An oil of sp.gr. 0.7 is flowing through a pipe of diameter 300 mm at the rate of 500 litres/s. Find the head lost due to friction and power required to maintain the floor for a length of 1000 m. Take v=0.29 stokes.

 An oil of sp.gr. 0.7 is flowing through a pipe of diameter 300 mm at the rate of 500 litres/s. Find the head lost due to friction and power required to maintain the floor for a length of 1000 m. Take v=0.29 stokes.

Solution: 

Given:-

s=0.7

Diameter of pipe, d=300mm=0.3m$d=300mm=0.3m$

Discharge, Q=500lit/s=0.5m3/s$Q=500lit/s=0.5m^3/s$

Length, L=1000m$L=1000m$

Step 1: Velocity

V=Qarea$V={\displaystyle \frac{Q}{\text{area}}}$

V=0.5π4×d2=0.5×4π×0.32=7.073m/s$V={\displaystyle \frac{0.5}{{\displaystyle \frac{\pi }{4}}\times d^2}}={\displaystyle \frac{0.5\times 4}{\pi \times {0.3}^2}}=7.073m/s$

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Step2: Reynold's Number

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

Re=7.073×0.30.29×10−4$R_e={\displaystyle \frac{7.073\times 0.3}{0.29\times {10}^{-4}}}$

Re=7.316×104$R_e=7.316\times {10}^4$

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Step 3: Coefficient of friction,

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

f=0.079(7.316×104)14$f={\displaystyle \frac{0.079}{(7.316\times {10}^4{)}^{{\displaystyle \frac{1}{4}}}}}$

f=0.0048$f=0.0048$

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Step 4: Head Lost

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

hf=4×0.0048×1000×7.07320.3×2×9.81$hf={\displaystyle \frac{4\times 0.0048\times 1000\times {7.073}^2}{0.3\times 2\times 9.81}}$

f=163.18m$f=163.18m$

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Step 5: Power required

=ρ.g.Q.hf1000$={\displaystyle \frac{\rho .g.Q.hf}{1000}}$

Density of oil, ρ=07×1000=700kg/m3$\rho =07\times 1000=700kg/m^3$

Power required =700×9.81×0.5×163181000$={\displaystyle \frac{700\times 9.81\times 0.5\times 16318}{1000}}$

Hence power required = 560.28kWa