Chapter 0 – Steady-State Hydraulics
Surge Analysis and the Wave Plan Method
Supplementary Material: Example Problems and Solutions
Chapter 0 – Problem 0.8
0.8 Suppose the pipeline shown in the previous question has non-uniform diameters between points A and B and carries a uniform flowrate. Compute the friction headloss (in meters of head) between A and B neglecting minor losses if the pressure, velocity, and elevation at point A are 5 bar, 1.5 m/s, and 200 m, respectively, and at point B are 4 bar, 2 m/s, and 180 m, respectively.
Solution:
The friction headloss between points A and B is the difference in the total energy per unit weight between points A and B, neglecting minor losses.
The total energy at point A is the sum of pressure head (p/γ), velocity head (V2/2g), and the elevation or datum head (Z):
EA = pA/γ + VA2/(2g) + ZA, where EA is the energy head in m at point A, pA is the pressure in Pa (pascals or N/m2) at point A, γ is the specific weight of water in N/m3, VA is the velocity at point A, g is the gravitational acceleration in m/s2, and ZA is the elevation at point A.
The total energy at point B is EB = pB/γ + VB2/(2g) + ZB, where pB is pressure in Pa, VB is the velocity, and ZB is the elevation at point B.
The friction headloss is the difference between the energy at points A and B:
ΔH = EA – EB = (pA/γ + VA2/(2g) + ZA) – (pB/γ + VB2/(2g) + ZB)
pA = 5 bar = 500000 Pa, pB = 4 bar = 400000 Pa
γ = 9810 N/m3, g = 9.81 m/s2
EA = 50.968 + 0.1146 + 200 = 251.083 m
EB = 40.775 + 0.2038 + 180 = 220.979 m
ΔH = 251.083 – 220.979 = 30.104 m
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