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

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?

• Why would $h$ change over time? Assume a stationary flow across the pipes. What happens to the water that is in the pipes going from time $t$ to time $t + \Delta t$? What's the variation of energy of this column of water? Commented Apr 27, 2017 at 22:35
• @Phoenix87 thanks for your answer. yes, but isn't each piece of the water going upwards? So, by definition, wouldn't dh/dt be equal to the velocity that each "piece of water" is going upward? Commented Apr 27, 2017 at 22:41
• Are you familiar with the open system (control volume) version of the first law of thermodynamics? Commented Apr 27, 2017 at 22:56
• @ChesterMiller yes, I am, and that's why I need your help, because I don't understand it very well. I know how to solve this problem correctly. But I don't know if P=dW/dt= d(KE+Pe)/dt, or if P=dE/dt=d(KE+PE+W)/dt.. I don't understand Commented Apr 28, 2017 at 0:00
• Are you saying that you don't understand the derivation of the open system version of the first law, or are you saying that, even though you understand its derivation, you are unable to apply it to problems? Commented Apr 28, 2017 at 0:44

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.
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: $$=g\cdot h \cdot \frac{dm}{dt}$$ $$\therefore \space =g \cdot h \cdot ρ\cdot \frac {dV}{dt}$$