# Manometer problem

1. Jun 25, 2014

### MrMechanic

1. The problem statement, all variables and given/known data
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

2. Relevant equations
Pressure = pgh

3. 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. Jun 25, 2014

### Staff: Mentor

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

3. Jun 25, 2014

### MrMechanic

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

4. Jun 25, 2014

### MrMechanic

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. Jun 25, 2014

### Staff: Mentor

The left column is filled with either air or water above the Hg. The right colunm is filled with air above the mercury.
Yes.
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

6. Jun 25, 2014

### MrMechanic

And I don't really know what a flow line is...

7. Jun 25, 2014

### Staff: Mentor

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

8. Jun 25, 2014

### MrMechanic

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. Jun 25, 2014

### MrMechanic

(a)
Py=14.7psi + 0.488(25) = 26.9
Px=26.9-(0.072lb/ft^3 / 12^3 in^3)(15) =26.9

(b)

10. Jun 25, 2014

### MrMechanic

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. Jun 25, 2014

### Staff: Mentor

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

12. Jun 25, 2014

### MrMechanic

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. Jun 25, 2014

### Staff: Mentor

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

14. Jun 25, 2014

### MrMechanic

Yes sir. But I couldn't find a way to start using the equations that I learned from my Professor.

15. Jun 25, 2014

### MrMechanic

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. Jun 25, 2014

### Staff: Mentor

In the units you are using, $ΔP=ρ\frac{g}{g_c}h$, 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 $\frac{lb_{mass}ft}{lb_{force}sec^2}$.

Chet

17. Jun 27, 2014

### MrMechanic

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: Jun 27, 2014
18. Jun 27, 2014

### Staff: Mentor

Don't forget to also do this for the Hg.

Chet

19. Jun 27, 2014

### MrMechanic

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: Jun 27, 2014
20. Jun 27, 2014

### MrMechanic

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