1. Proceed by first finding the set of servicing pipe segments unique to the critical consumer; this will
be his final service pipe only. Increase the diameter of that pipe segment until it reduces his pressure
loss to the same level as the consumer with the next highest pressure losses.
2. Now identify the pipe segments that these two consumers alone share and increase those pipe
diameters enough to reduce their pressure losses to the level of the next highest consumer. Note that
it may be that there are no shared pipe segments for these two consumers alone. In that event proceed
to the next step directly.
3. Again look for pipe segments shared by the three consumers with the highest pressure losses and
increase the diameters of those pipe segments enough to bring the pressure losses of these three
consumers to the level of the consumer with the fourth highest pressure loss. Once again, in the event
that no shared pipe segments exist, proceed directly to the next step.
4. Repeat this procedure until no consumers remain with pressure losses exceeding the constraints.
Figure 10. Method B.
lowest cost up to that point in the process.
To address the instances where more than one alternative is available at a
particular step in either of the processes outlined in Figures 9 and 10, we would like
a strategy that minimizes cost. Let's investigate the effect of pipe diameter to see if
it would be to our advantage to choose smaller or large pipes as candidates for the
First, we note that the capital cost Cpv,j is a linearly increasing function of pipe
diameter. Thus, an incremental increase in pipe diameter would have the same effect
regardless of the absolute value of the pipe diameter.
The cost of heat loss Chl,j is a somewhat complicated function of the pipe diameter.
It also includes an approximation introduced in Chapter 2. Within the range of
validity of the approximation (0.025 m ≤ d ≤ 1.0 m), we can see how the heat loss cost
behaves by examining its slope as shown in Figure 11.
The slope of the heat loss cost as plotted below in Figure 11 is essentially the first
term of eq 2-24 with the values of the parameters taken from the example of Chapter
3. From Figure 11 we see that the slope of the heat loss curve is always positive within
our range of interest. This tells us that whenever we increase the pipe diameter we
will increase heat losses, as we would expect. We also see that the slope is a
decreasing function of the diameter, except for pipe diameters over about 0.75 m,
where it becomes a slightly increasing function. For the portion of the range where
the slope is decreasing, we know that an incremental change in pipe diameter will
result in less increase in heat loss for larger diameters than for smaller ones.
The pressure loss as a function of pipe diameter is given by the sum of eq 5-8 and
5-9, which is our former eq 4-4
∆Ps&r = (a εb (4/π)2+c A6 m d2+c L d (5+b+c))j.
If we take the partial derivative of this pressure loss with respect to diameter, we
∆Ps&r/ d = (5 + b + c) (a εb (4/π)2+c A6 m d2+c L d (6+b+c))j.