Optimization
MECH 510
Design of Thermal Systems
American University of Beirut
Department of Mechanical Engineering
Design of Thermal Systems
1
What is Optimization?
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Optimization is the process of finding the conditions
that give maximum or minimum values of a
function
Optimization has always been an expected role of
engineers
Often a design is difficult to optimize because of its
complexity. In such cases, it may be possible to
optimize subsystems and then choose to optimize
a combination of them
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Design of Thermal Systems
Levels of Optimization
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There are two levels of optimization:
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Comparison of alternate concepts
Optimization within a concept
Optimization Methods
A complete optimization procedure consists of
proposing all reasonable alternate concepts,
optimizing the design of each concept, and then
choosing the best of the optimized designs.
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Design of Thermal Systems
Mathematical Representation
of Optimization Problems -1
The elements of the mathematical statement of
optimization are:
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Objective Function
Constraints
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Equality Constraints
Inequality Constraints
Optimization goal
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Maximizing (ex: cooling, production…)
Minimizing (ex: cost)
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Design of Thermal Systems
Mathematical Representation
of Optimization Problems -2
Objective Function
y  y ( x1, x2 ,......, xn )
(1)
Independent Variables
To Optimize
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Design of Thermal Systems
Mathematical Representation
of Optimization Problems -3

Equality Constraints
1  1( x1, x2 ,......., xn )  0
(2)
......................................
m  m ( x1, x2 ,......., xn )  0
(3)
Inequality Constraints
 1   1( x1, x2 ,......., xn )  L
(4)
......................................
 m   m ( x1, x2 ,......., xn )  L
(5)
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Design of Thermal Systems
Example #1
Determine the objective function for building a minimum cost
cylindrical refrigeration tank of volume 50 m3, if the circular ends
cost 10$/m2, the cylindrical wall costs 6$/mm2 and it costs
80$/m2 to refrigerate over the useful life.
Let x and L be respectively the diameter and length of the
cylinder:
L
4 V
200

  x2   x2
  x2
   x2

f  (10)(2)
 6    x  L  80   2 
  L
4
4


 45    x 2  86    x  L
Substituting for L we get
f  45    x 2 
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Design of Thermal Systems
17200
x
L
x
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Also it is very difficult to determine global minimum
but rather we will look for a strong local minimum.
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Design of Thermal Systems
Mathematical Representation
of Optimization Problems -4Additional properties of the optimum:
Additive constant appearing in the objective function does not
affect the independent variables at which the optimum occurs:
min [a  Y ( x1,...., xn )]  a  min [Y ( x1,...., xn )]
(6)
Maximum of a function appears at the same state point at
which the minimum of the negative of negative of the function
occurs:
max [ y ( x1,...., xn )]  min [ y ( x1,...., xn )]
(7)
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Design of Thermal Systems
Optimization Procedures -1
Calculus Methods: Lagrange Multipliers
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The basis of this method is to use derivatives to
indicate the optimum
Performs optimization where equality constraints
exists but the method can not directly
accommodate inequality constraints
A necessary requirement for using calculus
methods is the ability to extract derivatives of the
objective function and constraints
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Design of Thermal Systems
Optimization Procedures -2
Search Methods
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These methods involve examining a number of
combinations of values of the independent
variables and drawing conclusions from the
magnitude of the objective function at these
combinations
When applying search methods to continuous
functions, since only discrete points are
examined, the exact optimum can only be
approached, not reached, by a finite number of
trials
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Design of Thermal Systems
Setting Up the Mathematical
Statement of the Opt. Problem -1
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One of the first steps in performing an
optimization is to translate the physical
situation into a mathematical statement
[equations (1) to (5)]
In the optimization of thermal systems,
establishing the object function is often
simple and sometimes even a trivial task
The challenge is in writing the constraints
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Design of Thermal Systems
Setting Up the Mathematical
Statement of the Opt. Problem -2
Strategy for Writing Constraints
1.
Specify all the direct constraints (ex:
requirements of capacity, limitations of temperature and
pressure)
2.
3.
Describe in equation form the component
characteristics and properties of working
substances
Write mass and energy balances
Note: 1. and 3. usually provide a set of equations
containing more variables than exist in the objective
function. The set of constraint equations is then reduced in
number by eliminating variables that do not exist in the
objective function
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Design of Thermal Systems
Example
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Between two stages of air compression, the air is to be cooled
from 95 to 10°C. The facility to perform this cooling, shown in
the next figure, first cools the air in a pre-cooler and then in a
refrigeration unit. Water passes through the condenser of the
refrigeration unit, then into the pre-cooler, and finally to a
cooling tower, where heat is rejected to the atmosphere.
The flow rate of compressed air is 1.2 kg/s, and the specific
heat is 1.0 kJ/(kg K).
The flow rate of water is 2.3 kg/s, and its specific heat is 4.19
kJ/(kg K). The water leaves the cooling tower at a temperature
24°C.
The system is to be designed for minimum first cost, where
this first cost comprises the cost of the refrigeration unit, precooler, and cooling tower, designated x , x , & x
,
1
2
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respectively, in dollars.
Design of Thermal Systems
14
Example (continued)
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Where the costs are:
Refrigeration Unit:
x1  48q1
Precooler:
x2 
Cooling tower:
x3  25q3
(8)
50q2
( t3  t1)
t3  t1
(9)
(10)
And the q ‘s are rates of heat transfer in kilowatts, as designated
in the Figure. The compression power P kW required by the
refrigeration unit in O.25q1, and both q1 and the compression
power must be absorbed by the condenser cooling water
passing through the refrigeration unit.
Required:


Develop the objective function
Develop the constraint equations for an optimization to provide
minimum first cost
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Design of Thermal Systems
Example (Figure)
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Design of Thermal Systems
Example (Solution)
The goal of this example is only to set up the optimization
problem in the form of equations (1) to (5) and not to perform the
actual optimization.
y  y ( x1, x2 ,......, xn )  1   1( x1, x2 ,......., xn )  L
......................................
 m   m ( x1, x2 ,......., xn )  L
1  1( x1, x2 ,......., xn )  0
......................................
m  m ( x1, x2 ,......., xn )  0
Before proceeding, however, it would be instructive to examine
qualitatively the optimization features of this system. Since the
pre-cooler is a simple heat exchanger, under most operating
conditions it is less costly for a given heat- transfer rate than the
refrigeration unit. It would appear preferable, then, to do as
much cooling of the air as possible with the pre-cooler.
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Design of Thermal Systems
However, as the temperature t3 approaches the value of t1, the size of
the pre-cooler becomes very large. Some capacity is required of the
refrigeration unit in order to cool the air below 24°C. The cooling
tower must reject all the heat from the system, which includes the
heat from the air as well as the compression power to drive the
refrigeration unit. Shifting more cooling load to the refrigeration unit
increases the size and cost of the cooling tower moderately.
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Design of Thermal Systems
Example (Solution)
(a) The first assignment is to develop the expression for the objective
function. Since the total first cost is to be minimized, the objective
function will be the first cost in terms of the variables of
optimization. A choice must be made of these variables; the objective
function could conceivably be written in terms of the costs of the
individual components (the x ‘s), the energy flow rates (the q ‘s), or
even the temperatures (t1, t2, and t3). The most straightforward choice
is to use the component costs:
Total cost: y =?
(11)
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Design of Thermal Systems

Total cost: y =x1 + x2 + x3
(11)

