1 1. Introduction Storage is the art of keeping the quality of agricultural materials and preventing them from deterioration for specific period of time, beyond their normal shelf life. Cold storage Control ripening retards aging, softening, texture and color change, retards moisture loss, wilting, microbial activity, spoilage, sprouting and undesirable growth. Availability of proper cold storages are important for preserving perishable Commodities like milk, meat, eggs, vegetables, fruits, ornamental flowers and other floricultural goods. These cold storages give perishable food items a longer shelf life by preventing them from rotting due to humidity, high temperature and microorganisms. This results in a decrease in loss due to spoilage. 1.1. Principle of Refrigeration The cold storage like every other refrigerating systems of the same magnitude employs the vapour compression method of mechanical refrigeration. Fig.1 presents the T-s diagram of the vapour compression cycle, while the Fig.2 illustrates the processes of the refrigeration employed in the cold room, respectively. Fig. 1. T-s diagram Fig. 2. Processes of the refrigeration
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1
1. Introduction
Storage is the art of keeping the quality of agricultural materials and preventing them from
deterioration for specific period of time, beyond their normal shelf life. Cold storage
Control ripening retards aging, softening, texture and color change, retards moisture loss,
wilting, microbial activity, spoilage, sprouting and undesirable growth. Availability of
proper cold storages are important for preserving perishable
Commodities like milk, meat, eggs, vegetables, fruits, ornamental flowers and other
floricultural goods. These cold storages give perishable food items a longer shelf life by
preventing them from rotting due to humidity, high temperature and microorganisms. This
results in a decrease in loss due to spoilage.
1.1. Principle of Refrigeration
The cold storage like every other refrigerating systems of the same magnitude employs the
vapour compression method of mechanical refrigeration. Fig.1 presents the T-s diagram of
the vapour compression cycle, while the Fig.2 illustrates the processes of the refrigeration
employed in the cold room, respectively.
Fig. 1. T-s diagram Fig. 2. Processes of the refrigeration
2
2. Design Procedure:
Heat load factors normally considered in a cold storage design
Wall, floor and ceiling heat gains from solar radiation due to conduction.
Load due to ingression of air by frequent door openings and during fresh air
charge.
Product load from incoming goods and heat of respiration from stored product.
Heat from workers working in the room.
Cooler fan load, light load, aging of equipment.
Miscellaneous loads, if any.
2.1. Heat transmission through walls:
H T outside air (W/K π2 ) 80
