# ECONOMIC DISPATCH IN POWER SYSTEM* *

# 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.

# EXERCISE:

# 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 ; P_{1} in MW

dF2/dP2 = 0.0067P2 + 4.5 ; P_{2} in MW

The operating limits of both the units are given by

100<=P_{1}, P_{2}<=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 P_{1 }and P_{2} 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 = (P*_{D} + b_{1}/2a_{1 }+b_{2}/2a_{2})/(1/2a_{1}+1/2a_{2})

*For P*_{D }= 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*

*P*_{G1}^{* }= *l-b*_{1}/2a_{1} = (6.77691-5.6)/0.0056 = 210.626 MW

*P*_{G2}^{*} = *l-b*_{2}/2a_{2} = (6.77691-4.5)/0.0067 = 339.8374 MW

*Fuel Cost:*

*F*_{C1} = 0.0028 P_{G1}2 + 5.6 P_{G1 }+ C_{1 }Rs/Hr

*F*_{C2} = 0.00335 P_{G1}2 + 4.5 P_{G1} + C_{2} Rs/Hr

*Where C*_{1} & C_{2} are unknown constants

*Fuel Cost for optimal schedule (P*_{G1}^{*}, P_{G2}^{*})

*F*_{C1} = 0.0028(210.1626) 2+5.6(210.1626) + C_{1} Rs/Hr

*F*_{C2} = 0.00335(339.8374) 2 + 4.5(339.8374) + C_{2} Rs/Hr

*F*_{C}^{*}=F_{C1}+F_{C2}=3216.74+C_{1}C_{2} Rs/Hr

*Fuel Cost for equal Sharing: [P*_{G1} = 275 = P_{G2}]

*F*_{C1} = 0.0028(275) 2+5.6(275) + C_{1} Rs/Hr

*F*_{C2} = 0.00335(275) 2 + 4.5(275) + C_{2} Rs/Hr

*F*_{C1} + F_{C2} = 3242.59 + C_{1}C_{2} Rs/Hr

*Total Saving:*

*F*_{C }= F_{C} – F_{C}^{* }= 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:**

### B_{11} = 8.38*10^{-3}

B_{12} = -0.049*10^{-3}

B_{21} = 0

B_{22} = 5.96*10^{-3}

P_{L} = B_{11}P_{1}^{2} + P_{2}^{2}B_{22} + P_{1}P_{2}B_{12}

P_{D }= P_{1} + P_{2} – P_{L}

P_{D} = 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 P_{2} = 0

P_{1} = [1-(b_{1}/l)-(2B_{12}P_{2})]/[(2a_{1}/l)+2B_{11}]

### P_{1} = [1-(5.6/6.776)-0]/[(0.0056/6.776)+0.01676]

### P_{1} = 9.8686 MW

P_{2} = [1-(b_{2}/l)-(2B_{21}P_{1})]/[(2a_{2}/l)+2B_{22}]

### P_{2} = [1-(4.5/6.776)-0]/[(0.0067/6.776)+0.01192]

### P_{2} = 23.02 MW

#### P_{L} = P_{1}^{2}B_{11} + P_{2}^{2}B_{22} + P_{1}P_{2}B_{12}

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

### P_{L} = 4.8387 MW

#### P_{D} = P_{1} + P_{2} – P_{L}

= 9.8686 + 26.02 – 4.838

### P_{D} = 31.04 MW

: 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:

F_{1} = 392.7 + 5.544P_{1} + 0.001093P_{1}^{2} ; P_{1} in MW

F_{2} = 217.0 + 5.495P_{2 }+ 0.001358P_{2}^{2} ; P_{2} in MW

F_{3} = 65.5 + 6.695P_{3} + 0.004049P_{3}^{2 } ; P_{3} in MW

Generation Limit:

150<=P_{1}<=600 MW

100<=P_{2}<=400 MW

50<=P_{3}<=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