The fuel inputs per hour of two generators delivering 200 MW are as follows:
$F_{1} = 20 P_{1} + 0.05 P_{1}^{2} + 50$ Rs per hour,
$F_{2} = 16 P_{2} + 0.1 P_{2}^{2} + 40$ Rs per hour
The limits of generators are $20 \leq P_{2} \leq 60 M W;$
$40 \leq P_{1} \leq 150 M W$
For economic generation $P_{1}$ and $P_{2}$ should be

P_{1} = 120 M W, P_{2} = 80 M W

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P_{1} = 145 M W, P_{2} = 55 M W

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P_{1} = 140 M W, P_{2} = 60 M W

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P_{1} = 100 M W, P_{2} = 100 M W

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Solution

The correct option is C $P_{1} = 140 M W, P_{2} = 60 M W$
$\frac{d F_{1}}{d P_{1}} = 20 + 0.1 P_{1},$

$\frac{d F_{2}}{d P_{2}} = 16 + 0.2 P_{2},$

For economic operation

$20 + 0.1 P_{1} = 16 + 0.2 P_{2}$

$4 = - 0.1 P_{1} + 0.2 P_{2}$ ...(i)

$P_{1} + P_{2} = 200 M W$ ...(ii)

From equation (i) and (ii), we get

$P_{2} = 80 M W$

and $P_{1} = 120 M W$

But the limit of generator 2 is

$20 \leq P_{2} \leq 60 M W$

$P_{1} = 140 M W$ , $P_{2} = 60 M W$

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Similar questions

\frac{d C_{1}}{d P_{G 1}} = 0.10 P_{G 1} + 20.0; \frac{d C_{2}}{d P_{G 2}} = 0.12 P_{G 2} + 15

P_{G}

_{D}

_{1}

_{2}

_{1}

_{G 1}

_{2}

_{G 2}

_{l o s s (p u)}

_{G 1}^{2} (p u)

Unit	Rating	Speed drop R (per unit on unit base)
1	400	0.04
2	800	0.05

P_{1} = 200 M W, P_{2} = 500 M W

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