Solution Manual to Principles of Heating Ventilating and Air Conditioning 6th Edition
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PRINCIPLES OF HEATING VENTILATING AND
AIR CONDITIONING SOLUTIONS MANUAL
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ABOUT THE AUTHORS Ronald H. Howell, PhD, PE, Fellow ASHRAE, retired as professor and chair of mechanical engineering at the University of South Florida and is also professor emeritus of the University of Missouri-Rolla. For 45 years he taught courses in refrigeration, heating and air conditioning, thermal analysis, and related areas. He has been the principal or co-principal investigator of 12 ASHRAE-funded research projects. His industrial and consulting engineering experience ranges from ventilation and condensation problems to the development and implementation of a complete air curtain test program. William J. Coad, PE, Fellow ASHRAE, was ASHRAE president in 2001-2002. He has been with McClure Engineering Associates, St. Louis, Mo., for 45 years and is currently a consulting principal. He is also president of Coad Engineering Enterprises. He has served as a consultant to the Missouri state government and was a lecturer in mechanical engineering for 12 years and an affiliate professor in the graduate program for 17 years at Washington University, St. Louis. He is the author of Energy Engineering and Management for Building Systems (Van Nostrand Reinhold). Harry J. Sauer, Jr., PhD, PE, Fellow ASHRAE, was a professor of mechanical and aerospace engineering at the University of Missouri-Rolla. He taught courses in air conditioning, refrigeration, environmental quality analysis and control, and related areas. His research ranged from experimental boiling/condensing heat transfer and energy recovery equipment for HVAC systems to computer simulations of building energy use and actual monitoring of residential energy use. He served as an advisor to the Missouri state government and has conducted energy auditor training programs for the US Department of Energy. Dr. Sauer passed away in June 2008.
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PRINCIPLES OF HEATING VENTILATING AND
AIR CONDITIONING 6th Edition
Ronald H. Howell
William J. Coad
Harry J. Sauer, Jr.
American Society of Heating, Refrigerating and Air-Conditioning Engineers, Inc.
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SOLUTIONS MANUAL
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ISBN 978-1-933742-70-0 ©2009 American Society of Heating, Refrigerating and Air-Conditioning Engineers, Inc. 1791 Tullie Circle, N.E. Atlanta, GA 30329 www.ashrae.org All rights reserved. Printed in the United States of America
ASHRAE has compiled this publication with care, but ASHRAE has not investigated, and ASHRAE expressly disclaims any duty to investigate, any product, service, process, procedure, design, or the like that may be described herein. The appearance of any technical data or editorial material in this publication does not constitute endorsement, warranty, or guaranty by ASHRAE of any product, service, process, procedure, design, or the like. ASHRAE does not warrant that the information in the publication is free of errors, and ASHRAE does not necessarily agree with any statement or opinion in this publication. The entire risk of the use of any information in this publication is assumed by the user. No part of this publication may be reproduced without permission in writing from ASHRAE, except by a reviewer who may quote brief passages or reproduce illustrations in a review with appropriate credit, nor may any part of this publication be reproduced, stored in a retrieval system, or transmitted in any way or by any means—electronic, photocopying, recording, or other—without permission in writing from ASHRAE. Requests for permission should be submitted at www.ashrae.org/permissions.
SPECIAL PUBLICATIONS Mark Owen Editor/Group Manager of Handbook and Special Publications Cindy Sheffield Michaels Managing Editor James Madison Walker Associate Editor Amelia Sanders Assistant Editor Elisabeth Parrish Assistant Editor Michshell Phillips Editorial Coordinator
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Notes to Instructors This manual contains solutions to most of the problems in the textbook, Principles of Heating, Ventilating, and Air Conditioning, which is based on the 2009 ASHRAE Handbook—Fundamentals. Some of these problems require the use of tables, figures, or equations in the 2009 Handbook that may not be found in Principles of Heating, Ventilating, and Air Conditioning. The solutions in this manual are generally presented in abbreviated form, with some intermediate computations omitted. Answers and solutions are included for the majority of the problems. The remaining problems are either those requiring discussion or those whose solutions depend on arbitrary assumptions or data selected by the instructor.
R.H. Howell W.J. Coad
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CONTENTS Solutions to Chapter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 Chapter 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 Chapter 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 Chapter 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 Chapter 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61 Chapter 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77 Chapter 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 Chapter 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 Chapter 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111 Chapter 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133 Chapter 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139 Chapter 12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153 Chapter 13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165 Chapter 14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171 Chapter 15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177 Chapter 16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181 Chapter 17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185 Chapter 18 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195 Chapter 20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219
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Chapter 19 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203
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Solutions to
Chapter 1 BACKGROUND
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Chapter 1—Background⏐3
1.1 Estimate whether ice will form on a clear night when ambient air temperature is 45°F (7.2°C), if the water is placed in a shallow pan in a sheltered location where the convective heat transfer coefficient is 0.5 Btu/h⋅ft2⋅°F (2.8 W/m2⋅K). Heat in by convection = Heat out by radiation to space (Assume space @ 0°R, Assume water is black body, ε = 1 4
hA ( T air – T water ) = σεAT water Tw 4 ( 0.5 ) ( 505 – T w ) = 0.1714 ⎛ ---------⎞ → by trial and error ⎝ 100⎠ T w ≈ 410°R = – 50°F ∴ will freeze
1.4 Estimate the size of cooling and heating equipment that is needed for a new bank building in middle America that is 140 × 220 × 12 ft high (42.7 × 67 × 3.7 m high). Be conservative. Floor area = ( 140 ) ( 220 ) = 30,800 ft
2
Volume = ( 140 ) ( 220 ) ( 12 ) = 370,000 ft
3
2
From Table 1.1: 250 ft ⁄ ton and 3.0 Btu/h ⋅ ft
3
2
30, 800 ft - = 123 tons ∴Cooling: ---------------------------2 250 ft ⁄ ton 3
3
Heating: ( 370,000 ft ) ( 3.0 Btu/h ⋅ ft ) = 1,110,000 Btu/hr or 1110 Mbh
2
3
From Table 1.1: 700 ft ⁄ ton and 3.0 Btu ⁄ h ⋅ ft ( 28 ) ( 78 ) Cooling: ---------------------- = 3.12 tons or 3.12 × 12,000 = 37,400 Btu/h ( 700 ) 3
3
Heating: [ ( 28 ) ( 78 ) ( 8 )ft ] ( 3.0 Btu/h ⋅ ft ) = 52,400 Btu ⁄ h
1.6 Estimate the initial cost of the complete HVAC system (heating, cooling, and air moving) for an office building, 40 × 150 × 10 ft high (12.2 × 45.7 × 3.1 m high). 40 × 150 Cooling unit: ---------------------------= 17 tons 3 350 ft ⁄ ton 3
Heating unit: ( 40 ) ( 150 ) ( 10 ) ( 3 Btu/h ⋅ ft ) = 180,000 Btu/h 2
Air movement: 17 tons × 400 cfm/ton = 6900 cfm or [1.2 cfm/ft × 40 × 150 = 7200 cfm] Costs: Cooling system ($1500/ton) × 17 tons = $25,500 Heating system ($2.50/cfm) × 6900 [ 7200 ] cfm = $17,250 [$18,000] Fans/ducting ( $7.50 ) × 6900 [7200] cfm = $51,750 [$54,000] Total = $94,500 [$97,500]
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1.5 Estimate the size of heating and cooling equipment that will be needed for a residence in middle America that is 28 × 78 × 8 ft high (8.5 × 23.8 × 2.4 m high).
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4⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
1.7 Estimate the annual operating cost for the building in Problem 1.6 if it is all-electric. 2
From Table 1.2: 30.5 kWh / ft ⋅ yr Energy = ( 40 × 150 ) ( 30.5 ) = 183,000 kWh Cost = $0.08 ( 183,000 ) = $14,640
1.8 Open-ended design problem.
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Solutions to
Chapter 2 THERMODYNAMICS AND PSYCHROMETRICS
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Chapter 2—Thermodynamics and Psychrometrics⏐7
2.6 Two pounds of air contained in a cylinder expand without friction against a piston. The pressure on the back side of the piston is constant at 200 psia. The air initially occupies a volume of 0.50 ft3. What is the work done by the air in ft-lbf if the expansion continues until the temperature of the air reaches 100°F? PV = mRT W =
∫ P dv
mRT 2 ( 2 ) ( 53.3 ) ( 560 ) 3 V 2 = -------------- = ------------------------------------- = 2.073 ft P2 ( 200 ) ( 144 )
2
= P ∫ dv = P ( v 2 – v 1 ) = ( 200 ) ( 144 ) ( 2.073 – 0.5 ) = 45,300 ft ⋅ lb f 1
2.7 Determine the specific volume, enthalpy, and entropy of 1 kg of R-134a at a saturation temperature of –5°C and a quality of 14%. from R-134a tables: 1 -⎞ 3 ν = 0.14 ( 0.083 ) + 0.86 ⎛ ----------= 0.0123 m ⁄ kg ⎝ 1310⎠ h = 0.14 ( 395.8 ) + 0.86 ( 192.9 ) = 221.3 kJ/kg s = 0.14 ( 1.7306 ) + 0.86 ( 0.976 ) = 1.082 kJ/kg ⋅ K
2.8 Saturated R-134a vapor at 42°C is superheated at constant pressure to a final temperature of 72°C. What is the pressure? What are the changes in specific volume, enthalpy, entropy, and internal energy? from R-134a tables:
S g = 1.7108 kJ/kg ⋅ K
h g = 420.44 kJ/kg 6
u = h – Pν = 420.44 – 1.0721 ( 10 ) ( 0.0189 ⁄ 1000 )
u = 400.2 kJ/kg 1 2. 72°C = 345 K: P = 1.0037 MPa; ----- = 42.0; ν2
h 2 = 453, s 2 = 1.81
u 2 = 608.3 3 1 ν 2 – ν 1 = ⎛ ------ – 0.0189⎞ = 0.0049 m ⁄ kg; ⎝ 42 ⎠
s 2 – s 1 = 1.81 – 1.71 = 0.10 kJ/kg ⋅ k;
h 2 – h 1 = 453 – 420.44 = 32.6 kJ/kg u 2 – u 1 = 427.5 – 400.2 = 27.3 kJ/kg
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3
1. 42°C = 315 K: P = 1.0721 MPa; v g = 0.0189 m ⁄ kg ;
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8⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual 2.9 A tank having a volume of 200 ft3 contains saturated vapor (steam) at a pressure of 20 psia. Attached to this tank is a line in which vapor at 100 psia, 400°F flows. Steam from this line enters the vessel until the pressure is 100 psia. If there is no heat transfer from the tank and the heat capacity of the tank is neglected, calculate the mass of steam that enters the tank. Steam line: P = 100 psi ; t = 400°F Tank: P 1 = 20 psi, sat vapor P 2 = 100 psi 3 200 ν 1 = 20.09 ft ⁄ lb m ⇒ m 1 = ------------- = 9.955 lb m 20.09
V2 m 2 = -----ν2 Try
and T 2 = 550°F
mi hi = m2 u2 – m1 u1 u 2 = 1195
by trial and error
200 m 2 = --------- = 33.9 lbm 5.9
? ( 33.9 – 9.955 ) ( 1228 ) = ( 33.9 ) ( 1195 ) – ( 9.955 ) ( 1082 ) 29400 ≈ 29700 ∴ m i = 33.9 – 9.955 = 23.95 lb m
2.10 Determine the heat required to vaporize 50 kg of water at a saturation temperature of 100°C.
2.11 The temperature of 150 kg of water is raised from 15°C to 85°C by the addition of heat. How much heat is supplied? kJ Q = mc p ( Δt ) = ( 150 kg ) ⎛⎝ 4.180 --------------⎞⎠ ( 85 – 15 ) K = 43,890 kJ kg ⋅ K
2.12 Three cubic meters per second of water are cooled from 30°C to 2°C. Compute the rate of heat transfer in kilojoules per second (kilowatts). 3 ⎛ ⎞ V· 3 m ⁄s Q = mc p Δt = --- c p Δt = ⎜ ----------------------------------------⎟ ( 4.18 kJ ⁄ kg ⋅ K ) ( 30 – 2 ) K 3 ν ⎝ 0.001004 m ⁄ kg⎠
Q = 350,000 kW or
3 Q = m Δh = ⎛⎝ ----------------------⎞⎠ ( 125.72 – 8.39 ) = 351,000 kW 0.001004
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Q = mh fg = ( 50 kg ) ( 2256.28 ) kJ ⁄ kg = 112,800 kJ
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Chapter 2—Thermodynamics and Psychrometrics⏐9
2.13 Consider 10 lbm of air that is initially at 14.7 psia, 100°F. Heat is transferred to the air until the temperature reaches 500°F. Determine the change of internal energy, the change in enthalpy, the heat transfer, and the work done for: a. a constant-volume process b. a constant-pressure process. a. Closed system, constant volume process, perfect gas ΔU = mc v Δt = ( 10 ) ( 0.171 ) ( 500 – 100 ) = 684 Btu ΔH = mc p Δt = ( 10 ) ( 0.240 ) ( 500 – 100 ) = 960 Btu Q – W = ΔU (closed system); w = 0 since constant volume Q = ΔU = 684 Btu b. Closed system, constant pressure process, perfect gas ΔU = 684 Btu ; ΔH = 960 Btu p = c: Q = mc p Δt = ΔH = 960 Btu Q – W = ΔU (closed system) ; W = Q – Δu = 960 – 684 = 276 Btu
2.14 The discharge of a pump is 10 ft above the inlet. Water enters at a pressure of 20 psia and leaves at a pressure of 200 psia. The specific volume of the water is 0.016 ft3/lb. If there is no heat transfer and no change in kinetic or internal energy, what is the work per pound? From 1st Law: P1 ν1 – P2 ν2 g z1 – z2 w = ------------------------------ + ----- --------------J gc J
2.15 The discharge of a pump is 3 m above the inlet. Water enters at a pressure of 138 kPa and leaves at a pressure of 1380 kPa. The specific volume of the water is 0.001 m3/kg. If there is no heat transfer and no change in kinetic or internal energy, what is the work per unit mass? 2
2
ν1 ν1 ⎛ ⎞ ⎛ ⎞ m ⎜ u 1 + P 1 ν 1 + ----- + gz 1⎟ – ⎜ u 2 + P 2 ν 2 + ----- + gz 2⎟ + Q – W = 0 2 2 ⎝ ⎠ ⎝ ⎠ ( 138 ) ( 0.001 ) – ( 1380 ) ( 0.001 ) – ( 3 ) ( 9.806 ) – W = 0 W = – 30.66 J ( Note 1 J (Joule) = 1 N ⋅ m )
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( 0.016 ) ( 20 – 200 ) ( 144 ) ( 32.2 ) ( – 10 ) = ---------------------------------------------------------- + --------------- -------------778 ( 32.2 ) ( 778 ) = – 0.533 – 0.013 = – 0.546 Btu/lb m
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10⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
2.16 Air is compressed in a reversible, isothermal, steady-flow process from 15 psia, 100°F to 100 psia. Calculate the work of compression per pound, the change of entropy, and the heat transfer per pound of air compressed. P 1 = 15 psi , t 1 = 100°F
P 2 = 100 psi , t 2 = 100°F
RT 1 3 ( 53.3 ) ( 373 ) ν 1 = --------- = ----------------------------- = 9.21 ft ⁄ lb m P1 ( 15 ) ( 144 ) P1 15 3 ν 2 = ν 1 ⎛⎝ ------⎞⎠ = ( 9.21 ) ⎛⎝ ---------⎞⎠ = 1.38 ft ⁄ lb m P2 100 P1 ν1 Pν = constant ⇒ p = -----------ν w =
∫ P dν
2 ν2 = P 1 ν 1 ∫ dν ------ = P 1 ν 1 ln ⎛⎝ -----⎞⎠ ν1 1 ν
1.38 w = ( 15 ) ( 144 ) ( 9.21 ) ln ⎛⎝ ----------⎞⎠ 9.21
= – 37,800 ft ⋅ lb f ⁄ lb m
P2 T2 100 373 ΔS = C p ln ⎛ -----⎞ – R ln ⎛ ------⎞ = ( 0.24 ) ln ⎛ ---------⎞ – ( 53.3 ) ln ⎛ ---------⎞ ⎝ 15 ⎠ ⎝ 373⎠ ⎝ T 1⎠ ⎝ P 1⎠ = 0 – 101.2 = – 101.2 ft lb f ⁄ lb m ⋅ °R = – 0.13 Btu ⁄ lb m ⋅ °R
3
V = 3 ft m = 44.5 lb m
t = – 240°F = 220°R
3
⇒
ν f = 0.02613 ft ⁄ lb m 3
ν g = 0.0750 ft ⁄ lb m 3
ν = V ⁄ m = 3 ⁄ 44.5 = 0.0674 ft ⁄ lb m ν = νf + χ ( νg – νf ) ν – νf ( 0.0674 – 0.02613 ) x = ---------------- = ---------------------------------------------- = 0.8445 ( 0.0750 – 0.02613 ) νg – νf m v = mx = ( 44.5 ) ( 0.8445 ) = 37.58 lb m vapor m L = m ( 1 – x ) = ( 44.5 ) ( 1 – 0.8445 ) = 6.92 lb m liquid
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2.17 Liquid nitrogen at a temperature of –240°F exists in a container, and both the liquid and vapor phases are present. The volume of the container is 3 ft3 and the mass of nitrogen in the container has been determined as 44.5 lbm. What is the mass of liquid and the mass of vapor present in the container?
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Chapter 2—Thermodynamics and Psychrometrics⏐11
2.18 A fan in an air-conditioning system is drawing 1.25 hp at 1760 rpm. The capacity through the fan is 0.85 m3/s of 24°C air and the inlet and outlet ducts are 0.31 m in diameter. What is the temperature rise of the air due to this fan? 2
2
v 1⎞ ⎛ V 2⎞ ⎛ m ⎜ h1 + -----⎟ – ⎜ h 2 + ------⎟ – w = 0 2 2⎠ ⎝ ⎠ ⎝ m ( h1 – h2 ) – w = 0 m Cp ( t1 – t2 ) – w = 0
2
2 πD A = ---------- = 0.0755 m 4
V· 0.85 V 1 = --- = ---------------- = 11.26 m/s A 0.0755 V 2 = V 1 since small ΔT and ΔP
( 1.25 ) ( 0.746 ) t 1 – t 2 = ---------------------------------- = 0.91 K ( 1.02 ) ( 1.005 )
P 1 V· 1 = m· 1 RT 1 ( 101.3 ) ( 1000 ) ( 0.85 ) m· = --------------------------------------------------- = 1.02 kg/s ( 287 ) ( 294 )
2.19 Air is contained in a cylinder. Initially, the cylinder contains 1.5 m3 of air at 150 kPa, 20°C. The air is then compressed reversibly according to the relationship pvn = constant until the final pressure is 600 kPa, at which point the temperature is 120°C. For this process determine: a. the polytropic exponent n b. the final volume of the air c. the work done on the air and the heat transfer
V2 V1 1.5 b. m = ------ = ------------- = 2.674 kg = ------ ; ν 1 0.561 ν2
V 2 = ( 2.674 ) ( 0.188 ) = 0.503 m
P2 V2 – P1 V1 R ( T2 – T1 ) dv- ------------------------------ = -------------------------- = – 106.3 kJ c. W = ∫ P dv = c ∫ ---= n 1–n 1–n v Q = W + m ( U 2 – U 1 ) = – 106.3 + ( 2.674 ) ( 0.718 ) ( 120 – 20 ) = 85.7 kJ
2.20 Water at 20°C is pumped from ground level to an elevated storage tank above ground level; the volume of the tank is 50 m3. Initially, the tank contains air at 100 kPa, 20°C, and the tank is closed so that the air is compressed as the water enters the bottom of the tank. The pump is operated until the tank is three-quarters full. The temperature of the air and water remain constant at 20°C. Determine the work input to the pump.
3
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( 287 ) ( 393 ) 3 ( 287 ) ( 294 ) 3 ν 2 = ------------------------------- = 0.1880 m ⁄ kg ν 1 = ------------------------------- = 0.561 m ⁄ kg ( 600 ) ( 1000 ) ( 150 ) ( 1000 ) n ν2 n P1 n ⎛ 0.188 -------------⎞ = 150 --------a. ⎛ -----⎞ = ------ ; n = 1.27 0.335 = 0.25 ; ⎝ ν 1⎠ ⎝ ⎠ P2 0.561 600
P 2 ν 2 = RT 2 ;
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12⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
W against
air
=
∫ Pdv
v v = mRT ∫ dv ----- = mRT ln ----2- = P 1 v 1 ln ----2v1 v1 v
= ( 100 ) ( 1000 ) ( 50 ) ln ( 0.25 ) = – 6930 kJ ( 50 ) W to elevate = mg Δz = ( 3 ⁄ 4 ) --------------------------- ( 9.8 ) ( 40 ) = 14670 kJ ( 0.001002 ) W total from pump = 6930 + 14670 = 21,600 kJ
2.21 A centrifugal pump delivers liquid oxygen to a rocket engine at the rate of 100 lbm/s. The oxygen enters the pump as liquid at 15 psia and the discharge pressure is 500 psia. The density of liquid oxygen is 66.7 lbm/ft3. Determine the minimum size motor (in horsepower) to drive this pump. · w· = m· ∫ ν dP = m· ν ( P 2 – P 1 ) 1 8 = ( 100 ) ( 3600 ) ⎛ ----------⎞ ( 500 – 15 ) ( 144 ) = 3.769 × 10 ft ⋅ lb f ⎝ 66.7⎠ 8
( 3.769 × 10 ) w· HP = --------------------------------- = 190.4 Hp ( 778 ) ( 2545 )
P 1 = 200 psi P2 a. T 2 = T 1 ⎛⎝ ------⎞⎠ P1
t 1 = 1500°F K – 1-----------K
20 = ( 1960 ) ⎛⎝ ---------⎞⎠ 200
P 2 = 20 psi 1.4 – 1--------------1.4
= 1015.2 R
RT 2 ( 53.3 ) ( 1051.2 ) 3 b. ν 2 = --------- = ------------------------------------- = 19.469 ft ⁄ lb m P2 ( 20 ) ( 144 ) c. Δu = c u ΔT = ( 0.171 ) ( 1015.2 – 1960 ) = – 161.6 Btu/lb m d. ΔH = c p ΔT = ( 0.240 ) ( 1015.2 – 1960 ) = – 266.8 Btu/lb m e. δq – δw = δu = – 161.6 Btu/lb m
2.23 Air undergoes a steady-flow, reversible adiabatic process. The initial state is 1400 kPa, 815°C, and the final pressure is 140 kPa. Changes in kinetic and potential energy are negligible. Determine a. final temperature b. final specific volume c. change in specific internal energy d. change in specific enthalpy e. specific work
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2.22 Air undergoes a steady-flow, reversible adiabatic process. The initial state is 200 psia, 1500°F, and the final pressure is 20 psia. Changes in kinetic and potential energy are negligible. Determine a. final temperature b. final specific volume c. change in internal energy per lbm d. change in enthalpy per lbm e. work per lbm
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Chapter 2—Thermodynamics and Psychrometrics⏐13
P2 a. T 2 = T 1 ⎛⎝ ------⎞⎠ P1
k---------– 1k
140 0.286 = 1088 ⎛⎝ ------------⎞⎠ = 563 K = 290°C 1400
b. P 2 ν 2 = RT 2 , R air = 0.0287 N ⋅ m ⁄ g ⋅ K 3
( 140 ) ( 1000 )ν 2 = ( 0.0287 ) ( 563 ) ⇒ ν 2 = 0.115 m ⁄ kg c. u 2 – u 1 = c v ( T 2 – T 1 ) = ( 0.718 ) ( 290 – 815 ) = – 377 kJ/kg d. h 2 – h 1 = c p ( T 2 – T 1 ) = ( 1.005 ) ( 290 – 815 ) = – 528 kJ/kg e. h 1 – h 2 – w = 0 ; w = 528 kJ/kg
2.24 A fan provides fresh air to the welding area in an industrial plant. The fan takes in outside air at 80°F and 14.7 psia at the rate of 1200 cfm with negligible inlet velocity. In the 10 ft2 duct leaving the fan, air pressure is 1 psig. If the process is assumed to be reversible and adiabatic, determine the size motor needed to drive the fan. [Ans: W = 5.1 hp] ( 14.7 ) ( 144 ) ( 1200 ) Pv· m· = ------- = ---------------------------------------------- = 88.2 lb m /min RT ( 53.3 ) ( 540 ) P2 T 2 = T 1 ⎛⎝ ------⎞⎠ P1
k---------– 1k
15.7 = 540 ⎛⎝ ----------⎞⎠ 14.7
1.4 – 1--------------1.4
= 550.25°R
RT 2 ( 53.3 ) ( 550.25 ) 3 ν 2 = --------- = ------------------------------------- = 12.98 ft ⁄ lb m P2 ( 15.7 ) ( 14.4 ) mν 2 ( 88.7 ) ( 12.98 ) V 2 = ---------- = ---------------------------------- = 144.5 ft/min A2 ( 10 ) 2
V2 = m c p ( T 1 – T 2 ) + -----------2gcJ 2
( 144.5 ⁄ 60 ) = ( 88.2 ) ( 60 ) 0.24 ( 540 – 550.25 ) – ------------------------------------( 2 ) ( 32.2 ) ( 778 ) = – 13,020 Btu/h = 5.12 hp
2.25 If the fan in the previous problem has an efficiency of 64% and is driven by a motor having an efficiency of 78%, determine the required power, kW. HP Ideal ( 5.12 ) HP actual = ------------------------------------- = ------------------------------- = 10.3 hp ( η fan ) ( η motor ) ( 0.64 ) ( 0.78 )
2.26 A fan provides fresh air to the welding area in an industrial plant. The fan takes in outside air at 32.2°C and 101.4 kPa at the rate of 566 L/s with negligible inlet velocity. In the 0.93 m2 duct leaving the fan, air pressure is 102 kPa. Determine the minimum size motor needed to drive the fan.
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2
v2 ⎞ ⎛ W = m h 1 – ⎜ h 2 + ------------⎟ 2gcJ ⎝ ⎠
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14⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual P 1 V· 1 ( 101.4 ) ( 1000 ) ( 0.566 ) m = ------------ = ------------------------------------------------------ = 0.655 kg/s RT 1 ( 287 ) ( 305.2 ) P2 T 2 = T 1 ⎛ ------⎞ ⎝ P 1⎠
k – 1---------k
102 = ( 305.2 ) ⎛ -------------⎞ ⎝ 101.4⎠
1.4 – 1 ---------------1.4
= 305.7 K
RT 2 ( 287 ) ( 305.7 ) 3 ν 2 = --------- = -------------------------------- = 0.860 m ⁄ kg P2 ( 102 ) ( 1000 ) mν 2 ( 0.655 ) ( 0.860 ) V· 2 = ---------- = ------------------------------------- = 0.606 m/s (small, neglect kinetic energy) A2 ( 0.93 ) W = m ( h 1 – h 2 ) = mc p ΔT = ( 0.655 ) ( 0.718 ) ( 305.2 – 305.7 ) = 0.24 kW
2.27 In an insulated feedwater heater, steam condenses at a constant temperature of 220°F. The feedwater is heated from 60 to 150°F at constant pressure. a. Assuming the specific heat at constant pressure of the feedwater is unity, how many Btu are absorbed by each pound in its passage through the heater? [Ans: 90 Btu/lb] b. What is the change in entropy of the condensing steam per pound of feedwater heated? [Ans: −0.1324 Btu/lb m⋅R] c. What is the change in entropy of 1 lb of feedwater as it passes through the heater? [Ans: +0.1595 Btu/lb m⋅R] d. What is the change in entropy of the combined system? Does this violate the second law? Explain. [Ans: +0.0271 Btu/lb m⋅R, No] steam
= Q again
F.W.
Q = m F.W. C p ΔT ; ⎛ ----⎞ = C p ΔT ⎝ m⎠ FW
⎛Q ----⎞ ⎝ m⎠ F.W. = ( 1.0 ) ( 90 ) = 90 Btu ⁄ lb m F.W. Q b. ( m Δh ) steam = m FW ⎛ ----⎞ ⎝ m⎠ F.W.
Assume 1.0 lb m feed water
m FW ( Q ⁄ m ) FW ( 1.0 ) ( 90 ) m steam = ------------------------------------ = ----------------------- = 0.0932 lb m steam Δh ( 965.3 ) ΔS steam = m s s fg = ( 0.0932 ) ( 1.4201 ) = 0.1324 Btu ⁄ lb m c. ΔS FW = m FW ( s f
150°F
– sf
60°F
FW
⋅R
) = 1.0 [ ( 0.2150 ) – ( 0.0555 ) ]
= 0.1595 Btu ⁄ lbm ⋅ R d. Δs T = Δs FW – Δs steam = ( 0.1595 ) – ( 0.1324 ) = 0.0271 Btu ⁄ lb m ⋅ R Does not violate 2nd Law; ΔS is positive.
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a. Q loss
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Chapter 2—Thermodynamics and Psychrometrics⏐15
2.28 Steam at 124 kPa and 96% quality enters a radiator. The steam is condensed as it flows through the radiator and leaves as condensate at 88°C. If the radiator is to have a heating capacity of 1.85 kW, how many kilograms per hour of steam must be supplied to the radiator? Steam 1. 124 Kpa, 0.96 quality: h 1 = 444 + 0.96 ( 2241 ) = 2595 2. 88°C liquid: h 2 = 368.6 Q = m ( h 1 – h 2 ) = 1.85 = m ( 2595 – 368.6 ) m = 0.00083 kg/s = 6.6 lb m ⁄ h
2.29 Solve the following: a. Air at 50 psia and 90°F flows through a restriction in a 2 in. ID pipe. The velocity of the air upstream from the restriction is 450 fpm. If 58°F air is desired, what must the velocity downstream of the restriction be? Comment on this as a method of cooling. b. Air at 50 psia and 90°F flows through an insulated turbine at the rate of 1.6 lbm/s. If the air delivers 11.5 hp to the turbine blades, at what temperature does the air leave the turbine? c. Air at 50 psia and 90°F flows through an insulated turbine at the rate of 1.6 lbm/s to an exit pressure of 14.7 psia. What is the lowest temperature attainable at exit?
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16⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
2
2
V1 ⎞ ⎛ V2 ⎞ ⎛ a. ⎜ h 1 + ------------⎟ – ⎜ h 2 + ------------⎟ = 0 2gcJ 2gcJ ⎝ ⎠ ⎝ ⎠ 2
2
2 V1 V2 ( 450 ⁄ 60 ) C p + ( T 1 – T 2 ) + ------------ = ------------ = ( 0.24 ) ( 90 – 58 ) + ------------------------------------2gcJ 2gcJ ( 2 ) ( 32.2 ) ( 778 ) V 2 = 620 fps = 37,200 fpm
Not cooling since temperature will increase as fluid slows down. b. m ( h 1 – h 2 ) – W = 0 mC p ( T 1 – T 2 ) – W = 0 ; ( 1.6 ) ( 3600 ) ( 0.24 ) ( 90 – T 2 ) = 11.5 ( 2545 ) t 2 = 68.8°F c. Rev. Adiabatic for Minimum = Isentropic P2 T 2 = T 1 ⎛⎝ ------⎞⎠ P1
K – 1-----------K
14.7 = ( 550 ) ⎛⎝ ----------⎞⎠ 50
0.286
= 388R = – 72°F
2.30 Liquid water at a pressure of 10 psia and a temperature of 80°F enters a 1 in. diameter tube at the rate of 0.8 ft3/min. Heat is transferred to the water so that it leaves as saturated vapor at 9 psia. Determine the heat transfer per minute. [Ans: 95,800 Btu/min] P 1 = 10 psi
3 V· 1 = 0.8 ft /min
t 1 = 80°F
P 2 = 9 psi
Sat. vapor @ 2 2
ν1 ≅ νf h1 ≅ hf
3
80°F
80°F
= 0.01607 ft /lb m
= 48.05 Btu/lb m
2 3
v 2 = 42.367 ft /lb m h 2 = 1141.215 Btu/lb m
V· 1 ( 0.8 ) U 1 = ------ = ------------------------ = 146.8 ft/min A ( 0.00545 ) ν2 42.367 U 2 = U 1 ⎛ -----⎞ = 146.8 ⎛ -------------------⎞ = 387,024 ft/min ⎝ 0.01607⎠ ⎝ v1⎠ U1 A ( 146.8 ) ( 0.00545 ) m = ---------- = ------------------------------------------- = 49.78 lb m /min ν1 ( 0.01607 ) 2
2
2 2 U 2 – U 1⎞ ⎛ ( 387024 ) – ( 1468 ) Q· = m· ⎜ h 2 – h 1 + -------------------⎟ = ( 49.78 ) 1141.25 – 48.05 + ------------------------------------------------------( 2 ) ( 32.2 ) ( 3600 ) ( 778 ) 2gc ⎠ ⎝
= ( 49.78 ) [ 1093.165 + 830.43 ] = 95,800 Btu/min
2.31 A refrigerator uses R-134a as the refrigerant and handles 200 lbm/h. Condensing temperature is 110°F and evaporating temperature is 5°F. For a cooling effect of 11,000 Btu/h, determine the minimum size motor (hp) required to drive the compressor.
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Dia. = 1 in. ⇒ A = 0.7854 in. = 0.00545 ft
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Chapter 2—Thermodynamics and Psychrometrics⏐17
@ x = 1.0 t = 5°F h = 103.745, s = 0.22470 @ p = 161.05 psi, s = 0.22470 h ≅ 120
W· = m· ( h 1 – h 4 ) = 200 ( 120 – 103.7 ) = 3260 Btu/h 3260 W· = ------------ = 1.28 Hp [Minimum for mechanical vapor compression cycle.] 2545 1 1 COP = ------------------------- = ------------------------------- = 4.4286 TR ⁄ TA – 1 570 ⁄ 465 – 1 11,000 W· = ---------------- = 2483.9 Btu/h = 0.976 Hp [Minimum for reversed Carnot cycle] 4.4286
2.32 A heat pump is used in place of a furnace for heating a house. In winter, when the outside air temperature is 10°F, the heat loss from the house is 60,000 Btu/h if the inside is maintained at 70°F. Determine the minimum electric power required to operate the heat pump (in kW).
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18⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
For minimum consider reversible Carnot cycle. QR QR 1 1 COP heating = ------- = -------------------- = --------------------------- = -------------------------W QR – QA 1 – QA ⁄ QR 1 – TL ⁄ TH 1 = ------------------------------- = 8.833 1 – 470 ⁄ 530 QR ( 60,000 ) W = ----------------------------- = ----------------------------------- = 1.99 kW COP heating ( 8.833 ) ( 3413 )
2.33 A heat pump is used in place of a furnace for heating a house. In winter, when the outside air temperature is − 10°C, the heat loss from the house is 200 kW if the inside is maintained at 21°C. Determine the minimum electric power required to operate the heat pump. [Ans: 21.1 kW] 1 1 COP h ( max ) = ------------------------- = ------------------------------- = 9.48 1 – 263 ⁄ 298 1 – TL ⁄ Th QR 200 W min = --------------------- = ---------- = 21.1 kW COP max 9.48
2.34 Refrigerant-134a vapor enters a compressor at 25 psia, 40°F, and the mass rate of flow is 5 lbm/min. What is the smallest diameter tubing that can be used if the velocity of refrigerant must not exceed 20 ft/s? 3 3 1 25 psi, 40°F ⇒ ρ = 0.449 lb/ft ; v = ------------- = 2.23 ft /lb 0.449 2
2
d = 1.70 ⇒ d = 1.30 in.
2.35 An R-134a refrigerating system is operating with a condensing temperature of 86°F and evaporating temperature of 25°F. If the liquid line from the condenser is soldered to the suction line from the evaporator to form a simple heat exchanger, and if as a result of this, saturated liquid leaving the condenser is subcooled 6°F, how many degrees will the saturated vapor leaving the evaporator be superheated? (Use tables.)
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( 20 ) V πd 5 ⁄ 60 = A --- = ----------------- --------------4 ( 144 ) ( 2.23 ) v
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Chapter 2—Thermodynamics and Psychrometrics⏐19
h 1 – h 2 = h 4 – h 3 ⇒ ( 40.5 ) – ( 38 ) = h 4 – ( 106.6 ) h 4 = 109.1 Btu/lb m ⎫ ⎬t = 44°F P 4 = 36.8 psi ⎭ 4 °SH = 44 – 25 = 19°
2.36 Ammonia is heated in the evaporator of a refrigeration system from inlet conditions of 10°F, 10% quality, to saturated vapor. The pressure remains constant during the process. For each pound, determine the changes in enthalpy and volume. [Ans: 505 Btu/lb; 6.55 ft3/lb] t 1 = 10°F
x 1 = 10% = 0.10
Sat. vapor @ 2
P 1 = P 2 = constant
v f1 = 0.02446
v g1 = 7.304 3
v 1 = ( 0.02446 ) – ( 0.10 ) ( 7.304 – 0.02446 ) = 0.7524 ft /lb m 3
Δv = 7.304 – 0.7524 = 6.552 ft /lb m h f1 = 53.8
h g = 614.9
h 1 = 53.8 + ( 0.10 ) ( 614.9 – 53.8 ) = 109.9 Btu/lb m Δh = 614.9 – 109.9 = 505 Btu/lb m
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2.37 For a compressor using R-134a with an evaporator temperature of 20°F and a condensing temperature of 80°F, calculate per ton of refrigeration a. displacement b. mass flow c. horsepower required
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20⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
qL 12000 b. m· = ----------------- = ------------------------------------ = 176.7 = 2.94 lb m /min ton h1 – h4 105.9 – 37.978 3 3 a. PD = m· v 1 = ( 176.7 ) ( 1.4101 ) = 249 ft /h ⋅ ton = 415 ft /min ton c. w· = m· ( h 2 – h 1 ) = 176.7 [ 117 – 105.9 ] = 1961 Btu/h ⋅ ton
= 0.771 Hp/ton
2.38 For a compressor using an R-22 system operating between 100°F condensing temperature and −10°F evaporator temperature, calculate per ton a. displacement b. mass flow c. horsepower required qL 12000 b. m· = ----------------- = ------------------------------------ = 186.9 lb m ⁄ h ⋅ ton = 3.12 lb m /min ton h1 – h4 103.45 – 39.27 a. PD = m· v 1 = ( 186.9 ) ( 1.6825 ) = 314 ft /h ⋅ ton = 5.24 ft /min ton c. w· = m· ( h – h ) = 186.9 [ 125 – 103.46 ] = 4018.4 Btu/h ⋅ ton 3
2
3
1
= 1.58 Hp/ton
3
v ≅ 1.0 ft ⁄ lbm
q L = 110 – 44.9 = 65.1 Btu/lbm 2
compressor:
1 π3 V· ideal = ( 800 ) ⎛ ---------⎞ ( 4 ) ⎛ ------------⎞ ⎝ 1728⎠ ⎝ 4 ⎠ 3
= 13.09 ft ⁄ lbm 3 V· actual = 4 ( 13.09 ) ( 0.70 ) = 36.65 ft ⁄ min m· q L [ ( 36.65 ) ( 60 ) ] ( 65.1 ) capacity = --------------- = -------------------------------------------------- ≈ 12 tons 12000 ( 0.80 ) ( 12000 )
2.40 A mechanical refrigeration system with R-134a is operating under such conditions that the evaporator pressure is 160 kPa and the liquid approaching the refrigerant control valve is at a temperature of 41°C. If the system has
a capacity of 15 kW, determine: a. the refrigerating effect per kilogram of refrigerant circulated
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2.39 An industrial plant has available a 4 cylinder, 3 in. bore by 4 in. stroke, 800 rpm, single-acting compressor for use with R-134a. Proposed operating conditions for the compressor are 100°F condensing temperature and 40°F evaporating temperature. It is estimated that the refrigerant will enter the expansion valve as a saturated liquid, that vapor will leave the evaporator at a temperature of 45°F, and that vapor will enter the compressor at a temperature of 55°F. Assume a compressor-volumetric efficiency of 70% and frictionless flow. Calculate the refrigerating capacity in tons for a system equipped with this compressor. Plot the cycle on the p-h diagram. [Ans: 12 tons]
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Chapter 2—Thermodynamics and Psychrometrics⏐21
b. c. d. e.
the mass flow rate in kilograms per second per kilowatt the volume flow rate in liters per second per kilowatt at the compressor inlet the total mass flow rate in kilograms per second the total volume flow rate in liters per second at the compressor inlet
a)
q e = h 1 – h 4 = 389 – 258 = 131 kJ/kg
b)
1 m = --------- = 0.0075 kg/s 131 V 1 = mv 1 = ( 0.0076 ) ( 0.126 ) ( 1000 ) = 0.97 l/s
c) d) e)
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2.41 A vapor-compression R-22 refrigeration system is being designed to provide 50 kW of cooling when operating between evaporating and condensing temperatures of 0°C and 34°C, respectively. The refrigerant leaving the condenser is subcooled 3 degrees and the vapor leaving the evaporator is superheated 5 degrees. Determine a. ideal compressor discharge temperature, °C b. refrigerant flow rate, kg/s c. compressor motor size, kW d. COP for cooling e. compressor discharge temperature if compression efficiency is 60%
15 m = --------- = 0.115 kg/s 131 v· = mv 1 = ( 0.115 ) ( 0.126 ) ( 1000 ) = 14.4 l/s
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22⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
h 1 = 409
s 1 = 1.76
h 2 = 434
h 3 = h 4 = 237.7
a)
t 2i = 55°C
b)
50 m = --------------------------------- = 0.292 kg/s ( 409 – 237.7 ) P = 0.292 ( 434 – 409 ) = 7.30 kW 50 COP c = ------- = 6.85 7.3 ( 434 – 409 ) h 2a = 409 + ---------------------------- = 451 ⇒ t 2a ≅ 75°C 0.60
c) d) e)
3 1 h 1 = 252, s 1 = 10.4, v 1 = ------- = 0.303 m ⁄ kg 3.3 1 h 2i = 700, v 2i = ------- = 0.123 t 2i = 95°C 8.1
700 – 525 1 h 2a = 525 + ⎛ ------------------------⎞ = 738, v 2a = ------- = 0.123 ⎝ 0.82 ⎠ 8.1 d) 1.35 η av = 94 – 6.1 ⎛⎝ ----------⎞⎠ = 74.85% 0.43
t 2a ≈ 110°C 2
3 2π ( 0.092 ) PD = --------------------------- ( 0.74 ) ( 28 ) = 0.0275 m ⁄ s 4
b)
( 0.7485 ) ( 0.0275 ) m = ------------------------------------------- = 0.0675 kg/s 0.303 Q e = m ( h 1 – h 4 ) = 0.0675 [ 525 – ( – 620.7 ) ] = 77.4 kW
c)
W = m ( h 2 – h 1 ) = ( 0.0675 ) [ 738 – 525 ] = 14.4 kW
e)
Qe 77.5 COP = ------ = ---------- = 5.4 W 14.4
a)
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2.42 For a line of ammonia compressors, the actual volumetric efficiency is given by: ηva = 94 – 6.1(pd/ps), % The compression efficiency is fairly constant at 82%. A compressor in this line has two cylinders, each having a 92 mm bore and a 74 mm stroke. The compressor has 4.5% clearance and operates at 28 r/s. The system is being selected for an air-conditioning unit and will therefore operate between an evaporating temperature of 0°C and a condensing temperature of 35°C. There is 5°C of subcooling in the condenser and 10°C of superheating in the evaporator. Sketch and label the system, including appropriate values for the thermodynamic properties, starting with state 1 at the compressor inlet. Determine a. refrigerant flow rate b. refrigerating capacity c. compressor motor size d. compressor discharge temperature e. COPc
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Chapter 2—Thermodynamics and Psychrometrics⏐23
2.43 An ammonia refrigerating system is operating with a condensing temperature of 30°C and an evaporating temperature of −4°C. For the ideal standard vapor compression cycle, determine a. refrigerating effect b. COP Sketch and label a p-h diagram showing values.
h 1 = h g @ – 4°C = 494 h 2 = 650 h 3 = h 4 = – 621.4 a) b)
Q e = ( h 1 – h 4 ) = 494 – ( – 621.4 ) = 1115.4 kJ/kg h1 – h4 Qe ( 1115.4 ) COP = ------ = ----------------- = ---------------------------- = 7.15 h2 – h1 W ( 650 – 494 )
1.
h 1 = 408.3, s 1 = 1.7349, v 1 = 0.03462
3.
1 s 2 = 1.7349, h 2 ≅ 432, v 2 = ------ = 0.0147 68 h 3 = 256.1
4.
h 4 = 256.1
2.
Estimated volumetric efficienty based on re-expansion of TRAPPED GAS. v1 0.03462 η v = 100 – cv ⎛ ----- – 1⎞ = 100 – 4 ⎛ ------------------- – 1⎞ = 94.6% ⎝ v2 ⎠ ⎝ 0.0147 ⎠ 2
4 1725 3 π(5) Piston Displacement = ------------------2- ⎛⎝ ---------⎞⎠ ⎛⎝ ------------⎞⎠ = 0.002258 m ⁄ s 100 60 4 ( 100 ) ( 0.946 ) ( 0.002258 ) Flow Rate = ---------------------------------------------- = 0.062 kg/s ( 0.03462 ) Q e = 0.062 ( 408.03 – 256.1 ) = 9.42 kW W = 0.062 ( 432 – 408.03 ) ⁄ 0.746 = 1.99 Hp
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2.44 A single-cylinder R-22 compressor has a 50 mm bore, a 40 mm stroke, and operates at 1725 rpm. Clearance volume is 4%. Determine as close as possible the actual refrigerating capacity, kW, and the required motor size, in hp, if the compressor is used in a system operating between 10°C and 40°C, evaporating and condensing temperatures, respectively.
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24⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
2.45 For the lithium-bromide/water absorption refrigeration system shown below, determine a. heat required at generator per ton of cooling b. COP c. heat rejection ratio, (Qabsorber + Qcondenser)/Qevaporator
State
P
t
x
h
m
1.
10 mm
90
0.50
–70
9.85
2.
50 mm
150
0
1128
0.197
3.
50 mm
150
0.51
–41
9.65
4.
50 mm
101
0
68.96
0.197
5.
10 mm
52
0
1084.5
0.197
0.50 m 3 = ⎛⎝ ----------⎞⎠ mi = 0.98m 1 0.51 m 1 = 0.197 + 0.98m 1 , m 1 = 9.85 m 3 = 9.65 a)
Q g = 60 [ ( 0.197 ) ( 1128 ) + ( 9.65 ) ( – 41 ) – ( 9.85 ) ( – 70 ) ]
b)
= 30964 Btu/h = 516 Btu/min 12000 COP = --------------- = 0.39 30964 Q a = 60 [ 9.85 ( – 70 ) – 9.65 ( – 41 ) – 0.197 ( 1084.5 ) ]
c)
= 30450 Btu/h = – 508 Btu/min Q c = 60 ( 0.197 ) ( 68.96 – 1128 ) = – 12520 Btu/h = – 209 Btu/m QR 30450 + 12520 )------- = (---------------------------------------= 3.58 Qe ( 12000 )
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12000 m 2 = m 4 = m 5 = -------------------------------------------60 ( 1084.5 – 68.0 ) = 0.197
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Chapter 2—Thermodynamics and Psychrometrics⏐25
2.46 In the basic lithium-bromide water absorption system, the generator operates at 170°F while the evaporator is at 47°F. The absorbing temperature is 75°F and the condensing temperature is 88°F. Calculate the heat rejection ratio for these conditions.
State
P
t
x
h
1.
8.2 mm
75°F
0.47
–73
2.
33.9 mm 170°F
0
1138
3.
33.9 mm 170°F
0.62
–38
4.
--
88°F
0
56.04
5.
