Answer:
[tex]Q=12.59\ ft^3/s[/tex]
Explanation:
Given that
Diameter of pipe = 10 in
diameter of throat = 5 in
Pressure in the pipe = 8 pounds per square inch
Pressure in the throat = 6 pounds per square inch
We know that discharge given as follows
[tex]Q=A_1A_2\dfrac{\dfrac{\sqrt{\Delta P}}{\sqrt{\rho}}}{\sqrt{A_1^2-A_2^2}}[/tex]
For water
[tex]\rho=0.036\ lb/in^3[/tex]
[tex]A_1=\dfrac{\pi}{4}\times 10^2\ in^2[/tex]
[tex]A_1=78.53\ in^2[/tex]
[tex]A_2=\dfrac{\pi}{4}\times 5^2\ in^2[/tex]
[tex]A_2=19.63\ in^2[/tex]
So
[tex]Q=78.53\times 19.63\dfrac{\dfrac{\sqrt{2}}{\sqrt{0.036}}}{\sqrt{78.53^2-19.63^2}}[/tex]
[tex]Q=151.11\ in^3/s[/tex]
[tex]Q=12.59\ ft^3/s[/tex]
