A converging elbow (see Fig. P12.20) turns water through an angle of 135 degree

Question

A converging elbow (see Fig. P12.20) turns water through an angle of 135 degree in a vertical plane. The flow cross-sectional diameter is 400 mm at the elbow inlet, section (1), and 200 mm at the elbow outlet, section (2). The elbow flow passage volume is 0.2 m3 between sections (1) and (2). The water volumetric flow rate is 0.4 m3/s and the elbow inlet and outlet pressures are 150 kPa and 90 kPa. The elbow mass is 12 kg. Calculate the horizontal (x direction) and vertical (z direction) anchoring forces required to hold the elbow in place. Figure PI 2.20

Explanation / Answer

Velocity V1 = Q/A1 = 0.4/(3.14/4*0.4^2) = 3.185 m/s

Velocity V2 = Q/A2 = 0.4/(3.14/4*0.2^2) = 12.739 m/s

Mass flow rate m = density*volume flow rate = 1000*0.4 = 400 kg/s

Fx = Momentum loss in x-direction + Force due to pressure = [m*V1 - m*(V2*Cos135)] + (P1*A1 - P2*A2*Cos135)

Fx = [400*3.185 - 400*(12.739*Cos135)] + 150*10^3 *(3.14/4*0.4^2) - 90*10^3 *(3.14/4*0.2^2) *Cos135

Fx = 25715.4 N = 25.715 kN

Hence Fx will be = -25.715 kN

Fz = Momentum loss in z-direction + Force due to pressure + weight of fluid in elbow + weight of elbow

Fz = m*V2*Sin135 + P2*A2*Sin135 - (0.2*1000 + 12)*9.81

Fz = 400*(12.739*sin135) + 90*10^3 *(3.14/4*0.2^2) *Sin135 - (0.2*1000 + 12)*9.81

Fz = 3521.7 N

Fz = -3.521 kN

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