Why does potential energy time-derivative depend only on the mass flow?

Question

This is a homework exercise, but I don't want specifically to tell me how to resolve the exercise. I want you to help me what's wrong here, and why I can't do it.

Water is pumped from a lake to a storage tank 15 m above at a rate of 70 L/s while consuming 15.4 kW of electric power. Disregarding any frictional losses in the pipes and any changes in kinetic energy, determine the overall efficiency of the pump–motor unit.

$$ \begin{align} P&=\frac{dE}{dt}\\ &=\frac{d(KE+PE)}{dt}\\ &=\frac{d(KE)+d(PE)}{dt}\\ &=\frac{d(PE)}{dt}\\ &=\frac{d(mgh)}{dt}\\ &=m\cdot g \cdot \frac {dh}{dt} + g \cdot h \cdot \frac {dm}{dt}\\ &=m\cdot g \cdot \frac {dh}{dt} + g \cdot h \cdot \frac {dm}{dt}\\ &=m\cdot g \cdot \frac {dh}{dt} + g \cdot h \cdot ρ\cdot \frac {dV}{dt}\\ \end{align}$$

It is the chain rule, if they didn't say the change of kinetic energy is zero, I would do it too. I just want to know why can't be done like this. The $\frac {dh}{dt}$ would be velocity that water goes upward, and $\frac {dm}{dt}$ the mass flow.

Bonus question: Can power be defined like that? Because I only see it defined as the amount of work done. I know that work is energy, but energy isn't only on form of work, so, can it be specified like I did?

Answer 1

For your problem, $$\dot{W_P}=\dot{m}[(g z)_{top}-(gz)_{bottom}]$$where z is the elevation and $\dot{m}$ is the mass flow rate being pumped. $\dot{W_P}$ is the rate at which the pump does shaft work on the fluid.

Answer 2

First of all, about power. It is the rate of doing work, and more complicated than defining it as a rate of total energy.

Now, your working was not off. All is technically correct but $h$ does not actually change since the water behaves as a stationary column of water. If $\frac{dh}{dt}\neq0$ then this would mean that the height difference is changing.

Remember that $dh$ is similar to $\Delta h$ in that it represents a change in height of the water (in this case the height difference). We know that this is a constant at $15 \space m$, so there is no change over time. Thus $\frac{dh}{dt}=0$.

We can plug this in your last step: $$P=m\cdot g \cdot \frac {dh}{dt} + g \cdot h \cdot ρ\cdot \frac {dV}{dt}$$ $$=m\cdot g \cdot 0 + g \cdot h \cdot ρ\cdot \frac {dV}{dt}$$ $$\therefore \space =g \cdot h \cdot ρ\cdot \frac {dV}{dt}$$

Once realized that $h$ is a constant, the chain rule is no longer necessary and can get to this result from an earlier step: $$\frac{d(mgh)}{dt}$$ Factor out constants: $$=g\cdot h \cdot \frac{dm}{dt}$$ $$\therefore \space =g \cdot h \cdot ρ\cdot \frac {dV}{dt}$$