Basic Optimization Theory
• LP (Linear Programming)
• NLP (Non-Linear Programming)
• IP (Integer Programming)
• MIP (Mixed Integer Programming)
• MINLP (Mixed Integer Non-Linear Programming)
Analysis, Modelling and Simulation of Energy Systems, SEE-T9
Mads Pagh Nielsen
Types Of Optimization
• Parameter Optimization
• Configuration Optimization
• Operational Optimization
• Topology Optimization
Analysis, Modelling and Simulation of Energy Systems, SEE-T9
Mads Pagh Nielsen
Topology Optimization
Analysis, Modelling and Simulation of Energy Systems, SEE-T9
Mads Pagh Nielsen
The General Non-Linear Problem
Objective Function
max. or min. f ( x , x
subject to:
g (x , x
Constraints
g (x ,x
1
2
,..., x n )
1
1
2
,..., x n )  b1
2
1
2
L
g (x , x
m
1
,..., x n )  b 2
L
2
,..., x n )  b m
Design Space
Analysis, Modelling and Simulation of Energy Systems, SEE-T9
Mads Pagh Nielsen
Extrema: Ordering Situation
$
1000
800
Total Cost
600
400
Transport Cost
200
Ordering Cost
OPT.
0
0
10
20
30
40
50
Order Quantity
Analysis, Modelling and Simulation of Energy Systems, SEE-T9
Mads Pagh Nielsen
Extrema: Heat Exchanger
$
1000
800
Total Cost
600
400
Material Cost
200
Energy Cost
OPT.
0
0
100
200
300
400
500
Heat Exchanger Area
Analysis, Modelling and Simulation of Energy Systems, SEE-T9
Mads Pagh Nielsen
Local vs. Global Extrema
f(x)=sin(30x)/x
0,4
0,2
0
-0,2
0
2
4
6
8
10
12
14
-0,4
-0,6
-0,8
-1
-1,2
global
max
stationary
point
f(x)
local
max
local
min
local
min
x
Analysis, Modelling and Simulation of Energy Systems, SEE-T9
Mads Pagh Nielsen
Convexity
f(x)
f(x)
Concave Function
Convex Function
f(x2)
 f(x1)+(1- )f(x2)

f(x1)
 f(x1 +(1- )x2)
x1
x1+(1-)x2
x2 x
f(x) is a convex function if and only if for any given two points x1 and x2 in the
function domain and for any constant 0    1
f(x1 +(1- )x2)   f(x1)+(1- )f(x2)
Analysis, Modelling and Simulation of Energy Systems, SEE-T9
Mads Pagh Nielsen
x
The Hessian
• The gradient vector of f at x
 f

 x1
 f

f ( x )   x2
 ...

 f
 x
 n











• The Hessian Matrix of f at x
 2 f

 x1x1
 2 f ( x)   ...

 2 f

 xn x1
...
...
...
2 f 

x1xn 

...

2
 f 

xn xn 
Analysis, Modelling and Simulation of Energy Systems, SEE-T9
Mads Pagh Nielsen
Conditions for convexity
How can we use Hessian to determine whether or not
f(x) is convex?
• H(x) is positive semi-definite if and only if xTHx ≥ 0 for all x
and there exists and x  0 such that xTHx = 0. => Convexity
• H(x) is positive definite if and only if xTHx > 0 for all x  0.
• H(x) is indefinite if and only if xTHx > 0 for some x, and xTHx
< 0 for some other x.
Analysis, Modelling and Simulation of Energy Systems, SEE-T9
Mads Pagh Nielsen
Possible Solutions To Convex Problems
objective function
level curve
objective function
level curve
optimal solution
optimal solution
Feasible
Region
Feasible
Region
nonlinear objective,
linear constraints
linear objective,
nonlinear constraints
objective function
level curve
objective function
level curves
optimal solution
Feasible
Region
nonlinear objective,
nonlinear constraints
optimal solution
Feasible
Region
nonlinear objective,
linear constraints
Analysis, Modelling and Simulation of Energy Systems, SEE-T9
Mads Pagh Nielsen
Line Search
Line search techniques are in essence optimization algorithms for one
dimensional minimization problems.
They are often regarded as the backbones of nonlinear optimization
algorithms.
Typically, these techniques search a bracketed interval.
Often, unimodality is assumed.
a
x*
b
Exhaustive search requires N = (b-a)/ + 1
calculations to search the above interval, where
 is the resolution.
Analysis, Modelling and Simulation of Energy Systems, SEE-T9
Mads Pagh Nielsen
Bracketing Algorithm

