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GATE 2015 EE - SET 2 - Question 40

Solution :

\frac{dc_{1}}{dp_{1}} = 0.2P_{1} + 50

\frac{dc_{1}}{dp_{1}} = 76\frac{Rs}{MWh}

0.2P_{1} + 50 = 76
P_{1} = 130 MW

\frac{dc_{2}}{dP_{2}} = 0.24P_{2} + 40

\frac{dc_{2}}{dP_{2}} = 68.8\frac{Rs}{MWh}

0.24P_{2} + 40 = 68.8
P_{2} = 120 MW

P_{1} + P_{2} = 250 Mw   ---------------(1)

0.2P_{1} + 50 = 0.24 P_{2} + 40

0.2P_{1}  - 0.24 P_{2} = -10 ---------(2)

P_{1} = 113.636 MW

P_{2} = 136.36 MW

 

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