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with a discharge of 0.057 m 3 /sec flowing in the sudden expansion shown in the

ID: 1828578 • Letter: W

Question

with a discharge of 0.057 m3/sec flowing in the sudden expansion shown in the figure, calculate the change of pressure which occurs as the fluid flows from the small pipe to the large pipe?. The free body is chosen as shown, and the pressure acting across section 1 is assumed to be the pressure which exists in the 15.24 cm pipe.

with a discharge of 0.057 m3/sec flowing in the sudden expansion shown in the figure, calculate the change of pressure which occurs as the fluid flows from the small pipe to the large pipe?. The free body is chosen as shown, and the pressure acting across section 1 is assumed to be the pressure which exists in the 15.24 cm pipe.

Explanation / Answer

General Equation for Energy conservation: P1 / (Rho) + V1^2 / 2g = P2 /(rho) + V2^2 /2g (P1 - P2) = (rho) * (V2^2 - V1^2) / 2g Discharge = A1 * V1 0.057 = (3.1416/4)* (15.24/100)^2 * V1 V1 = 3.124 m/sec 0.057 = (3.1416/4)* (30.48/100)^2 * V2 V2 = 0.781 1/2 * ( V2^2 - V1^2 )/g = 4.574/9.81=0.466 if the fluid is water (generally considered )(rho =1000 Kg/m^3) 1/2 *(rho) * ( V2^2 - V1^2 )/g = 4.574*1000 =466.25 Kg /m2 =466.25 pascal = 0.047 metres of head