A pump installed in a l00-m pipeline to lift water at 20degreeC from reservoir.

Question

A pump installed in a l00-m pipeline to lift water at 20degreeC from reservoir. A to reservoir B. The pipe is rough concrete (epsilon = 0.6 mm) with a diameter of 80 cm. The design discharge is 5 m3 /sec. Determine the maximum distance from reservoir, A that the pump could be installed without encountering cavitation problems. Assume an entrance minor loss coefficient (KL) of 0.5 and neglect all other minor losses. Plot the energy line ( EL) and hydraulic grade line (HGL) along the pipe system. The x-axis should start at 0 and end at 100 m (total length of pipe). Assume datum al pump elevation.

Explanation / Answer

Pipe cross-section area A = (pi / 4) * d^2 = (3.14 / 4)* 0.8^2 = 0.5024 m^2

Velocity V = Q / A = 5 / 0.5024 = 9.95 m/s

Kinematic viscosity of water at 20 deg C is v = 1.005*10^-6 m^2 /s

Reynolds number Re = V*d / v

Re = 9.95*0.8 / (1.005*10^-6) = 7.92*10^6

Relative roughness e = epsilon / d = 0.6 / 800 = 0.00075

From Moody chart, for Re = 7.92*10^6 and e = 0.00075 we get friction factor f = 0.019

Head loss h_f = f*(L/d)*(V^2 / (2g))

h_f = 0.019*(100 / 0.8)*9.95^2 / (2*9.81)

h_f = 11.984 m

Assume distance between reservoir a and pump = x.

Total head at pump entrance = 4 - K*V^2 / (2g) - (x / 100)*11.984

= 4 - 0.5*9.95^2 / (2*9.81) - (x / 100)*11.984

= 1.477 - (x / 100)*11.984

Vapor pressure of water = 2340 Pa

In terms of head, 2340 / (1000*9.81) = 0.239 m

Thus, to avoid cavitation, 0.239 = 1.477 - (x / 100)*11.984

x = 10.33 m

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