# Pump Laws: When Do They Apply?

**URL:** <https://boards.straightdope.com/t/pump-laws-when-do-they-apply/134630>\
**Category:** Factual Questions\
**Created:** [October 31, 2002, 2:28am UTC](https://boards.straightdope.com/t/pump-laws-when-do-they-apply/134630 "2002-10-31T02:28:52Z")\
**Posts on this page:** 2\
**Page:** 1

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**Author:** ![Jinx](https://avatars.discourse-cdn.com/v4/letter/j/c6cbf5/32.png) [@Jinx](https://boards.straightdope.com/u/Jinx)\
**Post date:** [October 31, 2002, 2:28am UTC](https://boards.straightdope.com/t/pump-laws-when-do-they-apply/134630/1 "2002-10-31T02:28:52Z")

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I can’t get the pump laws to work for me! It’s supposed to be that head decreases as flow increases for a centrifugal pump. But, Cameron Handbook states Q1/Q2 = H1^(0.5)/H2^(0.5). If you play with this, you’ll find that head decreases as flow decreases. Also, there is no accounting for impeller size.

Also, a college fluids mechanics book states Q1/Q2 = wD1^2/wD2^2 where w is rpm, but we can assume rpm is constant. I can’t get this equation to work, either. In my case, Q1 = Q2, yet the impeller diameters are different, so the feet of head (H) must be different.

In short, I have this:  
A 10" impeller delivers 176 gpm @ 350 ft. It was trimmed to some smaller diameter where it now delivers 176 gpm @ 257 ft. Ultimately, I need to know how much feet of head will this new impeller size deliver at 75 gpm based on the info here. I can add that the original 10" impeller woudl deliver 75 gpm @ 372.5 ft of head.

Can anyone help me figure this out?  
Thanks,  
Jinx

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<div class="post-metadata">

**Author:** ![Jinx](https://avatars.discourse-cdn.com/v4/letter/j/c6cbf5/32.png) [@Jinx](https://boards.straightdope.com/u/Jinx)\
**Post date:** [October 31, 2002, 2:57am UTC](https://boards.straightdope.com/t/pump-laws-when-do-they-apply/134630/2 "2002-10-31T02:57:24Z")

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Correction: The above equation should read:  
Q1/ (w1\*D1^3) = Q2/ (w2 \* D2^3)

- Jinx