ICE PLANT AND COLD STORAGE DESIGN.docx

ICE PLANT DESIGN Capacity of plant = 114 Tons of Ice (TOI) per day Tons of Refrigeration Approximation From Kent’s Mechanical Engineers’ Handbook: TOI TR ≤ 0.61 TR ≤

114 0.61 TR ≤

186.88

Size of Can From Kent’s Mechanical Engineers’ Handbook, p. 11-50: Size of can = 11 x 22 x 44 in

Thickness of Can Material From Macintire’s Handbook of Mechanical Refrigeration, p. 380, Table 88: U.S. Standard Gage Sides: Bottom:

No. 16 No. 16

Weight of Ice block From Kent’s Mechanical Engineers’ Handbook, p. 11-50, Table 1: For 11 x 22x 44 in. ice can: W = 300 lb

Temperature of Brine From Kent’s Mechanical Engineers’ Handbook, p. 11-50, Table 1: For 11 x 22x 44 in. ice can: T = 15 ⁰F

Time of Freezing From Kent’s Mechanical Engineers’ Handbook, p. 11-50: Time of freezing may be calculated using the following formula: 2

7a x= 32−t where: x = time of freezing, hr a = thickness of cake, in t = temperature of brine, ⁰F

x=

7(11)2 32−15

x=49.82 hours ≈50 hours

Number of Cans Required From Kent’s Mechanical Engineers’ Handbook, p. 11-50: The number of cans may be computed using the formula: N=

WH 24 C

where: N = no. of cans W = weight of ice, lb H = freezing time, hr C = weight of one ice block of ice, lb

N=

(114 x 2000)(50) 24 (300)

N=1583.33 cans ≈1600 cans

Volume of Freezing Tank VFT = Vice + Vbrine where: VFT = volume of freezing tank Vice = volume of ice Vice = L x W x H of ice can Vice = 11 x 22 x 44 Vice = 10648 cu.in per block x 1600 blocks Vice = 9859.26 cu.ft

Vbrine = volume of brine Based from Kent’s Mechanical Engineers’ Handbook, p. 11-51: 60 cu. Ft of brine per TR of ice is required. Vbrine = 60 cu. ft per TR x 186.88 TR

Vbrine = 11212.8 cu. ft VFT = 9859.26 + 11212.8 VFT = 21072.06 cu. ft

Size of Freezing Tank For two freezing tanks of 800 cans each:

Ice can orientation:

Length of Tank

L = 2Cs + Nct+ Ci(Nc -1) where: Cs = wall clearance Cs = 13” (includes additional 18’’ for brine agitator space) Nc = number of cans on a side t = thickness of cake Ci = ice can spacing Ci = 3’’ L = 2(13) + 50(11) + 3(50-1) L = 723 inches

Width of Tank W = 2Cs + Ncw+ Ci(Nc -1) Cs = 8’’ w = 22’’ W = 2(8) + 16(22) 3(16-1) W = 413 inches

Height of Freezing Tank For two freezing tanks: VFT = 2 ( L x W x H ) 12∈ ¿ 1 ft )3 = 2 ( 723 in x 413 in x H ) 21072.06 cu. ft x ( ¿ H = 60.972 in

≈ 61 inches

Dimensions of One Freezing Tank Size = 723 x 413 x 61 inches

Cooling Load Calculations Heat Rejected from Water Raw water is at 77 ⁰F. Qice = mcpwΔtw + mhf + mcpiΔti Qice = (300 lb)(1 BTU/lb-⁰F)(77 – 32) ⁰F + (300 lb)(144 BTU/lb) + (300 lb)(0.5 BTU/lb-⁰F)(32-15) ⁰F Qice = 1185 BTU/hr per block Qice = 1185 1.055 kJ 1 BTU

x

1 hr 3600 s

BTU hr−block

(50 hours)

x 1600 blocks x

24 hrs 50 hrs

x

1 TR x 3.516 kW

Qice = 75.85 TR

Heat Infiltration through Insulation From Trane’s Air Conditioning Manual, p. 361, Table 3-3 and Kent’s Mechanical Engineers’ Handbook, p. 11-37, Table 19:

