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 ∆*P*hp must be the

difference between *P*hp,s and *P*hp,r to eliminate *P*hp,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 3*nn *+ 2*n *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

(5-12)

where ∆*P*cvs,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 ∆*P*cv,i unknowns in eq 4-2, thus

reducing both the number of unknowns and equations by *n *to 3*nn + 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 ∆*P*hp that appears in all of the constraints of eq 4-2. So, then our

total number of unknowns would be (*n *+ 4*nn * 1).

With (*n *+ 4*nn * 1) unknowns and only (*n *+ 3*nn*) 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*)

^

.

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