Thread: Pumps and heat
View Single Post
Unread 07-18-2002, 02:57 PM   #62
bigben2k
Responsible for 2%
of all the posts here.
 
bigben2k's Avatar
 
Join Date: May 2002
Location: Texas, U.S.A.
Posts: 8,302
Default

Quote:
Originally posted by myv65
As the old commercial says, "Sorry Charlie". You can't get there from here. You need delta-P.

It may help to take a gander at the units in their base form. Energy has units of force * length. Power has units of force * length / time. Flow has units of length^3 / time.

In order to get power based on flow, the units need to come out right. What are the missing units? Well, divide power by flow and you have (force * length / time) / (length^3 / time) = force / length^2. Recognize that? What if I said lbf / in^2 (psi)?

To calculate flow power you require flow rate and pressure rise. No two ways around it.
Of course, and I've even said that myself, in another thread, where I wrote something along the lines of 100 gph at 10 psi, isn't the same as 100 gph at 100 psi.

So I need to know the pressure drop between the pump inlet and outlet, as well as the effective flow rate.
bigben2k is offline   Reply With Quote