Answer to Question #159643 in Chemical Engineering for Adeleke olumide

Question

A pipeline connecting two tanks contains four standard elbows, a plug valve that is fully open and a gate valve that is half open. The line is commercial steel pipe, 25 mm internal diameter, length 120 m. The properties of the fluid are: viscosity 0.99 mNM_2 s, density 998 kg/m3. Calculate the total pressure drop due to friction when the flow rate is 3500 kg/h

Answer 1

Velocity of flow V = Q/A = (3500 kg / (3600 s x 998 kg/m³)) / ($\pi$/4 x (0.025 m)²) = 1.986 m/s

Reynolds number, Re = $\rho$VD/$\mu$ = (998 kg/m³ x 1.986 m/s x 0.025 m) / 0.099Ns/m² = 500

The Reynolds number is less than 2000. So the flow is laminar.

For laminar flow, Friction factor f = 64/Re = 64/500 = 0.128

Head loss, H_L = f(L/D)(V²/2g) = 0.128 x (120 m / 0.025 m) x ((1.986 m/s)² / 2 x 9.81 m/s²) = 121.4 m

Pressure drop = $\gamma$ H_L = 998 kg/m³ x 9.81 m/s² x 121.4 m = 11.9 x 10⁵ Ns/m² = 12 bar