## Demonstrative Video

### Problem-1

The fuel inputs per hour of plants 1 and 2 are given as \[\begin{aligned} &F_{1}=0.2 P_{1}^{2}+40 P_{1}+120 \text { Rs per hr } \\ &F_{2}=0.25 P_{2}^{2}+30 P_{2}+150 \text { Rs per hr } \end{aligned}\]

Determine the economic operating schedule and the corresponding cost of generation if the maximum and minimum loading on each unit is \(100 \mathrm{MW}\) and \(25 \mathrm{MW}\), the demand is \(180 \mathrm{MW}\), and transmission losses are neglected.

If the load is equally shared by both the units, determine the saving obtained by loading the units as per equal incremental production cost.

### Solution-1

Given: \[\begin{aligned} &F_{1}=0.2 P_{1}^{2}+40 P_{1}+120 \text { Rs per hr } \\ &F_{2}=0.25 P_{2}^{2}+30 P_{2}+150 \text { Rs per hr } \end{aligned}\]

The incremental production cost of both units: \[\begin{aligned} \dfrac{dF_1}{dP_1} & = 0.4P_1+40~\mathrm{Rs./MWhr} \\ \dfrac{dF_2}{dP_2} & = 0.5P_2+30~ \mathrm{Rs./MWhr} \end{aligned}\]

For economic operation of the units:

\[\begin{aligned} \lambda & = \dfrac{dF_1}{dP_1} = \dfrac{dF_2}{dP_2} \nonumber\\ & 0.4P_1+40 = 0.5P_2+30\\ & P_1+P_2 = 180 ~\mathrm{(Given)} \end{aligned}\]

On solving: \(P_1 = 88.89~\mathrm{MW} \qquad P_2 = 91.11~\mathrm{MW}\)

The cost of generation \(=F_{1}+F_{2}\) \[\begin{aligned} &F_{1}=0.2 P_{1}^{2}+40 P_{1}+120=\mathrm{Rs.}~5255.88/\mathrm{hr}. \\ &F_{2}=0.25 P_{2}^{2}+30 P_{2}+150=\mathrm{Rs.}~4958.55/\mathrm{hr} \end{aligned}\]

Total cost \(=\) Rs. \(10214.43 / \mathrm{hr}\).

If the load on each unit is 90 MW, the cost of generation will be \[\begin{aligned} F_1 & = \mathrm{Rs.}~5340/\mathrm{hr} \\ F_2 & = \mathrm{Rs.}~4875/\mathrm{hr} \end{aligned}\]

Total cost = Rs. 10215/hr

Saving will be = Rs. 0.57/hr.

### Problem-2

Determine the incremental cost of received power and the penalty factor of the plant shown in Fig., if the incremental cost of production is \[\begin{aligned} \dfrac{dF_1}{dP_1} & = 0.1P_1+3.0~\mathrm{Rs./MWhr} \end{aligned}\]

### Solution-2

\[\begin{aligned} \dfrac{dF_1}{dP_1} & = 0.1P_1+3.0~\mathrm{Rs./MWhr} \end{aligned}\]

Penalty factor = \(L_n = \dfrac{1}{1-\dfrac{dP_L}{dP_n}}= \dfrac{1}{1-\dfrac{2}{10}}=\dfrac{10}{8}\)

Cost of received power = \(\dfrac{dF_1}{dP_1}L_1=\lambda\) \[= (0.1\times 10+3)\times \dfrac{10}{8} = \mathrm{Rs. 5/MWh}\]

### Problem-3

A two-bus system is shown in Fig. If a load of 125 MW is transmitted from plant 1 to the load, a loss of 15.625 MW is incurred. Determine the generation schedule and the load demand if the cost of received power is Rs. 24/MWhr. Solve the problem using coordination equations and the penalty factor method approach. The incremental production costs of the plants are

\[\begin{aligned} \dfrac{dF_1}{dP_1} & = 0.025P_1+15 \\ \dfrac{dF_2}{dP_2} & = 0.05P_2+20 \end{aligned}\]

### Solution-3

\[\begin{aligned} \dfrac{dF_1}{dP_1} & = 0.025P_1+15 \\ \dfrac{dF_2}{dP_2} & = 0.05P_2+20 \end{aligned}\]

Since load is at Bus-2 alone, the loss in line will not be affected by Generator at Plant-2