8.2 mm
47°F
0
1081.74
m 1 x 1 = m 2 x 2 + ( m 1 – m 2 )x 3 m 1 ( 0.47 ) = ( m 1 – 1.0 ) ( 0.62 ) m 1 = 4.16, m 3 = 3.16 Q e = ( 1 ) [ ( 1081.74 ) – ( 56.04 ) ] = 1025.7 Btu/lb
Q c = – 12659 Btu/hr
Q a = ( 4.16 ) ( – 73 ) – ( 3.16 ) ( – 38 ) – ( 1 ) ( 1081.74 ) = – 1265.34 Btu/lb
Q a = – 14800 Btu/hr
Q g = ( 1 ) ( 1138 ) + ( 3.16 ) ( – 38 ) – ( 4.16 ) ( – 73 ) = 1321.6 Btu/lb
Q e = 15460 Btu/hr
Qa + Qc ------------------- = 2.28 Qe
2.47 For the aqua-ammonia absorption refrigeration system shown in the sketch below, complete the table of properties.
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Per ton of refrigeration 12000 m 2 = m 4 = m 5 = ---------------- = 11.7 lbm/h 1025.7 Q c = ( 1 ) [ ( 56.04 ) – ( 1138 ) ] = – 1081.96 Btu/lb
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26⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
3
200
260
0.85
735
4
200
160
0.54
40
5
200
160
0.992
606
6
200
97
0.992
75
7
25
20
0.999
550
2.48 Solar energy is to be used to warm a large collector plate. This energy will, in turn, be transferred as heat to a fluid within a heat engine, and the engine will reject energy as heat to the atmosphere. Experiments indicate that about 200 Btu/h·ft2 of energy can be collected when the plate is operating at 190°F. Estimate the minimum collector area that will be required for a plant producing 1 kW of useful shaft power, when the atmospheric temperature is 70°F.
Point
p, psia
t, °F
x, lb NH3/lb mix
h, Btu/lb
1
25
80
0.39
–48
2
200
260
0.26
165
For minimum area, maximum efficiency, USE CARNOT CYCLE QA 200 Δs = ------- = --------- = 0.308 TH 650 W = Q A – Q R = ( 200 ) – ( 163.24 ) = 36.8 Btu/h ⋅ ft
2
3413 2 area = ------------ = 92.74 ft ⁄ kW 36.8
2.50 A 20 by 12 by 8 ft (6.1 by 3.6 by 2.4 m) room contains an air-water vapor mixture at 80°F (26.7°C). The barometric pressure is standard and the partial pressure of the water vapor is measured to be 0.2 psia (1.38 kPa). Calculate a. relative humidity b. humidity ratio
c. d.
dew-point temperature mass of water vapor contained in the room
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Q R = ToΔs = ( 530 ) ( 0.308 ) = 163.24
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Chapter 2—Thermodynamics and Psychrometrics⏐27
a)
pw 0.2 φ = -------- = ---------------- = 0.394 = 39.4% p ws 0.5073
b)
pw 0.2 w = 0.62198 --------------- = ( 0.62198 ) ⎛ ----------------------⎞ = 0.0086 lb m /lb air ⎝ 14.7 – 12⎠ p – pw
c)
t d = t sat @ 0.2 PSI = 53.15°F
d)
Pw V ( 0.2 ) ( 144 ) ( 1920 ) m w = ----------------------- = ----------------------------------------------------- = 1.19 lb m ( R ⁄ m w )T ( 1545 ⁄ 18 ) ( 80 + 460 )
2.51 Given room conditions of 75°F (23.9°C) dry bulb and 60% RH, determine for the air vapor mixture without using the psychrometric charts a. humidity ratio b. enthalpy c. dew-point temperature d. specific volume e. degree of saturation a)
pw 0.258 w = 0.62198 ------ = ( 0.62198 ) ⎛⎝ ------------------------------⎞⎠ = 0.0111 lb m /lb air pa 14.7 – 0.258
b)
h = 0.24t + w ( 1061 + 0.45t ) = ( 0.24 ) ( 75 ) + ( 0.0111 ) [ ( 1061 ) + 0.45 ( 75 ) ] = 30.2 Btu/lb m
c)
t d = t sat @ 0.258 = 60.185°F
d)
RaT ( 53.3 ) ( 535 ) 3 v = ---------- = ------------------------------------------------- = 13.72 ft /lbm Pa ( 14.7 – 0.258 ) ( 144 )
e)
0.43 w s = 0.62198 ⎛⎝ ---------------------------⎞⎠ = 0.0187 14.7 – 0.43 ( p w = φp s = ( 0.6 ) ( 0.43 ) = 0.258 )
2.52 For the conditions of Problem 2.51 (above), using the ASHRAE Psychrometric Chart, find a. wet-bulb temperature [Ans: 65.2°F (18.4°C)] b. enthalpy [Ans: 30.2 Btu/lbm (70.2 J/g)] c. humidity ratio [Ans: 0.0112 lb/lb (0.0112 kg/kg)] a)
t wb = 65.2°F
b)
h = 30.2 Btu/lb m
c)
w = 0.0112 lb m /lb air
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w 0.0111 μ = ------ = ---------------- = 0.593 ws 0.0187
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28⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
2.53 Using the ASHRAE Psychrometric Chart, complete the following table. Enthalpy h, Relative Humidity φ, Specific Volume v, Btu/lbair % ft3/lbair 26.1 22 13.85
Dry Bulb, °F
Wet Bulb, °F
Dew Point, °F
Humidity W, lb/lbair
85
60
41
0.0054
75
60
50
0.0076
26.4
42
13.65
74.5
65
59.8
0.011
30
60
13.75
89.5
70
61
0.01143
34
38
14.1
99
85.5
82
0.0238
50
58
14.6
2.54 Using the ASHRAE Psychrometric Chart complete the following table: Dry Bulb, °F
Wet Bulb, °F
Dew Point, °F
Humidity Ratio, Relative Humidity, lbv/lba %
80
63.5
53.7
0.0088
40
28.8
13.8
70
55
43
0.0058
38
23.1
13.47
Enthalpy, Btu/lbair
Specific Volume, ft3/lbair
78
70
0.016
39
42
14.47
97
77
68
0.0157
40
40
14.3
79
65
57
0.01
46
30
13.8
86
60
40
0.0052
20
26.4
13.86
40
29
11
0.001
20
10.3
12.6
74
65
60
0.011
60
30
13.7
85
70
62
0.012
47
33.8
14.0
80
80
80
0.0224
100
43.8
14.1
Enthalpy, kJ/kg
Specific Volume, m3/kg
2.55 Complete the following table using the Psychrometric Chart. Dry Bulb, °C
Wet Bulb, °C
Dew Point, °C
Humidity Ratio, Relative Humidity, kg/kg %
26.5
17.3
12.2
0.0087
41
49
0.86
21
13
7.6
0.006
40
36.7
0.84
38
25.4
21
0.0155
38
78.2
0.905
41.7
29.2
25.2
0.0207
40
95
0.92
22.2
17
14.2
0.01
60
48
0.85
32
16
4
0.005
18
44.5
0.87
4
–2
–10
0.001
20
6.5
0.78
39.8
23.4
16
0.0115
26
70
0.904
30
21
17
0.012
45
61
0.875
27
27
27
0.0227
100
85.5
0.88
2.56 Complete the following table. Dry Bulb, °C
Wet Bulb, °C
Dew Point, °C
Humidity Ratio, Relative Humidity, kg/kg %
32
24
20.7
0.0155
40
26.3
21.2
0.0160
38.8
24.2
18
0.0130
33.8
28.1
26.4
0.022
7
7
7
0.0063
Enthalpy, kJ/kg
Specific Volume, m3/kg
52
72.2
0.886
34
81
0.91
30
72.5
0.902
66
90.5
0.9
100
23
0.801
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100
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Chapter 2—Thermodynamics and Psychrometrics⏐29
2.57 Without using the psychrometric chart, determine the humidity ratio and relative humidity of an air-water vapor mixture with a dry-bulb temperature of 90°F and thermodynamic wet-bulb temperature of 78°F. The barometric pressure is 14.7 psia. Check your result using the psychrometric chart. [Ans: W = 0.018 lb/lb, relative humidity φ = 59%] @ 90F
P g = 0.6989 PSI
@ 78F
P g = 0.475 PSI
P ws 0.4748 W* = 0.622 ------------------ = ( 0.622 ) ⎛ ---------------------------------⎞ = 0.02077 ⎝ 14.7 – 0.4748⎠ P – P ws ( 1093 – 0.556t* )W* – 0.24 ( t – t* ) W = -----------------------------------------------------------------------------------(t∗ = 78, W∗ = 0.02077) 1093 – 0.44t – t* W = 0.018 lb m /lb air (P )( W) P w = ------------------------ = 0.622 + W Pw φ = --------- = P ws
( 14.7 ) ( 0.0179 )---------------------------------------= 0.4119 psia ( 0.622 + 0.0179 ) 0.4119 ---------------- = 0.59 or 59% 0.6989
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Solutions to
Chapter 3 BASIC HVAC SYSTEM CALCULATIONS
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Chapter 3—Basic HVAC System Calculations⏐33
3.1 One of the many methods used for drying air is to cool the air below the dew point so that condensation or freezing of the moisture takes place. To what temperature must atmospheric air be cooled in order to have a humidity ratio of 0.000017 lb/lb (0.000017 kg/kg)? To what temperature must this air be cooled if its pressure is 10 atm?
Pw W W = 0.62198 ⎛⎝ -----------------⎞⎠ ⇒ Pw = ------------------- ( P – Pw ) P – Pw 0.62198 0.000017 Pw = ---------------------- ( 14.7 – Pw ) 0.62198 Pw = 0.000402 – 0.0000273Pw Pw = 4.02 × 10
–4
PSI ⇒ t sat = – 63°F
[Table 3, Chap 1 HF]
0.000017 10 ATM = 147 PSI ⇒ Pw = ---------------------- ( 147 – Pw ) 0.62198 Pw = 4.02 × 10
–3
PSI ⇒ t sat = – 27°F
Pg φ = 100% → Pw = Pg : W = 0.0001 = 0.622 ⎛⎝ --------------------⎞⎠ 0.1 – Pg Pg = 0.0000161 kPa : T < – 60°C
3.3 A room of dimensions 4 m by 6 m by 2.4 m contains an air-water vapor mixture at a total pressure of 100 kPa and a temperature of 25°C. The partial pressure of the water vapor is 1.4 kPa. Calculate: a. humidity ratio b. dew point c. total mass of water vapor in the room 3
V = ( 4 ) ( 6 ) ( 2.4 ) = 57.6 m ; P = 100 kPa ; Pw = 1.4 kPa Pw ( 0.622 ) ( 1.4 ) a) W = 0.622 ----------------- = ------------------------------- = 0.0088 kg ⁄ kg air P – Pw ( 100 – 1.4 ) b) @ Pw = 1.4 kPa t sat = Dew point = 11.8°C c)
m a = PaV ⁄ RaT = ( 98.6 ) ( 57.6 ) ⁄ ( 0.287 ) ( 298.2 ) = 66.4 kg m w = Wm a = ( 0.0088 ) ( 66.4 ) = 0.584 kg
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3.2 One method of removing moisture from atmospheric air is to cool the air so that the moisture condenses or freezes out. Suppose an experiment requires a humidity ratio of 0.0001. To what temperature must the air be cooled at a pressure of 0.1 kPa in order to achieve this humidity?
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34⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
3.4 The air conditions at the intake of an air compressor are 70°F (21.1°C), 50% RH, and 14.7 psia (101.3 kPa). The air is compressed to 50 psia (344.7 kPa), then sent to an intercooler. If condensation of water vapor from the air is to be prevented, what is the lowest temperature to which the air can be cooled in the intercooler?
@ 1 P 1 = 14.7, t 1 = 70°F, φ = 50% Pws 1 = 0.3632 Pw 1 = φPws = 0.1816 2 P 2 = 50 psi 3 P 3 = 50 psi, φ 3 = 100%, t 3 = ? Pw 1 0.1816 W 1 = 0.622 ------------- = 0.622 ⎛⎝ ---------------------------------⎞⎠ = 0.00778 lb/lb Pws 1 14.7 – 0.1816 x W 2 = W 1 = 0.622 ⎛⎝ --------------⎞⎠ = 0.00778 → x = 0.618 psia 50 – x φ 3 = Pw 3 ⁄ Pws 3 = 100 ⇒ Pws 3 = Pw 3 = 0.618 → t 3 = 86°F
a)
m· a = 287 lb m ⁄ min
hw 3 = 22 Btu ⁄ lb min
ha 1 = 21.6 Btu ⁄ lb m
ha 2 = 13.2 Btu ⁄ lb m
hw 1 = 1100 Btu ⁄ lb m
hw 2 = 1085 Btu ⁄ lb m
W 1 = 0.02 lb m ⁄ lb air
W 2 = 0.009 lb m ⁄ lb air
m· 3 = m· w – m· w = ( W 1 – W 2 )m· a = ( 0.02 – 0.005 )287 1
2
m· 3 = 3.16 lb m ⁄ min ° [ ( h – h ) + ( W h – W h ) + ( W – W )h ] b) Q° = m a a a 2 w 1 w 1 2 w 2
1
2
1
3
= 287 [ ( 13.2 – 21.6 ) + ( 0.009 ) ( 1085 ) – ( 0.02 ) ( 1100 ) + ( 0.02 – 0.009 ) ( 22 ) ] Q° = – 5860 Btu ⁄ min
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3.5 Humid air enters a dehumidifier with an enthalpy of 21.6 Btu/lbm of dry air and 1100 Btu/lbm of water vapor. There are 0.02 lbm of vapor per pound of dry air at entrance and 0.009 lbm of vapor per pound of dry air at exit. The dry air at exit has an enthalpy of 13.2 Btu/lbm, and the vapor at exit has an enthalpy of 1085 Btu/lbm. Condensate leaves with an enthalpy of 22 Btu/lbm. The rate of flow of dry air is 287 lbm/min. Determine: a. the amount of moisture removed from the air (lbm/ min) b. the rate of heat removal required
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Chapter 3—Basic HVAC System Calculations⏐35
3.6 Air is supplied to a room from the outside, where the temperature is 20°F (−6.7°C) and the relative humidity is 60%. The room is to be maintained at 70°F (21.1°C) and 50% RH. How many pounds of water must be supplied per pound of air supplied to the room?
t 1 = 20°F
t 2 = 70°F
φ 1 = 60%
φ 2 = 50%
W 2 = 0.0078 lb m ⁄ lb air Pw 1 = φPg 1 = ( 0.6 ) ( 0.0505 ) = 0.0303 P1 ( 0.0303 ) W 1 = 0.62198 --------- = ( 0.62198 ) ------------------------------------- = 0.00128 Pg 1 ( 14.7 – 0.0303 ) mw m a w 1 + m w = m a w 2 → ------- = w 2 – w 1 1 2 ma m ------w- = 0.0078 – 0.0013 = 0.0065 lb m ⁄ ( lb air ) ma
a)
φ 1 = 49%
b)
t i * = 41°F
c)
W = 0.0054 lb m ⁄ lb air
d)
h i = 20.3 Btu ⁄ lb m,
e)
h e = 25.2 Btu ⁄ lb m
f) q = 4.9 Btu ⁄ lb g) φ f = 25%
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3.7 Air is heated to 80°F (26.7°C) without adding water, from 60°F (15.6°C) dry-bulb and 50°F (10°C) wet-bulb temperature. Use the psychrometric chart to find: a. relative humidity of the original mixture b. original dew-point temperature c. original humidity ratio d. initial enthalpy e. final enthalpy f. the heat added g. final relative humidity
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36⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
3.8 Saturated air at 40°F (4.4°C) is first preheated and then saturated adiabatically. This saturated air is then heated to a final condition of 105°F (40.6°C) and 28% RH. To what temperature must the air initially be heated in the preheat coil?
t 2 = 101°F
3.9 Atmospheric air at 100°F (37.8°C) dry-bulb and 65°F (18.3°C) wet-bulb temperature is humidified adiabatically with steam. The supply steam contains 10% moisture and is at 16 psia (110.3 kPa). What is the dry-bulb temperature of the humidified air if enough steam is added to bring the air to 70% RH? 1. Air t = 100F t* = 65F h = 29.8 W = 0.0052
2
3. Air φ = 70%
h f = 184.5
m· s = m· a ( W 3 – W 1 )
2
m· a ( h 3 – h 1 ) = m· s ( h s ) = m· a ( W 3 – W 1 ) ( h s ) 2
2
h3 – h1 Δh h s = --------------------- = --------- = ( 184.5 ) + ( 0.9 ) ( 1152.1 – 184.5 ) W3 – W1 ΔW Δh-------= 1055 Btu ⁄ lb m ΔW
Using Psychrometric Chart
Δh through Pt. 1. along --------- = 1055 and 70% ΔW t db = 96°F
3.10 The conditions on a day in New Orleans, Louisiana, are 95°F (35°C) dry-bulb and 80°F (26.7°C) wet-bulb temperature. In Tucson, Arizona, the air conditions are 105°F (40.6°C) dry-bulb and 72°F (22.2°C) wet-bulb temperature. What is the lowest air temperature that could theoretically be attained in an evaporative cooler at these conditions in these two cities? New Orleans
80°Fwb
Tucson
72°Fwb
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· ma 1 h 1 + m· s h s = m· a h 3
2. Steam x = 0.90 P = 16 psia h g = 1152.1
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Chapter 3—Basic HVAC System Calculations⏐37
3.11 Air at 29.92 in. Hg enters an adiabatic saturator at 80°F dry-bulb and 66°F wet-bulb temperature. Water is supplied at 66°F. Find (without using the psychrometric chart) the humidity ratio, degree of saturation, enthalpy, and specific volume of entering air. Pw 1 s * W s = 0.62198 ⎛⎝ --------------------⎞⎠ P – Pws
Pw 1 s @ t = 66F = 0.31636 PSI
( 0.62198 ) ( 0.31636 ) * W s = ------------------------------------------------- = 0.01368 lb m ⁄ lb aire ( 14.696 – 0.31636 ) *
( 1093 + 0.556t* )W s – 0.24 ( t – t* ) W = ------------------------------------------------------------------------------------1093 + 0.444t – t* (------------------------------------------------------------------------------------------------------------------------1093 + ( 0.556 ) ( 66 ) ) ( 0.01368 ) – ( 0.24 ) ( 80 – 66 -) W = = 0.01044 1093 + ( 0.444 ) ( 80 ) – 66 W 0.01044 μ = ------- = ------------------- = 0.467 Ws 0.01368 h = 0.240t + W ( 1061 + 0.444t ) = ( 0.24 ) ( 80 ) + ( 0.01044 ) ( 1061 + ( 0.444 ) ( 80 ) ) = 30.65 Btu ⁄ lb RaT ( 53.3 ) ( 540 ) v = ---------- ( 1 + 1.6078W ) = ----------------------------- ( 1 + ( 1.6078 ) ( 0.01044 ) ) P ( 14.7 ) ( 144 ) 3
v = 13.832 ft ⁄ lb m
3.12 An air-water vapor mixture enters an air-conditioning unit at a pressure of 150 kPa, a temperature of 30°C, and a relative humidity of 80%. The mass flow of dry air entering is 1 kg/s. The air-vapor mixture leaves the airconditioning unit at 125 kPa, 10°C, 100% RH. The moisture condensed leaves at 10°C. Determine the heat transfer rate for the process.
3.4 W 1 = 0.622 ⎛ ----------------------⎞ = 0.0144 ⎝ 150 – 3.4⎠ 1.228 W 2 = 0.622 ⎛ ----------------------------⎞ = 0.0062 ⎝ 125 – 1.228⎠ m a ha 1 + m a W 1 h w – m a ha 2 – m a W 2 h w – m a ( W 1 – W 2 )h f + Q = 0 1
2
3
m a [ ( ha 1 – ha 2 ) + W 1 hg 1 – W 2 hg 2 – ( W 1 – W 2 )h f ] + Q = 0 3
Q = – 1 [ ( 1.0035 ) ( 20 ) + ( 0.0144 ) ( 2556.3 ) – ( 0.0062 ) ( 2519.8 ) – ( 0.0144 – 0.0062 ) ( 42.01 ) ] Q = – 41.0 kJ/s = – 41.0 W
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Pw 1 = φPg 1 = ( 0.8 ) ( 4.25 ) = 3.4 kPa
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38⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
3.13 Air at 40°C, 300 kPa, with a relative humidity of 35% is to be expanded in a reversible adiabatic nozzle. How low a pressure can the gas be expanded to if no condensation is to take place? What is the exit velocity at this condition?
S1 = S2
Isentreopic
Pw 1 = φ 1 Pg 1 = ( 0.35 ) ( 7.384 ) = 2.584 kPa 2.584 w 1 = 0.622 ⎛ ----------------------------⎞ = 0.0054 ⎝ 300 – 2.584⎠ Pg 2 W 1 = W 2 if no condensate. ⇒ 0.0054 = 0.622 ⎛ ---------------------⎞ ⎝ P 2 – Pg 2⎠ P 2 = 116.18Pg 2 P2 S 2 = S 1 ; T 2 = T 1 ⎛ ------⎞ ⎝ P 1⎠
k---------– 1k
116.18Pg 2 0.286 = 313.2 ⎛ --------------------------⎞ ⎝ 300 ⎠
Trial and Error → Try T 2 = 17.2°C = 290.4 K ; Pg 2 = 1.984 ? ( 116.18 ) ( 1.984 ) 0.286 290.4 = 313.2 ⎛⎝ ----------------------------------------⎞⎠ = 290.5 300 P 2 = ( 116.18 ) ( 1.984 ) = 230.5kPa 2
2
2
V2 V1 ⎞ ⎛ V2 ⎞ ⎛ ⎜ h 1 + --------⎟ – ⎜ h 2 + --------⎟ = 0 ; -------------------- = 1.0035 ( 313.2 – 290.4 ) 2g 2g 2 ( 1000 ) ⎝ c⎠ ⎝ c⎠ V 2 = 214 m/s
v = va + μ ( vs – va ) v = ( 1 – μ )v a + μv s W W = ⎛⎝ 1 – -------⎞⎠ v a + ⎛⎝ -------⎞⎠ ( v s ) Ws Ws
Ra T ⎧ ⎫ v a = --------⎪ ⎪ P ⎪ ⎪ ⎪ ⎪ Ra T ⎪ v s = -----------------⎪ P – P ws ⎪ ⎪ ⎨ ⎬ W⎪ ⎪ -----= μ ⎪ ⎪ Ws ⎪ ⎪ ⎪ ⎪ P ⎪ w = 0.622 ---------------- ⎪ P – P ⎩ w⎭
P – Pw Ra T Ra T P – P ws P – P ws Pw Pw W Ra T W = 1 – ⎛ ----------------⎞ ⎛ ----------------⎞ --------- + ------- ------------------ = ------------------ 1 – ⎛ ----------------⎞ ⎛ ------------------⎞ ------------------ + ------⎝ P – P w⎠ ⎝ P ws ⎠ P ⎝ P – P w⎠ ⎝ P ws ⎠ W s P – P ws P – P ws P Ws Pw P – P ws P – P ws Pw R a T P – P ws = ------------------ ------------------ – ⎛⎝ -------------⎞⎠ ⎛⎝ ------------------⎞⎠ ( P – P ws ) + ⎛⎝ ----------------⎞⎠ ⎛⎝ ------------------⎞⎠ P – Pws P P ws P ws P P – P w P – Pw 2
R a T ( P – Pws ) ( P ws ) ( P – P w ) – P w ( P – P ws ) + P w ( P – P ws )P = ------------------ --------------------------------------------------------------------------------------------------------------------------------------------P – P ws P ( P ws ) ( P – P w ) Ra T R a T PP ws ( P – Pws ) = ------------------ ------------------------------------- = ---------------P – P ws PP ws ( P – P w ) P – Pw
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3.14 By using basic definitions and Dalton’s Law of partial pressure, show that v = RaT/(p – pw)
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Chapter 3—Basic HVAC System Calculations⏐39
3.15 In an air-conditioning unit, 71,000 cfm at 80°F dry bulb, 60% RH, and standard atmospheric pressure, enter the unit. The air leaves the unit at 57°F dry bulb and 90% relative humidity. Calculate the following: a. cooling capacity of the air-conditioning unit, in Btu/h b. rate of water removal from the unit c. sensible heat load on the conditioner, in Btu/h d. latent heat load on the conditioner, in Btu/h e. the dew point of the air leaving the conditioner 1.
2.
t = 80F
t = 51F
3. h ≈ 25
φ = 60%
φ = 90% v = 13.2
( h f @ 57°F )
71000 CFM v = 13.9
h = 23.5
h = 33.7
W = 0.009
W = 0.0132
a)
b) c)
13.33 SCFM = 71000 ⎛⎝ -------------⎞⎠ = 67900 h c = 29.0 3 13.9 V· ( 71000 ) ( 60 ) m· a = --- = ------------------------------- = 306475 lbm air ⁄ h v 13.9 m· a [ ( h 1 – h 2 ) – ( W 1 – W 2 )h 3 ] = – Q – Q = ( 306475 ) [ ( 33.7 – 23.5 ) – ( 0.0132 – 0.009 )25 ] Q = – 3,093,900 Btu/h = 257.8 tons · m w = m· a ( W 1 – W 2 ) = ( 306,475 ) ( 0.0132 – 0.009 ) = 1287 lb/h Q = m· c ( t – t ) = ( 306,475 ) ( 0.244 ) ( 80 – 57 ) s
a p
1
2
= 1,719,900 Btu/h = 143.3 tons - or -
e)
- or · = 4840 ( SCFM ) ( Δw ) = ( 4840 ) ( 67,900 ) ( 0.0132 – 0.009 ) = 1,380,000 Btu/h = 115.0 tons td = 54.2°F
3.16 Four pounds of air at 80°F (26.7°C) dry bulb and 50% RH are mixed with one pound of air at 50°F (15.6°C) and 50% RH. Determine a. relative humidity of the mixture b. dew-point temperature of the mixture
by graphical solution φ mix = 52% t dp = 55.5°F
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d)
( 1.1 ) ( 67,900 ) ( 23 ) = 1.1 ( SCFM ) ( t 1 – t 2 ) = --------------------------------------------- = 143.2 tons 12000 Q L = m· w ( h fg ) = ( 1287 ) ( 1076 ) = 1,385,000 = 115.4 tons
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40⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
3.17 Air is compressed from 85°F, 60% RH, 14.7 psia to 60 psia and then cooled in an intercooler before entering a second stage of compression. What is the minimum temperature to which the air can be cooled so that condensation does not take place? Pw 2 W 1 = W 2 = 0.62198 -------------------P – Pw 2
P w = φP g = ( 0.6 ) ( 0.596 ) = 0.3576
( 0.62198 ) ( 0.3576 ) W = ---------------------------------------------- = 0.0155 lb/lb ( 14.696 – 0.3576 ) Ps 0.0155 = 0.62198 ⎛ ------------------⎞ ⇒ P s = 1.459 psia ⎝ 60 – Ps⎠ ⇒ t d = 114.5°F
3.18 An air-water vapor mixture flowing at a rate of 4000 cfm (1890 L/s) enters a perfect refrigeration coil at 84°F (28.9°C) and 70°F (21.1°C) wet-bulb temperature. The air leaves the coil at 53°F (11.7°C). How many Btu/h of refrigeration are required? 1. t = t* = v = h = W =
2. t = 53F saturated h = 22 h f = 21.1
84F 70F 13.97 34 0.01266
W = 0.00857
2
( 4000 ) ( 60 ) Q = ---------------------------- [ ( 34 – 22 ) – ( 0.01266 – 0.00857 ) ( 21.1 ) ] ( 13.97 ) = 204700 Btu/h = 17.1 tons
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m· a [ – ( h 1 – h 2 ) – ( W 1 – W 2 ) h f ] = Q
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Chapter 3—Basic HVAC System Calculations⏐41
3.19 Air at 40°F dry bulb and 35°F wet bulb is mixed with air at 100°F dry bulb and 77°F wet bulb in the ratio of 2 lb of cool air to 1 lb of warm air. Compute the resultant humidity ratio and enthalpy of the mixed air. t 1 = 40F *
t 2 = 100F *
t 1 = 35F m· 1 = 2m·
t 2 = 77F m· 2 = m·
W 1 = 0.0031
W 2 = 0.0148
h 1 = 13.1
h 2 = 40.5
m· 1 h 1 + m· 2 h 2 = m· h 3 → 2m· h 1 + m· h 2 ( 13.1 ) + 40.5- = 2---------------------------------= 22.23 Btu/lb h 3 = ----------------------------3m· 3 · m· 1 W 1 + m· 2 W 2 = m 3 W 3 → 2m· W 1 + m· W 2 ( 0.0031 ) + ( 0.0148 -) - = 2--------------------------------------------------w 3 = --------------------------------= 0.0070 lb/lb air 3m· 3
m· o h o + m· R h R = ( m· o + m· R )h mix ( 1000 ) ( 41.4 ) + ( 5000 ) ( 28.5 ) h mix = ----------------------------------------------------------------------- = 30.65 Btu/lb 6000 · · m o W o + m R W R = ( m· o + m· R )W mix ( 1000 ) ( 0.018 ) + ( 5000 ) ( 0.0096 ) w mix = -------------------------------------------------------------------------------- = 0.011 lb/lb air 6000 *
From Psych. Chart → t mix = 78.3°F ; t mix = 65.8°F
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3.20 Outdoor air at 90°F (32.2°C) and 78°F (25.6°C) wet bulb is mixed with return air at 75°F (23.9°C) and 52% RH. There are 1000 lb (454 kg) of outdoor air for every 5000 lb (2265 kg) of return air. What are the dry- and wetbulb temperatures for the mixed airstream?
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42⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
3.21 In a mixing process of two streams of air, 10,000 cfm of air at 75°F and 50% RH mix with 4000 cfm of air at 98°F dry-bulb and 78°F wet-bulb temperature. Calculate the following conditions after mixing at atmospheric pressure: a. dry-bulb temperature b. humidity ratio c. relative humidity d. enthalpy e. dew-point temperature From Psych. Chart as graphical solution a) t dB = 81.5°F b) W = 0.0112 lb/lb c) φ = 48.0% d) h = 32.0 Btu/lb e) t DP = 60.5°F
Pw 2 = Pg 2 = 3.169 kPa
a)
ma [ ( h1 – h2 ) + w1 hw
1
0.622 ( 3.169 ) W 2 = --------------------------------- = 0.0204 ( 100 – 3.169 ) – w 2 h w + ( W 2 – W 1 )h f ] = 0 2
2
c p ( T 2 – T 1 ) + W 2 ( hg 2 – h f ) 2 W 1 = ------------------------------------------------------------------hg 1 – h f 2
( 1.0035 ) ( 25 – 30 ) + 0.0204 ( 2442.3 ) w 1 = ----------------------------------------------------------------------------------------- = 0.0183 kg/kg 2556.3 – 104.89 Pw 1 0.0183 = 0.62189 ----------------------⇒ P w = 2.86 kPa 1 100 – P w 1
Pw 2.86 φ = --------1 = ------------- = 0.67 ⇒ 67% Pg 4.246 1
b)
φ = 67% ; W = 0.018 kg/kg ( for 101.3 kPa )
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3.22 Solve the following: a. Determine the humidity ratio and relative humidity of an air-water vapor mixture that has a dry-bulb temperature of 30°C, an adiabatic saturation temperature of 25°C, and a pressure of 100 kPa. b. Use the psychrometric chart to determine the humidity ratio and relative humidity of an air-water vapor mixture that has a dry-bulb temperature of 30°C, a wet-bulb temperature of 25°C, and a pressure of 100 kPa.
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Chapter 3—Basic HVAC System Calculations⏐43
3.23 An air-water vapor mixture at 100 kPa, 35°C, and 70% RH is contained in a 0.5 m3 closed tank. The tank is cooled until the water just begins to condense. Determine the temperature at which condensation begins and the heat transfer for the process. P w = φP g = 0.7 ( 5.628 ) = 3.94 kPa 1
1
@ dew point : P w = P g ; W = constant ; v = constant 2
2
Pw v m w RT 2 3.94T 2 2 ----------- = ----------------; P w = P g = ---------------2 2 m w RT 1 308.2 P w1 v = 0.0128T 2 Trial and Error Try T 2 = 25°C : 4.246 ≠ 0.0128 ( 298 ) = 3.81 T 2 = 30°C : 3.169 ≠ 0.0128 ( 303 ) = 3.88 By Interpolation T 2 = 28.2°C 0.622 ( 3.94 ) W 2 = W 1 = ------------------------------ = 0.0255 kg/kg ( 100 – 3.94 ) Pa V ( 100 – 3.94 ) ( 0.5 ) m a = ---------- = ------------------------------------------- = 0.543 kg Ra T ( 0.287 ) ( 308.2 ) Q = m ( u2 – u1 ) = ma cv ( T2 – T1 ) + ma W2 ug – ma W1 ug 2
1
= ( 0.543 ) [ 0.7165 ( 28.2 – 35 ) + 0.0255 ( 2414.2 – 2423.4 ) ] = – 277 kJ
Qs + Σm· w h w Δh100000 ) + 35 ( 1100 -) ------- = (--------------------------------------------------= -----------------------------Δw 35 Σm· w = 3957 Btu/lb m From Psychrometric Chart φ supply = 90% t dp = 36°F m· a h 1 + Q s + m· w h w = m· a h 2 Qs + mw hw ( 100000 ) + ( 35 ) ( 1100 ) m· a = --------------------------- = --------------------------------------------------------- = 11260 lb/hr ( 26.6 – 14.3 ) h2 – h1
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3.24 A room is to be maintained at 76°F and 40% RH. Supply air at 39°F is to absorb 100,000 Btu sensible heat and 35 lb of moisture per hour. Assume the moisture has an enthalpy of 1100 Btu/lb. How many pounds of dry air per hour are required? What should the dew-point temperature and relative humidity of the supply air be?
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44⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
3.25 Moist air enters a chamber at 40°F dry-bulb and 36°F wet-bulb temperature at a rate of 3000 cfm. In passing through the chamber, the air absorbs sensible heat at a rate of 116,000 Btu/h and picks up 83 lb/h of saturated steam at 230°F. Determine the dry-bulb and wet-bulb temperatures of the leaving air. V· a = 3000 cfm Q s + m· w h w 116000 ) + 83 ( 1157 -) Δh- = (--------------------------------------------------------= -------------------------m· w 83 Δw
m· a
1
= 2555 Btu/lbm V· 1 3000 = ------ = ------------ = 236.2 lb/min v1 12.7
83 m· a = m a + m w = 236.2 + ------ = 237.6 lb/min 2 1 1 60 116000 Qs = m· c p ΔT ⇒ Δt = ------------------------------------------------ = 34°F ( 0.240 ) ( 257.6 ) ( 60 ) t 2 = 40 + 34 = 74°F From Psych. Chart t wb = 64°F
a)
b)
Q s = m· a c p ( t R – t s )
350,000 m· a = ----------------------------------------- = 143,443 lb/h ( 0.244 ) ( 77 – 67 ) m· w 1,000,000 W s = W R – ------- = 0.0109 – ----------------------------------------- = 0.0099 lb/lb ma ( 7000 ) ( 143,443 ) φ = 70% ; t dp = 57°F
c)
( 1,000,000 ) Q L = m· w h w = ---------------------------- ( 1100 ) ≈ h g ( 7000 ) d)
90°F (people)
= 157,100 Btu/h
( 350,000 ) ( SHR ) = -------------------------------------------------- = 0.69 ( 350,000 ) + 157,000
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3.26 In an auditorium maintained at a temperature not to exceed 77°F, and a relative humidity not to exceed 55%, a sensible-heat load of 350,000 Btu and 1,000,000 grains of moisture per hour must be removed. Air is supplied to the auditorium at 67°F. a. How much air must be supplied, in lb/h? b. What is the dew-point temperature of the entering air, and what is its relative humidity? c. How much latent heat is picked up in the auditorium? d. What is the sensible heat ratio?
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Chapter 3—Basic HVAC System Calculations⏐45
3.27 A meeting hall is maintained at 75°F dry bulb and 65°F wet bulb. The barometric pressure is 29.92 in. Hg. The space has a load of 200,000 Btu/h sensible, and 200,000 Btu/h latent. The temperature of the supply air to the space cannot be lower than 65°F dry bulb. a. How much air must be supplied, in lb/h? b. What is the required wet-bulb temperature of the supply air? c. What is the sensible heat ratio? a) b)
Qs 200,000 m· a = ---------------- = ----------------------------------------- = 81,967 lb/h c p ( Δt ) ( 0.244 ) ( 75 – 65 ) h v ≅ h g @ 85-95°F = 1100 Btu/lb m· w ( 200,000 ) - = 0.011 – ------------------------------------= 0.0088 lb/lb W s = W R – -----m· ( 1100 ) ( 81,967 ) a
c)
From Psych. Chart → t = 65°F ; t* = 58°F ( 200,000 ) SHR = -------------------------------------------------- = 0.50 ( 200,000 + 200,000 )
3.28 A structure to be air conditioned has a sensible heat load of 20,000 Btu/h at a time when the total load is 100,000 Btu/h. If the inside state is to be at 80°F, 50% RH, is it possible to meet the load conditions by supplying air to the room at 100°F and 60% RH? If not, discuss the direction in which the inside state would be expected to move if such air were supplied.
3.29 A flow rate of 30,000 lb/h of conditioned air at 60°F and 85% RH is added to a space that has a sensible load of 120,000 Btu/h and a latent load of 30,000 Btu/h. a. What are the dry- and wet-bulb temperatures in the space? a)
Q s = m· c p ( t R – t s )
b.
If a mixture of 50% return air and 50% outdoor air at 98°F dry bulb and 77°F wet bulb enters the air conditioner, what is the refrigeration load?
Q L = m· ( w R – w s ) ( 1100 )
120,000 = ( 30,000 ) ( 0.24 ) ( t R – 60 ) t R = 76.6°F
30,000 = ( 30,000 ) ( w R – 0.0094 ) ( 1100 ) W R = 0.01031
⇒ h R = 29.7Btu/lb t wb = 64.5°F
m· a h a + m· R h R – ( m· a + m· R )h m ( 40.4 ) + ( 29.7 ) h m = ------------------------------------- = 35.05 Btu/lb m 2 ≈0 m· a ( h m – h s ) – m· a ( w m – w s )h f + Q = 0 h s = 24.7 Btu/lb m 60°F
Q = – ( 30,000 ) ( 35.05 – 24.7 ) = – 310,500 Btu/h = 25.9 tons
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20,000 SHR = ------------------- = 0.2 100,000 Not possible to meet conditions with supply since SHR doesn't pass through both room and supply conditions. If used room becomes warmer and/or less humid.
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46⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
3.30 An air-water vapor mixture enters a heater-humidifier unit at 5°C, 100 kPa, 50% RH. The flow rate of dry air is 0.1 kg/s. Liquid water at 10°C is sprayed into the mixture at the rate of 0.0022 kg/s. The mixture leaves the unit at 30°C, 100 kPa. Calculate: a. the relative humidity at the outlet b. the rate of heat transfer to the unit a)
b)
mw 0.0022 W 2 = W 1 + ------- = 0.0027 + ---------------- = 0.0247 m a ( 5°C,50% ) 0.1 and 30°C φ 2 = 91%
h 1 = 12.7 kJ/kg
m a [ h 1 – h 2 + ( W 2 – W 1 )h f ] + Q = 0 3
h 2 = 93.0 kJ/kg
Q = – 0.1 [ 12.7 – 93 + ( 0.0022 )42 ] = 7.94 kJ/s = 7.94 kW
h f = 42.0 kJ/kg 3
3.31 A room is being maintained at 75°F and 50% RH. The outside air conditions are 40°F and 50% RH at this time. Return air from the room is cooled and dehumidified by mixing it with fresh ventilation air from the outside. The total airflow to the room is 60% outdoor and 40% return air by mass. Determine the temperature, relative humidity, and humidity content of the mixed air going to the room.
( 0.4 ) ( 28.2 ) + ( 0.6 ) ( 12.5 ) h m = -------------------------------------------------------------(1) h m = 18.78Btu/lb ( 0.4 ) ( 0.0092 ) + ( 0.6 ) ( 0.0026 ) W m = -------------------------------------------------------------------------(1) = 0.00524 lb/lb
m· R W R + m· OA W OA = ( m· R + m· OA )W m ⇒ Wm
From Psychrometric Chart
t m = 54.5°F
; φ m = 58%
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Return Outside Air t = 75°F t = 40°F φ = 50% φ = 50% h = 28.2 h = 12.5 W = 0.0092 W = 0.0026 m· R h R + m· OA h OA = ( m· R + m· OA )h m
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Chapter 3—Basic HVAC System Calculations⏐47
3.32 A room with a sensible load of 20,000 Btu/h is maintained at 75°F and 50% RH. Outdoor air at 95°F and 80°F wet bulb was mixed with the room return air. The outdoor air, which is mixed, is 25% by mass of the total flow going to the conditioner. This air is then cooled and dehumidified by a coil and leaves the coil saturated at 50°F, which is on the condition line for the room. The air is then mixed with some room return air so that the temperature of the air entering the room is at 60°F. Find the following: a. the air-conditioning processes on the psychrometric chart b. ratio of latent to sensible load c. airflow rate d. the percent by mass of room return air mixed with air leaving the cooling coil
b)
Qs from Load line ------------------- = 0.78 Qs + QL
c)
d)
Room:
mR h3 + QT = mR hR
20000 + 5640 m R = --------------------------------- = 5455 lb m /h ( 28.2 – 23.5 ) 2 to 3: ( m R – m 1 )h 2 + ( m 1 )h R = ( m R )h 3
h2 – h3 20.3 – 23.5 m 1 = ----------------- m R = --------------------------- m R = 0.405 m R h2 – hR 20.3 – 28.2 m 1 = 40.5% of m R
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Q s – 0.78Q s 20,000 Q L = ----------------------------- = ---------------- – 20,000 = 5640 0.78 0.78 QL 5640 ------- = --------------= 0.282 Qs 20,000
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48⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
3.33 An air-water vapor mixture at 14.7 psia (101.5 kPa), 85°F (29.4°C), and 50% RH is contained within a 15 ft3 (0.425 m3) tank. At what temperature will condensation begin? If the tank and mixture are cooled an additional 15°F (8.3°C), how much water will condense from the mixture? v 1 = 15 ft
3
v 2 = 15ft
3
t 1 = 85°F
t2 = ?
φ 1 = 50%
φ 2 = 100%
P 1 = 14.7 psia
P2 = ?
p s1 = ( 50 ) ( 0.596 ) = 0.288psi
p s2 = p g
2
m w2 m w1 W 1 = W 2 ⇒ ---------- = ---------- ⇒ m w1 = m w2 m a1 m a2 pg ps 0.298 p s v = m w R w T ⇒ ----- 1 = ------------- = 0.000547 = -------2 T1 T2 545 Trial and error produces t 2 = 63.2°F ; p g = 0.287 psia 2
t 3 = 63.2 – 15.0 = 48.2°F ⇒ p g = 0.0167 psia s
0.298 W 1 = 0.622 ⎛⎝ ------------------------------⎞⎠ = 0.0129 lb/lb 14.7 – 0.298
T3 508.2 p a = p a ----- = 14.4 ⎛⎝ -------------⎞⎠ = 13.42 psia 3 1T 545 1
0.167 W 3 = 0.622 ⎛ -------------⎞ = 0.00774 lb/lb ⎝ 13.42⎠ pa v1 ( 14.4 ) ( 144 ) ( 15 ) 1 m a = -----------= ---------------------------------------- = 1.070 lb T1 ( 53.3 ) ( 545 )
m cond = 0.00552 lb m
V1 1000 m· a = ------ = ---------------- = 70.08 lb/min v1 14.268 · Q· = m a [ ( ha 2 – ha 1 ) + ( W 2 hw 2 – W 1 hw 1 ) + ( W 1 – W 2 )h ws ] = ( 70.08 ) ( 60 ) [ [ 0.24 ( 40 – 90 ) ] – [ 0.0183 ( 1100 ) – 0.0052 ( 1074 ) ] + [ ( 0.0052 – 0.0183 ) ( 8.04 ) ] ] = 111,840 Btu/h = 9.3 tons
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3.34 Air flowing at 1000 cfm and at 14.7 psia, 90°F, and 60% RH passes over a coil with a mean surface temperature of 40°F. A spray on the coil assures that the leaving air is saturated at the coil temperature. What is the required cooling capacity of the coil?
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Chapter 3—Basic HVAC System Calculations⏐49
3.35 An air-vapor mixture at 100°F (37.8°C) dry bulb contains 0.02 lb water vapor per pound of dry air (20 g/kg). The barometric pressure is 28.561 in. Hg (96.7 kPa). Calculate the relative humidity, dew-point temperature, and degree of saturation. pw W = 0.622 ⎛ ---------------⎞ = 0.2 ⇒ p w = 0.435 psi ⎝ p – p w⎠ From Chapter 1
p w = 0.949 psi , W s = 0.0432 s
pw 0.435 φ = ------- = ------------- = 45.8% pw 0.949 s
@ p w = 0.435
t dp = 75.3°F
0.02 W μ = ------- = ---------------- = 0.463 0.0432 Ws
3.36 Air enters a space at 20°F and 80% RH. Within the space, sensible heat is added at the rate of 45,000 Btu/h and latent heat is added at the rate of 20,000 Btu/h. The conditions to be maintained inside the space are 50°F and 75% RH. What must the air exhaust rate (lb/h) from the space be to maintain a 50°F temperature? What must the air exhaust rate (lb/h) from the space be to maintain a 75% RH? Discuss the difference. Qs = Gs [ cp ( t 0 – t1 ) ]
b)
45,000 G s = ------------------------ = 6250 lb/h 0.245 ( 30 ) Q L = G L ( ΔW ) ( 1060 )
20,000 G L = ------------------------------ = 4740 lb/hr 1060 ( 0.004 ) c) Supply conditions cannot maintain design for given Q s and Q L
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a)
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50⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
3.37 Moist air at a low pressure of 11 psia is flowing through a duct at a low velocity of 200 fpm. The duct is 1 ft in diameter and has negligible heat transfer to the surroundings. The dry-bulb temperature is 85°F and the wet-bulb temperature is 70°F. Calculate the following: a. humidity ratio, lb/lb b. dew-point temperature, °F c. relative humidity, %
a)
t = 85 t* = 70 ( 1093 – 0.556t* )W s – 0.24 ( t – t* ) W = --------------------------------------------------------------------------------- = 0.0177 lb/lb 1093 + 0.444t – t* pw ( 0.739 ) ( 0.491 ) * W s = 0.622 --------------*- = -------------------------------------------- = 0.0212 psia 11 – 0.739 ( 0.491 ) p – pw
b)
pw W = 0.622 ------------------ = 0.0177 ⇒ p w = 0.1915 psi = 0.39 in. Hg 11 – p w From Table 2-1:
c)
t dp = 52°F
pw 0.39 φ = -------- = ------------- = 32% p ws 1.213
3.38 If an air compressor takes in moist air (at about 90% RH) at room temperature and pressure and compresses this to 120 psig (827 kPa) (and slightly higher temperature), would you expect some condensation to occur? Why? If yes, where would the condensation form? How would you remove it?
condensation occurs in discharge pipe when exiting discharge valve. Remove moisture in an aftercooler.
3.39 Does a sling psychrometer give an accurate reading of the adiabatic saturation temperature? Explain. Normally within 1°F with a shielded thermometer in air-water vapor mixtures, because Lewis Relation is approximately equal to one.
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Yes. The air is heavily saturated @ Inlet, the Compression Process would lower t dewpoint ,
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Chapter 3—Basic HVAC System Calculations⏐51
3.40 An air processor handles 2000 cfm of air with initial conditions of 50°F and 50% RH. The air is heated with a finned heat exchanger with 78 ft2 of heat transfer surface area and a UA value of 210 Btu/h·°F. Also, a steam spray system adds moisture to the air from saturated steam at 16 psia. The outlet air is at 100°F and 50% RH. Do the following: a.
Show the processes on the psychrometric chart.
b. c. d.
Calculate the mass flow rate, lb/min. Calculate the pounds per minute of steam required. Calculate the heat added by the coil, Btu/min.