a
x1
x2
b
Two point search (dichotomous search) for finding the solution to minimizing ƒ(x):
0) assume an interval [a,b]
1) Find x1 = a + (b-a)/2 - /2 and x2 = a+(b-a)/2 + /2 where  is the resolution.
2) Compare ƒ(x1) and ƒ(x2)
3) If ƒ(x1) < ƒ(x2) then eliminate x > x2 and set b = x2
If ƒ(x1) > ƒ(x2) then eliminate x < x1 and set a = x1
If ƒ(x1) = ƒ(x2) then pick another pair of points
4) Continue placing point pairs until interval < 2 
Analysis, Modelling and Simulation of Energy Systems, SEE-T9
Mads Pagh Nielsen
Golden Section Search
b
a
Discard
a-b
a
b
In Golden Section, you try to have b/(a-b) = a/b
which implies b*b = a*a - ab
Solving this gives a = (b ± b* sqrt(5)) / 2
a/b = -0.618 or 1.618 (Golden Section ratio)
Note that 1/1.618 = 0.618
Analysis, Modelling and Simulation of Energy Systems, SEE-T9
Mads Pagh Nielsen
Golden Section Search Algorithm
{Initialize}
x1 = a + (b-a)*0.382
x2 = a + (b-a)*0.618
f1 = ƒ(x1)
f2 = ƒ(x2)
a
x1
x2
b
{Loop}
if f1 > f2 then
a = x1; x1 = x2; f1 = f2
x2 = a + (b-a)*0.618
f2 = ƒ(x2)
else
b = x2; x2 = x1; f2 = f1
x1 = a + (b-a)*0.382
f1 = ƒ(x1)
end
Analysis, Modelling and Simulation of Energy Systems, SEE-T9
Mads Pagh Nielsen
The 2D case
X2
Local optimal solution
C
E
Feasible Region
B
F
Local and global
optimal solution
G
A
D
X1
Analysis, Modelling and Simulation of Energy Systems, SEE-T9
Mads Pagh Nielsen
The 2D case
(From John Rasmussen, 1999)
Analysis, Modelling and Simulation of Energy Systems, SEE-T9
Mads Pagh Nielsen
Example: Heat Exchanger
Problem: Find the radius of tubes in a heat exchanger to
maximize the total surface area.
(The magnitude of pressure drops are not considered)
Analysis, Modelling and Simulation of Energy Systems, SEE-T9
Mads Pagh Nielsen
Setting up the problem:
What is the best value of r ?
What if we added a maximum allowable pressure drop?
Analysis, Modelling and Simulation of Energy Systems, SEE-T9
Mads Pagh Nielsen
Ideal Rankine Cycle Balance
•
•
Assumptions: steady flow process, no generation, neglect KE and PE changes for all four devices,
First Law: 0 = (net heat transfer in) - (net work out) + (net energy flow in)

