Question #207942

A 40cm diameter pipe conveying water, branches into two pipes of diameters 20cm and 10cm respectively if the average velocity in 40cm diameter pipe is 2m, find velocity in 20cm diameter pipe, if velocity in 10cm diameter pipe is 1.5 m/s.


1
Expert's answer
2021-06-18T02:28:32-0400


Diameter of pipe 1 d1=40 cm=0.4 md_1=40\ cm=0.4\ m


C/S Area of pipe 1 A1=π(d1)24=0.04π m2A_1 =\dfrac{\pi(d_1)^2}{4}=0.04\pi\ m^2


Diameter of pipe 2 d2=20 cm=0.2 md_2=20\ cm=0.2\ m


C/S Area of pipe 2 A2=π(d1)24=0.01π m2A_2 =\dfrac{\pi(d_1)^2}{4}=0.01\pi\ m^2


Diameter of pipe 3 d3=10 cm=0.1 md_3=10\ cm=0.1\ m


C/S Area of pipe 3 A3=π(d3)24=0.0025π m2A_3 =\dfrac{\pi(d_3)^2}{4}=0.0025\pi\ m^2


Discharge through pipe 1 

Q1=A1v1=0.04π m2(2 m/s)Q_1=A_1v_1=0.04\pi\ m^2 (2\ m/s)




=0.08π m3/s=0.08\pi\ m^3/s

From continuity Q1=Q2+Q3Q_1=Q_2+Q_3


Discharge through pipe 3 

Q3=A3v3=0.0025π m2(1.5 m/s)Q_3=A_3v_3=0.0025\pi\ m^2 (1.5\ m/s)




=0.00375π m3/s=0.00375\pi\ m^3/s

Discharge through pipe 2


Q2=Q1Q3=0.08π m3/s0.00375π m3/sQ_2=Q_1-Q_3=0.08\pi\ m^3/s-0.00375\pi\ m^3/s




=0.07625π m3/s=0.07625\pi\ m^3/s



Velocity at pipe 2


v2=Q2A2=0.07625π m3/s0.01π m2=7.625 m/sv_2=\dfrac{Q_2}{A_2}=\dfrac{0.07625\pi\ m^3/s}{0.01\pi\ m^2 }=7.625\ m/s

v2=7.625 m/sv_2=7.625\ m/s



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