Find Pressure at Point A in the Manometer

  • #1
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Homework Statement


Determine the pressure of the water in pipe A shown in the figure (p2.56) if the gage pressure of the air in the tank is 2 psi.

I have attached the image


Homework Equations





The Attempt at a Solution



Figure P2.56

I drew a jumpline at 2ft above point A because everything below point A has the same density. From here I assigned P1 to account for the pressure above the jumpline in the tank and p2 to account for the pressure in the pipe adjacent to it.

P1 = P + ρH2O*3ft

P2 = P1 + ρH2O* 4 ft

PA = P2 + ρSG*4ft

PA = ρSG*4ft + ρH2O*4ft + P + ρH2O*3ft

γH2O = 62.4 lb/ft3 taken from textbook

SG = ρfluid[email protected]°

ρSG = 0.9*1.940 slug/ft3

PA = 4ft(.9*1.940 slugs/ft3) + 62.4 lb/ft3 + 2lb/in[SUP2[/SUP]*144in2/1ft2 + 62.4lb/ft3*3ft

PA = 731.784 lb/ft2

My answer should be 525 lb/ft2

I'm not sure where I've gone wrong. I realize that I have slugs in my calculation witht he pressure of the fluid with the given SG but I'm not sure how to deal with them. Any suggestions would be appreciated.
 

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Answers and Replies

  • #2
gneill
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There's unequal lengths of the mystery fluid in the "U", so I don't think you can count on the fluid pressure at the bottom of the tank being the same as the pressure at point A, even though they are at the same elevation and the same fluid is below your "jumpline" in each.

Instead, start at a point of known pressure and "walk" the path to point A. The surface of the water in the tank would be a good place to start.

attachment.php?attachmentid=56950&stc=1&d=1363875841.gif


Note that you can count on the pressure being the same across the top of the "U" at the height of the water surface, since the same fluid is continuous along the path from one side of the "U" to the other.
 

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