APPENDIX 0 – Pre-requisite: Steady State Hydraulics
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Surge Analysis and the Wave Plan Method
Supplementary Material: Example Problems and Solutions
Appendix 0 – Problem 38
A0.38 Suppose the pipeline shown in the previous question has non-uniform diameters between points A and B and carries uniform flowrate. Compute the friction headloss between A and B neglecting the minor losses if pressure, velocity, and elevation at point A are 5 bars, 1.5 m/s, and 200m, respectively, and at point B are 4 bars, 2 m/s, and 180m, respectively.
Friction headloss between points A and B is the difference in total energy per unit weight at points A and B, neglecting the minor losses.
Total energy at point A is the sum of pressure head (p/ γ), velocity head (V 2 /2g), and elevation or datum head (Z).
E A = p A /γ + V A 2 /(2g) + Z A , where E A is energy head in m at point A, p A is pressure in Pa (pascals or N/m 2 ) at point A, γ is specific weight of water in N/m 3 , V A is velocity at point A, g is gravitational acceleration in m/s 2 , and Z A is elevation at point A.
Total energy at point B: E B = p B /γ + V B 2 /(2g) + Z B , where p B is pressure in Pa, V B is velocity, and Z B is elevation at point B.
Friction headloss between points A and B:
ΔH = E A – E B = (p A /γ + V A 2 /(2g) + Z A ) – (p B /γ + V B 2 /(2g) + Z B )
p A = 5 bars = 500000 Pa, p B = 4 bars = 400000 Pa
γ = 9810 N/m 3 , g = 9.81 m/s 2
E A = 50.968 + 0.1146 + 200 = 251.083m
E B = 40.775 + 0.2038 + 180 = 220.979m
ΔH = 251.083 – 220.979 = 30.104m
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