Side Walls

Material

Thickne ss (inches)

Thermal Conductivity, k BTU −¿ ¿ ( hr−ft 2−F

Capacitance, c BTU ( ) hr−ft 2−F

Corkboard Cement mortar Gypsum lightweight

3 2 -

0.27 5.0 -

3.12

Cover Flooring Still air

aggregate (1/2 in) Oak or maple wood Cement mortar Still air

0.75 5.0 -

1.15 5.0 -

Qside walls = UAΔt Asw = 2(L x H) + 2(W x H) x 2 tanks Asw = 2(723 x 61) + 2(413 x 61) x 2 Asw = 1924

U=

ft 2

1 1 x 1 x2 1 + + + fo k 1 k 2 c 1

U=

1 3 2 1 + + + 1.65 0.27 5 3.12

U = 0.080401

Qsw = (0.080401

BTU hr−ft 2−F

BTU 2 hr−ft −F ) (1924

Qsw = 10152.38 BTU/hr

Qfloor = UAΔt Af = L x W x 2 tanks Af = 723 x 413 x 2 Af = 4147.208

ft 2

ft 2 ) (80.6 – 15)⁰F

1.65

U=

k x

U=

5 8

U = 0.6

BTU 2 hr−ft −F

BTU hr−ft 2−F ) (4147.208

Qfloor = (0.6

ft 2 ¿

(80.6 – 15)⁰F

Qfloor = 163234.12 BTU/hr

Qcover = UAΔt Acover = Afloor = 4147.208

ft 2

1 U=

1 x + fo k 1

U=

1 0.75 + 1.65 1.15

U = 0.7948

Qcover = (0.7948

BTU hr−ft 2−F

BTU hr−ft 2−F

) (4147.208

ft

2

)(80.6 – 15)⁰F

Qcover = 216230.7976 BTU/hr

Qinsulation = Qsw + Qfloor + Qcover Qinsulation = 10152.38 BTU/hr + 163234.12 BTU/hr + 216230.7976 BTU/hr Qinsulation = 389617.2976

BTU hr

x

1.055 kJ 1 BTU

x

1 hr 3600 s

x

1 TR 3.516 kW

Qinsulation = 32.474 TR

Total Cooling Load Qtotal = Qice + Qinsulation Qtotal = 75.85 TR + 32.474 TR Qtotal = 108.324 TR

For other heat loads and non-computable loads, add 15% of total load. Qtotal = 1.15(108.324 TR) Qtotal = 124.5726 TR

≈

130 TR

From previous approximation of TR: TR ≤ 186.88 TR 130 ≤ 186.88 TR

(satisfied)

COLD STORAGE DESIGN For a 24-hour storage of frozen meat products: Based from Kent’s Mechanical Engineers’ Handbook, p. 11-40, Table 21:

Product

Beef Hams and Loins Lamb Poultry (fresh)

30 – 32

84%

27

98

Specifc Heat Above Freezing , (BTU/lbhr-F) 0.75

28 – 30

80%

27

86.5

0.68

0.38

28 – 30

85%

29

83.5

0.67

0.30

28 – 30

84%

27

106

0.79

0.37

Temperatu re Range, ⁰F

Optimu m Freezing Relative Temperatur Humidit e, ⁰F y

Latent Heat of Fusion, BTU/lbhr

Specifc Heat Below Freezing , (BTU/lbhr-F) 0.40

Heat Load Calculations Raw meat is at 59 ⁰F and taken from the storage at 0 ⁰F (frozen meat).

