# Example Problem with Complete Solution

5C-8 : Pump Horsepower Requirment 6 pts
The pump shown below increases the pressure in liquid water from 100 kPa to 6000 kPa. What is the minimum horsepower motor required to drive the pump for a flow rate of 25 L/s ? Assume the liquid water is incompressible and that its specific volume is equal to that of saturated liquid at 25oC.

Read : The minimum horsepower that a motor must supply to this pump is the value of Ws that we can determine by applying the 1st Law.  We can assume that heat transfer and changes in potential energy are negligible.  We still have a problem because without any temperature data, we cannot look up the properties of the water in the Steam Tables.  However, if we assume that the liquid water is incompressible, the problem gets much simpler.  For starters, the volumetric flow rate in and out of the pump must be equal at steady-state.  If we also assume that the temperature of the water does not change significantly in the process, then DU = 0 because U of an incompressible liquid is a function of temperature only.  These assumptions and the relationship between mass flow rate, volumetric flow rate and specific volume will dramatically simplify the 1st Law and allow us to evaluate Ws.  Unfortunately, we will still need to assume a value for the specific volume.
Diagram: The diagram in the problem statement is adequate.
Given: P1 100 kPa D1 4 cm
P2 6000 kPa 0.040 m
Vdot 25 L/s D2 6 cm
0.025 m3/s 0.060 m
Find: Ws ??? hP
Assumptions: 1 - The pump operates adiabatically and nearly isothermally.
2 - Changes in potential energy are negligible.
3- Water behaves as an incompressible fluid in this process.
Equations / Data / Solve:
Write the 1st Law for the pump, assuming that changes in potential energy are negligible.  This makes sense because we have no elevation data.  Also, assume the pump is adiabatic, Qpump = 0. Eqn 1 Eqn 2
The problem is that we cannot lookup the specific enthalpy and we do not know the mass flow rate.
We can use the definition of enthalpy to work around this : Eqn 3
For an incompressible liquid, U = fxn(T) only.  Since we assumed T1 = T2, DU = 0.
Also, specific volume is a constant for an incompressible liquid at constant temperature, so V pops out of the D brackets in the last term of Egn 3. Eqn 4
Eqn 4 is a pretty simple result, but we still cannot evaluate the specific volume.
Now, let's consider the kinetic energy term : Eqn 5
Velocity is related to the volumetric flow rate and the cross-sectional area for flow by : Eqn 6 where : Eqn 7
We can evaluate A1 and A2 : A1 0.0012566 m2
A2 0.0028274 m2
Now, we can use Eqn 6 and then Eqn 5 to determine the velocities and the change in the specific kinetic energy :
v1 19.89 m/s
v2 8.84 m/s DEkin -158.803 J/kg
Now, we need to think about the relationship between mass flow rate, volumetric flow rate and the specific volume. Eqn 8 or : Eqn 9
So, for an isothermal, adiabatic pump working on an incompressible fluid, with negligible changes in potential energy, the 1st Law simplifies from Eqn 2 to : Eqn 10
Unfortunately, in the end we still need to assume a value for the specific volume.
We will use the value of specific volume of saturated liquid water at 25oC :
V 0.001003 m3/kg
The, we can use Eqn 9 to determine mdot : mdot 24.93 kg/s
Finally, we can plug numbers into Eqn 10 to evaluate Ws : Ws -143.54 kW
Now, all we need to to do is convert to units of horsepower, hP : 1 hp = 745.7 W
Therefore : Ws -192.49 hP
Verify: None of the assumptions made in this problem solution can be verified. 