H T inside air (W/K π2) 3.8
T amb( oC) 38
Thickness Conductivity
Units m W/K m
Plaster 0.015 0.29
Brick 0.37 0.15
PUF 0.1 0.021
Plaster 0.015 0.29
Properties Length height breadth Temp Temp diff Area Q
Units M m m
W
Ice cream 7 10 5 -20 58 240 1829.73
Fish 7 10 5 -8 46 240 1457.17
Meat 7 10 5 -5 43 240 1356.52
Apple 7 10 5 2 36 240 1135.69
Carrot 7 10 5 0 38 240 1198.79
Banana 7 10 5 13 25 240 788.68
Q net 7760.58 Total H T C 0.131
ππΆ ππΆ π2
3
Q = U A βT
U = 1
1
βππ+
π₯1π1
+π₯2π2
+π₯3π3
+π₯4π4
+1
βππ’π‘
U = 1
1
80+
0.015
0.29+
0.37
0.15+
0.1
0.021+
0.015
0.29+
1
3.8
U = 0.131 W/ π2k
A = 2 x (L x H + H x W)
A = 2 x (7 x 10+10 x 5)
A = 240 π2
1) For ice cream unit
Q = 0.131 x 240 x (38 - (-20))
= 1829.73 W
2) For fish unit
Q = 0.131 x 240 x (38 - (-8))
= 1451.17 W
3) For meat unit
Q = 0.131 x 240 x (38 - (-5))
= 1356.52 W
4) For Apple unit
Q = 0.131 x 240x (38 - (2))
= 1135.69 W
5) For Carrot unit
Q = 0.131 x 240 x (38 - (0))
= 1198.79W
6) For Banana Unit
Q = 0.131 x 240 x (38 - 13)
= 788.68 W
4
2.2. Heat transmission through ceiling
Thickness Conductivity
Units M W/k m
Slab 0.150 0.300
PUF 0.021 0.021
Asbestos 0.015 0.750
Length breadth Temp Temp
diff
Area Q
Units M m
W
Ice
cream
7.000 5.000 -20 58.000 35.000 1130.5
Fish 7.000 5.000 -8 46.000 35.000 896.61
Meat 7.000 5.000 -5 43.000 35.000 838.13
Apple 7.000 5.000 2 36.000 35.000 701.69
Carrot 7.000 5.000 0 38.000 35.000 740.68
Banana 7.000 5.000 13 25.000 35.000 487.29
U = 1
1
βππ+
π₯1π1
+π₯2π2
+π₯3π3
+1
βππ’π‘
U = 1
1
80+
0.15
0.3+
0.021
0.021+
0.015
0.750+
1
3.8
U = 0.557W/ π2k
A = L x W
= 7 x 5
= 35 π2
H T outside air (W/K π2 ) 80
H T inside air (W/K π2 ) 3.8
T amb( oC) 38
Q net 4794.90 Total H T C 0.557
ππΆ
ππΆ π2
5
1) For ice cream unit
Q = 0.557x 35 x (38 - (-20))
= 1130.5 W
2) For fish unit
Q = 0.557x 35 x (38 - (-8))
= 896.61 W
3) For meat unit
Q = 0.557x 35 x (38 - (-5))
= 838.13 W
4) For Apple unit
Q = 0.557x 35 x (38 - (2))
= 701.69 W
5) For Carrot unit
Q = 0.557x 35 x (38 - (0))
= 740.68 W
6) For Banana Unit
Q = 0.557x 35 x (38 - 13)
= 487.29 W
2.3. Heat transmission through floor: -
Thickness Conductivity
Units M W/k m
Sand 0.060 0.700
Rubble 0.100 10.700
Concrete 0.080 0.300
PUF 0.100 0.021
Plaster 0.015 0.290
H T outside air (W/Kπ2 ) 80
H T inside air (W/K π2 ) 3.8
T amb( oC) 38
6
Length breadth Temp Temp
Diff
Area Q
Units M m oC
m2 W
Ice cream 7.000 5.000 -20 58.000 35.000 364.823
Fish 7.000 5.000 -8 46.000 35.000 289.342
Meat 7.000 5.000 -5 43.000 35.000 270.472
Apple 7.000 5.000 2 36.000 35.000 226.442
Carrot 7.000 5.000 0 38.000 35.000 239.022
Banana 7.000 5.000 13 25.000 35.000 157.251
U = 1
1
βππ+
π₯1π1
+π₯2π2
+π₯3π3
+π₯4π4
+π₯5π5
+1
βππ’π‘
U = 1
1
80+
0.06
0.7+
0.1
10.7+
0.08
0.3+
0.1
0.021+
0.015
0.29+
1
3.8
U = 0.180W/ π2k
A = L x W
= 7 x 5
= 35 π2
1) For ice cream unit
Q = 0.180x 35 x (38 - (-20))
= 364.823 W
2) For fish unit
Q = 0.180x 35 x (38 - (-8))
