# Thread: kW equals BTUs...or does it?

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An electric motor inside an AHU which is inside the airstream typically raises the air temperature 2 to 2.5 degrees F. This tells you that inside the unit most of the KW input is converting the power input to mechanical energy. What happens after that can be argued but the main thing to remember is in this application is the KW input is directly related to motor horse power.

2. "Heat Loss from an Electric Driven Device to a Room
The heat transferred to a room or enclosure depends on the location of the motor and the driven device related to the room.

1) motor and driven device in the room - all energy to the motor is at the end transferred to the room (heat loss from the motor and frictional energy from the driven device)
2) motor in the room and device outside the room - only heat loss from the motor is transferred to the room - frictional energy from the driven device is lost outside the room
3) motor outside the room and device inside the room - frictional loss from the driven device is transferred to the room - energy loss from the electric motor is lost outside the room"

There was a similar thread awhile back as someone mentioned.

https://www.engineeringtoolbox.com/e...oss-d_898.html

3. Really? I have been using 3.41 BTU's Per Watt for many years. Was I wrong all that time or has physics changed more recently?

PHM
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Originally Posted by Nuclrchiller
One BTU = 3.515 kW.

That's from my pocket reference book.
Those are just two different measurements of energy.
If you want the science behind converting one to the other, try Googling it in the form a question.
It'd probably be an interesting conversation, but be ready for some math, lol.

4. Originally Posted by Poodle Head Mikey
Really? I have been using 3.41 BTU's Per Watt for many years. Was I wrong all that time or has physics changed more recently?

PHM
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He he, good catch!

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5. No; the work also eventually becomes heat again at some point.

Unless that conversion happens outside the conditioned space the cooling system would have to deal with it.

PHM
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Originally Posted by OHDataCenterGuy
I agree that the only real way to know is to seal the bulb and the motor in separated insulated cases and compare the respective heat gains. The statements about the filament size vs the motor don't work for me, since the power draw vs the heat put into the environment is all I'm comparing. It doesn't matter if the heat is coming from a tiny filament or a giant motor, as long as it's trapped in the environment (and thus, must be removed with the cooling system).

I'm not implying that anything magical is happening that might contravene the law of conservation of energy. I'm just wondering if changing the energy into physical work RATHER THAN heat changes the amount of energy the cooling system has to remove. The more I consider it, the more convinced I am that it does...because if ALL the energy can be accounted for just from the heat coming off the motor, where would the energy come from that's doing the work? In THAT case, we're magically CREATING extra energy.

6. Originally Posted by WAYNE3298
An electric motor inside an AHU which is inside the airstream typically raises the air temperature 2 to 2.5 degrees F. This tells you that inside the unit most of the KW input is converting the power input to mechanical energy. What happens after that can be argued but the main thing to remember is in this application is the KW input is directly related to motor horse power.
Wayne, while you're on here, I think you said on another thread that even if a motor is outside of the heated or cooled space, (say it has a long shaft and is driving a fan inside the box) , that all the energy consumed by that motor should still be calculated into the cooling load of that box.
How can that be when so much of the heat is dissipated outside of the box? I realize that there is energy dissipated Inside the Box because of the fan but ...could you explain?

I'm sorry if this is a hijack, I don't mean it that way, it seems like it pertains to the subject.

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Originally Posted by OHDataCenterGuy
I agree that the only real way to know is to seal the bulb and the motor in separated insulated cases and compare the respective heat gains. The statements about the filament size vs the motor don't work for me, since the power draw vs the heat put into the environment is all I'm comparing. It doesn't matter if the heat is coming from a tiny filament or a giant motor, as long as it's trapped in the environment (and thus, must be removed with the cooling system).

I'm not implying that anything magical is happening that might contravene the law of conservation of energy. I'm just wondering if changing the energy into physical work RATHER THAN heat changes the amount of energy the cooling system has to remove. The more I consider it, the more convinced I am that it does...because if ALL the energy can be accounted for just from the heat coming off the motor, where would the energy come from that's doing the work? In THAT case, we're magically CREATING extra energy.
In your original example of lifting a weight, it is true that some of the energy is converted to potential energy because the height relative to the center of the Earth increased - the increase is mass X force of gravity X height change. If the weight never comes down that potential energy will never become heat. But when it does come down that energy will be converted to another form of energy and ultimately heat. For example, if you gently lower it using the brake on your winch the friction the brake is the heat from the potential energy.

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Originally Posted by Poodle Head Mikey
Really? I have been using 3.41 BTU's Per Watt for many years. Was I wrong all that time or has physics changed more recently?

