Can you help me understand this modification to a formula?

[renobueno]

Homework Statement

Its not exactly a homework. And the other threads weren't as suitable for the question.

Relevant Equations

https://imgur.com/a/I80VcFO and https://imgur.com/a/mfC62Io

Hello guys, I did a efficiency analysis with two different propellers. For that I build up an experiment to make measurements.
The experiment is build up like this:

You can see a prop inside the tube which after turning on will pump water from one side to the other. After a certain duration while pumping a new water level will arise on one side of the tube. I think the new water level is explained by that at some point the hydro static pressure and water weight is to big for the pump to overcome.
To calculate the efficiency I found a equation for water pumps. But the equation was designed for pumps which pump water in two different basins in a fixed height. The formula and a sketch:

But my experiment is in one system and not in two different that is why I went to my teacher for help. He modified the equation by dividing H by two:

He told me that in this experiment H (Head) and Q (flowrate) are proportional that is why we need to divide H by 2.
And here is the problem. I dont get why we can just cut a physical size in halve because of proportionality.

I hope you can help me. Thx in advance!
Just ask if you need more information about something.
Btw Im new to this forum so pls tell me if I did something wrong and what I can do better :)

 

[haruspex]

Surely your difficulty with both equations is that in your setup Q=0, giving η=0.
This implies that those equations, as a way of determining efficiency, are context dependent, i.e. they tell you how efficient the pump is at a specific task. If you give any pump a sufficiently hard task it will achieve nothing, so have zero efficiency.
What you need is a context-free way of defining the efficiency of a pump, but I'm not sure there is one. Different pumps in your setup would achieve different heights of water on the output side, but I see no way to express that as efficiency.

 

[renobueno]

Surely your difficulty with both equations is that in your setup Q=0, giving η=0.
This implies that those equations, as a way of determining efficiency, are context dependent, i.e. they tell you how efficient the pump is at a specific task. If you give any pump a sufficiently hard task it will achieve nothing, so have zero efficiency.
What you need is a context-free way of defining the efficiency of a pump, but I'm not sure there is one. Different pumps in your setup would achieve different heights of water on the output side, but I see no way to express that as efficiency.

Thank you for your answer. I think I understand what you are saying, but the thing is that my teacher and I have already agreed on using this specific modified formula. The main problem is now for me to understand his changes to it. I need to be able to explain why anyone would halve the H, but I feel like I dont have the expertise in math/physics to see why he did what he did. :(

 

[Steve4Physics]

Hello @renobueno. Welcome to PF.

Can I add to what @haruspex has said.

Your equation applies to the continuous flow of liquid being pumped upwards at a steady rate (Q) for some time (t). But that’s not what your diagram/description show. I’m guessing that you haven’t described the system correctly.

Maybe something like this is going on:

You have water (say) with initial height h=0 (e.g. the surface of a very large lake).

The water is pumped into a tall vertical cylinder next to the lake. Inside the cylinder the water fills the space between h=0 and h=H. We then have a vertical column of water of height H.

If the total mass of water in the cylinder is m, then the potential energy supplied to raise the water is mg(H/2), not mgH.

That’s because the centre of mass of the water has been raised to height H/2. Half of the water in the cylinder is above H/2 and half is below H/2.

Does that help?

 

[erobz]

Sorry, but has already been stated... the formula is nonsense. There is no power output. efficiency is:

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

Why your teacher halved the $H$ is because the center of mass of the column is at $H/2$. It doesn't make it good though.

 

[Steve4Physics]

If the column height changes the mass changes too?

Yes. For a column of liquid, density $\rho$, cross-sectional area $A$ and height $H$:

$m = \rho A H$

$PE = mg(\frac H2) = \rho A H g \frac H2 = \frac 12 \rho A H ^2.$

EDIT; Or as a function of mass: $PE = \frac {m^2g}{2 \rho A}$

But I don’t think any of this relates directly to the OP’s problem.

(BTW, w.r.t. Post #5, it's OK too define efficiency here as a ratio of energies - it doesn't have to be a ratio of powers. The OP's formula - if used in the correct context - is the ratio of PE-gained to electrical-energy-supplied-to-the-motor. This is the efficiency of the process.)

