The figure shows a two-generator system supplying a load

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The figure shows a two-generator system supplying a load of P_D = 40 MW, connected at bus 2.

The fuel cost of generators G₁ and G₂ are :
C₁ (P_G1) = 10,000 Rs/MWh and C₂ (P_G2) = 12,500 Rs/MWh
and the loss in the line is P_loss(pu) = 0.5 P_G1(pu) ²,
where the loss coefficient is specified in pu on a 100 M VA base. The most economic power generation schedule in MW is

1. P_G1 = 20, P_G2 = 22
2. P_G1 = 22, P_G2 = 20
3. P_G1 = 20, P_G2 = 20
4. P_G1 = 0, P_G2 = 40

Correct Option: B

P_L = 0.5 P_G1²
PL is given by,
PL = B₁₁ P_G1² + 2B₁₂ P_G1 P_G2 + P_G2² B₂₂
where B₁₁, B₁₂ & B₂₂ are loss - coefficients.
Since load P_D is connected to Bus 2,
B₂₂ = B₁₂ = 0.

Now ,	∂P_L	= 2 × 0.5 P_G1 = P_G1
	∂P_G1

and	∂P_L	= 0
	∂P_G2

Penality factors, L₁ =	1		=	1
	1 -	1		1 - P_G1
		∂P_G1

and L₂ =	1		= 1
	1 -	1
		∂P_G1

Now, L₁ .C₁ (P_G1) = L₂ C₂ (P_G2) = λ

⇒	1	× 10000 = 1 × (12500)
	1 - P_G1

⇒ P_G1 = 0.2 P.U.
⇒ P_G1 = 20 MW
As, P_G1 + P_G2 = P_D + P_L = P_D + 0.5 P_G1²
⇒ 20 + P_G1 = 40 + 0.5 × (0.2)² × 100
⇒ P_G2 = 22 MW