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

Answer:

Q = 0.030 m^3/sec

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 $1 inch = 2.54\times 10^{-4} m$

$1 feet = 12\times 2.54\times 10^{-4} m$

from hazen poiseville equation we have following relation for P

$\Delta P = \frac{32\mu Vl}{d^2}$

solving for v

$v = \frac{11.30\times 10^{11} (6\times 2.54\times 10^{-4})^2}{32\times 0.650\times 25\times 12\times 2.54\times 10^{-2}}$

v = 1.655 m/s

we know flow rate is given as

Q = AREA * VELOCITY

$Q = \frac{\pi}{4} (6\times 2.54\times 10^{-2})^2\times 1.625$

Q = 0.030 m^3/sec