Problem 4.5.4:

Water flows through a galvanized iron pipe at a rate of 0.08 cubic foot per second. Pressure drop through a 105 foot long is 5 psi. Pipe roughness is 0.0005 foot. Density is 62.36 pounds per cubic foot and viscosity is 7.56 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 11.55 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. If V is replaced by group 4q over pi d², then pipe diameter, d, is given by the relationship as:

d⁵ = 32 L q² f over pi² g_c h_f

Assume friction factor to be 0.005.

Diameter, d, can be calculated from the relationship, (32 L q² f over pi² g_c h_f )^0.2 as 0.124 foot. Velocity can be calculated from the relationship that is given as 4q over pi d² as 6.62 feet per second. Reynolds number is equal to V d rho over mu and is equal to 6.8 times 10⁴. Relative roughness, RR is calculated as 0.00373. With these values of Reynolds number and relative roughness, revised value of the friction factor, f_c, is 0.0075. Repeat these calculations till results converge.

Corrected diameter, d_c = d (f_c over f)^0.2 = 0.134 foot

Corrected velocity, V_c = 4 q over pi d_c² = 5.63 feet per second.

Reynolds number = Re_c = V_c d_c rho over mu = 6.25 times 10⁴.

Relative roughness, RR_c = epsilon over d_c = 0.00372

Corrected friction factor, f_c = 0.0074

Pipe diameter = 0.134 foot or 1.6 inches.