Answer to Question #149092 in Math for Usman

Question #149092
The diameter of a pipe at the section 1 and 2 are 15 cm and 20 cm respectively. Find the discharge through the pipe if velocity of water at section 1 is 4 m/s. Determine also the velocity at section 2.
1
Expert's answer
2020-12-08T06:39:13-0500

Diameter of pipe 1 d1=15cm=0.15 md_1=15 cm=0.15\ m

C/S Area of pipe 1 A1=π(d1)24=0.005625π m2A_1 =\dfrac{\pi(d_1)^2}{4}=0.005625\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


Velocity at pipe 1 v1=4 m/sv_1=4\ m/s

Velocity at pipe 2 v2=?v_2=?


Discharge through pipe 1 Q1=A1v1=0.005625π m2(4 m/s)Q_1=A_1v_1=0.005625\pi\ m^2 (4\ m/s)

=0.0225π m3/s=0.0707 m3/s=0.0225\pi\ m^3/s=0.0707\ m^3/s

Q1=0.0707 m3/sQ_1=0.0707\ m^3/s


From continuity Q1=Q2+Q3Q_1=Q_2+Q_3


Discharge through pipe 2 Q2=A2v2=0.01π m2(v2)Q_2=A_2v_2=0.01\pi\ m^2 (v_2)

=Q1=0.0225π m3/s=Q_1=0.0225\pi\ m^3/s


Velocity at pipe 2 v2=Q2A2=0.0225π m3/s0.01π m2=2.25 m/sv_2=\dfrac{Q_2}{A_2}=\dfrac{0.0225\pi\ m^3/s}{0.01\pi\ m^2 }=2.25\ m/s

v2=2.25 m/sv_2=2.25\ m/s



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