Modeling and Simulation
EEE621
Dr. Muhammad Naeem
1

Examples: Linear Programming
 Let the generator marginal costs be:
– Bus 1: 10 $ / MWhr; Range = 0 to 400 MW,
– Bus 2: 12 $ / MWhr; Range = 0 to 400 MW,
– Bus 3: 20 $ / MWhr; Range = 0 to 400 MW,
 Assume a single 180 MW load
 The combined load of Generator 1 and 2 must be less that 100MW and in this load 2/3 load will be from generator 1.
2

min
P P P
3
Examples: Linear Programming
 Let the generator marginal costs be:
10 P
1
 12 P
2
 20 P
3
– Bus 1: 10 $ / MWhr; Range = 0 to 400 MW,
– Bus 2: 12 $ / MWhr; Range = 0 to 400 MW,
P
1
 P
2
 P
3
 180
0.66
P
1
 0.33
P
2
 100
0  P P P
1 2 3
 400
– Bus 3: 20 $ / MWhr; Range = 0 to 400 MW,
 Assume a single 180 MW load
 The combined load of Generator 1 and 2 must be less that 100MW and in this load 2/3 load will be from generator 1.
3

min
P P P
3 f = [10 12 20];
Examples: Linear Programming
 Let the generator marginal costs be:
10 P
1
 12 P
2
 20 P
3
– Bus 1: 10 $ / MWhr; Range = 0 to 400 MW,
– Bus 2: 12 $ / MWhr; Range = 0 to 400 MW,
P
1
 P
2
 P
3
 180
0.66
P
1
 0.33
P
2
 100
0  P P P
1 2 3
 400
– Bus 3: 20 $ / MWhr; Range = 0 to 400 MW,
 Assume a single 180 MW load
 The combined load of Generator 1 and 2 must be less that 100MW and in this load 2/3 load will be from generator 1.
%inequality constraint
A = [0.66 0.33 0]; b = [100];
%equality constraint
Aeq = [1 1 1]; beq = [180];
%lower and bounds lb = zeros(1,3); ub = [400 400 400]; x =
123.0303
56.9697
0.0000
fval = 1914
[x,fval,exitflag] = linprog(f,A,b,Aeq,beq,lb,ub);
4

1
0
-2
-3
Examples: Linear Programming exitflag :Integer identifying the reason the algorithm terminated. The following lists the values of exitflag and the corresponding reasons the algorithm terminated.
Function converged to a solution x.
Number of iterations exceeded options.MaxIter.
No feasible point was found.
Problem is unbounded.
f = [10 12 20];
%inequality constraint
A = [0.66 0.33 0]; b = [100];
%equality constraint
Aeq = [1 1 1]; beq = [180];
%lower and bounds lb = zeros(1,3); ub = [400 400 400];
[x,fval,exitflag] = linprog(f,A,b,Aeq,beq,lb,ub);
5

 A good business practice is the one in which the production cost is minimized without sacrificing the quality. This is not any different in the power sector as well. The main aim here is to reduce the production cost while maintaining the voltage magnitudes at each bus.
 A power plant has to cater to load conditions all throughout the day, come summer or winter. It is therefore illogical to assume that the same level of power must be generated at all time.
 The power generation must vary according to the load pattern, which may in turn vary with season. Therefore the economic operation must take into account the load condition at all times. Moreover once the economic generation condition has been calculated, the turbine-governor must be controlled in such a way that this generation condition is maintained.
 In an early attempt at economic operation it was decided to supply power from the most efficient plant at light load conditions. As the load increased, the power was supplied by this most efficient plant till the point of maximum efficiency of this plant was reached. With further increase in load, the next most efficient plant would supply power till its maximum efficiency is reached. In this way the power would be supplied by the most efficient to the least efficient plant to reach the peak demand.
Unfortunately however, this method failed to minimize the total cost of electricity generation. We must therefore search for alternative method which takes into account
6 the total cost generation of all the units of a plant that is supplying a load.

