Answer:
friction factor is 0.01819
average speed v = 1.69 m/s
Reynolds number = 98786.7
and flow is turbulent
Explanation:
given data
diameter d = 0.052 m
roughness ε=0.0015 mm
length l = 200 m
ΔP = 100 kPa
to find out
friction factor f, Reynolds number Re and average speed v
solution
we solve this by hit and try assume flow is turbulent
so we can say pipe flow
...............1
here f is friction factor and l is length and d is diameter and ρ is density
so
![\frac{100*10^3}{10^3 g} = \frac{f*200*v^2}{2*g*0.052}](https://tex.z-dn.net/?f=%5Cfrac%7B100%2A10%5E3%7D%7B10%5E3%20g%7D%20%20%3D%20%5Cfrac%7Bf%2A200%2Av%5E2%7D%7B2%2Ag%2A0.052%7D)
f v² = 0.052 ................2
we know Reynolds number = ![\frac{\rho v d}{u}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Crho%20v%20d%7D%7Bu%7D)
here u is viscosity of water that is 8.9 ×
Reynolds number = ![\frac{10^3 v 0.052}{8.9*10^{-4}}](https://tex.z-dn.net/?f=%5Cfrac%7B10%5E3%20v%200.052%7D%7B8.9%2A10%5E%7B-4%7D%7D)
put here v from equation 2
Reynolds number =
× ![\sqrt{\frac{0.052}{f} }](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cfrac%7B0.052%7D%7Bf%7D%20%7D)
Reynolds number =
.................3
we know that 1/√f is
![\frac{1}{\sqrt{f}} = -2 log( \frac{2.51}{Re\sqrt{f} } + \frac{E}{3.7d})](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5Csqrt%7Bf%7D%7D%20%3D%20-2%20log%28%20%5Cfrac%7B2.51%7D%7BRe%5Csqrt%7Bf%7D%20%7D%20%2B%20%5Cfrac%7BE%7D%7B3.7d%7D%29)
![\frac{1}{\sqrt{f}} = -2 log( \frac{2.51}{13323.398 } + \frac{0.0015*10^{-3}}{3.7*0.052})](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5Csqrt%7Bf%7D%7D%20%3D%20-2%20log%28%20%5Cfrac%7B2.51%7D%7B13323.398%20%7D%20%2B%20%5Cfrac%7B0.0015%2A10%5E%7B-3%7D%7D%7B3.7%2A0.052%7D%29)
![\frac{1}{\sqrt{f}} = 7.41466](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5Csqrt%7Bf%7D%7D%20%3D%207.41466)
f = 0.01819
friction factor is 0.01819
and from equation 2
f v² = 0.052
0.01819 v² = 0.052
average speed v = 1.69 m/s
and from equation 3
Reynolds number = ![\frac{13323.398}{\sqrt{f} }](https://tex.z-dn.net/?f=%5Cfrac%7B13323.398%7D%7B%5Csqrt%7Bf%7D%20%7D)
Reynolds number = ![\frac{13323.398}{\sqrt{0.01819} }](https://tex.z-dn.net/?f=%5Cfrac%7B13323.398%7D%7B%5Csqrt%7B0.01819%7D%20%7D)
Reynolds number = 98786.7
so flow is turbulent