How to Calculate Heat Dissipation in an Electric Motor?

[newbphysic]

Consider an electric motor with a shaft power output of
20 kW and an efficiency of 88 percent. Determine the rate at
which the motor dissipates heat to the room it is in when the
motor operates at full load. In winter, this room is normally
heated by a 2-kW resistance heater. Determine if it is necessary
to turn the heater on when the motor runs at full load.

My question is how can i find heat dissipate ?

1. Homework Statement
power output = 20 kW
efficiency = 0.88
operates at full load
power input = 20/0.88 = 22.72 kW

The Attempt at a Solution

Method 1 (only inefficiency causes heat) : (1-0.88)* (22.72) = 2.72 kW
Method 2 (inefficiency causes heat & energy from shaft movement cause increase in air kinetic energy which in the end become thermal energy) = power input = 22.72 kW
Method 3 (from solution manual which i don't understand) = (1-0.88) * power output = 0.12 * 20 = 2.4 kW

Which one should i use and why ?

 

[russ_watters]

I don't like any part of this question.

What happens to the mechanical energy is critical to answering the question. Since they don't tell you IMO it is reasonable (usually but not quite always correct)to assume that the mechanical energy is all converted to heat in the room. So I would have answered #2.

Your answer #1 is correct based on its assumption.

Answer #3, from the solution manual, is not correct: The efficiency of any device is output divided by input. To see that their answer is gibberish, consider what the other side of the coin is: they (supposedly) calculated the inefficiency loss. So what's left? The output power. So if (1-.88)*output = loss, then .88*output = ...? Gibberish.

Please tell me this isn't an engineering thermodynamics problem?!

 

[newbphysic]

I don't like any part of this question.

What happens to the mechanical energy is critical to answering the question. Since they don't tell you IMO it is reasonable (usually but not quite always correct)to assume that the mechanical energy is all converted to heat in the room. So I would have answered #2.

Your answer #1 is correct based on its assumption.

Answer #3, from the solution manual, is not correct: The efficiency of any device is output divided by input. To see that their answer is gibberish, consider what the other side of the coin is: they (supposedly) calculated the inefficiency loss. So what's left? The output power. So if (1-.88)*output = loss, then .88*output = ...? Gibberish.

Please tell me this isn't an engineering thermodynamics problem?!

thanks russ,
yes it is engineering thermodynamics problem from cengel book, why ?

 

[russ_watters]

yes it is engineering thermodynamics problem from cengel book, why ?

Poorly written, poorly solved. Very disappointing for an engineering textbook.

 

[newbphysic]

Since they don't tell you IMO it is reasonable (usually but not quite always correct)to assume that the mechanical energy is all converted to heat in the room.

When i think about this, i think all mechanical work will increase air kinetic energy which will become heat.Can you give me an example (situation) where not all mechanical energy converted to heat ?

Also this maybe a little off topic but can you recommend good textbook for mechanical engineer ?
I searched amazon and found yunus cengel book as a recommended book

 

[russ_watters]

When i think about this, i think all mechanical work will increase air kinetic energy which will become heat.Can you give me an example (situation) where not all mechanical energy converted to heat ?

A mine conveyor. Much of the energy goes to lifting material out of the mine, increasing its potential energy.

Also note the phrase "in the room". If the motor does work that is exported from the room (turns a shaft through the wall, for example), the eventual heat doesn't stay in the room. [edit] A better example is an externally driven exhaust fan. Those are very common.

Also this maybe a little off topic but can you recommend good textbook for mechanical engineer ?
I searched amazon and found yunus cengel book as a recommended book

I keep mine at work and will check which it is. That book got good reviews though, so it is possible that's just a bad problem in an otherwise good book.
[Edit]
Heh - I have the same book. 6th edition, 2008. What edition is yours and what is the chapter/problem #?

 

[newbphysic]

8th edition, 2-58