Beef: m = 4000 lb QB = m [ cpAFΔt + LHF + cpBF ] QB = 4000 [ 0.75(59 – 27) + 98 + 0.4(27 – 0) ] QB = 531200 BTU/hr

Hams and Loins: m = 4000 lb QHL = m [ cpAFΔt + LHF + cpBF ] QHL = 4000 [ 0.68(59 – 27) + 86.5 + 0.38(27 – 0) ] QHL = 474080 BTU/hr

Lamb: m = 4000 lb QL = m [ cpAFΔt + LHF + cpBF ] QL = 4000 [ 0.67(59 – 29) + 83.5 + 0.3(29 – 0) ] QL = 449200 BTU/hr

Poultry (Fresh): m = 4000 lb QP = m [ cpAFΔt + LHF + cpBF ] QP = 4000 [ 0.79(59 -27) + 106 + 0.37(27 – 0) ] QP = 565080 BTU/hr

Product Load Qproduct = QB + QHL + QL + QP Qproduct = 531200 + 474080 + 449200 + 565080 Qproduct = 2019560 BTU/hr

Space Required for Refrigerated Goods From Kents’ Mechanical Engineers’ Handbook, p. 11-43, Table 24: For 700 lb beef: Space occupied = 108 ft3 Floor height = 12 ft Floor space = 9 ft2 Using the same requirements for other meat products: Mass total = 4 x 4000 lb MT = 16000 lb

3

Space required = 16000 lb x

108 ft 700 lb

Space required = 2468.6 ft3

Floor space = 16000 lb x

9 ft 2 700 lb

Floor space = 205.71 ft2

Height of pile of meat = 12 ft

For easy handling of storing meat products, the height of pile of meat is halved and the floor space is doubled (same volume of space required). Therefore: Height of pile of meat = 6 ft Floor space = 411.43 ft2

From Macintire’s Handbook of Mechanical Refrigeration, p. 529: 25 to 40% of storage room is used for storing products. For 30% storage space: Total floor space = 1371.43 ft2

Also, from Macintire’s Handbook of Mechanical Refrigeration, p. 529: Allowance to receiving and shipping floors amounts from 4 to 5% of total floor space. For 5%: Total Floor Space Required = 1443.61 ft2 ft2

≈

1500

Dimensions of Cold Storage Size = 50 ft x 30 ft x 12 ft

Wall Heat Gain Load Cold storage is designed based from the hottest temperature of Manila/Luzon Philippines. From ASHRAE: Month of May Dry bulb temperature = 34.5 ⁰C or 94.1 ⁰F Wet bulb temperature = 28.1 ⁰C or 82.58 ⁰F

Designed insulations: From Kents’ Mechanical Engineers’ Handbook:

Material

Side Walls

Flooring

Brick (common) Corkboard Concrete (cinder aggregate) Gypsum plaster Concrete (cinder aggregate)

Thickne ss, (inches)

Thermal Conductivity, k BTU −¿ ¿ ( hr−ft 2−F

Capacitance, c BTU ( ) hr−ft 2−F

8 4

4.8 0.32

-

1.5

4.9

-

3

4.9

4.4 -

Ceiling Still air Outside air

Corkboard Concrete (cinder aggregate) Corkboard Still air Outside air

5

0.32

-

4

4.9

-

6 -

0.32 -

1.65 6.0

Qside walls = UAΔt Asw = 2(L x H) + 2(W x H) Asw = 2(50 x 12) + 2(30 x 12) Asw = 1920 ft2

