# How do I find the number of floors in a building using power efficiency?

#### Chemist116

The problem is as follows:

An electrical motor of $80%$ of power efficiency requires $6\,kW$ to propel a centrifuge pump of $73.5%$ of efficiency, which the latter use it to raise the water from the basement to a water tank situated in the rooftop of the building at a rate of $0.54 \frac{m^3}{min}$. Find the number of floors that the building has if each of them has $2.5\,m$ of height. You may use $g=9.8\,\frac{m}{s^2}$.
The alternatives given in my book are as follows:

$\begin{array}{ll} 1.&14\,floors\\ 2.&16\,floors\\ 3.&12\,floors\\ 4.&10\,floors\\ 5.&10\,floors\\ \end{array}$

I'm still stuck on this problem as I don't know how should I use the information of the efficiency. The only thing which comes to my mind is this formula:

$\eta=\frac{P_{out}}{P_{in}}$

I'm assuming that "out" is the power which is produced by the process and "in" the power put in the process. But I'm confused on where should I put the percentages given and what should I do with the speed in cubic meters per minute. Can somebody help me?. Supposedly the answer is the second option. But I don't know how to get there.

#### DarnItJimImAnEngineer

$(0.80)(0.735)(6~kW)$ of power is being delivered to the water. Hydraulic power is pressure times volumetric flow rate*, where the pressure at the exit of the pump will be roughly $\rho gh$.
Don't forget to include your units and use unit conversions where necessary.

*Why is this power?
$\dot{W} = \frac{d}{dt}\int F dx = \frac{d}{dt}\int \frac{F}{A} Adx = \frac{d}{dt} \int PdV = P\dot{V}$ (steady-state)

#### skipjack

Forum Staff
Surely alternatives 4 and 5 aren't the same in the book.

#### Chemist116

$(0.80)(0.735)(6~kW)$ of power is being delivered to the water. Hydraulic power is pressure times volumetric flow rate*, where the pressure at the exit of the pump will be roughly $\rho gh$.
Don't forget to include your units and use unit conversions where necessary.

*Why is this power?
$\dot{W} = \frac{d}{dt}\int F dx = \frac{d}{dt}\int \frac{F}{A} Adx = \frac{d}{dt} \int PdV = P\dot{V}$ (steady-state)
Okay doing what you mentioned I got this:

Since what you mentioned that the power is the pressure times the volumetric flow rate I can find the height.

This would be as follows:

$(0.80)(0.735)(6\times 10^3\,W)=\rho \times g \times h \times \dot{V}$

$(0.80)(0.735)(6\times 10^3\,W)=\rho \times g \times h \times \dot{V}$

$(0.80)(0.735)(6\times 10^3\,W)=\left(1000\frac{kg}{m^3}\right) \times 9.8 \frac{m}{s^2} \times 0.54 \frac{m^3}{min}\times \frac{1\,min}{60\,s} \times h$

$h=40\,m$

Since it mentions that each floor has $2.5\,m$ then:

$40\,m\times\frac{1\,floor}{2.5\,m}=16\,floors$

which correspond to alternative two and it checks according the answers sheet.

Now I have a couple of questions:

What's the meaning of the little dot over the $W$?

Perhaps do you mean $P=\dfrac{dW}{dt}=\dot{W}$

Keep in mind I'm not very familiar with these notations of derivatives (its been a while since I taken calculus) so I'm just guessing the intended meaning.

why the little "d" is outside the integral here?, what's its meaning? (in words please )

$P=\dfrac{d}{dt}\int{Fdx}$

Should I understand

$Adx=dV$ ?

I'm assuming this is due area times the lenght, three dimentions but why $dV$ and not just $V$ is it because the $x$ is changing so the $V$ will change? (sorry if this assertion sounds dumb or obvius but I'm at this point doing my best at trying to understand).

Then I'm stuck here:

$\dfrac{d}{dt}\int{P dV}=P\dot{V}$

The pressure is constant it can go outside the integral, but when you apply the integral, this cancels the derivative for the volume,

so it will show like this

$\dfrac{d}{dt}\times P \times V$

Therefore

$P\dfrac{dV}{dt}=P\dot{V}$

Is this the justification?.

Again, I'm just confused about that little "d" at the beginning and its meaning:

I'm assuming since Power is defined as the Work per unit of time and Work is the change in mechanical energy (am I right with this?) then what the numerator is telling me, the infinitesimal change in the mechanical energy (work) with respect the infinitesimal change in time, therefore the little "d" represents that. Again did I got the clear picture?. Can you help me to clear out these concepts?

There's one thing: In this problem I was given two percentages of efficiency, you multiplied them, is this because are they additive? (I don't know if this is the right mathematical term, but should I proceed in the same way if I am given a series of succesive percentages in efficiency?).

Why
did you added the words (steady state) at the end of $P\dot{V}$. I'm assuming that is it an equilibrium?, the same quantity that does in is also out?. Can you help me to clear out these concepts as well?

Surely alternatives 4 and 5 aren't the same in the book.
I'm sorry I copied it down in a rush. It should have said

$\begin{array}{ll} 1.&14\,floors\\ 2.&16\,floors\\ 3.&12\,floors\\ 4.&10\,floors\\ 5.&18\,floors\\ \end{array}$

While for the power, it should had displayed:

electrical motor of $80\%$ of power efficiency

centrifuge pump of $73.5\%$ of efficiency

I forgot to include a character to let mathcode understand it as a normal letter and not a special character.

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#### DarnItJimImAnEngineer

Yes $\dot{W}=dW/dt$ is power. I tend to use $\dot{W}$ because $P$ is already being used for pressure.

$\frac{d}{dt} \Theta$ is the same as $\frac{d\Theta}{dt}$ ($\Theta$ is a generic placeholder here). Power is the rate of work done over time elapsed, and work is $\int Fdx$.

If the water flows by infinitesimal distance $dx$ in a pipe of cross-sectional area $A$, then this corresponds to an infinitesimal volume $dV=Adx$ of water flowing. Hence flow work = $\int PdV$. If it is steady-state (nothing changing with time), then pressure is constant at the pump, so flow work becomes $W=P\int dV = P\Delta V$, or the pressure times the volume of water pushed. Take the time derivative (again, pressure being constant), and you get $\dot{W} = P \frac{dV}{dt}$, or power is the pressure times the volumetric flowrate.

$\eta_{motor}=80\%$ means 80 % of the electrical power (6 kW) being supplied to the motor is transferred to the shaft.
$\eta_P=73.5\%$ means 73.5 % of this power from the shaft is transferred to moving the water.