A two bus system is shown below,

If 150 MW is transmitted from plant 2 to the load, a transmission loss of 15 MW is incurred. The incremental fuel costs of the two plants are:

$\frac{d C_{1}}{d P_{G 1}} = 0.04 P_{G 1} + 20 R s / M W h$

$\frac{d C_{2}}{d P_{G 2}} = 0.02 P_{G 2} + 16 R s / M W h$

The power received by load is ____ MW. (Assume system $λ$ = Rs. 25/MWh)
274.75

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Solution

The correct option is A 274.75
We know that,

Transmission loss, $P_{L} = B_{22} P_{G 2}^{2}$

$15 = B_{22} (150)^{2}$

$B_{22} = 6.6667 \times 10^{- 4} (M W^{- 1})$

$I C = λ (1 - \frac{\partial P_{L}}{\partial P_{1}})$

Now, $0.04 P_{G 1} + 20 = λ = 25$

$P_{G 1} = 125 M W$

and $0.02 P_{G 2} + 16 = λ (1 - 2 \times 6.6667 \times 10^{- 4} \times P_{G 2})$

$P_{G 2} = 168.75 M W$

$P_{L} = 6.6667 \times 10^{- 4} \times (168.75)^{2} ≅ 19 M W$

$P_{D 1} = P_{G 1} + P_{G 2} - P_{L}$

$= 125 + 168.75 - 19$

$P_{D 1} = 274.75 M W$

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