Chapter 0 – Steady-State Hydraulics
Main – Previous – Next
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.
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 (V 2 /2g), and the elevation or datum head (Z):
E A = p A /γ + V A 2 /(2g) + Z A , where E A is the energy head in m at point A, p A is the pressure in Pa (pascals or N/m 2 ) at point A, γ is the specific weight of water in N/m 3 , V A is the velocity at point A, g is the gravitational acceleration in m/s 2 , and Z A is the elevation at point A.
The total energy at point B is E B = p B /γ + V B 2 /(2g) + Z B , where p B is pressure in Pa, V B is the velocity, and Z B is the elevation at point B.
The friction headloss is the difference between the energy at 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 bar = 500000 Pa, p B = 4 bar = 400000 Pa
γ = 9810 N/m 3 , g = 9.81 m/s 2
E A = 50.968 + 0.1146 + 200 = 251.083 m
E B = 40.775 + 0.2038 + 180 = 220.979 m
ΔH = 251.083 – 220.979 = 30.104 m
Summary of revisions to this page: