Respuesta :
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