U=

1 1 x 1 x2 1 1 + + + + fo k 1 k 2 c fi 1

U=

1 8 4 1 1 + + + + 6 4.8 0.32 4.4 1.65

U = 0.06463

Qside walls = (0.06463

BTU 2 hr−ft −F

BTU 2 hr−ft 2−F ) (1920 ft ) ( 94.1 – 0)⁰F

Qside walls = 11676.83 BTU/hr

Qfloor = UAΔt Afloor = L x W

Afloor = 50 x 30 Afloor = 1500 ft2

U=

1 x1 x 2 1 + + k 1 k 2 fi

U=

1 3 5 1 + + 4.9 0.32 1.65

U = 0.05937

Qfloor = (0.05937

BTU hr−ft 2−F

BTU 2 hr−ft 2−F ) (1500 ft ) (94.1 – 0)⁰F

Qfloor = 8380.07 BTU/hr

Qceiling = UAΔt Aceiling = Afloor = 1500 ft2

U=

1 1 x 1 x2 1 + + + fo k 1 k 2 fi 1

U=

1 6 4 1 + + + 6 0.32 4.9 1.65

U = 0.049166

BTU hr−ft 2−F

Qceiling = (0.049166

BTU 2 2 hr−ft −F ) (1500 ft ) (94.1 – 0)⁰F

Qceiling = 6939.78 BTU/hr

Total Wall Gain Load Qwall = Qside walls + Qfloor + Qceiling Qwall = 11676.83 + 8380.07 + 6939.78 Qwall = 26996.68 BTU/hr

Air Change Load

Computations for air properties outside the cold storage: Pv = Pvm -

(Pt−Pvm)( DB −WB) 1546.622−1.44(WB )

Pv = 3.8046 -

(101.325−3.8046)(34.5−28.1) 1546.622−1.44(28.1)

Pv = 3.3902 kPa

wo =

0.622 Pv Pt −Pv

wo =

0.622(3.3902) 101.325−3.3902

wo = 0.02153 kg/kg

ho = 1.0062t + whfg ho = 1.0062 (34.5) + 0.02153(2564.4) ho = 89.925 kJ/kg = 38.656 BTU/lb

Computations for air properties inside the cold storage: tdb = 0 ⁰F = - 17.778 ⁰C RH = 80%

Pv Psat

RH =

0.80 =

Pv 0.12807

; Psat @ -17.778 ⁰C = 0.12807 kPa

Pv = 0.10245 kPa

From RenewAire Energy Recovery Ventilators Psychrometric Chart for Low Temperatures: hi = 0.7 BTU/lb

From Dossat’s Principles of Refrigeration, air change load may be computed using the formula: Qv = W (ho - hi)

For W: From Kent’s Mechanical Engineers’ Handbook, p. 11-48: 1 ( W = V v) V = volume of air per hour of ventilated air V = volume of room x air change rate For warehouses:

ACR = 2 to 3 hr-1

(ASHRAE) v = specifc volume of air specifed either outside space or as supplied space v=

0.287085(−17.778+ 273) 101.325−0.10245

v = 0.7238 m3/kg = 11.58 ft3/lb Therefore: 1 W = 2(30 x 50 x 12) ( 11.58 ) W = 3106.905 lb/hr Qv = 3106.905 (38.656 – 0.7) Qv = 117926.6 BTU/hr

Heat Sources in Space Heat from People/Workers

From Dossat’s Principle of Refrigeration, Table 10-14: Qpeople = 0.378 kW/person x no. of workers For 0 ⁰F or -17.778 ⁰C, use -15 ⁰C as reference: Qpeople = 0.378 kW/person x 6 workers Qpeople = 2.268 kW x

1 BTU 1.055 kJ

x

[assumed]

3600 s 1 hr

Qpeople = 7739.15 BTU/hr Heat from Lightings From Kent’s Mechanical Engineers’ Handbook Qlights = 3.41 x total wattage of lights Assume 40 fluorescent lamps with 15 W each: Qlights = 3.41 x (40 x 15) Qlights = 2046 BTU/hr

Heat from motors and equipment From Kent’s Mechanical Engineers’ Handbook: Qmo = 2950 BTU/hr per hp for 3-hp and above motors Assume maximum of 5 hp motors inside the cold storage: Qmo = 2950 BTU/hr x 5 hp Qmo = 14750 BTU/hr