Economic Distribution of Loads between the Units of a Plant
To determine the economic distribution of a load amongst the different units of a plant, the variable operating costs of each unit must be expressed in terms of its power output. The fuel cost is the main cost in a thermal or nuclear unit. Then the fuel cost must be expressed in terms of the power output. Other costs, such as the operation and maintenance costs, can also be expressed in terms of the power output. Fixed costs, such as the capital cost, depreciation etc., are not included in the fuel cost.
The fuel requirement of each generator is given in terms of the Rupees/hour. Let us define the input cost of an uniti , f i in Rs./h and the power output of the unit as be expressed in terms of the power output as
P i
. Then the input cost can f i
 a i
2
P i
2
 b P i i
 c i
/
7

•
•
•
A B C
L
8

400
0.4P
2
+10P+25
0.35P
2
+6P+20
350
P = 0:1:20; % power in mega watts
%a, b, and care the coeffi cients of the input - output characteristic a = [0.8 0.7]; b = [10 6]; c = [25 20];
300
250
200
150 i = 1; f(i,:) = a(i)/2 .* P.^2 + b(i) .* P + c(i); i = 2; f(i,:) = a(i)/2 * P.^2 + b(i) * P + c(i);
100
50
0
0 2 4 6 8 10
P (MW)
12 plot(P,f(1,:), 'r' , 'LineWidth' ,2); hold on ; plot(P,f(2,:), 'b' , 'LineWidth' ,2); xlabel( 'P (MW)' ); ylabel( 'Rs/h' ); legend( '0.4P^2+10P+25' , '0.35P^2+6P+20' )
14 16 18 f i
 a i
2
P i
2
 b P i i
 c i
/
20
9

min
P
K k

 1 f k
K k

 1
P k
 P
T
P k min
 P k
 P k max
, where f k
 a k
2 a = [0.8 0.7 0.95]; b = [10 5 15]; c = [25 20 35];
%------------------------------------------
Pmin = [30 30 30];
Pmax = [500 500 250];
%------------------------------------------
%inline function f = @(x) (a(1)/2 * x(1)^2 + b(1) * x(1) + c(1) ...
+a(2)/2 * x(2)^2 + b(2) * x(2) + c(2) ...
+a(3)/2 * x(3)^2 + b(3) * x(3) + c(3));
%-----------------------------------------lb = Pmin; ub = Pmax;
Aeq = [1 1 1]; beq = [1000]; x0 = [lb]; options = optimset( 'Display' , 'off' , 'Algorithm' , 'active-set' );
[x,fval,exitflag] = fmincon(f,x0,[],[],Aeq,beq,lb,ub,[],options);
P k
2
 b P k k
 c k
10
 k

a = [0.8 0.7 0.95]; b = [10 5 15]; c = [25 20 35];
%------------------------------------------
Pmin = [30 30 30];
Pmax = [500 500 250];
%-----------------------------------------lb = Pmin; ub = Pmax;
Aeq = [1 1 1]; beq = [1000]; x0 = [lb]; options = optimset( 'Display' , 'off' , 'Algorithm' , 'active-set' );
[x,fval,exitflag] = fmincon(@(x)ObjFunction(x,a,b,c),x0,[],[],Aeq,beq,lb,ub,[],options); min
P where f k
 a k
2 k
K

 1
P k min f k
K k

 1
P k
 P
T
 P k
 P k max
,
P k
2
 b P k k
 c k
 k function f = ObjFunction(x,a,b,c)
%Method 1
% f = (a(1)/2 * x(1)^2 + b(1) * x(1) + c(1)...
% +a(2)/2 * x(2)^2 + b(2) * x(2) + c(2)...
% +a(3)/2 * x(3)^2 + b(3) * x(3) + c(3));
%
%Alternate Method
Sum = 0; for i = 1:3
Sum = Sum + a(i)/2 * x(i)^2 + b(i) * x(i) + c(i); end f = Sum; 11

min
P
K k

 1 f k
K k

 1
P k
 P
T
P k min
 P k
 P k max
, where f k
 a k
2
P k
2
 b P k k
 c k
 k
St_Data.a = [0.8 0.7 0.95];
St_Data.b = [10 5 15];
St_Data.c = [25 20 35];
%------------------------------------------
Pmin = [30 30 30];
Pmax = [500 500 250];
%-----------------------------------------lb = Pmin; ub = Pmax;
Aeq = [1 1 1]; beq = [1000]; x0 = [lb]; options = optimset( 'Display' , 'off' , 'Algorithm' , 'active-set' );
[x,fval,exitflag] = fmincon(@(x)ObjFunctionStruct(x,St_Data),x0,[],[],Aeq,beq,lb,u b,[],options); function f = ObjFunctionStruct(x,St_Data) a = St_Data.a; b = St_Data.b; c = St_Data.c;
%Method 1
% f = (a(1)/2 * x(1)^2 + b(1) * x(1) + c(1)...
% +a(2)/2 * x(2)^2 + b(2) * x(2) + c(2)...
% +a(3)/2 * x(3)^2 + b(3) * x(3) + c(3));
%
%Alternate Method
Sum = 0; for i = 1:3
Sum = Sum + a(i)/2 * x(i)^2 + b(i) * x(i) + c(i); end f = Sum; 12

