$(0.80)(0.735)(6~kW)$ 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:

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

Using the adequate units:

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

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

$h=40\,m$

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.