Chapter 3 – What Causes Surge?
Surge Analysis and the Wave Plan Method
Supplementary Material: Example Problems and Solutions
Chapter 3 – Problem 3.3
3.3 Consider a pipe filling operation shown in Figures 3.7 and 3.8. Find the air slam pressure, if any, associated with this operation for the following conditions?
a. The pipe between isolations valves A and B is made of mild steel, diameter = 300 mm, length = 450m, with a wave speed = 1050 m/s. Valve B remains closed during the filling operation. There is a kinetic air valve at the location of valve B which has a 100 mm effective diameter for both its inlet and outlet orifices. Valve A (a butterfly valve) is opened rapidly letting water into the pipeline and the flow reaches a velocity of 2.5 m/s before all the air is expelled from the pipe.
b. All parameters remain the same as in case a except the air valve at location B is a kinetic air valve with a 35 mm effective outflow diameter. Because of the increased resistance to the outflow of air created by this smaller diameter orifice, water flow in the pipeline reaches a velocity of 1.0 m/s before all the air is expelled.
c. All parameters remain the same as in case a, except the air valve at location B is a non-slam air valve with a secondary outflow orifice size of 13 mm thereby further retarding the flow velocity to 0.3 m/s before all the air is expelled.
d. All parameters remain the same as in case c except the pipeline is made of PVC with an effective wave speed of 250 m/s.
Solution:
a. The pipe between isolations valves A and B is made of mild steel, diameter = 300 mm, length = 450m, with a wave speed = 1050 m/s. Valve B remains closed during the filling operation. There is a kinetic air valve at the location of valve B which has a 100 mm effective diameter for both its inlet and outlet orifices. Valve A (a butterfly valve) is opened rapidly letting water into the pipeline and the flow reaches a velocity of 2.5 m/s before all the air is expelled from the pipe.
When all the air is expelled from the pipe, the air valve float closes instantaneously forcing the flow velocity to 0 m/s. Use the Joukowsky relation (Eq. 1.22) to compute the air slam pressure. Change in velocity ∆V = initial velocity – final velocity = 2.5 – 0 = 2.5 m/s. Change in pressure ∆H representing air slam pressure = (1050/9.81) * (2.5) = 267.58m or 26 bar.
b. All parameters remain the same as in case a except the air valve at location B is a kinetic air valve with a 35 mm effective outflow diameter. Because of the increased resistance to the outflow of air created by this smaller diameter orifice, water flow in the pipeline reaches a velocity of 1.0 m/s before all the air is expelled.
∆V = 1.5 m/s,
air slam pressure ∆H = (1050/9.81) * 1.0 = 107.03 or 10.5 bar
c. All parameters remain the same as in case a, except the air valve at location B is a non-slam air valve with a secondary outflow orifice size of 13 mm thereby further retarding the flow velocity to 0.3 m/s before all the air is expelled.
∆V = 0.3 m/s,
air slam pressure ∆H = (1050/9.81) * 0.3 = 32.1m or 3.15 bar
d. All parameters remain the same as in case c except the pipeline is made of PVC with an effective wave speed of 250 m/s.
∆V = 0.3 m/s,
air slam pressure ∆H = (250/9.81) * 0.3 = 7.64m or 0.75 bar
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