V· 2000 m· = --- = ------------ = 155 lb/min v 12.9 m a ( w 1 ) + m w = m a w 2 ⇒ m w = m a ( ΔW ) b)
c)
m w = 155 ( 0.0208 – 0.0038 ) = 2.65 lb/min d)
Q s = m a ( h 2 – h 1 ) – m w ( h w ) = 155 ( 475 – 16.4 ) – 2.65 ( 1152 ) Q s = 1900 Btu/min
@ 5000 ft @ Sea Level
%RH = 53 %RH = 51.5
h = 34.9 Btu/lb h = 31.65 Btu/lb
3.42 The average person gives off sensible heat at the rate of 250 Btu/h and perspires and respires about 0.27 lb/h of moisture. Estimate the sensible and latent load for a room with 25 people in it (the lights give off 9000 Btu/h). If the room conditions are to be 78°F and 50% RH, what flow rate of air would be required if the supply air came in at 63°F? What would be the supply air relative humidity? Qs
people
= ( 250 ) ( 25 ) = 6250 Btu/h
Qs
total
= 6250 + 9000 = 15250
QL
people
= ( 0.27 ) ( 1100 ) ( 25 ) = 7430 Btu/h
QL
total
= 7430
Q s = m· c p Δt
Q L = m· h fg ( W R – W s )
15250 m· = ----------------------------- = 69.5 lb/min ( 0.244 ) ( 15 )
7430 = 69.5 ( 60 ) ( 1100 ) ( 0.0102 – W s ) w s = 0.00850 lb/lb @ 63°F φ = RH = 70%
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3.41 At an altitude of 5000 ft (1500 m), a sling psychrometer reads 80°F (26.7°C) and 67°F (19.4°C) wet bulb. Determine correct values of relative humidity and enthalpy from the chart. Compare these to the corresponding values for the same readings at sea level.
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52⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
3.43 A space in an industrial building has a winter sensible heat loss of 200,000 Btu/h and a negligible latent heat load (latent losses to outside are made up by latent gains within the space). The space is to be maintained precisely at 75°F and 50% RH. Due to the nature of the process, 100% outside air is required for ventilation. The outdoor air conditions can be taken as saturated air at 20°F. The amount of ventilation air is 7000 scfm and the air is to be preheated, humidified with an adiabatic saturator to the desired humidity, and then reheated. The temperature out of the adiabatic saturator is to be maintained at 60°F dry bulb. Determine the following: a.
temperature of air entering the space to be heated, °F
b. c. d.
heat supplied to preheat coil, Btu/h heat supplied to reheat coil, Btu/h amount of water required for humidification, gpm
Qs = 200,000 t3 = 60°F W3 = w4 = 0.0093 t2 = 91.8°F
c)
Q RH = 1.1 ( CFM ) ( Δt ) = ( 1.1 ) ( 7000 ) ( 101 – 60 ) = 315,700 Btu/h
d)
7000 0.0093 – 0.002152 m w = m a ΔW = ⎛ -------------⎞ ⎛ ---------------------------------------------⎞ = 0.45 gal/min ⎝ 13.33⎠ ⎝ ⎠ 8.33
3.44 Using the SI psychrometric chart at standard atmospheric pressure, find a.
dew point and humidity ratio for air at 28°C dry bulb and 22°C wet bulb
b.
enthalpy and specific volume a) W = 0.01403 kg/kg b) h = 64.7 kJ/kg
t dp = 19.5°C 3 v = 0.872 m /kg
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b)
200,000 Q s = 1.1 ( CFM )Δt ⇒ t 4 = ------------------------ + 75 = 101°F 1.1 ( 7000 ) Q PH = 1.1 ( CFM ) ( t 2 – t 1 ) = ( 1.1 ) ( 7000 ) ( 91.8 – 20 ) = 552,900 Btu/hr
a)
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Chapter 3—Basic HVAC System Calculations⏐53
3.45 Using the SI chart, find: a.
moisture that must be removed in cooling air from 24°C dry bulb, 21°C wet bulb to 13°C dry bulb, saturated
b.
total, sensible, and latent heat removal for the process.
1.
24°C db 21°C wb
2.
W 1 = 0.0145 kg/kg h 1 = 61 kJ/kg
Assume Saturated 13°C db W 2 = 0.00903 kg/kg h 2 = 36.8 kJ/kg
a.
Wa = W2 ta = t1 h a = 48 kJ/kg
mm ------- = ( W 1 – W 2 ) = ( 0.0145 – 0.00903 ) = 0.00547 kg/kg ma q t = h 1 – h 2 = 61 – 36.8 = 24.2 kJ/kg q s = h a – h 2 = 48 – 36.8 = 11.2 kJ/kg q L = h 1 – h a = 61 – 48 = 13 kJ/kg
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54⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
3.46 An air-conditioned space has a sensible heat load of 200,000 Btu/h, a latent load of 50,000 Btu/h, and is maintained at 78°F dry bulb and 60% RH. On a mass basis, 25% outside air is mixed with return air. Outside air is at 95°F dry bulb and 76°F wet bulb. Conditioned air leaves the apparatus and enters the room at 60°F dry bulb. The fan must produce a pressure increase of 3.5 in. water to overcome the system pressure loss. Fan efficiency is estimated as 55%. a. Draw and label the schematic flow diagram for the complete system. (Hint: See Fig. 3-1) b. Complete the table below. c. Plot and draw all processes on a psychrometric chart. d. Specify the fan size, scfm, and fan motor rating, HP. e. Determine the size refrigeration unit needed, in Btu/h and tons. f. What percent of the required refrigeration is for (1) sensible cooling and (2) for dehumidification? g. What percent of the required refrigeration is due to the outside air load?
Dry Bulb t, °F
φ, %
Enthalpy h, Btu/lb
W, lb/lb
ma, lb/h
scfm
acfm
vact
OA
95
42
39.4
0.0150
11380
2528
2710
14.3
r
78
60
32.2
0.0123
34160
7590
7860
13.8
m
82.3
55
34.0
0.0130
45540
10120
10550
13.9
f
84.6
50
34.6
0.0130
45540
10120
10630
14.0
s
60
100
26.7
0.0113
45540
10120
10130
13.35
b) 0.75 ( 32.2 ) + 0.25 ( 39.4 ) = 34.0 0.75 ( 0.0123 ) + 0.25 ( 0.0150 ) = 0.0130 0.75 ( 78 ) + 0.25 ( 95 ) = 82.3 mw 50,000 ⁄ 1100 W s = W r – ------- = 0.0123 – -------------------------------- = 0.0113 ma 45540 200,000 m a = ----------------------------------------- = 45,540 lb/h ⇒ 10120 cfm ( 0.244 ) ( 78 – 60 ) 10120 ( 3.5 ) d) w f = --------------------------- = 10.1 HP 0.55 ( 6350 ) 10.1 ( 2545 ) h f = 34 + --------------------------- = 34.6 ; W f = 0.0130 ; t f = 84.6 45540 e) q = 45540 [ 34.6 – 26.7 – ( 0.0130 – 0.0113 )28 ] = 357,600 Btuh = 29.8 tons 45540 ( 0.244 ) ( 84.6 – 60 ) f) % sensible = ------------------------------------------------------------- × 100 = 76.4% 357,600 2528 [ 1.10 ( 95 – 78 ) + 4840 ( 0.015 – 0.0123 ) ] g) % due to OA = -------------------------------------------------------------------------------------------------------------- × 100 = 22.5% 357,600
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Point
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Solutions to
Chapter 4 DESIGN CONDITIONS
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Chapter 4—Design Conditions⏐57
4.8 For the person in Problem 4.7 (5 ft, 5 in., 120 lb), compute the body surface area (ft2). A D = 0.108m
Eq. (4-1): W = 120 lb
0.425 0.725
l
l = 65 in.
A = 0.108 ( 120 )
0.425
( 65 )
0.725
= 17.1 ft
2
4.12 The living room in a home is occupied by adults at rest wearing medium clothing. The mean radiant temperature is 18°C (64°F). Determine the air temperature necessary for comfort. [Ans: 30°C (86°F)] t air + MRT t o ≈ 75°F ≅ --------------------------2
( Std. 55: t o = at a + ( 1 – a )t r ) a = 0.5
t air = 2 ( 75 ) – 64 = 86°F = 30°C
4.13 A room has a net outside wall area of 275 ft2 with a surface temperature of 54°F, 45 ft2 of glass with a surface temperature of 20°F, 540 ft2 of ceiling with a surface temperature of 60°F, 670 ft2 of partitions with a surface temperature of 70°F, and 540 ft2 of floor with a surface temperature of 70°F. If the air movement is 20 fpm and light clothing is being worn, determine the air temperature necessary for comfort. Assume occupants have equal view (i.e., all angle factors are identical) of all surfaces.
∑ A i = 2070
i =1
275 45 540 670 540 MRT = t r = ------------ ( 54 ) + ------------ ( 20 ) + ------------ ( 60 ) + ------------ 70 + ------------ ( 70 ) = 64.1 2070 2070 2070 2070 2070 t o = 75°F = 0.5t a + 0.5t r = 0.5t a + 0.5 ( 64.1 ) t a = 86°F
4.14 Workers on an assembly line making electronic equipment dissipate 700 Btu/h, of which 310 Btu/h is latent heat. When the MRT for the area is 69°F, what air temperature must the heating system maintain for comfort of the workers if the air movement is 40 fpm? 700 Btu/h= 2 met Activity level = ----------------------2 19.5 ft 89 + 69 t o = ------------------ = 79°F 2 t a = 79 – 5.4 ( 1 + 0.75 ) ( 2 – 1.2 ) = 71°F
Assuming light clothing (0.75 clo): Correcting for activity level:
Fig. 4-3;
t a = 89°F;
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N
Std. 55: t r = ∑ Fρ – i t i ;
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58⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual 4.15 A room has 1000 ft2 of surface, of which 120 ft2 is to be heated, and the balance has an average surface temperature of 60°F. The air temperature in the room is 68°F. The room is occupied by light clothed adults at rest. Determine the surface temperature of the heated panel necessary to produce comfort if the air velocity is 20 fpm. [Ans: 243°F? unfeasible] Assume uniform view of surfaces:
880 120 t r = ------------ ( 60 ) + ------------ ( t p ) 1000 1000
t o = 75 = 0.5t a + 0.5t r 880 120 t r = 150 – 68 = 82°F = ------------ ( 60 ) + ------------ t p 1000 1000 t p = 243°F
4.16 Assume that in Problem 4.15, the maximum allowable panel temperature is 120°F. The average temperature of other surfaces in the room remains at 60°F and the air temperature is still 68°F. What panel area will be required if the room is occupied by adults at rest? ( 1000 – P ) ( 60 ) + P ( 120 ) t r = 82 = -------------------------------------------------------------1000 P = 367 ft
2
a.
b.
390 Q s = 150 ( 225 ) ⎛⎝ --------- ⎞⎠ = 1.1 ( CFM ) ( 72 – 57 ) 330
[ Chap. 7 ]
CFM = 2417 Supply air must have sufficiently low humidity ratio since latent load actually represents moisture added, QL M H O ≅ -----------2 1100
4.18 Two hundred people attend a theater matinee. Air is supplied at 60°F. Determine the required flow rate (lb/h) to handle the heat gain from the occupants if the return air temperature is not to exceed 75°F. [Ans: 11,475 lb/h] q s = 210 Btu/person (Chap. 7) q s = m· C p ( t r – t s ) = 200 ( 210 ) = m· ( 0.244 ) ( 75 – 60 ) · V 11 ,475 · m· = 11 ,475 lb/h = --V = ---------------- ( 13.33 ) = 2550 cfm ν 60
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4.17 In an auditorium, 17,150 students are watching slides. The MRT is 80°F and the average room air temperature is 72°F. Air enters the room at 57°F. Assume the lights are out and no heat gain or loss occurs through the walls, floor, and ceiling. a. How much air (cfm) should be supplied to remove the sensible heat? b. Explain what must be done to remove the latent heat.
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Chapter 4—Design Conditions⏐59
4.19 Determine the increase in humidity ratio due to 80 people in a dance hall if air is circulated at the rate of 0.64 m3/s (1350 cfm). 43 ,600 M w = ---------------- = 39.64 lb/h 1100
Q L = 80 ( 545 ) = 43 ,600 Btu/h
Mw 39.64 ΔW = -------- = ---------------------------------------------- = 0.0065 lb/lb air Ma ( 1350 ) ( 60 ) ( 0.075 )
4.20 Specify the MRT for comfort in a space where the air temperature is 68°F and the relative velocity is 20 fpm, for sedentary activity and light clothing. [Ans: 92°F] From Fig. 4-3: MRT = 92°F [Assume medium activity is sedentary]
4.21 Specify completely a suitable set of indoor and outdoor design conditions for each of the following cases: a. winter; apartment building; St. Louis, Missouri b. summer; apartment building; St. Louis, Missouri c. winter; factory (medium activity); Rochester, Minnesota a.
Inside: ti = 72°F, 30% RH;
Outside: to = 4.1°F; Φo = 100%
b.
Inside: ti = 78°F, 60% RH;
Outside: 93.1 db/76.1°F wb
c.
Inside: Assume 2 met, 1 clo ⇒ ti = 75 – 5.4(1 + 1)(2 – 1.2) = 66°F, 30% RH Outside: to = –15.2°F; Φo = 100%
4.22 The mean radiant temperature in a bus is 6°C lower in winter than the air temperature. For passengers seated without coats, determine the desired air temperature if the relative air velocity is 0.2 m/s. Trial and Error with Fig. 4-3 t a ≅ 28°C MRT ≈ 22°C
4.23 For Atlanta, Georgia, specify the normal indoor design conditions listed below for a. Winter: Dry bulb = ____°C; W = _______ kg/kg b. Summer: Dry bulb = ____°C; W = _______ kg/kg a. b.
Winter: Summer:
Dry bulb = 22°C; Dry bulb = 25°C;
W = 0.004 kg/kg W = 0.012 kg/kg
4.24 Specify completely indoor and outdoor design conditions for winter for a clean room in Kansas City, Missouri, having a 1.2 by 1.2 m radiant panel at 49°C on each of the four walls. The room is 6 m by 4 by 3 m high and the other surfaces are all at 22°C. Assume very little activity and light clothing. Assume uniform view for each surface:
( ∑ A = 108 )
4 ( 1.2 ) ( 1.2 )49 [ 108 – 4 ( 1.2 ) ( 1.2 ) ]22 t r = ---------------------------------- + ------------------------------------------------------ = 23.4°C 108 108 Fig. 4-3 SI: for v = 0.2 m/s, MRT = 23.4 ⇒ t a = 28°C t o = – 16.3°C
φ o = 100%
φ a = 50%
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Assuming Icl = 0.5 with ta = MRT + 6
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Solutions to
Chapter 5 LOAD ESTIMATING FUNDAMENTALS
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Chapter 5—Load Estimating Fundamentals⏐63
NOTE: For the problems in this chapter the answers may vary depending on which tables are used for the R and k values as well as the assumptions in the selection of the tabulated listings.
5.1 With an 11.2 m/s wind blowing uniformly against one face of a building, what pressure differential would be used to calculate the air leakage into the building? 11.2 ⁄ 0.447 = 25 mph
11.2 m/s wind P v = 0.000482 at Standard ρ
V2 2
P v = ( 0.000482 ) ( 25 ) = 0.302 in. H2 O C s = 0.6
[ Fig. 5-5 at 0° ]
ΔP in = 0.6 ( 0.302 ) = 0.18 in. H 2 O
5.3 A double door has a 1/8 in. crack on all sides except between the two doors, which has a 1/4 in. crack. What would be the leakage rate for the building of Problem 5.1? From Problem 5.1, assume 6 × 7 ft doors. ΔP = 0.18 in. H 2 O
Average crack width is 1/8 in.
Fig. 14, Chap. 27, HBF → 300 cfm/door 0.18 CFM = 600 ⎛ ---------- ⎞ ⎝ 0.30 ⎠
0.55
Two doors at 0.3 in. H 2 O
= 453 cfm
5.4 Determine the heat loss due to infiltration of 236 L/s of outdoor air at 9°C when the indoor air is 24°C.
or
Q s = 1.1 ( SCFM ) Δt = 1.1 ( 500 ) ( 75 – 48 ) = 14 ,850 Btu/h
*There is also a latent load.
5.5 Give an expression for (a) the sensible load due to infiltration, and (b) the latent load due to infiltration. a.
Sensible
CFH Q s = 1.1 ( SCFM ) Δt = 1.1 ⎛ -----------⎞ Δt = 0.018 CFH ( t i – t o ), Btu/h ⎝ 60 ⎠
b.
Latent
CFH Q L = 4840 ( SCFM ) ( ΔW ) = 1.1 ⎛ -----------⎞ ( ΔW ) = 80.6 (CFH ) ( W i – W o ), Btu/h ⎝ 60 ⎠
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236 L/s at 9°C with 24°C inside 500 cfm 48°F with 75°F Q s = 1.23 ( 236 ) ( 24 – 9 ) = 4350 watts
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64⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
5.6 A building is 75 ft wide by 100 ft long and 10 ft high. The indoor conditions are 75°F and 24% RH, and the outside conditions are 0°F and saturated. The infiltration rate is estimated to be 0.75 ach. Calculate the sensible and latent heat loss. Volume = 75 × 100 × 10 = 75 ,000 ft @ in W i = 0.0046 @ out W o = 0.0007875
3
NV ( 0.75 ) ( 75 ,000 ) CFM = -------- = ------------------------------------- = 937.5 60 60 Q s = 1.1 ( CFM ) Δt = 1.1 ( 937.5 ) ( 75 – 0 ) Q s = 77 ,300 Btu/h
Q L = 4840 ( CFM ) ( ΔW ) = 4840 ( 937.5 ) ( 0.0046 – 0.0007875 ) = 17 ,000 Btu/h
5.8 A 3 by 3 ft ventilation opening is in a wall facing in the prevalent wind direction. There are adequate openings in the roof for the passage of exhaust air. Estimate the ventilation rate for a 25 mph wind. 3 × 3 ft opening Q = C 4 C v AU
25 mph
2005 HBF, Equations (27) – (29)
Q = 88 ( 0.55 ) ( 9 ) ( 25 ) = 10 ,890cfm
3
a.
Volume = ( 20 ) ( 40 ) ( 9 ) = 7200 ft ( 7200 ) ( 3 ⁄ 4 ) SCFM = NV = -------------------------------- = 90 60 Q s = 1.1 CFM ( Δt ) = 1.1 ( 90 ) ( 70 ) = 6930 Btu/h Q L = 4840 ( 90 ) ( 0.0055 – 0.001 ) = 1960 Btu/h Q t = 6390 + 1960 = 8890 Btu/h
b.
m w = Q L ⁄ 1100 = 1960 ⁄ 1100 = 1.8 lb/h
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5.9 A building 20 by 40 by 9 ft has an anticipated infiltration rate of 0.75 air changes per hour. Indoor design conditions are 75°F, 30% RH minimum. Outdoor design temperature is 5°F. a. Determine sensible, latent, and total heat loads (Btu/h) due to infiltration. b. Specify the necessary humidifier size (lb/h).
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Chapter 5—Load Estimating Fundamentals⏐65
5.10 A small factory with a 10 ft high ceiling is shown. There are 22 employees normally in the shop area and 4 employees in the office area. On a winter day when the outside temperature is 0°F, the office is maintained at 75°F, 25% RH, and the shop is kept at 68°F with no humidity control. Determine for each area a. infiltration, cfm b. minimum required outside air, cfm c. sensible heat loss due to infiltration, Btu/h d. latent heat loss due to infiltration, Btu/h 2
Office: 700 ft , 7000 ft 2
3
Shop: 2900 ft , 29 ,000 ft
CFM supply = 700 = V pz 3
o
CFM supply = 3600 = V pz
s
a.
INFILTRATION Office: 1/2 ach × 7000 ⁄ 60 = 58 cfm (more if traffic) Shop: 1 1/2 ach × 29 ,000 ⁄ 60 = 725 cfm
b.
VENTILATION ( Std. 62.1-2004 )
V bz = R p R z + R a A z
Office:
Rp = 5
R a = 0.06
V bz = 5 × 4 + 0.06 ( 700 ) = 62 cfm
Shop:
R P = 10
R a = 0.18
V bz = 10 × 22 + 0.18 ( 2900 ) = 742 cfm V oz = V bz ⁄ E z
Assuming ceiling supply (floor return): E z = 1.0 V oz
office
= 62;
V oz
shop
= 742
Assume single system for both spaces: V oz 62 742 Z P = ⎛ -------- ⎞ = --------- = 0.08 Z P = ------------ = 0.21 = max. Z p ⇒ E V = 0.9 ⎝ V pz ⎠o o s 700 3600
c.
d.
V ou 804 V ot = -------- = --------- = 893 cfm EV 0.9
q s = 1.10 × CFM × Δt qs
office
qs
shop
= 1.10 × 58 × ( 75 – 0 ) = 4785 Btu/h = 1.10 × 725 × ( 68 – 0 ) = 54 ,230 Btu/h
q L = 4840 × CFM × ΔW qL
offic
qL
shop
@ 75°F, 25% }
W i = 0.0046;
= 4840 × 58 × ( 0.0046 – 0.00079 ) = 1070 Btu/h = 0 no humidifier
5.11 Specify an acceptable amount of outside air for ventilation of the following: a. 12 by 12 ft private office with 8 ft high walls b. department store with 20,000 ft2 of floor area
Table 5-9 2
a.
Use 5 cfm/person + 0.06 cfm/ft ∴ OA = 5 × 1 + 0.06 ( 12 × 12 ) = 5 + 8.6 = 13.6 cfm
b.
Use 7.5 cfm/person & 15 p/1000 ft + 0.12 cfm/ft 20 ,000 ∴ OA = 7.5 × ---------------- × 15 + 0.12 ( 20 ,000 ) = 4650 cfm 1000
2
2
0°F W o = 0.00079
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V ou = Σ = 62 + 742 = 804 cfm;
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66⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
5.12 A wall consists of 4 in. of face brick, 1/2 in. of cement mortar, 8 in. hollow clay tile, an airspace 1 5/8 in. wide, and wood lath and plaster totaling 3/4 in. Find the U-factor for both winter and summer. R Outside air Face brick, 4 in. Cement, 1/2 in. Hollow clay tile, 8 in. Airspace, 1 5/8 in. Metal lathe and plaster, 3/4 in. Inside air ΣR = U = 1/ΣR – 0.215
Winter
Summer
0.17 0.44 0.10 1.85 0.95 0.47 0.68 4.66
0.25 0.44 0.10 1.85 0.95 0.47 0.68 4.74
0.211 Btu/h/ft2 ·°F
5.13 The ceiling of a house is 3/4 in. (19 mm) acoustical tile on furring strips with highly reflective aluminum foil across the top of the ceiling joists. Determine the U-factor for cooling load calculations. R Attic air 4.55 Air gap (top) 4.55 Air gap (bottom) 0.92 Acoustical tile 1.89 Room air 0.92 ΣR = 12.83
U = 1/ΣR = 0.078 Btu/h/ft2 ·°F
5.14 Calculate the winter U-factor for a wall consisting of 4 in. (100 mm) face brick, 4 in. (100 mm) common brick, and 1/2 in. (13 mm) of gypsum plaster (sand aggregate). 0.17 0.44 0.80 0.09 0.68 2.18
U = 1/ΣR = 0.459 Btu/h/ft2 ·°F
5.15 Find the overall coefficient of heat transmission U for a wall consisting of 4 in. of face brick, 1/2 in. of cement mortar, 8 in. of stone, and 3/4 in. of gypsum plaster. The outside air velocity is 15 mph and the inside air is still. R Outside air Face brick, 4 in. Cement mortar, 1/2 in. Stone, 8 in. (1/R = 0.08) Gypsum plaster, 3/4 in. Inside air ΣR =
0.17 0.44 0.10 0.64 0.47 0.68 2.50
(assume lightweight aggregate) U = 1/ΣR = 0.40 Btu/h/ft2 ·°F
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R Outside air Face brick, 4 in. Common brick, 4 in. Gypsum (sand) plaster, 1/2 in. Inside air ΣR =
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Chapter 5—Load Estimating Fundamentals⏐67 5.16 A wall has an overall coefficient U = 1.31 W/(m2 ·K). What is the conductance of the wall when its outside surface is exposed to a wind velocity of 6.7 m/s and the inside air is still? 11 1 1 --= ∑ R = ---- + ---- + ----U hi C ho 11 1 1 --------= ---------- + ---- + ------------1.31 8.29 C 34.08
2
C = 1.63 ( W/(m ⋅ K )
5.17 Compute the U-factor for a wall of frame construction consisting of 1/2 by 8 bevel siding, permeable felt building paper, 25/32 in. wood fiber sheathing, 2 by 4 studding on 16 in. centers, and 3/4 in. metal lath and sand plaster. Outside wind velocity is 15 mph. R Outside air Siding Felt paper Sheathing Airspace Lathe and plaster Inside air ΣR =
0.17 0.81 0.06 2.06 0.97 0.10 0.68 4.35
Neglecting Studs
U = 1/4.85 = 0.206 Btu/h·ft2 ·°F
5.18 For the wall of Problem 5.17, determine U if the space between the studs is filled with fiberglass blanket insulation. Neglect the effect of the studs.
5.19 Rework Problem 5.18 including the effect of the studs. U i = 0.067;
R s = 4.85 – 0.97 + 4.35 = 8.23;
1.5 S = ⎛ ------- ⎞ ( 100 ) = 9.4%; ⎝ 16 ⎠
U s = 0.12
U av = ( 0.094 ) ( 0.12 ) + ( 0.906 ) ( 0.067 )
2
U av = 0.072 Btu/h· ft ·°F
5.20 A concrete wall 250 mm thick is exposed to outside air at −15°C with a velocity of 6.7 m/s. Inside air temperature is 15.6°C. Determine the heat flow through 14.9 m2 of this wall. 2
R o = 0.030 ( m ⋅ K ) ⁄ W
2
R i = 0.120 ( m ⋅ K ) ⁄ W
2
R con = ( 0.25 ) ( 0.55 ) = 0.138 ( m ⋅ K ) ⁄ W
1 2 2 1 U = --------- = ----------------------------------------------- = 3.47 W ⁄ ( m ⋅ K ) = 3.47 W ⁄ ( m ⋅ °C ) ∑ R 0.03 + 0.12 + 0.138 Q = U ( A ) ( Δt ) = 3.47 ( 14.9 ) [ 15.6 – ( – 15 ) ] = 1540 W
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R corrected for insulation 4.85 – 0.97 + 11.0 = 14.88 U = 0.067 Btu/h·ft2 ·°F
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68⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
5.21 Find the overall coefficient of heat transfer and the total thermal resistance for the following exterior wall exposed to a 25 mph wind: face brick veneer, 25/32 in. insulating board sheathing, 3 in. fiberglass insulation in stud space, and 1/4 in. walnut veneer plywood panels for the interior. Neglecting Studs: R Outside air 0.09 Face brick 0.44 Insulating board sheathing 2.06 Airspace, 3/4 in. 1.01 Fiberglass, 3 in. 11.0 Plywood, 1 1/4 in. 0.31 Inside air 0.68 ΣR = 15.59
⇒ HBF
U = 1/ΣR = 0.064 Btu/h·ft2 ·°F
5.22 What is the thermal resistance of 12.1 cm (4 3/4 in.) thick precast concrete (stone aggregate, oven dried)? 2
R = ( 0.121 m ) [ 0.76 (m· K)/W ] = 0.092 ( m ⋅ K ) ⁄ W or 2
R = ( 4.75 ) ( 0.11 ) = 0.52 h· ft ·°F/Btu
R Outside air (R1) Face brick (R2) Fiberglass, (R3) Gypsum board (R4) Inside air (R5)
0.17 0.44 7.00 0.32 0.68
RT = ΣR =
8.61
U = 1/ΣR = 1/8.61 = 0.116 Ti = 75°F, To = –10°F q = UAΔT = (0.116)(85) = 9.87 Btu/h·ft2
T 1 = – 10°F R1 0.17 T 2 = T 1 + ------ ( ΔT ) = – 10 + ---------- ( 85 ) = – 8.3°F RT 8.61 R2 0.44 T 3 = T 2 + ------ ( ΔT ) = – 8.3 + ---------- ( 85 ) = – 3.9°F RT 8.61 R3 7.0 T 4 = T 3 + ------ ( ΔT ) = – 3.9 + ---------- ( 85 ) = 65.2°F RT 8.61 R4 0.32 T 5 = T 4 + ------ ( ΔT ) = – 65.2 + ---------- ( 85 ) = 68.3°F RT 8.61 T 6 = 75°F
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5.23 A composite wall structure experiences a –10°F air temperature on the outside and a 75°F air temperature on the inside. The wall consists of a 4 in. thick outer facebrick, a 2 in. batt of fiberglass insulation, and a 3/8 in. sheet of gypsum board. Determine the U-factor and the heat flow rate, per ft2. Plot the steady-state temperature profile across the wall.
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Chapter 5—Load Estimating Fundamentals⏐69
5.24 Find the overall heat transmission coefficient for a floor-ceiling sandwich (heat flow up) having the following construction. R Upper air Concrete, 2 1/2 in. Airspace, upper Airspace, lower Acoustical tile Inside air ΣR =
0.61 0.20 0.61 0.61 1.89 0.61 4.53
U = 0.22 Btu/h·ft2 ·°F
5.25 The exterior windows of a house are of double insulating glass with 1/4 in. airspace and have metal sashes. Determine the design U-factor for heating. U = 0.87 Btu/h·ft2 ·°F (Table 5-16) 5.26 In designing a house, the total heat loss is calculated as 17.9 kW. The heat loss through the outside walls is 28% of this total when the overall coefficient for the outside walls is 1.4 W/(m2 ·K). If 50 mm organic bonded fiberglass is added to the wall in the stud space, determine the new total heat loss for the house. q walls = 0.28 ( 17.9 ) = 5.01 kW; 1
- = 0.714; ∑ R = -----1.4 with fiberglass
= 0.714 + 0.05 ( 27.76 ) = 2.1
∑ R with fiberglass
0.714 = ------------- ( 5.01 ) = 1.70 kW 2.1
q total = 12.89 + 1.70 = 14.6 kW
5.27 The top floor ceiling of a building 30 by 36 ft is constructed of 2 by 4 in. joists on 18 in. centers. On the underside is metal lath with plaster, 3/4 in. thick. On top of the joists there are only scattered walking planks, but the space between the joists is filled with rock wool. The air temperature at the ceiling in the room is 78°F and the attic temperature is 25°F. Find the overall coefficient of heat transfer for the ceiling. R Room air 0.61 Lathe and plaster 0.13 Rock wool 11.0 Attic air 0.61 ΣR = 12.35
U = 1/ΣR = 0.0.81 Btu/h·ft2 ·°F
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q walls
q other = 17.9 – 5.01 = 12.89
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70⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
5.28 Determine the U-factor and the temperature at each point of change of material for the flat roof shown below. The roof has 3/8 in. built-up roofing, 1 1/2 in. roof insulation, 2 in. thick, 80 lb/ft3 lightweight aggregate concrete on corrugated metal over steel joists, with a metal lath and 3/4 in. (sand) plaster ceiling. Omit correction for framing. R Outside air Built-up roofing, 3/8 in. Roof insulation, 1 1/2 in. Concrete (80)#, 2 in. Airspace (top) Airspace (bottom) Lather and plaster Inside air
T1 T2 T3 T4 T5 T6 T7
ΣR =
0.17 0.33 4.17 0.54 0.61 0.61 0.13 0.61 7.17
0.17 T 1 = – 5 + ---------- ( 80 ) = – 3.1°F 7.17 0.5 T 2 = – 5 + ---------- ( 80 ) = 0.6°F 7.17 5.82 T 5 = – 5 + ---------- ( 80 ) = 59.9°F 7.17 6.50 T 7 = – 5 + ---------- ( 80 ) = 68.2°F 7.17 2
U = 0.139 Btu/h· ft ·°F
R Outside air Built-up roofing, 3/8 in. Cellular glass insulation, 1 in. Concrete, 4 in. Acoustical tile, 3/4 in. Inside air ΣR =
0.17 0.33 2.50 0.32 1.89 0.61 5.82
A = 100 ft2 to = 10°F ti = 70°F U = 1/ΣR = 0.172
q = UA ( t i – t o ) = 0.172 ( 100 ) ( 70 – 10 ) = 1030 Btu/h
5.30 Calculate the heat loss through 100 ft2 (9.29 m2) of 1/4 in. (6.5 mm) plate glass with inside and outside air temperatures of 70 and 10°F (21.1 and –12.2°C), respectively. 2
2
U glass = 5.91 W/ ( m ⋅ K ) { Table 5-15 } or 1.04 Btu/h· ft ·°F q = UA ( Δt ) = ( 5.91 ) ( 9.29 ) [ 21.2 – ( – 12.2 ) ] = 1833 W q = 1.04 ( 100 ) ( 70 – 10 ) = 6240 Bth/h
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5.29 Calculate the heat loss through a roof of 100 ft2 area where the inside air temperature is to be 70°F, outside air 10°F, and the composition from outside to inside: 3/8 in. built-up roofing, 1 in. cellular glass insulation, 4 in. concrete slab, and 3/4 in. acoustical tile.
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Chapter 5—Load Estimating Fundamentals⏐71
5.31 A building has single glass windows and an indoor temperature of 75°F. The outside air temperature is 40°F. With a 15 mph outside wind, still air inside, and after sundown, what can the maximum relative humidity of the inside air be without condensation forming on the glass? h i ( 75 – t s ) = U ( 75 – 40 );
1.46 ( 75 – t s ) = 1.04 ( 75 – 40 )
t s = 50.1°F = dew point;
dry bulb = 75°F ∴φ = 40% max.
5.32 Repeat Problem 5.31 for a double glass window with a 1/2 in. airspace. h i ( 75 – t s ) = U ( 75 – 40 );
1.46 ( 75 – t s ) = 0.59 ( 75 – 40 )
t s = 60.9°F = dew point;
dry bulb = 75°F ⇒ φ = 61% max.
5.33 A wall is constructed of 4 in. face brick, pressed fiber board sheathing (k = 0.44), 3 1/2 in. airspace, and 1/2 in. lightweight gypsum plaster on 1/2 in. plasterboard. When the inside air temperature is 70°F and the outside temperature is –15°F, how thick must the sheathing be to prevent water pipes in the stud space from freezing? R
0.17 + 0.44 + x ⁄ 0.44 32 – ( – 15 ) -------------------------------------------------= -------------------------3.0 + x ⁄ 0.44 70 – ( – 15 ) x = 1.0 in.
5.34 The roof of a rapid transit car is constructed of 3/8 in. plywood (C = 2.12), a vapor seal having negligible thermal resistance, expanded polystyrene insulation (k = 0.24), 3/4 in. airspace, 1/16 in. steel with welded joints and aluminum paint. If the car is traveling at 60 mph (film coefficient is 20.0 Btu/h·ft2 ·°F) when the ambient temperature is –20°F, what thickness of insulation is necessary to prevent condensation when the inside conditions in the car are 72°F dry bulb and 55% RH? 72°F db; Inside:
55% rh ⇒ 55°F dew point E = 0.6
h i = 0.81
2 --q- = 0.81 ( 72 – 55 ) = 13.77 Btu/h· ft ·°F A
R Outside air 0.05 Plywood 0.47 Polystyrene insulation x/0.24 Airspace 0.78 Inside air 0 ΣR = 1.3 + x/0.24
55 – ( – 20 ) q --- = 13.77 = ----------------------------------( 1.3 + x ⁄ 0.24 ) A x = 1.0 in.
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Outside air 0.17 Face brick 0.44 Sheathing x/0.44 Airspace 0.94 Plaster 0.45 Plasterboard 0.32 Inside air 0.68 ΣR = 3 + x/0.44
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72⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
5.35 A roof is constructed of 2 in. wood decking, insulation on top of deck, and 3/8 in. built-up roofing. It has no ceiling. Assuming that the insulation forms a perfect vapor barrier, determine the required resistance of the insulation to prevent condensation from occurring at the deck insulation interface when indoor conditions are 70°F and 40% RH, and the outside temperature is 20°F. R film Δt film ------------= --------------R total Δt total t i = 70°F
40% RH
dew point = 44.5°F
t o = 20°F
R Inside air Wood deck Outside air Insulation
– 44.50.61 + 2.50 -------------------------------------------------------- = 70 --------------------70 – 20 0.61 + 2.5 + 0.17 + R ins
0.61 2.5 0.17 Rins
R ins = 2.48
a.
b.
Ri
0.25 0.44 0.93 5.23
U = 1/ΣR = 0.259
0.129
Ceiling Attic air Plasterboard Inside air
Rs
Outside air 0.25 Face brick 0.44 Sheathing, 3/4 in. 0.93 Studs — Air { 0.68 0.68 Paneling 0.20 Inside air 0.68 ΣR = 3.86
Ui = 0.259; Us = 0.107 Uav = 0.15(0.129) + 0.85(0.259) = 0.24 Btu/h·ft2 ·°F (1.36 W/m2 ·K)
— 0.20 0.68 7.73
Roof 0.92 0.45 0.92
Outside air Shingles Plywood Attic Air
ΣR =
2.29 U = 0.437
ΣR =
0.25 0.44 0.93 0.76 2.38
u = 0.420
c.
U = 0.81 Btu/h·ft2 ·°F
d.
U = 0.46 Btu/h·ft2 ·°F (Table 5-17)
Ro,c = 2.29 + 2.38/(2717/1980) = 4.024 Uo,c = 1/4.024 = 0.25 Btu//h·ft2 ·°F
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5.36 Determine the summer U-factor for each of the following: a. building wall consisting of face brick veneer, 3/4 in. plywood sheathing, 2 by 6 studs on 24 in. centers, no insulation, and 5/32 in. plywood paneling b. ceiling/roof where the ceiling is composed of 1/2 in. plasterboard nailed to 2 by 6 joists on 16 in. centers and the roof consists of asphalt shingles on 3/4 in. plywood on 2 by 4 rafters on 16 in. centers. The roof area is 2717 ft2 while the ceiling area is 1980 ft2. c. sliding patio door with insulating glass (double) having a 0.50 in. airspace in a metal frame d. a 2 in. solid wood door with a wood storm door
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Chapter 5—Load Estimating Fundamentals⏐73
5.37 A prefabricated commercial building has exterior walls constructed of 2 in. expanded polyurethane bonded between 1/8 in. aluminum sheet and 1/4 in. veneer plywood. Design conditions include 105°F outside air temperature, 72°F indoor air, and 7.5 mph wind. Determine: a. overall thermal resistance b. value of U c. heat gain per ft2 R Outside air 0.25 Aluminum, 1/8 in. –0 Polyurethane (2 × 6.25) 12.50 Plywood, 1/4 in. 0.31 Inside air 0.68 ΣR = 13.74
b.
1 2 U w = ------- = 0.0728 Btu/h· ft ·°F Rw
c.
q = 0.0728 ( 105 – 72 ) = 2.40 Btu/h· ft
2
5.38 An outside wall consists of 4 in. face brick, 25/32 in. insulating board sheathing 2 in. mineral fiber batt between the 2 by 4 studs, and 1/2 in. plasterboard. Determine the winter U-factor. R
U = 1/ΣR = 0.085 Btu/h·ft2 ·°F
5.39 Solve the following: a. Compute the winter U-factor for the wall of Problem 5.38 if the wind velocity is 30 mph. b. Compute the summer U-factor for the wall of Problem 5.38. c. If full wall insulation is used, compute the summer U-value for the wall of Problem 5.38. a.
∑ R = 11.81 – 0.17 + 1 ⁄ 0.26 = 11.72;
b.
∑ R = 11.81 – 0.17 + 0.25 = 11.89;
c.
∑ R = 11.81 – 1.01 – 7.0 + 11.0 = 14.88;
U = 0.085 U = 0.084 U = 0.067
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Outside air 0.17 Face brick 0.44 Insulating board sheathing 2.06 Airspace 1.01 Fiber batt 7.0 Plasterboard 0.45 Inside air 0.68 ΣR = 11.81
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74⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
5.40 An exterior wall contains a 3 by 7 ft solid wood door, 1 3/8 in. thick, and a 6 by 7 ft sliding patio door with double insulating glass having a 1/2 in. airspace and metal frame. Determine the summer U-factor for each door. 2
1 3 ⁄ 8 in. wood dome:
U = 0.46 Btu/h· ft ·°F
Sliding patio door:
U = 0.76 Btu/h· ft ·°F
2
5.41 If the doors of Problem 5.40 are between the residence and a completely enclosed swimming pool area, determine the U-factor for each door. 1 3 ⁄ 8 in. wood door:
R inside = 1 ⁄ 0.39 – 0.25 + 0.68 = 2.99
Sliding patio door:
U = 1 ⁄ R = 0.33 R inside = 1 ⁄ 0.81 – 0.25 + 0.68 = 1.66 U = 1 ⁄ R = 0.60
5.42 Determine the winter U-factor in W/(m2 ·K) for the wall of a building which has the following construction: face brick, 4 in.; airspace, 3/4 in.; concrete, 9 in.; cellular glass board insulation, 1 in.; plywood paneling, 1/4 in. 1 1 U = --------- = -----------------------------------------------------------------------------------------------------------0.17 + 0.44 + 1.1 + 0.72 + 2.50 + 0.31 + 0.68 R ∑ 2
2
U = 0.169 Btu/h· ft ·°F [ 0.960 W ⁄ ( m ·K ) ]
a.
1 1 U i = ---------- = ----------------------------------------------------------------------------------------------- = 0.0706 Btu/h ft 2 °F ∑ R i 0.25 + 0.79 + 1.14 + 11.0 + 0.31 + 0.68 1 1 U s = ----------- = ----------------------------------------------------------------------------------------------- = 0.133 Btu/h ft 2 °F ∑ R s 0.25 + 0.79 + 1.14 + 4.35 + 0.31 + 0.68
b.
1.5 S = ------- ( 100 ) = 9.5% 16 U w = 0.27; R w = 3.703; U s = 0.264
R s = 3.703 – 0.17 + 0.25 = 3.784
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5.43 Determine the summer U-factor for the following building components a. Wall: wood drop siding, 1 by 8 in.; 1/2 in. nail-base insulating board sheathing; 2 by 4 studs (16 in. oc) with full wall fiberglass insulation; 1/4 in. paneling b. Door: solid wood, 1 1/2 in. thick, with wood and glass storm door
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Chapter 5—Load Estimating Fundamentals⏐75
area to ceiling area is 1:3. The attic is unvented in winter. 5.44 Determine the combined ceiling and roof winter U-value for the following construction: The ceiling consists of 3/8 in. gypsum board on 2 by 6 in. ceiling joists. Six inches of fiberglass (mineral/glass wool) insulation fills the space between the joists. The pitched roof has asphalt shingles on 25/32 in. solid wood sheathing with no insulation between the rafters. The ratio of roof Roof:
1 1 U roof = ------------------ = ---------------------------------------------------------------- = 0.463 0.17 + 0.44 + 0.934 + 0.62 R ∑ roof 1 1 U ceiling = ------------------------ = ---------------------------------------------------------------- = 0.0487 0.61 + 19.02 + 0.32 + 0.61 R ∑ ceiling
Ceiling:
1 1 2.16 R t = ------ + ---------- = 20.54 + ---------- = 22.20; U c nU R 1.3
2
U o ,c = 0.045 Btu/h· ft ·°F
5.45 The west wall of a residence is 70 ft long by 8 ft high. The wall contains four 3 by 5 ft wood sash 80% glass single pane windows each with a storm window; one double-glazed (1/2 in. airspace) picture window, 5 1/2 by 10 ft; and one 1 3/4 in. thick solid wood door, 3 by 7 ft. The wall itself has the construction of Problem 5.21. Specify the U-factor and corresponding area for each of the various parts of the wall with normal winter air velocities. 4 Windows:
U sin gle
pane
= 0.89
A w = 4 × 3 × 5 = 60 ft
2
A p 5.5 × 10 = 55 ft w
A wallproper = ( 70 × 8 ) – 60 – 55 – 21 = 424 ft
2
2
A d = 3 × 7 = 21 ft U wall = 0.064
5.46 A wall is 20 by 3 m, which includes 14% doubleinsulating glass windows with a 6 mm airspace. The wall proper consists of one layer of face brick backed by 250 mm of concrete with 12 mm of gypsum plaster on the inside. For indoor and outdoor design temperatures of 22°C and –15°C, respectively, determine the heat loss through this wall, kW. R
Wall Outside air Face brick Concrete, 250 mm Plaster, 12 mm Inside air
0.03 0.078 0.138 0.08 0.12
ΣR = 0.446 ;
Window: (Assume A1 frame operable) U = 4.93 W/(m2 ·K) Awall = (0.86)(20 × 3) = 51.6 m2 Awindow = (0.14)(20 × 3) = 8.4 m2 U = 2.24 W/(m2 ·K)
q = UA Δt = ( 2.24 ) ( 51.6 ) ( 37 ) + ( 4.93 ) ( 8.4 ) ( 37 ) q = 5800 W = 5.8 kW
2
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Picture window: U = 0.50 Door: U = 0.46
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Solutions to
Chapter 6 RESIDENTIAL COOLING AND HEATING LOAD CALCULATIONS
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Chapter 6—Residential Cooling and Heating Load Calculations⏐79 6.1 Determine which of the following walls of 150 ft2 gross area will have the greatest heat loss: a. b.
c.
Wall of 25% single glass and the remainder brick veneer (U = 0.25 Btu/h·ft2 ·°F) Wall of 50% double-glazed windows with the remainder of the wall brick veneer (U = 0.25 Btu/h·ft2 ·°F) Wall of 10% single pane glass and 90% of 6 in. poured concrete with ho = 6.0 and hi = 1.6 Btu/h·ft2 a.
q a = 0.25 ( 0.75 ) ( 150 ) ΔT + 1.13 ( 0.25 ) ( 150 ) ΔT q a = 70.5 ΔT
b.
q b = 0.25 ( 0.5 ) ( 150 ) ΔT + 0.69 ( 0.5 ) ( 150 ) ( ΔT ) q b = 70.5 ΔT
c.
1 q c = 1.13 ( 0.1 ) ( 150 ) ΔT + ----------------------------------------------- ( 0.9 ) ( 150 ) ΔT 0.167 + 0.48 + 0.62 q c = 123 ΔT
Wall (c) has greatest heat loss.
6.2 A house has a pitched roof with an area of 159 m2 and a U of 1.6 W/(m2 ·K). The ceiling beneath the roof has an area of 133 m2 and a U of 0.42 W/(m2 ·K). The attic is unvented in winter for which the design conditions are –19°C outside and 22°C inside. Determine the heat loss through the ceiling. [Ans: 1.88 kW (6400 Btu/h)]
1 1 q = UA ΔT U = --- = ------- = 0.345 W/(m 3 ·K) R 2.9 q = ( 0.345 ) ( 133 ) ( 41 ) = 1880 W ( 6420 Btu/h )
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1 1 R = ---------- + --------------------------------------- = 2.38 + 0.52 = 2.90 0.42 ( 159 ⁄ 133 ) ( 1.6 )
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80⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
6.3 Determine the design winter heat loss through each of the following components of a building located in Minneapolis, Minnesota: a.
b.
c.
