Calculation of refrigeration load

Calculation of refrigeration load

1. Calculation of Refrigeration Load
OBT554
Unit 4
Dr K.Geetha
Associate Professor, Dept of Biotechnology,
KamarajCollege of Engg& Tech, Madurai

2. Introduction
The major contributors to refrigeration loads are:
(1)the heat transmission through the roof, floor, and
walls
(2)infiltration through open doorways
(3)internal loads from lights, people, motors, and lift
trucks
(4)defrost heat and
(5)the product load cooling, freezing, and maintaining
the temperature of products.

3. The components of refrigeration load include
Transmission (heat gained through structure),
Product,
Internal (lights and motors),
Infiltration of air, and
Related equipment.

4. Where does all the heat come from that we need to remove?
Typically 5-15% is through transmission loads.

5. Product loads which account for typically 55-75% of the cooling
load.

6. Internal loads which account for around 10-20%.

7. Refrigeration equipment in the room which will account for
around 1-10% of the total cooling load.

8. Infiltration which adds 1-10% to the cooling load.

9. Cooling load calculation –Cold room
worked example
1. Transmission load
•The dimensions of the cold store are 6m long,
5m wide and 4m high.
•The ambient air is 30°C at 50% RH, The
internal air is 1°Cat 95% RH
•The walls, roof and floor are all insulated with
80mm polyurethane with a U value of
0.28W/m
2
.K
•The ground temperature is 10°C.

10. 1. To calculate the transmission load
the following formula may be used:

13. 2a. Product load –Product exchange
•Example: We’ll be storing Apples, we can look
up the specific heat capacity of the apples. in
this example we’re just cooling the Apples.
•There are 4,000kg of new apples arriving each
day at a temperature of 5°C and a specific
heat capacity of 3.65kJ/kg.°C.

14. We can then use the formula:

15. 2b.Product load –Product respiration
•heat generated by living products such as fruit
and vegetables.
•Example: 1.9kJ/kg per day as an average but
this rate changes over time and with
temperature.
•In this example thestore maintains a hold of
20,000kg of apples.

16. To calculate this we’ll use the formula

17. 3a. Internal heat load –People
•people working in the cold room
•Example: estimate 2 people working in the
store for 4 hours a day and at this
temperature they will give off around 270
Watts of heat per hour inside.

18. We’ll use the formula:

19. 3b. Internal heat load –Lighting

20. 4a. Equipment load –fan motors
The heat generation of the fan motors in the evaporator.

21. 4b. Equipment load –fan motors
The heat load caused by defrosting the evaporator.

22. 5. Infiltration load
•heat load from air infiltration.

23. Total cooling load
•To calculate the total cooling load we will just
sum all the values calculated
Transmission load:23.8kWh/day
Product load:26.5 kWh/day
Internal load:3.36kWh/day
Equipment load:8.94 kWh/day
Infiltration load: 9.67kWh/day
Total = 72.27 kWh/day

24. THANK YOU