Suppose the top surface of the vessel in the figure is subjected to an external

Question

Suppose the top surface of the vessel in the figure is subjected to an external gauge pressure P2.

Suppose the top surface of the vessel in the figure is subjected to an external gauge pressure P2. Derive a formula for the speed, v1, at which the liquid flows from the opening at the bottom into atmospheric pressure, PA. Assume the velocity of the liquid surface, v2, is approximately zero. (Use PA for PA, P2 for P2, v2 for v2, y1 for y1, y2 for y2, rho for ? and g for the acceleration due to gravity, as necessary.)

Explanation / Answer

Assuming you are allowed to take Bernoulli's theorem as given you can write for the liquid surface and the liquid as it flows out of the vessel

v1^2/2 + P1/Rho1 + y1g = v2^2/2 + P2/Rho2 + y2g

Rho is the density and is unchanged for water so Rho1 = Rho2 = Rho.
We are told that the velocity of the liquid surface v2 can be neglected, so v2 = 0.
Also the outlet is a gauge pressure and the pressure at the surface of the water is a certain gauge pressure. Because the density, Rho, is constant we can write gauge pressure for P1 and P2 and P1 = 0.

The Bernoulli expression thus simplifies to:
v1^2/2 + y1g = P2/Rho + y2g

Rearranging for v1:

v1 = [(y2-y1)g + P2/Rho]^(1/2)

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