Calculating power of a pump

Feb 2020
2
1
Sweden
Having some troubles understanding physical work is calculated in hydraulics.

Consider this problem:

Assume no friction and similar. Also use \(\displaystyle g = 10 \, [m/s^2] \)

A pump takes water from an underground reservoir (100 kPa pressure), pumps it through a pipe up to a fountain. The flow speed in the pipe (\(\displaystyle c\)) is 1 m/s, the pipe's diameter is 1 dm and the absolute pressure (\(\displaystyle p_{pipe}\)) is 200 kPa. The fountain sprouts the water 4 m into the air \(\displaystyle (h)\) from the nozzle and the absolute pressure here \(\displaystyle p_{atm}\) is 100 kPa. What is the power of the pump?

problem.png

Solution 1: Using conservation of energy we can figure out the speed of the water as it leaves the fountain's nozzle: \(\displaystyle mgh=\frac{mv^2}{2} \Rightarrow v=\sqrt{2gh} \approx \sqrt{2\cdot 10\cdot 4}=\sqrt{80}\quad [m/s]\)
We can use Bernoulli's equation to figure out how far the top of the water spout is above the reservoir level: \(\displaystyle p_{pipe}+\frac{\rho c^2 }{2}+g\rho z_{pipe}=p_{atm}+\frac{\rho v^2 }{2}+g\rho z_{nozzle}\)
Here \(\displaystyle z_{pipe}\) is taken to be zero, \(\displaystyle z_{nozzle}\) is the height level of the nozzle (from the pipe level), and \(\displaystyle \rho= 1000 \, [kg/m^2]\) being the density of water.
We get: \(\displaystyle 200000+\frac{1000 \cdot 1^2 }{2}=100000+\frac{1000 \cdot \sqrt{80}^2}{2}+10\cdot 1000\cdot\rho z_{nozzle} \Rightarrow z_{nozzle}= 6.05 \, [m]\)
We conclude that the pump is pumping water a total distance of: \(\displaystyle z_{nozzle}+h = 6.05+4 = 10.05 \, [m]\)
Now calculating the flow rate: \(\displaystyle Q=c\cdot \frac{D^2}{4}\cdot \pi\)
Since the pump essentially just moves the water from the reservoir and drops it 10.05 m in the air the power is given by: \(\displaystyle P=Q\cdot\rho g\cdot 10.05 \approx 789 \, [W] \)

Solution 2: Consider a volume of water \(\displaystyle dV\) that flows through the pump each second. This cylindrical volume of water has crossarea \(\displaystyle A\) and length \(\displaystyle L\) and thus \(\displaystyle dV=c\cdot A\cdot L\)
problem1.png
The work done on this volume of water is given by: \(\displaystyle \Delta W = \Delta W_{pump}+p_{atm}\cdot A\cdot L-p_{pipe}\cdot A \cdot L\)
We also know that: \(\displaystyle \Delta W = \Delta E_k + \Delta E_p\)
Here since the water wasn't moving before the pump: \(\displaystyle \Delta E_k = \frac{dV\cdot \rho c^2}{2}-0\)
and since the pipe doesn't change in height before and after the pump: \(\displaystyle \Delta E_p = 0\)
Taking this together we obtain: \(\displaystyle \Delta W_{pump}+p_{atm}\cdot A\cdot L-p_{pipe}\cdot A \cdot L =\frac{dV\cdot \rho c^2}{2} \equiv \Delta W_{pump}= A\cdot L \left(\frac{\rho c^2}{2}+p_{pipe}-p_{atm}\right)\approx 789 \, [W] \)

Solution 3: Using the formula for hydraulic power we obtain: \(\displaystyle P=Q\cdot (p_{pipe-}-p_{atm}) \approx 785 \, [W]\)


Since solution 3 is most likely the correct one (just applying a known formula) there must be something wrong in solution 1 and 2. What is the error?
 
Oct 2017
676
348
Glasgow
I think solution 3 neglects the dynamic pressure.

Including the dynamic pressure:
\(\displaystyle P = Q \cdot (\frac{\rho c^2}{2} + p_{pipe} - p_{atm})\)
\(\displaystyle = 0.0025 \pi (500 + 200000 - 100000)\)
\(\displaystyle = 789.325\) W
 
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Feb 2020
2
1
Sweden
I think solution 3 neglects the dynamic pressure.

Including the dynamic pressure:
\(\displaystyle P = Q \cdot (\frac{\rho c^2}{2} + p_{pipe} - p_{atm})\)
\(\displaystyle = 0.0025 \pi (500 + 200000 - 100000)\)
\(\displaystyle = 789.325\) W
I think I get it. That formula with \(\displaystyle P=Q\dot \Delta_{pressure}\) (which you can find on almost every page where hydraulic power is discussed) uses the total pressure difference.

This is great. Thank you!
 
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