Cold storage design
1. Selection of site
2. Orientation and building form
3. Size
4. Space requirement
5. Design of building
6. Thermal insulation
7. Heat load calculation
8. Refrigeration system for cold
store
Cold storage design 2. Orientation and building form
3. Size
1. Selection of site 4. Space
Sources of product not very far from cold store site requirement
Easy road access 5. Design of
building
Easy access of electricity
6. Thermal
insulation
7. Heat load
calculation
8. Refrigeration system for
cold store
e ratio less
Cold storage design
3. Size
2. Orientation and building form4. Space
N-S Direction requirement
W-E wall have good plantation 5. Design of
building
Surface to volume ratio less
6. Thermal
insulation
7. Heat load
calculation
8. Refrigeration system for
cold store
e ratio less
Cold storage design
3. Size 4. Space
Sufficient Space for product and container requirement
Sufficient space for walkway or machine way 5. Design of
building
Surface to volume ratio less
6. Thermal
insulation
7. Heat load
calculation
8. Refrigeration system for
cold store
e ratio less
Cold storage design
4. Space requirement
3.4 m3 /ton
Chamber height 3 to 10 m 5. Design of
building
Rack to rack distance 0.75 m
6. Thermal
insulation
7. Heat load
calculation
8. Refrigeration system for
cold store
e ratio less
Cold storage design
5. Design of
building
Roof
Ceiling
Floor
6. Thermal
Doors insulation
Rack 7. Heat load
calculation
8. Refrigeration system for
cold store
e ratio less
Cold storage design
6. Thermal
insulation
7. Heat load
calculation
8. Refrigeration system for
cold store
e ratio less
Cold storage design
7. Heat load calculation
8. Refrigeration system for
cold store
Cold Storage Design
Types of Loads in Cold Storages
1.Transmission load
2.Air change (infiltration) load
3.Miscellaneous loads, such as electric motors, lighting,
people, etc.
4.Product load, sensible
5.Product load, respiration
Determining the size of cold storage
Capacity of cold store = 500 MT
Number of Chamber = 2
Per day loading capacity @4% of total Number of column required = 250/ 18 = 27.7=14
load = 20 MT / day Cold room Height = 750 x 4 +800 = 3800 mm = 3.8 m
Store the potatoes in pallet box
Cold room Length = 20000 + 20000 = 20m + 20 m
Pallet loading capacity = 500 kg
Volume of pallet = 1.2x 1.0 x 0.75 = 0.9 m 3
Pallet size = 100 x 120 75 cm
Total volume of Pallets = 0.9 x 14 x 9 x 4 = 453 m 3
No. of pallet stack = 4
Volume of cold room = 20 x 14 x 3.4 = 1064 m 3
Width of cold room = 14 m
Free space required for passage = 2 m for fork lift
Free space required for wall side = 30 cm
Free space required for pallet to pallet = 15 cm
Total length required for product = 14000 -2000-300 = 11700
Total number of pallets in column wise = 11700/1350 = 8.6 =9
Total number of pallets in column wise with 4 stack = 9 x 4 =36
Total capacity in one column = 36x 500 = 18000 kg = 18 MT
20000 mm PLANT AREA 20000 mm
14000 mm
OFFICE PROCESSING AREA
20000 mm
14000 mm 6000 mm 14000 mm
Cooling load Estimation (Transmission load)
Weather data
DBT = 33 o C
RH = 70 %
Wall gain load
Wall material PUF (Polyurethane Foam)
Wall material thickness = 100 mm
South wall load calculation (Panel Shaded area)
Length = 20 m
Height = 3.8 m
Area = 76 m 2
U value of Panel = 0.22 W/(m2 *K )
Temperature of cold room = 2 o C
Heat loss from wall = A x U x (To –Ti)
76 x .22 x (33 – 2)
518.32 W =1768.58 BTU/hour
Cooling load Estimation (Transmission load)
Weather data
DBT = 33 o C
RH = 70 %
Wall gain load
Wall material PUF (Polyurethane Foam)
Wall material thickness = 100 mm
North wall load calculation (Panel Shaded area)
Length = 20 m
Height = 3.8 m
Area = 76 m 2
U value of Panel = 0.22 W/(m2 *K )
Temperature of cold room = 2 o C
Heat loss from wall = A x U x (To –Ti)
76 x .22 x (33 – 2)
518.32 W =1768.58 BTU/hour
Cooling load Estimation (Transmission load)
Weather data
DBT = 33 o C
RH = 70 %
Wall gain load
Wall material PUF (Polyurethane Foam)
Wall material thickness = 100 mm
East wall load calculation
Length = 14 m
Height = 3.8 m
Area = 53.2 m 2
U value of Panel = 0.22 W/(m2 *K )
