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Hey there, guys. Paul here from TheEngineeringMindset.com. In this video, we're going to

be learning how to calculate the cooling load for a

cold room storage unit. Coming up, we'll look

at what is a cold room and what are the sources of

heat that we need to remove. Then we'll calculate a worked

example with a safety factor, and we'll finish up sizing

the refrigeration unit to suit this cooling load. Before we dive in, I just

want to thank our partners over at Danfoss for sponsoring this video. They have an extensive

collection of solutions for cold rooms that can help

you meet future refrigerant and energy regulations without

compromising on performance. Their free Coolselector

tool will also help you put these principles I'm

sharing today into practice. You can download it for free

at Coolselector.Danfoss.com. So first of all, what is a

cold room and how does it work? 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 we need to cool down the food by removing the

heat and to remove the heat we use a refrigeration

system as this allows us to accurately and automatically

control the temperature to preserve the goods

for as long as possible.

So where does the heat come

from that we need to remove? Well, typically five to 15%

is through transmission loads. This is the thermal energy

transferred through the roof, the walls, and the floor

into the cold room. Now 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

temperature difference. 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 to 75% of the cooling load. This accounts for the

heat that is introduced into the cold room when

new products enter.

It's 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, but if you're freezing,

then you need to account for the latent heat also as

a phase change occurs. So there is energy used,

but you will not see a temperature change

while the product changes between a state of liquid and ice. Then there is an

additional energy required to further chill this food

down below the freezing point which is again a sensible heat load. You also need to account for the packaging as this will inherently be called also. Lastly, if you're cooling

fruit and vegetables, then these products are alive, and they will generate some heat. So you need to account

for removal of this also.

The next thing to consider

is the internal loads which account for around 10 to 20%. Now this is the heat given

off by people working in the cold room as well as

the lighting and equipment such as forklift trucks, et cetera. 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

generate, and how long will they and the equipment be in

the store for per day. Then we need to consider

the refrigeration equipment in the room which will

account for around one to 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 due to

defrosting the evaporator.

The last thing we need to

consider is air infiltration which again adds around one

to 10% to the cooling load. This occurs when the door's

open and 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 also if it's used. So let's now run a simplified example of a cooling load calculation. 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. First of all, let's start

with the transmission load. The dimensions of our cold

store are six meters long, five meters wide, and four meters high. The ambient air is thirty degrees Celsius at 50% relative humidity,

and the internal air is one degree Celsius at 95

degrees relative humidity. The walls, roof, and

floor are all insulated with 80 millimeters of polyurethane with a U value of 0.28 watts

per meter squared per Kelvin, and the ground temperature

is 10 degrees Celsius.

Now just to note that 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'll be using the formula Q equals U multiplied by A multiplied by the temperature

out minus temperature in multiplied by 24 and

then divided by 1,000. The U value we already know. A is just the surface area

for the walls, the roof, and the floor, and we'll

calculate that in just a second. The temperature in and

temperature out we already know. 24 is just how many

hours there are in a day, and we divide by 1,000 as this formula will calculate the value in watts, but we want to know the

answer in kilowatts, and so we just divide it by 1,000. To calculate the area A is fairly easy. It's just the size of each

of the internal walls. So drop the numbers in to

find the area for each wall, the roof, as well as the floor.

Then we can run these

numbers in the formula that we saw earlier. You'll need to calculate

the floor separately from the walls and the roof

as the temperature difference is different under the

floor so the heat transfer will therefore also be different. If the floor isn't insulated,

then you will need to use a different formula

based on empirical data. We can combine the walls

and the roof to see that the daily heat gain is

22 kilowatt hours per day, and then we can also

calculate the floor to be 1.8 kilowatt hours per day. Therefore the total

transmission cooling load is 23.8 kilowatt hours per day. And remember, if your cold

room is in direct sunlight, then you'll need to account

for the sun's energy also. Next we will calculate the cooling load from the product exchange,

that being the heat pulled into the cold room from

new products entering which are at a higher temperature.

For this example, we'll be storing apples. Now you 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 and then for

freezing and then for sub-cooling. So you'll need to account for this also, but in this example we're

just cooling apples. There are 4,000 kilograms of

new apples arriving each day at a temperature of five

degrees Celsius, and the store maintains a hold of 20,000

kilograms of apples. We can then use the formula

Q equals M multiplied by CP multiplied by the temp

enter minus temp store divided by 3,600, where Q

equals kilowatt hours per day. M equals the mass of

new products each day. Temp enter is the entering

temperature of the products. Temp store is the

temperature within the store, and 3,600 is simply to convert

kilojoules to kilowatt hours. We can then drop the numbers

in and see that it equates to 16 kilowatt hours per day. Next, we calculate the

product respiration. Now this is the heat

generated by living products such as fruit and vegetables. These will all generate heat

as they are still alive.