70
4500
4000
Cost Function
60
P
1
P
2
P
3
50
3500
40
3000
30 2500
2000
20
1500
80 90 100 110 120
P
T
(MW)
130 140 150 10
80 min
P
K k

 1 f k
K k

 1
P k
 P
T
P k min
 P k
 P k max
, where f k
 a k
2
P k
2  b P k
 c k
 k
90 100 110
P
T
(MW)
120 130 140 150
13

St_Data.a = [0.8 0.7 0.95];
St_Data.b = [10 5 15];
St_Data.c = [25 20 35]; min
P k
K

 1 f k
%------------------------------------------
Pmin = [30 10 30];
Pmax = [50 60 50];
PT = 80:10:150;
%------------------------------------------
K k

 1
P k
 P
T lb = Pmin; ub = Pmax;
Aeq = [1 1 1]; x0 = [lb]; options = optimset( 'Display' , 'off' , 'Algorithm' , 'active-set' );
P where f k k min

 a k
2
P
P k k
2


P k max b P k
,
 c
 k k for i = 1:length(PT) beq = PT(i);
[x,fval,exitflag] = fmincon(@(x)ObjFunctionStruct(x,St_Data),x0,[],[],Aeq,beq,lb,ub,[],options);
P(i,:) = x;
Cost(i) = fval;
ExitFlag(i) = exitflag; end
%==============================================
%Now do the plot plot(PT,P(:,1), '-or' , 'LineWidth' ,LINEWIDTH); hold on ; plot(PT,P(:,2), '-og' , 'LineWidth' ,LINEWIDTH); hold on ; plot(PT,P(:,3), '-ob' , 'LineWidth' ,LINEWIDTH); hold on ; xlabel( 'P_T (MW)' ) ylabel( 'Price (Rs/h)' ) legend( 'P_1' , 'P_2' , 'P_3' ) a = axis; axis([a(1) a(2) 10 70]) grid
14

4000 130
Cost
120
110
3500
3000
100
90
80
2500
2000
70
60
50
40
30
Load Demand
P
1
P
2
P
3
1500
2 4 6 8 10 16 18 20 22 24
20
2 4 6 8 10
12 14
Time
12 14 16 18 20 22 24
Time min
P
K k

 1 f k
K k

 1
P k
 P
T
P k min
 P k
 P k max
,  k
Pmin = [20 40 10];
Pmax = [50 45 50]; where f k
 a
2 k P k
2
 b P k
 c k
PT = [80 80 90 90 100 100 90 90 120 120 120 120 130 130 100 100 80 80 80 80 80 80 80 80]
Time = 1:24; 15

St_Data.a = [0.8 0.7 0.95];
St_Data.b = [10 5 15];
St_Data.c = [25 20 35];
%------------------------------------------
Pmin = [20 40 10];
Pmax = [50 45 50];
PT = [80 80 90 90 100 100 90 90 120 120 120 120 130 130 100 100 80 80 80 80 80 80 80 80]
Time = 1:24;
%-----------------------------------------lb = Pmin; ub = Pmax;
Aeq = [1 1 1]; x0 = [lb]; options = optimset( 'Display' , 'off' , 'Algorithm' , 'active-set' ); min
P k

 1
K k

 1
K f k
P k
 P
T
P k min
 P k
 P k max
,  k for i = 1:length(PT) beq = PT(i);
[x,fval,exitflag] = where f k
 a k fmincon(@(x)ObjFunctionStruct(x,St_Data),x0,[],[],Aeq,beq,lb,ub,[],options);
2
P k
2  b P k k
 c k
P(i,:) = x;
Cost(i) = fval;
ExitFlag(i) = exitflag; end stairs(Time,PT, '-m' , 'LineWidth' ,LINEWIDTH); hold on ; stairs(Time,P(:,1), '-r' , 'LineWidth' ,LINEWIDTH); hold on ; stairs(Time,P(:,2), '-g' , 'LineWidth' ,LINEWIDTH); hold on ; stairs(Time,P(:,3), '-b' , 'LineWidth' ,LINEWIDTH); hold on ; xlabel( 'Time' ) ylabel( 'MWatt' ) legend( 'Load Demand' , 'P_1' , 'P_2' , 'P_3' ) a = axis; axis([1 24 15 130]) grid
16

