Cooling Load Calculation for cold rooms. In this article we’ll be looking at how to calculate the cooling load for a cold room. We’ll first look at the heat sources and then we’ll look at a worked example of how to perform a cold room cooling load calculation in a simplified example. **Scroll to the bottom to watch the video tutorial.**

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## What is a cold room?

A cold room is used to store perishable goods such as meat and vegetables to slow down their deterioration and preserve them as fresh as possible for as long as possible. Heat accelerates their deterioration so the products are cooled down by removing the heat.

To remove the heat we use a refrigeration system as this allows accurate and automatic control of the temperature to preserve the goods for as long as possible.

To remove the heat we need to know what the cooling load will be. The cooling load varies throughout the day so in most cases the average cooling load is calculated and the refrigeration capacity is calculated to suit this.

## Cold Room Heat Sources

Where does all the heat come from that we need to remove?

Typically 5-15% is through transmission loads. This is the thermal energy transferred through the roof, walls and floor into the cold room. Heat always flows from hot to cold and the interior of the cold room is obviously a lot colder than its surroundings, so heat is always trying to enter the space because of that difference in temperature. If the cold store is exposed to direct sunlight then the heat transfer will be higher so an additional correction will need to be applied to allow for this.

Then we have Product loads which account for typically 55-75% of the cooling load. This accounts for the heat that is introduced into the cold room when new products enter. Its also the energy required to cool, freeze and further cool after freezing. If you’re just cooling the products then you only need to consider the sensible heat load. If you’re freezing the product then you need to account for the latent heat also as a phase change occurs. During this time energy is used but you will not see a temperature change while the product changes between a state of liquid and ice. There is additional energy required to further chill this food down below the freezing point which is again sensible heat. You also need to account for the packaging as this will inherently be cooled also. Lastly if you’re cooling fruit and vegetables then these products are alive and they will generate some heat so you’ll need to account for the removal of this too.

The next thing to consider is the internal loads which account for around 10-20%. This is the heat given off by people working in the cold room, the lighting and equipment such as fork lifts trucks etc. So for this you’ll need to consider what equipment will be used by the staff members in order to move the products in and out of the store, how much heat will they and the equipment give off and the daily duration.

Then we need to consider the refrigeration equipment in the room which will account for around 1-10% of the total cooling load. For this we want to know the rating of the fan motors and estimate how long they will run for each day, then we want to also account for any heat transferred into the space from defrosting the evaporator.

The last thing we need to consider is infiltration which again adds 1-10% to the cooling load. This occurs when the door opens so there is a transfer of heat into the space through the air. The other consideration is ventilation. Fruit and vegetables give off carbon dioxide so some stores will require a ventilation fan, this air needs to be cooled down so you must account for this if it’s used.

## Cooling load calculation – Cold room worked example

Lets consider a simplified example of a cooling load calculation for a cold room. Now If you’re doing this for a real world example then I recommend you use a design software such as the Danfoss coolselector app for speed and accuracy. Download here –> http://bit.ly/2Ars6yF

### Transmission load

- The dimensions of our cold store are 6m long, 5m wide and 4m high.
- The ambient air is 30
**°**c at 50% RH, The internal air is 1**°**C at 95% RH - The walls, roof and floor are all insulated with 80mm polyurethane with a U value of 0.28W/m
^{2}.K - The ground temperature is 10
**°**C.

*Just to note the manufacturer should tell you what the u value is for the insulation panels, if not, then you will need to calculate this.*

To calculate the transmission load we will be using the formula

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

- Q= kWh/day heat load
- U = U value of insulation (we already know this value) (W/m
^{2}.K) - A = surface area of walls roof and floor (we will calculate this) (m
^{2}) - Temp in = The air temperature inside the room (
**°**C) - Temp out = The ambient external air temperature (
**°**C) - 24 = Hours in a day
- 1000 = conversion from Watts to kW.

To calculate “A” is fairly easy, its just the size of each internal walls, so drop the numbers in to find the area of each wall, roof and floor.

Side 1 = 6m x 4m = 24m^{2}

Side 2 = 6m x 4m = 24m^{2}

Side 3 = 5m x 4m = 20m^{2}

Side 4 = 5m x 4m = 20m^{2}

Roof = 5m x 6m = 30m^{2}

Floor = 5m x 6m = 30m^{2}

Then we can run these numbers in the formula we saw earlier, you’ll need to calculate the floor separately to the walls and roof as the temperature difference is different under the floor so the heat transfer will therefore be different.

**Walls and roof**

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

Q = 0.28W/m^{2}.K x 113m^{2} x (30°C – 1°C) x 24 ÷ 1000

Q = 22 kWh/day

[113m^{2} = 24m^{2 }+ 24m^{2 }+ 20m^{2 }+ 20m^{2} + 30m^{2 }+ 30m^{2 }]

**Floor**

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

Q = 0.28W/m^{2}.K x 30m^{2} x (10°C – 1°C) x 24 ÷ 1000

Q = 1.8 kWh/day

*If the floor isn’t insulated then you will need to use a different formula based on empirical data*.

