the constraints from the sets of eq 4-11, 4-22 and 4-23, since at this point we have no
way of knowing which of these constraints will be active. For each of these latter sets
of constraints, we have one member for each node in the system. We can use the
constraint of either eq 4-23, 4-24 or 4-25 coupled with the fact that ∆Php must be the
difference between Php,s and Php,r to eliminate Php,s from eq 4-11, 4-22 and 4-23. Once
the parameter values for the problem are known, the choice of which equation to use
will be determined by the dominance argument given earlier. Thus, if we have nn
nodes in the system and n consumers, our result would be 3nn + 2n simultaneous
equations.
To introduce the inequality constraints of eq 4-5, 4-11, 4-22 and 4-23 directly into
a set of simultaneous equations, we would need to introduce a "slack" variable for
each inequality constraint. The slack variable allows us to convert the inequality
constraint to an equality constraint. For example, eq 4-5 would be converted into an
equality constraint of the form
Pcv,i = Pcvm,i + Pcvs,i
(5-12)
where ∆Pcvs,i is equal to the slack variable for consumer control valve pressure losses
(N/m2).
Equation 5-12 would introduce n slack variables into the problem. It could,
however, be used immediately to eliminate the ∆Pcv,i unknowns in eq 4-2, thus
reducing both the number of unknowns and equations by n to 3nn + n.
The constraints of eq 4-11 and 4-22 would each introduce nn slack variables as
well. The constraint of eq 4-23 would introduce one less slack variable than these
constraints, since there will be no slack variable at the heating plant where the
pressure level is determined by the constraint dominance arguments discussed
earlier. Thus, eq 4-23 will result in (nn 1) additional slack variables.
In addition to the slack variables, we would still have the diameters of our pipe
segments as unknowns as well. The number of pipe segments will always be one less
than the number of nodes. This result is easily shown if we consider the process of
building the network from one node to the next. The first two nodes introduced into
the system will require one pipe. Any subsequent nodes introduced will require one
pipe for each node, since one node will already be an existing connected node. The
only case where this would not be true is if we had a looped network, rather than the
pure branched networks to which we will limit our discussion. We have one
additional unknown ∆Php that appears in all of the constraints of eq 4-2. So, then our
total number of unknowns would be (n + 4nn 1).
With (n + 4nn 1) unknowns and only (n + 3nn) equations, we have no unique
solution. Recall, however, that (nn 1) of the unknowns are the pipe diameters,
which must take on discrete values. If the pipe diameters were to be considered as
continuous, we would have an infinite number of solutions. It's actually fortunate
that they are discrete because this limits the number of possible solutions. The
number of possibilities can still be quite large for a system of any significant size. For
example, if we were to consider only 3 possible pipe sizes for each pipe segment we
would have 3(nn 1) possible solutions. For our system discussed earlier with 125
nodes (124 pipe segments), we would have 3124 ≅ 1.46 1059, a combinatorial
problem of staggering proportions by any measure.
Notice that by applying monotonicity analysis to this problem we were able
reduce it to one of solving for the variables using the constraint set, which has been
reduced somewhat. The constraint set is linear in all the variables except the pipe
diameters and the pipe diameters only appear in one set of constraints. We could
make the problem linear by making the transformation for pipe diameters of
(5+b+c)
^
dj = dj
.
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