×
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS

Log In

Come Join Us!

Are you an
Engineering professional?
Join Eng-Tips Forums!
  • Talk With Other Members
  • Be Notified Of Responses
    To Your Posts
  • Keyword Search
  • One-Click Access To Your
    Favorite Forums
  • Automated Signatures
    On Your Posts
  • Best Of All, It's Free!
  • Students Click Here

*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail.

Posting Guidelines

Promoting, selling, recruiting, coursework and thesis posting is forbidden.

Students Click Here

Jobs

Heat Loss

Heat Loss

Heat Loss

(OP)
Is it possible to assess the heat loss from a 6Mw 4p 6kv motor with an efficiency of 97% (FL) and assumming the secondary cooling circuit is routed away ( air or water ), what heat would be rejected into the room.  

RE: Heat Loss

If there is a machine coupled to the motor, I think that it will produce so much heat that the motor itself will not be a problem.

If the driven machine is a pump with cool water so that not much heat is emitted from the machine (pump, for instance). Then the heat emitted from the motor becomes significant. It can be found from surface temperature, total surface and emittivity. You can use the same formula that you use to compute emission from a wall into a room.

In the second case, you can (in theory) force so much coolant through motor that the external surface is around room temperature. Then, you will not have any heating in the room.

Gunnar Englund
www.gke.org

RE: Heat Loss

You can also subtract the heat removed by water cooling from the 3% total heat loss. Heat removed by water heating can me calculated from flow and outlet temperature minus inlet temperature.

RE: Heat Loss

(OP)
CJCPE, Thanks for your input, how would calculate the heat removed by water cooling from flow and outlet temperature.

RE: Heat Loss

You need to determine from tables how much heat (BTUs per pound of mass or Joules per gram of mass) is contained in the fluid at the inlet temperature and pressure vs. the outlet temperature and pressure. Calculate the heat removed per unit mass and multiply by the mass flow rate. That will give you the joules/second (watts) carried away by the fluid.

This is starting to sound like a school question.

Red Flag This Post

Please let us know here why this post is inappropriate. Reasons such as off-topic, duplicates, flames, illegal, vulgar, or students posting their homework.

Red Flag Submitted

Thank you for helping keep Eng-Tips Forums free from inappropriate posts.
The Eng-Tips staff will check this out and take appropriate action.

Reply To This Thread

Posting in the Eng-Tips forums is a member-only feature.

Click Here to join Eng-Tips and talk with other members!


Resources