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