P
K k

 1 f k
St_Data.a = [0.8 0.7 0.95];
St_Data.b = [10 5 15];
St_Data.c = [25 20 35]; where f k
K k

 1
P k
 P
T f k
 F k
,  k f
1
 2 f
2
P k min
 P k
 P k max
,
 a k
2
 k
P k
2
 b P k
 c k
St_Data.F = 5e5 * ones(1,3);
%------------------------------------------
Pmin = [0 0 0];
Pmax = [500 500 250];
%-----------------------------------------lb = Pmin; ub = Pmax;
Aeq = [1 1 1]; beq = [1000]; x0 = [lb]; options = optimset( 'Display' , 'off' , 'Algorithm' , 'active-set' );
[x,fval,exitflag] = fmincon(@(x)ObjFunctionStruct(x,St_Data),x0,[],[],Aeq,beq,lb,ub,@
(x)Confun(x,St_Data),options); function f = ObjFunctionStruct(x,St_Data) a = St_Data.a; b = St_Data.b; c = St_Data.c;
%Method 1
% f = (a(1)/2 * x(1)^2 + b(1) * x(1) + c(1)...
% +a(2)/2 * x(2)^2 + b(2) * x(2) + c(2)...
% +a(3)/2 * x(3)^2 + b(3) * x(3) + c(3));
%
%Alternate Method
Sum = 0; for i = 1:3
Sum = Sum + a(i)/2 * x(i)^2 + b(i) * x(i) + c(i); end f = Sum; function [c, ceq] = Confun(x,St_Data) a = St_Data.a; b = St_Data.b; c = St_Data.c;
% Nonlinear inequality constraints
ConsIneq = zeros(1,3); for i = 1:3 f(i) = a(i)/2 * x(i)^2 + b(i) * x(i) + c(i);
ConsIneq(i) = f(i) - St_Data.F(i); end c = ConsIneq';
% Nonlinear equality constraints ceq = f(1)-2*f(2);
%ceq = [];
17

•
•
•
A B C
L
18

•
•
•
G
?
G
?
G
?
Load Profile
19

 Unit 1: P
Min
C
1
= 100MW, P
Max
= 400MW
= 510.0 + 7.9 P
1
+ 0.172 P
1
2
 Unit 2: P
Min
= 50MW, P
Max
= 400 MW
C
2
= 310.0 + 7.85 P
2
+ 0.194 P
2
2
$/h
$/h
 Unit 3: P
Min
= 150 MW, P
Max
= 500 MW
C
3
= 78.0 + 9.56 P
3
 Unit 4: P
Min
= 200 MW, P
+ 0.694 P
Max
3
2 $/h
= 400 MW
C
4
= 210.0 + 5.85 P
4
+ 0.5 P
4
2 $/h
 Unit 5: P
C
Min
5
= 150 MW, P
Max
= 50.0 + 1.56 P
5
= 500 MW
+ 0.8 P
5
2 $/h
 What combination of units 1, 2,3,4 and 5 will produce 550 MW at minimum cost?
 How much should each unit in that combination generate?
20

 Unit 1: P
Min
= 100MW, P
Max
= 400MW
C
1
= 510.0 + 7.9 P
1
+ 0.172 P
1
2 $/h
 Unit 2: P
Min
= 50MW, P
Max
= 400 MW
C
2
= 310.0 + 7.85 P
2
+ 0.194 P
2
2 $/h
 Unit 3: P
Min
= 150 MW, P
Max
= 500 MW
C
3
= 78.0 + 9.56 P
3
+ 0.694 P
3
2 $/h
 Unit 4: P
Min
= 200 MW, P
Max
= 400 MW
C
4
= 210.0 + 5.85 P
4
+ 0.5 P
4
2 $/h
 Unit 5: P
Min
= 150 MW, P
Max
= 500 MW
C
5
= 50.0 + 1.56 P
5
+ 0.8 P
5
2 $/h min
X,P
K k

 1 x f k k
K k

 1
P k
 P
T
P k
 x P k k max
, 
P k min x k
 P k
,  k k where f k
 a k
2
P k
2
 b P k k
 c k
 What combination of units 1, 2,3,4 and 5 will produce ---- MW at minimum cost?
 How much should each unit in that combination generate?
21

