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posted by  EngAve on 9/13/2008 12:49:31 PM  |  status: Closed  

Fluid Pressure - Mamometer

Course Textbook Chapter Problem
Fluid Mechanics N/A N/A N/A
Question Details:
In the following figure, the specific weight of liquid A is 8.4 kN/m3, and the specific weight of liquid B is 12.4 kN/m3. Manometer liquid M is mercury (g = 133.02 kN/m3). If the pressure at B is 207 kPa, find the pressure at A.
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Sage
Karma Points: 4,373
(IIT Delhi)
posted by higgs boson on 9/13/2008 1:17:18 PM  |  status: Live
Asker's Rating: Somewhat Helpful   
EngAve's comment:
"thanks, but are you assuming that the fluids in A & B have the same specific weight?"
Response Details:
Pressures at the same horizontal level must be equal, by pascals principle.
PB=PA+pressure due to liquid B between horizontal levels c and d

207 kPa=PA+hρg=PA+3*12.4 kPa (ρg=12.4 kN/m3 is given)
So, PA=169.8 kPa
Η α ρ ι
Sage
Karma Points: 4,373
(IIT Delhi)
posted by higgs boson on 9/15/2008 4:22:56 AM  |  status: Live
Asker's Rating: N/A   
EngAve's comment:
"you're ignoring the specific weight of A in your calculations - it has to be involved in there to get the right answer - I will find out the correct answer once my homework's returned - if you're right, I will reinstane the lifesaver rating."
Response Details:
Well, I did not make any such assumption..
I dont thin you dont need specific gravity of A, if I understand the question correctly..

You checked the answer? Is it correct?
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Sage
Karma Points: 4,373
(IIT Delhi)
posted by higgs boson on 9/16/2008 1:53:31 PM  |  status: Live
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EngAve's comment:
"Will do."
Response Details:
Got your papers back?
If this answer is wrong, do let me know!
Η α ρ ι
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posted by smily on 9/18/2008 1:53:27 AM  |  status: Live
Asker's Rating: Lifesaver   
EngAve's comment:
"that is correct - thank you"
Response Details:
P(A)= P(B)-γ(B)*h-γ(h)+γ(A)*h
P(A)= 207kPa-(12.4kN/m^3)*(5m)-(133.02kN/m^3)*(.4m)+(8.4kN/m^3)*(2.4m)
P(A)=111.95kPa

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