Parallel pipes

Text Box:  
Figure 9-2
The most simple system is that of parallel pipes. I have drawn such a system in figure 9-2 where two reservoirs with a difference of surface level of  metres are joined by a sequence of pipes. Pipe A is followed by parallel pipes B and C and finally pipe D. There is no reason why B should be the same length as C. This has the feeling of civil engineering about it rather than mechanical engineering but it will serve my purpose in looking at the calculations. If the dimensions of the pipes and the friction factors are known it is possible to find the net flow between the reservoirs and the flow in each pipe. Two conditions must be satisfied. First it is evident that, as there can be only one pressure at a junction, the head lost to friction in B must equal that in C and therefore that the friction loss in A plus the friction loss in either B or C plus the friction loss in D equals . The second condition is that the volume flow in A must equal that in D and also equal the sum of the volume flows in B and C.


Putting this in symbols the head loss in a pipe is  with a further subscript to indicate the pipe putting  with an identifying subscript for the volume flow I can write down the conditions that must be satisfied :-

                                                  and then that:-


Then we also have :-



It is obvious that the next step must be to have some way of evaluating the losses in the separate pipes. I think that the best way is to ignore the minor losses, unless this is really inappropriate, and to express the loss in the form :-

                                              in SI units.

There is only one way to get to grips with this problem and that is through a worked example. I have selected suitable dimensions in table 9-1. In a practical problem something would be known about the condition of the internal surfaces of the pipes. Then a good estimate of the value of the friction factor could be made using Darcy and Moody even if it involves iteration. For my purposes here it is sufficient to use one typical value of  and I will use 0.005 for all the pipes. Then I can write:-

Text Box: Pipe	Diameter mm	Length m
A	500	100
B	750	150
C	500	150
D	500	150
Table 9-1


Using the energy equation:-


As the loss in parallel pipes B must be equal to loss in C :-


Using continuity :-

We have  and then    and

                                     from which



This is a straightforward calculation although it could be refined by using a better choice of friction factor for each pipe. Only the context will show whether this is justified. The method can be extended to lots of parallel pipes but it may become a protracted calculation.