Total Heat Load Qtotal = Qproduct + Qwall + Qv + Qpeople + Qlights + Qmo Qtotal = 2019560 + 26996.68 + 117926.6 + 7739.15 + 2046 + 14750

Qtotal = 2189018.476 BTU/hr For miscellaneous loads, a safety factor of 10% of total heat load is added based from Dossat’s Principles of Refrigeration. Qtotal = 1.1 (2189018.476) Qtotal = 2407920.324 BTU/hr

Tons of Refrigeration Required From Kent’s Mechanical Engineers’ Handbook: Tons =

Heat Load 12000

24 hr x hr

x

x = hours of operation x = 16 for a system operating above 32 ⁰F x = 20 for a system operating below 32 ⁰F Tons =

2407920.324 12000

Tons = 240.729 TR

Refrigerating Unit

24 hr 20 hr

x ≈

245 TR

Schematic Diagram of Refrigerating Unit

Condenser 40 ⁰C

B

Evaporator 2 130 TR 5 ⁰F

Evaporator 1 245 TR -10 ⁰F

A

As suggested by Kent’s Mechanical Engineers’ Handbook, for ice plants, evaporator temperature is at 5 ⁰F. As suggested by Macintire’s Handbook of Mechanical Refrigeration, there’s a 10 degree difference between the temperatures of the coil and the room. Condenser temperature is assumed based on the country’s air condition.

For Ammonia as refrigerant: Pressures: Pressure at condenser = Psat @ 40 ⁰C Pcon = 1557 kPa Pressure at evaporator for ice plant = Psat @ -15 ⁰C Pevap2 = 237.09 kPa Pressure at evaporator for cold storage = Psat @ -23.33 ⁰C Pevap2 = 164.283 kPa Enthalpies: h1 = hg @ -23.33 ⁰C h1 = 1432.58 kJ/kg h2 = h @ 237.09 kPa and s1 = s2

h2 = 1480 kJ/kg h3 = hg @ -15 ⁰C h3 = 1443.9 kJ/kg h4 = h @ 1557 kPa and s3 = s4 h4 = 1742 kJ/kg h5 = h6 = hf @ 40 ⁰C h5 = h6 = 390.587 kJ/kg h7 = h8 = hf @ -15 ⁰C h7 = h8 = 131.2745 kJ/kg

Mass Flow Rates m1 =

245 x 3.516 h 1−h 8

m1 =

245 x 3.516 1432.58−131.2745

m1 = m2 = m7 = m8 = 0.662 kg/s

m6 =

130 x 3.516 h 3−h 6

m6 =

130 x 3.516 1443.9−390.587

m6 = 0.4339 kg/s

Heat Balance about Intercooler mh3

mh6

m2h2

m7h7 mh6 + m2h2 = mh3 + m7h7 m(390.587) + 0.662(1480) = m(1443.9) + 0.622(131.2745) m = 0.8477 kg/s

Mass flow rate through high-stage compressor m3 = m + m6 m3 = 0.8477 + 0.4339 m3 = 1.2816 kg/s

Compressor Power For Low Stage Power P = m1 (h2 – h1) P = 0.662(1480 – 1437.23) P = 31.39 kW

For High Stage Power P = m3 (h4 – h3) P = 1.2816(1742 – 1443.9) P = 382 kW

*For high stage power, use 4 compressors with 96 kW power each. Refrigerant Piping From Mark’s Standard Handbook, velocities for most refrigerants: Liquid Lines:

3.3 to 22 m/s

Suction Lines:

38 to 250 m/s

Discharge Lines:

55 to 275 m/s

Suction Lines 

At point 1 V1 = m1v1 m1 = 0.662 kg/s v1 = vg @ -23.33 ⁰C v1 = 0.7168 m3/kg V1 = 0.662(0.7168) V1 = 0.4745 m3/s For d = 3 in.: V = Av 0.4745 =

Π 4

v = 104 m/s



(3 x

1 39.37

)2 (v)

[satisfied the range]

