kW equals BTUs...or does it?

[OHDataCenterGuy]

We all know that, working with resistive heating coils, kW has a direct relationship to BTU's. I'm currently working with a company that refers to their chiller load in kW rather than tons, which was new to me. I came back to my own facility and, using our very accurately measured power consumption, derived the total heat produced in BTU's/tons for my server room. I was surprised and pleased to see that it pretty much exactly equaled my chiller load in tons (calculated from flow vs deltaT). What I want to know is...why?

It seems to me that if you compare a 1 kW metal halide bulb to a 1 kW motor rotating a pulley that's lifting a weight, the amount of BTU's put into the space by the bulb would be exponentially higher than the amount created by the motor, because the motor is transforming most of the energy into work. Sure, it might get warm, and there'd be some heat generated by whatever mechanical linkages are used to lift the weight...but nothing like the meat-broiling temperature of a metal halide bulb.

So I'm confused. Can anyone enlighten me? Does this have anything to do with the difference between resistive and inductive loads, but then why does my calculation work for my server room, then? There are inductive AND resistive loads there.

[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.

[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
-------

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.

[icy78]

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
-------

He he, good catch!

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[Nuclrchiller]

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
-------

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...

[icy78]

I probably should not Venture into this stuff but I'm going to anyhow LOL. I guess my first thought is that the temperature of the metal halide bulb is much higher than the temperature of the motor and bearings and pulley friction and all . But the total HEAT of each would be the same?

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[Guyinfield]

There was a thread a while back where someone (I think it was Wayne) was making the case that all mechanical work is being converted to heat. Any belt/ sheave friction but also airflow friction against duct walls/diffusers and turbulance. I think it was discussing how to factor a motor in a load calc and their point was that the full operating kw needs to be factored in not just the heat gain across the motor. I dont want to speak for them.

In your example I'd say by lifting you'd be generating a lot of potential energy too.

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[thatguy]

Looking at a bulb and a motor operating side by side isn’t really a fair comparison. The bulb has a rather small filament surface area where all of the work is being performed and that is contained inside a sealed glass housing that contains its heat within, while the motor uses the same energy but it’s work is spread over the entire surface area of the windings and is cooled with a fan while it operates in open air. The heat output for those very reasons are going to artificially look very different when observed from a distance.

A more fair comparison or test would be to seal each one individually inside a thermal insulated box of similar dimensions and let them both operate at the same time under the same conditions for the same duration. The motor for this example could be loaded with a fly wheel of matched size to operate the motor at exactly 1 Kw. I believe the temperature rise within the sealed boxes would be very similar if not identical.

In other words, make a 1Kw motor the same size as the filament inside a 1Kw bulb and seal that motor inside a glass tube and I bet it will look very similar to that bulb when it operates.

As I understand it some parts of the world use Kw instead of BTUH. Some manufacturers here that cater to data rooms are now starting to rate their units in Kw and use that as their standard measure for capacity. It seems logical. A lot of the equipment that we cool is rated in Kw so it makes it easy to calculate heat load and capacity without needing conversions.


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[coolcoil]

Looking at a bulb and a motor operating side by side isn’t really a fair comparison. The bulb has a rather small filament surface area where all of the work is being performed and that is contained inside a sealed glass housing that contains its heat within, while the motor uses the same energy but it’s work is spread over the entire surface area of the windings and is cooled with a fan while it operates in open air. The heat output for those very reasons are going to artificially look very different when observed from a distance.

A more fair comparison or test would be to seal each one individually inside a thermal insulated box of similar dimensions and let them both operate at the same time under the same conditions for the same duration. The motor for this example could be loaded with a fly wheel of matched size to operate the motor at exactly 1 Kw. I believe the temperature rise within the sealed boxes would be very similar if not identical.

In other words, make a 1Kw motor the same size as the filament inside a 1Kw bulb and seal that motor inside a glass tube and I bet it will look very similar to that bulb when it operates.