Total daily transmission heat gain = 22kWh/day + 1.8kWh/day = 23.8kWh/day

*Remember if your cold room is in direct sunlight you’ll need to account for the suns energy also.*

### Product load – Product exchange

Next we will calculate the cooling load from the product exchange, that being the heat brought into the cold room from new products which are at a higher temperature.

For this example we’ll be storing apples, we can look up the specific heat capacity of the apples but do remember if you’re freezing products then the products will have a different specific heat when cooling, freezing and sub cooling so you’ll need to account for this and calculate this separately, but in this example we’re just cooling.

There are 4,000kg of new apples arriving each day at a temperature of 5°C and a specific heat capacity of 3.65kJ/kg.°C.

We can then use the formula

**Q = m x Cp x (Temp enter – Temp store) / 3600. **

- Q = kWh/day
- CP = Specific Heat Capacity of product (kJ/kg.°C)
- m = the mass of new products each day (kg)
- Temp enter = the entering temperature of the products (°C)
- Temp store = the temperature within the store (°C)
- 3600 = convert from kJ to kWh.

Calculation

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

Q = 4,000kg x 3.65kJ/kg.°C x (5°C – 1°C) / 3600.

Q = 16kWh/day

### Product load – Product respiration

Next we calculate the product respiration, this is the heat generated by living products such as fruit and vegetables. These will generate heat as they are still alive, that’s why we’re cooling them to slow them down their deterioration and preserve them for longer.

For this example I’ve used 1.9kJ/kg per day as an average but this rate changes over time and with temperature. In this example we’re using a rules of thumb value just to simplify the calculation since this cooling load is not considered critical. If you were to calculate for a critical load you should use greater precision. In this example the store maintains a hold of 20,000kg of apples.

To calculate this we’ll use the formula

**Q = m x resp / 3600**

- Q = kWh/day
- m = mass of product in storage (kg)
- resp = the respiration heat of the product (1.9kJ/kg)
- 3600 = converts the kJ to kWh.

Q = m x resp / 3600

Q = 20,000kg x 1.9kJ/kg / 3600

Q = 10.5kWh/day

For the product section we’ll sum together the product exchange of 16kWh/day and respiration load of 10.5kWh/day to get a total product load of 26.5 kWh/day.

### Internal heat load – People

Next we’ll calculate the internal loads from people working in the cold room, as people generate heat and we need to account for this.

We’ll estimate 2 people working in the store for 4 hours a day and we can look up and see at this temperature they will give off around 270 Watts of heat per hour inside.

We’ll use the formula:

**Q = people x time x heat / 1000**

- Q = kWh/day
- people = how many people inside
- time = length of time they spend inside each day per person (Hours)
- heat = heat loss per person per hour (Watts)
- 1,000 just converts the watts into kW

Calculation:

Q = people x time x heat / 1000

Q = 2 x 4 hours x 270 Watts / 1000

Q = 2.16 kWh/day

### Internal heat load – Lighting

Then we can calculate the heat generated by the lighting, this is fairly simple to do and we can use the formula

**Q= lamps x time x wattage / 1000 **

- Q = kWh/day,
- lamps = number of lamps within the cold room
- time = hours of use per day
- wattage = power rating of the lamps
- 1000 = converts the Watts to kW.

If we have 3 lamps at 100W each, running for 4 hours a day, the calculation would be:

Q= lamps x time x wattage / 1000

Q= 3 x 4 hours x 100W / 1000

Q= 1.2kWh/day

For the total internal load we then just sum the people load (2.16 kWh/day) and lighting load (1.2kWh/day) to get a value of 3.36kWh/day.

### Equipment load – fan motors

Now we can calculate the heat generation of the fan motors in the evaporator. For this we can the use the formula of:

**Q = fans x time x wattage / 1000**

- Q = kWh/day
- fans = the number of fans
- time = fan daily run hours (hours)
- wattage = the rated power of the fan motors (Watts)
- 1000 = convert from watts to kw.

In this cold room evaporator we’ll be using 3 fans rated at 200W each and estimate that they will be running for 14 hours per day.

Calculation:

Q = fans x time x wattage / 1000

Q = 3 x 14 hours x 200W / 1000

Q = 8.4kWh/day

### Equipment load – fan motors

Now we will calculate the heat load caused by defrosting the evaporator. To calculate this we’ll use the formula:

**Q = power x time x cycles x efficiency**

- Q = kWh/day,
- power = power rating of the heating element (kW)
- time = defrost run time (Hours)
- cycles = how many times per day will the defrost cycle occur
- efficiency = what % of the heat will be transferred into the space.

In this example our cold room uses an electric heating element rated at 1.2kW, it runs for 30 minutes 3 times per day and the estimate that 30% of all the energy it consumes is just transferred into the cold room.

