design excluding the cost of heat losses. And, since we have found the optimum
design (lowest cost) neglecting heat losses, we now know that no design can achieve
a lower cost when heat losses are included. This simple result can be very useful. It
may be possible to find a design, not necessarily known to be optimal, whose cost
including heat losses is acceptably close to that of the optimal design for the lower
bounding problem.
The solution to the complete problem including heat losses is slightly more
complicated, but is easily obtained. To find the extremum of the total variable cost
′
function Ct , we simply take its partial derivative with respect to d and set the result
to zero. Before proceeding to do so, however, we must take note of the value of A10
in the heat loss term being a function of the pipe diameter. This results from the outer
diameter of the pipe being a function of pipe diameter and insulation thickness. The
appropriate insulation thickness is determined by a separate optimization proce-
dure that would consider the insulation and jacket material costs and the cost of heat
loss. As a result of this separate "sub-optimization," the insulation thickness
becomes a function of the pipe diameter.
For a given set of operating conditions and economic data, the optimal insulation
thickness can be found as a function of the pipe diameter. Here, for the sake of
simplicity, we will assume that the insulation thickness is fixed. We then find an
A10 = (d + 2∆xi)1γ (4Hp)γ
(2-21)
where ∆xi is the insulation thickness (m). In turn we approximate this expression by
one of the following form
A10 = (d1γ + (2∆xi)1γ) (4Hp)γ.
(2-22)
For a typical set of parameter values
ki = 0.030 W/m C
ks = 1.3 W/m C
∆xi = 0.050 m.
This approximation is within 2% for values of d from 0.025 to 1.0 m. Using this
′
approximation for A10, we obtain the following equation for Ct
[
]
Ct = I1/ln((4Hp)γ (dγ + (2∆xi)1γ d1)) + I3d(5+b+c) + A9d.
′
(2-23)
′
If we take the partial derivative of Ct and set the result to zero, we have
{
]2}{1((1γ)/(1+(2∆xi)1γ d γ1))}
[
0 = I1/d ln((4Hp)γ(dγ+ (2∆xi)1γ d1))
(5 + b + c)I3d(6+b+c) + A9.
(2-24)
This equation can not be solved explicitly for d. A solution can be obtained by
using a root-finder technique. An initial estimate needed for the solution can be
found by using the value of d obtained from the solution to the lower bounding
problem, which neglects heat losses. The cost associated with this optimal design
that neglects heat losses also provides us with a global lower bound on the actual
cost. In the next section, we consider the solution to a representative case.
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