Economic of power generation 6

 Question = A generating station has the following data : installed capacity = 300 MW , capacity factor = 50% or 0.5 , annual load factor = 60% or 0.6 , annual cost of fuel , oil etc (R) =  RS 9×10⁷ , capital cost (F) = RS 10⁹ , annual interest and depreciation (F) = 10% , 

• Calculate (1) the minimum reserve capacity of the station and (2) the cost per kWh generated ?

Solution = reserve capacity = installed capacity - maximum demand , 

» is given , Installed capacity = 300 MW , 

» M.D. = ( capacity factor × installed capacity ) ÷ Load factor , 

» capacity factor ( c f. ) = Average demand (A.D.) ÷ Installed capacity (I C.) ,  equations (1) AND , 

Load factor = average demand ( A.D.) ÷ maximum demand ( M.D.) , equations (2)

» Dividing 1 and 2 we get , 

» C.F. ÷ L.F. = ( A.D. ÷ I.C. ) ÷ ( A.D. ÷ M.D.) ,

» after calculate = C.F. ÷ L.F. = M.D. ÷ I.C. ,

» So , Maximum demand = ( C.F. × I.C. ) ÷ L.F. , 

» M D.= ( 0.5 × 300 MW ) ÷ 0.6 , = 250 MW , 

» So , reserve capacity = 300 - 250 = 50 MW , answer ✓ 

(2) Cost per unit generated = total annual charges ÷ unit generated per annum , 

» so , unit generated = M D. × L.F. × hours in a years ,  = 250 × 10³ MW × 0.6 × 8760 , = 1314 × 10⁶ kWh , 

( 24 hours × 365 days = 8760 days ) , 

» total annual charges = annual fixed charges + annual running charges , 

» annual fixed charges = annual interest and depreciation × capital cost , = 0.1 × 10⁹ ,= ( 1 ÷ 10 ) × 10⁹ , = 10⁸ RS , 

» is given , annual running charges = 9 × 10⁷ RS , 

» So , total annual charges = ( 10⁸ + 9 × 10⁷ ) RS , or  ( 10 × 10⁷ + 9 × 10⁷ ) = 19 × 10⁷ RS ,

» Cost per kWh = ( 19 × 10⁷ RS) ÷ (1314 × 10⁶ ) = 0.14 RS or (×100) , 14 paise , answer ✓

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