Transient Maths

Transient Maths

Transient analysis is an involved and complicated process, and care must be taken to assess the geometry of the distribution system and it's operating regieme to understand it fully. However here we give a short introduction and show that with a couple of assumptions we can get some useful information.

First as was stated in here the pressure transients move around the system at the speed of sound of the fluid in the pipes. If we had a large container of water the speed of sound would be about 1,484 m/s. However the pipe walls are always slightly elastic and deform as a pressure wave passes. This deformation has the effect of slowing the wave so we are left with the following expression for the wave speed in a pipe:

If we call the upstream or downstream length of pipe before a reflecting surface (reservoir, valve, junction etc.), l, then if the operation that causes the transient (valve opening closing, pump startup trip, etc.) occurs in a time shorter than 2l/c, then the pressure wave will not have had time to pass along the length and reflect back to the original site. This type of operation is then termed a 'fast' operation and will produce the largest transients. If the operation takes longer than 2l/c to occur then the returning pressure wave will interact with the wave generated by the operation and will reduce the magnitude of the pressure wave. In this case a more complicated analysis is required to determine the actual finall pressure rise.

The pressure rise in a fast operation is given by the Joukowski Equation:

Example

If we have a valve in the middle of a straight section of pipe of total length 1000 m. The pipe is connected to a pump at the upstream end and a reservoir at the downstream end and the flow rate is 4 l/s and runs at a steady state head of 20m. If we assume that the pipe is made of MDPE (E = 2e9 Pa) and has diameter and wall thickness of 100mm and 10mm respectively, we shall calculate the pressure rise of we close the valve in 1 s:

Putting the pipe properties into the equation for the wave speed in a pipe gives:

We now calculate whether the valve operation time is less than the reflection time: Given the total length of the pipe is 1000m and hte valve is in the middle l = 1000/2, therefore the time for a reflection to return to the valve is:

which is less than the valve operation time of 1 s, which means the valve operation is 'fast' and the magnitude of the transient can be calculated from the Joukowski Equation. We are assuming the valve completely closes during the operation so the flow rate drops from 4 l/s to zero. The initial velocity in the pipe is U = Q/A, where U is the flow velocity, Q the flow rate and A the cross sectional area of the pipe.

so the change in velocity during the operation, dU, is 0.51 m/s, which we can put into the Joukowski Equation:

the initial pressure in the pipe was 20m so upstream of the valve closure the pressure change will be positive giving a maximum pressure of 42.19 m downstream the pressure change is negative and the transient produces a minimum pressure of -2.19 m. This valve operation produces a negative transient and thus a risk of contaminant intrusion if there exists a pathway into the distribution system.