Hence, \[\begin{aligned} P_L & = B_{11}P_1^2 \qquad B_{12}=B_{21}=0~\text{and}~B_{22} = 0\\ 15.625 & = B_{11} \times 125^2 \Rightarrow B_{11} = 0.001 \end{aligned}\]

Now, coordination equation: \[\begin{aligned} \dfrac{dF_1}{dP_1}+\lambda \dfrac{dP_L}{dP_1} & = \lambda \end{aligned}\]

Since, \(P_L=0.001P_1^2 \Rightarrow \dfrac{dP_L}{dP_1} = 0.002P_1\)

Substituting in the coordination equation for plant \[\begin{aligned} & 0.025 P_{1}+15+\lambda 0.002 P_{1}=\lambda \\ \Rightarrow & 0.025 P_{1}+0.048 P_{1}+15=24 \\ \Rightarrow & 0.073 P_{1}=9 \\ \Rightarrow & P_{1}=123.28 \mathrm{MW} \end{aligned}\]

The coordination equation for plant 2, \[\begin{gathered} 0.05 P_{2}+20=24 \\ \Rightarrow P_{2}=80 \mathrm{MW} . \end{gathered}\]

Transmission loss \(P_{L}=0.001 \times(123.28)^{2}=15.19 \mathrm{MW}\)

Load, \(P_{D}=123.28+80-15.19=188.1 \mathrm{MW}\).

Penalty factor of Plant-1 is \[\begin{aligned} & \frac{1}{1-\frac{\partial P_{L}}{d P_{1}}}=\frac{1}{1-\left(0.002 P_{1}\right)} \\ & \frac{d F_{1}}{d P_{1}}\left(\frac{1}{1-0.002 P_{1}}\right)=24 \\ \Rightarrow & \frac{0.025 P_{1}+15}{1-0.002 P_{1}}=24 \\ \therefore & P_{1}=123.28 \mathrm{MW} \end{aligned}\]

Since \(\dfrac{dP_L}{dP_2}=0 \Rightarrow L_2 = \text{unity}\)

The incremental cost of received power equals the incremental cost of production \[\begin{aligned} 0.05P_2+20& = 24\\ \Rightarrow P_2 & = 80~\mathrm{MW} \end{aligned}\]

### Problem-4

Two generating stations A and B have full load capacities of 200 MW and 75 MW respectively. The inter-connector connecting the two stations has an induction motor/synchronous generator (plant C) of full load capacity 25 MW. Percentage changes of speeds of A, B and C are 5, 4 and 3 respectively. The loads on bus bars A and B are 75 MW and 30 MW respectively. Determine the load taken by the set C and indicate the direction in which the energy is flowing.

### Solution-4

Given data:

F.L. MW 200, 75, 25

% speed change 5, 4, & 3

Loads 75 MW and 30 MW

Assume that \(x\) MW is flowing from A to B

Load at station-A = \((75+x)\)

% drop in speed = \(\dfrac{5}{200}(75+x)\)

Load at station-B = \((30-x)\)

% drop in speed = \(\dfrac{4}{75}(30-x)\)

Since, the output of A corresponding to the power flow through the inter-connector is the input to C

Therefore, there is further reduction in frequency approximately by \(\dfrac{3x}{25}\)

The total reduction in frequency from both A and C and the reduction of frequency from B when referred to Bus-B should be same \[\begin{aligned} & \dfrac{5}{200}(75+x)+\dfrac{3x}{25} = \dfrac{4}{75}(30-x) \\ & \Rightarrow x = -1.35~\mathrm{MW} \end{aligned}\]

Hence, 1.35 MW of power will flow from B to A

### Problem-5

Two turbo-alternators rated for 110 MW and 210 MW have governor drop characteristics of 5 per cent from no load to full load. They are connected in parallel to share a load of 250 MW. Determine the load shared by each machine assuming free governor action.

### Solution-5

Since two-machines are working in parallel, the % drop in frequency from both the machines due to different loadings must be same.

Let \(x\) be the power supplied by 110 MW unit.

The % drop in speed = \(\dfrac{5x}{110}\)

Similarly, % drop in speed of 210 MW unit will be \(\dfrac{5x}{210}(250-x)\)

Then \[\begin{aligned} & \dfrac{5x}{100} = \dfrac{5x}{210}(250-x) \Rightarrow ~ x = 85.93~\mathrm{MW} \end{aligned}\]

Power shared by 210 MW unit is \((250-85.93)=164.07\) MW

Power supplied by 110 MW unit will be 85.93 MW