PHM
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Lol!
Yep, like Icy said, good catch. Book actually says One Ton Of Refrigeration = 3.515 kW.

When you don't type and have to finger peck, it's so slow that the disconnect between brain and finger is magnified. That's my story...

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Originally Posted by WAYNE3298
An electric motor inside an AHU which is inside the airstream typically raises the air temperature 2 to 2.5 degrees F. This tells you that inside the unit most of the KW input is converting the power input to mechanical energy. What happens after that can be argued but the main thing to remember is in this application is the KW input is directly related to motor horse power.
It is not always true that most of the kW input of a fan is converted to mechanical energy. Wire-to-air efficiency of a small motor and fan is very low - 25% is not unusual (eg. a fan coil with a 75% efficient motor and 33% efficient fan). In that case 75% of the electrical power becomes heat. I know a lot of people are surprised that the efficiency of small fans is so low, but that's just physics. When I first started in VAV and fan coil design I was surprised myself.

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icy I remember a conversation on another thread like you said but don't remember posting what you said. Someone on that thread did argue that all energy is converted to heat.

Coolcoil I was referring to the average temperature rise of most AHU'S. I am aware it can vary.

3412 and 3413 are both acceptable BTU per KW. Both can be found in reference books. I use 3412.

11. Originally Posted by WAYNE3298
icy I remember a conversation on another thread like you said but don't remember posting what you said. Someone on that thread did argue that all energy is converted to heat.

Coolcoil I was referring to the average temperature rise of most AHU'S. I am aware it can vary.

3412 and 3413 are both acceptable BTU per KW. Both can be found in reference books. I use 3412.

"If you Google the engineering tool box heat gain of electric motors in continuous operation there is a table that breaks down the distribution of the heat in kw. Notice with the motor and fan in the same enclosure all kw (heat) is dumped inside the enclosure. The table is set up to tell you how much heat is put into the airstream when the motor and fan are in separate enclosures. The heat of producing the energy needed is subtracted from the total motor input kw when calculating the heat added to the airstream when the motor and fan are in separate enclosures. I'm not going to start another subject by getting into that break down but with everything inside the fan enclosure the motor kw input is the total heat added to the air."

I am reposting this again from engineering toolbox

"Heat Loss from an Electric Driven Device to a Room
The heat transferred to a room or enclosure depends on the location of the motor and the driven device related to the room.

1) motor and driven device in the room - all energy to the motor is at the end transferred to the room (heat loss from the motor and frictional energy from the driven device)
2) motor in the room and device outside the room - only heat loss from the motor is transferred to the room - frictional energy from the driven device is lost outside the room
3) motor outside the room and device inside the room - frictional loss from the driven device is transferred to the room - energy loss from the electric motor is lost outside the room"

https://www.engineeringtoolbox.com/e...oss-d_898.html

12. Originally Posted by pageyjim

"If you Google the engineering tool box heat gain of electric motors in continuous operation there is a table that breaks down the distribution of the heat in kw. Notice with the motor and fan in the same enclosure all kw (heat) is dumped inside the enclosure. The table is set up to tell you how much heat is put into the airstream when the motor and fan are in separate enclosures. The heat of producing the energy needed is subtracted from the total motor input kw when calculating the heat added to the airstream when the motor and fan are in separate enclosures. I'm not going to start another subject by getting into that break down but with everything inside the fan enclosure the motor kw input is the total heat added to the air."

I am reposting this again from engineering toolbox

"Heat Loss from an Electric Driven Device to a Room
The heat transferred to a room or enclosure depends on the location of the motor and the driven device related to the room.

1) motor and driven device in the room - all energy to the motor is at the end transferred to the room (heat loss from the motor and frictional energy from the driven device)
2) motor in the room and device outside the room - only heat loss from the motor is transferred to the room - frictional energy from the driven device is lost outside the room
3) motor outside the room and device inside the room - frictional loss from the driven device is transferred to the room - energy loss from the electric motor is lost outside the room"

https://www.engineeringtoolbox.com/e...oss-d_898.html
Thanks pagey!
I mis-remembered again

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Thanks pageyjim.
I had forgotten about that post. I was actually thinking of another thread where we discussed conservation of energy. When I said most of the motor energy input was converted to mechanical energy I was hoping to get different comments than I got. You guys are too clever for that bit of bait. My out was going to be remember the motor KW.
If you have 800 CFM being moved by a 1/2 horsepower motor the air temperature rise calculates to right at 2 F. That indicates all the motor KW input results in heat.

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