Question

A 6 nominal schedule 40 galvanized steel pipe is 25 ft long. It is to convey castor oil. The available pump can provide a pressure drop of 1.64 psi. Determine the expected flow rate of castor oil in the pipe.

Answer 1

Q = 0.030 m^3/sec

Step-by-step explanation:

Given data:

`\mu` for castor oil is given as 0.650 N s/m^2

P = 1.64 psi

we know 1 psi = 6.894 kPa, so

1.64 psi = 11.30 Kpa

w know
`1 inch = 2.54* 10^(-4) m`

`1 feet = 12* 2.54* 10^(-4) m`

from hazen poiseville equation we have following relation for P

`\Delta P = (32\mu Vl)/(d^2)`

solving for v

`v = (11.30* 10^(11) (6* 2.54* 10^(-4))^2)/(32* 0.650* 25* 12* 2.54* 10^(-2))`

v = 1.655 m/s

we know flow rate is given as

Q = AREA * VELOCITY

`Q = (\pi)/(4) (6* 2.54* 10^(-2))^2* 1.625`

Q = 0.030 m^3/sec