APPENDIX 0 – Pre-requisite: Steady State Hydraulics
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.
Solution:
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 (V2/2g), and elevation or datum head (Z).
EA = pA/γ + VA2/(2g) + ZA, where EA is energy head in m at point A, pA is pressure in Pa (pascals or N/m2) at point A, γ is specific weight of water in N/m3, VA is velocity at point A, g is gravitational acceleration in m/s2, and ZA is elevation at point A.
Total energy at point B: EB = pB/γ + VB2/(2g) + ZB, where pB is pressure in Pa, VB is velocity, and ZB is elevation at point B.
Friction headloss between points A and B:
ΔH = EA – EB = (pA/γ + VA2/(2g) + ZA) – (pB/γ + VB2/(2g) + ZB)
pA = 5 bars = 500000 Pa, pB = 4 bars = 400000 Pa
γ = 9810 N/m3, g = 9.81 m/s2
EA = 50.968 + 0.1146 + 200 = 251.083m
EB = 40.775 + 0.2038 + 180 = 220.979m
ΔH = 251.083 – 220.979 = 30.104m
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