Find the head lost due to friction in a pipe of diameter 300mm and length 50m through which water is flowing at a velocity of 3m/s using (i) Darcy weisbach (ii) Chezy's Formula (c=60) (Take V=0.01 stoke (for water))

  Find the head lost due to friction in a pipe of diameter 300mm and length 50m through which water is flowing at a velocity of 3m/s using (i) Darcy weisbach (ii) Chezy's Formula (c=60)

(Take V=0.01 stoke (for water))

Solution:

Given: Diameter of the pipe, d = 300mm (divide by '1000' to convert it from 'mm' to 'm')

So, d=0.3 m

Length, L= 50m

Velocity, v=3m/s

Kinematic Viscosity, V=0.01stoke

V=0.01cm2/s$V=0.01c{m}^2/s$

V=0.01×10−4m2/s$V=0.01\times {10}^{-4}{m}^2/s$

∴$\therefore$ By Darcy's Formula,

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

All the values are given except 'f'

'f' = coefficient of friction which is the function of Reynolds number 'Re$R_e$'

To find: 'f'

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

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

=3.0×0.300.01×10−4$={\displaystyle \frac{3.0\times 0.30}{0.01\times {10}^{-4}}}$

Re=9×105$R_e=9\times {10}^5$

∴f=0.079(9×105)14$\therefore f={\displaystyle \frac{0.079}{(9\times {10}^5{)}^{{\displaystyle \frac{1}{4}}}}}$

f=2.56×10−3$f=2.56\times {10}^{-3}$

or f=0.00256$f=0.00256$

Therefore, Head lost, hf=4×0.00256×50×320.3×2.0×9.81$hf={\displaystyle \frac{4\times 0.00256\times 50\times 3^2}{0.3\times 2.0\times 9.81}}$

hf=782.87×10−3$hf=782.87\times {10}^{-3}$

hf=0.7828m$hf=0.7828m$

Now by using Chezy's formula:-

V=cmi−−−√$V=c\sqrt{mi}$

where, c=60,m=d4=0.304=0.075m$c=60,m={\displaystyle \frac{d}{4}}={\displaystyle \frac{0.30}{4}}=0.075m$

∴3=600.075×i−−−−−−−√.................(i)$\therefore 3=60\sqrt{0.075\times i}.................(i)$

or, ∴,i=(360)2×10.075...................(ii)$\therefore ,i={\left({\displaystyle \frac{3}{60}}\right)}^2\times {\displaystyle \frac{1}{0.075}}...................(ii)$

By equating, we get

i=0..333$i=\mathrm{0..333}$

But, i=hfL$i={\displaystyle \frac{hf}{L}}$

∴,0.333=hf50$\therefore ,0.333={\displaystyle \frac{hf}{50}}$

∴hf=1.665m