calculation of load for meat cold storage

1. Design of cold Storage
Submitted By : Akash Gangwar
(MEAT Storage)

2. Contents
1)Introduction
2) Principle of Operation of Refrigeration Room
3)Parameters considered for the design
4)Calculation of Refrigeration Loads
5)Insulated Material Selection
6)Calculation of Cold Storage
7)References

3. Introduction:Cold Storage Facility For Meat Products
1) The refrigeration room of 2m x 2m x 2m was successfully designed having a total refrigeration load of 0.52 TR.
3) The moist, warm surface of the freshly cut meat provides ideal conditions for the growth of micro-organisms and
this can result into spoilage of the meat.
4) The surface of meat also may dry out due to lack of moisture while refrigeration. The color of the meat is also
influenced by the rate of cooling. Slower cooling rate will result in lighter colored meat.
5) Slower cooling also avoids the problem of cold shortening. But slower cooling produces decrease of shell life,
weight loss, dull meat color, and bigger crystals are formed which destroys the tissues.
6) Rapid chilling increases the shell life, reduces the bacterial growth, retains color and weight and also produces
smaller crystals preventing tissue damage.
7) The lower the freezing temperature, higher will be the life of meat.

4. Principle of Operation of Refrigeration Room
The refrigeration room operates on basic vaporcompression refrigeration cycle.
The evaporator,compressor, condenser, expansion valve are the key components of the refrigeration c
The low pressure low temperature liquid refrigerant absorbs and removes heat from the inside cabine
This low pressure refrigerant from evaporator turns into saturated vapor which is then compressed to
pressure and passed to the condenser. The heat is rejected in the condenser. Then the saturated liquid
enters the expansion valve, where it is undergoes pressure reduction.
After expansion valve the refrigerant again enters the evaporator and the cycle continues. A fan is use
the warm air over the evaporator coils. The condenser may be water or air cooled type.

6. Parameters considered for the design
 Site Selection:
The refrigeration room should be in a shady areas having good wind flow.
 Storage temperature:
In many meat types greater the frozen temperature higher is the life of meat.
Fig.1: As per the table the temperature for
refrigeration should be -18℃ to -24℃.
Therefore, this parameter is crucial for the
design.

7.  Rate of Freezing
If meat is frozen slowly, large ice crystals are formed and their size breaks down the tissues. When
defrosted for use, the food spoils rapidly and taste is ruined.
Fast freezing, from 0℃ to -15℃, forms small ice crystals which do not damage the meat tissues.
Faster freezing also tends to produce a lighter colored product as the small crystals scatter the light
more than the large crystals and enhances the appearance of the meat products.
 Relative Humidity
Relative humidity is the second most important condition for storage of meat products. Using proper
temperature and relative humidity, the life of meat products can be extended.
Relative humidity in a storage space is affected by many variables, such as system running tim
Infiltration, condition and amount of product surface exposed, air motion,outside air condition
system control, etc.

8. Calculation of Refrigeration Loads
The most accurate means of determining the refrigeration load is by calculating each of the factors contributing
to the load.
 Product Load
Mass = 200 kg
Initial temp of meat = 24℃
Freezing temp = -2℃
Final / storage temp. = -18℃
Specific heat of meat before freezing = 3.23 KJ/kgK
Specific heat of meat after freezing = 1.68 KJ/kgK
Latent heat = 233 KJ/kg
Solution:
To cool from 24℃ to -2℃ = 200 x 3.23(24 - (-2))
= 16796 KJ
To freeze = 200 x 233 = 46600 KJ
To cool from -2℃ to -18℃ = 200 x 1.68(-2 - (-18))
= 5376 KJ
Total = 16796 KJ + 46600 KJ + 5376 KJ
= 68772 KJ/24hr
= 800 W or 0.8 KW

9.  Heat Transfer due to opening and closing of Door
Hc = V x Ac x He
= (2 x 2 x 2 x) 34 x 1
= 3.148 x 10-3 KW
 Heat load due to Lamps
HL = (No. of lamps) x (Lamp power rating) x (Functioning time)
= 2 x 40W/hr x 8hrs
= 0.64 KW
 Heat Load due to occupancy
At -18℃ heat released by a person is 1372 KJ/hr or 0.3811 KW
Ho = (No. of persons working) x (Heat released) x (Working time)
= 1 x 0.3811 x 1
= 0.3811 KW

10. Therefore,
Total heat load = 0.8 KW + 3.148 x 10-3 KW + 0.64 KW + 0.3811 KW
= 1.8242 KW
From the table of refrigerant 134a
H1= 386.7 KJ/kg, H3 = H4 = 234.3 KJ/kg
And calculating H2 = 420 KJ/kg
Therefore,
Compressor work = H2 – H1
= 420 – 386.7
= 33.3042 KJ/kg
Refrigerating effect = H1 – H4
= 386.7 – 234.3
= 152.4 KJ/kg

11. Coefficient of performance
COP = Refrigerating effect / Compressor work
= 152.4 / 33.3042
COP = 4.57
(COP)Actual = (1/3) x COP
= 4.57 / 3
(COP)Actual = 1.52
Fig.2: Pressure-Enthalpy diagram

12. Material Selection
Insulating material
 Cold storage insulation material has several types: one is that it can be processed into fixed sha
and specification which has fixed length, width and thickness.
 High and medium temperature cold storage rooms generally need 10cm thick panel whereas low
temperature cold storage and freezing cold storage rooms need 12cm or 15cm thick panels.
 Heat insulation material includes PU and PHB.
 Modern structures are developing fabricated cold storage rooms which make moisture proof
layers and insulating layers in advance and assembling is done on site.
 Therefore as per the temperature considerations and also considering the economics and durabi
Urethane is best suited insulating material.

13. Design of cold Storage
Refrigerant contains mainly
1) Compressor
2) Evaporator
3) Condensor design
 Compressor design
Mass flow rate of refrigerant (M)
= Cooling load / H1 – H4
= 1.8242 / 152.4
= 0.01196 kg/s
Compressor capacity = M x compressor work
= 0.01196 x 33.3042
= 0.3986 kW or 0.5345 HP
= 1 HP
Compressor rated power = compressor capacity / (COP)Actual
= 0.3986 / 1.52
= 0.2622 kW or 0.35HP

14.  Evaporator design
Evaporator capacity = M x Refrigerating effect
= 0.01196 x 152.4
= 1.82 kW or 3 HP
Rated evaporated capacity = evaporator capacity / (COP)Actual
= 1.82 / 1.52
= 1.2 kW or 1.6HP
 Condenser design
Condenser capacity = M x condenser work
= 0.01196 x (420 – 234.3)
= 2.22 kW or 2.97 HP
= 3 HP
Rated condenser power = condenser capacity / (COP)Actual
= 2.22 / 1.52
= 0.2622 kW or 2HP

15. REFRENCES:
 Imperial journal of discipilinary Research
 Authors: S J James Bruce James, Meat Refrigeration,
1st Edition.
 American Society of Heating, Refrigerating and Air-
Conditioning Engineers, Inc., 2010 ASHRAE Handbook -
Refrigeration (SI Edition).

16. THANK
YOU