Question #159643

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

Expert's answer

Velocity of flow V = Q/A = (3500 kg / (3600 s x 998 kg/m^{3})) / ("\\pi"/4 x (0.025 m)^{2}) = 1.986 m/s

Reynolds number, Re = "\\rho"VD/"\\mu" = (998 kg/m^{3} x 1.986 m/s x 0.025 m) / 0.099Ns/m^{2} = 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)(V^{2}/2g) = 0.128 x (120 m / 0.025 m) x ((1.986 m/s)^{2} / 2 x 9.81 m/s^{2}) = 121.4 m

Pressure drop = "\\gamma" H_{L} = 998 kg/m^{3} x 9.81 m/s^{2} x 121.4 m = 11.9 x 10^{5} Ns/m^{2} = 12 bar

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godzo sampson07.02.24, 13:40this was awesome

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