A bent pipe is attached at point B to a support. The pipe does not move. Liquid

Question

A bent pipe is attached at point B to a support. The pipe does not move. Liquid enters at port 1 at a pressure pi and velocity Vi, and exits at port 2 with pressure p2 and velocity V2. The pipe cross-sectional area is A1 at port 1 and A2 at port 2. The fluid density ? is constant. Distances hand hare given, as 6 shown in the figure. The process is at steady state. Assume properties are uniform across the pipe cross- section. There are no body forces. Derive an expression for the torque T provided by the support wall at point B in terms of above-given parameters. Also assume that the external pressure surrounding the pipe is zero, and that no torques are transmitted within the pipe walls at ports 1 and 2. Show all work. P1 P2 2

Explanation / Answer

Let us assume that the fluid is incompressible and no accumulation of fluid inside the pipe.

So, Flow rate, q = A1V1 = A2V2

Let assume that the bent pipe with fluid is rotating at B.

The resultant torque acting on a rotating fluid = rate of change of moment of momentum

Moment of momentum per second at section 1 = mass*velocity*h1 = (?*q)* V1* h1 (clockwise)

Moment of momentum per second at section 2 = mass*velocity*h2 = (?*q)* V2* h2 (clockwise)

Hence, the resultant torque = Rate of change of moment of momentum = (?*q)* V1* h1 + (?*q)* V2* h2

= (?*q)* (V1* h1 + V2* h2) = (?* A1*V1)* (V1* h1 + V2* h2) (clockwise)

But the pipe with fluid is not rotating.

Hence, the support B will provide a resisting torque, T = (?* A1*V1)* (V1* h1 + V2* h2) (anti clockwise)

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