First off, I'm a CompE so I have no idea why I let myself get talked into taking this class but now that I am I might as well try to do decent. I have a homework due tomorrow and I think I have most of the answers correct but this problem is really chapping my ass.

"Determine the minimum size motor (HP) required to steadily pump 3 liters/sec of 20 C water from an lnlet pressure of 100kpa to a discharge pressure of 1000kpa, neglecting changes in kinetic and potential energy."

I can generally figure things out for myself if I find an example somewhere but it seems that I can't find one for this particular problem. Any help would be greatly appreciated. Thanks!

try finding the enthalpy value at the first and second state, then use that big equation, i forget if its the first or second law.

Ein-Eout= Q-W + Sum(E ins) -Sum (E outs)

Its steady state, so your E's turn into zeros, and has no heat transfer, so Q is zero; you are left with:

Sum(E ins) = Sum(E outs) -W

turns out to be
M(h1+v^2/2+gz) = M(h2 + V^2/2 + gz) - W
where M is the mass flow rate (M dot)
considering neglecting kinetic and potential, your big terms turn just into the h1 and h2.

So finally you are left with: W= M(h1)-M(h2)

hope that helps, thats how im looking at solving it.
Also, if its constant specific heats, since the input and output M(flow rate) is the same, you can use M(Cp)(change in T) to replace the m(h1)-m(h2). terms. (but prolly not in this problem cause you were given pressures)

man i am sorry but that looks really hard, good luck

That's basically how I did it. Just convert your L/sec to kg/sec and then use that value for your "M". Then, instead of "W" you will get "W/time" which is power in kW. also, h = u + Pv. They give you the "P's" and you can assume v and u stay constant since the water never vaporizes (pressures are too high).

Thanks a lot guys!

although i didnt help you are welcome