Problem 4.5.3:

Pressure drop through a galvanized iron pipe (1.38 inches ID) is 100 psi, Equivalent length of the pipe is 2000 feet. Pipe roughness is 0.0005 foot. Density is 62.36 pounds per cubic foot and viscosity is 7.556 pound per foot per second.

Solution: For given situation, Z₁ = Z₂ = 0. (Points are at same elevation).

V₁ = V₂ (Points are within the pipe having constant diameter).

Pump work, w_p, = 0. The Bernoulli’s equation reduces to P₁ over rho is equal to P₂ over rho plus h_f, where h_f is form friction loss. h_f is calculated as 230.92 foot pound-force per pound-mass. With no fitting losses, h_fs is the same as h_f.

h_fs is given by 4 f L over d V² over 2 g_c. Reynolds number is V d rho over mu or V is Re mu over d rho. Expressing velocity in terms of Reynolds number and rearranging one can obtain the value of Re sqrt f as rho over mu sqrt (d³ g_c h_f over 2 L). Figure 5-11 shows a plot of friction factor f versus a dimensionless parameter, Re sqrt f.

For Re sqrt f value of 4.387 times 10³, friction factor is 0.0078. For the given relative roughness, epsilon over d, is 0.00435, Figure 5-10, gives value of Reynolds number as 4.975 times 10⁴. Velocity of fluid is Re mu over d rho and is found to be 5.24 foot per second.