Text Box: Fig 17-19 Effect of friction on the pressure/time trace

We are now in a position to explain the ramps on the tops of the wave in figure 17-3.

In figure 17-19 I have shown the steady state flow of water through a horizontal pipe. The pressure at inlet will be  as before, but now, at velocity , there will be a friction loss in the pipe and the pressure at the valve will be . Now we have to think out what happens when the valve is suddenly closed. We know that the pressure at the valve will rise instantly by  and a wave front of this magnitude will be propagated along the pipe towards the inlet end bringing to rest the water between the wave front and the valve.

 

When we dealt with this before we ignored friction and the wave advanced into water all at the same pressure. Now, as the wave advances it meets water at an ever-increasing pressure. This must affect what goes on at the wave front.

Text Box: Fig 17-20

In figure 17-20 I have drawn a wave front in the pipe just after the valve has been closed. The wave will have moved a short distance through the pipe and brought the water between the wave front and the valve to rest. In figure 17-20 I have shown the pressures on the two sides of the wave. Pressure  must equal the pressure in the pipe at the position of the wave plus the rise in pressure caused by the destruction of the momentum of the water brought to rest from velocity . As the wave moves along the pipe that pressure will rise to a maximum value of  when it started at . We made the following statement in deriving the expression for the rise in pressure through the wave. The mass of water that has been brought to rest in the short time  has given up momentum as a result of a force equal to  acting on it for a time . So :-

                                         or,

                                                  .

 

Text Box: Fig 17-21 The continual change in upstream pressure has disturbed the original equality because  are all unchanged yet the pressure rise through the wave is not equal to . The water solves this problem by propagating a wave of continuously increasing magnitude towards the valve increasing the pressure in the water at rest and allowing a small additional compression of the water and a small increase in diameter of the pipe to increase very slightly the volume of water at rest over that which would be at rest if there were to be no friction. The reverse wave will travel at the normal speed  and, as a consequence in the time that the main wave front takes to travel from end to end of the pipe and back the pressure at the valve will rise by the original pressure loss. The net effect of the returning pressure wave is to increase the pressure at the valve at a rate that is one half of the rate that the pressure ahead of the wave front is increasing due to the existing pipe friction.

 

We can draw a plot of the pressure distribution when the wave is say half way between the valve and the open end. It is figure 17-21.

Text Box: Fig 17-22

The important area of this diagram in figure 17-22 is the area depicting these returning waves that add to become a distributed wave. I have enlarged this small area to give figure 17-22. We can see that the net effect of the procession of returning waves generated by moving into a region of rising pressure resulting from the gradient produced by friction is the distributed wave shown in green and the effect of that wave at the valve is to give a steady rise in pressure.

 

Text Box: Fig 17-23 We can move on the show the pressure distribution when the wave front reaches the tank. It is figure 17-33 and we can see that the pressure at the valve has risen by one half of the pressure drop due to friction.

 

At the instant shown in figure 17-23 the wave front reverses just as it did in figure 17-9. As the wave front progresses towards the valve the distributed wave goes on travelling unchanged towards the valve and we can see a part way position in figure 17-24.

 

 

 

 

 

 

 

 

 

 

Text Box: Fig 17-24 In figure 17-24 the wave front is half way back. The pressure at the valve has risen by ¾ of the pressure drop due to friction. There is now a very important new line in blue. As the wave front moves towards the valve the water flows out of the pipe and into the tank and in doing so reverses the pressure gradient due to friction. It is this pressure gradient that is shown in blue.

 

Clearly the arrival of the wave front at the valve will have established this reverse pressure gradient from end to end with some loss due to friction. It will also have raised the pressure at the valve by the original pressure drop due to friction. When this is transferred to figure 17-18 it gives the ramps on the top of the square wave shown originally in figure 17-3.

 

We can also see that the reversal of the friction gradient gives the upside down ramps on the negative going waves in 1figure 17-3. We now know enough to understand the pressure-time trace of figure 17-3 including the attenuation of the wave as a result of the steady loss of energy to friction in the water in the pipe with no energy input from the tank.

 

This is an analysis that I think can be called physics although others might see it as an application of physics to engineering. However it is viewed, an engineer involved with some application involving unsteady flow will, in the first instance, need to understand this behaviour even if it is difficult to quantify. Let me give an example.

 

Suppose that you were in charge of a water turbine that ran under a high head and the supply pipe was 1000 metres long. If the machine were to be delivering its full output, when the velocity in the pipe might be 5 metres/sec, a sudden reduction of the load of  25% would have to be met by a rapid reduction in the output of the machine. Let us now suppose that the machine was a Pelton wheel where the water emerges into the machine to interact with the moving blades as a jet at atmospheric pressure. The jet is created in an adjustable nozzle. Closing the nozzle is not an option because that will produce a rapid, very large, increase in pressure in the pipe and the whole gamut of unsteady flow pressure waves running back and forth in the pipe with a transit time of about 1.4 seconds. Furthermore the rise in pressure will increase the velocity of the jet which is the opposite to what is required. One of two solutions must be used; the jet must be broken up by making it swirl or the valve must close as required to reduce the load and a by pass valve must open to release surplus water to waste. That sounds simple but there is now a problem of what to do with the bypass water. It is my understanding that the jet d’eau in Geneva is a replacement for a fountain in the lake that was the by pass flow from a Pelton wheel in a factory. It would have been very attractive with its continual random fluctuation of height. The importance of this example lies in the fact that only a qualitative knowledge of unsteady flow is needed to devise a relatively simple mechanical solution. Computers would not help even if the full analysis of the unsteady flow would be quite interesting academically.