Cooling Load Calculation - Cold Room hvac
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Date: 2023-11-17
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Comments and reviews: 27
kdmq
Does your factor of energy per cubic meter of air (2 kJ m3 / deg C) account for latent cooling as well as sensible cooling? The value of the heat capacity times temperature difference (sensible cooling) rarely accounts for the full effect of air infiltration. If you want to calculate based on both sensible and latent cooling, a psychrometric chart is the way to go.
When I say latent cooling, this is not to be confused with the freezing of the fruits, which I am aware is not present in this calculation. I am referring to the humidity difference between the air outside and the air inside. Basically the refrigeration system will have to condense large amounts of water that enter the unit as the door is opened, which costs energy.
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Does your factor of energy per cubic meter of air (2 kJ m3 / deg C) account for latent cooling as well as sensible cooling? The value of the heat capacity times temperature difference (sensible cooling) rarely accounts for the full effect of air infiltration. If you want to calculate based on both sensible and latent cooling, a psychrometric chart is the way to go.
When I say latent cooling, this is not to be confused with the freezing of the fruits, which I am aware is not present in this calculation. I am referring to the humidity difference between the air outside and the air inside. Basically the refrigeration system will have to condense large amounts of water that enter the unit as the door is opened, which costs energy.
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Nightbow9
Great video, nicely done: )
However, the fans' heat load calculation is not correct. You count the 200 W of total electricity consumption as heat. This is true for a heater: 200 W electricity will turn into 200 W heat. That's 100% efficiency, as heating is the purpose of a heater.
The electric motor in the fan has an efficiency of ca. 70-80%, so 30-20% of the electricity intake dissipates as heat. Therefore in the formula the 200 W needs to be multiplied by 0. 2-0. 3, so the 200 W fan will generate only ca. 40-60 W heat.
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Great video, nicely done: )
However, the fans' heat load calculation is not correct. You count the 200 W of total electricity consumption as heat. This is true for a heater: 200 W electricity will turn into 200 W heat. That's 100% efficiency, as heating is the purpose of a heater.
The electric motor in the fan has an efficiency of ca. 70-80%, so 30-20% of the electricity intake dissipates as heat. Therefore in the formula the 200 W needs to be multiplied by 0. 2-0. 3, so the 200 W fan will generate only ca. 40-60 W heat.
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Zaher
why we take into account the rated power of the equipment?
Like the power of the fan is the responsible for moving the fan by a specific torque
so, i see that the energy (from the power given by time) to move the fan cannot all be converted to a heat load (or heat generation) as the fan would not have the energy to move
I believe that there is some of heating in the process but with a ratio
please give me an answer as it is so confusing to me
I almost forgot, thank you for this great video
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why we take into account the rated power of the equipment?
Like the power of the fan is the responsible for moving the fan by a specific torque
so, i see that the energy (from the power given by time) to move the fan cannot all be converted to a heat load (or heat generation) as the fan would not have the energy to move
I believe that there is some of heating in the process but with a ratio
please give me an answer as it is so confusing to me
I almost forgot, thank you for this great video
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education
Brother can you please upload video of cooling load estimation on banana ripening chamber having capacity of 7. 5 mt of 3 chamber having 201210 feet chamber each temp is 17-20 deg celcius having water droplet (fog) humidifier 3kg/hr capacity. It is my mtech final year project currently im working on topic PRACTICAL INVESTIGATION AND STUDY OF BANANA RIPENING CHAMBER so it is very useful for me im waiting for ur response.
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Brother can you please upload video of cooling load estimation on banana ripening chamber having capacity of 7. 5 mt of 3 chamber having 201210 feet chamber each temp is 17-20 deg celcius having water droplet (fog) humidifier 3kg/hr capacity. It is my mtech final year project currently im working on topic PRACTICAL INVESTIGATION AND STUDY OF BANANA RIPENING CHAMBER so it is very useful for me im waiting for ur response.
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Mostafa
excellent explanation, based on this which of the AC system components contributes most to the thermal capacity of the system, in other words, say we need to upgrade a standard car SUV factory AC, to be able to handle blowing 42 degree F during extended exploration in extreme middle east desert heat in excess of 130 F, what do we upgrade in order of priority and, how do we measure the overall additional capacity?
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excellent explanation, based on this which of the AC system components contributes most to the thermal capacity of the system, in other words, say we need to upgrade a standard car SUV factory AC, to be able to handle blowing 42 degree F during extended exploration in extreme middle east desert heat in excess of 130 F, what do we upgrade in order of priority and, how do we measure the overall additional capacity?
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kamozoren
Very helpful. One remark: for Internal load - people calculation, it should be 2 people x 4 hours x 270W / 1000.
not 270W/h which is a meaningless notion. You have either Watt(power) or Watt x h(energy.
Try dimension analysis on your equation: people x hours/day x w/hour = w/d. Which is again, meaningless.
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Very helpful. One remark: for Internal load - people calculation, it should be 2 people x 4 hours x 270W / 1000.
not 270W/h which is a meaningless notion. You have either Watt(power) or Watt x h(energy.
Try dimension analysis on your equation: people x hours/day x w/hour = w/d. Which is again, meaningless.
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Shubham
In this example we assumed that 4000 kg of apple changes daily from a total of 20K so 16K still be there untouched. In such case this cold storage should work 24 hours per day but in the end you took 14 hours/day working. Is it just assumed or there is reason which i could not understand
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In this example we assumed that 4000 kg of apple changes daily from a total of 20K so 16K still be there untouched. In such case this cold storage should work 24 hours per day but in the end you took 14 hours/day working. Is it just assumed or there is reason which i could not understand
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Arisztid
At the time stamp 5: 48 your calculation for walls and roof are off. When you add up the walls and roof you get 118m sq not 113m sq. It is only a difference of. 97Wh/d when you figure the equation. Would that small of a difference make any issues arise down the line?