Wall having 648 ft2 of area and construction of 4 in. face brick; 3/4 in. plywood sheathing; 2-1/2 in. glass fiber insulation in 2 by 4 stud space (16 in. on centers); 1/2 in. plasterboard interior wall. A 2185 ft2 ceiling topped by a 2622 ft2 hip roof. The ceiling consists of 1/2 in. acoustical tile with R-19 insulation between the 2 by 6 (16 in. on centers) ceiling joists. The roof has asphalt shingles on 3/4 in. plywood sheathing on the roof rafters. The attic is unvented in winter. Two 4 by 6 ft single pane glass windows with storm windows. t i = 72°F (selected) ; t c = – 13.4°F ( Fig. 4-4 ) Wall
Ceiling Ri 0.17 0.44 1.08 6.7 1.01 – 0.45
Rs 0.17 0.44 1.08 – – 4.35 0.45
Inside air
0.68
0.68
10.53
7.17
ER
Attic air Insulation Acoustical tile, 1/2 in. Inside air ER
w
Rs 0.61 6.25 1.19 0.61 8.16
Ui = Us = 0.047, 0.122 S = 10%
Ui = 0.095, Us = 0.139 S ≅ 15% for 2x on 16 in. Centers U avg = ( 0.15 ) ( 0.139 ) + ( 0.85 ) ( 0.095 ) = 0.10
Ri 0.61 19.0 1.19 0.61 21.41
R 0.17 0.44 1.08 0.62 2.31
UR = 0.433
Uav (0.1)(0.122) + (0.9)(0.047) = 0.054 1 U C/R = -----------------------------------------------------------------------------1 ⁄ 0.054 + ( 2622 ⁄ 2195 ) ( 0.433 )
With storm window, treat as double glazing Wwindow = 0.55. Assume wood frame. ΔT design = [ 72 – ( – 13.4 ) ] = 85.4°F Selected ⇑
Outside air Shingles Plywood Attic Air ER
⇑ 99.6 % Value
Wall loss = ( 0.1 ) ( 648 ) ( 85.4 ) = 5533 Btu/h Ceiling loss = ( 0.0525 ) ( 2185 ) ( 85.4 ) = 9796 Btu/h Window loss = ( 0.55 ) ( 48 ) ( 85.4 ) = 2254 Btu/h
U C/R = 0.0525
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Outside air Face brick, 4 in. Plywood sheathing, 3/4 in. Insulation, 2 1/2 in. Airspace Studs Plasterboard
Roof
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Chapter 6—Residential Cooling and Heating Load Calculations⏐81
6.4 If the building of Problem 6.3 is a residence having a volume of 17,480 ft3 and is equipped with a humidifier set for 25% RH, determine a. b.
Sensible heat load due to infiltration Latent heat load due to infiltration
⇓ Estimated
⎛ 17 ,480 ft 3⎞ Infiltration ≈ ( 0.75 AC/h ) ⎜ -------------------------⎟ = 219 cfm ⎝ 60 min/h ⎠ q s = 1.10 (cfm) q l = 4840 (cfm)
( Δt ) = 1.1 ( 219 ) ( 88 ) = 21 ,200 Btu/h
( ΔW ) = 4840 ( 219 ) ( 0.0042 – 0.00038 ) = 4049 Btu/h
72°F, 25% ⇑
⇑ –13.4°F, 100%
6.5 For a frame building with design conditions of 72°F indoor and 12°F outdoor, determine the heat loss through each of the following components: a. b.
c. d.
Slab floor, 56 by 28 ft, on grade without perimeter insulation [Ans: 12,100 Btu/h] Single glass double-hung window, 3 by 5 ft, with storm window in common metal frame [Ans: 780 Btu/h] 1 3/8 in. thick solid wood door, 3 by 7 ft, with 25% single glazing [Ans: 490 Btu/h] Sliding patio door, 6 by 7 ft, metal frame with double insulating glass having 1/4 in. air space [Ans: 2192 Btu/h] q = F p P ( t i – t o ) = ( 1.20 ) [ 2 ( 56 + 28 ) ] ( 72 – 12 ) = 12 ,100 Btu/h (Table 6-18) b.
q = UA ( t i – t o ) = ( 0.81 ) ( 3 × 5 ) ( 72 – 12 ) = 729 Btu/h (Table 5-16)
c.
q = UA ( t i – t o ) = ( 0.58 ) ( 3 × 7 ) ( 72 – 12 ) = 731 Btu/h (Table 5-17)
d.
q = UA ( t i – t o ) = ( 0.81 ) ( 6 × 7 ) ( 72 – 12 ) = 2041 Btu/h (Table 5-16)
6.6 Determine the heat loss for a basement in Chicago, Illinois, which is 8 by 12 by 2.1 m high, of standard concrete construction, and entirely below grade. Wall area = ( 2 ) ( 8 + 12 ) ( 2.1 ) = 84 m 2 ; Floor area = 8 × 12 = 96 m 2 U w ,av = 0.980 W/(m 2 ·°C) U f ,av = 0.164 W/(m 2 ·°C) t i = 21°C ( selected ) ; t g = t w ,av – A = 2 – 12 = – 10°C (Table 4-8) (Fig. 6-1) Δt = t i – t g = 31 q = ( UA Δt ) walls + ( UA Δt ) floor q = ( 0.980 ) ( 84 ) ( 31 ) + ( 0.164 ) ( 96 ) ( 31 ) = 3040 W
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a.
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82⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
6.7 A residence located in Chicago, Illinois, has a total ceiling area of 1960 ft2 and consists of 3/8 in. gypsum board on 2 by 6 ceiling joists. Six inches of fiberglass (mineral/glass wool) insulation fills the space between the joists. The pitched roof has asphalt shingles on 25/32 in. solid wood sheathing with no insulation between the rafters. The ratio of roof area to ceiling area is 1.3. The attic is unvented in winter. For winter design conditions, including a 72°F inside dry bulb at the 5 ft line, determine a. b. c. d.
Outside design temperature, °F Appropriate temperature difference, °F Appropriate overall coefficient U, Btu/h·ft2 ·°F Ceiling heat loss q, Btu/h t o = – 4°F, Chicago CO
a.
99.6% value
b.
72°F inside – ( – 4°F ) outside = 76°F 1 U c = -------------------------------------------------------- = 0.049 0.61 + 19 + 0.34 + 0.61 1 U R = ------------------------------------------------------------- = 0.43 0.17 + 0.44 + 1.08 + 0.62 1 1 1 1 R t = ------ + ---------- = ------------- + ---------------------------- = 21.8 U c nU R 0.049 ( 1.3 ) ( 0.43 )
c.
d.
1 U T = ---------- = 0.046 21.8 q = U T A Δt = ( 0.046 ) ( 1960 ) ( 76 ) = 6852 Btu/h
t o = – 6.9°F [ Table 4-7 ] ; t a = 35.5°F ; A = 23 [Table 4-8, Fig. 6-1] Walls:
U av = 0.157 (use conservative 8 ft value) [Table 6-16]
Floor:
U f = 0.026 (conservative) [Table 6-17]
Q walls = U av A ( t i – t o ) = ( 0.157 ) ( 260 × 9 ) [ 72 – ( 35.5 – 23 ) ] = 21 ,900 Btu/h Q floor = U f A ( t i – t o ) = ( 0.026 ) ( 3000 ) [ 72 – ( 35.5 – 23 ) ] = 10 ,300 Btu/h -------------------------------Total = 32 ,200 Btu/h
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6.8 A residential building, 30 by 100 ft, located in Des Moines, Iowa, has a conditioned space which extends 9 ft below grade level. Determine the design heat loss from the uninsulated below grade concrete walls and floor.
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Chapter 6—Residential Cooling and Heating Load Calculations⏐83
6.9 Determine the heating load and specify the furnace for the following residence (located in St. Louis, Missouri) with a. 1 in. fiberglass wall insulation and 2 in. fiberglass ceiling insulation b. Full wall fiberglass insulation and 4 in. fiberglass ceiling insulation
Diagram for Problem 6.9
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Basic Plan Wall construction: Face brick, 25/32 in. insulating board sheathing, 2 by 4 studs on 16 in. centers, 3/8 in. gypsum board interior Ceiling: 2 by 6 ceiling joists, 16 in. on center, no flooring above, 3/8 in. gypsum board ceiling Roof: Asphalt shingles on solid wood sheathing, 2 by 6 rafters, no insulation between rafters, no ceiling applied to rafters, 1:4 pitch, 1 ft overhang on eaves, no overhang on gables Full basement: Heated, 10 in. concrete walls, all below grade, 4 in. concrete floor over 4 in. gravel Fireplaces: One in living room on first floor Garage: Attached but unheated Windows: W1: 3 by 5 ft singles glazed, double-hung wood sash, weather stripped with storm window W2: 10 by 5 1/2 ft picture window, double glazed, 1/2 in. airspace W3: 5 by 3 ft wood sash casement, double glazed, 1/2 in. airspace W4: 3 by 3 ft wood sash casement, double glazed, 1/2 in. airspace Doors: D1: 3 by 6 ft-8 in, 1 3/4 in. solid with glass storm door D2: Sliding glass door, two section, each 3 by 6 ft, 8 in. double glazed, 1/2 in. airspace, aluminum frame [Ans: (b) 52,000 Btu/h (15 kW)]
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84⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
Problem 6.9 continued t a ≅ 44°F A ≅ 22°F t i = 72°F
t o = 4.1°F
W i = 0.0042
Surface W1 windows (6) W2 windows W3 windows (2) W4 windows (2) D1 door (West) D1 door (South) D2 door Wall (N, W, E) Wall (S) Ceiling/Roof
A 90 55 30 18 20 20 40 1091 204 1960
W o = 0.00097 U 0.51 0.50 0.51 0.51 0.26 0.26 0.81 0.067 0.068 0.067
Δt = 67.9 67.9 67.9 67.9 67.9 67.9 67.9 67.9 67.9 67.9
t g = 44 – 22 = 22°F
q 3114 1872 1038 621 349 349 2202 4963 890 8916 24366 Btu/h
Δt = 72 – 22 = 50°F
Basement wall:
U av = 0.157 Btu/h· ft 2 ·°F (Table 6-16) Basement floor:
q = ( 0.157 ) [ ( 2 ) ( 70 + 28 ) ( 8 ) ] ( 50 ) = 12 ,300 Btu/h U f = 0.026 (Table 6-17)
q = ( 0.026 ) ( 70 × 28 ) ( 50 ) = 2550 Btu/h Infiltration: (Fig. 5-7) ACH ≅ 0.5 q s = [ 0.5 ( 70 × 28 × 8 ) ⁄ 60 ] ( 1.10 ) ( 72 – 4.1 ) = 9758 Btu/h q L = [ 0.5 ( 70 × 28 × 8 ) ⁄ 60 ] ( 4840 ) ( 0.0042 – 0.00097 ) = 2036 Btu/h q T = 24,366 + 12 ,300 + 2550 + 9758 + 2036 = 51, 010 Btu/h
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NOTE: U or R values may differ depending on which table is used.
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Chapter 6—Residential Cooling and Heating Load Calculations⏐85
6.10 Determine the total conductance loss through the wall panel as shown below. The window has a wooden sill and the plate glass (U = 1.06) covers 85% of the window area. [Ans: 9640 Btu/h (2.77 kW)]
q = UAΔt Wall: q = (0.39)(520 – 192– 21)(70 – 40) Door: q = (0.64)(21)(70 – 40) Window: q = (1.06)(192)(70 – 40)
= = =
3592 403 6105 10,100 Btu/h
6.11 Calculate, for design purposes, the heat losses from a room of a building as shown in the diagram, if the outside ambient is 0°F. [Ans: 62,820 Btu/h (18.4 kW)]
Volume = ( 24 ) ( 40 ) ( 13 ) = 12 ,480 ft 3
Glass = ( 13.75 ) ( 9 ) + ( 13.5 ) ( 9 ) = 245 ft 2
Net wall = ( 24 ) ( 13 ) + ( 40 ) ( 13 ) – 245 = 587 ft 2 Floor/Ceiling = ( 24 ) ( 40 ) = 960 ft 2 1 U wall = -------------------------------------------------------------------------------------------------- = 0.381 ( 1 ⁄ 1.65 ) + ( 16 ⁄ 9 ) + ( 0.5 ⁄ 5 ) + ( 1 ⁄ 7.2 ) 1 U glass = 0.89 U ceiling = -------------------------------------------------------------------------------------------------------------- = 0.143 ( 2 ⁄ 1.65 ) + ( 0.5 ⁄ 5 ) + ( 1 ⁄ 0.26 ) + 0.98 + 0.85 Floor: F P = 0.68 , P = 64 ft Infiltration:
Assume 1/2 ACH ⇒ CFM = 1.2 ( 12 ,480 ) ⁄ 60 = 104
Heat Losses: qglass = (0.89)(245)(70 – 0) = 15, 260 Btu/h qwalls = (0.381)(587)(70 – 0) = 15, 660 Btu/h qceiling = (0.143)(960)(70 – 0) = 4,800 Btu/h qfloor = (0.68)(64)(70 – 0) = 3,050 Btu/h qinfil = (1.10)(104)(70 – 0) = 8,000 Btu/h Total Loss = 46,800 Btu/h (13.7 kW)
NOTE: U or R values may differ depending on which tables are used.
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Diagram for Problem 6.11
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86⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
6.12 A room has three 760 by 1520 mm well-fitted double hung windows. For design conditions of –1°C and 21°C, calculate: a. b.
heating load from air leakage heating load from transmission through the windows. Assume single story frame house, 6.7 m/s wind. Leakage area = ( 3 + 0.3 ) ( 3 ) ( 0.76 ) ( 1.52 ) = 11.4 cm 2 = L 1/4 2 1/2 · V = [( A ) ( Δt ) + ( B ) (V )] L = [ ( 0.00188 ) ( 39 ) + ( 0.00413 ) ( 6.7 ) 2 ] 11.4 = 5.8 m 3 /h
a.
q s = ( 1.232 ) ( 5800 ⁄ 3600 ) ( 39 ) = 73.4 W
b.
q tran = AU ( Δt ) = ( 6.2 ) ( 3 ) ( 0.76 ) ( 1.52 ) ( 39 ) = 838 W
6.13 A residence has a total ceiling area of 1960 ft2 and consists of 3/8 in. gypsum board on 2 by 6 in. ceiling joists. Six inches of fiberglass (mineral/glass wool) insulation fills the space between the joists. The effect of the joists themselves can be neglected. The pitched roof has asphalt shingles on 5/8 in. plywood with no insulation between the rafters. The ratio of roof area to ceiling area is 1:3. The attic contains louvers which remain open all year. The residence is located in Louisville, Kentucky. For winter design conditions, determine: a. b. c.
(a) appropriate temperature difference Δt overall coefficient U ceiling heat loss. Louisville:
1 U R = ------------------------------------------------------------- = 0.362 0.17 + 0.44 + 0.77 + 0.62
a.
1 U ci = -------------------------------------------------------- = 0.049 ⎫ 0.61 + 0.32 + 19 + 0.61 ⎪ ⎬ U av = ( 0.1 ) ( 0.115 ) + ( 0.9 ) ( 0.049 ) = 0.055 1 U cs = ------------------------------------------------------------- = 0.115 ⎪ ⎭ 0.61 + 0.32 + 7.14 + 0.61 ( 1960 ) ( 0.055 ) ( 75 ) + 6 [ ( 1.08 )1960 ( 0.5 ) + ( 2548 ) ( 0.362 ) ] t a = ------------------------------------------------------------------------------------------------------------------------------------------------ = 9.6°F 1960 [ 0.055 + 1.08 ( 0.5 ) ] + ( 2548 ) ( 0.362 ) Δt = 75 – 7.1 = 67.9°F
b.
U c = 0.055 Btu/h· ft 2 ·°F
c.
q = U c A c ( t c – t a ) = ( 0.055 ) ( 1960 ) ( 67.9 ) = 7319 Btu/h ( 2.15 kW )
6.14 Estimate the heat loss from the uninsulated slab floor of a frame house having dimensions of 18 by 38 m. The house is maintained at 22°C. Outdoor design temperature is –15°C in a region with 5400 degree kelvin days. [Ans: 8.6 kW] q = FP P ( t i – t o ) = ( 2.07 ) ( 112 ) [ 22 – ( – 15 ) ] = 8580 W
NOTE: U or R values may differ depending on which tables are used.
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t o = 7.1°F ; t i = 75°F ; A c = 1960 ; A R ⁄ A c = 1.3 ; A R = 2548 · A c U c t c + t o ( 1.08A c Vc + A R U R ) · t a = --------------------------------------------------------------------------Vc ≅ 0.5 cfm/ft 2 · A c ( U c + 1.08Vc ) + A R U R
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Chapter 6—Residential Cooling and Heating Load Calculations⏐87
6.15 Repeat Problem 6.14 for the case where insulation [R = 0.9 (m2 ·K)/W] is applied to the slab edge and extended below grade to the frost line. q = FP P ( t i – t o ) = ( 0.92 ) ( 112 ) ( 22 + 15 ) = 3800 W
6.16 To preclude attic condensation, an attic ventilation rate of 59 L/s is provided with outside air at –13°C. The roof area is 244 m 2 and Uroof = 2.7 W/(m2 ·K). The ceiling area is 203 m 2 and Uclg = 0.30 W/(m2 ·K). Inside design temperature is 22°C. Determine the ceiling heat loss W with ventilation and compare to the loss if there had been no ventilation. · A c U c t c + t o ( 1200A c Vc + A R U R ) t a = ----------------------------------------------------------------------------· A c ( U c + 1200Vc ) + A R U R
( Eq. 4-6 )
( 203 ) ( 0.3 ) ( 22 ) + ( – 13 ) [ ( 1200 ) ( 203 ) ( 0.059 ) + ( 244 ) ( 2.7 ) ] = -------------------------------------------------------------------------------------------------------------------------------------------------203 ( 0.3 + 1200 ( 0.059 ) ) + ( 244 ) ( 2.7 ) t a = – 12.9°C ; q = ( 0.3 ) ( 203 ) ( 22 + 12.9 ) = 2125 W ( 7250 Btu/h ) 1 U o ,c = --------------------------------------------------------------------------- = 0.275 ( 1 ⁄ 0.3 ) + [ ( 203 ) ⁄ ( 244 ) ( 2.7 ) ] q = ( 0.275 ) ( 203 ) ( 22 + 13 ) = 1954 W ( 6670 Btu/h )
If uninsulated:
t i = 72°F
Ceiling: 1 1 U c = -------------------------------------------- = ---------- = 0.649 0.61 + 0.32 + 0.61 1.54
t o = 14.2°F
Roof: 1 1 U R = ------------------------------------------------------------- = ---------- = 0.543 0.17 + 0.44 + 0.62 + 0.61 1.84 ( 2300 ) ( 0.649 ) ( 72 ) + ( 14.2 ) ( 1.08 ) ( 325 ) + ( 2950 ) ( 0.543 ) t a = -------------------------------------------------------------------------------------------------------------------------------------------- = 33.1°F ( 2300 ) ( 0.649 ) + ( 1.08 ) ( 325 ) + ( 2950 ) ( 0.543 )
NOTE: U or R values may differ depending on which tables are used.
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6.17 For a residence in Roanoke, Virginia, the hip roof consisting of asphalt shingles on 1/2 in. plywood has an area of 2950 ft2. The 2300 ft2 ceiling consists of 3/8 in. plasterboard on 2 by 6 joists on 24 in. centers. The attic has forced ventilation at the rate of 325 cfm. Determine the attic air temperature at winter design conditions.
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88⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
6.18 Solve the following: a.
b.
A 115 by 10 ft high wall in Minneapolis, Minnesota, consists of face brick, a 3/4 in. air gap, 8 in. cinder aggregate concrete blocks, 1 in. organic bonded glass fiber insulation, and 4 in. clay tile interior. Determine the design heat loss through the wall in winter, Btu/h. If the wall of Part (a) is converted to 60% single glazed glass, what is the winter design heat loss through the total wall, Btu/h? ti = 72°F; a.
to = –13.4°F
R 0.17 0.44 ~ 1.0 1.72 4.0 1.11 068 = 9.12 ΣR
Outside air Face brick Air gap, 3/4 in. Cinder block, 8 in. Insulation, 1 in. Clay tile, 4 in. Inside air
1 U = --------- = 0.110 Btu/h· ft 2 ·°F ∑R b.
Q = ( 0.11 ) ( 115 ) ( 10 ) ( 72 + 13.4 ) = 10, 803 Btu/h Q = ( 0.60 ) ( 115 ) ( 10 ) ( 1.13 ) ( 85.4 ) + ( 0.40 ) ( 115 ) ( 10 ) ( 0.11 ) ( 85.4 ) = 70, 911 Btu/h U glass ¡
NOTE: U or R values may differ depending on which tables are used.
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Chapter 6—Residential Cooling and Heating Load Calculations⏐89
6.19 Determine the design heating load for a residence, 30 by 100 by 10 ft, to be located in Windsor Locks, Connecticut, which has an uninsulated slab on grade concrete floor. The construction consists of Walls: 4 in. face brick, 3/4 in. plywood sheathing, 4 in. cellular glass insulation, and 1/2 in. plasterboard Ceiling/roof: 3 in. lightweight concrete deck, built-up roofing, 2 in. of rigid, expanded rubber insulation, and a drop ceiling of 1/2 in. acoustical tiles, some 18 in. below the roof. Windows: 45% of each wall is double pane, nonoperable, metal-framed glass (1/4 in. air gap). Doors: Two 3 by 7 ft, 1.75 in. thick, solid wood doors are located in each wall. NOTE: U or R values may differ depending on which tables are used. Walls Outside air Face brick, 4 in. Plywood sheathing, 3/4 in. Cellular glass insulation, 4 in. Plasterboard, 1/2 in. Inside air ΣR = Uw =
R 0.17 0.44 0.93 12.12 0.45 0.68 14.79 0.0676
Ceiling/Roof Outside air Built-up roofing Lightweight concrete, 3 in. Rubber insulation, 2 in. Air Air Acoustical tiles, 1/2 in. Inside air
~
ΣR = Uc =
Windows: U g = 0.69
Floor: F P ≈ 0.84 1 30 × 100 × 10 Infiltration: @ 1/2 ACH --- ⎛⎝--------------------------------- ⎞⎠ = 250 CFM 2 60
Design Values: t i = 72°F ; φ i = 30% rh ; W i = 0.005 t o = 3°F ; φ o = 100% rh ; W o = 0.00092 Heat Losses: Walls: Roof: Doors: Windows: Floor: Infiltration:
Q Q Q Q Q Qs QL
= (0.0676)(1262)(69) = 5,880 Btu/h = (0.0763)(3000)(69) = 15,790 Btu/h = (0.46)(168)(69) = 5,330 Btu/h = (0.69)(1170)(69) = 55,700 Btu/h = (0.84)(260)(69) = 15,070 Btu/h = (1.10)(250)(69) = 18,975 Btu/h = (4840)(250)(0.005 – 0.00092) = 4,940 Btu/h Total Loss = 121,700 Btu/h
Problems 6.20, 6.21, and 6.22 are essentially openended but should be solved using the RLF method, illustrated in Section 6.6.
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Doors: U d = 0.46
R 0.17 0.33 0.42 9.10 0.61 0.61 1.25 0.61 13.10 0.0763
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Solutions to
Chapter 7 NONRESIDENTIAL COOLING AND HEATING LOAD CALCULATIONS
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Chapter 7—Nonresidential Cooling and Heating Load Calculations⏐93
7.1 The exterior windows are of double insulating glass with 0.25 in. (6 mm) airspace and have metal sashes. Determine the design U-factor for cooling for the window. From Table 5-6, ID #4: U = 0.81 Btu/h· ft 2 ·°F, if operable, or U = 0.68 Btu/h· ft 2 ·°F, if fixed
7.2 A store in Lafayette, Indiana, is on the northeast corner of an intersection with one street running due north. The bottom of the show windows are 2 ft 6 in. above the sidewalk; the show windows are 7 ft high. An aluminum awning with a 3 in. rise per horizontal foot is to be hung with the bottom strut at the window header. Both south and west awnings are to have the same dimensions. a.
b. c. d.
What minimum distance should the strut extend from the building to keep the shade line on the windows at 3 P.M. sun time? Which face of the building governs the awning dimensions? Where will the shade line be at 3 P.M. on the other face of the building? What is the elevation of the top of the awnings above the sidewalk? P = S H cot Ω
West, Sept.: Ω = 37° P south = 7 × cot 50° = 7 ( 0.839 ) = 5.9 ft P west = 7 × cot 37° = 7 ( 1.327 ) = 9.3 ft
b. West c. 9.3 = SH ( 0.839 ) ⇒ S H = 11.1 ft, shadow to ground d.
3 9.3 ⎛ ------ ⎞ + 9.5 = 11.8 ft above ground ⎝ 12 ⎠
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a.
South, Sept.: Ω = 50°
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94⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
7.3 Calculate the solar radiation entering through clear glass as shown below. [Ans 692 Btu/h]
Mullion: M = P tan γ Tansom: T = P tan β⁄ cos γ 40° N latitude, July 21, 1 P . M ., south γ ( ψ = 0° from south ) = φ = 37°; β = 66° 17 17 M = ------ tan 37 = 1.1 ft T = ------ tan 66 ⁄ cos 37 = 4 ft 12 12 Area in Sun = ( width – M ) ( height – T ) A s = [ 3 – ( 1.1 – 0.83 ) ] [ 4 – ( 4 – 0.83 ) ] = 2.27 ft 2 Total glass area = 3 × 4 = 12 ft 2 q = As ( E D )SHGC ( θ )IAC + AT ( E d + E r )SHGCD IAC A s = 2.27 ;
IAC = 1 ;
A T = 12
A E DN = ---------------------------------- ; A = 1093 W/m 2 exp ( E ⁄ sin β ) B = 0.186 , C = 0.138 1093 E DN = ---------------------------------------------- = 892 W/m 2 = 283 Btu/h· ft 2 exp ( 0.186 ⁄ sin 66 ) E D = E DN cos θ = 283 ⋅ cos 71 = 92 Btu/h· ft 2 2
E d = CYE DN ; Y = 0.55 + 0.437 cos θ + 0.313 cos θ Y = 0.725 E d = ( 0.138 ) ( 0.725 ) ( 283 ) = 28 Btu/h· ft 2 E r = E DN ( C + sin β )ρg ( 1 – cos Σ ) ⁄ 2 ; cos Σ = 0 for vertical = 283 ( 0.138 + sin 71 ) ( 0.2 ) ⁄ 2
ρ g = 0.2 (typical)
= 31 Btu/h· ft 2 ·°F q = ( 2.27 ) ( 92 ) ( 0.67 ) ( 1 ) + ( 12 ) ( 28 + 31 ) ( 0.78 ) ( 1 ) q = 139.9 + 552.2 = 692 Btu/h
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Table 7-13, θ = 71° ; Table 7-4, SHGC ( 71 ) = 0.67 Table 7-4, SHGCD ( 71 ) = 0.78
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Chapter 7—Nonresidential Cooling and Heating Load Calculations⏐95
7.4 Solve the following: a.
b.
Determine the solar angle of incidence for a vertical wall facing 15° west of south when the sun has an azimuth of 79.2° west of south and an altitude of 75.7°. [Ans: 83.8°] Find the solar incident angle (for direct solar radiation) for a vertical surface facing southeast at 8:30 A.M. CST on October 22 at 32° N latitude and 95° W longitude. [Ans: 28.4°] a.
cos θ v = cos β cos γ ; ψ = 15 ; φ = 79.2 ; β = 75.7 γ = φ – ψ = 64.2
–1
θ v = cos [ ( cos 75.7 ) ( cos 64.2 ) ] = 83.8°
LCT = 8:30 + 4 ( 90 – 95 ) = 8:10 A.M. Equation of time = 15.4 AST = 8:10 + :15 = 8:25 A.M. ( 215 min. before noon ) H = 0.25 ( 215 ) = 53.8° δ ≅ – 10.5° sin β = cos L cos δ cos H + sin L sin δ ⇒ sin β = 0.396 ⇒ β = 23.3° cos φ = ( sin β sin L – sin δ ) ⁄ ( cos β cos L ) ⇒ cos φ = 0.503 ⇒ φ = 60° cos θ v = cos β cos γ = 0.8872 ⇒ θ v = 28.4°
b.
7.5 What environmental factors affect the solar intensity reaching the earth’s surface? Angle through which the solar radiation passes through the atmosphere increases or decreases the quantity of air mass.
2.
Gas molecules, ozone, and water vapor
3.
Dust and other contaminants
4.
Height of ground upon which the structure is to be built where the elevation is substantial.
7.6 Determine the heat being dissipated by 50 pendant mounted fluorescent luminaires with four 40 W lamps in each luminaire. 4-40 Watt lamps in each of 50 luminares. q = ( wattage ) ( use factor ) ( spec. allow. factor ) ( 3.413 ) = ( 50 ) ( 4 ) ( 40 ) ( 1 ) ( 1.15 ) ( 3.413 ) = 31 ,400 Btu/h
7.7 How much sensible, latent, and total heat is contributed by 50 customers shopping in a drugstore? Sensible: Latent:
(50)(250) = (50)(250) = Qtotal =
12,500 Btu/h 12,500 Btu/h 25,000 Btu/h
(Table 7-14)
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1.
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96⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
7.8 A 1 hp motor driving a pump is located in a space to be air conditioned. Determine heat dissipated to the space from the motor and pump. [Ans: 3390 Btu/h] Table 7-16 1 Hp Motor Assume both in the space
3390 Btu/h
7.9 Calculate the heat gain to a room from an open deep fat fryer if (a) hooded and (b) nonhooded. Table 7-18 (a) Hooded: (b)
q s = 47, 800 Btu/h
Nonhooded: NOT RECOMMENDED
7.10 Calculate the maximum heat gain through the floor for a room directly over a boiler room. The air temperature at the underside of the floor is 100°F, and the room air temperature desired close to the floor is 70°F. The floor is 4 in. concrete with vinyl tile finish. [Ans: 18.87 Btu/h·ft2 (18.9 W/m2)] 1 1 U = --------------------------------------------------- = ------------------------------------------------------------- = 0.70 1 1 0.61 + 0.31 + 0.51 + 0.61 ---- + R con + R tile + ---hi hi q = ( U ) ( A ) ( Δt ) = ( 0.7 ) ( 1 ) ( 100 – 70 ) = 21 Btu/h· ft 2 Description General office Director’s room Conference room 5 private offices
Size 25 by 50 ft 25 by 25 ft 10 by 25 ft 10 by 10 ft
Occupancy 75 ft2 per person 16 people Plush furnishings Smoking permitted
Values from Table 5-9: General Office: Directors Room: Conference Room: Private Offices:
25 × 50 ft cfm- + 0.06 -------5 CFM/person × ------------------------------( 25 × 50 ) = 158 cfm 75 ft 2 /person ft 2 16 people × 5 + 0.06 ( 25 × 25 ) = 118 cfm 50 people ----------------------- ( 10 × 25 )5 CFM/person + 0.06 ( 10 × 25 ) = 78 cfm 1000 ft 2 Smoking areas not covered by ASHRAE Standards; This area should have separate system.
Total = 354 cfm
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7.11 An air-conditioning unit serves an office having the following areas: What quantity of outdoor air must be brought into the air-conditioning unit for ventilation?
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Chapter 7—Nonresidential Cooling and Heating Load Calculations⏐97
7.12 Suppose the fan of the air-conditioning unit in Problem 7.11 supplies 3200 cfm to the ductwork. a. b. c.
How many air changes per hour are being used? (Assume a ceiling height of 9 ft.) What is the percentage of outdoor air? Suppose the minimum recommended quantities of total air and outside air are used, what will be the percentage of outdoor air? [Ans: 10, 11%, 15%] Assume smoking spaces are on separate dedicated system. Total volume of spaces: [(25 × 50) + (25 × 25) + (10 × 25)]9 = 19,100 ft3 a. b. c.
3200 ( 60 ) Supply air changes/h = ----------------------- = 10 ach (high) 19 ,000 158 + 118 + 78 % OA = ------------------------------------ ( Problem 7.11 ) × 100 = 11% 3200 For proper air distribution, a “typical” number of 6 ach is sometimes used. If so, supply air, cfm = 6 × 19,100 ft3/60 min/h = 1910 cfm 158 + 118 + 78 % OA = ------------------------------------ × 100 = 15% 1910
7.13 A small parts assembly area in a factory has a working force of 25 men and occupies a space 27.4 by 9.1 m with a 3 m ceiling. Smoking is not allowed. Determine Sensible heat load from the occupants Latent heat load from the occupants Moisture added from the occupants The minimum volume of outdoor air for ventilation Suitable summer design inside dry-bulb temperature a.
q s = 25 100 --------- ( 235 ) = 2554 W 230
b.
q L = 25 130 --------- ( 235 ) = 3320 W 230
c. d. e.
q L ( Btu/h ) 3320 ( 3.413 ) Moisture added = --------------------------- = ------------------------------ = 10.3 lb/h ( 0.00128 kg/s ) 1100 Btu/h 1100 Occupancy (Table 5-9 for metal shop) ( 5 × 25 ) + 0.9 ( 27.4 × 9.1 ) = 349 L/s 20 to 22°C
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a. b. c. d. e.
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98⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
Problems 7.14 through 7.17. No manual (hand) cooling load method is currently recommended by ASHRAE. Cooling loads for each of these buildings/spaces should be determined using the RTS software available on the CD provided with “Principles.” 7.18 Solve the following: a.
b.
A 115 by 10 ft high wall in Minneapolis, Minnesota consists of face brick, a 3/4 in. air gap, 8 in. cinder aggregate concrete blocks, 1 in. organic bonded glass fiber insulation, and 4 in. clay tile interior. Determine the design heat loss through the wall in winter in Btu/h. If the wall of Part (a) is converted to 60% single glazed glass, what is the winter design heat loss through the total wall in Btu/h? ti = 72°F, to = –13.4°F a.
R 0.17 0.44 1.0 1.72 4.0 1.11 0.68 9.12
1 U = --------- = 0.110 Btu/h· ft 2 ·°F ∑R b.
q = ( 0.11 ) ( 115 ) ( 10 ) ( 72 + 13.4 ) = 10 ,800 Btu/h q = ( 0.60 ) ( 115 ) ( 10 ) ( 1.12 ) ( 85.4 ) + ( 0.40 ) ( 115 ) ( 10 ) ( 0.11 ) ( 85.4 ) U glass ¡ q = 70, 300 Btu/h
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Outside air Face brick Air gap, 3/4 in. ~ Cinder block, 8 in. Insulation, 1 in. Clay tile, 4 in. Inside air ΣR =
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Solutions to
Chapter 8 ENERGY ESTIMATING METHODS
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Chapter 8—Energy Estimating Methods⏐101
8.1 The total design heating load on a residence in New York City is 32.8 kW (112,000 Btu/h) for an indoor temperature of 72°F. The furnace is off from June through September. Estimate: a. the annual energy requirement for heating b. the annual heating cost if electric heat is used with the single rate of 16¢/kWh, $/yr c. the maximum savings effected if the thermostat is set back to 65°F between 10 P.M. and 6 A.M., $/yr New York City: DD = 4848 to = 13; ti = 72; η = 1 CD = 0.77; V = 1 a.
32.8 ⎛ ( 4848 ) ( 24 ) ⎞ E = ---------------------- ⎜ ---------------------------- ⎟ ( 0.77 ) = 49 ,800 kWh ( 72 – 13 ) ⎝ ( 1.0 ) ( 1 ) ⎠
b.
Cost = ( 49 ,800 ) ( 0.16 ) = $7969 00
c.
816 ⎛ ----⎞ -----( t i – t o ) with ⎝ 24 ( 65 ) + 24 ( 72 ) ⎠ – 42.5 % with Setback ---------------------------------- ( 100 ) = ---------------------------------------------------------------------- = 92% or 8% Savings ( t i – t o ) without ( 72 – 42.8 )
Cleveland: ti = 72°F; to = 1°F; January 1159 DD DD = (Days in Period)(65 – to,av) = 1159 = 31(65 – to, av) to,av = 27.6°F for January
Without Setback:
( q L ) ( DD ) ( 24 ) ( 112 ,000 ) ( 1159 ) ( 24 ) E = ----------------------------------- ( C D ) = --------------------------------------------------------- ( 1 ) ( 72 – 1 ) ( 0.7 ) ( 140 ,000 ) ( Δt ) ( η ) (V )
E = 448 gal 8 16 t i ,av = ------ ( 68 ) + ------ ( 72 ) = 70.67°F 24 24 ( 72 – 27.6 ) – ( 70.67 – 27.6 ) % Savings = -------------------------------------------------------------------- ( 100 ) = 3.0% ( 72 – 27.6 ) Fuel Savings = ( 448 ) ( 0.03 ) = 13.4 gal ( 50 L ) With Setback:
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8.3 A home is located in Cleveland, Ohio, and has a design heat loss of 112,000 Btu/h at an inside design temperature of 72°F and an outside design temperature of 0°F. The home has an oil-fired furnace. Find the savings in gallons of fuel oil if the owner lowers the temperature in the home to 68°F between 10 P.M. and 6 A.M. every day during January.
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102⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
8.4 The total design heating load on a residence in Kansas City, Missouri, is 32.8 kW (112,000 Btu/h). The furnace is off during June through September. Estimate: a. Annual energy requirement for heating b. Annual heating cost if No. 2 fuel oil is used in a furnace with an efficiency of 80% (assume fuel oil costs 68¢/L) c. Maximum savings effected if the thermostat is set back from 22.2 to 18.3°C (72 to 65°F) between 10 P.M. and 6 A.M. in $/yr Kansas City: HL = 32.8 kW (112,000 Btu/h); to = –1°F; ti = 72°F DD = 5161 a. b.
c.
CD = 0.77
( q L ) ( DD ) ( 24 ) ( 112 ,000 ) ( 5161 ) ( 24 ) E = ----------------------------------- ( C D ) = ----------------------------------------------------- ( 0.77 ) = 1.46 ×108 Btu ( Δt ) ( 73 ) ( 1.46 ×108 ) E F = ----------- = ---------------------------------------- = 1300 gal ( 4950 L ) ( 0.80 ) ( 140 ,000 ) η⋅V Cost = ( 4950 ) ( 0.68 ) = $3366 t o ,av = 43.9°F 8 16 t i ,av = ------ ( 65 ) + ------ ( 72 ) = 70°F 24 24 ( 72 – 43.9 ) – ( 70 – 43.9 ) % of Savings = ------------------------------------------------------------ ( 100 ) = 7.1% Savings ( 72 – 43.9 ) Cost Savings = ( 0.071 ) ( 2300 ) = $165
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Chapter 8—Energy Estimating Methods⏐103
8.5 A residence located in Tulsa, Oklahoma has a design heating load of 20 kW and a design cooling load of 9.4 kW. Determine the following: a. Heating energy requirements, kWh b. Litres of No. 2 fuel oil per season if used as heating fuel c. Litres of natural gas per season if used as heating fuel d. kWh of electric energy if used as heating fuel with baseboard units e. kWh of electric energy if used for air-conditioning system having COPseasonal = 3.4 f. Total airflow rate in L/s if a warm air system is used g. Total steam flow in kg/s if a steam system is used Tulsa, OK: DD = 3680; to = 9°F; ti = 72°F CDD = 1949; to = 97; ti = 78 ( H L ) ( DD ) ( 24 ) ( 20 ) ( 3413 ) ( 3680 ) ( 24 ) E = ------------------------------------- ( C D ) = -------------------------------------------------------- ( 0.77 ) = 7.37 ×107 Btu ( 21 590 kWh ) ( Δt ) ( η ) (V ) ( 72 – 9 )
b.
7.37 ×107 F = ------------------------------------- = 658 gal ( 2500 L ) ( 140 ,000 ) ( 0.8 )
c.
7.37 ×107 - = 8770 ft 3 ( 2.48 ×106 L ) F = -------------------------------------------------( 1050 Btu/ft 3 ) ( 0.80 )
d.
7.37 ×107 F = ------------------------- = 21 600 kW ⋅ h ( 3413 ) ( 1 )
e.
Q c ⎛ ( CDD ) ( 24 )⎞ ( 9.4 ) ( 1949 ) ( 24 ) E c = ------ ⎜ ----------------------------⎟ = ---------------------------------------- = 6800 kWh COP ⎠ ( 97 – 78 ) ( 3.4 ) Δt ⎝
g.
¡
f.
Typical · · Q s = ( 1.2 ) ( V ) ( t supply – t return ) ; 20 ,000 = ( 1.2 ) (V ) + ( 54.4 – 22.2 ) · V = 518 L/s Q s = m· ( h fg ) ; 20 kJ/s = m· ( 232.8 kJ/kg ); m· = 0.0086 kg/s
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a.
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104⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
8.6 Estimate the annual energy costs for heating and cooling a residence located in Cleveland, Ohio, having design loads of 65,000 Btu/h (heating) and 30,000 Btu/h (cooling) based on a 75°F indoor temperature. In winter the thermostat is set back to 60°F for 10 h each night. The furnace is on from October 1 through May 31. Electric baseboard heat is used. The air conditioner has an SEER of 7.3 (Btu/h)/W. Electricity costs 0.0725 $/kWh year round. Cleveland: HDD = 6154; CDD = 613; SEER = 7.3 Average Winter Temperature = 37.2°F; CD = 0.77 Design Temperatures: 1°F Winter ; 86 °F Summer 65 ,000 Winter: ER = ---------------- ( 6154 ) ( 24 ) ( 0.77 ) = 99 ,900 ,000 Btu without Setback ( 75 ) ( 75 – 37.2 ) – ( 68.8 – 37.2 ) 14 10 Setback Savings: t i , av = ------ ( 75 ) ÷ ------ ( 60 ) = 68.8°F Savings = ----------------------------------------------------------------- ( 100 ) = 16.4% 24 24 ( 75 – 37.2 ) ER = ( 99 ,900 ,000 ) ( 1 – 0.164 ) = 83 ,500 ,000 Btu with Setback 83 ,500 ,000 Cost: Heating → --------------------------- ( 0.0725 ) = $1774 3413 30 ,000 613 ( 24 ) Summer: ER = ------------------- ----------------------------- = 7560 kWh ( 8 – 78 ) ( 7.3 ) ( 1000 ) Cost: ( 7560 ) ( 0.0725 ) = $548 Annual Cost: 1774 + 548 = $2322
a. b. c. c. e. f. g. h.
c. d. e. f. g. h.
Cubic feet of natural gas/yr kWh Total airflow rate in cfm if a warm air system is used Total steam flow in lb/h if a steam system is used Total water flow rate in gpm if a hydronic system is used Total electric power in kW if electric heating is used
q 68 ,000 ER = -------------------- ( DD ) ( 24 )C D = ------------------- ( 5435 ) ( 24 ) ( 0.77 ) = 94 ,900 ,000 Btu/yr ( ti – to )d ( 75 – 3 ) 94 ,900 ,000 gal, Fuel Oil = ---------------------------------------- = 966 gal/yr ( 140 ,000 ) ( 0.70 ) 94 ,900 ,000 ft 3 , Natural Gas = -------------------------------- = 120 ,800 ft 3 /yr ( 1050 ) ( 0.75 ) 94 ,900 ,000 kWh, Electricity = -------------------------------- = 29 ,300 kWh/yr ( 3413 ) ( 0.95 ) 68 ,000 ( cfm ) air = ----------------------------------------- = 1120 cfm ( 1.10 ) ( 130 – 75 ) 68 ,000 ( lb/h ) steam = ---------------- = 68 lb/h 1000 68 ,000 ( gpm ) water = ------------------------------------------ = 6.8 gal/min ( 1 ) ( 20 ) ( 60 ) ( 8.3 ) 68 ,000 kW = ---------------- ≅ 20 kW 3413
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8.7 A residence in St. Joseph, Missouri, has a design heating load of 68,000 Btu/h when design indoor and outdoor temperatures are 75°F and 3°F, respectively. The furnace is off from June through September. Determine the fuel and energy requirements for heating in: a. Btu b. Gallons of No. 2 fuel oil/yr
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Chapter 8—Energy Estimating Methods⏐105
8.8 For a residence located in New Orleans, Louisiana, the design cooling load is 12 kW (41,000 Btu/h). Determine: a. Annual energy requirements for cooling, kWh b. Cost of this energy if the electric rate is 6.5¢/kWh New Orleans:
q c = 12 kW ( 41 ,000 Btu )
Summer Design: 92 ⁄ 78°F ( Inside = 78°F ) CDD = 2706 Assume Air Conditioner SEER = 13 41 ,000 ( 2706 ) ( 24 ) a. ER = ---------------------- ---------------------------- = 14 ,600 kWh ( 92 – 78 ) 13 ( 1000 ) b. Cost = ( 0.065 ) ( 14 ,600 ) = $951
8.9 An office building located in Springfield, Missouri, has a heat loss of 2,160,000 Btu/h for design condition of 75°F inside and 10°F outside. The heating system is operational between October 1 and April 30. Determine: a. Annual energy usage for heating b. Estimated fuel cost if No. 2 fuel oil is used having a heating value of 140,000 Btu/gal and costing $2.50/ gal Springfield: DD = 4570; CD = 0.77
a.
2 ,160 ,000 ER = ------------------------ ( 4570 ) ( 24 ) ( 0.77 ) = 2.81 ×109 Btu ( 75 – 10 )
b.
2.81 ×109 Fuel Cost ≈ ---------------------------------------- ( 2.50 ) = $71, 800/yr ( 140 ,000 ) ( 0.70 )
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Average Winter Temperature = 44.5°F
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106⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
8.10 A small football promotion office is being designed for Jacksonville, Florida. The design heating and cooling loads are 61,200 and 55,400 Btu/h, respectively, based on 99.6% and 1% outdoor design dry bulb temperatures. Balance point has been estimated as 65°F. a. Select an appropriate heat pump from the XYZ Corporation models listed on the next page and estimate the energy costs for summer and winter if electricity is 8¢/kWh. b. Compare the heating energy cost for the heat pump to that for a condensing gas furnace with natural gas costing $1.20 per therm. Jacksonville: Winter: 32°F; HDD = 1327 Summer: 93/77°F; CDD = 2596 a.
Heat Pump A060 for 55,400 cooling need Winter: ( Energy = 5946 kWh see BIN Sheet ) Cost = 0.08 × 5946 = $476 55 ,400 ( 2596 ) ( 24 ) Summer: Cost = ---------------------- -------------------------------- ( 0.08 ) = $1753 ( 93 – 78 ) ( 10.5 ) ( 1000 ) 61 ,200 ER = ---------------------- ( 1327 ) ( 24 ) ( 0.77 ) = 37 ,500 ,000 Btu ( 72 – 32 ) Chapter 19: AFUE = 0.925 = η 37 ,500 ,000 Cost = ------------------------------------------- ( $1.20 ) = $487 ( 100 ,000 ) ( 0.925 )
b.
A
B
Calculation of Annual Heating Energy Consumption
House C
D
Heat Pump E
F
G
H
Supplemental I
Jd
Ke
Lf
Mg
Nh
Heat Seasonal SuppleTotal Heat Pump Cycling Pump Heat Pump mental Electric Weather Rated Heat Integrated Capacity Adjusted Temp. Temp. Heating Energy Data Loss Heating Adjust- Heat Pump Electric Operating Supplied Elec. Con- Space Bin, Diff. Load, Required, ConsumpHeating, sumption, Bin, Capacity, Input, Time Rate, Capacity, ment °F tbal − tbin hours 1000 Btu/h 1000 Btu/h Factora 1000 Btu/hb 106 Btuf kWhe kWhg tionh kW Fractionc 106 Btud 62 3 879 4.59 81.3 0.76 61.8 6.79 0.07 3.80 418 4.03 0 418 57 8 692 12.24 76.3 0.79 60.3 6.56 0.20 8.35 908 8.47 0 908 52 13 530 19.89 71.2 0.82 58.4 6.32 0.34 10.52 1139 10.54 0 1139 47 18 355 27.54 66.0 0.85 56.1 6.05 0.49 9.76 1052 9.78 0 1052 42 23 288 35.19 61.0 0.89 54.3 5.81 0.64 10.01 1071 10.13 0 1071 37 28 154 42.84 56.0 0.94 52.6 5.56 0.81 6.56 694 6.60 0 694 32 33 83 50.49 51.0 1 51.0 5.30 0.99 4.19 436 4.19 0 436 27 38 24 58.14 46.3 1 46.3 5.05 1 1.11 121 1.40 85 206 22 43 2 65.79 41.7 1 41.7 4.81 1 0.08 10 0.13 12 22 17 12 7 2 –3 TOTALS: a Cycling
Capacity Adjustment Factor = 1 − Cd(1 − x), where Cd = degradation coefficient (default = 0.25 unless part load factor is known) and x = building heat loss per unit capacity at temperature bin. Cycling capacity = 1 at the balance point and below. b Col G = Col E × Col F c Operating Time Factor equals smaller of 1 or Col D/Col G
5849
d Col J = (Col I × Col G × Col C)/1000 e Col K = Col I × Col H × Col C f Col L = Col C × Col D/1000 g Col M = (Col L – Col J) × 106/3413 h Col N = Col K + Col M
97
5946
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Problem 8-10 Climate
Chapter 8—Energy Estimating Methods⏐107 8.11 A 1980 ft2 residence located in Cincinnati, Ohio, has design heating and cooling loads of 74,000 Btu/h and 35,000 Btu/h, respectively. Determine a. Heating energy requirements, Btu b. Gallons of No. 2 fuel oil if 75% efficient oil-fired warm air system is used c. Therms of natural gas if 88% efficient gas-fired warm air system is used d. kWh of electricity if 98% efficient baseboard units are used e. Required airflow, cfm, for warm air systems f. kWh of electricity if heat pump system (WA-36 specifications follow) including supplementary electric resistance heat is used g. kWh of electricity for cooling for air conditioner with SEER of 8.5 using degree-day estimation h. Required airflow, cfm, for air conditioning
Performance Data for Model WA-36 Heat Pump Air Temperature, °F
Heat Pump Output, Heat Pump Input, 1000 Btu kW
62 57 52 47 42 37 32 27 22 17 12 7 2 −3
44 43 41 39 36 33 30 27 24 22 19 17 15 13
4.5 4.4 4.3 4.1 4.0 3.9 3.7 3.6 3.5 3.3 3.2 3.1 2.9 2.8
Cincinnati: Winter: 5°F; Summer: 90°F; DD = 5070; CDD = 1080; CD = 0.77 a.
