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About ValonaBrau

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  • Birthday 12/03/1981

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  1. Ah, I wasn't familiar with the term "specific weight" = rho * g. So yes, I follow your equation now.
  2. 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 job at the local wastewater plant as an instrument technician (better pay, less responsibility), but they require a PE for any engineering positions, even in the controls department so I decided to bite the bullet and finally get mine. I originally wanted to take the CSE, but after trying and giving up in despair on the practice exam I found the thermal fluid systems much easier. I'm signed up for the October exam, so I might hit this board up for some tips. I've got 14 years experience with engines, wastewater, PLC's, software development, database work, instrumentation and much more, so I'd be glad to help anywhere I can. -Mike Takacs
  3. I think the word "water" was left in by mistake and that the author intended it to read as 60 ft of pump head.
  4. 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 = 3.03.
  5. 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, H= Hw / SG = rhoW * Hw / rho -> Wdot = rhoW *g *Hw * Q
  6. 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 water" is a unit of pressure (eg. see this calculator: The actual head on the pump is 60 / .7 = 85.7 ft. ME_VT, that equation is incorrect. Power = Force * Velocity. Force = pressure * area, Velocity is Volumetric flow /area. Therefore Power = Pressure * Volumetric Flow. Mass does not appear in the equation.
  7. 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 to 26 psi. Pump work is pressure times volumetric flow rate, so specific gravity doesn't enter in to the equation. The book answer, however, multiplies the head by specific gravity to get 2.1. If they had specified the head was simply 60 ft, I would agree, but in specifying the head in feet water they are specifying the pressure regardless of the fluid density. Anyone have an opinion?
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