Q = power x time x cycles x efficiency

Q = 1.2kW x 0.5hours x 3 x 0.3

Q = 0.54kWh/day

The total equipment load is then the fan heat load (8.4kWh/day) plus the defrost heat load (0.54kWh/day) which therefore equals 8.94 kWh/day

### Infiltration load

Now we need to calculate the heat load from air infiltration. I’m going to use a simplified equation but depending on how critical your calculation is then you may need to use other more comprehensive formulas to achieve greater precision. We will use the formula:

**Q = changes x volume x energy x (Temp out – Temp in ) / 3600**

- Q = kWh/d
- changes = number of volume changes per day
- volume = the volume of the cold store
- energy = energy per cubic meter per degree Celsius
- Temp out is the air temperature outside
- Temp in is the air temperature inside
- 3600 is just to convert from kJ to kWh.

We’ll estimate that there will be 5 volume air changes per day due to the door being open, the volume is calculated at 120m^{3}, each cubic meter of new air provides 2kJ/°C, the air outside is 30°C and the air inside is 1°C

Q = changes x volume x energy x (Temp out – Temp in ) / 3600

Q = 5 x 120m^{3} 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 just sum all the values calculated

Transmission load: 23.8kWh/day

Product load: 26.5 kWh/day

Internal load: 3.36kWh/day

Equipment load: 8.94 kWh/day

Infiltration load: 9.67 kWh/day

**Total = 72.27 kWh/day**

### Safety Factor

We should also then apply a safety factor to the calculation to account for errors and variations from design. Its typical to add 10 to 30 percent onto the calculation to cover this, I’ve gone with 20% in this example so well just multiply the cooling load by a safety factor of 1.2 to give us our total cooling load of 86.7 kWh/day

### Refrigeration cooling capacity sizing

The last thing we need to do is calculate the refrigeration capacity to handle this load, a common approach is to average the total daily cooling load by the run time of the refrigeration unit. For this I’m estimating the unit to run 14 hours per day which is fairly typical for this size and type of store. Therefore our total cooling load of 86.7kWh/day divided by 14 hours means our refrigeration unit needs to have a capacity of 6.2kW to sufficiently meet this cooling load.

hello,i am HVAC student:

Thermostatic expansion valve is a throttling device used for flow control of refrigerant in the refrigeration system. Throttling is done to reduce the boiling point of the liquid from the condenser. this is achieved by reducing the presure of the refrigerant as it is metered through the small orifice of the throttling device. As the pressure reduces, the boiling temperature of the refrigerant also reduces

Heat pumps

When a friend of mine brought up learning about load calculations I became interested too. She wants to make sure that this service will help her get to a temperature she needs. I will let her know that cooling products only means accounting for the sensible heat load.

Hi

For the sample calculation, you have estimated that the unit runs 14 hours per day. Can I find out how this basis is derived? The refrigeration cooling should be running on 24hours less the defrosting duration (ie duration that the cooler is not running) and in your example, you have used 3 defrost cycles in a day of 30mins per cycle. Shouldn’t the unit be running for (24 – 3*30mins) 22.5hrs instead of 14 or are there any other factors that may have made the unit run lesser hours that I am unaware of?

The load is the time that the compressor will actually be running to remove the heat content. There is never a 100% load on the unit otherwise the compressor would never shut down and any extra load(because the load varies from hour to hour and day to day) would mean the unit can not do it”s job

For the Transmission load

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

Temp in and out must be in Kelvin (K) as you have (K) in the value of U that you have used (0.28W/m2.K)

Doesn’t matter difference between temperatures remains the same

For the Transmission load

Temp in and out must be in Kelvin (K) as you have (K) in the value of U that you have used (0.28W/m2.K)

HOW WE WILL CALCULATE THE COOLING LOAD OF A ROOM IF THE ROOM IS SURROUNDED BY CONDITIONED AIR ,WHAT WE WILL GIVE TO THE OUTSIDE THE OUTSIDE DRY BULB TEMP VALUE

This calculation does not stop here. You will have to select the equipment’s ( Evaporator & Condensing Unit) and then rate them to 72.27kWh/day

Correction,

Actually rate to 86.7 kWh/day and not 72.27 kWh/day as stated earlier

Side 1 = 6m x 4m = 24m2

Side 2 = 6m x 4m = 24m2

Side 3 = 5m x 4m = 20m2

Side 4 = 5m x 4m = 20m2

Roof = 5m x 6m = 30m2

Floor = 5m x 6m = 30m2

Then we can run these numbers in the formula we saw earlier, you’ll need to calculate the floor separately to the walls and roof as the temperature difference is different under the floor so the heat transfer will therefore be different.

Walls and roof

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

Q = 0.28W/m2.K x 113m2 x (30°C – 1°C) x 24 ÷ 1000

Q = 22 kWh/day

[113m2 = 24m2 + 24m2 + 20m2 + 20m2 + 30m2 + 30m2 ]

I would check your calcs if i were you and at this point its not the floor added either.

Nice explanation buddy, thanks for sharing

A/c:for 1000sq/ft we need one ton=12000 BTUH

For refrigeration we’ve to calculate that how much stuff we need to cool down or freeze

For heating total area of the building or house

Area X 30=total BTUH we need to warmup a building or house. How solved this question

will there be any air changes

Why each cubic meter of new air provides 2kJ/°C ?

Specific heat of air at 25°C is 1.005kj/kg°C