What is a cold room?
Cold rooms are used to slow the deterioration of perishable products like fruits, vegetables, and meat, keeping them fresh for as long as possible. Heat accelerates spoilage. Therefore, cold rooms remove heat from the environment, allowing the products to cool.
To remove heat from the environment and ensure that stored products can be preserved for as long as possible, cooling systems that allow accurate and automatic temperature control are needed.
To remove heat from the environment and achieve proper cooling, the cooling load must be calculated. Cooling load varies throughout the day depending on the product stored, the location of the warehouse, and how the warehouse is used. Therefore, in most cases, the average cooling load is calculated and the cooling capacity of the system to be installed is calculated accordingly.
Cold Room Heat Sources
Heat gain in cold rooms typically comes from conduction loads, accounting for 5-15% of the heat loss. This means that thermal energy flows into the cold room from the roof, walls, and floor of the warehouse.
Heat always flows from hot to cold, and the interior of a cold room is much colder than the surrounding environment, so heat is always trying to enter the space due to this temperature difference. If the cold room is exposed to direct sunlight, heat transfer will be higher.
Another factor affecting the cooling load is the load from the products stored in the warehouse. Cooling loads from the products account for 55-75% of the total heat gain. Furthermore, if the product is subjected to a second freezing, freezing, or post-cooling process after cooling, this will also create an additional cooling load.
When calculating heat gain, packaging must also be taken into account, as the product's packaging will also be subject to cooling in the warehouse.
Additionally, if fruits and vegetables are to be cooled, the heat gain from these products must be taken into account, as these products are living and produce some heat in the environment through respiration.
The next thing to consider is the internal load, which accounts for approximately 10-20%. This includes heat given off by people working in the cold room, as well as heat gain from equipment like forklifts and lighting. Therefore, factors such as the equipment used to load and unload products, the number of warehouse personnel, and the time spent loading and unloading should be included in the heat gain calculation.
In addition, we need to consider the cooling equipment in the room, which will account for approximately 1-10% of the total cooling load. This requires knowing the fan motor temperature and the approximate duration of operation each day, as well as factoring in the heat the evaporator will release into the space during defrosting.
The final consideration is air leaking into the room, which again accounts for 1-10% of the cooling load. This occurs when the cold room door is opened. Another consideration is ventilation. Because fruits and vegetables release carbon dioxide, some warehouses require ventilation fans. Because this air also needs to be cooled, it must be included in the heat gain calculation.
Cooling load calculation example
Let's do a simplified example of cooling load calculation for a cold room:
Transmission load calculation
The dimensions of our cold storage are 6m long, 5m wide and 4m high.
Ambient air (location where the warehouse is located) 30°C with 50% relative humidity, Indoor air (desired air condition inside the warehouse) 1°C with 95% relative humidity.
Walls, roof and floors are insulated with 80 mm polyurethane with a U value of 0.28W/m2.K.
The ground temperature is 10°C.
To calculate the transmission load we will use the following formula:
Q = U x A x (Outside Temperature – Inside Temperature) x 24 ÷ 1000
Q = kWh / g heat load
U = U insulation value (we already know this value) (W / m2.K)
A = Surface area of walls, roof and floor (we will calculate this) (m2)
Indoor Temperature = Air temperature inside the room (°C)
Outside Temperature = Ambient outside air temperature (°C)
24 = Number of hours in a day
1000 = Watt to kW conversion.
Calculating “A” is quite easy:
Wall = 6m x 4m = 24m2
Wall = 6m x 4m = 24m2
Wall = 5m x 4m = 20m2
Wall = 5m x 4m = 20m2
Roof = 5m x 6m = 30m2
Floor = 5m x 6m = 30m2
You need to calculate the floor separately from the walls and ceiling, because the temperature difference is different under the floor, so the heat transfer will also be different.
Walls and Roof
Q = U x A x (Outside temperature – Inside temperature) x 24 ÷ 1000
Q = 0.28W / m2.K x 148m2 x (30°C – 1°C) x 24 ÷ 1000
Q = 28.8 kWh / day
[148m2 = 24m2 + 24m2 + 20m2 + 20m2 + 30m2 + 30m2]
Ground
Q = U x A x (Outside temperature – Inside temperature) x 24 ÷ 1000
Q = 0.28W / m2.K x 30m2 x (10°C – 1°C) x 24 ÷ 1000
Q = 1.8 kWh / day
Total daily transmission heat gain = 28.8kWh/day + 1.8kWh/day = 30.6 kWh/day
Product Loading – Cooling Load Calculation from Product Change
In the next step, we will calculate the cooling load based on the heat from the newly arriving product change in the cold room.
For this example, we'll be storing apples. If you're going to be chilling the products, freezing them, or performing post-cooling, you'll need to calculate their heat gain separately. In this example, we're only chilling.
Every day, 4,000 kg of new apples arrive at the warehouse at a temperature of 5°C and a heat capacity of 3.65 kJ/kg.
For this calculation we can use the following formula:
Q = mx Cp x (Product inlet temperature – Temperature inside the warehouse) / 3600
Q = kWh / day
CP = Specific Heat Capacity of the Product (kJ/kg.°C)
m = Mass of newly added products (kg)
Product inlet temperature = Products inlet temperature (°C)
Temperature inside the warehouse = Temperature inside the warehouse (°C)
3600 = kJ to kWh conversion
Calculation
Q = mx Cp x (Product inlet temperature – Temperature inside the warehouse) / 3600
Q = 4,000 kg x 3.65 kJ / kg°C x (5°C – 1°C) / 3600
Q = 16kWh / day
Cooling Load Calculation from Product Respiration
The next step is to calculate the cooling load from product respiration. In this example, let's use an average heat of product respiration of 1.9kJ/kg per day, but this rate varies with time and temperature. In this example, this cooling load is not considered critical, so we use a single value simply to simplify the calculation. In this example, we're storing 20,000kg of apples in storage.
