Answer:
Q = 0.030 m^3/sec
Explanation:
Given data:
[tex]\mu[/tex] 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 [tex]1 inch = 2.54\times 10^{-4} m[/tex]
[tex]1 feet = 12\times 2.54\times 10^{-4} m[/tex]
from hazen poiseville equation we have following relation for P
[tex]\Delta P = \frac{32\mu Vl}{d^2}[/tex]
solving for v
[tex]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}}[/tex]
v = 1.655 m/s
we know flow rate is given as
Q = AREA * VELOCITY
[tex]Q = \frac{\pi}{4} (6\times 2.54\times 10^{-2})^2\times 1.625[/tex]
Q = 0.030 m^3/sec