 

[erobz]

So In the case of the OP, the power output of the pump ( ignoring rate of change of kinetic energy over time period $T$

$$ \dot W_p = (P_2 -P_1)Q $$

$P_2 = \rho g ( H + h)$
$P_1 = \rho g ( H - h)$
$Q = A \dot h$

Thus;

$$ \dot W_p = 2 \rho A g h \dot h $$

It follows that for this set up:

$$W_p = \rho A g h^2 $$

And the efficiency over this transient period is a function of time:

$$ \eta = \frac{W_p}{W_{in}} = \frac{\rho A g h^2}{ \int_0^T V(t)I(t) dt } $$

(BTW, w.r.t. Post #5, it's OK too define efficiency here as a ratio of energies - it doesn't have to be a ratio of powers. The OP's formula - if used in the correct context - is the ratio of PE-gained to electrical-energy-supplied-to-the-motor. This is the efficiency of the process.)

I agree, over the (short) period of time that the process occurs. As time goes on efficiency tends to 0.

 

[Steve4Physics]

View attachment 337904
So In the case of the OP, the power output of the pump ( ignoring rate of change of kinetic energy over time period $T$
$$ \dot W_p = (P_2 -P_1)Q $$
$P_2 = \rho g ( H + h)$
$P_1 = \rho g ( H - h)$
$Q = A \dot h$
Thus;
$$ \dot W_p = 2 \rho A g h \dot h $$
It follows that for this set up:
$$W_p = \rho A g h^2 $$
And the efficiency over this transient period is a function of time:
$$ \eta = \frac{W_p}{W_{in}} = \frac{\rho A g h^2}{ \int_0^T V(t)I(t) dt } $$
I agree, over the (short) period of time that the process occurs. As time goes on efficiency tends to 0.

I'm still not clear if the OP is asking about pumping liquid along a U-tube, or about something else. It seems a peculiar experiment.

@renobueno, if you are still reading this thread, can you clarify?

 

[erobz]

I'm still not clear if the OP is asking about pumping liquid along a U-tube, or about something else. It seems a peculiar experiment.

@renobueno, if you are still reading this thread, can you clarify?

The diagram (and supporting text) seems clear enough to me, however peculiar it may seem.

 

[haruspex]

BTW, w.r.t. Post #5, it's OK too define efficiency here as a ratio of energies - it doesn't have to be a ratio of powers. The OP's formula - if used in the correct context - is the ratio of PE-gained to electrical-energy-supplied-to-the-motor. This is the efficiency of the process.

If I understand post #1 correctly, the pump, continuing to run, just maintains the height H on the output side, with Q=0. So no specific quantity of energy has been used - it depends how long the pump has been running.
Ok, you could judge how long it takes to reach height H, but, clearly, the efficiency (measured using Q) has been declining to that point.

Can you measure the efficiency of an electric motor by stalling it? That only tells you its max torque.

 

[Steve4Physics]

If I understand post #1 correctly, the pump, continuing to run, just maintains the height H on the output side, with Q=0. So no specific quantity of energy has been used - it depends how long the pump has been running.
Ok, you could judge how long it takes to reach height H, but, clearly, the efficiency (measured using Q) has been declining to that point.

Can you measure the efficiency of an electric motor by stalling it? That only tells you its max torque.

It appears that the OP is trying to measure propellor 'efficiency' with an inappropriate method/formula.

 

[erobz]

It appears that the OP is trying to measure propellor 'efficiency' with an inappropriate method/formula.

Perhaps an interesting theoretical problem, but yeah...surely a poorly designed experiment.

 

[haruspex]

It appears that the OP is trying to measure propellor 'efficiency' with an inappropriate method/formula.

But what is an appropriate method?
Suppose the ascending water can escape at a height h<H, and a flow speed v is achieved.
By Bernoulli, $\frac 12v^2=\frac{\Delta p}\rho-gh$, where $\Delta p$ is the pressure difference created by the pump.
The useful power is $P_{out}=\Delta p Av$, where A is the cross-sectional area.
$=\Delta p A\sqrt{2\frac{\Delta p}\rho-2gh}$.
So the measured efficiency is a function of h.

Ok, I have treated $\Delta p$ as independent of v, which it probably isn't, but you see the problem.

 

[Steve4Physics]

But what is an appropriate method?
Suppose the ascending water can escape at a height h<H, and a flow speed v is achieved.
By Bernoulli, $\frac 12v^2=\frac{\Delta p}\rho-gh$, where $\Delta p$ is the pressure difference created by the pump.
The useful power is $P_{out}=\Delta p Av$, where A is the cross-sectional area.
$=\Delta p A\sqrt{2\frac{\Delta p}\rho-2gh}$.
So the measured efficiency is a function of h.

Yes. Your approach uses a steady-state, so that power in and out are constant. The Post #1 method fails on that count.

 

[renobueno]

Oh my day THANK YOU ALL! I didn't expect to receive so much help!