# How do I find the minimum power of a motor required to pump water from a well?

#### Chemist116

The problem is as follows:

A pump propels water from a well which has a depth of $20\,m$ so that will be poured to a canal with a speed of $8\,\frac{m}{s}$. The volume which is to be pumped is $0.5$ Liters per second. Find the minimum power of the motor.

The alternatives given in my book are as follows:

$\begin{array}{ll} 1.&32\,W\\ 2.&100\,W\\ 3.&76\,W\\ 4.&80\,W\\ 5.&50\,W\\ \end{array}$

I'm confused about how to find the least power. What I thought to solve this problem was should I use the density of water so as to find the mass and from then calculate the power?. Can somebody help me with this question?.

#### skeeter

Math Team
1 liter of water has a mass of 1 kg $\implies$ 0.5 L/sec is equivalent to a mass flow rate of 0.5 kg/sec

$P = \dfrac{W}{t} = \dfrac{mg \Delta h}{t} = \dfrac{m}{t} \cdot g \cdot \Delta h = (0.5 \text{ kg/sec}) \cdot (10 \text{ m/sec}^2) \cdot (20 \text{m}) = 100 \text{ W}$

#### DarnItJimImAnEngineer

Wait, the pump is imparting both gravitational potential energy and kinetic energy to the water. Shouldn't the power be...
$\dot{W}=\dot{m} \left[ g\Delta h + \frac{v^2}{2}\right]=(0.5~kg/s)[(9.81~m/s^2)(20~m)+(0.5)(8~m/s)^2]=114~W$?

#### skeeter

Math Team
Thought that at first also, but a you-tube video I viewed on the subject said relatively low flow rates can ignore KE and 114W isn't one of the choices ... go figure.

#### DarnItJimImAnEngineer

You can ignore low velocities and shallow heights in comparison to most changes in temperature because the values are so much smaller. There are no temperature changes here. All we have is a low velocity and a shallow height; we can't automatically ignore either.

My only other thought is if they meant the canal was moving at 8 m/s, but not the flow from the well. That would bring us back to the ~100 W assuming the well channel were sufficiently wide.

#### skeeter

Math Team
To be honest, I haven't dealt with flow rate problems in quite a long time; that's why I had to research it a bit before posting a possible solution. Glad that you're the engineer ... #### DarnItJimImAnEngineer

It doesn't hurt that I taught thermodynamics for years.

Raising something ten stories (30 m) takes $\Delta pe = gh = 0.29 ~\frac{kJ}{kg}$ of energy.
Accelerating something to highway speeds (30 m/s) takes $\Delta ke = \frac{v^2}{2} = 0.45 ~\frac{kJ}{kg}$ of energy.
Heating water by 1 °C takes $\Delta h = C_p \Delta T = 4.19 \frac{kJ}{kg}$ of energy.
Hence why we often (but not always) can neglect kinetic and potential energy terms.