Question #149096
A 30 cm diameter pipe carries oil of sp. gr. 0.8 at a velocity of 2 m/s. At another section the diameter is 20 cm. Find the velocity at this section and also mass rate of flow of oil.
1
Expert's answer
2020-12-08T07:47:45-0500

Diameter of pipe 1 d1=30cm=0.3 md_1=30 cm=0.3\ m

C/S Area of pipe 1 A1=π(d1)24=0.0225π m2A_1 =\dfrac{\pi(d_1)^2}{4}=0.0225\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(d2)24=0.01π m2A_2 =\dfrac{\pi(d_2)^2}{4}=0.01\pi\ m^2


Velocity at pipe 1 v1=2 m/sv_1=2\ m/s

Velocity at pipe 2 v2=?v_2=?


Discharge through pipe 1 Q1=A1v1=0.0225π m2(2 m/s)Q_1=A_1v_1=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


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=0.045π m3/s=Q_1=0.045\pi\ m^3/s


Velocity at pipe 2 v2=Q2A2=0.045π m3/s0.01π m2=4.5 m/sv_2=\dfrac{Q_2}{A_2}=\dfrac{0.045\pi\ m^3/s}{0.01\pi\ m^2 }=4.5\ m/s

v2=4.5 m/sv_2=4.5\ m/s

Mass rate of flow 


ΔmΔt=ρAv=0.81000kg/m3(0.045π m3/s)\dfrac{\Delta m}{\Delta t}=\rho Av=0.8\cdot1000kg/m^3(0.045\pi\ m^3/s)

=113.097kg/s=113.097 kg/s

ΔmΔt=113.097kg/s\dfrac{\Delta m}{\Delta t}=113.097 kg/s


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