78 ,000 = ------------------- ( 5070 ) ( 24 ) ( 0.77 ) = 103.5 ×106 Btu ( 72 – 5 )
g.
103.5 ×106 F = ---------------------------------------- = 1175 therms ( 100 ,000 ) ( 0.88 )
h.
c. d.
103.5 ×106 F = -------------------------------- = 30 ,900 kWh ( 3413 ) ( 0.98 )
A
B
Calculation of Annual Heating Energy Consumption
House C
D
Heat Pump E
F
74,000/(72 – 5) × 0.77 = 850 (0.85)
1
1
1
1
1
1
1
0.93 0.72 0.54 0.39 0.27 0.16 0.60
Heat Pump Weather Heat Integrated Data Loss Temp. Temp. Heating Bin, Rate, Bin, Capacity, Diff. °F tbal − tbin hours 1000 Btu/h 1000 Btu/h × 62 3 726 2.55 44 57 8 639 6.80 43 52 13 611 11.05 41 47 18 599 15.30 39 42 23 627 19.55 36 37 28 698 23.80 33 32 33 711 28.05 30 27 38 460 32.30 27 22 43 249 36.55 24 17 48 131 40.80 22 12 53 68 45.05 19 7 58 44 49.30 17 2 63 18 53.55 15 –3 68 8 57.80 13
G
H
Supplemental Jd
I
Lf
Mg
Nh
Heat Seasonal SuppleTotal Cycling Rated Pump Heat Pump mental Electric Capacity Adjusted Heating Energy Adjust- Heat Pump Electric Operating Supplied Elec. Con- Space Capacity, Input, Load, Required, ConsumpHeating, sumption, Time ment a b c 6 d e 6 f g kW 10 Btu kWh kWh tionh Fraction 10 Btu Factor 1000 Btu/h 0.765 0.790 0.818 0.848 0.885 0.930 0.983 1.0 1.0 1.0 1.0 1.0 1.0 1.0
33.7 34.0 33.5 34.8 31.9 30.7 29.5 27 24 22 19 17 15 13
4.5 4.4 4.3 4.1 4.0 3.9 3.7 3.6 3.5 3.3 3.2 3.1 2.9 2.8
0.08 0.20 0.33 0.44 0.61 0.78 0.95 1 1 1 1 1 1 1
12.42 5.98 2.88 1.29 0.75 0.27 0.10
TOTALS: a Cycling
Ke
Capacity Adjustment Factor = 1 − Cd(1 − x), where Cd = degradation coefficient (default = 0.25 unless part load factor is known) and x = building heat loss per unit capacity at temperature bin. Cycling capacity = 1 at the balance point and below. b Col G = Col E × Col F c Operating Time Factor equals smaller of 1 or Col D/Col G
261 562 867 1081 1530 2123 2499 1656 872 432 218 136 52 22 12,311
d Col J = (Col I × Col G × Col C)/1000 e Col K = Col I × Col H × Col C f Col L = Col C × Col D/1000 g Col M = (Col L – Col J) × 106/3413 h Col N = Col K + Col M
14.86 9.10 5.34 3.06 2.17 0.96 0.46
715 914 721 519 416 202 105 3592
15,903
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103.5 ×106 F = ---------------------------------------- = 985 gal ( 100 ,000 ) ( 0.75 )
Problem 8-11
74 ,000 cfm = -------------------------------------- = 1160 cfm ( 1.1 ) ( 130 – 72 ) kWh = 15 ,903 kWh → See table below. 35 ,000 ( 1080 ) ( 24 ) kWh = ---------------------------------------------------- = 8900 kWh ( 90 – 78 ) ( 8.5 ) ( 1000 ) ( 35 ,000 ⁄ 1.3 ) cfm = ------------------------------------- = 1220 cfm ( 1.1 ) ( 78 – 5.8 ) (assumed 30% latent)
f.
b.
Climate
Q s = ( 1.10 ) ( cfm ) ( Δt )
e.
qL ER = -------------------- ( DD ) ( 24 ) ( C D ) ( ti – to )d
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108⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
8.12 A small commercial building located in Oklahoma City, Oklahoma, has design loads of 245,000 Btu/h, heating, and 162,000 Btu/h, cooling. Balance point for the building has been estimated at 65°F. Determine: a. Annual energy requirements for heating, Btu b. Fuel cost using LPG at $2.50/gal, $ c. Fuel cost using electric baseboard units with electricity at 6.7¢/kWh, $ d. Savings if setback to 55°F is effected between 10 P.M. and 6 A.M., Monday through Saturday, and all day Sunday, % e. Cooling season energy cost using cooling degreedays if conditioner has SEER of 11.5 and electricity is 7¢/kWh Oklahoma City: Winter: 10°F Summer: 96°F HDD = 3695; CD = 0.77 Average: 48.3°F; CDD = 1876 a.
245 ,000 ER = ---------------------- ( 3695 ) ( 24 ) ( 0.77 ) = 2.69 ×108 Btu ( 72 – 10 )
b.
2.69 ×108 C f = ------------------------------------- ( $2.50 ) = $9,963 ( 90 ,000 ) ( 0.75 )
c.
64.7 – 48.3 % Savings = 1 – --------------------------- = 0.31 or 31% 72 – 48.3 e.
162 ,000 ( 1876 ) ( 24 ) C e = ---------------------- -------------------------------- × 0.07 = $2467 ( 96 – 78 ) ( 11.5 ) ( 1000 )
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d.
2.69 ×108 C f = ----------------------------- ( 0.067 ) = $5280 ( 3413 ) ( 1.0 ) t i ,av = [ ( 6 ) ( 16 ) ( 72 ) + ( 168 – ( 6 ) ( 16 ) ( 55 ) ] ⁄ 168 = 64.7°F
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Chapter 8—Energy Estimating Methods⏐109
8.13 A small commercial building in Indianapolis, Indiana, has design heating and cooling loads of 98,000 Btu/ h and 48,000 Btu/h, respectively. Internal heat gains throughout the winter are relatively steady at 4.5 kW. Electricity costs 7.1¢/kWh. Estimate: a. Annual heating cost if baseboard electric resistance units are used.
b.
Annual cooling cost with a conventional vapor compression air-cooled unit, using your choice of method.
Select a heat pump system for the building from the XYZ Corporation models. Determine the a. b.
Annual heating cost and Annual cooling cost.
Indianapolis: Winter: –3°F; Average: 39.6°F; DD = 5577; CD = 0.77 Summer: 88°F; CDD = 974 a.
98 ,000 ER = --------------------------- ( 5577 ) ( 24 ) ( 0.77 ) = 135 ×106 Btu [ 72 – ( – 3 ) ]
b.
135 ×106 Cost = -------------------------------- ( 0.071 ) = $2970 ( 3413 ) ( 0.95 ) Cooling Unit: Assume SEER = 11 Btu/Wh 48 ,000 CDD = ---------------------- ( 974 ) ( 24 ) = 112 ×106 Btu ( 88 – 78 ) 112 ×106 Cost = ---------------------------- ( 0.07 ) = $714 ( 11 ) ( 1000 ) ( 4.5 ) ( 3413 ) t bal = 72 – ------------------------------------------------ = 60°F 98 ,000 ⁄ [ 72 – ( – 3 ) ] Heat Pump: Model A-048; SEER = 10.5 ; Watts = 5270 a. Energy Input = 13 ,220 + 5976 = 19 ,196 kWh → See table below. Balance Point:
Cost = 19196 (0.071) = $1363
Problem 8-13 Climate A
B
112 ×106 ------------------------------- ( 0.071 ) = $757 ( 10.5 ) ( 1000 )
CDD Method:
Calculation of Annual Heating Energy Consumption
House C
D
Heat Pump E
F
G
H
Supplemental I
Jd
Ke
Lf
Mg
Nh
98,000/[72 – (–3)] = 1306 (1.3)
Heat Seasonal SuppleTotal Heat Pump Cycling Weather Heat Rated Pump Heat Pump mental Electric Integrated Capacity Adjusted Data Loss Temp. Heating Energy Heating Adjust- Heat Pump Electric Operating Supplied Elec. Con- Space Bin, Rate, Capacity, Input, Bin, Load, Required, ConsumpHeating, sumption, Capacity, Time ment 60 a b c 6 d e 6 f g kW °F tbal − tbin hours 1000 Btu/h 1000 Btu/h Factor 1000 Btu/h 10 Btu kWh kWh tionh Fraction 10 Btu 62 — 57 3 585 3.9 59.2 0.766 45.3 4.99 0.09 263 52 8 586 10.4 55.4 0.797 44.2 4.81 0.24 676 47 13 579 16.9 51.0 0.833 42.5 4.60 0.40 1065 42 18 605 23.4 48.0 0.872 41.9 4.46 0.56 1511 37 23 712 29.9 44.4 0.918 40.8 4.28 0.73 2225 32 28 791 36.4 40.8 0.973 39.7 4.10 0.92 2984 27 33 551 42.9 37.3 1 37.3 3.93 1 20.55 2165 23.64 905 22 38 293 49.4 33.8 1 33.8 3.76 1 9.90 1102 14.47 1339 17 43 152 55.9 30.0 1 30.0 3.58 1 4.56 544 8.50 1154 12 48 97 62.4 27.3 1 27.3 3.45 1 2.65 335 6.05 996 7 53 60 68.9 24.2 1 24.2 3.31 1 1.45 199 4.13 785 2 58 35 75.4 21.2 1 21.2 3.18 1 0.74 111 2.64 557 –3 63 13 81.9 18.4 1 18.4 3.06 1 0.24 40 1.06 240 Temp. Diff.
TOTALS: a Cycling
Capacity Adjustment Factor = 1 − Cd(1 − x), where Cd = degradation coefficient (default = 0.25 unless part load factor is known) and x = building heat loss per unit capacity at temperature bin. Cycling capacity = 1 at the balance point and below. b Col G = Col E × Col F c Operating Time Factor equals smaller of 1 or Col D/Col G
13,220
d Col J = (Col I × Col G × Col C)/1000 e Col K = Col I × Col H × Col C f Col L = Col C × Col D/1000 g Col M = (Col L – Col J) × 106/3413 h Col N = Col K + Col M
5976
19,196
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b.
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110⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual 8.14 A small (2200 ft2) food mart store located in Charlotte, NC, has design heating and cooling loads of 94,500 Btu/h and 57,400 Btu/h, respectively, based on inside design temperatures of 72°F (winter) and 78°F (summer). The store is open 24 h a day and has a relatively constant internal load due to lights, food cases, people, etc., of 3.3 W/ft2. Select a suitable heat pump for the XYZ Corporation and estimate its operating energy costs for both summer and winter if the price of electricity is 7.4¢/kWh. Charlotte: Winter: to = 18°F Summer: to = 91°F CDD = 1596 HL 94 ,500 ------- = ---------------------- = 1480 Δt ( 72 – 18 )
Q int = ( 3.3 ) ( 2200 ) ( 3.413 ) = 24 ,780
⎛ 24 ,780⎞ t balance = 72 – ⎜ ----------------⎟ = 55°F ⎝ 1480 ⎠ Select Model 060JA Heat Pump: Total Watts = 6250 SEER = 10.5 Winter: 7187 kWh × 0.074 = $532 Summer:
Climate A
B
( 1596 ) ( 24 ) -------------------------------- ( 0.074 ) = $1192 ( 10.5 ) ( 1000 )
Calculation of Annual Heating Energy Consumption
House C
D
Heat Pump E
F
G
H
Supplemental I
Jd
Ke
Lf
Mg
Nh
98,000/(72 – 18) = 1480 (1.48)
Heat Seasonal SuppleTotal Heat Pump Cycling Weather Heat Rated Pump Heat Pump mental Electric Integrated Capacity Adjusted Temp. Data Loss Heating Energy Heating Adjust- Heat Pump Electric Operating Supplied Elec. Con- Space Bin, Load, Required, ConsumpBin, Rate, Capacity, Input, Time Heating, sumption, Capacity, ment 55 a b c 6 d e 6 f g °F tbal − tbin hours 1000 Btu/h 1000 Btu/h Factor 1000 Btu/h 10 Btu kW Fraction 10 Btu kWh kWh tionh 62 57 52 3 730 4.4 71.2 0.765 54.5 6.32 0.08 369 47 8 684 11.8 66.0 0.795 52.5 6.05 0.22 910 42 13 634 19.2 61.0 0.829 50.6 5.81 0.38 1400 37 18 515 26.6 56.0 0.869 48.7 5.56 0.55 1575 32 23 360 34.0 51.0 0.917 46.8 5.30 0.73 1393 27 28 166 41.4 46.3 0.974 45.1 5.05 0.92 771 22 33 64 48.8 41.9 1 41.9 4.81 1 2.68 308 3.12 129 17 38 23 56.2 37.0 1 37.0 4.61 1 0.85 106 1.29 129 12 43 5 63.6 33.2 1 33.2 4.35 1 0.17 22 0.32 44 7 48 2 71.0 29.4 1 29.4 4.13 1 0.06 8 0.14 23 2 –3 Temp. Diff.
TOTALS: a Cycling
Capacity Adjustment Factor = 1 − Cd(1 − x), where Cd = degradation coefficient (default = 0.25 unless part load factor is known) and x = building heat loss per unit capacity at temperature bin. Cycling capacity = 1 at the balance point and below. b Col G = Col E × Col F c Operating Time Factor equals smaller of 1 or Col D/Col G
6862
d Col J = (Col I × Col G × Col C)/1000 e Col K = Col I × Col H × Col C f Col L = Col C × Col D/1000 g Col M = (Col L – Col J) × 106/3413 h Col N = Col K + Col M
325
7187
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Problem 8-14
57 ,400 ---------------------( 91 – 78 )
Solutions to
Chapter 9 DUCT AND PIPE SIZING
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Chapter 9—Duct and Pipe Sizing⏐113
9.1 The air velocity in a human occupied zone should not exceed (a) 10 to 25 ft/min, (b) 25 to 40 ft/min, (c) 40 to 64 ft/min, (d) None of the above. c.
40 to 60 ft/min
9.2 You are to select the type of outlets for a home to be constructed in Houston, Texas. Discuss your selection of outlets and locations for each of the following combinations: (a) Group A or Group C, (b) Group B or Group E. a.
Group A:
These outlets would be satisfactory for Houston where cooling is predominant and heating is minimum.
Group C:
These outlets would not be completely satisfactory where cooling is predominant.
b.
Group B:
These outlets would be satisfactory but probably not as good as Group A.
Group E:
These outlets would be satisfactory if properly designed and selected.
9.3 What velocity of air is necessary at a location in a room such that most people will feel neither cool nor warm? Assume that the local temperature is equal to the control temperature of 24.4°C. 24.4°C = 76°F Δt = ( t x – 76 ) – 0.07 ( V x – 30 ) for t x = 76°F and Δt = 0 V x = 30 fpm ( 0.15 m/s )
a.
P v = P T – P S = 0.54 – 0.50 = 0.04 in. H 2 O V = 4005 P v = 4005 0.04 = 801 fpm
b.
Q = AV = 801 ( 1 ) ( 2 ) = 1600 cfm ΔP 0.2 ------------= ------- ( 100 ) = 0.33 in. H 2 O ⁄ 100 ft 100 ft 60 From Fig. 9-2 V = 1650 cfm ⇒ 1300 cfm
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9.4 Solve the following problems. a. Find the airflow through a 12 in. by 24 in. (305 mm by 610 mm) duct if the static pressure is measured at 0.5 in. of water (125 Pa) and total pressure is measured at 0.54 in. of water (135 Pa) b. The pressure difference available to a 60 ft (18.3 m) length of circular duct is 0.2 in. of water (50 Pa). The duct has an ID of 12 in. (305 mm). What rate of airflow is expected?
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114⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
9.5 For a residential air-conditioning system, one branch duct must supply 207 cfm to one of the rooms. The branch duct has a run of 16 ft. (a) Determine the branch duct size and the pressure drop from the main duct to the room, and (b) specify the supply and return grille sizes for the room. a.
207 cfm – 16 ft branch run
V recomm = 600 fpm
From Fig. 9-2 @ 600 fpm,
207 cfm ⇒ D = 8 in.
16 ΔP = 0.08 ⎛ --------- ⎞ ⎝ 100 ⎠
ΔP ------------= 0.08 in. H 2 O 100 ft \
= 0.0128 in. H 2 O b.
return grille – 600 fpm supply grille ≈ 6 × 9 in. or 5 × 11 in.
9.6 How large of a duct is required to carry 20,000 cfm (9400 L/s) of air if the velocity is not to exceed 1600 fpm (8.1 m/s)? For 20,000 cfm @ 1600 fpm, use 48 in. or larger. Fig. 9.2
9.7 Given the duct system shown below, plot pv, Δps, and Δpt for the flow through the system.
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Chapter 9—Duct and Pipe Sizing⏐115
9.8 Solve the following. a. For Problem 9.7, find the frictional pressure loss between points (1) and (2). b. How can the static pressure be increased in a duct system as the air moves away from the fan? a.
H loss = H 1 – H 2 = P1 – P2 + HV – HV 1
b.
2
= ( 0.32 – 0.05 ) + ( 0.12 – 0.0 ) = 0.39 in. w.g. loss Increase cross-sectional area
9.9 Determine the dynamic loss of total pressure that occurs in an abrupt expansion from a 1 ft2 (0.093 m2) duct to a 2 ft2 (0.186 m2) duct carrying 1000 cfm (470 L/s) of air. A 1 = 1 ft 2
A 2 = 2 ft 2 Q V = ---A
Q = AV
1000 V 1 = ------------ = 1000 fpm 1 2
V1 ⎛ A 1 ⎞ H L = ------- ⎜1 – ------ ⎟ 2g ⎝ A 2 ⎠
2
2
⎛ V1 ⎞ 1 2 = ⎜ ------------ ⎟ ⎛⎝ 1 – ---- ⎞⎠ = 0.0156 in. w.g. 2 ⎝ 4005 ⎠
From Fig. 9-2 Loss = 0.25 in. H2O/100 ft (150 ft) = 0.375 in. H2O
9.11 Find the equivalent rectangular duct for equal friction and capacity for the duct in Problem 9.6, one side is 26 in.
From Table 9-1 80 in.
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9.10 Determine the friction loss when circulating 20,000 cfm (9430 L/s) of air at 75°F (23.9°C) through 150 ft (45.7 m) of 36 in. (0.914 m) diameter galvanized steel duct.
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116⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
9.12 Find the pressure loss between points A and D for the 12 by 12 in. duct shown below. Air at standard conditions is being supplied at the rate of 2000 cfm in galvanized duct of average construction. Elbows No. 1 and No. 2 have center line radii of 13 and 24 in., respectively.
Elbow #1
Elbow #2
H12 ---= ------ = 1.0 From Table 9-4 W 12 ⇒ C D = 0.21 V 13 ----- = ------ = 1.08 W 12 H 12 ----- = ------ = 1.0 W 12 V24 ---= ------ = 2.0 W 12
Table 9-1
Fig. 9-2
D c = 13.1 in.
⇒ C D = ( 0.05 ) ( 0.6 ) = 0.03
V = 2000 fpm
V 2 P v = ⎛ ------------ ⎞ = 0.249 in. H 2 O ⎝ 4005 ⎠ Loss straight = 0.45 in. w.g. ⁄ 100 ft ⎛ 0.45 in. H 2 O⎞ ΔP loss = ( 10 + 20 + 30 ) ⎜ -------------------------------⎟ + ( 0.21 + 0.03 ) ( 0.249 ) 100 ft ⎠ ⎝
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= 0.27 + 0.059 = 0.33 in. w.g.
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Chapter 9—Duct and Pipe Sizing⏐117
9.13 Analyze the air-handling system shown in the following diagram. Determine if a damper is needed in either section (d) or (b), and if so, in what section. There is a damper located in section (u) so that the proper static pressure can be maintained in section (u). If a damper is needed, what is the pressure loss across the damper? f = 0.02 ρ = 0.075 lb/ft3 D = 12 in.
(R/D)elbows = 1 Q u = 3000 cfm
V u = 3825 fpm
H v = 0.915 in. H 2 O
Q b = 1000 cfm
V b = 1275 fpm
H v = 0.102 in. H 2 O
Q d = 2000 cfm
V d = 2550 fpm
H v = 0.407 in. H 2 O
u
b
d
V--= 1.0 D
Section d → 2 elbows
C D = 0.22
Vd ------ = 2550 ------------ = 0.666 Vu 3825
Straight Section
Table 9-4
C D = 0.04
ΔP loss = 0.037 + 2 ( 0.22 ) ( 0.407 ) + ( 50 + 40 + 10 ) ( 0.02 ) ( 0.407 ) = 1.03 in. H 2 O V -----b- = 0.33 Vu
Section b → Diverted Flow fitting
ΔP fitting = ( 1.1 ) ( 0.415 ) = 1.01 in. H 2 O
C D = 1.1 Table 9-4
10 ΔP loss = 1.01 + 0.02 ⎛ ------ ⎞ ( 0.102 ) = 1.03 in. H 2 O ⎝1⎠ ΔP loss – ΔP loss = 0 d
No damper needed.
b
R = 0.8 Table 9-4 Q1 2000 V 1 = ------ = ------------ = 667 fpm A1 3×1 1500 V 2 = ------------ = 750 fpm 2×1
P v = 0.0277 in. w.g. 1
P v = 0.0351 in. w.g. 2
a.
500 V b = ------------ = 500 fpm 1×1 P v = R ( P v – P v ) = 0.8 ( 0.0277 – 0.0351 ) = – 0.006 ≈ 0 in. H2 O
b.
ΔP branch = ( 1 – R ) ( P v – P v ) = 0.2 ( 0.0277 – 0.0351 ) ≅ – 0.0015 ≅ 0 in. H2 O
c.
V -----L- = 500 --------- = 0.75 V1 667
Assume Branch 1 × 1 ft 1
2
1
2
C = 0.28
Table 9-4
ΔP branch = ( 0.28 ) ( 0.0277 ) = 0.008 in. H 2 O ≅ 0 in. H 2 O
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9.14 A 1 ft high by 3 ft wide main duct carries 2000 cfm of air to a branch where 1500 cfm continues in the 1 ft by 2 ft straight through section and 500 cfm goes into the branch. Find the actual static pressure regain and the total pressure loss in the straight through section if the static regain coefficient is 0.80. If the branch take off is a 45° cylindrical Y, find the static pressure loss in this section.
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118⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
9.15 The supply ductwork for an office space is shown in the following diagram. Size the ductwork by the equalfriction method and calculate the pressure drop. Assume a maximum duct depth of 12 in. and that all duct take-offs are straight rectangular take-offs.
Total cfm = 800 + 900 + 1200 = 2900 cfm Deq = 20 in.
V
AB
12 × 30 in.
Table 9-1
ΔP --------- = 0.13 in. H O 2 100
( Table 9-4 )
V b 1060 ------ = ------------ = 0.76 V 1400 u
First branch take-off
= 1400 fpm
C = 1.0
ΔP loss = 1.0 ( 0.122 ) = 0.122 in. H 2 O Straight through loss ≅ 0 in. H 2 O Section
cfm
ΔP/100 ft
Deq.
Velocity
A-B C-D C-E C-H B-F
2900 2000 1200 800 900
0.13 0.13 0.13 0.13 0.13
20 in. 17 in. 14 in. 12 in. 12.5 in.
1400 1300 1130 1010 1060
12 × 30 12 × 21 12 × 14 12 × 10 12 × 11 H12 ----= ------ = W 14
Elbows Radius Ratio = 1.5
Run D Table 9-4
Rectangular Size, in. Length
ΔP
C = 0.18
Run G Radius Ratio = 1.5
loss
0.065 0.0754 0.167 0.211 0.183
⎛ 1130⎞2 = 0.18 ⎜ ------------⎟ = 0.014 in. H O 2 ⎝ 4005⎠ Table 9-4
⎛ 1010⎞ = 0.09 ⎜ ------------⎟ ⎝ 4005⎠
V b 1010 ------ = ------------ = 0.776 V u 1300
Divided Flow Fitting
Cum. Loss
C = 0.09
2 = 0.005 in. H O 2
Table 9-3 C = 1.0
1300 2 ΔP loss = 1.0 ⎛⎝ ------------⎞⎠ = 0.105 in. H 2 O 4005 Through loss ≅ 0 Total Pressure Drop ( No Outlet Grille ) = 0.211 in. H O 2 Damper in A-E and A-F or reduce size appropriately.
9.16 The following duct system contains circular, galvanized duct. The velocity in the ducts is to be 2000 fpm, and each outlet is to handle 2000 cfm. Each outlet grille has a pressure loss of 0.12 in. of water. Estimate the required pressure increase of the fan. cfm
fpm
D
ΔP/100 ft
ΔP
Pv
Length
4000 2000 2000
2000 2000 2000
19.2 13.5 13.5
0.27 0.4 0.4
0.081 0.20 0.36
0.25 0.25 0.25
30 ft 50 ft 90 ft
Table 9-4
r ⁄ D = 1.0 C = 1.34
Longest Run:
ΔP elbow ≅ 0.054 in. H 2 O each ΔP div, flow ≅ 0.35 in. H 2 O
ΔP loss = 0.081 + 0.36 + 0.35 + 0.054 + 0.054 + 0.12 = 1.02 in. H 2 O
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loss
Loss 0.065 0.014 0.078 + 0.014 0.105 + 0.026 + 0.0054 0.013 + 0.105
0.86
H12 ----= ------ = 1 W 12 ΔP
50 ft 8 ft 60 ft 20 ft 10 ft
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Chapter 9—Duct and Pipe Sizing⏐119
9.17 Select a fan for the following system. The radius ratio of the elbows is 1.0 and the elbows are 3 piece. The pipe is circular. Calculate the frictional pressure loss for the system and the total capacity required by the fan.
Q total = 300 cfm 100 cfm branch ΔP through ≅ 0.10 200 cfm branch
20 300 cfm at 1000 fpm → ΔP = 0.24 --------- = 0.046 in. 100 20 100 cfm at 1000 fpm → ΔP = 0.46 --------- = 0.092 in. 100 ΔP loss = 0.138 in. straight
300 cfm at 1000 fpm → ΔP = 0.046 Branch Loss (Table 9-3) ΔP = 0.08 Elbows (Table 9-3) ΔP ≅ 0.07 Straight ΔP = 0.087 ----------------------------------------------------------ΔP loss = 0.283 in. H 2 O
Fan must supply 0.283 in. H 2 O static for 300 cfm Damper the 100 cfm branch
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120⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
9.18 a. Estimate the total pressure loss between points (1) and (2) and between (1) and (3) in the following take-off. when: Q1 = 1510 L/s V1 = 8.12 m/s Q2 = 1227 L/s V2 = 6.1 m/s Q3 = 283 L/s V3 = 3.05 m/s The duct is rectangular, of commercial fabrication, and has mastic tape joints. b.
Estimate the static pressure at (3) if the static pressure at (1) is 1.0 in. of water. V 1 = 1600 fpm
Q 1 = 3200 cfm
V 2 = 1200 fpm
Q 2 = 2600 cfm
V 3 = 600 fpm
Q 3 = 600 cfm
Table 9-4
a. –
V3 ------ = 0.375 V1
C = 0.54
–
P V = 0.16
V -----2- = 0.75 V1
1
1600 ΔP loss = 0.54 ⎛⎝ ------------⎞⎠ 4000 A2 ------ = 1.08 A1
2
= 0.086 in. H 2 O
C = 0.04
P V = 0.09 2
ΔP loss = ( 0.04 ) ( 0.16 ) = 0.0064 in. H 2 O b. –
P T – P T = 0.086 in. H 2 O 1
3
P S + P V – P S – P V = 0.086 1
1
3
P S = 1.05 in. H 2 O 3
3
; P S = 1.0 + 0.16 – 0.023 – 0.086 3
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Chapter 9—Duct and Pipe Sizing⏐121
9.19 Solve the following problems: a. What is the expected approximate frictional pressure from (1) to (2) in the below length of duct. Assume round ducts, clean sheet metal, and air at standard temperature and pressure.
b.
If the static pressure at (3) is 0.350 in. of water, what friction drop will be required of a damper at 68°F? What size duct would be required (for ducts C and D) if the damper is eliminated? What is the velocity in the line? Assume a temperature of 68°F. Assume that the static pressure at (3) is still 0.35 in. of water and that the R/D of the elbow is 2. Also assume that the grille loss is linear with velocity.
c.
Elbow Radius = 36 in. Grille Loss = 0.1 in w.g. at 600 fpm Duct
cfm
Velocity, fpm
Length, ft
A B C D
2000 1000 1000 1000
1000
40 — 30 30
600 600 600
Grille loss = 0.10 in H20 at 600 fpm
a. Section A
D = 19 in.
ΔP = 0.071 in. H 2 O/100 ft
ΔP = ( 40 ⁄ 100 ) ( 0.071 ) = 0.028 in w.g. Branch cylindrical tee Table 9-4 V c ⁄ V a = 0.6
Section C Elbow
D = 17.5 in.
Section D
= 0.075 in w.g.
ΔP = ( 0.031 ) ( 30 ⁄ 100 ) = 0.0093 in. H 2 O
R ⁄ D = 36 ⁄ 17.5 = 2.05 ΔP loss
C = 1.2
C = 0.13
6000 2 = 0.13 ⎛ ------------⎞ = 0.003 in w.g. ⎝ 4000⎠
ΔP = ( 0.031 ) ( 30 ⁄ 100 ) = 0.0093 in. H 2 O
Total loss = 0.028 + 0.075 + 0.0093 + 0.003 + 0.0093 + 0.10 = 0.225 in. H 2 O b. at
P S = 0.35 in. H 2 O and P v = P v 2
3
ΔP friction = 0.0093 + 0.003 + 0.0093 + 0.10 = 0.122 in w.g. ΔP damper = 0.35 – 0.122 = 0.228 in. H2 O c.
P S at = 0.35 in. H 2 O
R⁄D = 2
C = 0.13
V ΔP = 0.13 ⎛ ------------⎞ ⎝ 4003⎠
2
ΔP V 2 ΔP L = 0.35 = ( 30 + 3 ) --------- + 0.13 ⎛⎝ ------------⎞⎠ + ΔP grille 100 4000 Trial and Error Solution: Assume a velocity, using 1000 cfm Find a D, thus → V = 1300 fpm D = 12 in.
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1600 ΔP loss = 1.2 ⎛⎝ ------------⎞⎠ 4000
2
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122⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
9.20 The pressure (energy) loss between A-B, A-C, A-D, and A-E must be equal if a proper air balance is to be achieved. The static pressure required in the duct at points B, C, D, and E to produce the proper flow from the air diffusing terminal units (ceiling diffusers) is assumed to be uniform and to be 0.10 in. of water. Ignoring interference losses due to terminal unit take off: 1.
Calculate the total pressure loss between A and D.
2.
Size the ducts between A and B.
3.
Size the ducts between A and C.
4.
Size the ducts between A and E.
5.
What would the static pressure be at F?
Note: In practical application, minor static pressure imbalances up to 0.05 in. of water can be absorbed by adjustment of dampers installed in air-diffusing terminal units and further minor adjustments of diverting dampers, in branch fittings. For greater imbalances, duct sizing must be modified and/or butterfly dampers installed in the branch ducts.
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This is an open-ended design problem with a number of satisfactory solutions.
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Chapter 9—Duct and Pipe Sizing⏐123
9.21 Determine the equivalent feet of pipe for a 2 in. (50 mm) open gate valve and a 2 in. (50 mm) open globe valve at a flow velocity of 5 fps (1.5 m/s). (See Table 11 in Chap. 22, 2009 HBF) a.
at 5 fps – 2 in. pipe → L eq 1 in. elbow = 5.9 ft open gate valve → 0.5 elbow L eq = 0.5 ( 5.9 ft ) = 2.95 ft
b.
at 5 fps – 2 in. pipe → L eq = 5.9 ft open globe valve → 12 L eq = 12 ( 5.9 ft ) = 70.8 ft
9.22 A convector unit is rated at 1.8 gpm (0.113 L/s) and has a 3.4 ft (10 kPa) pressure loss at rated flow. Estimate the pressure loss with a flow of 2.3 gpm through the convector. ⎛W ⎞ H 2 = H 1 ⎜ ------2- ⎟ ⎝ W1 ⎠
2
⎛2.3 ⎞ -⎟ = 3.4 ⎜-----⎝1.8 ⎠
2
= 5.6 ft
9.23 Size the system shown in Example 9.5 for a 10°F temperature drop. (See Chap. 22, 2009 HBF) A-B-C
Supply Side I-II at II-III at III-IV at Conv. A at Conv. B at Conv. C at V-VIII at VIII-IX at IX-X at Conv. D at Conv. E at Conv. F at Conv. G at
14,400 Btu/h (14.4 gpm) 54,000 Btu/h (5.4 gpm) 32,000 Btu/h (3.2 gpm) 22,000 Btu/h (2.2 gpm) 18,000 Btu/h (1.8 gpm) 14,000 Btu/h (1.4 gpm) 90,000 Btu/h (9 gpm) 54,000 Btu/h (5.4 gpm) 30,000 Btu/h (3 gpm) 36,000 Btu/h (3.6 gpm) 14,000 Btu/h (1.4 gpm) 16,000 Btu/h (1.6 gpm) 24,000 Btu/h (2.4 gpm)
Size, in. 1 1/2 1 1 3/4 3/4 5/8 1 1/4 1 1 1 5/8 5/8 3/4
Return Side V-VI 4 gpm VI-VII 5.4 gpm XI-XII 3 gpm XII-XIII 6.6 gpm XIII-VII 9.0 gpm VII-I 14.4 gpm
Size, in. 1 1 1 1 1/4 1 1/4 1 1/2
Assume run containing Conv. E is longest run. This should be verified later.
The head loss for this straight pipe in this run from Fig. 2 I-II II-VIII VIII-IX IX-X X-Conv. Conv.-X XI-XII XII-XIII XIII-VII VII-I
(280)(2) = (280)(16) = (280)(7) = (100)(11) = (200)(3) = (200)(12) = (100)(11) = (150)(10) = (250)(14) = (280)(25) =
560 milli-inch (4.67 ft/100 ft) 4480 milli-inch (37.3 ft/100 ft) 1960 milli-inch (16.3 ft/100 ft) 1100 milli-inch (9.2 ft/100 ft) 600 milli-inch (5 ft/100 ft) 2400 milli-inch (20 ft/100 ft) 1100 milli-inch (9.2 ft/100 ft) 1500 milli-inch (12.5 ft/100 ft) 4750 milli-inch (39.6 ft/100 ft) 7000 milli-inch (58.3 ft/100 ft) Total = 212.1 ft/100 ft Convector Losses = 11.9 ft/100 ft
From Table 4 and Fig. 2 friction loss for elbow is 35 ft/100 ft. From Fig. 4, Fig. 2 the friction loss for Tees is 158 ft/100 ft. Total friction loss for this run = 4.17 ft Pump must supply at least this head at 14.7 gpm. See Notes in Example for this design.
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27 ,000 gpm = ------------------------- = 5.51 gpm ( 480 ) ( 10 ) 45 ,000 D-E-F-G gpm = ------------------------- = 9.18 gpm ( 480 ) ( 10 ) Assume design friction loss = 2.5 ft/100 ft Multiply heat required at convectors by ( 20 ⁄ 10 ) = 2 and use Fig. 2. 10° ΔT
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124⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
9.24 Size the system shown in Example 9.5 for 30°F temperature drop. (See chapter 22, 2009 ASHRAE Handbook—Fundamentals) 27 ,000 gpm = ------------------------- = 1.84 gpm ( 490 ) ( 30 ) 45 ,000 D-E-F-G gpm = ------------------------- = 3.06 gpm ( 490 ) ( 30 ) Assume design friction loss = 2.5 ft/100 ft Multiply heat required at convectors by ( 20 ⁄ 30 ) = 0.667 and Fig. 2 to get pipe size. 30° ΔT
A-B-C
Supply Side I-II at II-III at III-IV at Conv. A at Conv. B at Conv. C at V-VIII at VIII-IX at IX-X at Conv. D at Conv. E at Conv. F at Conv. G at
48,000 Btu/h (4.8 gpm) 18,000 Btu/h (1.8 gpm) 10,670 Btu/h (1.1 gpm) 7340 Btu/h (0.7 gpm) 6000 Btu/h (0.6 gpm) 4670 Btu/h (0.5 gpm) 30,000 Btu/h (3.0 gpm) 18,000 Btu/h (1.8 gpm) 10,000 Btu/h (1.0 gpm) 12,000 Btu/h (1.2 gpm) 4670 Btu/h (0.5 gpm) 5340 Btu/h (0.5 gpm) 8000 Btu/h (0.8 gpm)
Pipe Size, in. 1 3/4 5/8 1/2 1/2 1/2 1 3/4 5/8 5/8 1/2 1/2 1/2
Return Side V-VI 1.33 gpm VI-VII 1.8 gpm XI-XII 1.0 gpm XII-XIII 2.2 gpm XIII-VII 3.0 gpm VII-I 4.8 gpm
Pipe Size, in. 5/8 3/4 5/8 3/4 1 1
Assume Conv. E is longest run. Verify after various calculations.
The head loss for this straight pipe from Fig. 2 (240)(2) = (100)(16) = (170)(7) = (130)(11) = (85)(3) = (85)(12) = (130)(11) = (220)(10) = (100)(19) = (220)(25) =
480 milli-inch (4 ft/100 ft) 1600 milli-inch (13.3 ft/100 ft) 1190 milli-inch (9.9 ft/100 ft) 1430 milli-inch (11.9 ft/100 ft) 225 milli-inch (2.1 ft/100 ft) 1020 milli-inch (8.5 ft/100 ft) 1430 milli-inch (11.9 ft/100 ft) 2200 milli-inch (18.3 ft/100 ft) 1900 milli-inch (15.8 ft/100 ft) 5500 milli-inch (45.8 ft/100 ft) Total = 142 ft/100 ft
Convector Losses = 11.9 ft/100 ft
From Table 4, Fig. 2; loss for elbows = 10.5ft/ 100 ft. From Fig. 4; Fig. 2 the total friction loss for Tees = 79.6 ft/100 ft. Convector loss = 11.9 ft/100 ft Total friction loss for this run = 2.43 ft Pump must supply at least this head at 4.9 gpm of flow. See Notes in Example 4 in Chap. 9 PHVAC.
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I-II II-VIII VIII-IX IX-X X-Conv. Conv.-XI XI-XII XII-XIII XIII-VII VII-I
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Chapter 9—Duct and Pipe Sizing⏐125
9.25 Size the system shown for iron pipe. The water leaves the boiler at 200°F and has a 20°F temperature drop. The convectors have a loss given by the equation: Loss (milli-inches) = 0.3 (Btu/h output). Note: 1 milliinch = 0.001 in. of water. Assume a 3 ft rise is needed to get to the convectors and then a 3 ft drop to return to the boiler. What head must be developed by the pump and what flow rate (gpm) is required? (See chapter 22, 2009 ASHRAE Handbook—Fundamentals)
A
B
Iron pipe 20°Δt
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24,000 27 ,500 gpm = ------------------------- = 2.45 ; Loop B gpm = ------------------------- = 2.81 ( 490 ) ( 70 ) ( 490 ) ( 70 ) Total gpm = 5.25 Assume Loop B will have longest pipe. Assume 300 milli-inch/ft friction loss. For Loop B Measured length = 49 ft From Fig. 1 Loss = ( 225 ) ( 49 ) = 11 ,025 milli-inch Table 11, HBF, elbows Loss = ( 2 ) ( 1.8 ) ( 2.25 ) = 810 milli-inch Convectors Loss = ( 0.3 ) ( 27 ,500 ) = 8250 milli-inch Fig. 7, HBF, supply tee Loss = ( 3.7 ) ( 225 ) ( 1.8 ) = 1500 milli-inch Loss = ( 3.7 ) ( 225 ) ( 1.8 ) = 1500 milli-inch return tee -----------------------------------------------------------------------------------------------------Total loss = 23 ,085 milli-inch Boiler Circuit: L = 15 ft – Fig. 1 – 1 in. pipe Loss = ( 180 ) ( 15 ) = 2700 milli-inch Table 11, HBF, elbows Loss = ( 2 ) ( 2.4 ) ( 180 ) = 864 milli-inch 6 ft riser and return Loss = ( 6 ) ( 180 ) = 1080 milli-inch Loss = ( 3 ) ( 2.4 ) ( 180 ) = 1296 milli-inch Table 11, HBF, boilers -------------------------------------------------------------------------------------------------Total = 5940 milli-inch Total friction loss = 29 ,025 milli-inches for Loop B. Loop A
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126⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
9.26 Rework Problem 9.25 using Type L copper tubing. Use copper tubes. See Figure 5, chapter 22, 2009 ASHRAE Handbook—Fundamentals. gpm = 5.25 gpm; Loop A = 2.8 gpm; 2.5 ft/100 ft; 3/4 in. tube 44 Pipe loss = 2.5 × --------- = 1.22 ft 100 Elbows 2 × ( 1.8 ) ( 2.5 ) = 0.09 ft Convectors ( 0.3 ) ( 2.29 ) = 0.69 ft 2 ( 3.7 ) ( 2.5 ) ( 1.8 ) = 0.33 ft Tees ---------------------------------------------------------------Total = 2.33 ft Boiler circuit 1.41 ( 21 ) + 2 ( 2.4 ) ( 1.41 ) + 4 ( 2.4 ) ( 1.41 ) = 0.5 ft Total friction = 2.83 ft for Loop A including Boiler
9.27 A steam system requires 15,000 lb/h of steam at an initial pressure of 150 psig. The design pressure drop is to be 6 psi per 100 ft. Determine the size of schedule 40 pipe required and the velocity in the steam pipe.
15,000 lb/h
Pinitial = 150 psig (Fig. 13D, chapter 22, 2009 ASHRAE Handbook—Fundamentals) 6 psi/100 ft – Schedule 40 Select 3 1/2 in. pipe Velocity at 150 psig = 10,000 fpm
This is an open ended design problem; there are many solutions.
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9.28 Determine the pipe sizes for the refrigeration systems shown in the following figure. a. Refrigerant lines using R-22 b. Condenser water lines
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Chapter 9—Duct and Pipe Sizing⏐127
9.29 A gas appliance has an input rating of 80,000 Btu/h and is operated with natural gas (specific gravity = 0.60) having a heating value of 1050 Btu/ft3. What size of supply pipe is necessary when the equivalent length is 70 ft and a pressure loss of 0.3 in. of water is allowable? (See chapter 22, Table 26, 2009 ASHRAE Handbook—Fundamentals.) Leq = 70 ft; 0.3 in. H2O 80,000/1050 = 76.19 ft3/h
3/4 in. pipe
9.30 A fan operating at 1200 rpm has been delivering 6500 cfm against a static head of 3.25 in. of water and a total head of 5.25 in. of water. The air temperature is 130°F, gage temperature, 90°F, and the input power is 6.1 kW. During a time of power difficulty, the operator notices that the static head is now 2.36 in. of water. There has been no change in the system. Find: a. New capacity in cfm b. New power input in kW c. Original efficiency of the fan (%) a.
n 2⎞ 2 h 2 2.38 ⎛ ---- = ----- = ---------- ; ⎝ n 1⎠ h 1 3.25
h 2 = 0.855 ( 1200 ) = 1028 rpm
b. c.
3
kW 2 = ( 6.10 ) ( 0.855 ) = 3.82 kW QP t ( 144 ) ( 0.746 ) ( 6500 ) ( 5.25 ) ( 0.036 ) ( 144 ) ( 0.746 ) AHP Eff = ------------ = ------------------------------------------ = -----------------------------------------------------------------------------------( 33 ,000 ) ( 6.1 ) ( 33 ,000 ) ( 6.1 ) BHP EFF = 65.5%
9.31 A certain damper design introduces a head loss of 0.5 velocity heads when wide open. A damper of this design is to be installed in a 12 by 30 in. duct that handles 3000 cfm. The pressure drop in the undampered system is 1.5 in. of water. If the pressure drop through the damper when wide open is to be 5% of the total system resistance, how much cross-sectional area in the duct should the damper occupy? ΔP T = ΔP duct + ΔP damper = 1.5 + 0.05 ΔP T ;
ΔP T = 1.58
ΔP damper = 1.58 – 1.5 = 0.08 in. w.g. wide open 12 ( 0.075 )V 2 12 Pa V 2 CFM 2 HV ( in. w.g. ) = --------------------- = ---------------------------------- = ⎛ ----------------⎞ ⎝ 4005A⎠ 2 g ρw 2 ( 32.2 ) ( 62.4 ) ΔP damper
open
3000 = 0.08 = 0.5 ⎛ ----------------⎞ ⎝ 4005A⎠
2
A = 1.87 ft
2
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Q 2 = ( 6500 ) ( 0.855 ) = 5570 cfm
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128⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
9.32 What effect on the following parameters does a variation in air density have for a fan operating in a system: a. flow rate b. developed head c. horsepower Air Density Variation: a. Flow rate; Volume – Remains same; Mass – Increases with increases. b. Developed head – Increases directly with density increase. c. Horsepower – Increases directly with density increase.
9.33 A fan delivers 1500 cfm (708 L/s) of dry air at 65°F (18.3°C) against a static pressure of 0.20 in. of water (50 Pa) and requires 0.10 BHP. Find the volume circulated, the static pressure, and the BHP required to deliver the same weight of air when the air temperature is increased to 165°F (73.9°C). (Note: Atmospheric pressure is constant.) 1500 cfm, 65°F, P s = 0.20 in. H 2 O, 0.10 bph for air temperature of 165°F T ρ2 460 + 65----- = -----2 = ----------------------= 0.81 ρ1 T1 460 + 165
CFM 2 = CFM 1 = 1500 cfm
ρ2 H P = H P ⎛ ----- ⎞ = 0.10 ( 0.84 ) = 0.084 BHP 2 1⎝ ρ ⎠ 1
9.34 Should fans be placed before or after air heaters? Why? For the same mass flow rate through the air heater, the fan law gives: ρ BHP 1 -------------- = -----2 BHP 2 ρ1
ρ2 BHP 2 = ⎛⎝ ----- ⎞⎠ BHP 1 ρ1
if ρ 1 is cold, ρ 2 is warm then BHP 2 > BHP 1 – Place fan before heater.
9.35 A 40 in. by 24 in. rectangular duct conveying
12,000 cfm of standard air divides into 3 branches (see
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ρ2 P S = P S ⎛ ----- ⎞ = 0.20 ( 0.84 ) = 0.168 in. H 2 O 2 1⎝ ρ ⎠ 1
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Chapter 9—Duct and Pipe Sizing⏐129
figure). Branch A carries 6000 cfm for 100 ft, B carries 4000 cfm for 150 ft, and C carries 2000 for 35 ft. (a) Size each branch for equal total friction of 0.15 in. of water. Do not exceed upper velocity limit of 2000 fpm. (b) What is the total friction loss if the same quantity of air, 12,000 cfm of air at 150°F and 14.0 psia, is passed through the same system as part (a)? (c) For a fan selected for part (a), at what percentage of the speed in part (a) must the fan run to satisfy part (b)?
