Smart questions
Smart answers
Smart people
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS

Member Login




Remember Me
Forgot Password?
Join Us!

Come Join Us!

Are you an
Engineering professional?
Join Eng-Tips now!
  • 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!

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

Donate Today!

Do you enjoy these
technical forums?
Donate Today! Click Here

Posting Guidelines

Promoting, selling, recruiting, coursework and thesis posting is forbidden.
Jobs from Indeed

Link To This Forum!

Partner Button
Add Stickiness To Your Site By Linking To This Professionally Managed Technical Forum.
Just copy and paste the
code below into your site.

cryotechnic (Chemical)
21 May 08 14:06
I have the following situation:
3 coolingwaterpumps, 3 different coolingwater users with gascoolers, coolingwaterpressure reading, coolingwaterflowreading.

Can I use the formula as mentioned below for calculation of the pump power?
We don't have an accurate powerreading on the pumpmotors, but we want to have an idea of the pumppower when changing the flow, pressure and when we do other adjustments in the coolingwater system.

This is the formula:

Ph =(q*rho*g*h)/3.6*10^6
Pshaft = Ph/motorefficiency

unit's are in SI

Thank you in advance,
Cryotechnic

"Math is the ruler of your potential succes...."  

BigInch (Petroleum)
21 May 08 14:28
Brake Power = motor_efficiency should be pump_efficiency
(read pump's efficiency at its flowrate off the pump curve)
Power Input to Motor = Brake Power/Motor Efficiency
 

http://virtualpipeline.spaces.msn.com

"What gets us into trouble is not what we don't know, its what we know for sure" - Mark Twain

Artisi (Mechanical)
21 May 08 20:47
The easiest formula to use in metric units is:

l/sec x h(m)x S.G. / 102 x pump% = Kw (pump shaft)
 
25362 (Chemical)
22 May 08 11:50

(L/s)(m)(kg/L)(m/s2) = kg.m2/s3 = W

Q → L/s
H → m
ρ → kg/L
g → 9.8 m/s2

If you divide by 1000 and by the system's efficiency as a decimal, you get the equation given by Artisi (expressed in kW)
cryotechnic (Chemical)
22 May 08 14:14
Thanks,

in my formula
Ph =(q*rho*g*h)/3.6*10^6
Pshaft = Ph/motorefficiency

q = m3/h
rho = kg/m3
g = 9,81
h = m

 

"Math is the ruler of your potential succes...."  

BigInch (Petroleum)
22 May 08 18:14
No it is not.  If it says so, its a mistake.
 

http://virtualpipeline.spaces.msn.com

"What gets us into trouble is not what we don't know, its what we know for sure" - Mark Twain

25362 (Chemical)
23 May 08 0:40

The formula for liquid horsepower (Ph) expressed in kW, is OK.

Pshaft (aka brake horsepower) is wrong. The efficiency should be pump's effy. not motor's.

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!

Back To Forum

Close Box

Join Eng-Tips® Today!

Join your peers on the Internet's largest technical engineering professional community.
It's easy to join and it's free.

Here's Why Members Love Eng-Tips Forums:

Register now while it's still free!

Already a member? Close this window and log in.

Join Us             Close