To calculate this we will use the following formula:
Q = mx resp / 3600
Q = kWh / day
m = Product quantity in the warehouse (kg)
resp = heat of respiration of the product (1.9kJ / kg)
3600 = converts kJ to kWh.
Q = mx resp / 3600
Q = 20.000kg x 1.9kJ/kg / 3600
Q = 10.5 kWh / day
So, when we calculate the cooling load from the new product entering the warehouse and the cooling load due to the respiration of the product, we get a total cooling load of 26.5 kWh/day.
Internal Heat Load – Cooling Load Calculation from People
In the next step, we'll calculate the heat load from people working in the warehouse. Assuming there are two people working in the cold storage room for four hours a day, we can estimate that they can generate 270 watts of heat per hour.
We will use the following formula:
Q = number of workers x time x heat / 1000
Q = kWh / day
Number of employees = Number of people working in the warehouse
Time = Length of time spent in the warehouse per person (Hours)
Heat = Heat loss per person per hour (Watt)
1000 = Just converts watts to kW
Calculation:
Q = number of workers x time x heat / 1000
Q = 2 x 4 hours x 270 Watt / 1000
Q = 2.16 kWh/day
Internal heat load – Cooling Load Calculation from Lighting
In the next step, we'll calculate the heat generated by lighting. This is pretty straightforward and we can use the formula below.
Q = lamp x time x watts / 1000
Q = kWh / day,
Lamps = number of lamps in the cold room
Hours = hours of use per day
Watt = power rating of lamps
1000 = Converts Watts to kW.
If there are 3 lamps at 100W each, working 4 hours a day, the calculation would be:
Q = lamp x time x watts / 1000
Q = 3 x 4 hours x 100W / 1000
Q = 1.2kWh/day
Total internal load: Heat load from people (2.16 kWh/day) and heat load from lighting (1.2kWh/day) we get a total value of 3.36kWh/day.
Equipment load – Cooling Load Calculation from Fan Motors
Now, let's calculate the heat gain from the evaporator fan motors.
Q = fans x time x watts / 1000
Q = kWh/day
Fans = Number of fans
Time = Daily operating time of the fan (hours)
Watt = Nominal power of fan motors (Watt)
1000 = Convert from Watt to kW.
This cold room evaporator uses 3 fans, each rated at 200W, and we assume they will run for 14 hours a day.
Calculation:
Q = fans x time x watts / 1000
Q = 3 x 14 hours x 200 W / 1000
Q = 8.4kWh / day
Equipment load – Cooling Load Calculation from Fan Motor Defrost
Now we'll calculate the heat load from evaporator defrosting. To do this, we use the following formula:
Q = power x time x defrost cycle x efficiency
Q = kWh / day,
Power = Power rating of the heating element (kW)
Time = Defrost operating time (Hour)
Defrost cycle = How many times a day the defrost cycle occurs
Efficiency = Percentage of heat transferred to the environment
In this example, our cold room uses a 1.2kW electric heating element. It runs for 30 minutes, three times a day, and 30% of all the energy it consumes is transferred to the cold room.
Q = 1.2kW x 0.5h x 3 x 0.3
Q = 0.54kWh / day
The total equipment cooling load is the fan heat load (8.4kWh/day) plus the defrost heat load (0.54kWh/day), equal to 8.94 kWh/day.
Cooling Load Calculation from Infiltration
Now we need to calculate the heat load from air infiltration. Using a simplified formula:
Q = volume x energy x change x (Outside Temperature – Inside Temperature) / 3600
Q = kWh / d
Change = Number of volume changes during the day
Volume = Cold storage volume
Energy = energy per cubic meter degrees Celsius
Outside Temperature = Outside air temperature
Internal Temperature = Cold room temperature
3600= just converts from kJ to kWh.
If we assume that the door will create 5 volumes of air exchange per day due to product entry and exit into the warehouse, the volume is calculated as 120m3, each cubic meter of new air is 2kJ /°C, the outside air is 30°C and the air inside the warehouse is 1°C.
Q = change x volume x energy x (Outside Temperature – Inside Temperature) / 3600
Q = 5 x 120m3 x 2kJ /°C x (30°C – 1°C) / 3600
Q = 9.67 kWh / day
Total cooling load
To calculate the total cooling load, we will simply add up all the calculated values.
Transmission load: 28.8kWh/day
Product loading: 26.5 kWh / day
Internal load: 3.36kWh/day
Equipment load: 8.94 kWh / day
Leaking load: 9.67 kWh / day
Total = 77.27 kWh / day
Safety Factor
To account for design errors and variations, we must also apply a safety factor to the calculation. A 10% to 30% deviation can be added to account for this.
Let's use a 20% safety factor in this example. Therefore, if we multiply the cooling load by a 1.2 safety factor, we get a total cooling load of 92.7 kWh/day.
Cooling Capacity Calculation
The final step is to calculate the cooling capacity required to remove this heat gain from the environment. To do this, divide the calculated total cooling load by 14, assuming the unit operates 14 hours per day. This means the required capacity for our cooling unit is 92.7/14 = 6.6 kW.
K: Storage and Preparation of Horticultural Products for Market “Somtad Publications Textbook No:1” prepared by Prof. Dr. Rahmi TURK, Prof. Dr. Nurdan TUNA GUNES, Prof. Dr. Mustafa ERKAN, Prof. Dr. Mehmet Ali KOYUNCU / 2017