Temperature outside wall = To + 5 = 33 + 5 = 38
Temperature of cold room = 2 o C
Heat loss from wall = A x U x (To –Ti)
53.2 x .22 x (38 – 2)
421.34 W =1437.67 BTU/hour
Cooling load Estimation (Transmission load)
Weather data
DBT = 33 o C
RH = 70 %
Wall gain load
Wall material PUF (Polyurethane Foam)
Wall material thickness = 100 mm
West wall load calculation (Neighbor room as cold room)
Length = 14 m
Height = 3.8 m
Area = 53.2 m 2
U value of Panel = 0.22 W/(m2 *K )
Temperature outside wall = (To + Ti)/2 = (33 + 2)/2 =17.5
Temperature of cold room = 2 o C
Heat loss from wall = A x U x (To –Ti)
53.2 x .22 x (17.5 – 2)
181.41 W =618.9 BTU/hour
Cooling load Estimation (Transmission load)
Weather data
DBT = 33 o C
RH = 70 %
Wall gain load
Wall material PUF (Polyurethane Foam)
Wall material thickness = 100 mm
Ceiling (Shaded by corrugated roof)
Length = 20 m
Height = 14 m
Area = 280 m 2
U value of Panel = 0.22 W/(m2 *K )
Temperature outside wall = To + 5= 33 + 5=38
Temperature of cold room = 2 o C
Heat loss from wall = A x U x (To –Ti)
280 x .22 x (38 – 2) = 2217.6
2217.6 W =7566.76 BTU/hour
Cooling load Estimation (Transmission load)
Weather data
DBT = 33 o C
RH = 70 %
Wall gain load
Wall material PUF (Polyurethane Foam)
Wall material thickness = 100 mm
floor
Length = 20 m
Height = 14 m
Area = 280 m 2
U value of Panel = 0.41 W/(m2 *K )
Temperature outside wall = 10
Temperature of cold room = 2 o C
Heat loss from wall = A x U x (To –Ti)
280 x .41 x (10 – 2) = 2217.6
914.4 W =3120 BTU/hour
Cooling load Estimation (Transmission load)
Total Transmission load = South + North + East + West + Ceiling + Floor
Total Transmission load = 518.32 + 518.32 + 421.34 + 181.41 + 2217.6 +
914.4
Total Transmission load = 4771.39 W = 16280 BTU/hour
Cooling load Estimation (Product load)
Weather data
DBT = 33 o C
RH = 70 %
Sensible heat load (Field heat)
Potato main crop
Mass of potato = 4% of 250 ton = 10000 kg = 22046 lb
Entering temperature =25 o C
Final temperature = 2 o C
Specific heat of potato 3400 J/kg K
Total heat to be remove = 10000 x 3400 x (25 – 2)/(3600 x 24x1000)
Total Heat to be remove = 9.050 kW
Cooling load Estimation (Product load)
Weather data
DBT = 33 o C
RH = 70 %
Respiration heat load
Respiration Load Assuming that the remaining 250000 -10000 = 240000kg Potatoes are already in store,
and refrigeration load on the last day of loading is considered
The respiration load would be:
Respiration load = 240000 x 0.018 = 4320 W = 4.32 kW
Total product load would be Field heat + respiration heat = 9.05 +4.32 = 13.37 kW, say 14 kW
Cooling load Estimation (infiltration load)
Weather data
DBT = 33 o C
RH = 70 %
Infiltration load
Infiltration Load Based on 4 air changes/day,
outside enthalpy 99.173kJ/ kg at 45 0 C DB and 30 0C WB,
inside enthalpy 13.62 kJ/kg at 3 0C and 90% RH
Δh = 85.553.
(These values have been taken from Psychrometric property tables for moist air.)
Amount of ventilation air for 1064 (volume of room, m3 ) x 4 air changes ÷ (24x3.6) = 49.25 litre/second.
Using the standard formula for total heat load as 1.2 x l/s x (Δh), = 1.2x49.25 x 85.53/1000 = 5.055 kW.
With 70% recovery, it would be 3.53 kW
Cooling load Estimation (miscellaneous load)
Weather data
DBT = 33 o C
RH = 70 %
Internal Load due to Fan Motors
Assuming 3 coolers per room, each with 2 fans of 0.5 kW, total
motor power is 6 kW. Power contributed to heat load at the
rate of 400 W per motor = 8x.4 = 3.2 kW
Lighting Load At 2 W/sq m during loading = 280 sq ft floor area
x2 W/1000 = 0.56 kW
Occupancy Load Assuming 2 persons working inside the cold
room during loading, each person would be contributing 250
W. Total occupancy load = 250 Wx3 = 0.75 kW
Total Internal load = 3.2+ 0.56 +0.75 = 4.51 kW
Now Total Load = Transmission load + product load + infiltration load + fan load +light load +
occupant load
Now Total Load = 4.77 W + 14 + 3.53 + 3.2 + 0.56 + 0.75 = 26.81 kW
Safety Factor 10% = 26.81 + 26.81 x .1 = 29.49 kW
System running hour = 16 hour then:
System running hour = 29.49 x 24 /16 = 44.23 = 45 kW
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