Water rho 998 kgm3 and mu 0001002 Pa s flows from a large

Water (rho = 998 kg/m^3 and mu = 0.001002 Pa s) flows from a large tank through the pipe system shown in the diagram at a flow rate of 6.2 kg/s. The pressure above the liquid in the tank is maintained at 200 kPa(gauge). The pipe from the tank is 2-inch schedule 40 pipe made of commercial steel. A pump delivers 1000 W of power at an efficiency of 75%. The exit of the pipe is open to the atmosphere. To what height above the liquid level in the tank can the water be pumped (H in m)? Include all friction losses in your calculation. Assume that the entrance from the tank and the exit to the atmosphere are sharp.

Solution

For 2\" schedule 40 commercial steel pipe, Inner Dia D = 2.067\" = 0.172 ft = 0.0524 m

Cross-section area A = pi/4 * D^2 = 3.14 / 4 * 0.0524^2 = 0.002155 m²

Mass flow rate m = rho*A*V

6.2 = 998*0.002155*V

Velocity V = 2.88 m/s

Reynolds number Re = rho*V*D / meu

= 998*2.88*0.0524 / 0.001002

Re = 150309

Absolute roughness for commercial steel = 0.00015 ft

Relative roughness = 0.00015 / 0.172 = 0.000872

From Moody diagram, for Re = 6186 and rel roughness = 0.000872, we get friction factor f = 0.021

For 50% open gate valve, K = 2.1

For 45 deg elbow, K = 0.4

For threaded 90 deg elbow, K = 1.5

For sharp entrance, K = 0.5

For sharp exit, K = 1

Total K = 2.1 + 2*0.4 + 2*1.5 + 0.5 + 1 = 7.4

Total pipe length L = 15 + 10 + 10+ 20 + 30 + H = 85+H metre

P₁ - (fL/D + K)*rho*V² / 2 + (Power / Q) * Eff = rho*g*H

200*10³ - (0.021*(85+H) / 0.0524 + 7.4) * (998*2.88² / 2) + (1000 / (0.002155*2.88) * 0.75) = 998*9.81*H

200000- (0.021*(85+H) / 0.0524 + 7.4) * 4138.9 + 120843 = 998*9.81*H

Solving H = 13.03 m