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posted by  EngAve on 9/15/2008 2:15:45 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 (specific weight = 133.02 kN/m3). If the pressure at B is 207 kPa, find the pressure at A.
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posted by EEHW on 9/16/2008 12:33:47 AM  |  status: Live
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EngAve's comment:
"That's the answer I had too. Great! Thanks."
Response Details:
First, draw horizontal line(in red below) at the lowest interface of Mercury P1 and connect along the line to P2 , then write the equations: going up (-), going down(+)
 
P1 = PB-γwater(h) >>>P1 = 207kPa-12.4kn/m3(5m)
P2= PA-γwater(h)  + γHg (h) >>> P2= PA-8.4kn/m3(4m) + 133.02kn/m3(.4m)
P1= P2
207kPa-62kPa = PA-20.16kPa + 53.208kPa
Answer PA=111.95kPa
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