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 tempera

Question

3 years ago

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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 tempera

ture 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

Physics

1 answer:

KonstantinChe [14] · Answer 1 · 2022-04-05T16:07:29+03:00

Answer:

Q = 178.41 m^3 / s

Explanation:

Given:

Length of the pipe L = 0.5 km
Diameter D = 0.05 m
Pressure head @ A (P_a / γ )= 21.7 m
Pressure head @ B (P_b / γ )= 76.1 m
Elevation head Z_a = 115 m
Elevation head Z_b = 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
Velocity at cross section A and B: V_a = V_b m/s
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 = pi*V*D^2/4 m^3/s ??

Solution:

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

(P_a / γ ) - (P_b / γ ) + (V_a^2 - V_b^2) / 2*g + (Z_a - Z_b) = Major Losses

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

f*V^2 = 0.18897199

To find correction factor f which is a function of e / D = 0.05, and Reynold's number which is unknown. 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