Question

Two sections, A and B, are 0.5 km apart along a 0.05 m diameter rough concrete pipe. A is 115 m higher than B, the water temperature is 4 °C, and the pressure heads measured at A and B are 21.7 m and 76.1 m, respectively. Assume minor losses are negligible. Determine the flow rate through the pipe.

Answer 1

Answer:

Q = 178.41 m^3 / s

Explanation:

Given:

- Length of the pipe L = 0.5 km

- Diameter D = 0.05 m

- Pressure head at point A (Pa / γ )= 21.7 m

- Pressure head @ point B (Pb / γ )= 76.1 m

- Elevation head Za = 115 m

- Elevation head Zb = 0 m

- Minor Losses = 0 m

- Major Losses = f*L*V^2 / 2*D*g = (500/2*0.05*9.81) *f*V^2 = 509.684*f*V^2

- Roughness e = 2.5 mm

- Dynamic viscosity of water u = 8.9*10^-4 Pa-s

- Density of water p = 997 kg/m^3

Find:

Flow Rate Q ?

Solution:

- We will use the Head Balance as derived from Energy Balance:

(Pa / γ ) - (Pb / γ ) + (Va^2 - Vb^2) / 2*g + (Za - Zb) = Major Losses

- Velocity at cross section A and B: Va = Vb m/s. Hence,

                   21.7 - 76.1 + 0 + 115-0 = 509.684*f*V^2

                              f*V^2 = 0.18897199

- Correction factor f which is a function of e / D = 0.05, and Reynold's number. In such cases we will guess a value of f and perform iterations as follows:

- Guess: f_o = 0.072 (Moody's Chart @ e / D = 0.05 and most turbulent function).

V_o = sqrt(0.18897199 / 0.072) = 1.28504 m/s

Re_o = p*V_o*D / u =  997*1.28504*0.05 / 8.9*10^-4 = 71976.6

1st iteration

f_1 = g (Re_o , e/d) = 0.0718702 (Moody's Chart)

V_1 = sqrt(0.18897199 / 0.0718702) = 1.621528 m/s

Re_1 = p*V_1*D / u =  997*1.621528*0.05 / 8.9*10^-4 = 90823.81698

2nd iteration

f_2 = g (Re_1 , e/d) = 0.0718041 (Moody's Chart)

V_2 = sqrt(0.18897199 / 0.0718041) = 1.662273585 m/s

Re_2 = p*V_2*D / u =  997*1.662273585*0.05 / 8.9*10^-4 = 90865.54854

3rd iteration

f_3 = g (Re_2 , e/d) = 0.0718040  (Moody's Chart)

V_3 = sqrt(0.18897199 / 0.0718040) = 1.622274714 m/s

Re_3 = p*V_3*D / u =  997*1.622274714*0.05 / 8.9*10^-4 = 90865.61182

- V converges to 1.6222 m /s and any further iterations will be meaningless, so we will stop at 3rd iteration. Hence, the required velocity will be used to calculate the Flow rate Q:

Q =  pi*V*D^2/4 = pi*1.6222*0.05^2 / 4

Q = 178.41 m^3 / s