The next obvious combination of pipes is three pipes joining three reservoirs. I have shown an arrangement in figure 9-2. It looks harmless but, if the second reservoir is as I have shown it, the direction of flow in pipe B is by no means certain. Once more we need an example to see what happens and that means dimensions.
I have chosen suitable dimensions as given in the table 9-2. I shall take to be 0.005 as before.
There are at least two ways to tackle the problem of finding the flows in the three pipes. In one the energy equation can be applied for the paths 1 to 2 and 1 to 3 and then use continuity to relate the flows and evaluate by trial. In the other the head at the junction is regarded as a variable and related to the friction losses in the three pipes. Using continuity an equation is set up and solved by trial. Both methods are sound, it comes down to which makes the least work. I used the junction method.
From the energy equation:-
Using continuity a first attempt is:-
Clearly it is possible to substitute these expressions in j into the continuity equation and then solve. The most simple way is by plotting a graph of against to find out the head at the junction when this sum is zero.
Using Mathcad the graph in calculation 9-1. The graph passes through zero at j = 11.25m. This gives flows of :-
and these satisfy continuity.
I do not think that there is an algebraic method to avoid the trial solution but Mathcad takes the labour out of the trial.
There would obviously be a value of Z2 for which the flow in B is zero and a lower one for which the flow is into the reservoir 2. One must be careful when setting up the equations for graphs.
This leads us to consider the problem of having four reservoirs where the four pipes meet at a common junction.
Figure 9-3 shows such a system in very diagrammatic form. Once more we need some dimensions and values for the levels.
The next step is to apply continuity of flow but that is not certain because we do not know for certain the direction of flow in pipe B. However, a trial solution is inevitable, and I will suppose that A and B flow into C and D to see what transpires. Then the condition to be satisfied is that:-
This is a job for Mathcad and the outcome is given in calculation 9-2
The value of j when the equation is satisfied is about 10.25 m and the flows are then :-
We must look carefully at this result. It worked out for the dimensions that were chosen but it could so easily have been otherwise and then two trials would have been necessary.
I have been dealing with very simple systems and if there had been a short length of pipe between the junction of A and B and the junction of C and D the computation would have been very much more difficult. We are mechanical engineers and we are moving into the specialist field of pipe networks and the province of civil engineers and especially water engineers. We have a method for dealing with simple combinations of pipes and know that complicated systems are very difficult to analyse and that is probably adequate for mechanical engineers.
As a final note let me say that the use of these highly stylised diagrams is apt to give the wrong impression. Try drawing the pipe system to scale.