Answer:
The pump delivers 32.737 kilowatts to the water.
Explanation:
We can describe the system by applying the Principle of Energy Conservation and the Work-Energy Theorem, the pump system, which works at steady state and changes due to temperature are neglected, is represented by the following model:
(Eq. 2)
Where:
- Mass flow, measured in kilograms per second.
- Gravitational acceleration, measured in meters per square second.
, - Initial and final heights, measured in meters.
, - Initial and final flow speeds at pump nozzles, measured in meters per second.
, - Initial and final internal energies, measured in joules per kilogram.
, - Initial and final pressures, measured in pascals.
, - Initial and final specific volumes, measured in cubic meters per kilogram.
Then, we get this expression:
(Ec. 3)
We note that specific volume is the reciprocal of density:
(Ec. 4)
Where is the density of water, measured in kilograms per cubic meter.
The initial pressure of water (), measured in pascals, can be found by Hydrostatics:
(Ec. 5)
Where:
- Atmospheric pressure, measured in pascals.
- Depth of the entrance of the inlet pipe with respect to the limit of the water reservoir.
If we know that , , and , then:
And the specific volume of water (), measured in cubic meters per kilogram, is: ()
The power losses due to friction is found by this expression:
Where is the total friction head loss, measured in meters.
The mass flow is obtained by this:
(Ec. 6)
Where is the volumetric flow, measured in cubic meters per second.
If we know that and , then:
Then, the power loss due to friction is: ()
Now, we calculate the inlet and outlet speed by this formula:
(Ec. 7)
Inlet nozzle (, )
Oulet nozzle (, )
(, , , , , , , , )
The pump delivers 32.737 kilowatts to the water.