As I understand it some parts of the world use Kw instead of BTUH. Some manufacturers here that cater to data rooms are now starting to rate their units in Kw and use that as their standard measure for capacity. It seems logical. A lot of the equipment that we cool is rated in Kw so it makes it easy to calculate heat load and capacity without needing conversions.

A very good answer. I was going to write something similar.

The Law of Conservation of Energy https://en.wikipedia.org/wiki/Conservation_of_energy always applies, at least in the physical world in which we live. Pretty much all electric power put into a system eventually becomes heat. For example, if you input 1 kW of power into a motor to run a fan, that 1 kW will become heat through:

1. Motor inefficiency.
2. Fan inefficiency (the typical AHU fan heats the airstream by between 1.5 and 2.5 degF).
3. Friction of the air through the ducts
4. Friction of the air with other air molecules both in the ducts and in the occupied space.

Some of the energy goes to creating noise and vibration, but those ultimately become heat as the power of the vibration is absorbed by surrounding materials and dissipated through friction.

It is understandable that it seems that the light bulb is generating more heat, but the explanation that it is doing so on a small surface is correct. Here is a thought experiment: If you have a 1,000W light bulb and one hundred 10W light bulbs they are all use the same energy, but you can safely put your hand on the 10W bulb.

[icy78]

All the answers given all point to the same thing. Nuclrchillers answer puts it in a nutshell but basically energy in equals energy out.
So 1000 watts into a metal halide bulb is equal to 1000 watts into a motor it's just that the heat is spread out over a wider area, giving the illusion that 1 part of the experiment is producing more energy or using more energy than the other.


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[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.

[thatguy]

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).

What I was trying to drive home with the reference to the size variation between the bulb and the motor is that the perceived difference in heat output is different because the heat is dispersed through a larger surface area on the motor than it is with the bulb. Even though the heat output is the same they feel different when you touch them. Another example is to look at two resistive loads of similar power but different physical size. Like the he light bulbs mentioned above, you could also look at a heated blanket that you sleep with. A 1Kw blanket produce the same heat output as the bulb but you would never be able to sleep with the bulb against you.

I tried to contain both the work output vs heat output confusion by keeping all the work in the same box with the flywheel. What is stoping that flywheel from achieving the same rotational speed as the motor then staying at that speed without any further input from the motor? Friction. The byproduct of friction is heat. You can neither create energy or destroy energy, you can only convert it. Eventually you will have to remove it with cooling. Run a blender with room temperature water for an hour. You will see that water heats up exceptionally, probably close to boiling by that time. Some of that work that the motor is doing is transferring to the water through the blades friction against it. At the end of the day you will still need to cool that water.


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[OHDataCenterGuy]

First, thanks for everyone's comments. This is interesting.

I'm imagining a 1 kW metal halide bulb in an enclosed, insulated box (maybe the size of a refrigerator). And a 230vac motor that pulls ~5 amps at no load, in its own fridge. I've actually got that motor in the room next to me at the moment. It's about the size of a big toaster. Power them both up for an hour, the motor shaft connected to nothing. At the end of that hour, it seems unlikely that the temperature in the bulb fridge won't be substantially higher. The motor isn't creating any frictional heat, besides that coming from its own bearings and windings.

[Poodle Head Mikey]

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
-------

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.

[coolcoil]

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.

[icy78]

At the end of an hour the motor will have absorbed a lot of the heat that the light bulb would have given up freely. So it would take considerably longer for the motor heat to dissipate out into the air and raise that sensible temperature.

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[OHDataCenterGuy]

Now there's an interesting point. I've been all caught up in the intensity of the heat...the time component did not occur to me.

[Nuclrchiller]

At the end of an hour the motor will have absorbed a lot of the heat that the light bulb would have given up freely. So it would take considerably longer for the motor heat to dissipate out into the air and raise that sensible temperature.

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Yep, the mass of that motor will hold much more heat than the air will. In other words, the temperature of that mass will change less, but there is much more mass changing temperature.

[pageyjim]

"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

[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.