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At the time stamp 5: 48 your calculation for walls and roof are off. When you add up the walls and roof you get 118m sq not 113m sq. It is only a difference of. 97Wh/d when you figure the equation. Would that small of a difference make any issues arise down the line?
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Edward
Hey Paul,
Could you do a video on residential load calcs and go more in depth on radiation heating. I'd love to learn more of the math and standards behind doing something similar to a manual J without dedicated software.
Thanks for your informative videos!
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Hey Paul,
Could you do a video on residential load calcs and go more in depth on radiation heating. I'd love to learn more of the math and standards behind doing something similar to a manual J without dedicated software.
Thanks for your informative videos!
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Keep
how can I find the respiration heat of specific fruits, how are you using the values for respiration heat per day and heat of people and stuff, I mean like where are you taking these values from, just want to know that, thanks, very nice video indeed
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how can I find the respiration heat of specific fruits, how are you using the values for respiration heat per day and heat of people and stuff, I mean like where are you taking these values from, just want to know that, thanks, very nice video indeed
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YM
Thanks for the explanation of the previous question, but i got an other question about the calculation on the cooling load, is it not reasonable to also calculate cooling load of the air. I was thinking that also heat need to be extracted from the air.
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Thanks for the explanation of the previous question, but i got an other question about the calculation on the cooling load, is it not reasonable to also calculate cooling load of the air. I was thinking that also heat need to be extracted from the air.
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Dee
Medium temperature refrigeration maintaining box temperature of 35 to 38F degrees have evaporator fan motors that run 24hrs per day when set point is satisfied and compressor is off to help defrost evaporator and for air circulation.
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Medium temperature refrigeration maintaining box temperature of 35 to 38F degrees have evaporator fan motors that run 24hrs per day when set point is satisfied and compressor is off to help defrost evaporator and for air circulation.
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Furkan
great video, thanks! i have a question. the 20000 apples that are stored in the cold room requires a cooling load too in order to keep the apples cool. how can you calculate that?
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great video, thanks! i have a question. the 20000 apples that are stored in the cold room requires a cooling load too in order to keep the apples cool. how can you calculate that?
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Donald
Sir can you show us how to calculate to get what compressor and evaporator cooler and its capacity to use after you get the kw/h cooling load. Thank you very much
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Sir can you show us how to calculate to get what compressor and evaporator cooler and its capacity to use after you get the kw/h cooling load. Thank you very much
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Lucas
The best video that I've ever watched about cooling load calculation.
You did an amazing job, man! So clear and straightforward. Thank you so much for that
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The best video that I've ever watched about cooling load calculation.
You did an amazing job, man! So clear and straightforward. Thank you so much for that
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Yufanyi
It took me 5hours to learn everything you just explain. Very straight forward. And thank you for adding the calculation for the kW needed to power the unite
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It took me 5hours to learn everything you just explain. Very straight forward. And thank you for adding the calculation for the kW needed to power the unite
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education
will be watching all your videos on cooling loads great break down on the numbers and reasons why we need them to achieve the proper cooling required
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will be watching all your videos on cooling loads great break down on the numbers and reasons why we need them to achieve the proper cooling required
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Amma
What am excellent explanation Bro
We weren't taught like this on our bachelor's
Hats off bro
Thank you very much for the video
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What am excellent explanation Bro
We weren't taught like this on our bachelor's
Hats off bro
Thank you very much for the video
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AnimeDose
Great videos! Very helpful. I hope you could upload more videos specifically on cooling load estimation of a building. thanks
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Great videos! Very helpful. I hope you could upload more videos specifically on cooling load estimation of a building. thanks
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Engineering
I wonder how much time you would have spent on each of these. This definitely needs over lot of time and effort. Hats Off.
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I wonder how much time you would have spent on each of these. This definitely needs over lot of time and effort. Hats Off.
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Bullish
Being in Canada we are suppose to use metric system but use imperial because everything we do and use is from the states.
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Being in Canada we are suppose to use metric system but use imperial because everything we do and use is from the states.
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Jayke
How about plate evaporator freezer for freezing a fish. Fish is in rectangular aluminum tray. is this formula can apply? ?
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How about plate evaporator freezer for freezing a fish. Fish is in rectangular aluminum tray. is this formula can apply? ?
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Bilal
Loved the video but I believe the area summation for transmission load is incorrect. It should be 118m2 instead of 113m2
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Loved the video but I believe the area summation for transmission load is incorrect. It should be 118m2 instead of 113m2
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Zak
75kwh/ day seems insane. Are you sure it wasnt per month? 75 a day would mean a very hefty electricity bill. Please verify
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75kwh/ day seems insane. Are you sure it wasnt per month? 75 a day would mean a very hefty electricity bill. Please verify
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Donald
Sir, in Temp out - Temp in, how can we subtract if Temp out is 33celsius minus to -28celsius? Thank you very much
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Sir, in Temp out - Temp in, how can we subtract if Temp out is 33celsius minus to -28celsius? Thank you very much
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riyadh
dear- may i know how to calculate the load for cooling room with full fresh air --the room is 5x6x3 m --regards
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dear- may i know how to calculate the load for cooling room with full fresh air --the room is 5x6x3 m --regards
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John
Really helpful video, how would one use the cooling load calculation to design the refrigeration system?
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Really helpful video, how would one use the cooling load calculation to design the refrigeration system?
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