Question

A water pipe with a 5 cm inner diameter is designed for to have a flow rate of 75 Umin. What is the entrance length for this pipe (in cm)?

Answer 1

121.20 cm

Step-by-step 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=
`(V × D)/(v))` =
`(Q × D)/(π/4 × D²× v))`

Re=
`(4Q )/(π × D× v))`

here we know Q is flow rate and D is dia of pipe and v is kinematic viscosity that is 1.14 ×
`10^(-2)` cm ² / sec

so Re=
`(4Q )/(π × D× v))`

Re =
`(4×1.25× 10³ )/(π × 5 × 1.14 × 10^(-2)  ))`

So Re will be 27936 that is greater than 4000 thats why it is turbulent flow

and we know
`(Length)/(dia))` ≡ 4.4
`Re^(1/6)`

so
`(Length)/(dia))` ≡ 24.24

length will be 121.20 cm