Question

A water pump delivers 3 hp of shaft power when operating. If the pressure differential between the outlet and the inlet of the pump is measured to be 1.2 psi when the flow rate is 8 ft^3/s and the changes in velocity and elevation are negligible, determine the mechanical efficiency of this pump.

Answer 1

Answer:

Mechanical Efficiency =  83.51%

Explanation:

Given Data:

Pressure difference = ΔP=1.2 Psi

Flow rate = $V=8ft^3/s$

Power of Pump = 3 hp

Required:

Mechanical Efficiency

Solution:

We will first bring the change the units of given data into SI units.

$P=1.2*6.895 = 8.274KPa\\V=8*0.00283=0.226 m^3/s\\P=3*0.746=2.238KW$

Now we will find the change in energy.

Since it is mentioned in the statement that change in elevation (potential energy) and change in velocity (Kinetic Energy) are negligible.

Thus change in energy is

$=(Mass * change in P)/density\\= \frac{M*P}{p}$

As we know that Mass = Volume x density

substituting the value

Energy = Volume * density x ΔP / density

Change in energy = Volumetric flow x ΔP

Change in energy = 0.226 x 8.274 = 1.869 KW

Now mechanical efficiency = change in energy / work done by shaft

Efficiency = 1.869 / 2.238

Efficiency = 0.8351 = 83.51%