Branch A
Total length = 100 ft + 6 ft. Assume branch similar to elbow. with r ⁄ w = 1.5 ; H ⁄ W = 1.0 → C = 0.09 ΔP ΔP = 0.15 in. H 2 O = ( 100 + 6 ) --------- + 0.0225 100
ΔP ⁄ 100 ft = 0.12 in. w.g. ( Chap. 21, 2009 HBF ) Assume W = duct width = 24 in. for 600 cfm and D eg = 33.6 V = 1400 fpm Size = 24 in. × 24 in. ΔP = 0.15 in. w.g. = ( 6 + 150 ) ΔP′ ⁄ 100 ΔP′ = 0.096 in. w.g./100 ft with W = 24 in. V = 1350 fpm Size = 24 in. × 19.5 in. Branch C ΔP = 0.15 in. w.g. = ( 35 + 6 ) ΔP′ ⁄ 100 + 0.0225 ΔP′ = 0.31 in. H 2 O for 2000 cfm, D eg = 11 in. Branch B
b.
V = 1800 fpm ( Chap. 21, HBF )
Size = 24 in. × 7.5 in. K = 0.89 Q actual = 12 ,000 cfm
H o = ( 0.89 ) ( 0.15 ) = 0.133 in. H2 O c.
For Part a.
ΔP static, fan = 0.15 + 25 Δp = 0.183 in. H 2 O Q fan = 12 ,000 cfm
Using fan lows, the required speed for 150°F with 12,000 cfm is rpm2 = rpm1 × 1 × 1.
9.36 A centrifugal fan operating at 2400 rpm delivers 20,000 cfm of air through a 32 in. diameter duct against a static pressure of 4.8 in. of water. The air is 40°F. The barometer is 29.0 in. Hg. Determine the horsepower input
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2000 2 ΔP = 0.04 ⎛⎝ ------------⎞⎠ = 0.0225 in. w.g.; 4000
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130⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
if the efficiency is 70%. If the fan size, gas density, and duct system remain the same, calculate the horsepower required if operated at 3200 rpm.
π 2 π 32 2 2 A = --- D = --- ⎛ ------⎞ = 5.6 ft ; 4 4 ⎝ 12⎠
Q 20 ,000 V = ---- = ---------------- = 3570 fpm A 5.6
2
12ρ u V cfm 2 20 ,000 h v = ------------------ = ⎛ ---------------- ⎞ = ⎛ ----------------------------- ⎞ = 0.8 in. w.g. ⎝ 4005A ⎠ ⎝ ( 4005 ) ( 5.6 ) ⎠ 2gρ w h t = h s + h v = 4.8 + 0.8 = ( 5.6 in. w.g. ) ( 0.0361 ) = 0.202 psi ( 144 ) AHP = ( 0.202 ) ---------------- ( 20 ,000 ) = 17.6 hp 33 ,000 17.6 a. BHP = ---------- = 25.2 hp 0.70 b.
3200 2 BHP 2 = ( 25.2 ) ⎛ ------------ ⎞ = 59.7 hp ⎝ 2400 ⎠
9.37 Compute the efficiency of Fan 303 (Fig. 9-11b) when delivering 15,500 cfm at 4 in. static pressure (SP). From Fig. 9-116, fan BHP ≅ 13.4 CFM × ΔP ( in. w.g. ) 15 ,500 ( 4 ) Ideal BHP = ------------------------------------------------- = ------------------------- = 9.8 6350 6350 Wi 9.8 η f = ------- × 100 = ---------- × 100 = 74% Wa 13.4
a.
CFM × ΔP ( in. w.g. ) CFM × ΔP W i = m· ∫ ν dp ≅ m· ν ΔP = ------------------------------------------------- = -------------------------conversion factors 6350 Wi CFM × ΔP W a = ------ = -------------------------ηf 6350η f
b.
c.
W ( kWh ) = HP × 0.746 × kW ⁄ HP × time Motor input ( kWh ) = Fan input ⁄ η m CFM × 60 CFM × ΔP × 2545 W = ( 0.240 ) ( m· ) ( Δt ) = m ∫ ν dp ( 0.240 ) ------------------------ ( Δt ) = -------------------------------------------13.33 6350η f Δt = 0.371 ΔP ⁄ h f
d.
W = – m ∫ ν dp = – m ν dp ∼ CFM ⋅ ΔP V 2 CFM 2 L ΔP = ⎛⎝ f ---- + C o⎞⎠ ------ ∼ ⎛⎝ ------------ ⎞⎠ 2 A D 2
W ∼ CFM ( CFM ) ∼ ( CFM )
3
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9.38 Develop and explain the following relations for fan performance: (a) HP = CFM × ΔP/6350ηf (b) kWH = HP (0.746) Hours/ηm (c) Δtf = ΔP(0.371)/ηf (d) HP ~ CFM3
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Chapter 9—Duct and Pipe Sizing⏐131
9.39 A water pump develops a total head of 200 ft. The pump efficiency is 80% and the motor efficiency is 87.5%. If the power rate is 1.5¢ per kilowatt-hour, what is the power cost for pumping 1000 gal? Hm ( 200 ) ( 1000 ) ( 231 ⁄ 1728 ) ( 62.4 ) WHP = ---------------- = ----------------------------------------------------------------------------- = 50.2 hp 33 ,000 33 ,000 50.2 BHP = ------------------------------- = 72.5 hp ( 0.8 ) ( 0.875 ) Cost = ( 72.5 hp ) ( 2545 ) ( 1 ⁄ 3413 ) ( 0.015 ) = $0.81/h 0.81 ---------- = 1.35¢ for 1000 gal 60
9.40 For a certain system it is required to select a pump that will deliver 2400 gpm (150 L/s) at a total head of 360 ft (110 m), and a pump shaft speed of 2400 rpm. What type of pump would you suggest? n Q 2400 2400 - = ---------------------------- = 1540 η s = -----------3⁄4 3⁄4 H ( 360 ) Use a centrifugal pump.
( 180 ) ( 144 ) H d = ---------------------------- + 22 = 437 ft; 62.4
H S = +40 ft
231 1 3 Q = 1400 ⎛ ------------ ⎞ ⎛ ------ ⎞ = 3.11 ft ⁄ s ⎝ 1728 ⎠ ⎝ 60 ⎠ ( 3.11 ) ( 144 ) V d = ----------------------------- = 133 fps; 3.355
( 3.11 ) ( 144 ) V s = ----------------------------- = 35.2 fps 12.73
( 3.11 ) ( 60 ) ( 62.4 ) ( 654 ) H p = -------------------------------------------------------- = 385 hp ( 33 ,000 ) ( 0.60 ) kW = ( 385 ) ( 0.746 ) = 287 kW
2
H vd – H vs
2
Vd – Vs = ------------------ = 257 ft 2g c
H t = 437 – 40 + 257 = 654 ft
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9.41 A pump delivers 1400 gpm of water. The inlet pipe is 4 in. nominal and the outlet pipe is 2 in. nominal standard pipe. The water temperature is 40°F. The surface of the inlet supply is 40 ft higher than the pump centerline. The discharge gage, which is 22 ft above the pump centerline, reads 180 psi. If the pump and motor combined efficiency is 60%, calculate the necessary input to the motor in kilowatts.
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132⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
9.42 A pump is required to force 9250 lb/h (4200 kg/h) of water at 165°F (74°C) through a heating system against a total resistance of 82,300 milli-inches of water (20.5 kPa). If the mechanical efficiency of the pump is 65%, find the required horsepower input. 1 1 3 ρ = ------------ = ---------------- = 60.9 lb/ft ν 165° 0.0164 –3 Qρh m· h ( 9250 ) ( 82 ,300 ) ( 10 ) BHP = ----------- = ------- = ----------------------------------------------------------- = 0.05 hp η η ( 60 ) ( 12 ) ( 0.65 ) ( 33 ,000 )
9.43 How many horsepower are required to pump 66 gpm (4.16 L/s) against 60 ft (18.3 m) of head assuming 75% efficiency? Qρh ( 66 ) ( 231 ⁄ 1728 ) ( 62.4 ) ( 60 ) BHP = ----------- = -------------------------------------------------------------------- = 1.33 hp η 33 ,000 ( 0.75 )
h = 0.001 ( 300 ) = 90 ft H 2 O
b.
BHP = 8.35 hp P v = 4.74 psi
c.
P b = 14.741 psi
( 10 ) ( 144 ) H s = ------------------------- = 23.6 ft H2 O ( 61.01 ) η2 Q 1 = 300 Q 2 = Q 1 -----η1 Q 2 = 400
d.
( 300 ) ( 231 ⁄ 1728 ) ( 61.01 ) ( 90 ) BHP = -------------------------------------------------------------------------( 33 ,000 ) ( 0.80 )
2
a.
backward; yes
P s = 10 psi
η2 ------ = 400 --------- = 1.33 η1 300
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9.44 Solve the following problems: a. A certain system is found to have losses due to frictional effects according to the equation H = 0.001 (gpm)2 where H is in ft of water. The system is handling water at 160°F. For a design capacity of 300 gpm, what is the head developed by the pump and the BHP if the pump efficiency is 80%? b. What would be the theoretical maximum length of suction in order to prevent cavitation if the level of the supply tank is below the centerline of the pump? Assume atmospheric pressure to be 14.7 psi. c. If a capacity of 400 gpm is desired, what would be the speed ratio n2/n1 for the same pump, density of fluid, and system? d. Should a backward- or forward-curved blade pump be chosen? Would you make arrangements for a priming system for the pump?
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Solutions to
Chapter 10 LIFE-CYCLE COSTS
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Chapter 10—Life-Cycle Costs⏐135
10.1 If $1000 is invested at 8% interest, determine the value of this money in 10 years. Sum = Amount (1 + i)n = $1000 (1 + 0.08)10 Sum = $2159 10.2 Find the present worth of money that will have a value of $35,000 in 3 years with an interest rate of 9%. Sum = Amount (1 + i)n $35,000 = Amount (1 + 0.09)3 Amount = $27,026 10.3 $1000 is invested at the end of each year for 10 years. Interest is 11%. Find the amount accumulated. n
(1 + i) – 1 S = R --------------------------- = R ( CAF ) i 10
( 1 + 0.11 ) – 1 S = 1000 -------------------------------------0.11 S = $16,722
10.4 If $100,000 is invested at 8% interest, find the yearly withdrawal that will use up the money in 20 years.
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n
20
( 1 + 0.08 ) – 1 $100,000 = R -------------------------------------------20 ( 0.08 ) ( 1 + 0.08 ) R = $10,185
10.5 The cost of a new heat pump system is $3000 with an expected lifetime of 20 years. Neglect energy and maintenance costs. Find the annual cost if the salvage value is $0 and the interest rate is 8%. Series present worth n
1 ( 1 + i ) – 1P = R ⎛⎝ ------------⎞⎠ = R -------------------------n CRF i( 1 + i) 20
( 1 + 0.08 ) – 1$3000 = R --------------------------------------20 0.08 ( 1 + 0.08 ) R = $306
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(1 + i) – 1 1 -⎞ - = R ⎛ ----------P = R -------------------------n ⎝ CRF⎠ i( + i )
136⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
10.6 A new heating system has a cost of $15,000 and a salvage value of $5000, independent of age. The new system saves $1400 per year in fuel cost. Calculate the break-even point if i = 9%. Neglect maintenance costs. Initial cost – Salvage = Savings $15,000 ( CRF, 9%, N years ) – $5,000 ⁄ ( CAF, 9%, N years ) = $1,400 trial and error solution for N years yields N = 34 years (break even) (reasonable?)
10.7 A new high-efficiency cooling system costs $60,000 and saves $7500 in energy costs each year. The system has a salvage value of $10,000 in 20 years. Compute the rate of return. Neglect maintenance costs. Initial cost – Salvage = Savings $60,000 ( CRF, x%, 20 ) – $10,000 ⁄ ( CAF, x%, 20 ) = $7,500 trial and error solution for x% yields rate of return = 11.25%
10.8 The costs of two small heat pump units A and B are $1000 and $1200 and the annual operating costs are $110 and $100, respectively. The interest rate is 8% and the amortization is selected as 20 years. Compare the systems on the basis of present worth.
Present Worth
System B
$1000
$1200
Initial Cost CRF (8%, 20 yr.)
Operating Cost
= 0.10185
110 × 1/0.10185
$1080
100 × 1/0.10185 Present Worth
$982 $2080
$2182
System A is least costly to own.
10.9 Compare the units in Problem 10.8 on the basis of uniform annual costs.
Costs
System A
System B
Owning Cost Initial Cost × CRF $1000 × 0.10185
$101.85
$1200 × 0.10185 Operating Cost Uniform Annual O&O Cost
$122.22 $110
$100
$211.85
$222.22
System A is least costly to operate.
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System A
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Chapter 10—Life-Cycle Costs⏐137
10.10 An installation is going to require a 500 ton chiller. An annual energy analysis for this office building application shows that the required ton-hours over the year will be 2,100,000. The economic data is given below.
Average Chiller Efficiency
Chiller A
Chiller B
0.73 kW/ton
0.63 kW/ton
Initial Cost
$221,500
$240,500
Installation Cost
$19,000
$19,000
Electricity Cost
6¢/kWh
5.9¢/kWh
Maintenance Costs
$9,500
$10,000
Estimated Life
20 years
20 years
Perform a simple payback analysis for this option.
A
B
Energy Required:
0.73 × 2,100,000 1,533,000 kWh
0.63 × 2,100,000 1,323,000 kWh
Annual Operating Cost:
1,533,000 (0.06) +9500 $101,480
1,323,000 (0.059) +10000 $88,057
Initial Cost:
$221,500 + $19,000 $240,500
$240,500 + $19,000 $259,500
For the remaining time (18.6 yrs.) savings in operating cost would be 18.6 × 13423 = $250,000
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259,500 – 240,500 $19,000 Y pb = --------------------------------------------- = ------------------- = 1.41 yrs. 101,480 – 88,057 $13,423
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Solutions to
Chapter 11 AIR-CONDITIONING SYSTEM CONCEPTS
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Chapter 11—Air-Conditioning System Concepts⏐141
11.1 A room is to be cooled to a temperature of 75°F and a relative humidity of 50%. If there is negligible latent load within the space, what is the highest temperature at which the conditioned air can be supplied (ts)? Why? ts
max
= dew-point temperature of the space conditions P s at 75°F = 0.87567 in. Hg P s at 75°F/50% = 0.5 ( 0.87567 ) = 0.43784 t s at 0.43784 in. Hg = 55.12°F
11.2 A room has a total space cooling load of 20 tons and a sensible heat ratio of 0.90. If the conditioned air is to be supplied at 20°F less than the room temperature, how much air must be circulated? q s = q T ( SHR ) = 20 ( 12000 ) ( 0.9 ) q s = 216000 Btu/h q s = 1.1 ( CFM ) ( t r – t s ) qs 216000 CFM = ------------------ = -----------------1.1 ( Δt ) 1.1 ( 20 ) 3
CFM = 9820 ft ⁄ min
Heat - cool - off Dual stream Reheat Variable Air Volume
11.4 What are the 18 fundamental parameters that must be addressed in the selection and design of an HVAC system? Load dynamics Performance requirements Availability of equipment Capacity Spatial requirements First cost Energy Consumption Operating cost Simplicity
Flexibility Operations requirements Service ability Maintainability Availability of service Availability of replacement components Environmental requirements of space Environmental requirements of community Reliability
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11.3 What are the four generic types of air systems expressed by thermodynamic methods?
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142⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
11.5 Before designing a system, the cooling and heating load for each room in a building must be calculated? Why? Because the system must be designed to add heat or remove it in each room at the same rate at which the load occurs in order to maintain thermal equilibrium. Also, an understanding of the nature of the load is fundamental to the selection of a system type. 11.6 If outdoor air at 95°F dry bulb and 78°F wet bulb is cooled to 75°F dry bulb without any dehumidification, what will the relative humidity be?
90% RH
11.7 In the air handling unit of Figure 11-1, under design conditions the outdoor air temperature is 95°F dry bulb and 78°F wet bulb and the space temperature is 75°F and 50% RH. The supply fan handles 60,000 cfm of air at 55°F saturated (entering the fan). If the minimum outdoor air dampers are sized for 6000 cfm of ventilation air, what is the statepoint (dry-bulb and wet-bulb temperatures) of the mixed air?
From Psych. chart
11.8 A constant-flow air-handling system is designed to circulate 60,000 cfm of air at a total fan pressure rise of 6 in. w.g. The system is designed to operate continuously. The fan efficiency is 70% and the motor efficiency is 90%. a. How much power (hp) is required to drive the fan? b. What will be the annual fan energy consumption? a)
b)
CFM ( DP t ) Hp = ---------------------------6350η f 60000 ( 6 in. ) = ------------------------------6350 ( 0.7 ) Hp = 81 CFM ( ΔP f )θ kWh = ------------------------------( 8512 )η f η m 60000 ( 6 ) ( 8760 ) kWh = ---------------------------------------- = 588100 kWh 8512 ( 0.7 ) ( 0.9 )
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CFMO t m = t i + ------------------ ( t o – t i ) CFMT 6000 = 75 + --------------- ( 95 – 75 ) 60000 = 77°F db 64°F wb h m = 29.6 Btu/lb
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Chapter 11—Air-Conditioning System Concepts⏐143
11.9 If, in the above problem, the sensible space load were reduced by 25% by using a more energy effective building envelope and improved lighting system, and this change were accommodated by reducing the air flow rate at the same fan pressure and efficiencies, what would be the reduction in annual fan energy? q s = CFM ( 1.1 ) ( Δt ) q s ≈ CFM and kWh ≈ CFM kWh ≈ q s ∴Δ ( kWh ) = kWh (% reduction in load) = 588100 ( 0.25 ) = 147000 kWh
11.10 It is desired to transfer a given quantity of heat energy from one location to another location in a building. Two methods being considered are either by an air system operating at 4 in. of water total pressure or by a water system with a pump head of 40 ft. Calculate the ratio of fan power required for an air system to pump power required for a water system with the following system variables: Fan efficiency 70% Pump efficiency 80% Air Δt 20°F Water Δt 40°F
q = 1.1 ( CFM ) ( Δt air ) q CFM = -----------------------1.1 ( Δt air ) q ( ΔP t ) q(Δ) –5 Hp air = ----------------------------------------------- = ------------------------------------------------ = 4.09 × 10 q 6350 ( 1.1 ) ( Δt air )η f 6350 ( 1.1 ) ( 20 ) ( 0.7 ) GPM ( ΔH ) Hp w = --------------------------3960 ( η p ) q = GPM ( 500 ) ( Δt w ) GPM = 8 ⁄ ( 500 ) ( Δt w ) q ( 40 ) q ( ΔH ) –7 Hp w = ----------------------------------------------- = -------------------------------------------------- = 6.31 × 10 q 3960 ( 500 ) ( 40 ) ( 0.8 ) 3960 ( 500 )Δt w ( η p ) –5 Hp air × 10 q------------------------------------------- = 4.09 = 65 –7 Hp w 6.31 × 10 q
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CFM ( ΔP t ) Hp air = --------------------------6350η f
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144⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
11.11 A fan with a variable speed drive is selected to operate at 900 rpm, and it is installed on a spring isolator mount with 1 in. static deflection. Determine a. Transmissibility of the isolator b. Minimum speed that the unit can be operated at before the transmissibility is 0.50 1 TR = ------------------------2 ( f ⁄ fn ) – 1
a)
N 900 f = ------ = --------- = 15 Hz 60 60 1 f n = ------ g--2π y 1 386 f n = ------ --------- = 3.13 Hz 2π 1 1 - = 0.046 TR = -------------------------2 15 ⎛ ----------⎞ – 1 ⎝ 3.13⎠ 1 TR = ------------------------2 ( f ⁄ fn ) – 1
b)
2
( f ⁄ fn ) TR – TR = 1 2 TR + 1 0.5 + 1 ( f ⁄ f n ) = ---------------- = ---------------- = 3 TR 0.5
f = f =
3 ( 3.13 ) = 5.42 N min f = ----------60 = 60f = 60 ( 5.42 )
N min = 325 rpm
11.12 Specify typical temperatures for the following: a. Air leaving a gas-fired warm air furnace b. Air leaving a heat pump condenser c. Air leaving the cooling coil of a residential air conditioner d. Air leaving the cooling coil of a commercial air conditioner e. Hot water entering the convectors (radiators) of a hydronic system f. Hot water returning to the boiler from the convectors a. b. c. d. e. f.
135°F 105°F 58°F 55°F 190°F 170°F
(57°C) (41°C) (14°C) (13°C) (88°C) (77°C)
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N min
3f n
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Chapter 11—Air-Conditioning System Concepts⏐145
11.13 An air-conditioned room has a sensible heat load of 200,000 Btu/h, a latent heat load of 50,000 Btu/h, an occupancy of 20 people, and is maintained at 76°F dry bulb and 64°F wet bulb. Twenty-five percent of the air entering the room leaves through cracks and hoods. Outside air is assumed to be at design conditions of 95°F dry bulb and 76°F wet bulb. Conditioned air leaves the apparatus and enters the room at 60°F dry bulb. Use the following letters to designate state points: A Outside design conditions B Inside design conditions C Air entering apparatus (mixed air) D Air entering room (supply air) a. Complete the table provided. b. Calculate the room SHR. c. What air quantity must enter the room? d. What is the apparatus load in tons? e. What is the load of the outside air? In lb per hour? In cfm? f. Does the room load plus the outside air load equal the coil load? a)
Point
Dry Bulb
Wet Bulb
h
W
A
95
76
39.4
0.015
B
76
64
29.2
0.010
C
80.6
67.2
31.75
0.01125
D
60
56.8
24.3
0.0091
A: 95°F db, 76°F wb; h A = 39.4, W A = 0.015 B: 76°F db, 64°F wb; h B = 29.2, W B = 0.010 M A h A + M B h B = ( M A + M B )h c ; M A W A + M e W e = ( M A + M e )W c ( 0.25 ) ( 39.4 ) + ( 0.75 ) ( 29.2 ) = h c ; ( 0.25 ) ( 0.015 ) + ( 0.75 ) ( 0.010 ) = W c C: h c = 31.75; W c = 0.01125
q s = 200000 = M da C p ( t B – t D ) = M da ( 0.244 )(76 – 60) M da = 51230 lb/h MW ( 50000 ⁄ 1054 ) D: W D = W B – ---------- = 0.010 – ----------------------------------- = 0.0091 M da 51230 D: 60°F db, 56.8°F wb, h D = 24.3 b)
SHR = q s ⁄ ( q s + q L ) = 200000 ⁄ 250000 = 0.8
d)
3 51230 51230 lb/h; v D = 13.3 ft ⁄ lb; V· = --------------- ( 13.3 ) = 11356 cfm 60 M da h c – M da ( W c – W D )h f60°F – M da h D + q = 0
e)
51230 [ 31.75 – ( 0.01125 – 0.0091 )28 – 24.3 ] = – q q = – 378600 Btuh = 31.5 tons O.A. Load: q s = M OA C p ( t A – t B ) = 0.25 ( 51230 ) ( 0.244 ) ( 95 – 76 ) = 59376 Btu/h
c)
q L = M OA ( W A – W B )1054 = 0.25 ( 51230 ) ( 0.015 – 0.010 ) ( 1054 ) = 67495 Btu/h 126872 Btu/h f)
250000 + 126872 = 376872 ≈ 378600
YES
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C: 80.6°F db, 67.2°F wb Assuming occupancy is included in q s and q L as should be:
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146⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
11.14 A space has a sensible heat loss of 60,000 Btu/h and a latent loss of 20,000 Btu/h. The space is to be maintained at 70°F and 40% RH. The air that passes through the conditioner is 90% recirculated and 10% outdoor air at 40°F and 20% RH. The conditioner consists of an adiabatic saturator and a heating coil. Estimate the temperature and humidity ratio of the air entering the conditioned space. What is the flow rate in lb/h and cfm? How much heat is added by the coil to the air in Btu/h? How much water is added to the air by the adiabatic saturator (lb/h)? q L = 20000 Btuh q s = 60000 Btuh t r = 70°F t OA = 40°F r: OA:
t r = 70°F, θ r = 40%; W r = 0.0062; h r = 23.7 40°Fdb, 20% RH; W OA = 0.001; h OA = 10.7
m:
90% recirc.; 10% outside air 0.1 ( 0.001 ) + 0.9 ( 0.0062 ) = W m = 0.0057
m:
0.1 ( 10.7 ) + 0.9 ( 23.7 ) = h m = 22.4 ; t m = 67°F t wb, m = 53.7°F = t x
x:
t = 53.7°F, θ = 100%, W x = 0.0088
60000 SHR (space) = --------------- = 0.75 80000 From Psych. chart: at intersection of W = 0.0088 and space condition line (slope of SHR = 0.75 ): t s = 102°Fdb, W s = 0.0088, v s = 14.35
or 1853 cfm q HC = mCp ( t s – t x ) = 7748 ( 0.242 ) ( 102 – 53.7 ) = 90563 Btu/h M W, AS = M ( W s – W m ) = 7748 ( 0.0088 – 0.0057 ) = 24 lb/h
11.15 Air at the 800 ft3/min leaves a residential air conditioner at 65°F with 40% RH. The return air from the rooms has average dry- and wet-bulb temperatures of 75°F and 65°F, respectively. Determine a. Size of the unit in tons (12,000 Btu/h = 1 ton) b. Rate of dehumidification 75°Fdb 65°Fwb h 1 = 30
65°F, 40% RH, h 2 = 21.4, W 2 = 0.0053, v = 13.335 ( 800 ) ( 60 ) M da = ------------------------- = 3600 lb/h 13.335
w 1 = 0.0109 a)
q = M da [ h 1 – h 2 – ( W 1 – W 2 )h f ] = 3600 [ 30 – 21.4 – ( 0.0109 – 0.0053 )33 ] b)
= 30300 Btu/h = 2.52 tons · M w = M da ( W 1 – W 2 ) = ( 3600 ) ( 0.0109 – 0.0053 ) = 20.2 lb/h
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qs 60000 M da = ------------------------- = --------------------------------------- = 7748 lb/h Cp ( ts – tr ) 0.242 ( 102 – 70 )
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Chapter 11—Air-Conditioning System Concepts⏐147
11.16 In an air-conditioning unit 6000 cfm at 80°F dry bulb, 60% RH, and standard atmospheric pressure, enter the unit. The leaving condition of the air is 57°F dry bulb and 90% RH. Calculate
a)
a. b. c. d. e.
Cooling capacity of the air-conditioning unit, tons Rate of water removal from the unit, lb/h Sensible heat load on the conditioner, Btu/h Latent heat load on the conditioner, Btu/h Dew point of the air leaving the conditioner, °F
6000 × 60 M da = ------------------------ = 26100 lb/h 13.8 M da h 1 – M da h 2 – M da ( W 1 – W 2 )h 3 + q c = 0 26100 [ 33.8 – 23.4 – ( 0.0132 – 0.009 )25 ] = – q c q c = – 268700 Btu/h = 22.4 tons
b)
M c = M da ( W 1 – W 2 ) = 26100 ( 0.0132 – 0.009 ) = 109.6 lb/h
c)
q s = M da C p ( t 1 – t 2 ) = 26100 ( 0.244 ) ( 80 – 57 ) = 146470 Btu/h
d)
= 12.2 tons q L = M c ( 1076 ) = 109.6 ( 1076 ) = 117930 Btu/h = 9.8 tons
e)
Dew Point = 54°F
r: OA:
a. b. c. d. e. f. g.
Temperature of the air entering the preheater Temperature of the air entering the space to be heated Heat supplied to preheat coil, Btu/h Heat supplied to reheat coil, Btu/h Quantity of make-up water added to adiabatic saturator, gpm Temperature of the spray water Show the processes and label points on the psychrometric diagram
75°Fdb, 50% RH; W r = 0.0093, h r = 28.2 20°Fdb, 100% RH; W OA = 0.002152, h OA = 7.106
Leaving adiabatic saturator: t = 60°Fdb, W = 0.0093 ∴wb = 57°F Leaving preheater: W = 0.002152, WB = 57°F ∴t = 91°Fdb a) Entering preheater, t = 20°Fdb b) q s = 200000 = 1.1 cfm ( t s – t r ) = 1.1 ( 7000 ) ( t s – 75 ) t s = 101.5°F c) d)
q ≅ 1.1 ( 7000 ) ( 91 – 20 ) = 536760 Btu/h (preheat) q ≅ 1.1 ( 7000 ) ( 101.5 – 60 ) = 313740 Btu/h (reheat)
e)
700 lb gpm = ---------- ( 0.0093 – 0.0022 ) ⁄ ⎛ 8.33 -------⎞ = 0.44 gpm ⎝ 13.5 gal⎠
f)
57°F
g)
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11.17 A space in an industrial building has a winter sensible heat loss of 200,000 Btu/h and a negligible latent heat load (latent losses to outside are made up by latent gains within the space). The space is to be maintained at 75°F and 50% RH. Due to the nature of the process, 100% outdoor air is required for ventilation. The outdoor air conditions can be taken as saturated air at 20°F. The amount of ventilation air required is 7000 cfm and the air is to be preheated, humidified with an adiabatic saturator, and then reheated. The temperature out of the adiabatic saturator is to be maintained at 60°F dry bulb. Calculate:
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148⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
11.18 In winter, a meeting room with a large window is to be maintained at comfort conditions. The inside glass temperature on the design day is 40°F. Condensation on the window is highly undesirable. The room is to accommodate 18 adult males [250 Btu/h (sensible) and 200 Btu/h (latent) per person]. The heat loss through the walls, ceiling, and floor is 33,600 Btu/h. There are 640 watts of lights in the room. a. Determine the sensible heat loss or gain. b. Specify the desired interior dry-bulb temperature and relative humidity. c. If the heating system provides air at 95°F, determine the required airflow (cfm) and the maximum relative humidity permissible in the incoming air. a)
q s ( loss ) = 33600 – 640 ( 3.413 ) – 18 ( 250 ) = 26916 Btu/h
b)
t db = 75°F (selected), D.P. = 40°F, ∴φ = 28%; W r = 0.0052
c)
60 26916 = 1.08 cfm ( 95 – 75 ); cfm = 1246 × ---------- = 5497 lb/h 13.6
loss
Mw ( 18 ) ( 200 ) ( 1070 ) W s = W r – ---------- = 0.0052 – ------------------------------------------ = 0.0046 M da 5497 φ s = 14% max
b)
70000 SHR = ------------------------------------ = 0.70 70000 + 30000 Using protractor on Psych. chart: t = 49°F db, W = 0.0074 at 100% RH q s = M da ( 0.244 ) ( 77 – 49 ) = 70000; M da = 10246 lb/h
c)
r: 77°F db, 50% RH; W r = 0.010, h r = 29.4
a)
OA: 97°F db, 60% RH; W OA = 0.023, h OA = 48.7 m: W m = 0.5 ( 0.01 ) + 0.5 ( 0.023 ) ; h m = 0.5 ( 29.4 ) + 0.5 ( 48.7 ) W m = 0.0165 ; h m = 39.05 s: 49°F db, 100% RH; W s = 0.0074, h s = 19.8 M da [ h m – h s – ( W m – W s )h f49° ] = – q q = – 10246 [ 39.05 – 19.8 – ( 0.0165 – 0.0074 )17 ] = – 195650 Btu/h = – 57.3 kW
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11.19 A zone in a building has a sensible load of 20.5 kW (70,000 Btu/h) and a latent load of 8.8 kW (30,000 Btu/h). The zone is to be maintained at 25°C (77°F) and 50% RH. a. Calculate the conditions (t and W) of the entering air to the zone if the air leaves the coil saturated. b. What flow rate is required in order to maintain the space temperatures? c. If a mixture of 50% return air and 50% outdoor air at 36.1°C (97°F) and 60% RH enters the air conditioner, what is the refrigeration load?
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Chapter 11—Air-Conditioning System Concepts⏐149
11.20 Sketch (with line diagrams) and list the advantages, disadvantages, and typical uses of the following systems: a. Fan-coil units b. Terminal reheat system c. Multizone system d. Double-duct system e. Variable volume system f. Induction system a) b) c) d) e) f)
Article 12.8.1 Article 12.3.1 Article 12.5 Article 12.4 Article 12.3.2 Article 12.7.2
11.21 A general office building in St. Louis, Missouri, has a winter sensible space heating load of 1,150,000 Btu/h for design conditions of 75 and −5°F. The heating system operates with 25% outside air mixed with return air. a. Schematically draw the flow diagram and label, including temperatures and flow rates at each location. b. Specify the necessary furnace size. Note: different outside design conditions may be selected.
1150000 CFM s = --------------------------------- = 19000 scfm 1.1 ( 130 – 75 )
b) Qf = 19000(1.10)(130 – 55) = 1,570,000 Btu/h = 460 kW
11.22 For the building of Problem 11.21, determine: a. Annual energy requirements for heating, Btu b. Annual fuel cost using No. 2 fuel oil at $1.60/gal. a)
H L × DD × 24 9 1570000 × 4900 × 24 E = ---------------------------------- × C D = --------------------------------------------------- × 0.6 = 1.385 × 10 Btu Δt × k × v ( 75 – ( – 5 ) ) × 1 × 1 = 405800 kWh 9
b)
1.385 × 10 Btu Cost = ------------------------------------- × $1.60/gal = $17600 126000 Btu/gal
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a)
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150⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
11.23 To provide comfort conditions for a general office building, 38 ft by 80 ft by 8 ft, an air-treating unit consisting of cooling coil, heating coil, and humidifier is provided for this space with the flow diagram as shown. Ninety people are normally employed doing light work while seated. The building is in Kansas City, Missouri. Fan operation is constant all year long. Winter: Sensible space heat loss is 189,000 Btu/h at design conditions, latent load is negligible. Maximum supply air temperature is 155°F.
Summer: Sensible space heat gain is 101,200 Btu/h at design conditions. Latent load is due entirely to the occupancy. The minimum supply air temperature from the cooling coil is 58°F. a. Determine the fan size (scfm) needed to provide sufficient air b. Size the heating unit needed, Btu/h c. Size the cooling coil needed, Btu/h d. Size the humidifier, gal/h
Summer design Inside: 78°F, 60% RH (max); W rmax = 0.0124 Outside: 96°Fdb, 74°Fwb; 20° range; W O = 0.013 Inside: 72°F, 25% RH (min); W rmin = 0.0042
Winter design
Outside: 6°F, W O = 0.001074, h O = 2.583 Ventilation air a)
General office: 15 cfm/person × 90 = 1350 cfm
Q s, w = 189000 = 1.10 ( scfm ) ( 155 – 72 ) ⇒ SCFM = 2070 Q s, s = 101200 = 1.10 ( scfm ) ( 78 – 58 ) ⇒ SCFM = 4600
b)
Winter SCFM on T OA + SCFM r T r = SCFM T T m 1350 ( 6 ) + 3250 ( 72 ) = 4600 T m ; T m = 52.6°F SCFM O W O + SCFM ⋅ W r = SCFM T W m 1350 ( 0.001074 ) + 3250 ( 0.0042 ) = 4600 W m ; W m = 0.0033 Q s = 4600 ( 1.10 ) ( 109.4 – 52.6 ) = 287400 Btu/h Q L = 4600 ( 4840 ) ( 0.0042 – 0.0033 ) = 20000 Btu/h
c)
Furnace size = 307400 Btu/h ( say 300000 ) Summer s: T s = 58°F, φ s = 100%, W s = 0.0104, h s = 25.2 Ms 90 ( 255 ) ⁄ 1100 r: T r = 78°F, W r = W s + ------- = 0.0104 + ------------------------------------ = 0.0114 60 Ma 46000 × ---------13.3 m: 1350 ( 0.013 ) + 3250 ( 0.0114 ) = 4600 W m ;W m = 0.0119 1350 ( 96 ) + 3250 ( 78 ) = 4600 T m ;T m = 83.3°F M a [ h m – h s – ( W m – W s )h f ] + Q c = 0 4600 ------------ ( 60 ) [ 33.2 – 25.2 – ( 0.0119 – 0.0104 )26 ] = – Q c 13.3 Q c = – 165200 Btu/h A/C size
4600 60 1 M c = ------------ × ---------- ( 0.0042 – 0.0033 ) = 2.24 or 2 ------------13.3 8.33 4 gph * Note: different outside design conditions may be selected. d)
Humidifier
h m = 33.2
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use 4600 cfm Q s, w = 189000 = 1.10 ( 4600 ) ( T s – 72 ) ⇒ T s, w = 109.4°F
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Chapter 11—Air-Conditioning System Concepts⏐151
11.24 A view of the air-conditioning system for a building in Denver, Colorado (elevation = 5000 ft; barometric pressure = 12.23 psi), is given. Outside air at the rate of 2500 cfm is required for ventilation. Other conditions at summer design are Space Loads Sensible = 410,000 Btu/h Latent = 220,000 Btu/h Outside Air: 91°F, 30% RH For an indoor design temperature of 78°F, determine
1. 2.
1.
Supply airflow, lb/h
2.
Supply airflow, cfm
3.
Relative humidity at return, %
4.
Size of cooling unit, Btu/h
5.
Latent component of (4)
6.
Sensible component of (4)
7.
Sensible cooling load due to outside air, Btu/h
410000 Q s = m a c p ( t r – t s ) ⇒ m a = ------------------------------------ = 73060 lb/h 0.244 ( 78 – 55 ) 73060 v· = m a v = --------------- × 13.33 = 16230 cfm @ s: P w = P w, s = 0.2141 60 0.2141 W s = 0.622 ------------------------------ = 0.0110 12.28 – 2.41 h s = 0.24 ( 58 ) + 0.011 ( 1061 + 0.444 – 59 ) = 25.23
3.
Q L ⁄ 1100 220000 ⁄ 1100 W r = W s + ----------------------- = 0.0110 + ---------------------------------- = 0.01874 lb v ⁄ lb a ; P w, s = 0.47511 psia Ma 73060
h r = 0.240 ( 78 ) + 0.01374 ( 1061 + 0.444 × 78 ) = 33.74 4.
m O = 2500 × 60 ⁄ 13.33 = 11250 lb/h ; P w
1 s0
= 0.72113 ; P w
10
= 0.30 ( 0.72113 ) = 0.21634
0.21634 W O = 0.622 --------------------------------------- = 0.0112 ;h o = 0.240 ( 91 ) + 0.0112 ( 1061 + 0.444 × 91 ) = 34.18 12.23 – 0.21634 h m = [ ( 11250 ) ( 34.18 ) + ( 73060 – 11250 ) ( 33.74 ) ] ⁄ 73060 = 33.80 Btu/lb a W m = [ ( 11250 ) ( 0.0112 ) + ( 73060 – 11250 ) ( 0.01374 ) ] ⁄ 73060 = 0.1335 lb v ⁄ lb a m a [ h m – h s – ( W m – W s )h c ] + Q c = 0 73060 [ 33.80 – 25.23 – ( 0.01335 – 0.0110 ) ( 23.07 ) ] = – Q c = 622200 Btu/h 5.
Q L ≅ 4840 × CFM × ΔW = 4840 × 16230 × ( 0.01335 – 0.0110 ) = 184600 Btu/h
6.
33.80 – 0.240 ( t ) + 0.01335 ( 1061 + 0.444 ⋅ t ) ⇒ t m = 79.8°F
c
Q s ≅ 1.10 × CFM × Δt = 1.10 × 16230 × ( 79.8 – 55 ) = 443500 Btu/h c
or Q s = Q T – Q L = 622200 – 184600 = 437600 Btu/h c
7.
Q s ≅ 1.10 × 2500 × ( 91 – 78 ) = 35750 Btu/h o
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Pw Pw 0.26432 w = 0.622 ---------------- = 0.01374 = 0.622 -------------------------- ⇒ P w = 0.26432 psia ; φ = ------------------- × 110 = 55.6% P – Pw 12.23 – P w 0.47511
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Solutions to
Chapter 12 SYSTEM CONFIGURATIONS
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Chapter 12—System Configurations⏐155
12.1 From an energy consumption perspective, list the four fundamental psychrometric system types from least consumption to most consumption. 1. Variable - air - volume 2. Heat - cool - off 3. Dual stream 4. Reheat 12.2 In a VAV system with series fan powered terminals, why must all of the terminal fans be running prior to turning on the system fan? If the main fan is blowing air through the terminal fans, they will turn in the wrong direction. They, being single phase motors, will run in the direction they are turning when turned on, thus running backwards. 12.3 What is the advantage of a parallel fan powered terminal over a series fan-powered terminal? Less energy consumption, because the fan does not have to run continuously. 12.4 What is the purpose of using a fan powered terminal in a variable air volume system? Provides better mixing, ambient air circulation, and air distribution. Prevents dumping. All as the primary supply air throttles down.
a)
b)
1) The dual duct or reheat provides false loading, which keeps the airflow high enough to assure adequate ventilation and air circulation rate. 2) A VAV system alone cannot handle a heating load. 1) VAV system: As the space cooling load decreases, the space thermostat closes a supply air damper, reducing the flow of conditioned air to the space. Most systems have a lower limit setting to assume adequate ventilation and air circulation, below which there is no room temperature control. 2) VAV reheat system: Similar to (1) above except that when the load falls below the minimum airflow setting, the reheat valve opens and reheats the minimum volume flow air to prevent overcooling and maintain room temperature control at all times. 3) Dual-duct VAV system: Similar to (1) above except that when the load falls below the minimum airflow setting, the warm duct damper starts modulating open mixing warm air with the conditioned air in increasing amounts to prevent overcooling and maintaining temperature control at all times.
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12.5 a. Why do some VAV systems also use dual-duct or reheat features? b. In your own words, describe the operating sequence of the zone or terminal control of 1. A VAV system 2. A VAV reheat system 3. A dual-duct VAV system
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156⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
12.6 What is the primary advantage of a separate outdoor makeup air conditioning unit? The outdoor air make-up air-conditoning unit (primary air unit [12.6.1]) provides a constant volume of outdoor ventilating air and is used in conjunction with VAV systems. It also simplifies the design of the system for the space with no outside air entering the space conditioning unit there is danger of freezing, no ventilation control dampers, no return air fan, and the cooling coil usually operates dry, reducing the likelihood of microbial growth in the system. 12.7 Why is a high-pressure primary system fan required with an induction system? Because there is a high pressure required to create the high velocity flow through the induction nozzles. 12.8 Are fan coil units with connections to the outdoors recommended as an acceptable method for providing ventilation air? Why? No. Because when the room thermostat turns the fan coil unit off, untreated ventilation air can enter the sapce, causing discomfort and sometimes freeze damage or microbial growth. Also, the quantitiy of ventilation air is not controllable because of varying pressure differentials resulting from chimney effects and wind variations.
Vertical units can be mounted under windows in extremely cold climates, providing better heating performance, and horizontal models overhead and above ceilings can create problems related to condensate collection and disposal, mixing of return air from other rooms, leakage of pans causing damage to ceilings, difficulty of access for maintenance and service, and IAQ concerns. 12.10 Size the basic components and sketch the equipment arrangement if the HVAC system now under consideration for the building of Example 12.1 is a triple deck multizone (hot, cold, and neutral decks). This is a design problem and many solutions can be found. 12.11 Size the basic components and sketch the equipment arrangement if the HVAC system now under consideration for the building of Example 12.1 is a variable volume, dual fan, dual duct. This is a design problem and many solutions can be found.
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12.9 Why are vertical floor mounted fan coil units recommended in some applications in preference to horizontal ceiling mounted units? Explain.
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Chapter 12—System Configurations⏐157
12.12 A small single-zone classroom building is being designed for Knoxville, Tennessee, to use the HVAC system shown in the sketch. Minimum outside air for meeting the ventilation requirements of the anticipated 550 occupants will be maintained throughout the year. Fan speed will be changed between summer and winter. The duct system will be designed so that at summer air flow rate the pressure drop does not exceed 3.75 in. w.g. At winter design conditions, the air is heated to 130°F at which temperature it is supplied to the conditioned space. The winter conditioning unit includes both a heating coil and a humidifier supplied with city water at 60°F. The humidistat in the return air steam maintains the design relative humidity of 30% in winter. During summer operation, the cooling coil supplied air to the conditioned space at 58°F. The space design loads are Summer: 423,000 Btu/h sensible (gain) 139,000 Btu/h latent (gain) Winter: 645,000 Btu/h sensible (loss) negligible latent
Unit Physical Data (Approximate)
Size the following system components: 2
Cooling coil, Btu/h and ft of face area Chiller unit, Btu/h Heating coil, Btu/h and ft2 of face area Boiler, Btu/h Humidifier, gph
a. b. c. d. e.
Select an appropriate air handler from the following data. Winter Dry Bulb, °F
φ, %
OA
3
100
1.699
r
72
30
22.7
ma, lb/h
SCFM
.00092 37200 550×15=8250 * .005
8400
1850
m
16
94
5.57
.0017
45600
10100
s
130
5.3
36.8
.005
45600
10100
Design cfm
Face Area*, ft2
3 6 8 10 12 14 17 21 25 30 35 40 50 66 80 100
1,660 2,930 3,770 4,820 6,150 7,110 8,400 10,390 12,190 14,505 17,050 19,650 24,715 32,815 39,375 50,180
2.34 – 3.32 4.31 – 5.86 5.49 – 7.54 7.01 – 9.64 9.46 – 12.3 10.2 – 14.2 12.3 – 16.8 15.0 – 20.8 17.8 – 24.4 21.2 – 29.0 26.72 – 34.10 30.78 – 39.30 34.22 – 49.43 48.13 – 65.63 56.88 – 78.75 73.44 – 100.4
Max Unit Wt., lb
≤ 3,600
≤ 4,500 ≤ 6,000 (all modules)
* Actual face area varies with unit coil type.
* Other outdoor design conditions could be used. Pw 0.119 W s = 0.005 = 0.622 ----------------------- ; P w = 0.119 ; P w, s = 2.2256 ; φ = ---------------- × 100 = 5.3 14.7 – P w 2.2256 h s = 0.240 ( 130 ) + 0.005 [ 1061 + 0.444 ( 130 ) ] = 36.79 Q s = 645000 = 0.244M a ( 130 – 72 ) ; M a = 45600 lb/h = 10100 cfm 37200 ( 1.699 ) + 8400 ( 22.7 ) h m = ------------------------------------------------------------------- = 5.57 45600 Pw 37200 ( 0.00092 ) + 8400 ( 0.005 ) w m = ---------------------------------------------------------------------------- = 0.00167 = 0.622 ----------------------- ; P w = 0.039 14.7 – P w 45600 0.039 37200 ( 3 ) + 8400 ( 72 ) t m = ---------------------------------------------------- = 15.7°F; P w, s = 0.042 ; φ = ------------- 100 = 94% 0.042 45600 45600 [ 36.8 – 5.57 – ( 0.0017 – 0.005 ) ( 60 – 32 ) ] + Q h = 0 ; Q h = 1428000 Btu/h M h = ( 45600 ( 0.005 – 0.0017 ) ) ⁄ 8.33 = 18 gph
Problem 12.12 continued on next page.
