Solve Pressure at X & Y in Manometer Problem

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In summary, the pressure at points X and Y can be found by using the equation P = (0.072) x (30/32.2) (40/12) for air and P = (62.1) x (30/32.2) (40/12) for water, and then adding the pressures found to the atmospheric pressure of 14.7 psi. For part (c), where the local gravity is 30 ft/s^2, the pressure at points X and Y can be found by using the equation ΔP=(0.488)(30/32.2)x(25), and then adding this pressure to the atmospheric pressure of 14.7 psi. The resulting pressure at points X and
  • #1
MrMechanic
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Homework Statement


A simple mercury manometer connected into a flow line gives readings as shown in the figure. Local gravity is standard and the mercury density is 0.488lb/in^3. Find the pressure at points X and Y when the flow line and left leg contain (a) air whose density is 0.072lb/ft^3, (b) water whose density is 62.1lb/ft^3. (c) Answer (a) and (b) if the local gravity is g= 30ft/s^2.


ANSWER: (a) 26.90, 26.90, (b) 25.46, 26.90, (c) 26.10 psia


Homework Equations


Pressure = pgh


The Attempt at a Solution


I've done getting Py but I can't find Py at (a)
Also I've tried solving (b) but i can't solve it actually.
Here's my solution on Py at (a)
Py = 14.7psi + (0.072lb/ft^3 / 12^3 in^3)(15) + (0.488lb/in^3)(25) =26.90psi
 

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  • #2
Your equation for py at (a) for the left column filled with air is incorrect. The "air term" should not be there. If the left column is filled with water, the corresponding "water term" should not be there either. The air and water columns only come in when you are trying to get the pressure at point x...and then they should be subtracted.

Chet
 
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  • #3
Thanks for replying. The left column is filled water. While the right column is filled with air?
Do you mean that I should use Py = 14.7psi + 0.488(25) = 26.9
then how about Px?
 
  • #4
I'm thinking and re-reading the problem stated above. I just noticed this "Find the pressure at points X and Y WHEN THE FLOW LINE AND THE LEFT LEG CONTAIN" So does that mean ...The left leg & the right leg contain a specific condition each letter (a and b) right?
 
  • #5
MrMechanic said:
Thanks for replying. The left column is filled water. While the right column is filled with air?
The left column is filled with either air or water above the Hg. The right colunm is filled with air above the mercury.
Do you mean that I should use Py = 14.7psi + 0.488(25) = 26.9
Yes.
then how about Px?
For (a), Px=Py-(0.072lb/ft^3 / 12^3 in^3)(15)
For (b), Px=Py-(62.1lb/ft^3 / 12^3 in^3)(15)

Chet
 
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  • #6
And I don't really know what a flow line is...
 
  • #7
MrMechanic said:
And I don't really know what a flow line is...
I think that circle with the x inside it represents a pipe running into the paper. Water or air is flowing through that pipe.

Chet
 
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  • #8
I see. Wow. Now I know what you mean when you said that my Py is incorrect. Thanks.
and for (b) i don't really know where did they put the water. So I'm totally confused. I'm done with (a). Thanks for helping Sir Chet!
 
  • #9
(a)
Py=14.7psi + 0.488(25) = 26.9
Px=26.9-(0.072lb/ft^3 / 12^3 in^3)(15) =26.9

(b)
Py = 14.7 + (62.1/12^3)(15) + (0.488x25)?? <--- not really sure about this.
 
  • #10
at (b) i think i got Px but I'm not really sure about it.
Px = Py - (62.1/12^3 x 40) =25.46

I got Py using what we did above.
 
  • #11
MrMechanic said:
at (b) i think i got Px but I'm not really sure about it.
Px = Py - (62.1/12^3 x 40) =25.46

I got Py using what we did above.
Yes. It isn't clear from the diagram whether it should be a 40 or a 15. Visually, if the diagram is close to scale, the 40 is the correct number to use.

Chet
 
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  • #12
Thanks for clearing that up! Could you explain to me what should I do on (c). (a) and (b) if the local gravity is g= 30ft/s^2. I haven't use the local gravity yet...
 
