| 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 ? |
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| Assume the liquid water
is incompressible and
that its specific volume
is equal to that of saturated liquid at 25oC. |
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| 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. |
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| Diagram: |
The diagram in the
problem statement is adequate. |
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| Given: |
P1 |
100 |
kPa |
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D1 |
4 |
cm |
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P2 |
6000 |
kPa |
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0.040 |
m |
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Vdot |
25 |
L/s |
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D2 |
6 |
cm |
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0.025 |
m3/s |
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0.060 |
m |
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| Find: |
Ws |
??? |
hP |
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| Assumptions: |
1 - |
The pump operates adiabatically and nearly isothermally. |
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2 - |
Changes in potential energy are negligible. |
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3- |
Water behaves as an incompressible fluid in this process. |
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| Equations
/ Data / Solve: |
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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. |
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Eqn 1 |
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Eqn 2 |
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The problem is that we
cannot lookup the specific enthalpy and we do not know the mass flow rate. |
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We can use the definition of enthalpy to work around this : |
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Eqn 3 |
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For an incompressible liquid, U = fxn(T) only. Since we assumed T1 = T2, DU = 0. |
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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. |
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Eqn 4 |
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Eqn
4 is a pretty simple result, but we still cannot evaluate the specific volume. |
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Now, let's consider
the kinetic energy term : |
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Eqn 5 |
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Velocity is related to the volumetric flow
rate and the cross-sectional
area for flow by : |
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Eqn 6 |
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where : |
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Eqn 7 |
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We can evaluate A1 and A2 : |
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A1 |
0.0012566 |
m2 |
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A2 |
0.0028274 |
m2 |
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Now, we can use Eqn 6 and then Eqn 5 to determine the velocities and the change in the specific kinetic energy : |
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v1 |
19.89 |
m/s |
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v2 |
8.84 |
m/s |
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DEkin |
-158.803 |
J/kg |
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Now, we need to think
about the relationship between mass flow rate, volumetric flow rate and the specific volume. |
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Eqn 8 |
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or : |
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Eqn 9 |
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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 : |
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Eqn 10 |
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Unfortunately, in the
end we still need to assume a
value for the specific volume. |
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We will use the value
of specific volume of saturated liquid water at 25oC : |
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V |
0.001003 |
m3/kg |
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The, we can use Eqn 9 to determine mdot : |
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mdot |
24.93 |
kg/s |
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Finally, we can plug
numbers into Eqn 10 to
evaluate Ws : |
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Ws |
-143.54 |
kW |
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Now, all we need to to
do is convert to units of horsepower, hP : |
1 hp = |
745.7 |
W |
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Therefore : |
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Ws |
-192.49 |
hP |
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| Verify: |
None of the assumptions
made in this problem solution can be verified. |
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| Answers : |
Ws |
-192 |
hP |
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