# Cooling Load Calculation – Cold Room hvac

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
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
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.

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