At point 3 V3 = m3v3 m3 = 1.2816 kg/s v3 = vg @ -15 ⁰C v3 = 0.508013 m3/kg V3 = 1.2816(0.508013)

V3 = 0.651 m3/s For d = 3 in.: V = Av 0.651 =

Π 4

(3 x

1 39.37

v = 142.75 m/s

)2 (v)

[satisfied the range]

Discharge Lines 

At point 2 V2 = m2v2 m2 = 0.662 kg/s v2 = v @ 237.09 kPa and s1 = s2 v2 = 0.56 m3/kg V2 = 0.662(0.56) V2 = 0.3707 m3/s

For d = 2 ½ in: V = Av 0.3707 =

Π 4

v = 117 m/s



(2 ½ x

1 39.37

)2 (v)

[satisfied the range]

At point 4 V4 = m4v4 m4 = 1.2816 kg/s v4 = v @ 1557 kPa and s3 = s4 v4 = 0.13 m3/kg V4 = 1.2816(0.13) V4 = 0.165 m3/s For d = 1 ½ in:

V = Av 0.165 =

Π 4

(1 ½ x

v = 144.72 m/s

1 39.37

)2 (v)

[satisfied the range]

Liquid Lines 

At point 5 V5 = m5v5 m5 = 1.2816 kg/s v5 = vf @ 40 ⁰C v5 = 0.0017257 m3/kg V5 = 1.2816(0.0017257) V5 = 0.002212 m3/s For d = ½ in: V = Av 0.002212 =

Π 4

v = 17.46 m/s



(½ x

1 39.37

[satisfied the range]

At point 7 V7 = m7v7 m7 = 0.662 kg/s v7 = vf @ -15 ⁰C v7 = 0.0015185 m3/kg V7 = 0.662(0.0015185) V7 = 0.0009445 m3/s For d = 3/8 in: V = Av

)2 (v)

0.0009445 =

Π 4

(3/8 x

v = 13.26 m/s

1 39.37

)2 (v)

[satisfied the range]

Condenser Based from Stoecker and Jones’ Refrigeration and Air Conditioning: Designed Values for Condenser Temperature at condenser = 40 ⁰C Common temperature of water in and out the cooling tower: Temperature entering the condenser = 85 ⁰F or 29.44 ⁰C Temperature leaving the condenser = 95 ⁰F or 35 ⁰C Nominal Size of Pipe = 1 ½ in. Schedule 40 Outside Diameter = 1.9 in or 0.04826 m Inside Diameter = 1.61 in or 0.040894 Thickness = 0.145 in or 0.003683 Tubes per pass = 14 No. of Tubes = 112 tubes

No. of Passes = 112/14 = 8 passes

Rate of heat rejected at the condenser Qr = m4 (h4 – h5) Qr = 1.2816 (1742 – 390.587) Qr = 1731.97 kW

Condensing Coefficient hcond = 0.725 (

g ρ2 hfg k 3 μΔ tND

)1/4

For Ammonia at 40 ⁰C ρ

=

1 1.7257

= 0.5795 kg/L = 579.5 kg/m3

hfg = 1490.42 – 390.587 hfg = 1099.83 kJ/kg From Table 15.5, p. 300 μ

= 0.000122 Pa-s

k = 0.447 W/m-K Average number of tubes in a vertical row, N N=

384 48

N=8

hcond = 0.725 (

(9.81) ( 579.5 )2 (1099.83)(0.447)3 0.000122(5.56)(8)(0.04826)

hcond = 764.5443 W/m2-K

Resistance of Metal R=

x Ao kA m x = 0.003683 ksteel = 50 W/m-K

x Ao kA m

=

x Ao kA m

= 8.6335 x10-5 m2-K/W

(0.00383)(0.04826) (50)(0.044577)

Fouling Factor 1 hff

= 0.000176 m2-K/W

Water side coefficient hw D k

=

0.023(

ρ VD ) μ

0.8

(

cp μ 0.4 k )

