Problem 3.14

# Problem 3.14 - Problem 3.14 Given Find[Difficulty 3...

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Problem 3.14 [Difficulty: 3] Air H D Air H – y y y Given: Cylindrical cup lowered slowly beneath pool surface Find: Expression for y in terms of h and H. Plot y/H vs. h/H. Solution: Governing Equations: dp dh ρ g = (Hydrostatic Pressure - h is positive downwards) pV MR T = (Ideal Gas Equation) Assumptions: (1) Constant temperature compression of air inside cup (2) Static liquid (3) Incompressible liquid First we apply the ideal gas equation (at constant temperature) for the pressure of the air in the cup: constant = Therefore: p a π 4 D 2 H = p π 4 D 2 Hy () = and upon simplification: p a H pH y = Now we look at the hydrostatic pressure equation for the pressure exerted by the water. Since ρ is constant, we integrate: pp a ρ g hy = at the water-air interface in the cup. Since the cup is submerged to a depth of h, these pressures must be equal: p a H p a ρ g + = p a H p a y ρ g + = Explanding out the right hand side of this expression: 0p a y ρ g + = ρ g h H ρ g h y

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## This note was uploaded on 10/19/2011 for the course EGN 3353C taught by Professor Lear during the Fall '07 term at University of Florida.

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Problem 3.14 - Problem 3.14 Given Find[Difficulty 3...

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