A crude oil of kinematic viscosity 0.4 stoke is flowing through a pipe of diameter 300 mm at the rate of 300 litres per se. Find the head lost due to friction for a length of 50 m of the pipe.

 A crude oil of kinematic viscosity 0.4 stoke is flowing through a pipe of diameter 300 mm at the rate of 300 litres per se. Find the head lost due to friction for a length of 50 m of the pipe.

Solution:

Given:

V=0.4stoke=0.4cm2/s=0.4×10−4m2/s$V=0.4stoke=0.4c{m}^2/s=0.4\times {10}^{-4}{m}^2/s$

d=300mm=0.3m3/s$d=300mm=0.3m^3/s$

L=50m

 

V=DischargeArea$V={\displaystyle \frac{\text{Discharge}}{\text{Area}}}$

V=Qπ4d2$V={\displaystyle \frac{Q}{{\displaystyle \frac{\pi }{4}}{d}^2}}$

V=0.3×4π(0.3)2$V={\displaystyle \frac{0.3\times 4}{\pi (0.3{)}^2}}$

V=4.24m/s$V=4.24m/s$

Step 2: Reynold's Number

Re=V×dv$R_e={\displaystyle \frac{V\times d}{v}}$

Re=4.24×0.30.4×10−4$R_e={\displaystyle \frac{4.24\times 0.3}{0.4\times {10}^{-4}}}$

Re=3.18×104$R_e=3.18\times {10}^4$

As the value of Re$R_e$ lies between 4000 and 10000, the value of 'f' is given by

f=0.079R14e$f={\displaystyle \frac{0.079}{R_e^{{\displaystyle \frac{1}{4}}}}}$

f=0.079(3.18×104)14$f={\displaystyle \frac{0.079}{(3.18\times {10}^4{)}^{{\displaystyle \frac{1}{4}}}}}$

f=5.91×10−3$f=5.91\times {10}^{-3}$

or, f=0.00591

Now by Darcy's Formula

hf=4×f×L×V2d×2g$hf={\displaystyle \frac{4\times f\times L\times V^2}{d\times 2g}}$

hf=4×0.00591×50×4.2420.3×2×9.81$hf={\displaystyle \frac{4\times 0.00591\times 50\times {4.24}^2}{0.3\times 2\times 9.81}}$

∴$\therefore$ Head lost due to friction hf=3.61m$hf=3.61m$

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