An oil of viscosity 0.1NS/m20.1NS/m2 and relative density 0.9 is flowing through a circular pipe of diameter 50 mm and of length 300 m. The rate of flow of fluid through the pipe is 3.5 litres per second. Find the pressure drop in a length of 300 m and also the shear stress at the pipe wall.

An oil of viscosity 0.1NS/m2$0.1NS/m^2$ and relative density 0.9 is flowing through a circular pipe of diameter 50 mm and of length 300 m. The rate of flow of fluid through the pipe is 3.5 litres per second. Find the pressure drop in a length of 300 m and also the shear stress at the pipe wall.

Solution:-

 Given:-

Viscosity, μ=0.1NS/m2$\mu =0.1NS/m^2$

Relative density = 0.9

ρo=0.9×1000=900kg/m3${\rho }_o=0.9\times 1000=900kg/m^3$ (As density of water =1000kg/m3$1000kg/m^3$)

D=50mm=0.05m

L=300mQ=3.5lit/s=0.0035m3/s$L=300mQ=3.5lit/s=0.0035m^3/s$

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1) Pressure Drop (P1−P2$P_1-P_2$)

=32μu¯¯¯LD2$={\displaystyle \frac{32\mu \overline{u}L}{D^2}}$

∴u¯¯¯=QA$\therefore \overline{u}={\displaystyle \frac{Q}{A}}$

=0.0035π4D2$={\displaystyle \frac{0.0035}{{\displaystyle \frac{\pi }{4}}{D}^2}}$

=0.0035π4(0.05)2=1.782m/s$={\displaystyle \frac{0.0035}{{\displaystyle \frac{\pi }{4}}(0.05{)}^2}}=1.782m/s$

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Re=ρVDμ$R_e={\displaystyle \frac{\rho VD}{\mu }}$ (ρ=900kg/m3,V=u¯¯¯=1.782$\rho =900kg/m^3,V=\overline{u}=1.782$)

Re=900×1.782×0.050.1=801.9$R_e={\displaystyle \frac{900\times 1.782\times 0.05}{0.1}}=801.9$

As Re<2000$R_e<2000$, the flow is laminar,

∴P1−P2=32×0.1×1.782×300(0.05)2$\therefore P_1-P_2={\displaystyle \frac{32\times 0.1\times 1.782\times 300}{(0.05{)}^2}}$

=684288N/m2=684288×10−4N/cm2$=684288N/m^2=684288\times {10}^{-4}N/c{m}^2$

∴P1−P−2=68.43N/cm2$\therefore P_1-P-2=68.43N/c{m}^2$

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2) Shear stress at the pipe wall (τ0${\tau }_0$)

∴τ=−δPδx.r2$\therefore \tau =-{\displaystyle \frac{\delta P}{\delta x}}.{\displaystyle \frac{r}{2}}$ (r=R)

∴τ=−δPδx.R2$\therefore \tau =-{\displaystyle \frac{\delta P}{\delta x}}.{\displaystyle \frac{R}{2}}$

−δPδx=−(P2−P1)x2−x1$-{\displaystyle \frac{\delta P}{\delta x}}={\displaystyle \frac{-(P_2-P_1)}{x_2-x_1}}$

−δPδx=P1−P2x2−x1$-{\displaystyle \frac{\delta P}{\delta x}}={\displaystyle \frac{P_1-P_2}{x_2-x_1}}$

−δPδx=P1−P2L=684288300=2280.96N/m2$-{\displaystyle \frac{\delta P}{\delta x}}={\displaystyle \frac{P_1-P_2}{L}}={\displaystyle \frac{684288}{300}}=2280.96N/m^2$

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R=D2=0.052$R={\displaystyle \frac{D}{2}}={\displaystyle \frac{0.05}{2}}$

R=0.025m

∴τ0=2280.96×0.0252$\therefore {\tau }_0=2280.96\times {\displaystyle \frac{0.025}{2}}$

τ0=28.512N/m2