Question

A water main made out of galvanized iron has an inner pipe diameter of 0.1 meters, an inlet pressure of 1000 kPa, outlet pressure of 500 kPa, water flow rate of 5 m3/min, and 10 90 degree elbows. What length galvanized iron line should be used to reach the desired outlet pressure?

Answer 1

Step-by-step explanation:

The given data is as follows.

Diameter = 0.1 m,
`P_(1)` = 1000 kPa

`P_(2)` = 500 kPa

Change in pressure
`\Delta P` = 1000 kPa - 500 kPa = 500 kPa

Since, 1000 Pa = 1 kPa. So, 500 kPa will be equal to
`500 * 10^(3)`.

Q = 5
`m^(3)/min` =
`(5)/(60) m^(3)/sec` = 0.0833
`m^(3)/sec`

It is known that Q =
`A * V`

where, A = cross sectional area

V = speed of the fluid in that section

Hence, calculate V as follows.

V =
`(Q)/(A)`

=
`(Q * 4)/(\pi * d^(2))`

=
`(0.0833 * 4)/(3.14 * (0.1)^(2))`

= 10.61 m/sec

Also it is known that Reynold's number is as follows.

Re =
`(\rho * V * d)/(\mu)`

=
`(1000 * 10.61 * 0.1)/(10^(-3))`

= 1061032.954

As, it is given that the flow is turbulent so we cannot use the Hagen-Poiseuille equation as follows. Therefore, by using Blasius equation for turbulent flow as follows.

`\Delta P = (0.241 * \rho^(0.75) * \mu^(0.25) * L)/(D^(4.75)) * Q^(1.75)`

`500 * 10^(3) = (0.241 * (1000)^(0.75) * (10^(-3))^(0.25) * L)/((0.1)^(4.75)) * (0.0833)^(1.75)`

`500 * 10^(3) = 5535.36 * L`

L =
`(500 * 10^(3))/(5535.36)`

= 90.328 m

Thus, we can conclude that 90.328 m length galvanized iron line should be used to reach the desired outlet pressure.