Answer to Question #193825 in Math for Pavan

Question

9 Water flows through a pipe AB 1.2m diameter at 3m/s and then passes through a pipe BC of diameter 1.5m. At C, the pipe branches. Branch CD is 0.8m in diameter and carries one third of the flow in AB. The flow velocity in branch CE is 2.5m/s. find the volume rate of flow in AB, the velocity in BC, the velocity in CD and diameter of CE

Answer 1

a. Discharge in $AB:Q_{AB}=\nu_{AB }S_{AB}=3m/s\cdot (\dfrac{\pi(1.2m)^2}{4})$

$$=3.3929\ m^3/s$$

b. Velocity in $BC:\nu_{BC }=\dfrac{Q_{BC}}{S_{BC}}=\dfrac{Q_{AB}}{\dfrac{\pi(1.5m)^2}{4}}$

$$=\dfrac{3.3929\ m^3/s}{\dfrac{\pi(1.5m)^2}{4}}=1.92m/s$$

c. Velocity in $CD:\nu_{CD }=\dfrac{Q_{CD}}{S_{CD}}=\dfrac{\dfrac{1}{3}Q_{AB}}{\dfrac{\pi(0.8m)^2}{4}}$

$$=\dfrac{3.3929\ m^3/s}{3\cdot\dfrac{\pi(0.8m)^2}{4}}=2.25m/s$$

d. Diameter of $CE:\nu_{CE }=\dfrac{Q_{CE}}{S_{CE}}=\dfrac{\dfrac{2}{3}Q_{AB}}{\dfrac{\pi(D_{CE})^2}{4}}$

$$D_{CE}=\sqrt{\dfrac{\dfrac{8}{3}Q_{AB}}{\pi\nu_{CE}}}=\sqrt{\dfrac{\dfrac{8}{3}(3.3929\ m^3/s)}{\pi\cdot2.5m/s}}$$

$$=1.0733\ m$$