If the q’s are chosen as the variables of optimization, it
is necessary to
start with (11) and express the x’s in terms of the q’s.
Refrigeration Unit:
x1  48q1
Precooler:
x2 
Cooling tower:
x3  25q3
50q2
( t3  t1)
t3  t1
given
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Design of Thermal Systems
Example (Solution)
(b) The next task is to write the constraints, and this means developing
the set of equations in terms of the variables used in the objective function.
Establishing the objective function is usually a simple process; the major
challenge is setting up the constraints. The advice in the early part of this
section was to specify the direct constraints, the component
characteristics, and finally the energy and mass balances. This expanded
set of equations is then reduced by eliminating the variables that do not
appear in the objective function.
A direct constraint is the requirement that the airflow rate of 1.2 kg/s be
cooled from 95 to 10°C, This requirement can be expressed in two
equations:
q1  ?
(12)
q2  ?
(13)
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Design of Thermal Systems
Energy Balances
(12)
(13)
q1  (1.2 kg / s )[1.0 kJ /(kg.K )](t 3  10)
q2  (1.2 kg / s )[1.0 kJ /(kg.K )](95  t 3 )
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Design of Thermal Systems
Example (Solution)
Under the heading of component characteristics fall the expression for the
compression power equaling 0.25q1 and the relationships of the sizes
(costs) to the capacity, equations (8) to (10).
The final category includes energy and mass balances:
Refrigeration unit:
Pre-cooler:
Cooling tower:
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Design of Thermal Systems
Example (Solution)
(14)
The final category includes energy and mass balances:
(15)
Refrigeration unit:
q1 + P = (2.3 kg/s) [4.19 kJ/(kg K)] (t1 -24)
(16)
Pre-cooler:
(1.2)(1.0)(95 - t3) = (2.3)(4.19)(t2 - t1)
Cooling tower:
(2.3)(4.19)(t2-24) = q3
The complete set of constraint equations is ??
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Design of Thermal Systems
24
Example (Solution)
The complete set of constraint equations is
q1 = (1.2)(1.0)(t3 — 10)
q2 = (1 2)(1.0)(95 — t3)
P = 0.25q1
x1= 48q1
x2 
50q2
t3  t1
(17)
(18)
(19)
(20)
(21)
x3 = 25q3
q1 + P = (2.3 kg/s) [4.19 kJ/(kg K)] (t1 -24)
(1.2)(1.0)(95 - t3) = (2.3)(4.19)(t2 - t1)
(2.3)(4.19)(t2-24) = q3
(22)
(23)
(24)
(25)
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Design of Thermal Systems
Example (Solution)
There are nine equations in the set, equations (17) to (25) and ten
unknowns, q1, q2, q3, P, x1, x2, x3, t1, t2, and t3. The next operation is to
eliminate in this set of equations all but the variables of optimization, x1, x2,
and x3. As the elimination of variables and equations proceeds, there will
always be one more unknown than the number of equations, so when all
but the three x’s are eliminated, there should be two equations remaining.
These two constraint equations are equations (27) and (28), so the
complete mathematical statement of this optimization problem is as
follows:
Minimize
y =x1 + x2 + x3
(26)
subject to
0.01466 x1x2 - 14x2 + 1.042 x1 = 5100
7.69 x3 - x1 =19,615
(27)
(28)
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Design of Thermal Systems
Example (Discussion)
The objective of this chapter is to introduce procedures for setting up the
mathematical statement of the optimization problem. With one of the
methods in the subsequent chapters the execution of the optimization
process would show that the optimal values of x1, x2, and x3 are $1450,
$496, and $2738, respectively. Equation (9) [or (21)] was presented as
valid if t3 > t1, and this condition could legitimately be listed as one
of the constraints. If t3 < t1, x2 becomes negative, which is physically
impossible.
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Design of Thermal Systems
Example (Discussion)
The constraints are an integral part of the statement of the optimization
problem. The objective function without the constraints is
meaningless because the x ‘s could all shrink to zero and there would be
no cost for the system. The constraint in Eq. (27) requires a positive value
of which is the same as requiring the existence of a refrigeration unit. From
heat-transfer considerations the pre-cooler can cool the air no lower than a
temperature of 24°C. Substituting x1 = 0 in Eq. (27) makes x2 negative,
which is physically impossible. Equation (27) does permit x2 to be zero, in
which case all cooling is performed by the refrigeration unit.
The constraint equation (28) imposes a minimum value of the coolingtower size and cost x . As the size of the refrigeration unit and x1
increases, x3 also increases because of the compression power
associated with the refrigeration unit.
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