Question #149094
A 40 cm diameter pipe, conveying water, branches into two pipes of diameters 30 cm and 20 cm respectively. If the average velocity in the 40 cm diameter pipe is 3 m/s. Find the discharge in this pipe. Also determine the velocity in 20 cm pipe if the average velocity in 30 cm diameter pipe is 2 m/s.
1
Expert's answer
2020-12-08T06:37:36-0500


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=30 cm=0.3 md_2=30\ cm=0.3\ m

C/S Area of pipe 2 A2=π(d1)24=0.0225π m2A_2 =\dfrac{\pi(d_1)^2}{4}=0.0225\pi\ m^2


Diameter of pipe 3 d3=20 cm=0.2 md_3=20\ cm=0.2\ m

C/S Area of pipe 3 A3=π(d3)24=0.01π m2A_3 =\dfrac{\pi(d_3)^2}{4}=0.01\pi\ m^2


Discharge through pipe 1 Q1=A1v1=0.04π m2(3 m/s)Q_1=A_1v_1=0.04\pi\ m^2 (3\ m/s)

=0.12π m3/s=0.3770 m3/s=0.12\pi\ m^3/s=0.3770\ m^3/s

Q1=0.3770 m3/sQ_1=0.3770\ m^3/s


Velocity at pipe 2 v2=2 m/sv_2=2\ m/s

Velocity at pipe 3 v3=?v_3=?


From continuity Q1=Q2+Q3Q_1=Q_2+Q_3

Discharge through pipe 2 Q2=A2v2=0.0225π m2(2 m/s)Q_2=A_2v_2=0.0225\pi\ m^2 (2\ m/s)

=0.045π m3/s=0.1414 m3/s=0.045\pi\ m^3/s=0.1414\ m^3/s


Discharge through pipe 3 Q3=Q1Q2=0.12π m3/s0.045π m3/sQ_3=Q_1-Q_2=0.12\pi\ m^3/s-0.045\pi\ m^3/s

=0.075π m3/s=0.075\pi\ m^3/s


Velocity at pipe 3 v3=Q3A3=0.075π m3/s0.01π m2=7.5 m/sv_3=\dfrac{Q_3}{A_3}=\dfrac{0.075\pi\ m^3/s}{0.01\pi\ m^2 }=7.5\ m/s

v3=7.5 m/sv_3=7.5\ m/s



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