Fluid Mechanics - Pressure - U-Tube Manometer

In summary, a piston with a cross-sectional area of 0.07 m^2 is located in a cylinder filled with water and connected to an open u-tube manometer. For given heights of h1=60 mm and h=100 mm, the value of the applied force, P, acting on the piston can be calculated by solving the equation P1 + γH2O*h1 - γHg*h = 0 and then multiplying by the area of the piston. It is unclear if the air pressure on top of the mercury should be included in the calculation.
  • #1
Aerospace
39
0
A piston having a cross-sectional area of 0.07 m^2 is located in a cylinder containing water as shown in the figure. An open u-tube manometer is connected to the cylinder as shown. For h1=60 mm and h=100 mm, what is the value of the applied force, P, acting on the piston? The weight of the piston is negligible.


I'm confused because there is more than one "u"-shape in the figure, so, it's unclear to me how to proceed.

Help would be appreciated!

legend for figure :
p = piston
first weird looking h = h1
second weird looking h = h
piston is present on top of the water.
Force P downwards on piston.
 

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  • #2
The U on the right is the manometer. I've never seen a manometer filled with water on one side, but that's the way your diagram looks. Is there any air between the water and the mercury? It's not clear to me why the 60mm is given, but the height of the water relative to the top of the mercury is important
 
  • #3
Here's a scan of the piston I'm talking about.
 

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  • #4
This fact may help: Since the fluid is connected (and in hydrostatic equilibrium) the pressure at any height within the same fluid must be the same throughout.
 
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  • #5
Ok, so far this is what I've got. Let me know if I'm on the right track.

If you take the pressure at the piston as P1 then,
P1 + [tex]\gamma[/tex]H2O*h1 - [tex]\gamma[/tex]Hg*h = 0

Solve for p1 and then
P(Force)= p1*Area
 
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  • #6
Don't forget that the tube is open to the air. (But you are on the right track.)
 
  • #7
Hmmm. I did take that into account when I did -[tex]\gamma[/tex](Hg)*h
 
  • #8
Aerospace said:
Hmmm. I did take that into account when I did -[tex]\gamma[/tex](Hg)*h
That's the pressure due to the column of mercury. But what about the air on top of the mercury?
 
  • #9
well, isn't the pressure at the open end supposed to be equal to zero?
 
  • #10
Aerospace said:
well, isn't the pressure at the open end supposed to be equal to zero?
Only if it's in a vacuum.
 
  • #11
hmmm then what would it be? because my book says "at the open end the pressure is zero."
 
  • #12
Aerospace said:
hmmm then what would it be? because my book says "at the open end the pressure is zero."

This is a matter of interpretation. If the book says zero, they intend for you to find only the extra force applied to the piston that is added to the force of air pressure already acting. Air pressure is the same on both sides, so it can be neglected if all you need is the extra force.

Don't forget to calculate the force. If all you have done is caclulate the added pressure, you are not finished.
 
  • #13
so the force P = p1*A, correct?
I stated that above in my post...is this incomplete?

And in any case, if Pair was not neglected, how would you write it into the equation?
 
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  • #14
Sorry. I had missed your force calculation. That is OK.

If you had to include air pressure in the force you would need to look up the normal pressure of air (1 atmosphere) and multiply by the area of the piston to find the force the air exerts on the piston.
 
  • #15
Aerospace said:
hmmm then what would it be? because my book says "at the open end the pressure is zero."
I was interpreting the problem as the piston being enclosed but the tube open to air. (But there's no good reason to think that the piston is enclosed.) I think OlderDan is correct, that they just want you to ignore the equal air pressure on both sides.
 

1. What is fluid mechanics?

Fluid mechanics is the branch of physics that deals with the study of fluids (liquids and gases) and the forces that act upon them.

2. How does pressure affect fluid mechanics?

Pressure is a key factor in fluid mechanics as it determines the direction and magnitude of fluid flow. Changes in pressure can cause fluids to move or flow in different ways, such as through pipes or channels.

3. What is a U-tube manometer?

A U-tube manometer is a device used to measure pressure differences between two points in a fluid system. It consists of a U-shaped tube filled with a liquid (usually mercury) and attached to the system at two points. The difference in liquid levels indicates the pressure difference between the two points.

4. How does a U-tube manometer work?

The U-tube manometer works on the principle of balancing the pressure of the fluid in the system with the weight of the liquid in the manometer tube. When the pressure in the system changes, it causes a corresponding change in the liquid levels in the manometer tube, allowing for pressure to be measured.

5. What are some applications of U-tube manometers in fluid mechanics?

U-tube manometers are commonly used in various industries and fields, such as in plumbing, HVAC systems, chemical engineering, and aerospace engineering. They are also used in laboratory experiments to measure pressure in different fluid systems.

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