Pressure in a tank/pipe

Shlllink

Water flows steadily from an open tank. The elevation of point 1 is 10.0 m, and the elevation of points 2 and 3 is 2.00 m. The cross-sectional area at point 2 is .048 m^2; at point 3 it is .016 m^2. The area of the tank is very large compared with the cross-sectional area of the pipe.

I already got part A which tells us that the discharge rate is .2 m^3/s

Part B asks What is the gauge pressure at point 2?

I'm familiar with Bernoulli's equation and I thought I knew how to do the problem, but some how I got it wrong enough times to bring me down to one guess, so I want to be sure about what I'm doing.

Andrew Mason

Quote by Shlllink

Water flows steadily from an open tank. The elevation of point 1 is 10.0 m, and the elevation of points 2 and 3 is 2.00 m. The cross-sectional area at point 2 is .048 m^2; at point 3 it is .016 m^2. The area of the tank is very large compared with the cross-sectional area of the pipe.

I already got part A which tells us that the discharge rate is .2 m^3/s

Part B asks What is the gauge pressure at point 2?

I'm familiar with Bernoulli's equation and I thought I knew how to do the problem, but some how I got it wrong enough times to bring me down to one guess, so I want to be sure about what I'm doing.

You will have to describe what points 1, 2 and 3 are in relation to the tank and pipe and what the cross-sectional areas refer to (eg. tank, pipe).

AM

billiards

i don't think there's enough there to tell, but if that discharge rate is for point 2 you could say that the speed of the water is due to the pressure and equate the velocity and pressure terms from the bernoulli. then solve for pressure.

v=Q/A (dischargre rate over area)

(v^2)/2=p/rho

that might work but impossible to say without a better description of question/diagram.

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billiards	i don't think there's enough there to tell, but if that discharge rate is for point 2 you could say that the speed of the water is due to the pressure and equate the velocity and pressure terms from the bernoulli. then solve for pressure. v=Q/A (dischargre rate over area) (v^2)/2=p/rho that might work but impossible to say without a better description of question/diagram.