= 289.342 W
3) For meat unit
Q = 0.180x 35 x (38 - (-5))
= 270.472 W
4) For Apple unit
Q net 1547.351 Total H T C 0.180
ππΆ
7
Q = 0.180x 35 x (38 - (2))
= 226.442 W
5) For Carrot unit
Q = 0.180x 35 x (38 - (0))
= 239.022 W
6) For Banana Unit
Q = 0.180x35 x (38 - 13)
= 157.251 W
2.4. Heat transmission through door: -
Avg air change per hr 0.125
Air density(kg/m^3) 1.2
Temp Amb 38
Cp(kJ/kg k) 1.005
Length Breadth Height Vol Temp Enthalpy Diff Q
Units M M m
KJ/Kg W
Ice cream 7 5 10 350 -20 58.29 850.063
Fish 7 5 10 350 -8 46.23 674.188
Meat 7 5 10 350 -5 43.215 630.2190
Apple 7 5 10 350 2 36.18 527.625
Carrot 7 5 10 350 0 38.19 556.938
Banana 7 5 10 350 13 25.125 366.406
Heat gain, Q = room volume x air changes per hour x air density x enthalpy change
1) For ice cream unit
Q = 350 x 0.125x 1.2 x 58.29
= 850.063 W
2) For fish unit
Q = 350 x 0.125x 1.2 x 46.23
= 674.188 W
3) For meat unit
Q = 350 x 0.125x 1.2 x 40.2
Q net 3605.438
π3 ππΆ
8
= 630.219 W
4) For Apple unit
Q = 350 x 0.125x 1.2 x 36.18
= 527.625 W
5) For Carrot unit
Q = 350 x 0.125x 1.2 x 38.19
= 556.938 W
6) For Banana Unit
Q = 350 x 0.125x 1.2 x 24.14
= 366.406W
2.5. Equipment load:-
Nos of Tube lights 10
Ratings of Tube lights (W) 40
No of Tube Lights = 10
Q = No of tube lights x rating
= 10 x 40
= 400 W
2.6. Heat extracted:-
Q net (W) 400
Storeag
e Time
Load Cp
abv
freezin
g
Pre
cool
temp
Q abv
freezin
g
Laten
t Heat
Cp
below
freezin
g
Temp Q
below
freezin
g
Total Q
Units days Tons KJ/Kg
K
KJ/kg KJ/kg KJ/KgK
KJ/kg W
Ice
cream
7
666.667 3.10 -2.0 -6.220 186.0 2.57 -20.0 55.00 25879.63
Fish 6 666.667 3.68 6.0 22.08 255.0 2.17 -8.0 17.36 37865.22
Meat 8 666.667 3.25 6.0 19.5 194.0 2.24 -5.0 11.2 21672.45
Apple 9 666.667 3.81 10.0 38.1 280.0 1.98 2.0 -3.96 26932.44
Carrot 4 666.667 3.92 10.0 39.2 293.0 2.0 0.0 0.000 64081.79
Banana 6 666.667 3.56 16.0 56.96 248.0 2.03 13.0 -26.39 35824.33
Q net 319161.111
ππΆ ππΆ
9
Q =
π ( πΆπ(ππππ£π)π₯ (ππππβ 0)+πΏππ‘πππ‘+πΆπ(πππππ€)π₯ (0βππ))π₯1000
24 π₯ 3600 π₯ πππ¦π
1) For ice cream unit
Q = 50 x (3.1 x 38+190+1.67 x 20)x 1000x1000
24 x3600 x 7 =25879.63 W
2) For fish unit
Q = 60 x (3.18 x 38+276+1.67 x 8)x 1000x1000
24 x3600 x 6 =37865.226 W
3) For meat unit
Q = 30 x (3.52 x 38+200+1.51 x 2)x 1000x1000
24 x3600 x 8 =21672.454 W
4) For apple unit
Q = 50 x (3.64 x 38+280+1.76 x 1)x 1000x1000
24 x3600 x 9 =26932.442 W
5) For carrot unit
Q = 100 x (3.81 x 38+293+1.84 x 0)x 1000x1000
24 x3600 x 4 =64081.79 W
6) For banana unit
Q = 80 x (3.35 x 38+248+1.67 x(β14))x 1000x1000
24 x 3600 x 6 =35824.331 W
2.7. Human occupancy: -
Avg working hr per day per person 10
Nos of
occupants
Heat
dissipated
Q
Units NA KJ/hr W
Ice cream 2 1500 833.333
Fish 2 1500 833.333
Meat 2 1500 833.333
Apple 2 1500 833.333
Carrot 2 1500 833.333
Banana 2 1500 833.333 Q net 5000.000
10
Q = No of occupant x heat dissipated
For all individual units
