Question

The fuel inputs per hour of two generators delivering 200 MW are as follows:
$F_1=20P_1+0.05P_1^2+50$ Rs per hour,
$F_2=16P_2+0.1P_2^2+40$ Rs per hour
The limits of generators are $20\le P_2\le 60MW;$
$40\le P_1\le 150MW$
For economic generation $P_1$ and $P_2$ should be

• A. $P_1=120MW,P_2=80MW$
• B. $P_1=145MW,P_2=55MW$
• C. $P_1=140MW,P_2=60MW$
• D. $P_1=100MW,P_2=100MW$

Answer 1

The correct option is C $P_1=140MW,P_2=60MW$

$\frac{d{F}_1}{d{P}_1}=20+0.1P_1,$

$\frac{d{F}_2}{d{P}_2}=16+0.2P_2,$

For economic operation

$20+0.1P_1=16+0.2P_2$

$4=-0.1P_1+0.2P_2$ ...(i)

$P_1+P_2=200MW$ ...(ii)

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

$P_2=80MW$

and $P_1=120MW$

But the limit of generator 2 is

$20\le P_2\le 60MW$

$P_1=140MW$, $P_2=60MW$