  • #13
MrMechanic said:
Thanks for clearing that up! Could you explain to me what should I do on (c). (a) and (b) if the local gravity is g= 30ft/s^2. I haven't use the local gravity yet...
Well, normal gravity is 32.2 ft/sec^2. From your equations for the hydrostatic pressure variation, how would the hydrostatic contributions change if g were only 30? Have you learned how to work with the "american" system of units, and with using gc=32.2?

Chet
 
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  • #14
Yes sir. But I couldn't find a way to start using the equations that I learned from my Professor.
 
  • #15
How would I start (c) ... If the gravity changed to 30ft/sec^2 then what would be the starting equation to get 26.10psia.
 
  • #16
In the units you are using, [itex]ΔP=ρ\frac{g}{g_c}h[/itex], where ΔP is in lbforce/in2, ρ is the density in lbmass/in^3, g is the local acceleration of gravity (ft/sec2), h is the height of the column in inches, and g_c is the constant 32.2 [itex]\frac{lb_{mass}ft}{lb_{force}sec^2}[/itex].

Chet
 
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  • #17
Hmmm. Let me know if this is the correct equation sir...
P = (0.072) x (30/32.2) (40/12)
when i get that should I make another equation for Water?
P = (62.1) x (30/32.2) (40/12)
and sum what I got? is it the correct procedure?
Since,
Pabs = Patm + Pgage i think.
 
Last edited:
  • #18
MrMechanic said:
Hmmm. Let me know if this is the correct equation sir...
P = (0.072) x (30/32.2) (40/12)
when i get that should I make another equation for Water?
P = (62.1) x (30/32.2) (40/12)
and sum what I got? is it the correct procedure?
Since,
Pabs = Patm + Pgage i think.
Don't forget to also do this for the Hg.

Chet
 
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  • #19
Pabs = Pair + PH20 + PHg + Patm?

So i got this...

Pair = 0.02lb/ft^3 /12^3 x (30/32.2) (40) = 1/644 psi
PH20 = 62.1 / 12^3 x (30/32.2)(40)= 1.339psi
PHg = 0.488 x (30/32) (40) = 18.19psi

Is it wrong?
 
Last edited:
  • #20
Well, I kinda worked out the formula you gave me.
So, here is my solution to (c) ΔP=(0.488)(30/32.2)x(25) = 11.38 lb/in^2
Pgage = 11.38
Pabs = Patm + Pgage
Pabs = 14.7 + 11.38
Pabs = 26.08 psia
 
  • #21
MrMechanic said:
Well, I kinda worked out the formula you gave me.
So, here is my solution to (c) ΔP=(0.488)(30/32.2)x(25) = 11.38 lb/in^2
Pgage = 11.38
Pabs = Patm + Pgage
Pabs = 14.7 + 11.38
Pabs = 26.08 psia
Good job.

Chet
 
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1. What is a manometer and how does it work?

A manometer is a scientific instrument used to measure pressure. It works by comparing the pressure of a gas or liquid in a closed system to the atmospheric pressure. This is done by using a U-shaped tube filled with a liquid, typically mercury, and measuring the difference in liquid levels between the two sides of the tube.

2. How do you solve for pressure at a specific point in a manometer?

To solve for pressure at a specific point in a manometer, you will need to use the equation P = ρgh, where P is pressure, ρ is the density of the liquid in the manometer, g is the acceleration due to gravity, and h is the difference in liquid levels between the two sides of the manometer at the specific point you are solving for.

3. What are the units of measurement for pressure in a manometer?

The units of measurement for pressure in a manometer are typically in pascals (Pa) or millimeters of mercury (mmHg). However, the specific units will depend on the system of measurement being used.

4. Can a manometer be used to measure pressure in both gases and liquids?

Yes, a manometer can be used to measure pressure in both gases and liquids. The type of liquid used in the manometer may vary depending on the substance being measured, but the principle remains the same.

5. How does the position of the liquid in a manometer relate to the pressure in the system being measured?

The position of the liquid in a manometer is directly related to the pressure in the system being measured. The higher the liquid level on one side of the tube, the higher the pressure in that part of the system. The difference in liquid levels between the two sides of the manometer is used to calculate the pressure at a specific point in the system.

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