Q = 2 x 1500 x 1000
3600 = 833.333 W
Total TR:
Q
Units KW
Ice cream 31.29
Fish 42.01
Meat 25.6
Apple 30.36
Carrot 67.65
Banana 38.46
Q = Qwall + Qceiling + Qfloor + Qdoor + Qequipment + Qextract + Qhuman
1) For ice cream unit
Q = 31.29 kW
2) For fish unit
Q = 42.01 kW
3) For meat unit
Q = 25.6 kW
4) For apple unit
Q = 30.36 kW
5) For carrot unit
Q = 67.65 kW
6) For banana unit
Q = 38.46 kW
Qnet = 235.96 kW = 66.92 TR
Units KW TR
Net Q 235.36 66.92
11
COP
Fig 3.Actual P-h diagram
Enthalpy outlet @ Compressor(KJ/Kg) 1780
Enthalpy inlet
@ evaporator
Enthalpy outlet @
evaporator
Refrigerating
Effect
Work
Done
Mass flow
rate
Units KJ/Kg KJ/Kg KJ/Kg KJ/Kg kg/s
Ice cream 350 1450 1100 330 0.028
Fish 350 1452 1102 0.038
Meat 350 1458 1108 0.023
Apple 350 1460 1110 0.027
Carrot 350 1461 1111 0.061
Banana 350 1474 1124 0.034
COP 3.698
RE net 6655.000
Mas flow rate 0.212
Work net 330.000
12
Refrigeration Effect:-
R.E. = Enthalpy at evaporator outlet β Enthalpy at evaporator inlet
1) For Ice cream unit
R.E. = 1450 β 350 = 1100 KJ/Kg
2) For fish unit
R.E. = 1452 - 350 = 1102 KJ/Kg
3) For Meat unit
R.E. = 1458 β 350 = 1108 KJ/Kg
4) For apple unit
R.E. = 1460 β 350 = 1110 KJ/Kg
5) For carrot unit
R.E. = 1461 β 350 = 1111 KJ/Kg
6) For banana unit
R.E. = 1474β 350 = 1124 KJ/Kg
π . πΈπππ‘ = π . πΈπΌππ + π . πΈπππ β + π . πΈππππ‘ + π . πΈπππππ + π . πΈππππππ‘ = 6655ππ½/ππ
Mass Flow Rate:-
m = π
ββ
1) For Ice cream unit
m = 202.58
1100= 0.028 ππ/π
2) For fish unit
m = 288.79
1102= 0.038 ππ/π
3) For Meat unit
m = 120.46
1108= 0.023 ππ/π
13
4) For Apple unit
m = 246.56
1110= 0.027 ππ/π
5) For Carrot unit
m = 510.08
1111= 0.061 ππ/π
6) For Banana unit
m = 328.30
1124= 0.034 ππ/π
ππππ‘=ππππ+ππππ β+πππππ‘+ππππππ+πππππππ‘+πππππππ=0.212 ππ/π
ππππ‘ = β2 β β1= 330 kJ/kg
C O P = π πΈπππ‘
ππππ‘= 3.698
14
3. Equipment Selection
3.1. Compressor Selection [2]
R-717 refrigerant
Model Evaporator Temp Condenser Temp 40
KC42 -20 Q(kW)=242 P (kW)=86.8
3.2. Evaporative Condenser Selection [3]
R-717 refrigerant
Compressor evaporator capacity = 66TR
Compressor BHP = 116 HP
Condensing temperature = 104Β°F
Entering wet-bulb temperature = 80Β°F
Determine the total heat rejection:
Compressor evaporator capacity = 66 TR x 12 MBH/TR = 840 MBH
Compressor BHP input = 116 BHP x 2.545 MBH/BHP = 293 MBH
Total heat rejection = 1134 MBH
From Table 6, the heat rejection capacity factor for R-717 at 105Β°F condensing temperature and 80Β°F
entering wet-bulb temperature is 1.18
Multiply: 1134 MBH x 1.18 = 1340 MBH
From Tables 5 select a unit with a Base Heat Rejection equal to or greater than 1340 MBH:
Model VCL-096.
15
4. References
[1] ASHRAE hand Book (2010) chapter 24- Refrigerated-Facility Loads.
[2] Compressor manufacture catalogue of Kirloskar.
[3] Evaporative Condenser manufacture catalogue of Baltimore Air Coil.
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