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Enthalpy h, W, Btu/lb lb/lb
Point
Unit Coil
Unit Size
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158⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
Problem 12.12 continued. Summer Enthalpy h, W, Btu/lb lb/lb
ma, lb/h
Point
Dry Bulb, °F
φ, %
OA
91
48
38.5
.0151
37200
8250
r
78
58
32.0
.0120
49500
11050
m
83.6
53
34.8
.0133
86700
19300
f
85.9
49
35.4
.0133
86700
19300
s
58
100
25.1
.0105
86700
19300
SCFM
Q s = 423000 = 0.244M a ( 78 – 58 ) ;to M a = 86700 lb/h ( 139000 ⁄ 1100 ) W r = 0.0105 + -------------------------------------- = 0.0120 86700 W m = [ 37200 ( 0.0151 ) + 49500 ( 0.0120 ) ] ⁄ 86700 = 0.0133 lb/lb h m = [ 37200 ( 38.5 ) + 49500 ( 32 ) ] ⁄ 86700 = 34.8 Btu/lb t m = [ 37200 ( 91 ) + 49500 ( 78 ) ] ⁄ 86700 = 93.6°F W fan = 19300 ( 3.75 ) ( 0.0361 ) ( 144 )60 ⁄ 778 = 29000 Btu/h 29000 Assume R f = 60% W fan = --------------- = 48300 Btu/h = 19 hp or 0.56 Btu/lb 0.6 Δt = 2.3°F h f = 34.8 + 0.56 = 35.4 86700 [ 35.4 – 25.1 – ( 0.0133 – 0.0105 )26 ] + Q c = 0 ;Q c = – 887000 Btu/h Select 600 fpm face velocity 2 19300 A = --------------- = 32 ft 600
Unit Size
40
a) b)
112000 Winter: CFM 1 = ------------------------------------ = 1850 1.10 ( 130 – 75 ) 23500 CFM 2 = --------------------------------- = 1260 1.10 ( 75 – 58 ) 67000 Summer: CFM 1 = --------------------------------- = 3580 1.10 ( 75 – 58 ) 49000 CFM 2 = --------------------------------- = 2620 1.10 ( 75 – 58 ) 6200 ( 3.1 ) ( 62.4 ) ( 60 ) FAN: hp = v· ΔP = --------------------------------------------------- = 3 hp 6200 SCFM 3.1 in. Hg ( 12 )778 ( 2545 ) 1.10 ⋅ x 1 ( 130 – 75 ) = 112000 + 1.10 ( 3580 – x 1 ) ( 75 – 58 ) ;to 60.5x 1 = 112000 + 66950 – 18.7x 1 x 1 = 2259 cfm 1.10 ⋅ x 2 ( 130 – 75 ) = – 23500 + 1.10 ( 2620 – x 2 ) ( 75 – 58 ) ;to 60.5x 2 = – 23500 + 48994 – 18.7x 2 x 2 = 322 cfm CFM Hc = 2259 + 322 = 2581 4800 ( 75 ) + 1400 ( 0 ) 3 ( 2545 ) t m = ------------------------------------------------- = 58.2 ; t f = 58.2 + -------------------------------------------------------- = 58.3°F 6200 ( 0.24 )6200 ( 60 ) ⁄ 13.33 Q Hc = 1.10 × 2581 ⋅ ( 130 – 58.3 ) = 200000 Btu/h
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12.13 A double duct system is to be used for air conditioning of a two-zone building. At winter design outdoor temperature of 0°F, exterior SPACE 1 has a design sensible heat loss of 112,000 Btu/h while interior SPACE 2 has a net sensible heat gain of 23,500 Btu/h. At summer design outdoor conditions of 95°F db and 75°F wb, SPACE 1 has a design sensible heat gain of 67,000 Btu/h while SPACE 2 experiences a design sensible heat gain of 49,000 Btu/h. Interior design temperatures of both spaces is 75°F, all year long. Duct pressure drop is 3.1 in. water. Outside air requirement is 1400 cfm. Calculate the size of a. Fan (scfm, pressure, motor horsepower) b. Heating coil (Btu/h).
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Chapter 12—System Configurations⏐159
12.14 To maintain necessary close control of humidity and temperature required for a computer room, the reheat air-conditioning system shown in the sketch is used. Space loads for the computer room include a heat load of 85,000 Btu/h and a moisture load of 42 lb/h. The return air conditions from the space must be exactly 50% relative humidity and 78°F. After mixing of the outside ventilation air with return air, the mixed air is at 80°F dry bulb with a relative humidity of 0.0114 lb/lb. The air is then cooled to saturation at 50°F by the cooling coil. There is a 2°F temperature rise across the fan. Air flow is controlled by a humidistat in the return air duct. The thermostat controls the temperature leaving the reheater. Size the reheater (kW) and the cooling coil (Btu/h). From a manufacturer’s catalog, select an appropriate electric resistance reheater coil. From a manufacturer's catalog, select an appropriate chilled water cooling coil. RETURN AIR CONDITION: p v, s @ 78°F = 0.475 psia RH/100 = 0.5 = p v ⁄ p v, s = p v ⁄ 0.475 p v = 0.3375 W r = 0.622p v ⁄ ( p – p v ) = 0.622 ( 0.3375 ) ⁄ ( 14.7 – 0.3375 ) = 0.0102 lb v ⁄ lb a h r = 0.24t + Wh g = 0.240 + 0.0102 ( 1095.5 ) = 29.91 Btu/lb [ Note: Psychrometric properties could also have been obtained from the Psychrometric chart. ]
h m = 0.240 ( 80 ) + 0.0102 ( 1096.4 ) = 31.7 Btu/lb CONDITION OF AIR LEAVING COOLING COIL: 50°F, RH = 100%; p v = p v, s = 0.178 psia W cc = 0.622 ( 0.178 ) ⁄ ( 14.7 – 0.178 ) = 0.00762 lb v ⁄ lb a h cc = 0.240 ( 50 ) + 0.00762 ( 1083.3 ) = 20.25 Btu/lb SUPPLY AIR CONDITION: W s = W f = W cc = 0.00762 lb v ⁄ lb a m a = m s ⁄ ( W r – W s ) = 42 ⁄ ( 0.0102 – 0.00762 ) = 16300 lb/h t s = t r – Q ⁄ [ m a c p ], where c p = 0.244 Btu/lb °F for moist air 78 – 85000 ⁄ ( 0.244 ) ( 16300 ) = 56.6°F COIL SIZES (RATINGS) Q cc = – 16300 [ 31.7 – 20.25 – ( 0.0114 – 0.00762 ) ( 18 ) ] = – 185500 Btu/h Q rh = 16300 ( 0.244 ) ( 56.6 – 52 ) = 18300 Btu/hr = 5.4 kW (can probably get by with 5 kW heater)
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MIXED AIR CONDITION: 80°F, W m = 0.0114 lb v ⁄ lb a
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160⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
12.15 A small commercial building located in St. Louis, Missouri is to be conditioned using a variable air volume (VAV) system with reheat, as shown in the following sketch. At this stage of the process, preliminary sizing of the central cooling unit, of the reheaters, and of the fan (scfm) is to take place. There are four zones (separately thermostated spaces) in the building. Supply air from the cooling coil is maintained at 55°F during the summer and 58°F during the winter. Relative humidity off the coil is approximately 90% in both cases. Minimum outside air of 4000 scfm is maintained at all times (just don’t ask how). The VAV boxes are not to be cut back beyond 50% of rated flow. The design conditions and calculated design load for each zone are as follow: Zone 1 Winter inside temperature = 72°F Winter design heat loss = −55,000 Btu/h (a loss) Summer inside temperature = 78°F Summer design heat gains = 124,000 Btu/h (sensible) and 31,000 Btu/h (latent) 1.
Determine required airflow rates 124000 CFM 1 = --------------------------------- = 4901 1.10 ( 78 – 55 ) 220000 CFM 2 = --------------------------------- = 8696 1.10 ( 78 – 55 )
Zone 2 (an interior space) Winter inside temperature = 78°F Winter design heat loss = 40,000 Btu/h (a gain) Summer inside temperature = 78°F Summer design heat gains = 220,000 Btu/h (sensible) and 71,000 Btu/h (latent) Zone 3 (an interior space) Winter inside temperature = 78°F Winter design heat loss = 115,000 Btu/h (a gain) Summer inside temperature = 78°F Summer design heat gains = 140,000 Btu/h (sensible) and 42,000 Btu/h (latent) Zone 4 Winter inside temperature = 72°F Winter design heat loss = −180,000 Btu/h (a loss) Summer inside temperature = 78°F Summer design heat gains = 210,000 Btu/h (sensible) and 52,500 Btu/h (latent)
[ Q s = 1.10 × CFM s × Δt ]
140000 115000 CFM 3 = --------------------------------- = 5534 [check CFM 3 = --------------------------------- = 5288] 1.10 ( 78 – 55 ) 1.10 ( 78 – 58 )
2.
t r = 78°F ⇒ φ r = 48% [Plenty low, could reset t s to higher value] t m = [ 4000 ( 94 ) + ( 27431 – 4000 ) ( 78 ) ] ⁄ 27431 = 80.3 W m = [ 4000 ( 0.0144 ) + ( 27431 – 4000 ) ( 0.0097 ) ] ⁄ 27431 = 0.0104 Q cc = 27431 [ 1.10 ( 80.3 – 55 ) + 4840 ( 0.0104 – 0.0083 ) ] = 27431 [ 27.83 + 10.16 ] = 1042000 Btu/h Cooling coil → 1042 Mbh * Note:Other outdoor design conditions could be selected. 3. Winter reheater discharge temperature 55000 ( t m ) 1 = 72 + ----------------------------------------- = 92.4°F 1.10 ( 4901 × 0.5 ) – 40000 ( t m ) 2 = 78 + ----------------------------------------- = 69.6°F 1.10 ( 8696 × 0.5 ) – 115000 ( t m ) 3 = 78 + ----------------------------------------- = 40.2°F < 58 ∴ no reheating 1.10 ( 5534 × 0.5 ) 180000 ( t m ) 4 = 72 + ----------------------------------------- = 111.4°F 1.10 ( 8300 × 0.5 ) REHEATERS ( Q m ) 1 = 1.10 ( 4901 × 0.5 ) ( 92.4 – 58 ) = 93000 Btu/h ( Q m ) 2 = 1.10 ( 8696 × 0.5 ) ( 69.6 – 58 ) = 55700 Btu/h ( Qm )3 = 0
none needed
( Q m ) 4 = 1.10 ( 8300 × 0.5 ) ( 111.4 – 58 ) = 244000 Btu/h
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210000 CFM 4 = --------------------------------- = 8300 1.10 ( 78 – 55 ) FAN → 27431 SCFM ⇒ 123750 lb/h Summer (55°F, 90%) Σm w ( 31000 + 71000 + 42000 + 51500 ) ⁄ 1100 W r = W s + ----------- = 0.0083 + ---------------------------------------------------------------------------------------------------- = 0.0097 ma 123750
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Chapter 12—System Configurations⏐161
12.16 A commercial three zone office building is being designed for St. Louis, Missouri where summer outdoor design conditions are 94°F db and 75°F wb and winter outdoor design conditions are 3°F and 100% RH. Each zone is to contain 10,000 sq ft of floor space. A blow-thru multizone unit will be used with cold deck temperature maintained at 58°F all year long and with hot deck temperature varying from a maximum of 130°F at winter design to 85°F during the summer. The amount of outside air is to equal the recommended 20 cfm per person in accordance with ASHRAE Standard 62-1989. Design occupancy is to be 10 people per 1000 ft2 of floor area. The duct system will be designed so that the pressure drop does not exceed 2.0 in. w.g. Fan efficiency is estimated at 65%. In winter, the control humidistat in the common return air duct is set at 30% RH. Due to the building orientation and internal zoning, all spaces will experience their peak loads at the same time. The space design loads at indoor design temperatures of 78°F summer and 72°F winter are
Summer Zone 1: 116,000 Btu/h sensible, 43,000 Btu/h latent (gains) Zone 2: 290,000 Btu/h Sensible, 59,000 Btu/h Latent (gains) Zone 3: 190,000 Btu/h sensible, 39,000 Btu/h latent (gains) Winter Zone 1: −215,000 Btu/h sensible (loss), negligible latent Zone 2: 110,000 Btu/h sensible (gain), negligible latent Zone 3: −171,000 Btu/h sensible (loss), negligible latent Conduct the preliminary sizing of the fan (scfm and horsepower), cooling coil (scfm and Btu/h), heating coil (scfm and Btu/h), and humidifier (gal/h). Provide a completely labeled sketch of the system.
CFM 1s CFM 2s CFM 3s
215000 CFM 1w = ------------------------------------ = 3370 1.10 ( 130 – 72 ) 110000 CFM 2w = --------------------------------- = 5000 1.10 ( 72 – 58 ) 171000 CFM 3w = ------------------------------------ = 2620 1.10 ( 130 – 72 )
CFM = 5270 CFM = 13180 CFM = 8640 Fan CFM = 27090
Summer
27090 M a = --------------- 60 = 122000 lb/h 13.33 27090 ( 2 ) ( 0.0361 )144 ( 60 ) W = ----------------------------------------------------------------- = 13.1 Hp 778 ( 0.65 )2545 t o = 94 ;to W o = 0.0144 ;to t r = 78 ;to t s = 58, φ s = 100% (max) ⇒ W s = 0.0104 ( 43000 + 59000 + 39000 ) ⁄ 1100 W r = 0.0104 + ------------------------------------------------------------------------------ = 0.0115 and 78° ⇒ φ r ≅ 55% ok 122000 6000 ( 94 ) + 21090 ( 78 ) 6000 ( 0.0144 ) + 21090 ( 0.0115 ) t m = ------------------------------------------------------- = 81.5°F W m = ---------------------------------------------------------------------------- = 0.0121 27090 27090 13.1 ( 2545 ) W f = 0.0121 t f = 81.5 + ------------------------------------ = 82.6°F 122000 ( 0.244 ) CFM cc ( max ) = 27090 ( since all zones peak at same time ) Q cc = 27090 [ 1.10 ( 82.6 – 58 ) + 4840 ( 0.0121 – 0.0104 ) ] = 956000 Btu/h
Problem 12.16 continued on next page.
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USE 116000 = --------------------------------- = 5270 1.10 ( 78 – 58 ) 290000 = --------------------------------- = 13180 1.10 ( 78 – 58 ) 190000 = --------------------------------- = 8640 1.10 ( 78 – 58 )
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162⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
Problem 12.16 continued. Winter
zone 1 1.10CFMh1 ( 130 – 72 ) = 215000 + 1.10 ( 5270 – CFM h1 ) ( 72 – 58 ) ;
CFMh1 = 3739
zone 2 1.10CFMh2 ( 130 – 72 ) = 1.10 ( 13180 – CFM h2 ) ( 72 – 58 ) – 110000 ; CFM h2 = 1174 zone 3 1.10CFMh3 ( 130 – 72 ) = 171000 + 1.10 ( 8640 – CFM h3 ) ( 72 – 58 ) ;
CFMh3 = 3839 CFM h = 8752
6000 ( 3 ) + 21090 ( 72 ) t m = ---------------------------------------------------- = 56.7 ; t f = 56.7 + 1.1 = 57.8°F 27090 Humidification: W o = 0.00092 ( 3°, 100% ) ; W r = 0.005 ( 72 ;to 30% ) 6000 ( 60 ) 1 M c = ----------------------- ( 0.005 – 0.00092 ) = 100 lb/h ÷ 8 --- = 13gph 13.33 3 Q h = 110 ( 1076 ) + 1.10 ( 8752 ) ( 130 – 57.8 ) = 118400 + 695100 = 813500 Btu/h c
* Note: Other outdoor design conditions could be selected.
12.17 An air-conditioning unit takes in 2000 cfm of outside air at 95°F dry bulb and 76°F wet bulb, and 6000 cfm of return air at 78°F dry bulb and 50% RH. The conditioned air leaves the chilled water coil at 52°F dry bulb and 90% RH. a. What is the refrigeration load on the chiller in tons? b. Assume the conditioned air were reheated to 58.5°F dry bulb with electric heaters. What would be the operating cost of these heaters at 2.5 cents per kWh?
h m = 32.3, t m = 82°F a)
m da [ h m – h s – ( W m – W s )h f ] = q c c
34479 [ 32.3 – 20.6 – ( 0.0114 – 0.00745 ) ( 20 ) ] = q c q c = 400680 Btu/h = 33.4 tons b)
q rh = m da c p ( t rh – t s ) = 34479 ( 0.244 ) ( 58.5 – 52 ) = 52160 Btu/h = 15.3 kW Cost = 15.3 ( 0.025 ) = 38 cents/hour
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OA: 95°Fdb, 76°Fwb, 2000 cfm; W = 0.015, h = 39.4, v = 14.3 ; m· = 8392 lb/h r: 78°Fdb, 50% RH, 6000 cfm; W = 0.0102, h = 30.0, v = 13.8 ; m· = 26087 lb/h s: 52°Fdb, 90% RH, W = 0.00745, h = 20.6, v = 13.07 ; m· = 34479 lb/h m: 8392 ( 0.015 ) + 26087 ( 0.0102 ) = 34479 W m ; W m = 0.0114 lb/h
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Chapter 12—System Configurations⏐163
12.18 In Problem 12.17, assume 2000 cfm of return air bypasses the chilled water coil and is used for reheat. a. How does the final condition of the air compare with the reheated air in part (b) of 12.1? b. Comment on the ability of the leaving air to absorb latent load in the conditioned space.
(--------------------------4000 ) ( 60 )= 17391 lb/h 13.8 (--------------------------2000 ) ( 60 -) = 8696 lb/h 13.8 m:
8392 ( 0.015 ) + 17391 ( 0.0102 ) = 25783 W m ; W m = 0.0118 8392 ( 39.4 ) + 17391 ( 30 ) = 25783 h m ; h m = 33.06
rh:
8696 ( 30 ) + 25783 ( 20.6 ) = 34479 h rh ; h rh = 22.97 8696 ( 0.0102 ) + 25783 ( 0.00745 ) = 34479 W rh ; W rh = 0.0081
t rh = 58°F W = 0.0081 compared to 0.00745 for 2.11(b), ∴ can absorb less latent load in conditioned space.
12.19 For the building and reheat system shown below, determine a. Fan rating, scfm b. Return air relative humidity at summer design conditions, % c. Size cooling coil, Btu/h d. Size reheat coils, Btu/h and scfm for each
Latent design loads (moisture produced) Space 1: 38 lb/h Space 2: 26 lb/h Year-round: 10% by mass outside air required for ventilation. Conditions of cooling coil: 58°F, 90% RH.
Summer: Outdoor 95°F dry bulb, 78°F wet bulb; indoor 78°F. Sensible design cooling loads Space 1: 64,500 Btu/h Space 2: 55,000 Btu/h 64500 55000 SCFM 2 = --------------------------------- = 2500 SCFM 1 = --------------------------------- = 2932 1.10 ( 78 – 58 ) 1.10 ( 78 – 58 ) a. Fan SCFM = 2932 + 2500 = 5432 cfm ( = 24444 lb/h ) b. 24444 ( 0.0093 ) + 38 + 26 = 24444 W r ; W r = 0.0119 ; φ r, s ≅ 58%,h r, s = 31.8 av
c.
av
h m = 0.9 ( 31.8 ) + 0.1 ( 41.4 ) = 32.8 w m = 0.9 ( 0.0119 ) + 0.1 ( 0.0169 ) = 0.0124
t m = 80°F
5 ( 2545 ) t f = 80 + --------------------------- = 82.1°F; h f = 33.4 1.10 ( 5432 ) 24444 [ 33.4 – 24 – ( 0.0124 – 0.0093 )26 ] + Q c = 0 ;to Q c = 228000 Btu/h d.
162000 143000 t s1 = 72 + --------------------------- = 122F; t s2 = 72 + --------------------------- = 124°F 1.10 ( 2932 ) 1.10 ( 2500 ) Q R = 2500 ( 1.10 ) ( 124 – 58 ) Q R = 2932 ( 1.10 ) ( 122 – 58 ) ; 1
2
= 206400 Btu/h
= 181500 Btu/h
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Winter: Outside 6°F, W = 0.001; indoor 72°F, no humidity control. Sensible design heating loads Space 1: 162,000 Btu/h Space 2: 143,000 Btu/h
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164⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
12.20 A basic reheat system has been retrofitted with an improved control system. For the operating conditions shown in the sketch below and with all thermostats set at 78°F, for what cooling coil discharge temperature T should the logic system of the controller be calling if there is no humidity override?
#1
81000 = 3900 × 1.10 × ( 78 – t s ) ; t s = 59.1°F
#2
149600 = 8500 × 1.10 × ( 78 – t s ) ; t s = 62.0°F
#3
30100 = 2100 × 1.10 × ( 78 – t s ) ; t s = 65.0°F
1
2
3
∴ Coil discharge temperature = 59.1°F
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Solutions to
Chapter 13 HYDRONIC HEATING AND COOLING SYSTEM DESIGN
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Chapter 13—Hydronic Heating and Cooling System Design⏐167
13.1 What is the maximum temperature at which a heating water system can be operated if the boiler (hot water generator) is rated as low pressure by the ASME Boiler and Pressure Vessel Code? From 13.1.2, maximum temperature is 250°F. 13.2 Sketch the fundamental components for a chilledwater system with a single load and source, and a capacity of 100 tons of cooling. a. What is the water circulation rate (GPM) required if the temperature range of the water is 12°F. b. If the head loss in the system is 60 feet, and the pump is 80% efficient, what is the pump horsepower? Motor size? c. If the motor is 90% efficient and it operates for onethird of the total hours in the year, what is the annual energy consumption of the pump? a)
b)
q = GPM ( 500 ) ( Δt ) GPM = q ⁄ 500 ( Δt ) GPM = 100 ( 12000 ) ⁄ ( 500 × 12 ) = 200 gpm GPM ( ΔH ) 200 ( 60 ) Hp = -------------------------- = ------------------------ = 3.79 hp 3960 η r 3960 ( 0.8 ) Motor Hp = 5 Hp
c)
kW 1 Annual Energy (kWh) = Hp × hours × 0.746 -------- × ------Hp η m
13.3 Calculate the size of the expansion tank for a hot water heating system of 1,200,000 Btu/h heating capacity if the tank is a closed tank with an air/water interface and the following system parameters are known:
Supply water temperature Ambient temperature Fill pressure (at tank) Max. operating pressure (at tank) System water volume Steel piping system material 3
v 2 = v f @ 210°F = 0.01670 ft ⁄ lb 3
v 1 = v f @ 60°F = 0.01604 ft ⁄ lb –6
α = 6.5 × 10 in./in.°F Δt = 210 – 60 = 150°F Pa = 14.7 psia P 1 = 30 + 14.7 = 44.7 psia P 2 = 35 + 14.7 = 49.7 psia [ ( v 2 ⁄ v 1 ) – 1 ] – 3αΔt V t = V s ---------------------------------------------------Pa ⁄ P 1 – Pa ⁄ P2
(from Equation 13.12) –6
[ 0.01670 ⁄ 0.01604 – 1 ] – 3 × 6.5 × 10 × ( 150 ) V t = 6000 --------------------------------------------------------------------------------------------------------------------( 14.7 ⁄ 44.7 ) – ( 14.7 ⁄ 49.7 ) V t = 6930 gallons
210°F 60°F 30 psig 35 psig 6,000 gallons
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8760 kWh = 3.79 × ⎛⎝ ------------⎞⎠ ( 0.746 ) ( 1 ⁄ 0.9 ) = 9173 kWh 3
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168⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
13.4 What tank size diaphragm tank would be required for the above system? From Equation 13.14
[ ( v 2 ⁄ v 1 ) – 1 ] – 3αΔt V t = V s ---------------------------------------------------1 – P1 ⁄ P2 –6
[ ( 0.01670 ⁄ 0.01604 ) – 1 ] – 3 × 6.5 × 10 × 150 V t = --------------------------------------------------------------------------------------------------------------------1 – 44.7 ⁄ 49.7 V t = 2280 gallons
13.5 In a given chilled water system, the pump head required at 640 gpm is 80 ft. a. What is the system constant, Cs? b. Plot the system curve from 0 to 800 gpm.
Q = C s ΔH Q 640 gpm C s = ------------ = ---------- = 71.6 ---------ΔH 80 ft ΔH = ( Q ⁄ 71.6 )
2
ΔH
Q
ΔH
0
0
400
31.21
50
.49
450
39.50
100
1.95
500
48.77
150
4.39
550
59.01
200
7.80
600
70.22
250
12.19
650
82.01
300
17.56
700
95.58
350
23.90
750
108.72
800
124.84
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From Equation 13.16
Q
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Chapter 13—Hydronic Heating and Cooling System Design⏐169
13.6 In a chilled-water system, the pump is located in a basement equipment room with the expansion tank connected to the pump suction. The pump is the lowest point in the system and the highest point is a pipe in the penthouse, which is 115 feet above the pump. The dynamic head losses in the system are Piping and fittings 30 ft Chiller 20 ft Control valve 10 ft Cooling coil 10 ft When the system is filled (at 95°F ambient temperature) it is desired to have a pressure of 10 psig at the high-
est point in the system which will reduce to 5 psig when the water temperature reduces to 45°F. a. What operating pressures (p1, p2) should the expansion tank be designed for? b. What pump head is required? c. With the pump off and a cold (45°F) system, what is the pressure at the pump suction? The pump discharge? d. With the pump on and a cold (45°F) system, what is the pressure at the pump suction? The pump discharge?
P i = 10 psig @ 95°F P f = 5 psig @ 45°F P 2 = P t @ 95°F
w = 62.05 lb/ft
3
P 1 = P t @ 45°F
w = 62.42 lb/ft
3
62.42 ( 115 ) P 1 = 5 + --------------------------- = 54.85 psig ≈ 55 psig 144 b) H = ΣH = 30 + 20 + 10 + 10 = 70 ft c) Assume pressure at pump inlet equals the tank pressure.
(Pump off)
P x = P x = P 1 = 55 psig 2
d)
1
P x = P 1 = 55 psig 1
70 P x = P x + wH = P x + 62.4 ⎛⎝ ---------⎞⎠ 2 1 1 144 P x = 55 + 30.3 2
85 psig
13.7 In your own words, explain the difference between a three-way control valve and a two-way control valve as they affect the hydraulics of the system. Both vary the flow (gpm) through the controlled load as they modulate. However, from the system perspective, the three-way valve provides a constant flow variable Δt in the system as it modulates and the two-way valve provides a variable flow as it modulates.
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62.05 ( 115 ) P 2 = 10 + w ( 115 ) = 10 + --------------------------144 P 2 = 59.5 psig ≈ 60 psig
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170⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
13.8 A control valve is to be sized for a cooling coil with a capacity of 30 tons of cooling. The water temperature entering the coil is at 44°F with a 12°F Δt. It is determined that the valve should have a pressure drop of 5 psi. What is the required Cv of the valve? GPM = C v ΔP C v = GPM ⁄ ΔP q = GPM ( 500 ) ( Δt ) 8 360000 GPM = ------------------- = -------------------- = 60 gpm 500 ( Δt ) 500 ( 12 ) C v = 60 ⁄ Δ5 = 26.83
13.9 A section of 1 in. steel pipe in a 45°F chilled-water system at 50 psig is in a pipe chase and is isolated between two service valves. If the chase is at a temperature of 95°F and the pipe reaches thermal equilibrium with the chase, what will the final pressure in the pipe be? ( β – 3α )Δt ΔP = -----------------------------------------------5 ⁄ 4 ( D ⁄ ( EΔr ) ) + α solving for 1 in. steel pipe from Figure 13.28 ΔP @ 50°Δt = 1380 psi P 2 = P 1 + ΔP P 2 = 50 + 1380 P 2 = 1430 psig
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Solutions to
Chapter 14 UNITARY AND ROOM AIR CONDITIONERS
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Chapter 14—Unitary and Room Air Conditioners⏐173
14.1 An air-cooled packaged air conditioning unit with a hot water heating coil is to be used to condition a small office suite in a high-rise office building. The unit has a total cooling capacity of three tons of refrigeration, and the power requirement to the compressor is 1 kW per ton of cooling. How many cfm of air must be brought into the condenser from an ambient outdoor temperature of 95°F db and 78°F wb if the condensing temperature is to be 115°F with a 10°F approach to the leaving air temperature? Total heat rejected = q rej q rej = q ref + q motor = 36,000 + ( 3 ) ( 3413 ) q rej = 46,239Btu/h q rej = m· ( 0.241 ) ( 115 – 95 ) 46,239 m· = ------------------------------------------------------( 60 ) ( 0.241 ) ( 115 – 95 ) m· = 319.77 lb/min CFM = m × v v = 14.37 ft 3 /lb ( at 95°F db and 78°F wb ) CFM = ( 319.77 ) ( 14.37 ) CFM = 4595.12 ft 3 /min
a.)
A i = CFM i /v = 4595.12/800 A i = 5.74 ft 2
b.) In passing through the condenser coil, the air would be heated a a constant humidity ratio. Air at 95°F db and 78°F wb (w = 117.49 gr/lb) heated to 115°F db has a final specific volume (v) of 14.85 ft3/lb. CFM o = m· 2 = ( 319.77 ) ( 14.88 ) CFM o = 4758.18 ft 3 /min A o = CFM o /v = 4758.18/800 A o = 5.95ft 2
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14.2 If the ductwork supplying the air to and from the condenser section in Problem 14.1 were sized for a velocity of 800 ft/min, what would be the cross-sectional area of the ductwork? a. From the outdoors to the condenser? b. From the condenser back to the outdoors?
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174⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
14.3 If the packaged air-conditioning unit of Problem 14.1 were provided with a water-cooled condenser instead of an air-cooled unit, and 1) the water was supplied at 85°F, 2) the leaving water temperature was 95°F, and 3) the condensing temperature was 105°F, what would be a. The Carnot COP between 40°F suction temperature and the 90°F condensing temperature? b. The Carnot COP between the 40°F suction temperature and the 105°F condensing temperature of the Problem 14.1 air-cooled unit? a.) To COP a = --------------T – To T o = 40 + 460 = 500°R T = 105 + 460 = 565°R 500 COP a = -----------------------565 – 500 COP a = 7.69 (water-cooled)
b.)
T o = 40 + 460 = 500°R T = 105 + 460 = 575°R 500 COP a = -----------------------575 – 500 COP a = 6.67 (air-cooled)
Short Solution: ( 1 )COP a ( air-cooled ) kW/J on = ----------------------------------------------------COP a ( water-cooled )
kW/J on
6.67 = ---------7.69 = 0.867kW/Jon
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14.4 Assuming that the actual power requirement for the cooling cycles of Problems 14.1 and 14.3 were proportioned in the same relationship as the Carnot COPs of Problem 14.3, what would be the kW per ton for the water-cooled unit of Problem 14.3?
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Chapter 14—Unitary and Room Air Conditioners⏐175
14.5 How many gallons per minute of water would be required for the water-cooled unit of Problem 14.3? q r = Total heat rejection q r = Refrigeration capacity and compressor heat = 36,000 + 3 ( 0.867 ) ( 3413 ) = 44,881 q r = m w ( 1 ) ( 95 – 85 ) ( 60 ) Btu/h ( 44,881 ) ( 7.48 ) GPm = ---------------------------------------------( 1 ) ( 10 ) ( 60 ) ( 62.4 ) GPm = 8.97 gal/min
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Solutions to
Chapter 15 PANEL HEATING AND COOLING SYSTEMS
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Chapter 15—Panel Heating and Cooling Systems⏐179 15.1 A room has a net outside wall area of 300 ft2 that has a surface temperature of 55°F; 50 ft2 of glass with a surface temperature of 30°F; 560 ft2 of ceiling with a surface temperature of 70°F; and 560 ft2 with a surface temperature of 70°F. Estimate the average unheated surface temperature or the area-weighted mean radiant temperature. 300 ( 55 ) + 50 ( 30 ) + 560 ( 70 ) + 560 ( 70 ) MRT = AUST = ------------------------------------------------------------------------------------------------300 + 50 + 560 + 560 96400 MRT = --------------- = 65.6°F 1470
15.2 For the room in Problem 15.1, estimate the following: a. radiant output for a 100 ft2 heating panel with a panel surface temperature of 120°F b. natural convection output for the ceiling panel when the air temperature is 70°F a)
From Figure 1, chapter 6, 2008 Systems and Equipment Handbook
b)
- radiant 55 Btu/h ⋅ ft 2 Btu 55 ------------2- × 100 ft = 5500 Btu/h h ⋅ ft t p – t a = 120 – 70 = 50°F
2
15.3 A room has 1500 ft2 of surface area and 320 ft2 is to be heated. The average unheated surface temperature in the room is 67°F. The air temperature in the room is 75°F. The room is occupied by adults in light clothing at a sedentary activity. Determine the surface temperature of the heated panel necessary to produce comfort if the air velocity is 20 fpm. From Figure 4-3, chapter 4, PHVAC @ 20 fpm, 75°F air temp., MRT = 80.6°F ( 1500 – 320 )67 + 320 ( t ) 80.6 = -----------------------------------------------------------1500 t = 131°F
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From Figure 3, chapter 6, 2008 Systems and Equipment Handbook 2 Btu 17.5 ------------2- × 100 ft = 1750 Btu/h h ⋅ ft
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180⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
15.4 For Problem 15.3, determine the total heat transferred by the ceiling heating panel.
t p = 131°F, AUST = 67°F t p – t a = 131 – 75 = 56°F Q R = 63 Btu/h ⋅ ft
2
2
Q R = 63 ( 320 ft )
Figure 1, chapter 6, 2008 Systems and Equipment Handbook
Q R = 20160 Btu/h Q C = 20 Btu/h ⋅ ft
2
2
Q C = 20 ( 320 ft ) = 6400 Btu/h Q T = Q C + Q R = 20160 + 6400 Q T = 26560 Btu/h
Figure 3, chapter 6, 2008 Systems and Equipment Handbook
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Solutions to
Chapter 16 HEAT PUMP, COGENERATION, AND HEAT RECOVERY SYSTEMS
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Chapter 16—Heat Pump, Cogeneration, and Heat Recovery Systems⏐183
16.1 A heat pump is used in place of a furnace for heating a house. In winter, when the outside air temperature is 15°F, the heat loss from the house is 100,000 Btu/h if the inside is maintained at 70°F. (a) Determine the minimum electric power (Carnot COP) required to operate the heat pump. (b) Determine the actual electric power to operate the heat pump with a heating COP of 3. TH 530°R COP h, Carnot = ------------------- = ------------------------------------- = 9.64 TH – TL 530°R – 475°R
a.
Qh 100000 Btu/h W Carnot = -------------------------------- = --------------------------------COP h, Carnot 9.64 Btu = 10373 -------- = 3.0 kW h
b.
Qh 100000 Btu/h W = ------------ = --------------------------------COP 3 = 33333 Btu/h = 9.8 kW
16.2 An air-source heat pump is to be used for both air conditioning and heating of a residence, maintaining the interior at 80°F in summer with an outside air temperature of 95°F and a cooling load of 36,000 Btu/h. As a heat pump, it is to maintain 70°F in winter with an outside air temperature of 2°F and a heating load of 52,000 Btu/h. Select a heat pump from the table in Problem 8.13, sized for cooling. What size resistance heater is required at the winter design condition?
16.3 A 100,000 ft2 building design has a design electrical load of 5 W/ft2. A reciprocating natural gas engine cogeneration plant is to serve the building. The engine-generator is sized for the electrical load, with salvaged heat being used for heating and for driving a single-effect absorption chiller. The design heating load is 3,000,000 Btu/h. The design cooling load is 250 tons; the absorber requires 20,000 Btu/ton⋅h input. Calculate hourly design operating costs for heating and cooling. Any shortfall in heating from recovered heat
must be made up by a boiler. Any shortfall in cooling by the absorber with recovered heat must be made up by the boiler as input to the absorber. Compare design operating costs with hourly design operating costs using conventional equipment (purchased electricity for the building and for cooling with an electric chiller at 1.0 kW/ton, purchased gas for a boiler for heating). Use $1.00 per therm, boiler efficiency of 80% for fuel cost, $0.10/kWh for purchased electricity cost.
Solution on following page.
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From the table in Problem 8.13, a cooling load of 36000 Btu/h at 95°F outdoor requires an A036 heat pump. At 2°F outdoor, this heat pump has an output of 15100 Btu/h. The heating load is 52000 Btu/h. Supplemental heat = 52000 Btu/h – 15100 Btu/h = 36900 Btu/h = 10.8 kW
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184⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
W 2 Engine-generator output = ⎛ 5 -----2-⎞ ( 100000 ft ) = 500000 W ⎝ ⎠ ft = 500 kW = 1706500 Btu/h which is 33% of the fuel input Btu Fuel input = 1706500 Btu/hr ⁄ 0.33 = 5171000 -------- input h Salvaged heat = 70% (30% + 30%) = 42% of input Btu = 2172000 -------h Design heating = 3000000 Btu/h Btu Btu Supplemental fuel input = ⎛ 3000000 -------- – 2172000 --------⎞ ⁄ 80% ⎝ h h ⎠ Btu = 1035000 -------h Design cooling = 250T Btu Btu Absorption chiller input = ( 250T )20000 ------------- = 5000000 -------ton-h h Btu Btu Supplemental fuel input = ⎛⎝ 5000000 -------- – 2172000 --------⎞⎠ ⁄ 80% h h Btu = 3535000 -------h Heating Design Condition Cogeneration Plant Engine input Supplemental fuel input
Conventional Plant ⎛ 500 kWh -----------⎞ ( $0.10/kWh ) = $50.00/h ⎝ h ⎠
Electricity
Heating
⎛ 3000000 Btu --------⎞ ⎜ h ⎟ ⎜ ------------------------------⎟ ( $1.00/therm ) = $30.00/h 5 Btu ⎜ 10 -------------- ⎟ ⎝ therm ⎠ $80.00/h
Cooling Design Condition Cogeneration Plant Engine input Supplemental fuel input
$51.71/h Btu ⎛ 3535000 --------⎞ ⎜ h ⎟ ⎜ ------------------------------⎟ ( $1.00/therm ) = $35.35/h Btu - ⎟ ⎜ 10 5 ------------⎝ therm ⎠ $87.06/h
Conventional Plant Building electricity Chiller electricity
⎛ 500 kWh -----------⎞ ( $0.10/kWh ) = $50.00/h ⎝ h ⎠ kWh ( 250T ) ⎛⎝ 1.0 -------------⎞⎠ ( $0.10/kWh ) = $25.00/h ton-h $75.00/h
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Btu therms 5171000 -------- = 51.71 ---------------- = $51.71/h h h Btu therms 1035000 -------- = 10.35 ---------------- = $10.25/h h h $62.06/h
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Solutions to
Chapter 17 AIR-PROCESSING EQUIPMENT
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Chapter 17—Air-Processing Equipment⏐187
17.1 Air enters a coil at 95°F dry-bulb and 78°F wet-bulb temperature and leaves at 62°F dry-bulb and 60°F wetbulb temperature. The condensate is assumed to be at a temperature of 56°F. Find the total, latent, and sensible cooling loads on the coil with air at 14.7 psia.
1.
95 db, 78 wb : w 1 = 0.0168, h 1 = 41.4
2.
62 db, 60 wb : w 2 = 0.0107, h 2 = 26.5
3.
56°F
4.
95 db w = 0.0107 : h c = 34.6
h s = h f = 24 B/lb
[ h 1 – h 2 – ( w 1 – w 2 )h 3 ] = q c 41.4 – 26.5 – ( 0.0168 – 0.0107 )24 = q c q c = 14.75 Btu/lb q s ≅ h c – h 2 = 34.6 – 26.5 = 8.1 or q s ≈ 0.244 ( 95 – 62 ) = 8.05 Btu/lb q L ≅ h 1 – h c = 41.4 – 34.6 = 6.8 or q L ≈ ( 0.0168 – 0.0107 )1076 = 6.6 Btu/lb
17.2 Air enters a direct expansion coil at 29.4°C dry bulb and 21.1°C wet bulb and leaves at 16.7°C dry bulb and 90% RH. a. How much sensible heat and how much latent heat is removed from the air by the coil? b. How much condensate drains off the coil? 29.4°C = 85°F
16.7°C = 62°F 21.1°C = 70 wb, °F 85 db, 70 wb : w 1 = 0.0124 w 2 = 0.0107
2.
62 db, 90% RH :
a)
q s = 0.244 ( 85 – 62 ) = 5.61 B/lb
q L = ( 0.0124 – 0.0107 )1076 = 1.83 B/lb b)
m c = ( 0.0124 – 0.107 ) = 0.0017 lb/lb air
17.3 Air enters a direct-expansion coil at 90°F dry bulb (32.2°C) 60% RH and leaves the coil at 60°F dry bulb (15.6°C), 95% RH. Find a. heat removed from air b. moisture condensed from air c. SHR for the condition line 1. 2.
90 db, 60% RH 60 db, 95% RH
: :
h 1 = 42, w 1 = 0.0184 h 2 = 25.8, w 2 = 0.0105
q = 42 – 25.8 = 16.2 Btu/lb Δw = 0.0184 – 0.0105 = 0.0079 lb/lb air 42 – 33.4 SHR = q s ⁄ q T = ---------------------- = 0.53 16.2
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1.
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188⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
17.4 Water flowing at 60 lb/min and at 51°F is chilled in an evaporator to 40°F. The heat transfer area is 20 ft2 and the heat exchanger has an overall heat transfer coefficient of 60 Btu/h·ft2 ·°F. The direct-expansion evaporator uses R-12 and operates at 35°F. Find the evaporator effectiveness. t in – t out 51 – 40 Effectiveness = -------------------------------- = ------------------ = 0.687 51 – 35 t in – t out, max
17.5 Outside air at 35°F and 70% RH is supplied to an air-conditioning apparatus. Recirculated air is returned from the plant at 69°F dry bulb and 40% RH; 8100 cfm of outside air mixes with 18,900 cfm of recirculated air. The mixture is heated by a steam coil and humidified by a pan humidifier to final conditions of 115°F dry bulb and 20% RH. a. What steam flow, in pounds per hour, should be supplied to the heating coil? b. Estimate the steam consumption of the humidifier.
OA 35 db, 70% RH : w oA = 0.003, h OA = 11.8, v OA = 12.55 r : 65 db, 40% RH; w r = 0.006, h r = 23.2 = v r = 13.45 m r = ( 18900 ( 60 ) ) ⁄ 13.45 = 84132 lb/h m : W: 38725 ( 0.003 ) + 84312 ( 0.006 ) = ( 38725 + 84312 )w m w m = 0.005 h: 38725 ( 11.8 ) + 84312 ( 23.2 ) = ( 38725 + 84312 )h m ⇒ h m = 19.6 at w m and h m → t m = 59 db s : 115 db, 20% RH; w s = 0.0128, h s = 41.4 Assume humidifier is steam coil heated pan type and process is pure humidification. a)
Leaving heating coil: t = 115°F, w = 0.005, h = 33.2 q HC = m da ( h out – h m ) = ( 38725 + 84312 ) ( 33.2 – 14.6 )
q HC = 123037 ( 13.6 ) = 1673300 Btuh 1673300 q HC = 1673300 ≅ m s hf s : m s = --------------------- = 1673 lb/h 1000 b) q Hum = m da ( h out – h in ) – m da ( w out – w in )h f h f = 28
water @ 60°F
= 123037 [ ( 41.9 – 33.2 ) – ( 0.0128 – 0.005 )28 ] 1043500 = – 1043500 Btuh = m s h fg ⇒ m s = --------------------1000 m s = 1043 lb/h
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m OA = ( 8100 ( 60 ) ) ⁄ 12.58 = 38725 lb/h
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Chapter 17—Air-Processing Equipment⏐189
17.6 Outdoor air (8000 cfm) at 10°F dry bulb and 50% RH enters the central apparatus of a split heating system. It is tempered to 55°F dry bulb. Then, it flows through a spray humidifier where the leaving sump water is maintained at 50°F. The spray humidifier has a performance factor of 0.80. After leaving the humidifier, the air flows through a steam heating coil and is heated to 70°F dry bulb. a. What is the final relative humidity and humidity ratio of the air as it leaves the heating coil? b. Assume steam at 2 psig and 90% quality is supplied to the tempering coil, the sump water heat exchanger, and the heating coil. How many pounds of steam per hour should be supplied to each?
: 55°F, W 2 = 0.00064; h 2 = 14 14 – h 3 h2 – h3 E = -------------------------- = 0.80 = ------------------ ; h 3 = 17 h 2 – h f50°F 14 – 18 :
W 3 = 0.00628
(a) : 70°F, W = 0.00628; h = 23.6, = 40% RH (b) h steam = 1056 Btu/lb; Δh H O = 1056 – 18 = 1038 Btu/lb 2
60 Tempering Coil: 676.8 ( 14 – 3 ) = 7445 Btu/min × ------------ = 430 lb/h 1038 60 Sump: 676.8 ( 17 – 14 ) = 2030 Btu/min × ------------ = 117.4 lb/h 1038 60 Heat Coil: 676.8 ( 23.6 – 17 ) = 4462 × ------------ = 257.9 lb/h 1038
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at :10°F, 50%; W 1 = 0.00064, h 1 = 3, v 1 = 11.82
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190⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
17.7 The heat exchanger for the spray water in Problem 17.6 is out of service for maintenance. The split heating system is operating as specified except that the sump water is recirculated. Assume make-up water to the sump is 37°F and saturating effectiveness is equal to the performance factor. a. What is the final relative humidity and humidity ratio of the air leaving the heating coil? b. What is the steam rate (lb/h) for the tempering coils and for the heating coil?
h1 = 3 w 1 = 0.00064
h 2 = 14 w 2 = 0.00064
55 – t 3 E = 0.8 = ---------------------- ; t sat = 37.5 55 – 37.5 t 3 = 41°F w 3 = w 4 = 0.00386 ; t 4 = 70°F; RH 4 = 25% (b) Tempering Coils:to 430 lb/h from Problem 17.8 60 Heating Coil: 676.8 ( 21.2 – 14 ) = 4873 Btu/min × ------------ = 267 lb/h 1038
1.
Pv 105°F db, 75% RH; φ = ----- ; P v = 0.75 ( 1.1021 ) = 0.8266 Ps Pv w 1 = 0.622 ---------------------- = 0.0371 lb/lb 14.7 – P v
2.
45°F db, 100% RH; w 2 = 0.0063 lb/lb Δw = ( 0.0371 – 0.0063 ) = 0.0308 lb s /lb da × 7000 = 215.6 grains/lb da
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17.8 Air at 105°F dry bulb (40.6°C) and 75% RH passes through a chilled water spray. Air leaves the spray chamber at 45°F dry bulb (7.2°C) saturated. How many grains of moisture per pound of entering air are condensed?
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Chapter 17—Air-Processing Equipment⏐191
17.9 Air enters a coil at 35°C dry-bulb and 26°C wetbulb temperatures and leaves at 17°C dry-bulb and 16°C wet-bulb temperatures. The condensate is assumed to be at a temperature of 13.5°C. Find the total, latent, and sensible cooling loads on the coil with air at 101 kPa.
1.
35°C, 26°C wb : W 1 = 0.0177; h 1 = 80.2
2.
17°C, 16°C wb : W 2 = 0.0110, h 2 = 45
3.
13.5°C : h 3 = h f = 56.7 q T = h 1 – h 2 – ( w 1 – w 2 )h 3 = 80.2 – 45 – ( 0.0177 – 0.0110 )56.7 = 34.8 J q s ≅ c p Δt = 1.02 ( 35 – 17 ) = 18.4 J q w ≅ ( h g – h f )Δw = 2500 ( 0.0177 – 0.0110 ) = 16.8 J
17.10 A building space is to be maintained at 70°F and 35% RH when outdoor design temperature is 10°F. Design heat losses from the space are 250,000 Btu/h, sensible, and 45,000 Btu/h, latent. Ventilation requires that 1500 cfm of outdoor air be used. Supply air is to be at 120°F. Determine: a. the amount of supply air required, lb/h, and cfm
b.
the capacity of the heating coil, Btu/h, if 1.to the humidifier is a spray washer using recirculated spray water with makeup water provided at 60°F 2.to the humidifier is a steam humidifier using dry, saturated steam at 17.2 psia c. the capacity of the humidifier, lb/h. The conditioning equipment and nomenclature are shown in the following sketch.
t 2 = 135.2°F (b)2.
h 2 = 37.1 – ( 0.0074 – 0.00405 ) ( 1153.4 ) = 33.2 t 2 = 119°F
(b)1. Q = m ( h 2 – h 1 ) = 20492 ( 37 – 16.5 ) = 420000 Btuh (b)2. Q = m ( h 2 – h 1 ) = 20492 ( 33.2 – 16.5 ) = 342000 Btuh (c) H = m ( w 3 – w 2 ) = 20492 ( 0.0074 – 0.00405 ) = 68.6 lb/h
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250000 250000 CFM s = ------------------------------------ = 4545 cfm m a = --------------------------------------- = 20492 lb/h 1.10 ( 120 – 70 ) 0.244 ( 120 – 70 ) 1500 ( 3.803 ) + 3045 ( 22.8 ) 1500 ( 0.001315 ) + 3045 ( 0.0054 ) h 1 = ---------------------------------------------------------------- = 16.5 ; w 1 = ------------------------------------------------------------------------------- = 0.0040 ; t 1 ≅ 51°F 4545 4545 mw 45000 ⁄ 1100 w s = w r + ------- = 0.0054 + ------------------------------- = 0.0074 ; t s = 120 ; h s = 37.1 ma 20492 h 2 + ( w 3 – w 2 )h h – h 3 = 0 (b)1. h 2 = 37.1 – ( 0.0074 – 0.00405 ) ( 28 ) = 37.0
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192⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
17.11 A spray-type air washer is to be used for humidification as well as cleaning of 9000 SCFM of air. Inlet conditions to the washer are 75°F db and 48 F wb. Desired humidity ratio at outlet is 0.005 lbw/lba. Determine: (a) the necessary humidification efficiency of the washer, % (b) the make-up water requirements (humidifying capacity) of the unit, lbw/h.
