Optimization of Cold Room Utility Can Be Enhanced By Proper Cooling Load Calculation

What is a Cold Room?

The term cold room refers to a room that is used to store perishable goods such as meat and vegetables so that their deterioration can be slowed down and they can be preserved as fresh as possible and for a longer period as possible. The heat accelerates the product’s deterioration, so the products need to be cooled down by removing the heat. To take out the heat a refrigeration system is used so that it allows an accurate automatic control of the temperature so that it can preserve the products for a longer period.

Before removing the heat, we must know about the cooling load in detail or what is going to be the cooling load. The cooling load is different throughout the day so in most cases the average cooling load is calculated & the refrigeration capacity is calculated to suit this.

Cold Room Heat Sources:

The heat mainly comes from transmission loads (5% – 15%). The heat is transmitted into the cold room in the form of thermal energy & it is transferred through the roofs, walls & floors of the cold room. Heat always flows from hot areas to cold areas and the interior of the cold room is much colder than its surroundings, so the heat is trying to enter the space because of the difference in temperature. If the cold room is exposed to direct sunlight, then the heat transfer is going to be higher and so additional correctional measures need to be taken for this.

Then we have product loads which make up 55%-75% of the cooling load. This is accountable for the heat that is entering the cold room when the products are entering the cold room. We should also consider that is required to cool, freeze & further cooling after freezing. If you are doing only cooling, then you should take into consideration the sensible heat load but on the other hand if you are freezing a product you need to take latent heat into consideration as well.

Cooling Load Calculation- Cold Room Worked Example:

For Example, let’s take into consideration that the dimensions of the cold store are 6m long, 5m wide& 4m high. The ambient air is 30°c at 50% RH, the internal air is at 1°c at 95% RH. The walls, roof & the floors are all insulated with 80mm polyurethane with a U value of 0.28W/m². K. The ground temperature is 10°c.

Transmission Load

To calculate the transmission load we are going to take the help of this formula

Q= U x A x (Temp out – Temp in) x 24 ÷ 1000.

Here, Q= kWh/day heat load, U= U value insulation (W/m². K), A= surface area of walls, roof & floor (m²), Temp in= air temperature inside the room (°C), Temp out= ambient external air temperature (°C), 24= Hours in a day, 1000= conversions from Watts to kW.

Now to calculate “A”, we just need to do the following steps:

Side 1= 4m x 5m = 20m²

Side 2 =4m x 5m = 20m²

Side 3 =4m x 5m = 20m²

Side 4 =4m x 5m = 20m²

Roof = 4m x 5m = 20m²

Floor = 4m x 5m = 20m²

Therefore, A= 120m²

Walls & Roof Calculation:

Q= U x A x (Temp out – Temp in) x 24 ÷ 1000

Q= 0.028 W/m². K x 120m² x (25°c – I-20I°c) x 24 ÷ 1000

Q=3.6 kWh/day

Product Load- Product Exchange:

Now we are going to calculate the cooling load from the product exchange, this happens when the heat is brought into the cold room from new products which are at a higher temperature.

For Example, we are going to store apples, so that we can look up the specific heat capacity of the apples. Suppose we are storing 4000kg of apples every day at a temperature of 5°C and a specific heat capacity of 3.65kJ/kg. °C. Now we can apply the following formula.

Q= m x CP x (Temp enter – Temp store)/3600

Calculation:

Q= m x CP x (Temp enter – Temp store)/3600

Q= 4000kg x 3.65kJ/kg. °C x (5°C – 1°C) /3600

Q= 16kWh/day

There are formulae for Internal Heat Load, Equipment Load, Infiltration Load, Internal Heat Load-Lighting. Thus, it can be concluded that we should apply safety factors at the time of calculating cooling load of cold rooms. We should also calculate the refrigeration cooling capacity sizing of the cold rooms.