Manometer Oil Problem: How to Calculate Water Level Difference | Expert Help

In summary, you tried to calculate the change in the oil levels of a manometer with a pressure differential, but got an incorrect result because you did not use the given pipe diameters.
  • #1
*Alice*
26
0
Dear all,
would anyone be able to help me with the following problem

In the manometer shown in the figure, the reservoir diameters are 50 mm and the tube diameters are 10 mm. The upper fluid is an oil of density 850 kg/m3, and the lower fluid (hatched) is water. Initially the levels of oil in the two reservoirs are the same, and the levels of water in the two tubes are the same. If a pressure differential of 200 Pa is applied to the air above the oil in the reservoirs, what is then the difference in water levels?
PICTURE SEE ATTACHED FILE

Basically, I tried to calculate the change in height on the top rhs and then calculated the height difference at the water on the right hand side since the relation

h1 = h2 (d/D)^2

holds. due to the symmetry h on the lhs should be the same and hence dh=2h
which gives me the wrong result.

Can anyone please give me a hint? THat would be amazing! Thanks a lot!
 

Attachments

  • Manometer.jpg
    Manometer.jpg
    6 KB · Views: 510
Physics news on Phys.org
  • #2
You have not defined h, h1, h2, d, or D and they are not in your picture. You have not said anything about how you calculated the change in the oil levels. Be more explicit, and show what you did to get your answer.
 
  • #3
Set up 2 equations for the pressure on the two sides of the manometer.

One equation on top of the oil in the right arm - with the applied pressure differential. Call the height of the oil in the left arm [itex]h_o[/itex].

Set up another equation at the top of the water level in the right arm of the manometer. Call the water height in the left arm [itex]h_w[/itex]. Call the height of the oil in the right arm a and in the left arm b.

Find a relation between a,b,ho and hw. More exact you want a relation for a - b.
 
Last edited:
  • #4
Please see attached the definitions.

pressure at C = pressure at D

p(C) = p(D)

p(D) = 1000g(x-h1) + h(2)*g*1000 + 200Pa

p(C) = 850g*25h(1) + (h(2)-25h(1))g1000 + (x + h(1))*1000g

now p(C) = p(D)

and solve for h(1):

200 - 1000gh(1) = 1000gh(1) - 2500gh(1) + 2150gh(1)

h(1) = 1.09mm

therefore the difference at the water levels is

h(1)*25*2 = 54.5 mm

which is wrong

correct answer: 111mmm


Can anyone see the mistake? Thanks so much!

[Edit: sorry just saw the second reply with the definitions]
 

Attachments

  • Manometer definitions.jpg
    Manometer definitions.jpg
    20 KB · Views: 590
Last edited:
  • #5
Note slight change to previous post.
 
  • #6
Were are C and D located in the manometer diagram?
 
  • #7
C and D are at the equilibrium levels on the rhs and lhs respectively.
Basically I though that the level of the water below the equilibrium level on the rhs must be (50/10)^2=25 times the height difference h1 at the large diameter on the top (see attached file in the above post)
 
Last edited:
  • #8
Note change inpost #3 again (sorry it was a bit confusing). I do not understand why you multiply the heights with the radia of the reserviour for the oil. That will make the pressure equations invalid!
 
  • #9
sorry, I'm confused. What do you mean by height of the oil exactly, i.e. from which point to which?
Also shall I call the height of the oil on the left arm h0 or a (you ay two different things in your post).

I thought that it was 25 h1, since the volume change should be the same. Since the density is constant and only the diameter changes I thought that

h1 (top, rhs) = 5^2*h1 = h2(bottom rhs)

But anyway, I shall try your attempt - if you could let me know the exact definitions first that would be awesome and I will give it a go!
Thanks a lot!
 
  • #10
Are you trying to set up the equations that I suggested?
 
  • #11
The oil level in the left arm is a height [itex]h_o[/itex] above that in the right arm.

The water level in the left arm is a height [itex]h_w[/itex] above that in the right arm.

Both these heights are of cause with the applied pressure differential.
 
  • #12
The height of the oil above the water in the left arm is b and in the right arm a - again with the pressure differential in effect.
 
