Find the diameter of a pipe of length 2000m when the rate of flow of water through the pipe is 200 litres/s and the head lost due to friction is 4m. Take the value of c=50 in Chezy's Formula

  Find the diameter of a pipe of length 2000m when the rate of flow of water through the pipe is 200 litres/s and the head lost due to friction is 4m. Take the value of c=50 in Chezy's Formula

Solution:-

Given:

L=200m

Q=200lit/s=2001000=0.2m3/s$Q=200lit/s={\displaystyle \frac{200}{1000}}=0.2m^3/s$

hf=head lost due to friction=4m

c=50

Step 1: Velocity of flow

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

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

V=0.2×4πd2...............................(i)$V={\displaystyle \frac{0.2\times 4}{\pi d^2}}...............................(i)$

Step 2: Hydraulic mean depth

m=d4.........................(ii)$m={\displaystyle \frac{d}{4}}.........................(ii)$

Loss of head per unit length

i=hfL=42000$i={\displaystyle \frac{hf}{L}}={\displaystyle \frac{4}{2000}}$

i=0.002.....................(iii)$i=\mathrm{0.002.....................}(iii)$

Now substituting the value of V, m, iand c from the equation (i), (ii), (iii), we get

V=cmi−−−√$V=c\sqrt{mi}$

0.2×4πd2=50d4×0.002−−−−−−−−√${\displaystyle \frac{0.2\times 4}{\pi d^2}}=50\sqrt{{\displaystyle \frac{d}{4}}\times 0.002}$

0.2×4πd2×50=d4×0.002−−−−−−−−√${\displaystyle \frac{0.2\times 4}{\pi d^2\times 50}}=\sqrt{{\displaystyle \frac{d}{4}}\times 0.002}$

0.00509d2=d4×0.002−−−−−−−−√${\displaystyle \frac{0.00509}{d^2}}=\sqrt{{\displaystyle \frac{d}{4}}\times 0.002}$

Squaring on both sides, we get

(0.00509)2d4=d4×0.002${\displaystyle \frac{(0.00509{)}^2}{d^4}}={\displaystyle \frac{d}{4}}\times 0.002$

2.59×10−5d4=d4×0.002${\displaystyle \frac{2.59\times {10}^{-5}}{d^4}}={\displaystyle \frac{d}{4}}\times 0.002$

Rearranging, we get

d5=4×(2.59×10−5)0.002$d^5={\displaystyle \frac{4\times (2.59\times {10}^{-5})}{0.002}}$

d5=4×(0.0000259)0.002$d^5={\displaystyle \frac{4\times (0.0000259)}{0.002}}$

∴d5=0.0518$\therefore d^5=0.0518$

∴d=0.0518−−−−−√5$\therefore d=\sqrt[5]{0.0518}$

d=0.553m