# Thermodynamics Chapter 4 Solution Manual

August 3, 2017 | Author: Naser Fernandez | Category: British Thermal Unit, Gases, Statistical Mechanics, Heat, Mechanics

#### Short Description

Thermodynamics CHAPTER 4(1-7) by Hipolito Sta. Maria SOLUTION MANUAL...

#### Description

SOLUTION MANUAL OF THERMODYNAMICS By Hipolito Sta. Maria

Answered by: ENGR. NASER A. FERNANDEZ Published by: ‘I Think, Therefore I’m An Atheist’ Enterprises and Priority Development Fund (PDF)

CHAPTER 4 1.A perfect gas has a value of R = 58.8 ft.lbf/lbm - °R and k = 1.26. If 20 Btu are added to 5 lbm of this gas at constantvolume when the initial temperature is 90 °F, find (a) T 2, Change in H, Change in S, Change in U and (b) Work for a non flowprocess.

Given:

R = 58.8

Q = 20 BTU

k = 1.26 m = 5 lb

T1 = 90 F + 460 = 550 °R

Solution: (a) Q = mcv(T2 – T1) i = R = 58.8

x

= 0.0756

ii. cv = R/(k-1) = 0.0756/(1.26-1) cv = 0.29 iii. Q = mcv(T2 – T1) 20 = (5)(0.29)( T2- 550) T2 = 563.8 °R (b) i. cp = (kR)/(k-1) = (1.26)(0.0756)/(1.26-1) cp = 0.366 ii. ΔH = mcp(T2 – T1) = (5)(0.366)(563.8-550) ΔH = 25.25 BTU (c) ΔS = mcvln( ) = (5)(0.29)ln ΔS

)

= 0.036

(d) ΔU = mcv(T2 – T1) = (5)(0.29)(563.8-550) = 20.01 BTU

2. A reversible, non flow, constant volume process decreases the internal energy by 316.5 KJ for 2.268KG of a gas R=430 J/KG-K and k=1.35. for process determine: a.)the work ; b.) the heat and c.) the change in entropy if the initial temperature is 204.4 °C? Given: U = -316.5 kJ m = 2.268 kg R = 430 J/kg.K

k = 1.35 T1 = 204.4 +273 = 477.4 K

Solution: (a) Wn = __ pdv ; constant volume Wn = 0 (b) Q = U + Wn = -316.5 + 0 Q = -316.5 kJ (c) i. cv = R/(k-1) = 430/(1.35-1) cv = 1228.57 J/kg.K = 1.22857 kJ/kg.K ii. finding for T2 Q = mcv(T2-T1) -316.5 = (2.268)(1.22857)(T2-477.4) T2 = 363.81 K iii. ΔS = mcvln( ) = (2.268)(1.22857)ln ΔS

)

= -0.757 kJ/K

3. 10ft^3 vessel of hydrogen at a pressure of 305 psia is vigorously stirred by paddles until the pressure becomes 400 psia. determine ∆U and W. no heat is transferred, Cv = 2.434 btu / lb.R. Given: V1 = 10 ft3 cv= 2.434 BTU/lb.R P1 = 305 psia = 43920 lb/ft2 P2 = 400 psia = 57600 lb/ft2

Q=0

Solution: (a) i. R(hydrogen) = 765.9 lb.ft/lb.R cv = 2.434 BTU/lb.R ii. ∆U = mcv(T2-T1) = mcv( ∆U =

)

(p2-p1)

=

(57600-43920)

∆U = 434.75 BTU (b) Irreversible nonflow constant volume Q = U + Wn ;Q = 0 Wn = -434.75 BTU

4. Three pounds of a perfect gas with R = 38 ft.lb/lb.R and k = 1.667 have 300 Btu of heat added during the reversible nonflow constant pressure change of state. The initial temperature is 100 . Determine (a) final temperature, (b) ∆H, © W, (d) ∆U and (e) ∆S. Given: R = 38 lb.ft/lb.R k = 1.667 m = 3 lb