That's why we're cooling

them down to slow down their deterioration and

preserve them for longer. For this example, I've used

1.9 kilojoules per kilogram per day as an average, but this rate changes over

time and with temperature. Now you can use rules of

thumb such as this one just to simplify the calculation, but the precision of

your answer will reflect how critical the cooling load needs to be. To calculate this, we'll

use the formula Q equals M multiplied by resp divided by 3,600 where Q equals kilowatt hours per day. M equals the mass of product in storage. Resp equals the respiration

heat of the product. And 3,600 is just to convert the kilojoules to kilowatt hours.

If we drop those numbers in,

then we'll see it comes out at 10.5 kilowatt hours per day. For the product section,

we'll just sum the exchange in the respiration load together to get 26.5 kilowatt hours per day. 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 two people

working in the store for four hours a day,

and we can look up to see that at this temperature

they will give off around 270 watts of heat per

hour that they're inside. For this, we'll use the

formula Q equals people multiplied by time multiplied

by heat divided by 1,000 where Q equals kilowatt hours per day. People equals how many

people who are inside. Time equals the length of time

they spend inside each day. And heat equals the heat

loss per person per hour. 1,000 is just to convert

the watts to kilowatts. So dropping those numbers

in, we should see that we get 2.16 kilowatt hours per day. Then we can calculate the heat

generated by the lighting. Now this is fairly simple to

do, and we can use the formula Q equals lamps multiplied by

time multiplied by wattage divided by 1,000 where Q

equals kilowatt hours per day.

Lamps equals the number of lamps. Time equals the hours of use per day. Wattage is the power rating of the lamps, and 1,000 is to convert

the watts to kilowatts. So if we have three lamps

inside at 100 watts each, and they run for four hours

a day, then we should see that these would give off around 1.2 kilowatt hours of heat per day. For the total internal load,

we then just sum the people and lighting load together to get a value of 3.36 kilowatt hours per day. Now we can calculate the heat generation of the fan motors in the evaporator. In this cold room evaporator,

we'll be using three fans rated at 200 watts

each, and we'll estimate that they will be running

for 14 hours per day. We can use the formula Q

equals fans multiplied by time multiplied by wattage divided by 1,000 where Q equals kilowatt hours per day.

Fans equals the number of fans

within the evaporator unit. Time equals the daily run hours. Wattage equals the rated

power of the fan motors, and 1,000 is to simply convert

from watts to kilowatts. So if we drop these numbers in, we'll see that the answer comes out at

8.4 kilowatt hours per day. Now we will calculate the heat load caused by the defrosting

of the evaporator. In this example, our cold room uses an electric heating element

rated at 1.2 kilowatts. It runs for thirty

minutes three times a day, and the transfer efficiency is 30%. So 30% of all the energy it consumes is actually just transferred

into the cold room. We will then use the formula

power multiplied by time multiplied by cycles

multiplied by efficiency where Q equals kilowatt hours per day. Power equals a power rating

of the heating element. Time equals the defrost runtime. Cycles is how many times per day will the defrost cycle occur, and the efficiency is

which percent of the heat will be transferred into the space. So if we drop those numbers in, we'll see that the value comes out at

0.54 kilowatt hours per day.

The total equipment load

is then the fan heat load plus the defrost heat load which is 8.94 kilowatt hours per day. 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'll estimate that there will

be five volume air changes per day due to the door being opened.

The volume is calculated

at 120 cubic meters. Each cubic meter of air

provides around two kilojoules of heat per degree Celsius

of temperature difference, and the outside air is 30 degree Celsius, whereas the inside air

is of one degree Celsius. We can then use the

formula Q equals changes multiplied by volume multiplied by energy multiplied by temp out minus

temp in divided by 3,600 where Q equals the kilowatt hours per day. Changes equals the number

of volume changes per day. Volume equals the volume

of the cold store. Energy equals the energy per

cubic meter per degree Celsius. Temp out is the outside air temperature, and temp in is the inside air temperature.

And 3,600 is just to convert from kilojoules to kilowatt hours. So we'll drop those numbers

in, and we'll see that that comes out to 9.67

kilowatt hours per day. Now we need to calculate

the total cooling load. So for that we'll just need

to sum up the calculations that we've just performed,

and we'll see that that comes out to 72.27

kilowatt hours per day. We should also then apply a

safety factor to the calculation to account for errors and

variations from design. It's typical to add 10 to 30% onto the calculation to cover this. I've gone with 20%. So we'll just multiply the

cooling load by a safety factor of 1.2 to give us our total cooling load of 86.7 kilowatt hours per day. The last thing we need to do is calculate the refrigeration capacities

to handle this load. A common approach is to average

the total daily cooling load by the runtime 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 86.7 kilowatt hours per day divided by our 14 hours of runtime means that the refrigeration

unit needs to have a capacity of 6.2 kilowatts to sufficiently

meet our cooling load. Okay, that's it for this video. Before we wrap things up though, I just want to thank Danfoss one last time and remind you to give everything

you learned today a try by downloading their

free Coolselector tool at Coolselector.Danfoss.com. Thanks for watching.

I hope you've enjoyed

this and it's helped you. If so, then please like,

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This post may contain affiliate links which means I may receive a commission for purchases made through links. Learn more on my Private Policy page.