Some Applications of Knapsack Problem
Optimal Allocation of Renewable Energy (RE) Parks
The optimization problem we address is defined as follows.
Given the country of Pakistan with its data and constraints :
1) Pakistan map of populated areas,
2) Wind and Solar atlas,
3) electricity grid map,
4) current and forecast energy cost per Renewable Energy (RE) resource,
5) current and forecast energy demands per month,
6) a set of potential RE park locations ;
Determine the set of energy parks to be invested in today, the set of energy parks to be invested in the future, such that 20% of the current and forecast energy demand are covered for each month of the year, and the anticipated financial cost is minimized.
The cost is determined in terms of sum of total costs associated with each potential park : cost of connection to the grid, cost of installation, and cost of park maintenance.
22

Some Applications of Knapsack Problem
Optimal Allocation of Renewable Energy (RE) Parks
The following input data is provided by the decision maker to our system, but can also be preprocessed and entered automatically.
Input data
Ci = Cost of park i
Mi = Maximum area of park I
(xi ; yi) = Coordinates of park i
Gij = Watts/m2 produced by park i in month j
Dj = 20% Electricity demand in month j
Note that the cost Ci is the total cost for each potential location and is composed of, installation cost, maintenance cost, and the cost of building the transportation lines to connect and bring the electricity to the grid.
23

Some Applications of Knapsack Problem
Optimal Allocation of Renewable Energy (RE) Parks
The following input data is provided by the decision maker to our system, but can also be preprocessed and entered automatically.
Decision variables Since a Boolean model is used, the decision variables are associated to each potential park location :
Input data
Ci = Cost of park i
Mi = Maximum area of park I
(xi ; yi) = Coordinates of park i
Gij = Watts/m2 produced by park i in month j
Dj = 20% Electricity demand in month j x i
 1
 

0 if park i is chosen otherwise
Note that the cost Ci is the total cost for each potential location and is composed of, installation cost, maintenance cost, and the cost of building the transportation lines to connect and bring the electricity to the grid.
24

Some Applications of Knapsack Problem
Optimal Allocation of Renewable Energy (RE) Parks
The following input data is provided by the decision maker to our system, but can also be preprocessed and entered automatically.
Decision variables Since a Boolean model is used, the decision variables are associated to each potential park location :
Input data
Ci = Cost of park i
Mi = Maximum area of park I
(xi ; yi) = Coordinates of park i
Gij = Watts/m2 produced by park i in month j
Dj = 20% Electricity demand in month j
Note that the cost Ci is the total cost for each potential location and is composed of, installation cost, maintenance cost, and the cost of building the transportation lines to connect and bring the electricity to the grid.
x i
 1
 

0 if park i is chosen otherwise
Constraints The total electricity generated by the RE parks must satisfy 20% of the current demand.
Given the demand Dj for month j , we have : n i

 1
G x ij i
 D j
,    1, 2,...,12 
Objective Function We seek to minimize the current cost over all the potential parks : min n i

 1
C x i i
,
25

Some Applications of Knapsack Problem
Optimal Allocation of Renewable Energy (RE) Parks
The following input data is provided by the decision maker to our system, but can also be preprocessed and entered automatically.
Decision variables Since a Boolean model is used, the decision variables are associated to each potential park location :
Input data
Ci = Cost of park i
Mi = Maximum area of park I
(xi ; yi) = Coordinates of park i
Gij = Watts/m2 produced by park i in month j
Dj = 20% Electricity demand in month j
Note that the cost Ci is the total cost for each potential location and is composed of, installation cost, maintenance cost, and the cost of building the transportation lines to connect and bring the electricity to the grid.
x i
 1
 

0 if park i is chosen otherwise
Constraints The total electricity generated by the RE parks must satisfy 20% of the current demand.
Given the demand Dj for month j , we have : n i

 1
G x ij i
 D j
,    1, 2,...,12 
Objective Function We seek to minimize the current cost over all the potential parks : min n i

