Find the diameter of the pipe and efficiency of transmission if the pressure drop over the length of pipe is 100 N/cmo. Take f- 0.02 also find diameter if pipe corresponding to maximum efficiency of transmission

Length of the pipeline, L=2400 m

Power transmitted, P=115 kW

Pressure at inlet, p=$500N⁄cm^{2} =5000k N⁄m^2$

H=$\frac{p}{w}=\frac{5000}{9.81}=509.68 m$

w=specific weight of water.

Pressure drop=$100 N⁄cm^2=1000 kN⁄m^2 )$

Therefore loss of head, $h_f=\frac{1000}{9.81}$=101.94 m

Coefficient of friction, f=0.026.

i)Diameter of Pipe, D:

Head available at the end of the pipe, $H-h_f$=509.68-101.94=407.74 m

Now, power transmitted is given by : P=$wQ(H-h_f )$kW

115=9.81×Q×407.74

Q=$0.02875m^3⁄s$

But, Q=$\frac{π}{4} D^2 V$

0.02875=$\frac{π}{4} D^2 V$

V=$\frac{0.02875×4}{πD^2}=\frac{0.03661}{D^2 }$

The head lost due to friction,$h_f=\frac{4fLV^2}{2gD}$

But, $h_f$=101.94

$101.94=\frac{4×0.026×2400×(0.0366⁄D^2 )^2}{2×9.81×D}$

5981.87=$\frac{1}{D}^5$

D=$\sqrt{\frac{1}{5981.87}}$=0.1756m

**ii)Efficiency of transmission,$η$:** $η=\frac{H-h_f}{H}=\frac{509.68-101.94}{509.68}$=0.79=79% iii)Maximum efficiency: power transmitted through the pipe is maximum, when head lost due to friction in the pipe is equal to 1/3 of the total supply head. $h_f=\frac{H}{3}$ $H-3×\frac{4fLV^2}{2gD}$=0 509.68-$\frac{3×4×0.026×2400×(0.0366⁄D^2 )^2}{2×9.81×D}$=0 509.68-$\frac{0.0511}{D^{5}}$ =0

D=0.1586 m