Mass of flowing water mw =

Qr CpΔ t

)1/4

1731.97 4.187(35−29.444)

mw =

mw = 74.458 kg/s

Volume flow rate V=

m ρ

V=

74.458 1000

V = 0.07446 m3/s Water velocity v=

Volume ( no .of tubes per pass ) ( Area)

v=

0.07446 Π 2 96[ ( 0.040894 ) ] 4

v = 0.5905 m/s

To calculate the water-side heat transfer coefficient, Use water properties at 32 ⁰C: ρ

= 995 kg/m3

cp = 4190 J/kg-K μ

= 0.000773 Pa-s

k = 0.617 W/m-K

hw =

0.023(0.617) ( 995 ) ( 0.5905 ) ( 0.040894 ) [ ] 0.040894 0.000773

0.8

[

(4190)(0.000773) ]0.4 0.617 hw = 2644.97 W/m2-K

Overall Heat Transfer Coefficient 1 Uo

+

=

1 ho

x Ao kA m

+

+

Ao hff A i

A0 hi A i 1 Uo

=

1 764.5443

+ (8.6335 x10-5) +

(0.000176)(1.9) 1.61

1.9 2644.97 (1.61) Uo = 461.688 W/m2-K

Logarithmic Mean Temperature Difference

+

LMTD =

Δ t max Δt min ¿ ¿ ln ¿ Δ t max− Δ t min ¿

LMTD =

40−29.44 40−35 ¿ ¿ ln ¿ ( 40−29.44 )−(40−35) ¿

LMTD = 7.437 ⁰C

Heat Transfer Area Ao =

Qr U o LMTD

Ao =

1731.97 x 1000 461.688(7.437)

Ao = 504.421 m2

Length of Tube Ao = ΠDLN L=

504.421 Π (0.04826)(384)

L = 8.66 m

Fans Fan Capacity Based from Macintire’s Handbook of Mechanical Refrigeration, p. 535 Fan capacity ranges from 6 to 10 times the volume of the refrigerated space per hour. For 10 times the volume of cold storage: Fan Capacity = 10 (30 x 50 x 12) Fan Capacity = 180000 ft3/hr or 1.42 m3/s

Power Required P = ƴQTDH For a supply air temperature of -10 ⁰C: Since at very low temperatures, the value of Pv is relatively small, v=

0.287085(−10+ 273) 101.325

v = 0.745 m3/kg Then, ƴ=

g v

=

9.81 0.745

ƴ = 13.165 N/m3

Q = Fan Capacity = 1.42 m3/s Assume a total head of 15 m. TDH = 15 m

P = 13.165(1.42)(15) P = 280.15 W Assume 50% efficiency of fan. Power = 560.3 W

Brine Pumps Designed Temperature of Brine From Macintire’s Handbook of Mechanical Refrigeration, a 5 degree temperature difference between the evaporator temperature and the brine temperature is needed. Evaporator Temperature = 5 ⁰F Temperature at tank entrance = 10 ⁰F Temperature at tank exit = 15 ⁰F Mass Flow of Brine From Kent’s Mechanical Engineers’ Handbook, p.11-52 Mass Flow of Brine = 242.4 lb/min per ⁰F rise of brine temperature mb = 242.4 (15 – 10) mb = 1212 lb/min

Volume Flow of Brine From Kent’s Mechanical Engineers’ Handbook, p.11-52 Volume Flow of Brine = 25.2 GPM per ⁰F rise of brine temperature

Vb = 25.2 (15 – 10) Vb = 126 GPM

Brine Flow on Freezing Tank

Power Required For Six Pumps P = ƴQTDH Assume a total head of 10 m. SGbrine = 1.16 Q = 126 GPM or 0.00795 m3/s P = 1.16(9.81)(0.00795)(10) P = 904.5 W Assume 50% efficiency of pump. Power = 1.809 kW