This chapter is a clearing up exercise in that none of the topics justify a chapter but none can be ignored.
+ input power from a pump =
+ power output to a motor or turbine + loss of power to friction.
See worked examples at the end of this chapter for the use of this equation.
It seems to me that mechanical engineers need some acquaintance with the behaviour of pipes when joined to form networks. The pipe system that most people come into contact with is the simple waterfilled central heating system. I have drawn it diagrammatically in figure 91. The components are, the set of radiators, the heating coil for hot water, a boiler, a pump and the pipework. Radiators are of different sizes to give a heat output to suit the sizes of rooms. As a result the flow through each radiator should be proportional to the heat output that is required.
In the common arrangement two pipes, called the flow and return, of larger diameter than all the others are routed side by side through the structure of the building. Radiators and the coil are connected between the flow and return. The radiators and the coil are sited in the building often at the whim of the installer. The size of the pipes is chosen by some rule of thumb.
The routing of the pipework is determined either for minimum cost or to be least intrusive. As a result the lengths of pipe associated with any radiator is not selected to suit the hydraulic requirements of the system. Often the largest radiator is on the end of the longest pipe and vice versa. Valves are fitted at one end of each radiator with the idea of adjusting them to make the flow suit the intended output of each radiator.
We have seen just how nonlinear such valves are and now, any adjustment of a valve in one pipe affects the flow through all the other pipes. There is very little chance that the system will operate as is intended and add to this the need for some sort of timed control and we are looking at a system that a mechanical engineer would find wanting in design. The business is best left to plumbers and tolerant house owners.
In engineering there are other much less complicated systems of pipes that are required to work as planned and it is prudent to have some acquaintance with both their behaviour and their analysis.
It turns out that only the very simple systems can be analysed at all except with expensive computers and software. We can look at two simple but useful systems to assess the difficulty.
Parallel pipes
The most simple system is that of parallel pipes. I have drawn such a system in figure 92 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:
or
Then we also have :
and
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 91. 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:
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
Then
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.
The next obvious combination of pipes is three pipes joining three reservoirs. I have shown an arrangement in figure 92. 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 92. 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 91. 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 Z_{2} 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 93 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 92
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.
This is a traditional topic to appear in degree courses. It is really quite mundane but the need for examinations has to be taken into account and this does require the use of the energy equation and some proficiency in calculus. In practice, as distinct from examination conditions, software like Mathcad have removed the problem of integration leaving just the energy equation.
The only way to illustrate the method is by a worked example.
There is a model boating lake at Maldon in Essex in England. The lake lies alongside the tidal estuary of the river Chelmer from which it is separated by a sea wall. The lake is surrounded with banks and is effectively in a bowl. For over 100 years the lake has been filled from the river through an open pipe that let water in when the tide was high and let it out again when the tide was low. Somehow someone found a pipe size that kept the level steady within a few centimetres. During a recent redevelopment this arrangement was changed. The pipe was replaced by a much larger pipe set well below the required surface level in the lake and fitted with a gate valve. This permits the controlled filling of the lake and of course the controlled emptying. It also creates a problem when there are extra high tides that top the river wall and then the lake is filled by up to one metre above normal and there is no automatic outlet.
Someone has to open the valve fully and keep watch to ensure that the valve is closed again when the normal level is attained. That someone could go home for several hours if the time taken to reach the normal level is long enough. The effective size of the lake is 50 metres by 20 metres, the pipe is 10 metres long and metres in diameter set metres below the normal surface level. The top of the river bank is about 1 metre above normal level in the lake and metres above the outlet.
It is possible to get an idea of the time taken for the level to fall using the energy equation and integration. If I take the friction coefficient to be and the loss at entry to this pipe to be equal to I can make an estimate of the time taken for the level in the lake to fall from the level of the top of the bank to normal level.
Suppose that at some instant in time the level in the lake is above the outlet level of the pipe. Applying the energy equation to the free surface and the outlet gives : + the loss at inlet and to friction. This reduces to :
.
I can substitute in this to give : .
Then and the instantaneous flow is :
In a time of the outflow is m^{3}. This outflow will cause the level in the lake to drop by :
where is the area of the free surface.