0 = (qin - qout) - (Wout - Win) + (hin - hout)
• 1-2: Pump (q=0)
 Wpump = h2 - h1 = v(P2-P1)
3
T
• 2-3: Boiler (W=0)  qin = h3 - h2
2
1
• 3-4: Turbine (q=0)  Wout = h3 - h4
4
s
• 4-1: Condenser (W=0)  qout = h4 - h1
Thermal efficiency h = Wnet/qin = 1 - qout/qin = 1 - (h4-h1)/(h3-h2)
Wnet = Wout - Win = (h3-h4) - (h2-h1)
Analysis, Modelling and Simulation of Energy Systems, SEE-T9
Mads Pagh Nielsen
Possibility for improvement of thermal efficiency:
Thermal efficiency can be improved by manipulating the temperatures and/or pressures in
various components
(a) Lowering the condensing pressure (lowers TL, but decreases quality, x4 )
(b) Superheating the steam to a higher temperature (increases TH but requires higher
temp materials)
(c) Increasing the boiler pressure (increases TH but requires higher temp/press
materials)
T
3
(b) Superheating
1
3
T
2
2
T
(c) increase pressure
1
2
4
4
1
2
4
s
(a) lower pressure(temp)
1
Low quality
high moisture content
s
s
Red area = increase in W net
Blue area = decrease in W net
Analysis, Modelling and Simulation of Energy Systems, SEE-T9
Mads Pagh Nielsen
Reheating
•
The optimal way of increasing the boiler pressure without increasing the moisture
content in the exiting vapor is to reheat the vapor after it exits from a first-stage turbine
and redirect this reheated vapor into a second turbine.
3
high-P
turbine
T
Low-P
turbine
high-P
turbine
5
3
low-P
turbine
boiler
4
4
5
2
6
2
1
pump
4
6
1
condenser
Analysis, Modelling and Simulation of Energy Systems, SEE-T9
Mads Pagh Nielsen
s
Reheating
•
•
•
•
•
Reheating allows one to increase the boiler pressure without increasing the moisture
content in the vapor exiting from the turbine.
By reheating, the average temperature of the vapor entering the turbine is increased,
thus, it increases the thermal efficiency of the cycle.
Multistage reheating is possible but not practical. One major reason is because the
vapor exiting will be superheated vapor at higher temperature, thus, decrease the
thermal efficiency.
Energy analysis: Heat transfer and work output both change
qin = qprimary + qreheat = (h3-h2) + (h5-h4)
Wout = Wturbine1 + Wturbine2 = (h3-h4) + (h5-h6)
Analysis, Modelling and Simulation of Energy Systems, SEE-T9
Mads Pagh Nielsen
Regeneration
•
•
From 2-2’, the average temperature is very low, therefore, the heat addition process is at
a lower temperature and therefore, the thermal efficiency is lower.
Use a regenerator to heat the liquid (feedwater) leaving the pump before sending it to
the boiler. This increases the average temperature during heat addition in the boiler,
hence it increases efficiency.
Lower temp
heat addition
T
3
T
higher temp
heat addition
2’
5
4
6
2
3
2
1
4
1
s
7
s
Use regenerator to heat up the feedwater
Analysis, Modelling and Simulation of Energy Systems, SEE-T9
Mads Pagh Nielsen
Regeneration Cycle
•
•
Improve efficiency by increasing feedwater temperature before it enters the boiler.
Two Options:
– Open feedwater : Mix steam with the feedwater in a mixing chamber.
– Closed feedwater: No mixing.
Open FWH
5
T
6
(y)
Open
FWH
boiler
4
Pump 2
3
5
4
7 (1-y)
2
2
(y) 6
(1-y)
3
1
7
s
Pump 1
1
condenser
Analysis, Modelling and Simulation of Energy Systems, SEE-T9
Mads Pagh Nielsen
Analysis Of Regenerative Cycle
• Assume y percent of steam is extracted from the turbine and is directed into
open feedwater heater.
• Energy analysis:
qin = h5-h4,
qout = (1-y)(h7-h1),
Wturbine, out = (h5-h6) + (1-y)(h6-h7)
Wpump, in
= (1-y)Wpump1 + Wpump2
= (1-y)(h2-h1) + (h4-h3)
= (1-y)v1(P2-P1) + v3(P4-P3)
(Note: Ideal pumps)
• In general, more feedwater heaters result in higher cycle efficiencies.
• However this does not mean that it is necessary a practical optimal solution!
Analysis, Modelling and Simulation of Energy Systems, SEE-T9
Mads Pagh Nielsen