Q = 300 BTU T1 = 100 F + 460 = 560 °R

Solution: (a) i. cp = (kR)/(k-1) = (1.667)(38)/(1.667-1) = 94.97

x

= 0.1221 BTU/lb.R

ii. Q = mcp(T2-T1) 300 = (3)(0.1221)(T2 - 560) T2 = 1379 R or 919 °F (b) Q= mcp(T2-T1) = H ∆H = 300 BTU (c) Wn = p(V2-V1) = p(

-

= pmR( Wn = mR(T2-T1)

) ; p 1 = p2 )

= (3)(38)(1379-560) Wn = 120.008 BTU (d) i. cv = R/(k-1) = 38/(1.667-1) = 56.97

x

cv = 0.0732 BTU/lb.R ii. ∆U = mcv(T2-T1) = (3)(0.0732)(1379-560) ∆U = 179.85 BTU (e) ∆S = mcpln( ) = (3)(0.1221)ln(1379/560) ∆S = 0.3301 BTU/R 5. While the pressure remains constant at 689.5 kPa, the volume of a system of air changes from 0.567 m³ to 0.283 m³, what are a. Change in U b. Change in H c. Q d. Change in S e. if the process is non-flow and internally reversible, what is the work? Given: P = 689.5 kPa V1 = 0.567 m3 V2 = 0.283 m3 Solution: (a) i. ∆U = mcv(T2-T1) = mcv( ∆U =

)

(V2-V1)

=

x (0.283-0.567)

∆U = 490 kJ (b) ∆H = mcp(T2-T1) = mcp( ∆H =

)

(V2-V1)

=

(0.283-0.567)

∆H = -686.39 kJ (c) Q = mcp(T2-T1)= ∆H Q = -686.39 kJ

(d) ∆S = mcpln( ) = mcpln(

)

∆S = mcpln( ) = cpln( ) = (1.006)ln(0.283/0.567)

= -0.699 kJ/kgK (e) Wn = p(V2-V1) = (689.5)(0.283-0.567) Wn = -195.82 kJ 6. Four pounds of air gain 0.491 Btu/°R of entropy during a non-flow isothermal process. If P1 = 120 psia and V2 = 42.5 ft³, find a. V1 and T1 b. Wnf c. Q and d. Change in U. Given: m = 4lb ∆S = 0.491 BTU/R P1 = 120 psia V1 = 42.5 ft3

Rair = 53.34 lb.ft/lb.R

Solution: (a) i. ∆S = (0.491 BTU/R)(778 lb.ft/BTU) = 382 lb.ft/R Rair = 53.34 lb.ft/lb.R P1 = (120 lb/in2 )(144 in2/ft2)= 17280 lb/ft2 ii. ∆S= mRln(p1/p2) 382 = (4)(53.34)[ln(17280)]-ln(p2) - ln(17280)]-= -ln(p2) e7.9669 = eln(p2) p2 = 2883.91 lb/ft2 iii. p1V1 = p2V2 V1 = p2V2/p1 = (2883.91)(42.5)/(17280) V1 = 7.093 ft3 iv. p1V1 = mRT T = p1V1/mR = (17280)(7.093)/(4)(53.34) T = 574.46 R

(b) W = p1V1ln(V2/ V1) = (17280)(7.093)ln(42.5/7.093) = (219443.50 lb.ft)(

)

W = 282.06 BTU (c) Q = U + W; U= 0 Q = 282.06 BTU (d) U= 0 7. If 10 kg/min of air are compressed isothermally from P1=96kPa and V1=7.65m^3/min to P3=620kPa, find the work change of entropy and the heat for: a) nonflow process and b) steady flow process with v1=15m/s and v2=60m/s.

Given: m = 10 kg/min p1 = 96 kPa/kJ V1 = 7.65 m3 min P2 = 620 kPa/kJ Solution: (a) i. p1V1 = p2V2 V2 = p1V1/p2 = (96)(7.65)/620 V2 = 1.185 m3/min ii. Wn = p1V1ln(V2/ V1) = (96)(7.65)ln(

)

Wn = -1369.63 kJ/min iii. ∆S = mRln(p1/p2) ; Rair = 0.287 kJ/kg.K = (10)(0.287)ln(

)

20

∆S = -5.35 kJ/minK