So, . Then the time taken to fall from to above the outlet is . Mathcad gives this result in hours :
This result is much dependent on the value chosen for the friction coefficient. The range of value for friction coefficient is from about and this gives a range of about 9 hours to 18 hours. Regardless this is a long time and may involve supervision of this process during the night if the lake is to be available for use from say 10 am.
The calculation above is for a real situation and is probably quite typical. It is not very complicated mathematically but some ingenuity might be required in some cases. It can generate all sorts of examination questions.
1 The new form of the equation is a power equation. It differs from the energy equation by having all the terms multiplied by the mass flow. This permits the inclusion of devices like pumps and turbines to the equation. The diagram shows the elements of such a system.
If the power equation is applied to sections 1 and 2 of this system we can write :
+ the power supplied between 1 and 2
= + the power extracted between 1 and 2 + the power lost to internal friction between 1 and 2.
Clearly the new terms are the power input, the power output and the loss expressed not in energy per unit weight but in power lost to internal friction. The power input would come from a pump or compressor and the output would be through a motor or a turbine. The loss term will be the rate at which mechanical energy is absorbed into the flowing fluid as a result of internal friction arising from the flow in ducts and from the loss of unrecovered kinetic energy.
2 We have the power equation.
+ the power loss between 1 and 2.
.
Therefore power loss 1 to 2.
power loss 1 to 2.
power loss 1 to 2.
Now the power loss is included in the difference between the useful power input to the oil, that is 4,875 watts, and the input power to the shaft.
So the input to the shaft is watts.
3 Applying the energy equation to the turbine :
+ the power extracted from the water between A and B + the loss of power to friction between A and B.
.
Therefore :
power extracted + the loss =
= .
Of the power extracted 85% goes to the shaft and the rest to the loss into the fluid by friction. Therefore :
output power = kw.
4(i) Applying the energy equation to the pipe at the pressure gauge and the delivery point, that is, to A and B we get :
+ power to loss to friction.
Power to pipe loss = .
Therefore cancelling common term gives :
and
Therefore and
4(ii)
Now the system can be considered without reference to the pressure gauge.
Applying the energy equation to D (a point in the free surface in the reservoir) and to B gives:
power loss in pipes where P is the power input from the pump to the water. Then:
5 This question really contains a Venturimeter question as well.
If is the volume flow,
For a Venturimeter : and
.
Therefore and
That is :
.
Applying the energy equation to the section at the gauge 1 and to the exit plane of the 70 mm pipe :
+power to losses.
Therefore
.
Power to losses =
=
Cancelling
The gauge will read bar
6 (i) .
(ii) Using the energy equation applied to the section just before the nozzle 1 and to the jet 2 we get :
.
Therefore (Compare with 40m)
and bar
(iii) The friction loss in the pipe must be .
1 The diagram shows the tank and the gate valve. It also shows the instant at which the level in the tank is above the outlet flange of the gate valve. I have shown point 1 in the free surface and 2 in the water leaving the valve. The energy equation can be applied to 1 and to 2 to give :
+ loss of energy in the vicinity of the valve.
The worst case for this is for the loss at entry to be and at exit .
and then :
.
The instantaneous value of the volume flow is m^{3}/s if in metres.
In time the outflow is m^{3} and this will produce a drop in level :
. Then and Mathcad gives a time of 457 seconds
There is a question mark over the choice of loss at the valve. It is worth reworking using other decisions to see just how much difference they make.
2 Read the coordinates from the graph to give this table.
h 



0.5 
0.00035 
0.301 
3.45 
1 
0.0005 
0 
3.3 
1.5 
0.00061 
0.176 
3.22 
2 
0.0007 
0.301 
3.15 
2.5 
0.00079 
0.4 
3.1 
3 
0.00087 
0.477 
3.06 
From graph :
Slope and k=
.
In time the outflow will be and the drop in level = Rearranging and integrating using Mathcad gives 1,800 seconds.
3 (i) The volume of a cone is where is the base area and is the height. We are given that the radius is equal to the height then .
Therefore
(ii) We have the rate of flow . The outflow in a short time will be given by:
m^{3}
This will produce a fall in level of where A is the area of the free surface at depth . But so .
This can be rearranged for integration. .
(When this question was written programs were written to find the data to plot the graph. This was an opportunity to practice. It is not necessary now.) Mathcad gives: