# Fluid Mechanics: Determining Elevation in a Piezometric Tube

#### WhiteWolf98

1. Homework Statement

2. Homework Equations
$P=\rho gh$

3. The Attempt at a Solution
So this is the first time I'm doing fluid mechanics, and I'm trying my best to understand it. This was the first question, and truth to be told, I'm not very confident.

I know what gauge pressure is: the total pressure minus the atmospheric pressure. I don't know what it means in this question though, and how I'm supposed to use it.

This was my attempt at finding the elevation of the petrol:

$Pressure~due~to~air = \rho_{air} gh$
$=1.225 \times 9.81 \times 2$
$=24.0345~kPa$

$Pressure~due~to~petrol = \rho_{petrol} gh$
(The height I'll be using is the height considering everything above the tube)
$=750 \times 9.81 \times 0.75$
$=5518.125~kPa$

$Atmospheric~pressure=101.325~kPa$

$Total~pressure=24.0345+5518.125+101.325=5643.4845~kPa$

$5643.4845=\rho_{petrol} gh$
$5643.4845=750\times 9.81\times h$
$h=0.767~m~(to~3~s.f.)$

This would be the height from the base of the tube (just a little higher at the dotted line).

Like I said before, I don't know how to use that gauge value, or even if my calculations without it are correct.

Any help would be appreciated.

Thank you

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#### Doc Al

Mentor
I know what gauge pressure is: the total pressure minus the atmospheric pressure. I don't know what it means in this question though, and how I'm supposed to use it.
The cylinder is sealed and the air is under pressure. That gauge tells you what the pressure is. (If the gauge pressure were zero, what would be the pressure of the air inside the tank?)

This was my attempt at finding the elevation of the petrol:

$Pressure~due~to~air = \rho_{air} gh$
$=1.225 \times 9.81 \times 2$
$=24.0345~kPa$
(1) Your final units would be in Pa, not kPa.
(2) This ignores the given pressure of the air. (So don't do this!)

This calculation gives you the difference in pressure between the top and bottom of the air section, which is not significant here.

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#### WhiteWolf98

If the gauge read zero, then I'm guessing the pressure of the air inside the tank would also be zero.

Then, using what you've told me, I'll try again:

$Pressure~due~to~air=1.5~kPa$

$Pressure~due~to~petrol~(stays~the~same)=5.518125~kPa$

$Hence,~total~pressure=1.5+5.52=7.02~kPa~(to~3~s.f.)$

$7.02\times 10^3=750(9.81)(h)$

$h=0.954~m~(to~3~s.f.)$

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#### Doc Al

Mentor
If the gauge read zero, then I'm guessing the pressure of the air inside the tank would also be zero.
Careful! Remember that gauge pressure is total pressure minus atmospheric pressure.

Then, using what you've told me, I'll try again:

$Pressure~due~to~air=1.5~kPa$

$Pressure~due~to~petrol~(stays~the~same)=5.518125~kPa$

$Hence,~total~pressure=1.5+5.52=7.02~kPa~(to~3~s.f.)$

$7.02\times 10^3=750(9.81)(h)$

$h=0.954~m~(to~3~s.f.)$
You left off atmospheric pressure at both ends -- the tubes are open to the air, presumably -- so you should get the right answer despite using the wrong pressure from the air in the cylinder. (But you need to correct your understanding -- the atmospheric pressures will cancel out so you'll end up with the same final equation.)

#### WhiteWolf98

I understand that gauge pressure is total pressure minus atmospheric pressure. But since the tank is sealed at the top, isn't atmospheric pressure 0? And if atmospheric pressure is 0, then the total pressure would equal the gauge pressure

(As in, since it's sealed, the atmospheric pressure has no effect on the air)

#### Doc Al

Mentor
I understand that gauge pressure is total pressure minus atmospheric pressure. But since the tank is sealed at the top, isn't atmospheric pressure 0? And if atmospheric pressure is 0, then the total pressure would equal the gauge pressure

(As in, since it's sealed, the atmospheric pressure has no effect on the air)
Since the cylinder is sealed, the outside atmosphere has no effect on the air. But you can measure the pressure of the air inside the cylinder with respect to the outside atmosphere -- which is what gauge pressure is. And you are told that the pressure is greater than atmospheric pressure.

#### WhiteWolf98

I don't see how the atmospheric pressures cancel out. Including the atmospheric pressure now:

$Total~pressure=108~kPa~(to~3s.f.)$

I've added the atmospheric pressure to it because the pressure of the air given was gauge pressure. Why now do I have to subtract it again?

Wouldn't you have two lots of atmospheric pressure? One from the gauge pressure, and the other because the piezometric tube is open

#### Doc Al

Mentor
Wouldn't you have two lots of atmospheric pressure? One from the gauge pressure, and the other because the piezometric tube is open
Exactly. Which is why they cancel out. (You don't have to cancel them out, it just saves work in doing the arithmetic.)

#### WhiteWolf98

No no, as in they sum, rather than cancel out.

#### Doc Al

Mentor
No no, as in they sum, rather than cancel out.
Why would they sum?

You need to equate the pressure in the cylinder to the pressure in the tube (at the level of the bottom of the tube).

Pressure in cylinder = Atm Pressure + Gauge Pressure + ρgh1
Pressure in tube = Atm Pressure + ρgh2

When you set those equal, the Atm Pressure terms can be canceled.

#### WhiteWolf98

Ahh, I see now. I didn't know you had to equate them, that wasn't what I was doing. Thank you, I think with your help, I should be okay with the second part of the question too. You have my gratitude.