  • #13
Thank you. This is what I did:

relation 0:

b + h(w) = a + h(0)

equation 1:

200 = h(0)*850*g

This h(0) = 0.024

equation 2:

850g*a = h(w)*g*1000 + b*850*g

(a-b)*850g = h(w)*9810
a-b = h(w)*1.18


PUT IN RELATIONSHIP 0:

a-b = h(w) - h(0)
0.18h(w) = -h(0)
h(w) = -133mm


The correct result is 111mm - so is that just a matter of rounding or did I do something wrong? Also, why did I not have to use the given pipe diameters?
Again, cheers for your help!
 
  • #14
850g*a = h(w)*g*1000 + b*850*g

You forgot the pressure on top.

It also seems I misled you a bit...

The pressure in manometer is the same at the bottom of the manometer. As you move up in the two arms the pressures can be compared on the right and the left if you stay in the same fluid. So the equation above is fine (with the correction), but the comparison on top of the oil surface in the two arms is invalid. As soon as one move into different media up in the liquid the pressure cannot be compared any more due to the different densities of the media.

How to solve the problem then?

The volume of water moved in the tube will be the same as the volume of oil moved in the reserviour. This gives you another relation between [itex]h_o[/itex] and [itex]h_w[/itex].
 
Last edited:
  • #15
*Alice* said:
h(w) = -133mm

The correct result is 111mm - so is that just a matter of rounding or did I do something wrong? Also, why did I not have to use the given pipe diameters?
Again, cheers for your help!
You do need the diameters because they affect the total depth of oil on each side. The pressure difference between the top and bottom surface of the oil is proportional to the total depth (resevior plus tube)

andrevdh has made the crucial point I think you are missing.

andrevdh said:
The pressure in manometer is the same at the bottom of the manometer. As you move up in the two arms the pressures can be compared on the right and the left if you stay in the same fluid. So the equation above is fine (with the correction). As soon as one move into different media up in the liquid the pressure cannot be compared any more due to the different densities of the media.
The pressures are equal on both sides at any level that has water on both sides. The pressure difference between the top and bottom surface of oil on each side is proportional to the total depth of oil. The total oil depths are different on the two sides, and the left side has an additional depth of water.

You could use the top of the water on the right as a reference level where you know the pressures in both tubes are equal.

On the left side, this reference pressure is
Pleft = pressure change from depth of water + pressure change from depth of oil + pressure of gas on left

On the right side it is just
Pright = pressure change from depth of oil + pressure of gas on right

Pleft=Pright

I think you understand everything else you need.
 
Last edited:

1. How does a manometer measure pressure?

A manometer is a device used to measure pressure by comparing the pressure of a fluid (usually a liquid) to the pressure of a gas. It does this by using a U-shaped tube filled with the fluid and connected to the gas source. The difference in the levels of the fluid in the two sides of the tube indicates the pressure difference between the gas and the atmosphere.

2. What is the purpose of using oil in a manometer?

The oil in a manometer serves as the fluid that is used to measure pressure. It is typically used because it has a low volatility and does not evaporate easily, making it a reliable and stable fluid for pressure measurements.

3. What causes the oil in a manometer to rise or fall?

The oil in a manometer rises or falls due to changes in pressure. When the pressure in the gas source increases, it pushes down on the oil, causing it to rise on the other side of the tube. Similarly, when the pressure decreases, the oil on the other side of the tube will fall.

4. How can I determine the pressure using a manometer?

To determine the pressure using a manometer, you must measure the difference in height between the two sides of the U-shaped tube. This difference in height, known as the "manometer reading", can be converted to pressure using a simple equation that takes into account the density of the fluid and the acceleration due to gravity.

5. What are the limitations of using a manometer for pressure measurements?

While manometers are a simple and reliable method for measuring pressure, they do have some limitations. They are not suitable for measuring highly corrosive or toxic gases, as the fluid in the manometer may become contaminated. They also have a limited range of measurement and may not be accurate for very high or very low pressures.

Similar threads

  • Introductory Physics Homework Help
Replies
11
Views
5K
  • Classical Physics
Replies
24
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
5K
  • Introductory Physics Homework Help
Replies
15
Views
8K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
16
Views
5K
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
1K
Back
Top