# Calculating power of a pump

#### Johrnak

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? 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 $$\displaystyle dV=c\cdot A\cdot L$$ 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?

#### benit13

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

• Johrnak

#### Johrnak

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!

• benit13