Answer:
121.20 cm
Explanation:
Given data in question
inner dia = 5 cm
flow rate = 75 umin = 1.25 ×10³ cm³/sec
Solution
First we calculate Re by this formula
Re= [tex]\frac{V × D}{v)}[/tex] = [tex]\frac{Q × D}{π/4 × D²× v)}[/tex]
Re= [tex]\frac{4Q }{π × D× v)}[/tex]
here we know Q is flow rate and D is dia of pipe and v is kinematic viscosity that is 1.14 × [tex]10^{-2}[/tex] cm ² / sec
so Re= [tex]\frac{4Q }{π × D× v)}[/tex]
Re = [tex]\frac{4×1.25× 10³ }{π × 5 × 1.14 × 10^{-2} )}[/tex]
So Re will be 27936 that is greater than 4000 thats why it is turbulent flow
and we know [tex]\frac{Length}{dia)}[/tex] ≡ 4.4 [tex]Re^{1/6}[/tex]
so [tex]\frac{Length}{dia)}[/tex] ≡ 24.24
length will be 121.20 cm
