Sunday, April 15, 2007
HW #5, P6 - Pump Horsepower Requirment - 6 pts
The pump shown here increases the pressure in liquid water from 200 to 4000 kPa. What is the minimum horsepower motor required to drive the pump for a flow rate of 0.1 m3 /s ?
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8 comments:
what exactly is D?
D is the diameter of the of the cross sectional area through which the liquid water is flowing into the pump.
Does the temperature of water not matter for this problem? And if not, why not?
Anon & Sparticus:
D = diameter.
Thank you Sparticus.
Anon:
Please come to the problem sessions like the one on Friday.
Assume the water is an incompressible fluid. That means Vhat does not change with P. Since no T is given look at saturated liquid T = 25 degC for Vhat. Look at other temperatures as well...say 10 degC and 60 degC. Vhat doesn't change much. So, for this problem, just assume Vhat of the liquid is constant. Furthre, assume that T does not change significantly from the inlet to the outlet of the pump. These assumptions let you calculate the shaft work for the pump from Wsh = - Vhat * deltaP.
I can't decide which delta V to use. Because we do not know the temperature of the liquid, only that it is subcooled, we cannot accurately gauge what Vhat should be, only a range. Also, should we approach this with Wsh = -Vhat * deltaP or -mdot*delta Hhat? Any hints?
I can't get "Wsh = - Vhat * deltaP".
Wsh = - mdot*delta Hhat
= - Vdot / Vhat *(deltaU + deltaP*Vhat)
= -Vdot*deltaP if we assume delta T is small, and therefore U, as a function of T only, changes very little.
Is this correct?
Questioning:
Just use Vhat at 25 degC. Vhat doesn't change too much with T.
Changes in kinetic energy ARE important here !
deltaHhat = deltaUhat + delta(PVhat)
But deltaHhat = 0 if we assume Tin = Tout and incompressible fluid.
So:
deltaHhat = delta(PVhat)
But Vhat = constant for the incompressible fluid, so:
deltaHhat = Vhat * deltaP
So, your last post WOULD be correct if it included the deltaEkin term.
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