Fluid Mechanics: Determining Elevation in a Piezometric Tube

In summary, this student is trying to understand fluid mechanics, but is not very confident. They attempted to find the elevation of petrol using gauge pressure and failed. They asked for help and provided a summary of the conversation.
  • #1
WhiteWolf98
86
5

Homework Statement


Smart-Select-20190205-183326-Adobe-Acrobat.jpg
[/B]
p1XG6Qz

Homework Equations


##P=\rho gh##

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
 

Attachments

  • Smart-Select-20190205-183326-Adobe-Acrobat.jpg
    Smart-Select-20190205-183326-Adobe-Acrobat.jpg
    28.4 KB · Views: 1,061
Physics news on Phys.org
  • #2
WhiteWolf98 said:
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?)

WhiteWolf98 said:
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.
 
Last edited:
  • Like
Likes WhiteWolf98
  • #3
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.)##
 
Last edited:
  • #4
WhiteWolf98 said:
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.

WhiteWolf98 said:
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.)
 
  • Like
Likes WhiteWolf98
  • #5
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)
 
  • #6
WhiteWolf98 said:
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.
 
  • Like
Likes WhiteWolf98
  • #7
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
 
  • #8
WhiteWolf98 said:
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.)
 
  • Like
Likes WhiteWolf98
  • #9
No no, as in they sum, rather than cancel out.
 
  • #10
WhiteWolf98 said:
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.
 
  • Like
Likes WhiteWolf98
  • #11
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.
 

1. What is a piezometric tube?

A piezometric tube is a device used in fluid mechanics to measure the elevation of a fluid. It consists of a long, narrow tube that is inserted into a fluid and connected to a pressure gauge. The height of the fluid in the tube corresponds to the pressure of the fluid at that point, allowing for the determination of elevation.

2. How does a piezometric tube work?

A piezometric tube works based on the principle that the pressure of a fluid is directly proportional to its elevation. As the fluid rises in the tube, the pressure increases, and this pressure is measured by the pressure gauge. By knowing the density of the fluid, the elevation can be determined using the pressure reading.

3. What are the applications of piezometric tubes?

Piezometric tubes are commonly used in various engineering and scientific fields to measure the elevation of fluids. They are especially useful in hydrology, groundwater studies, and hydraulic engineering projects. They can also be used to measure the elevation of liquids in tanks and reservoirs.

4. Are there any limitations to using a piezometric tube?

One limitation of using a piezometric tube is that it can only be used for fluids with a relatively low viscosity, such as water. The tube may also become clogged or damaged if used with fluids containing solid particles. Additionally, the accuracy of the measurement may be affected by changes in temperature and atmospheric pressure.

5. How can the accuracy of a piezometric tube measurement be improved?

The accuracy of a piezometric tube measurement can be improved by using a well-designed and calibrated tube, ensuring the fluid is at a constant temperature, and taking multiple readings at different levels in the tube to account for any variations. Additionally, using a manometer instead of a pressure gauge can provide more precise measurements.

Similar threads

  • Advanced Physics Homework Help
Replies
5
Views
905
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
11
Views
5K
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
3K
  • Introductory Physics Homework Help
Replies
2
Views
954
  • Introductory Physics Homework Help
Replies
1
Views
912
  • Introductory Physics Homework Help
Replies
6
Views
1K
Back
Top