Pump Performance net s Part A Calculate the pump head A water pump is used to pu

Question

Pump Performance net s Part A Calculate the pump head A water pump is used to punp water from a reservoir into a tank. The pipe is L 100 mlong and dameter D=10 rthe flow rate is Q=01 m/s.what is the requred purp head? Usefs 2x 10-2 for the fnction factor. The minor losses total K 32·the elevaton change is 10 m . and the i and exit are at the same pressure Express your answer in m to three significant figures Learning Goal: To use a pump performance curve to calculate the low rate that will be delivered and the required pump power Pumps are fairly complex systems, so their performance is often measured by experiment and reported as a graph of the pump head versus the flow rate. As the pressure bebween the outlet and inlet changes, the flow rate produced by the pump will change There are very specific experiments for measuring the pump performance so thet diferent pumps can be compared more easily Hints To use pump performance curves for a system, we can use the energy equabion Submit My Answers Give Up For a constant diameter pipe, the velocity is constant if the inlet and outlet are both at atmosphenic pressure then those two terms can be eliminated Under these conditions, the equation reduces to Part B Calculate the performance A pump is being evaluated for the applicaion described in Part A The performance of the pump is gi by the graph below What flow rate will be produced by this pump? 1000 750 500 250 Recall that the head loss for a pipeishyf the loss "or attrgishl-K1 The actual pump power required to operate a pump is going to be larger than the ideal power that would be indicated by the fow rate, pump head, and specific weight of the fluid. This diference is reported as the pump efficiency 08 016 024

Explanation / Answer

Aim : Caluculate hpump

Answer:PART A

For the constant diameter pipe - velocity, inlet and oulet atmpressure are also constant then eqution required is:

zin +hpump = Zout + hL ........1(zin =zout)

hl (pipe) = f (L/D) v 2/ 2g.

=.02(100/0.1)v2 / 2g or fLQ2 / D3g.

= 0.02 *100 * (.1)2 /0.13 *9.8

= 2.04 m

From equation 1 we get,

  hpump = hl. (Elevation inlet = Elevation outlet)

hpump = 2.04 m

PART B

Flow rate = V / T (m3/sec)

From the graph it clears that for the 0.24m3 flowrate the pump head is 875m and performance of pump is gradually increased by flowrate.

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