The document provides calculations to determine the cooling load requirements for a cold storage facility. It outlines the following: 1) The structural heat gain through the walls, floor, and ceiling is calculated to be 44.82 kW. 2) The infiltration load from air changes is calculated to be 20.37 kW. 3) The product load from cooling the soybeans is calculated to be 982.5 kW. 4) Additional loads include lighting at 5 kW and occupant heat at 9.2 kW. 5) The total cooling load is calculated to be 1061.89 kW or 301.67 TR.

1. Apex Holdings Ltd.
Gaibanda project Cold stores design
Problem: A Chilling room Size 122’X56’x12’(37.19mX17mX3.65m) is used to chill 943200 kg
of fresh soybeans seed per day from an initial temperature 360 C to final temperature 180C
in 18 hours .The inside relative humidity is 80% . 40 persons work 24 hrs. In shift in the cold
room during the loading and unloading period and the lighting load is 5000 Watts (5KW).
The floor & the ceiling are located over and beneath the unconditioned space respectively.
The four walls are inside partitions adjacent to unconditioned spaces (360 C ,60% RH).
Use 1.5 w/m2 0 C as the overall heat transfer co-efficient for the walls, floor & Ceiling. Take
an air density value of 1.22 kg/m3. A chilling factor is .8. And the specific heat of soybean
seed is 4 KJ/Kg K at 12% Moisture Content .Allow 12 air changes per 24 hours for the door
usages & 230 watt of heat gain per occupant.
Design the cold storage (Calculate the cooling load in kW based on a 24 hours operating
time for the equipment.
Solutions: (a) Structural heat gain (Qs): It constituents the heat transmissions into the cold
store through the walls, ceiling & roof.
Qs= UA∆T
Walls: Western: 17X3.65X (36-18) X1.5=1675.35 Watt
Northern: 37.19X3.65X (36-18) X1.5= 3665.07 Watt
Eastern: 17X3.65X (36-18) X1.5= 1675.35 Watt
Southern: 37.19X3.65X (36-18) X1.5= 3665.07 Watt
Total heat gain through the wall (4 no’s) = 10.68 KW
Floor and ceiling: Heat gain through floor and ceiling = 2X37.19X17X (36-18) X1.5
= 34.14 KW
Therefor the total structural heat gain=44.82 KW
(B) Infiltration load(Ql): Ql = VXdXNacX(ho –hi )
= 2358X1.22x12X (95-44)/24X3600
= 20.37 KW
(C) : Product load(Qp): Qp = mp Xcp(Tp-Ts)/CF

2. = 943200X4X (36-18)/24X.8X3600
= 982.5 kW
(D) Heat of respiration (Qresp)=0
(E) Q appliance= 5 KW
(F) Occupancy load, Q Occupancy = 40X230(W)=9200 W=9.2 KW
Q Total
= Qs + Q l + Q p + Q resp + Q appliance + Q Occupancy
= 44.82 KW+ 20.37 KW+982.5 kW+0+5 KW+9.2 KW
= 1061.89 KW/3.52 TR
= 301.67 TR