Example-
One cold store dimensions is -10 m x 10 m × 5 m = 500 Sq mtr, External store surface area is 385.75 Sq mtr, Storage temperature is minus 20 Degree C, maximum ambient temperature is 35 Degree C. Insulation thickness is 250 mm. Consider two men are working inside the cold room. Calculate total calculated refrigeration load.
Calculation:
(1) Insulation heat leak through walls, floor and roof
Conductivity of polystyrene 0.033 kcal/h m C
Surface area of store = 385.75 Sq mtr
Temperature difference between ambient and store 35 Deg C and -20 Deg C = 55 Deg C
Thickness of polystyrene = 250 mm
Heat leak in Kcal/hr = 385.75 x 55 x 0.033/ 0.250 = 2800 kcal/hr
(2) Product load
5.5 kcal/kg for fresh fish load at an average temperature of -20 Deg C
Fish loaded per day 25 000 kg
Product load = 25000 x 5.5 / 24 = 5729 kcal/hr
(3) Fan load
2 x 200 W = 400 kcal/hr
(4) Air changes
Average of 2.5 air changes in 24 h
Store volume = 500 Cum
Heat gain (35 C and 60% R.H. air) 40 kcal/Cum
Air change heat gain = 500 x 2.5 x 40/ 24 = 2083 kcal/h
(5) Lights during working day
Light load during working day- 1000 W = 860 kcal/h
(6) Men working
1 man working at -20C gives off 370 kcal/hr
2 men working is equivalent = 740 kcal/hr
(7) Defrost heat
1 defrost of 8000 W for 1 h (recovered over 6 h) = 1142 kcal/hr
Total calculated refrigeration load
= sum of Items (1+2+3+4+5+6+7)
= (2800+5729+400+2083+860+740+1142)=13754 kcal/hr
One cold store dimensions is -10 m x 10 m × 5 m = 500 Sq mtr, External store surface area is 385.75 Sq mtr, Storage temperature is minus 20 Degree C, maximum ambient temperature is 35 Degree C. Insulation thickness is 250 mm. Consider two men are working inside the cold room. Calculate total calculated refrigeration load.
Calculation:
(1) Insulation heat leak through walls, floor and roof
Conductivity of polystyrene 0.033 kcal/h m C
Surface area of store = 385.75 Sq mtr
Temperature difference between ambient and store 35 Deg C and -20 Deg C = 55 Deg C
Thickness of polystyrene = 250 mm
Heat leak in Kcal/hr = 385.75 x 55 x 0.033/ 0.250 = 2800 kcal/hr
(2) Product load
5.5 kcal/kg for fresh fish load at an average temperature of -20 Deg C
Fish loaded per day 25 000 kg
Product load = 25000 x 5.5 / 24 = 5729 kcal/hr
(3) Fan load
2 x 200 W = 400 kcal/hr
(4) Air changes
Average of 2.5 air changes in 24 h
Store volume = 500 Cum
Heat gain (35 C and 60% R.H. air) 40 kcal/Cum
Air change heat gain = 500 x 2.5 x 40/ 24 = 2083 kcal/h
(5) Lights during working day
Light load during working day- 1000 W = 860 kcal/h
(6) Men working
1 man working at -20C gives off 370 kcal/hr
2 men working is equivalent = 740 kcal/hr
(7) Defrost heat
1 defrost of 8000 W for 1 h (recovered over 6 h) = 1142 kcal/hr
Total calculated refrigeration load
= sum of Items (1+2+3+4+5+6+7)
= (2800+5729+400+2083+860+740+1142)=13754 kcal/hr