Question #193825

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


Expert's answer

a. Discharge in AB:QAB=νABSAB=3m/s(π(1.2m)24)AB:Q_{AB}=\nu_{AB }S_{AB}=3m/s\cdot (\dfrac{\pi(1.2m)^2}{4})



=3.3929 m3/s=3.3929\ m^3/s

b. Velocity in BC:νBC=QBCSBC=QABπ(1.5m)24BC:\nu_{BC }=\dfrac{Q_{BC}}{S_{BC}}=\dfrac{Q_{AB}}{\dfrac{\pi(1.5m)^2}{4}}



=3.3929 m3/sπ(1.5m)24=1.92m/s=\dfrac{3.3929\ m^3/s}{\dfrac{\pi(1.5m)^2}{4}}=1.92m/s

c. Velocity in CD:νCD=QCDSCD=13QABπ(0.8m)24CD:\nu_{CD }=\dfrac{Q_{CD}}{S_{CD}}=\dfrac{\dfrac{1}{3}Q_{AB}}{\dfrac{\pi(0.8m)^2}{4}}



=3.3929 m3/s3π(0.8m)24=2.25m/s=\dfrac{3.3929\ m^3/s}{3\cdot\dfrac{\pi(0.8m)^2}{4}}=2.25m/s

d. Diameter of CE:νCE=QCESCE=23QABπ(DCE)24CE:\nu_{CE }=\dfrac{Q_{CE}}{S_{CE}}=\dfrac{\dfrac{2}{3}Q_{AB}}{\dfrac{\pi(D_{CE})^2}{4}}



DCE=83QABπνCE=83(3.3929 m3/s)π2.5m/sD_{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=1.0733\ m


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