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
The Attempt at a Solution
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/ρ[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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