×
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

Cooling Water Flow

Cooling Water Flow

Cooling Water Flow

(OP)
I'm trying to determine the amount of water flow I will need to cool a heated platen (350 to 70 deg) in 1 hour.  The platen is made of stainless steel and is 48" by 48" by 4" thick.  Water flows inside the platens.

Any help on equations that apply will be appreciated.

RE: Cooling Water Flow

Here is your answer:

Mass = Vol*Density = 48x48x4/144 (40.8 lbs/ft^2*in) = 2611 lbs

Energy used to heat steel
kW = m*Cp*dT/3412*Hrs = 2611 lbs(0.12 Btu/lb/F)(350-70)/(3412*1hr) = 25.7 kWhr = 87688 Btu/hr

Energy gained by water to cool steel to say 150F
m = q/(Cp*dT) = 87688 Btu/hr / (1 Btu/lb/F)(150-70)
m = 1096 lbs/hr = 18.27 lbs/min / 8.31 lbs/gal = 2.2 gal/min.
I am not sure of all your system variables but this should be close enough.

RE: Cooling Water Flow

(OP)
Thank you Dracula

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