 1
C x i i
,
26

Applications
Hydroelectric Power Systems Planning
ENGINEERING OPTIMIZATION Methods and Applications, A. Ravindran, K. M. Ragsdell, G. V. Reklaitis
An agency controls the operation of a system consisting of two water reservoirs with one hydroelectric power generation plant attached to each, as shown in Figure 4.1.
The planning horizon for the system is broken into two periods. When the reservoir is at full capacity, additional inflowing water is spilled over a spillway. In addition, water can also be released through a spillway as desired for flood protection purposes. Spilled water does not produce any electricity.
Assume that on an average 1 kilo-acre-foot (KAF) of water is converted to 400 megawatt hours (MWh) of electricity by power plant A and 200 MWh by power plant B. The capacities of power plants A and B are 60,000 and
35,000 MWh per period. During each period, up to 50,000 MWh of electricity can be sold at $20.00/MWh, and excess power above 50,000 MWh can only be sold for $14.00/MWh.
27

Applications
Hydroelectric Power Systems Planning
ENGINEERING OPTIMIZATION Methods and Applications, A. Ravindran, K. M. Ragsdell, G. V. Reklaitis
The following table gives additional data on the reservoir operation and inflow in kilo-acre-feet:
Conversion Rate per kilo-acre-foot (KAF)
Capacity of Power Plants
Capacity of Reservoir
Predicted Flow
Period 1
Period 2
Minimum Allowable Level
Level at the beginning of period 1
Plant / Reservoir A Plant / Reservoir B
400 MWh 200 MWh
60,000 MWh/Period 35,000 MWh/Period
2000 1500
200
130
1200
1900
40
15
800
850
28

Applications
Hydroelectric Power Systems Planning

∞
29

Applications
Hydroelectric Power Systems Planning
PH1
PL1
XA1
XB1
SA1
SB1
EA1
EB1
Power sold at $20/MWh
Power sold at $14/MWh
Water supplied to power plant A
MWh
MWh
KAF
Water supplied to power plant B
Spill water drained from reservoir A
Spill water drained from reservoir B
KAF
KAF
KAF
Reservoir A level at the end of period 1 KAF
Reservoir B level at the end of period 1 KAF
Conversion Rate per kilo-acre-foot (KAF)
Capacity of Power Plants
Capacity of Reservoir
Predicted Flow
Period 1
Period 2
Minimum Allowable Level
Level at the beginning of period 1
Plant / Reservoir A Plant / Reservoir B
400 MWh 200 MWh
60,000 MWh/Period 35,000 MWh/Period
2000 1500
200
130
1200
1900
40
15
800
850
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Applications
Hydroelectric Power Systems Planning
PH1
PL1
XA1
XB1
SA1
SB1
EA1
EB1
Power sold at $20/MWh
Power sold at $14/MWh
Water supplied to power plant A
MWh
MWh
KAF
Water supplied to power plant B
Spill water drained from reservoir A
Spill water drained from reservoir B
KAF
KAF
KAF
Reservoir A level at the end of period 1 KAF
Reservoir B level at the end of period 1 KAF
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Applications
32

