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Ah, I wasn't familiar with the term "specific weight" = rho * g. So yes, I follow your equation now.

I got my degree in ME from California Maritime back in 2004 and took the FE exam that fall. I worked for a company doing automation and emission research on large reciprocating engines. None of that industry had any need for PE licensing so I forgot about it. for a while  A few years ago I got a...

I think the word "water" was left in by mistake and that the author intended it to read as 60 ft of pump head.

SacMe24, that equation is correct and equivalent to mine, but my point is that you need to use the actual fluid head. The head pressure in the problem is given in feet of WATER. The actual fluid head is 60 / .7 = 85.7 ft which cancels out the SG term in the above equation so 200 gpm * 85.7 / 0.7...

I don't know where you are getting that equation. In your equation you would end up with units of Ft^4 / s. I have the FET supplemental handbook in front of me. It says Wdot = rho * g * H * Q / N. Rho * g * H = Pressure, thus Wdot = P * H /N. Or another way, H would be the actual feet of head...

Justin, If the problem had stated simply, "feet of head", then I would agree with you. The problem stated "Feet of water", thus the pressure across the pump is enough to support a column of water 60 ft high. The density of the fluid that is doing the pushing is irrelevant . The term "feet of...

Question 132: A centrifugal pump is sized to deliver 200 gpm of liquid with a specific gravity of 0.7 and a total differential head of 60 ft of water. The hydraulic horsepower required is most nearly: A 2.1 B 2.7 C 3.0 D 4.3 In my reasoning, the pump head is 60 ft OF WATER which is equal...