EE2404 ECONOMIC DISPATCH IN POWER SYSTEM - C C++ Java Programs - Examples

# C C++ Java Programs - Examples

C C++ Java Python Perl Programs Examples with Output -useful for Schools & College Students

# AIM:

To understand the basics of the problem of Economic Dispatch of optimally adjusting the generation schedules of thermal generating units to meet the system load which are required for unit commitment & Economic operation of power systems.
To understand the development of co-ordination equation (the mathematical model for ED) without and with losses & operating constraints and solution of these equations using direct and iterative methods.

# OBJECTIVE:

Ø  To write a program for solving ED problem without and with transmission losses for a given load condition/daily load cycle using
(a)Direct method
(b)Lambda-iteration method
Ø  To study the effect of reduction in operation cost resulting due to changing from simple load dispatch to economic load dispatch.
Ø  To study the effect of change in fuel cost on the economic dispatch for a given load.
Ø  To study the use of ED in finalizing the unit commitment for tomorrow’s operating conditions of power system.

# 1. The system load in a power system varies from 250MW to 1250MW. Two thermal units are operating at all times and meeting the system load. Incremental fuel cost in hundreds of rupees per Megawatt hour for the units are

dF1/dP1 = 0.0056P1 + 5.6   ;           P1 in MW
dF2/dP2 =  0.0067P2 + 4.5  ;           P2 in MW
The operating limits of both the units are given by
100<=P1, P2<=625MW
Assume that the transmission loss is negligible.
a)      Determine the economic (minimum fuel cost) generation schedule of each unit, the incremental fuel cost of each unit and the incremental cost of received power for different load levels from 250 to 1250MW in steps of 100MW.
b)      Draw the following characteristics from the results obtained in (a)
i.            Incremental cost of received power in hundreds of rupees per MW hr  versus system load in MW.
ii.            Unit outputs P1 and P2 in MW versus system load in MW.
c)      Determine the saving in fuel cost in hundreds of rupees per hour for the economic distribution of a total load of 550MW between the two units compared with equal distribution of that load between the two units.

# INPUT:

Economic Dispatch - Lambda Iteration Method Without Loss
AU Power lab
VII
0.001 0.05    10
2  \$/hr  \$/MWhr
0.0028 5.6 0 100 625
0.00335 4.5 0 100 625
1
250 1250 100

# OUTPUT:

Manual Calculation:
l = (PD + b1/2a1 +b2/2a2)/(1/2a1+1/2a2)
For PD = 550 MW
l= 550 + ((5.6/0.0056 + 4.5/0.0067)/(1/0.0056 + 1/0.0067))
Þ      l = 6.77691 Rs MW/Hr
PG1* = l-b1/2a1 = (6.77691-5.6)/0.0056 = 210.626 MW
PG2* = l-b2/2a2 = (6.77691-4.5)/0.0067 = 339.8374 MW
Fuel Cost:
FC1 = 0.0028 PG12 + 5.6 PG1 + C1 Rs/Hr
FC2 = 0.00335 PG12 + 4.5 PG1 + C2 Rs/Hr
Where C1 & C2 are unknown constants
Fuel Cost for optimal schedule (PG1*, PG2*)
FC1 = 0.0028(210.1626) 2+5.6(210.1626) + C1 Rs/Hr
FC2 = 0.00335(339.8374) 2 + 4.5(339.8374) + C2 Rs/Hr
FC*=FC1+FC2=3216.74+C1C2 Rs/Hr
Fuel Cost for equal Sharing: [PG1 = 275 = PG2]
FC1 = 0.0028(275) 2+5.6(275) + C1 Rs/Hr
FC2 = 0.00335(275) 2 + 4.5(275) + C2 Rs/Hr
FC1 + FC2 = 3242.59 + C1C2 Rs/Hr
Total Saving:
FC = FC – FC* = 3242.59 – 3216.74 = 25.85 Rs/Hr

### EXERCISE :

2. For the system in exercise 4.3 take into account the transmission loss.
a.       Determine the economic loading of each unit to meet a total customer load of 550MW, using the program developed in 4.2
b.      What is transmission loss of the system at the economic loading?
c.       Determine the penalty factor for each unit and the incremental fuel cost at each generating bus.
d.      Determine also the incremental cost of received power (or system l).Assume that the loss coefficient in per unit on a 100MVA base of customer load level of 550MW are given by

#### B11 = 0.008383183

B12 = B21 = -0.000049448
B10/2 = 0.000375082
B22 = 0.005963568
B20/2 = 0.000194971
B00 = 0.000090121

### INPUT:

Economic Dispatch - Lambda Iteration Method With Loss
AU Powerlab
2001399126
VII
0.01  0.05  10
2  \$/hr   \$/MWhr
0.0028 5.6 0 100 625
0.00335 4.5 0 100 625
0.008383183 -0.000049448
0.000049448 0.005963568
0.000375082 0.000194971
0.000090121 100
0
550OUTPUT :
Manual Calculation:

### B11 = 8.38*10-3

B12 = -0.049*10-3
B21 = 0
B22 = 5.96*10-3
PL = B11P12 + P22B22 + P1P2B12
PD = P1 + P2 – PL
PD = 550 MW
l = (550 + 5.6/0.0056 + 4.5/0.0067) / (1/0.0056 + 1/0.0067)
Þ l = 6.7767 Rs/MW-Hr
For First Iteration
Assume P2 = 0
P1 = [1-(b1/l)-(2B12P2)]/[(2a1/l)+2B11]

### P1 = 9.8686 MW

P2 = [1-(b2/l)-(2B21P1)]/[(2a2/l)+2B22]

### P2 = 23.02 MW

#### PL = P12B11 + P22B22 + P1P2B12

= [(9.8686)2*(0.00838) + (26.02)2*(0.00596) + 9.8686*26.02*-0.049*10-3]

### PL = 4.8387 MW

#### PD = P1 + P2 – PL

= 9.8686 + 26.02 – 4.838

: 3292.06  \$/hr

### EXERCISE :

3.      A power system with negligible transmission loss, the system load varies from a peak of 1200MW to a valley of 500MW. There are three thermal generating units which can be committed to take the system load. The fuel cost data and generation operation limit data are given below
In hundreds of rupees per hour:
F1 = 392.7 + 5.544P1 + 0.001093P12 ;           P1 in MW
F2 = 217.0 + 5.495P2 + 0.001358P22 ;           P2 in MW
F3 = 65.5 + 6.695P3 + 0.004049P32  ;             P3 in MW

Generation Limit:
150<=P1<=600 MW
100<=P2<=400 MW
50<=P3<=200 MW
There are no other constant on system operation. Obtain an optimum (minimum fuel cost) with commitment table for each load level taken in steps of 100 MW from 1200 to 500.Adopt “Brute force enumeration” technique. For each load level obtain economic schedule using economic dispatch program developed in ex 4.3 for each feasible combination of units and choose the lowest fuel cost schedule among the combination.
Show the details of economic schedule and the component and total cost of operation for each feasible combination of unit for the load level of 900MW.

### INPUT:

Economic Dispatch - Lambda Iteration Method
Without Loss
AU Power lab
2003557
VII
0.001 0.05    5
3             \$/hr      \$/MWhr
0.001093 5.544 392.7 150 600
0.001358 5.495 217.0 100 400
0.004049 6.695 65.50 50 200
1
500 1200 100

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