Applications
33

Applications
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Combined cooling, heat and power system planning
1) Jradi, M., and S. Riffat. "Tri-generation systems: Energy policies, prime movers, cooling technologies, configurations and operation strategies." Renewable and Sustainable Energy
Reviews 32 (2014): 396-415.
2) Gu, Wei, et al. "Modeling, planning and optimal energy management of combined cooling, heating and power microgrid: A review." International Journal of Electrical Power & Energy
Systems 54 (2014): 26-37.
3) Liu, Mingxi, Yang Shi, and Fang Fang. "Combined cooling, heating and power systems: A survey." Renewable and Sustainable Energy Reviews 35 (2014): 1-22.
4) Chandan, Vikas, et al. "Modeling and optimization of a combined cooling, heating and power plant system." American Control Conference (ACC), 2012. IEEE, 2012.
5) Zhengyi, Li, Huo Zhaoyi, and Y. Hongchao. "Optimization and Analysis of Operation Strategies for Combined Cooling, Heating and Power System." IEEE (2011).
6) Bischi, Aldo, et al. "A detailed MILP optimization model for combined cooling, heat and power system operation planning." Energy (2014).
7) http://www.midwestchptap.org/Archive/pdfs/080530_JC_AEE_St%20Louis.pdf
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Combined cooling, heat and power system planning
Conventional Energy System
Electricity :
 Either purchased from the local utility (regulated market) or purchased from the utility / competitive electric provider (de-regulated market)
 Power is generated at a central station power plant
 Normally generated at approx. 30% energy efficiency (10 units of fuel in, 3 units of electric power (kW) out)
 70% of the fuel’s energy is lost in the form of heat vented to the atmosphere
Thermal (heating)
 –Normally generated on-site with multiple boilers (hot water or steam loop)
 –Boiler energy efficiencies between 60% and 80%
Thermal (cooling)
 –Normally use electric chillers with chilled water loop
 –May use some absorption chillers in conjunction with electric chillers to offset peak electric demands
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Combined cooling, heat and power system planning
Conventional Energy System
 Central Station Electric Power (30% efficiency)
 Boilers at the Facility Provide Heat (60% to 80%)
 Chillers either using electricity produced at 30% or
 steam produced at 60% to 80% (Absorption)
 Conventional System Energy Efficiency
 Depends on heat/power ratio
 Typical system efficiencies range from 40% to 55%
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Combined cooling, heat and power system planning
Distributed Generation
DG is
 An Electric Generator
 Located At a Substation or
 Near a Building / Facility
 Generates at least a
 portion of the Electric Load
DG Technologies
 Solar Photovoltaic
 Wind Turbines
 Engine Generator Sets
 Turbine Generator Sets
 Combustion Turbines
 Micro-Turbines
 Steam T
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Combined cooling, heat and power system planning
Combined Heat & Power (CHP)
A Form of Distributed Generation CHP is …
An Integrated System
Located At or Near a Building/Facility
Provides at Least a Portion of the Electrical Load and Recycles the Thermal
Energy for
 Space Heating / Cooling
 Process Heating / Cooling
 Dehumidification
As in a cascade process, first primary energy (fuel) is converted into electric/mechanical power by a thermodynamic cycle, then the heat discharged by the thermodynamic cycle is used to satisfy the user’s heat demand and/or to generate refrigeration power with absorption chillers.
Simple thermodynamic considerations can prove that such cascade energy flow ensures a saving of primary energy with respect to notintegrated solutions.
39

Combined cooling, heat and power system planning
Companies and industries that would likely benefit from installation of their own Trigeneration plant include; (http://www.cogeneration.net/)
* Airports
* Agriculture
* Casinos
* Central Plants
* City centers
* Colleges & Universities
* Company campuses
* Dairies
* Data Centers
* District Heating & Cooling
* Electric utilities
* Food Processing Plants
* Government Buildings and Facilities
* Grocery Stores
* Hospitals
* Hotels
* Manufacturing Plants
* Military Bases
* Nursing Homes
* Office Buildings
* Refrigerated Warehouses
* Resorts
* Restaurants
* Schools
* Server Farms
* Shopping centers
* Supermarkets
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Combined cooling, heat and power system planning
One-degree of freedom units feature a single operative variable (i.e., decision variable related to their operation mode) which determines the amount of heat recovered at a given temperature as well as the electric power output and the energy consumption.
Examples of “one-degree of freedom” units are internal combustion engines and simple cycle gas turbines whose electric power output and cogenerated thermal power depend on the fuel mass flow rate (which is considered here as their operative variable). In “two-degree of freedom” units, the operation mode (i.e., the electric power output and the thermal power output) depends on two independent operative variables. Examples are gas turbines with post-firing, and extraction-condensing steam turbines with a single turbine bleed
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Combined cooling, heat and power system planning
The operation of CCHP systems poses a number of challenges due to the fact that
(1) in many instances electric power, heat and refrigeration power demands do not follow the same trend,
(2)different units can be used to satisfy the user demands.
As a consequence of (1), the prime mover load cannot be regulated to exactly match all the three power demands, and a compromise solution must be found. For example, if the location is not remote, electricity can be purchased from the electric grid, while heat surplus can be either wasted
(rejected to the environment) or stored in a heat storage tank to make it available during peak hours. Another solution to improve the CCHP system flexibility could be the downgrade of high temperature heat into low temperature heat. Due to the large number of decision variables and the necessity of determining trade-off solutions, the operation planning of
CCHP plants requires the development of specific optimization tools
42

Combined cooling, heat and power system planning
A general version of the short-term operation planning problem can be summarized as follows:
Given a set of production units comprising prime movers generating only electricity or electricity and heat (CHP), boilers, heat pumps, compression and absorption refrigeration units and optional heat storage systems,
Determine for each time period t which units must be switched on and the value of their operative variables in order to minimize a relevant objective function (such as the total operating cost) while satisfying the demands of thermal power, cooling load and electricity, and while dealing with the operational features and constraints of the units (e.g., performance curves at off-design conditions, minimum and maximum allowed loads, average start-up time, maximum number of startups per day, etc.).
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