(a)
w i = 0.005 ; w s = 0.007
0.005 – 0.001 E w = --------------------------------- x100 = 65.6% 0.007 – 0.001 9000 ( 60 ) (b) m w = ----------------------- ( 0.005 – 0.001 ) = 162 lb/h 13.33
17.12 A heat pipe air-to-air energy recovery device is being considered for a system requiring 9000 SCFM of outside air. Initially, a separate preheater was planned for bringing the outside air from its –2°F design ambient outdoor temperature to 40°F. Determine: (a) the rating (Btu/h) and (b) the sensible effectiveness (%) to specify for the heat pipe unit if it is to eliminate the need for the air preheater.
(b)
Q = 1.10 ( 9000 ) ( 40 – ( – 2 ) ) = 415800 Btu/h 40 – ( – 2 ) E s = ----------------------- x100 = 57% 72 – ( – 2 )
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(a)
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Chapter 17—Air-Processing Equipment⏐193
17.13 The HVAC system for a hospital operating room which requires 100% outside air is shown in the following figure and includes an air-to-air heat pipe energy recovery unit having a sensible effectiveness of 73%. The air leaving the cooling coil is maintained at 58°F, 90% RH, all year long. During winter operation, air leaves the heater at 130°F. Fan speed is changed between summer and winter operation. Design duct system pressure drop (summer) is 3.25 in. water. 1. At winter design conditions (Indoor: 72°F & 30% RH; Outdoor: 5°F & 100% RH) the space load is 235,000 Btu/h (sensible) with negligible latent load. Deter-
2.
mine: (a) the necessary size of heating unit (Btu/h) both with and without the energy recovery unit and (b) the humidifier size (gallons/ day). Neglect fan effects. At summer design conditions (Indoor: 78°F; Outdoor: 95°F db/76°F wb), the space cooling loads are 146,000 Btu/h (sensible) and 79,000 Btu/h (latent). Determine: (a) fan size (hp & scfm), (b) sensible coil load, Btu/h, (c) latent coil load, Btu/h, and (d) necessary size of cooling unit, Btu/h, both with and without the energy recovery unit. Include fan effects.
t o = 5°F, φ o = 100% ⇒ w o = 0.00102 t r = 72°F, φ r = 30% ⇒ w r = 0.005 t s = 58°F, φ s = 90% ⇒ w s = 0.0093 1.(a) without heat pipe unit: 235000 = 0.244m a ( 130 – 72 ); m a = 16600 lb/h m a [ 0.244 ( t s – t c ) ] + Q h = 0
Q s = 6650 ( 1.10 ) ( 96.2 – 58 ) = 279400 Btu/h with 3.4 ( 2545 ) t 1 = 95 – 0.73 ( 95 – 78 ) = 82.6 ; t 2 = 82.6 + --------------------------------- = 83.8°F 29900 ( 0.244 ) Q s = 6650 ( 1.10 ) ( 83.8 – 58 ) = 188700 Btu/h (c) with/without :w 2 = 0.015 Q L = 4840 ( 6650 ) ( 0.015 – 0.0093 ) = 183500 Btu/h (d) with: Q T = 188700 + 183500 = 372200 Btu/h without: Q T = 279400 + 183500 = 462400 Btu/h 19.5% savings
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1600 [ 0.244 ( 130 – 58 ) ] + Q h = 0; Q h = 292000 Btu/h same with heat pipe since cooling coil discharge is same (b) m h = m a ( w r – w o ) = 16600 ( 0.005 – 0.00102 ) = 66 lb/h ≈ 190 gal/day 13.33 2.(a) 146000 = 0.224m a ( 78 – 58 ); m a = 29900 lb/h × ------------- ⇒ 6650 scfm 60 ( 3.25 ) ( 0.036 ) ( 144 )60 W = V· ΔP = 6650 ------------------------------------------------------ = 3.4 HP 778 ( 2545 ) (b) without heat pipe unit: 3.4 ( 2545 ) t 1 = t o = 95°F; t f = 95 + ------------------------------------ = 96.2°F = t 2 29900 ( 0.0244 ) w 1 = w f = w 2 = w o = 0.015
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Solutions to
Chapter 18 REFRIGERATION EQUIPMENT
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Chapter 18—Refrigeration Equipment⏐197
18.1 A condenser used in a refrigeration system has a capacity of 10 tons at a 40°F evaporating temperature. When 20 gpm of cooling water enters at 75°F, the condensing temperature is 90°F. The manufacturer claims a U-factor of 95 Btu/h· ft2· °F, with a heat transfer area of 83 ft2. Are these claims reasonable? Why? Q = UA Δt m = m w C p ( tw o – tw i ) To 550 Ideally: COP = 1 ⁄ ⎛ ------ – 1⎞ = 1 ⁄ ⎛ --------- – 1⎞ = 10 ⎝ TR ⎠ ⎝ 500 ⎠ QA 10 tons COP = 10 = -------------------- = ------------------QR – QA Q R – 10 Q R = 11 tons ⇒ Q R = 11 × 12000 = 132000 Btuh 132000 t w,o = ---------------------------- + 75 = 88.2°F 20 ( 8.33 )60 ( 90 – 75 ) – ( 90 – 88.2 ) Δt m = -------------------------------------------------------- = 6.2°F 90 – 75 ln ---------------------90 – 88.2 132000 U = ------------------ = 257(needed) 83 ( 6.2 ) More than 95 inconsistent
18.2 Given a compressor using R-22 condensing at 80°F (26.7°C) and evaporating at 20°F (−6.7°C), find the enthalpy of the refrigerant when it enters the a. compressor b. condenser c. evaporator Find the power required for the compressor.
P 4 = P 1 = 57.795 psia
20°F; x = 1.0; s = 0.22415 h 1 = 106.53 Btu/lb
a) b) c)
158.33 psia; s = 0.22415i h 2 = 118 Btu/lb 80°F; h 3 = 33.342 = h 4
HP( 11.5 ⁄ 2545 ) -------= ---------------------------------- = 0.74 Ton ( 73.3 ⁄ 12000 )
h1 – h2 – w = 0 w = 106.53 – 118 = – 11.5 B/lb q e = h 1 – h 4 = 106.53 – 33.34 q e = 73.2 Btu/lb
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P 2 = P 3 = 158.33 psia
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198⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
18.3 What is the maximum theoretical COP of a refrigeration device operating between 0°F and 75°F (−17.8°C and 23.9°C). Why is this theoretical limit difficult to obtain? TL 460 Ideal COP = ------------------- = --------- = 6.14 TH – TL 75
18.4 A reference book on refrigeration indicates that a compressor using R-22 requires a displacement of 40.59 cfm per ton for evaporation at −100°F and condensing at −30°F. Is this correct? Substantiate your answer with calculations based on knowledge of R-22 for these conditions. Also, verify the mass flow rate in lb per min.
– 100°F, x = 1.0, h g = 93.37, v g = 18.43
3)
– 30°F, x = 0, h = 2.547 q e = 93.37 – 2.547 = 90.82 Btu/lb
12000 ⁄ 90.82 = 132.1 lb/h/ton = 2.2 lb/min v· = 132.1 ( 18.43 ) = 2435 cfh = 40.6 cfm
18.5 An R-134a refrigerating system develops 10 tons of refrigeration when operating at 100°F condensing and +10°F evaporating, with no liquid subcooling or vapor superheating. Determine the volume of the refrigerant leaving the expansion valve in cubic feet per minute.
3)
1 + 10°F; h f = 15.328, h g = 104.617, v f = ------------- , v g = 1.7357 83.29 100°F; x = 0, h 3 = 45.155
4)
h 4 = h 3 = 45.155 = 15.328 + x ( 104.617 – 15.328 ); → x = 0.334
1)
3 1 v 4 = 0.334 ( 1.7357 ) + 0.666 ⎛ -------------⎞ = 0.588 ft ⁄ lb ⎝ 83.29⎠
m = 120000 ⁄ ( 104.47 – 44.94 ) = 2018 lb/h = 33.6 lb/min v· = 33.6 ( 0.588 ) = 19.5 cfm
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1)
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Chapter 18—Refrigeration Equipment⏐199
18.6 An expansion device has a mass flow rate for R-134a given by m = 60 + 0.25Δp where m = flow rate in lb/min Δp = pressure drop across the valve in psi. For an evaporator temperature of 0°F and a condenser temperature of 100°F, estimate the piston displacement required for a compressor if C = 0.04 and the polytropic compression coefficient n = 1.1 for the compression process. P 1 = 21.16 psia P 2 = 138.83 psia v 1 = 2.1587 m = 60 + 0.25 ( 138.8 – 21.16 ) = 89.4 lb/min 1 ---
n
p1 v1 = p2 v2
n
P1 n ⇒ v 2 = v 1 ⎛⎝ ------⎞⎠ P2
1------
21.16 1.1 3 v 2 = ⎛ ----------------⎞ 2.1587 = 0.39 ft ⁄ lb m ⎝ 138.83⎠ v1 2.1587 n v = 1 + C – C ⎛ -----⎞ = 1 + 0.04 – 0.04 ⎛ ----------------⎞ ⎝ 0.39 ⎠ ⎝ v 2⎠ n v = 0.819 mv 1 89.4 ( 2.1587 ) PD = --------- = -------------------------------- = 235.7 cfm nv 0.819
h 1 = 107.32 1 h1
h3 = hf
P L = 40.76 psia
= 112 ( 40.76 psia, 50°F )
100F
= 44.94
P H = 138.83 1
( 112 – 107.32 ) = ( 44.94 – h 3 ) 1
h 3 = 40.26 Btu/lb 10 ( 12000 ) m = ----------------------------------------------- = 29.82 lb/min ( 107.32 – 40.26 )60
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18.7 A liquid-to-suction heat exchanger is installed in an R-134a system to cool liquid that comes from the condenser with vapor that flows from the evaporator. The evaporator generates 10 tons (35.17 kW) of refrigeration at 30°F (−1.1°C). Liquid leaves the condenser saturated at 100°F (37.8°C), vapor leaves the evaporator saturated, and vapor leaves the heat exchanger at a temperature of 50°F (10°C). What is the flow rate of the refrigerant?
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200⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
18.8 An eight-cylinder ammonia compressor is designed to operate at 800 rpm and deliver 30 tons of refrigeration. The evaporator is to operate at 10°F with a condensing temperature of 100°F. The vapor enters the compressor at 30°F. The ammonia leaves the condenser as saturated liquid. If the average piston speed is to be 600 ft/min and the actual volumetric efficiency at this condition is 83%, find the bore of the compressor. Assuming superheating to 30°F takes place in the evaporator 3
1)
30°F, 38.5 psi; h 1 = 627, v 1 = 7.7 ft ⁄ lb
3)
h 3 = h f = 155.2
30 ( 12000 ) m = --------------------------------------- = 12.7 lb/min 60 ( 627 – 155.2 ) v· = 12.7 ( 7.7 ) = 97.8 cfm 2
97.8 πD ---------- = 8 ( 800 ) ⎛ ------------------⎞ ( 0.375 ) ⎝ 4 × 144⎠ 0.83
PH = 211.9 psi, PL = 38.5 psi
D = 3.0 in bore x = vt 1 2L = 600 --------800 L = 0.375 ft = 4.5 in.
Assume an ideal Carnot cycle QA QA 1 1 COP = ------- = -------------------- = ---------------- = ------------------ = 4.7 570 W QR – QA TH --------- – 1 ------ – 1 470 TL QA 30 ( 12000 ) 4.7 = -------------------- = --------------------------------------- ⇒ Q R = 436,600 Btuh QR – QA Q R – 30 ( 12000 ) Assuming 85°F inlet → Q R = 436,600 = mCp ( t o – t i ) 436600 t o = 85 + ---------------------------- = 94.7 90 ( 8.33 )60 ( 110 – 85 ) – ( 110 – 94.7 ) Δt m = -------------------------------------------------------------- = 19.6°F 25 ln ---------15.3 436,600 = 130 × A × 19.6 ⇒ A = 171 ft
2
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18.9 A condenser is to be selected for a system that generates 30 tons (105.5 kW) of refrigeration at 10°F (−12.2°C). The condenser is to operate at 110°F (43.3°C) and is cooled with 90 gpm (5.68 L/s) of water at 85°F (29.4°C). If the expected U-factor of the condenser is 130 Btu/h⋅ft2·°F [738 W/(m2·K)], calculate the condensing area required.
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Chapter 18—Refrigeration Equipment⏐201
18.10 A cooling tower cools water by passing it through a stream of air. If 1000 cfm of air at 95°F dry bulb and 78°F wet bulb enters the tower and leaves saturated at 84°F, to what temperature can this air cool water that enters at 110°F with a flow of 80 lb/min? What is the makeup water rate? m da ( h out – h in ) air = mCp ( t ou – t off ) air:
in
water:
out on
1000 Cfm, 95°F db, 78°F wb, v = 14.35, h = 41.4, w = 0.0168 1000 m da = ------------- = 69.7 lb/min 14.35 84°F db, 100% RH, h = 48.2, w = 0.0256 t = 110°F, m = 80 lb/min, C p = 1.0 69.7 [ 48.2 – 41.4 ] = 80 ( 1 ) ( 110 – t off ) t off = 104°F
make-up = m da ( w out – w on ) check:
= 69.7 ( 0.0256 – 0.0168 ) = 0.61 lb/min 0.61 ( 1046 ) – 0.244 ( 95 – 84 ) = 635 = 80 ( 110 – t off ) t off = 102°F
18.11 Water flowing at 28 kg/min at 11°C is chilled in an evaporator to 4.5°C. The heat transfer area is 1.9 m2 and the heat exchanger has an overall heat transfer coefficient of 341 W/(m2· K). The direct-expansion evaporator uses R-12 and operates at 2°C. Find the evaporator effectiveness.
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Assuming refrigerant flow is not limiting, 11 – 4.5 Effectiveness = ------------------- × 100 = 72% 11 – 2
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Solutions to
Chapter 19 HEATING EQUIPMENT
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Chapter 19—Heating Equipment⏐205
19.1 Set up the necessary combustion equations and determine the mass of air required to burn 0.45 kg (1 lb) of pure carbon to equal masses of CO and CO2. x x- ----x x x ⎛ ----+ -⎞ C + ⎛ ------ + ------⎞ O 2 → ------ CO + CO 2 ⎝ 28 44⎠ ⎝ 56 44⎠ 28 x- ----x-⎞ ⎛ ----⎝ 28 + 44⎠ 12 = 0.45 kg C;
x = 0.642 kg CO, CO 2
0.642 0.642 wt. of air = ⎛ ------------- + -------------⎞ ( 4.76 ) ( 29 ) = 3.6 kg air ⎝ 56 44 ⎠
19.2 The gravimetric analysis of a gaseous mixture is: CO2 = 32%, O2 = 54.5%, and N2 = 11.5%. The mixture is at a pressure of 20.7 kPa (3 psia). Determine (a) the volumetric analysis and (b) the partial pressure of each component. a.
b.
lb/lbmix lb/mol mol/molmix CO2 0.32 ÷ 44 = 0.00727 O2 0.545 ÷ 32 = 0.01705 N2 0.115 ÷ 28 = 0.00411 0.02843
molmix/lbmix mol/molmix % Volume ÷ 0.02843 = 0.2555 ~ 25.55% ÷ 0.02843 = 0.5995 59.95% ÷ 0.02843 = 0.1447 14.47% molmix/lbmix 99.97%
⎛ n CO ⎞ P CO = P mix ⎜ -----------2⎟ = 3 ( 0.2555 ) = 0.7665 psia (5.28 kPa) 2 ⎝ n mix ⎠ P O = 3 ( 0.5995 ) = 1.7985 psia (12.39 kPa) 2
P N = 3 ( 0.1447 ) = 0.4341 psia (2.99 kPa) 2
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206⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
19.3 A liquid petroleum fuel, C2H6OH is burned in a space heater at atmospheric pressure. a. For combustion with 20% excess air, determine the air/fuel ratio by mass, the mass of water formed by combustion per pound of fuel, and the dew point of the combustion products. [Ans: 11.45, 1.34, 133.9°F] a.
Theoretical:
b.
For combustion with 80% theoretical air, determine the dry analysis of the exhaust gases in percentage by volume. [Ans: 5.94% C02, 11.04% CO, 83.02% N2]
C 2 H 6 OH + 3.25O 2 + ( 3.25 ) ( 3.76 )N 2 → 2CO 2 + 3.5H 2 O + ( 3.25 ) ( 3.76 )N 2
20% Excess (12.0% Theroretical): C 2 H 6 OH + ( 1.2 ) ( 3.25 )O 2 + ( 1.2 ) ( 3.25 ) ( 3.76 )N 2 → 2CO 2 + 3.5H 2 O + ( 1.2 ) ( 12.22 )N 2 + 0.65O 2
b.
( 1.2 ) ( 3.25 ) ( 4.76 ) ( 29 ) A --- = ------------------------------------------------------- = 11.45 lb air ⁄ lb fuel F 2 × 12 + 6 + 16 + 1 ( 3.5 ) ( 18 ) m H O = ----------------------- = 1.34 lb H O ⁄ lb fuel 2 2 47 3.5 P w = ----------------------------------------------------- ( 14.7 ) = 2.47 psia 2 + 3.5 + 14.66 + 0.65 DP = 133.9°F 80% Theoretical Air: C 2 H 6 OH + ( 0.8 ) ( 3.25 )O 2 + ( 0.8 ) ( 3.25 ) ( 3.76 )N 2 → 0.7CO 2 + 1.3CO + 3.5H 2 O + ( 8.0 ) ( 3.25 ) ( 3.76 )N 2 CO 2 0.7 ÷ 11.78 = 5.94% CO 2 CO 1.3 ÷ 11.78 = 11.04% CO 9.78 N2 ------------- ÷ 11.78 = 83.02% ------------------ N 2 11.78 100.0%
19.4 Find the air/fuel ratio by mass when benzene (C6H6) burns with theoretical air and determine the dew point at atmospheric pressure of the combustion products if an air/fuel ratio of 20:1 by mass is used. lb air lb air Mol Air × lb air ⁄ mol air ( 7.5 ) ( 32 ) + ( 7.5 ) ( 3.76 ) ( 28 ) ------------------------------------------------------------ = ----------- = --------------------------------------------------------------------- = 13.25 ----------Mol Fuel × lb fuel ⁄ mol fuel lb fuel lb fuel ( 1 ) ( 78 ) Mol Air × 29Mol Air 78 × 20 --------------------------------= 20, ∴---------------------- = ------------------ = 53.8 Mol Fuel × 78 Mol Fuel 29 Actual Mols Air ………………………53.8 Theorical Mols Air ( 7.5 + 7.5 × 3.76 )…35.7 ---------18.1 mols excess air C 6 H 6 + 53.8 Air → 6CO 2 + 3H 2 O + 28.2N 2 + 18.1 Air Mol H 2 O 3 ----------------------= ---------------------------------------------- = 0.054 Mol Exh. 6 + 3 + 28.2 + 18.1 P v = 0.054 ( 14.7 ) = 0.797 psia, ΔP = 94°F
19.5 A diesel engine uses 30 lbm of fuel per hour (3.8 g/s) when the brake output is 75 hp. If the heating value of the fuel is 19,600 Btu/lb (45 600 kJ/kg), what is the brake thermal efficiency of the engine? 75 hp × 2545 Btu/hp·h η 6 = --------------------------------------------------------- × 100 = 32.5% 30 lb/h × 19 ,600 Btu/lb
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C 6 H 6 + 7.5O 2 + ( 7.5 ) ( 3.76 )N 2 → 6CO 2 + 3H 2 O + ( 7.5 ) ( 3.76 )N 2
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Chapter 19—Heating Equipment⏐207
19.6 Methane (CH4) is burned with air at atmospheric pressure. The Orsat analysis of the flue gas gives: CO2 = 10.00%, O2 = 2.41%, CO = 0.52%, and N2 = 87.07%. Balance the combustion equation and determine the airfuel ratio, the percent theoretical air, and the percent excess air. [Ans: 10.48 (vol.), 18.89 (mass), 110.1%, 10.1%] aCH 4 + bO 2 + cN 2 → 10.0CO 2 + 0.52CO + 2.41O 2 + dH 2 O + 87.07N 2 N2 :
c = 87.07
82.07 --c- = 79 ------ = 3.76, b = ------------- = 23.16 b 21 3.76 C: a = 10 + 0.52 = 10.52 H: 4a = 2d = 4 ( 10.52 ) → d = 21.04 10.52CH 4 + 23.16O 2 + 87.07N 2 → 10.0CO 2 + 0.52CO + 2.41O 2 + 21.04H 2 O + 87.07N 2 Since nitrogen is from air:
moles air ft 3 air A ⁄ F ( by volume ) = ( 23.16 + 87.07 ) ⁄ 10.52 = 10.48 ---------------------- or ---------------mole fuel ft 3 fuel lb air A ⁄ F ( by mass ) = [ 23.16 ( 32 ) + 87.07 ( 28 ) ] ⁄ ( 10.52 ) ( 16 ) = 18.89 -----------lb fuel Theoretical: CH 4 + 2O 2 + 2 ( 3.76 )N 2 → CO 2 + 2H 2 O + 7.52 N 2 2 ( 32 ) + 7.52 ( 28 ) A ⁄ F ( by mass ) = ------------------------------------------ = 17.16 lb a ⁄ lb f 16 18.89 % Theoretical Air = ------------- × 100 = 110.1% 17.16 18.89 – 17.16 % Excess Air = --------------------------------- × 100 = 10.1% 17.16
a. b. c. d.
Moisture formed per kg of fuel; moisture formed per lb of fuel Partial pressure of the water vapor, kPa; water vapor psia Percentage of CO2 in the stack gases on an Orsat basis Volume of exhaust gases per unit mass of oil, if the gas is at 260°C (500°F) and 102 kPa (14.8 psia). C 16 H 32 + 24O 2 + ( 24 ) ( 3.76 )N 2 → 16CO 2 + 16H 2 O + 90.24N 2 1 mol 24 moles 90.24 moles 16 moles 224 lb or kg 768 lb 2527 lb 704 lb a. M ⁄ F = 288 ⁄ 224 = 1.286 lb m /lb fuel also 1.286 kg/kg b.
PH O 16 2 -----------= -------------------------------------- = 0.13; PT 16 + 16 + 90.24
PH
2O
16 moles 288 lb
90.24 moles 2527 lb
= 0.13 ( 14.8 ) = 1.924 psia ( 133 kPa )
[dew point = 125°F(52°C)] c. d.
16 % CO 2 (vol., dry) = ------------------------- × 100 = 15.1% 16 + 90.24 ( 16 + 16 + 90.24 ) ( 1544 ) ( 960 ) 2 3 PV = nRT; V = -------------------------------------------------------------------------- = 379.5 ft /lb (23.7 m /kg) ( 14.8 ) ( 144 ) ( 224 )
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19.7 Fuel oil composed of C16H32 is burned with the chemically correct air-fuel ratio. Find
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208⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
19.8 Determine the composition of a hydrocarbon fuel if the Orsat analysis gives: CO2 = 8.0%, CO = 1.0%, O2 = 8.7%, and N2 = 82.3%. C a H b + dO 2 + cN 2 → 8.0CO 2 + 1.0CO + 8.7O 2 + eH 2 O + 82.3N 2 N2 :
c = 82.3 --c- = 3.76, d
From composition of air:
d = 21.9
1.0 e O 2 : 21.9 = 8.0 + ------- + 8.7 + --- , e = 9.4 2 2 C: a = 8.0 + 1.0 = 9.0 H: b = 2e = 18.8 ∴ Composition of fuel: C 9 H 18.8 or C 45 H 94
19.9 Determine the air/fuel ratio by mass when a liquid fuel of 16% hydrogen and 84% carbon by mass is burned with 15% excess air. [Ans: 17.49] H2 :
C:
0.16 × 8 = 1.28 lb O ⁄ lb fuel 2
0.84 × 32 ⁄ 12 = 2.24 lb O ⁄ lb fuel 2
3.52 lb O ⁄ lb fuel 2
2H 2 + O 2 → 2H 2 O 4 32 36 1 8 9 C + O 2 → CO 2 12 32 44 1 32/12 44/12
A = 3.52 lb O ⁄ lb fuel ⁄ 0.2315 lbO ⁄ lb air = 15.20 lb air ⁄ lb fuel , theor. 2
2
A ⁄ F = 1.15 ( 15.20 ) = 17.49 lb air ⁄ lb fuel
C 8 H 18 + 12.50O 2 + 12.5 ( 3.76 )N 2 → 8.0CO 2 + 9.0H 2 O + 47.0N 2 0.85 ( 12.5 ) ( 2 ) = 21.25 oxygen atoms available – 9.00 for H 2 → H 2 O 12.25 – 8.00 for C → CO 4.25 for CO → CO2 8.00 – 4.25 =
2H 2 + O 2 → 2H 2 O C + 1 ⁄ 2O 2 → CO 2C + O 2 → 2CO C + O2 → CO 2 3.75 CO remaining
C 8 H 18 + ( 0.85 ) ( 12.5 )O 2 + ( 0.85 ) ( 47.0 )N 2 → 4.25CO 2 + 3.75CO + 9.0H 2 O + 39.9N 2 4.25 CO 2 = ---------- × 1 = 8.9%; 47.9
3.75 CO = ---------- × 100 = 7.8%; 47.9
39.9 N 2 = ---------- × 100 = 83.3% 47.9
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19.10 Compute the compositions of the flue gases on a percent by volume on dry basis (same as Orsat) resulting from the combustion of C8H18 with 85% theoretical air.
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Chapter 19—Heating Equipment⏐209
19.11 A liquid petroleum fuel having a hydrogen to carbon ratio of 0.169 by mass is burned in a heater with an air/fuel ratio of 17 by mass. Determine: a. b.
the volumetric analysis on both wet and dry bases of the exhaust gases the dew point of the exhaust gas. H H ---- = 0.169 ---- = 0.169 × 12 = 2.028 = 2.03 C C CH 2.03 + 1.5075O 2 + 5.67N 2 → CO 2 + 1.015H 2 O + 5.670N 2 x ( 1.5075 + 5.6700 )29 ----------------------------------------------------- = 17; x = 1.145 or 14.5% excess 1 ( 12 + 2.03 ) CH 2.03 + 1.725O 2 + 6.49N 2 → CO 2 + 1.015H 2 O + 0.2175O 2 + 6.59N 2 a.
b.
1 wet CO 2 = ---------------- = 8.7225 1.015 H 2 O = ---------------- = 8.7225 0.2175 O 2 = ---------------- = 8.7225 6.4900 N 2 = ---------------- = 8.7225 P w = 0.1165 ( 14.7 ) =
11.50%
1 dry CO 2 = ---------------- = 12.95% 7.7075
11.65% 0.2175 O 2 = ---------------- = 2.82% 7.7075 6.490 74.40% N 2 = ---------------- = 24.20% 7.7075 1.71 psia ( 11.8 kPa ); Dew Point = 120°F ( 49°C ) 2.49%
19.12 Compare the heating value for semianthracite coal as given in Table 8, chapter 28, 2009 ASHRAE Handbook—Fundamentals with the value predicted using the Dulong Formula. [Ans: 1.24% difference]
HV = 13 ,600 Btu/lb O2 = 5 ; H 2 = 3.9; C = 80.4; N 2 = 1.1; S = 1.1; Ash = 8.5 Dulong:
Difference:
HHV = 14 ,544C + 62 ,028 [ H – ( 0 ⁄ 8 ) ] + 4050S = 14 ,544 ( 0.804 ) + 62 ,028 [ 0.39 – ( 0.05 ⁄ 8 ) ] + 4050 ( 0.011 ) = 13 ,769 Btu/lb 13 ,769 – 13 ,600 --------------------------------------- × 100 = 1.24% 13 ,600
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From Table 8, pg. 18.8, 2009 HBF
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210⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
19.13 Natural gas with a volumetric composition of 93.32% methane, 4.17% ethane, 0.69% propane, 0.19% butane, 0.05% pentane, 0.98% carbon dioxide and 0.61% nitrogen burns with 30% excess air. Calculate the volume of dry air at 60°F, 30 in. Hg (15.6°C, 101.5 kPa) used to burn 1000 ft3 (28.3 mL) of gas at 68°F and 29.92 in. Hg (20°C and 101.4 kPa) and find the dew point of the combustion products. Table 1, pg. 28.2, 2009 HBF
Natural Gas; 30% Excess Air
Methane (CH4) 93.32; Ethane (C2H6) 4.17; Propane (C3H7) 0.69 Butanes (C4H10) 0.19; Pentanes (C5H12) 0.05; CO2 0.98; N2 0.61 A ⁄ F = ( 9.57 ) ( 0.9332 ) + ( 16.75 ) ( 0.0417 ) + ( 23.95 ) ( 0.0069 ) + ( 31.14 ) ( 0.0019 ) + 38.29 ( 0.0005 ) 3
3
3
3
= 9.88 ft air /ftgas at 68°F, 29.92 in. Hg = 9880 ft air /1000 ft gas 3
3
3
3
A ⁄ F at 30% excess air → A ⁄ Fa = 1.3 × 9.88 = 12.89ft air /ft gas or 12 ,890 ft air /1000 ftgas T1 P2 ( RT 1 ) ⁄ ( MP 1 ) V1 - = ---------------- = ---------------------------------V2 ( RT 2 ) ⁄ ( MP 2 ) T2 P1 3 3 520 29.92 V 60°F, 30 in. Hg = 12 ,890 ⎛ ---------⎞ ⎛ -------------⎞ = 12 ,660 ft air /1000 ft gas ⎝ 528⎠ ⎝ 30 ⎠
Methane:
→ ( 0.9332 )CO 2 + ( 0.9332 ) ( 2 )H 2 O + ( 0.9332 ) ( 2 ) ( 3.76 )N 2
Ethane:
→ ( 0.0417 ) ( 2 )CO 2 + ( 0.0417 ) ( 3 )H 2 O + ( 0.0417 ) ( 3.5 ) ( 3.76 )N 2
Propane:
→ ( 0.0069 ) ( 3 )CO 2 + ( 0.0069 ) ( 4 )H 2 O + ( 0.0069 ) ( 5 ) ( 3.76 )N 2
Butane, Pentane: small, neglect 0.0098CO 2 CO 2 : N2 :
2.018
1.047CO 2
------------------------2.019H 2 O 1.8% 2.019H 2 O
P w = 15.11 ( 30 in. Hg ) = 4.0 in. Hg;
0.0061N 2 ----------------------- = 10.71 Total 7.702N 2 71.5% 10.0N 2 + 2.047O 2 = 15.11 Total Dew Point = 126°F
19.14 The proximate analysis of a coal is: moisture = 4.33%, volatile matter = 40.21%, fixed carbon = 45.07%, and ash = 10.39%. The heating value was determined as 29 000 kJ/kg (12,490 Btu/lb). Find the ASTM rank of the coal. Prox.: M = 4.33 ; VM = 40.21 ; FC = 45.07 ; A = 10.39 ; HV = 12 ,490 Btu/lb Change to Dry: VM = 42% ; FC = 47.1% ; A = 10.9% Class II. Bituminous, Group 5 High-Volatile C Bituminous Coal From Table 7, pg. 28.8, 2009 HBF
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With 30% excess air
-----------------------1.047CO 2 9.7%
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Chapter 19—Heating Equipment⏐211
19.15 A fuel oil shows an API gravity of 36. Calculate the specific gravity at 60/60°F and the pounds per gallon of fuel. Estimate the ASTM Grade. [Ans: 0.845, 7.05, No. 2] Eq. (2), pg. 28.7, 2009 HBF 141.5 Degrees API = -------------------------------------- – 131.5 S.G. at 60/60°F 141.5 36 = -------------------------- – 131.5 ; S.G. 60/60°F = 0.845 S.G. 60/60°F at 60°F: ν H ρ fuel
oil
2O
= 0.01603 ft 2 /lb ; ρ = 62.383 lb/ft 3
= 0.845 ( 62.383 ) = 52.714 lb/ft 3
M = 52.714 lb/ft 3 × 0.13368 ft 3 /gal = 7.047 lb/gal Grade No. 2
19.16 A representative No. 4 fuel oil has a gravity of 25° API and the following composition: carbon = 87.4%, hydrogen = 10.7%, sulfur = 1.2%, nitrogen = 0.2, moisture = 0, and solids = 0.5%. a. b.
Estimate its higher heating value. Compute the mass of air required to burn, theoretically, 1 gallon of the fuel.
lb a ⁄ gal fuel = ( 13.82 ) ( 7.5 ) = 103.6lb a ⁄ galf
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141.5 25 = ------------------------ – 131.5 S.G./60°F Eq. (2), pg. 28.7, 2009 HBF S.G./60°F = 0.904 No. 4 fuel oil, 25°API C = 87.4; H = 10.7; S = 1.2; N = 0.2; Solids = 0.5 a. HHV, Btu/lb = 22 ,320 – 37 ( S.G. ) = 18 ,903 Btu/lb Eq. (3), pg. 28.7, 2009 HBF Table 6: No. 4 1 gal = 7.5 lb ; HV = 145 ,000 Btu/gal b. A ⁄ F = 0.0144 ( 8C + 24H + 3S – 30 ) Eq. (6), pg. 28.10, 2009 HBF = 0.0144 [ 8 ( 87.4 ) + 24 ( 10.7 ) + 3 ( 1.2 ) ] = 13.82 lb a /lb f
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212⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
19.17 The following data was taken from a test on an oilfired furnace: Fuel rate = 20 gal oil/h Specific gravity of fuel oil = 0.89% by mass Hydrogen in fuel = 14.7% Temperature of fuel for combustion = 80°F Temperature of entering combustion air = 80°F Relative humidity of entering air = 45% Temperature of flue gases leaving furnace = 550°F a. b.
Calculate the heat loss in water vapor in products formed by combustion. Calculate the heat loss in water vapor in the combustion air. [Ans: 1672.5 Btu/lb (3888 kJ/kg), 29.4 Btu/lb (68.3 kJ/kg)] a.
b.
9H 2 q 3 = ---------- ( h ) tg – ( h f ) ta Eq. (19), Chap. 28, 2009 HBF 100 ( 9 ) ( 14.7 ) = ----------------------- ( 1312.2 – 48.05 ) = 1672.5 Btu/lb f 100 q 4 = Mw a [ ( h ) tg – ( h f ) ta ]
Eq. (20), Chap. 28, 2009 HBF
= ( 0.0098 ) ( 13.89 ) ( 1312.2 – 1096.12 ) = 29.4 Btu/lb
19.18 An office building requires 2901 MJ (2.75 × 109 Btu) of heat for the winter season. Compute the seasonal heating costs, if the following fuel is used: a.
Assume that the conversion efficiency is 75% for the oil and 61% for the coal. a.
2 ,750 ,000 ,000 Btu ----------------------------------------------------------------------------------- × $70.00/ton = $11, 680 coal 13 ,500 Btu/lb × 2000 lb/ton × 0.61
b.
2 ,750 ,000 ,000----------------------------------× $2.75/gal = $73 ,067 oil 138 ,000 × 0.75
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b.
Bituminous coal; 31 380 kJ/kg (13,500 Btu/lb); $70.00 per ton No. 2 fuel oil, 38 500 kJ/L (138,000 Btu/gal); $2.75 per gallon.
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Chapter 19—Heating Equipment⏐213
19.19 Saturated air at 41°F dry bulb (5°C) enters a furnace; it leaves the furnace at 110°F dry bulb (43.3°C) and 0.00543 lbv /lba (0.00543 kg/kg) and circulates through a factory. Air leaves the factory at 65°F dry bulb (18.3°C) and 63°F wet bulb (17.2°C). a. b.
What is the sensible and latent heat change for the air passing through the factory? State whether the air gains or loses sensible and latent heat during each process. [Ans: qs = +6 Btu/lb (14.0 kJ/kg), ql = +7 Btu/lb (16.3 kJ/kg)] a.
q s = mc p Δt = 0.24 ( 110 – 65 ) = 10.8 Btu/lb loss q L = m ( Δw )h f = ( 0.012 – 0.0054 )1060 = 7 Btu/lb gain g
b.
Across space: loses sensible, gains latent, see (a) Across furnace: q s = 0.24 ( 110 – 41 ) = 16.6 Btu/lb gain by air q L = ( 0.0054 – 0.0054 )h f = 0 g
19.20 A plant is maintained at 70°F dry bulb, 60% RH, and has a low-pressure steam heating system. A makeup air system is being added to the plant and it has been decided that the input air should be 10,127 cfm. Outside design conditions are −1°F dry bulb, 50% RH. The plant is 250 by 560 ft and normally has 325 people working per shift. a.
Makeup air system for ventilation: OA: –1°F db; 50% RH; Pv = 0.5(1.754 × 10–2) = 0.009; ν = 11.6
r:
0.009 w = 0.622 ------------------------------ = 0.0004 lb ν /lb a 14.7 – 0.009 70°F db; 60% RH; w = 0.0094 m w = 325 ( 475 ) ⁄ 1100 = 140.3 lb w /h lightbench work m a = 10 ,127 ( 60 ) ⁄ 11.6 = 52 ,380 lb da /h
a. b.
140.3 w s = 0.0094 – ---------------- = 0.0067 lb v /lb a 52 ,380 q = 52 ,380 { 0.240 [ 70 – ( – 1 ) ] + 1060 ( 0.0067 – 0.0004 ) } = 1 ,242 ,300 Btu/h m H = 52 ,380 ( 0.0067 – 0.0004 ) = 330 lb/h
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b.
What are the total steam requirements for the heating coil and the humidifier? What capacity should the humidifier have in pounds of water per hour?
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214⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
19.21 A residence with a design heating load of 26 kW (89,000 Btu/h) is to use an oil-fired warm air system with forced circulation. Return air to the furnace is at 22.2°C (72°F). Specify the following: This problem requires catalog data. a. b. c.
Supply air temperature Airflow rate Make and catalog number of suitable furnace a. b.
135°F q s = 1.10 ( cfm ) ( t s – t r ) ; 89 ,000 = 1.10 ( cfm ) ( 135 – 72 ) Airflow = 1280 cfm
19.22 A residence with a design heating load of 16 kW (55,000 Btu/h) is to use a forced circulation hot-water baseboard radiator system. The baseboard units house copper tubing with aluminum fins and operate with the inlet air temperature at 18.3°C (65°F). Specify the following: This problem requires catalog data. a. b. c. d. e.
Hot water inlet temperature and outlet temperature Total water flow rate Total length of radiator panel for house Location of panels Make and catalog number of suitable hot water heater 190°F inlet; 170°F outlet q s = 490 ( gpm ) Δt ; 55 ,000 = 490 ( gpm )20
c.
gpm = 5.6 L ≅ 90 ft from catalog for t av = 180°F
19.23 For the residence of Problem 19-22, electric baseboard units replace the hot water system. Specify This problem requires catalog data. a. b.
Total rating of electric system, kW Total length of baseboard units 55 ,000 P = ---------------- = 16.1 = 16 kW 3413 b. from catalog: ∼ 250 W/ft = 852 Btu/h/ft 55 ,000 L ≅ ---------------- = 64.5 ft 852 a.
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a. b.
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Chapter 19—Heating Equipment⏐215
19.24 A large classroom has a winter design heat loss of 19.9 kW (68,000 Btu/h) with installed forced circulation hot-water baseboard radiators. The baseboard units house copper tubing with aluminum fins and operate with the inlet air temperature at 18.3°C (65°F). Specify the following: This problem requires catalog data. a. b. c.
Hot water inlet temperature and outlet temperature Water flow rate Length of radiator panel a. b.
190°F inlet; 170°F outlet q s = 490 ( gpm ) Δt ; 68 ,000 = 490 ( gpm )20
gpm = 6.95 ∼ 7 c. from catalog: at t av = 180°F; ∼ 600 Btu/h/ft 68 ,000 L = ---------------- = 113 ft 600
19.25 A large classroom has a winter design load of 26 kW (89,000 Btu/h). A forced circulation warm air system is to be used with return air at 23.3°C (74°F). Specify a. b.
Supply air temperature Airflow rate q s = 1.10 ( cfm ) ( t s – t r ) t s = 135°F (selected)
b.
89 ,000 = 1.10 ( cfm ) ( 135 – 74 ) Airflow rate = 1330 cfm
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a.
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216⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
19.26 For a heat loss from the space to be conditioned of Q, write the expression for determining a. b. c. d.
Amount of air L/s (cfm), which must be supplied if a hot air system is used Amount of hot water L/s (gpm), which must be supplied if a hydronic system is used Amount of steam kg/h (lb/h), which must be supplied if a steam heating system is used Size, in watts, of electric heaters required if electric heat is used a. Hot air:
qs · · q s = 1.2V ( t r – t s ); V ( l ⁄ s ) = -------------------------1.2 ( t s – t r )
b. Hydronic:
· q s = m w c p ( t in – t out ) w = ρ w V w c p ( t in – t out ) w w
· Vw
w
qs = ------------------------------------------ρ w c p ( t in – t out ) w w
c. Steam:
qs q s = m st h f ; m st = -----g hf
d. Electric:
P ( watt ) q s ( kW ) = ------------------- ; P = 1000 q s 1000
g
19.27 List the steps taken when designing a forcedcirculation hot water heating system. Compute the heat required for each room or space Sketch runs, boiler, and convector locations. Compute gpm for each circuit. Select pump. Find ΔP/ft capacity of pump. Select pipe sizes required.
19.28 Compute the increase in length of 28.3 m (93 ft) of steel steam pipe when the average steam temperature is 113°C (235°F) and the air is 21°C (70°F). The pipe was installed during a period when the temperature was 15.6°C (60°F). α = 6.5 ×10
–6
in./in. –6
ΔL = αL Δt = 6.5 ×10 ( 93 ) ( 235 – 60 ) = 0.106 ft = 1.27 in.
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1. 2. 3. 4. 5. 6.
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Chapter 19—Heating Equipment⏐217
19.29 The total mass of steel in the boiler and piping of a school’s heating system is 9080 kg (20,000 lb). The piping and boiler also contain 6810 kg (15,000 lb) of water. After a weekend shut-down, the temperature of the system is 10°C (50°F). The operating temperature is 93°C (200°F). a.
b.
Assuming the system should be warmed up in one hour, determine the required furnace size. [Ans: 764 kW (2,610,000 Btu/h)] For a furnace size of 146 kW (500,000 Btu/h) output, when should the furnace be started to be up to the operating temperature of 93°C (200°F) by 7:30 A.M. Monday morning? [Ans: 2:16 A.M. Sunday] q a γ = m H O C H O Δt + m s C s Δt = ( m H O C H 2
a. b.
2
2
2O
+ m s C s ) Δt
q ( 1 ) = ( 15 ,000 × 1 + 20 ,000 × 0.12 ) ( 200 – 50 ) = 2 ,610 ,000 Btu/h 500 ,000 ( γ ) = 2 ,610 ,000; γ = 5.22 h = 5 h – 13 min. 7:30 A.M. – 5:13 = 2:17 A.M.
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Solutions to
Chapter 20 HEAT EXCHANGE EQUIPMENT
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Chapter 20—Heat Exchange Equipment⏐221
NOTE: All of the problems in this chapter are open-ended design problems and require the reader to make certain design assumptions. There are a number of possible solutions. A possible solution for 20.5 is provided below to illustrate the open-ended nature of these design problems. 20.5 Design the evaporator/condenser for a cascade lowtemperature refrigeration system using R-410A in the high temperature loop and R-22 in the low temperature loop. The shell-and-tube heat exchanger will use standard size copper tubes with a steel pipe as the shell. R-22 at the rate of 0.130 kg/s is to be condensed from saturated vapor to saturated liquid at a pressure of 0.91 MPa as it flows through the tubes. R-410A surrounds the tubes and evaporates under pool boiling conditinos at a pressure of 1.1 MPa. The exterior of the heat exchanger shell is to be well insulated. Space limites the length of the exchanger to 2 m.
Q = m c ( h co – h ci ) = m h ( h hi – h ho ) FΔ m, cf Q = UAFΔt m, cf = ----------------ΣR
where: 1 U = -----------------------------------------------------------------------------------------------------------------------------------------do A ln ----A - AR di AR fo AR c A fi --------+ ---------- + ---------------- + ----------- + --------- + --------------------------------------------2πkl Ai Ao A o h o ( A plane + φA fin ) hi Ai Δt a – Δt b Δt m, cf = ---------------------Δta ln ⎛ ---------⎞ ⎝ Δtb⎠
Saturated vapor → saturated liquid at 0.91 MPa R-22 at 293 K condensing
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Governing Relations:
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222⏐Principles of Heating, Ventilating, and Air Conditioning—Solutions Manual
R-410a at 1.1 MPa evaporating at 283 K Q = m· h fg
R – 22
= 0.13 kg/s (187.6 kJ/kg) = 24.4 kW = 24,000 kW
Δta – Δtb Δt m = ------------------------ or Δt avg = 293 – 283 = 10°C Δta ln ⎛ ---------⎞ ⎝ Δtb⎠ m i = N T ρA T V
Select: 5--in. OD 8 ( D i = 13.4 mm ) ( D o = 15.9 mm ) πD i2 A T = ---------- = 0.00014 m 2 4
at x = 0.1 Select:
1 v x = 0.1 0.026 + 0.9 ------------ = 0.0034 1200 μ = [ 0.1 ( 12.4 + 0.9 ( 175 ) ) ]10 – 6 = 159 × 10 – 6
Inside: 40,000 × 159 × 10 – 6 ( 0.0034 ) V = -----------------------------------------------------------------------0.0134 V = 1.6 m/s 0.13 kg/s × ( 0.0034 ) N T = -------------------------------------------------0.00014 ( 1.6 ) N T = 1.97
Two tubes per pass.
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D i VP Rev = 40,000 = ------------μ
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Chapter 20—Heat Exchange Equipment⏐223
From Equation 20.33: ρ 1/2 i Di ⎛ h--------- = 0.26Pr l1 / 3 Re l + Rev ⎛ -----l ⎞ ⎝ ρ v⎠ ⎝ K
0.8
⎞ ⎠
h i = 30,000 W/ ( m 2 ⋅ K )
Table 20.7 → Rfi = 0.00018 Assume: ΔT = 5 h o = 14, 500 W/ ( m 2 ⋅ K )
( 30,000 ( 20 – T s ) = 14,500 ( T s – 10 ) ) → T s = 16.7°C ΔT = 6.7°C
From Table 20.7: h i = 26,000 W/ ( m 2 ⋅ K ) R fo = 0.00035
dirty
1 U clean = --------------------------------------- = 14,000 W/ ( m 2 ⋅ K ) 1 1 ---------------- + --------------30,000 26,000 U avg = 7800 W/ ( m 2 ⋅ K )
( Q = U o A o ΔT m = 24,000 = 7800 ( 2 )π ( 0.0159 ) × L T ( 10 )ΔT m ) L T = 3.2m 3.2 L max pass = ------- = 1.6 2
Use two passes where a pass equals 1.6 m; use two tubes for each pass where a tube is 5/8 in. OD.
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1 U dirty = ------------------------------------------------------------------------------------------1 1 ---------------- + 0.00035 + 0.00018 + --------------30,000 26,000 = 1660 w/m 2 °C U
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