Generating stations 4

 Question = A steam power station spends rs 30 lakhs annum for coal used in the station . The coal has a calorific value of 5000 kcal/kg and costs rs 300/ton . If the station has thermal efficiency of 33% and electrical efficiency of 90 % , find .the average load on the station .? 

Solution = formula , Average load on the station = unit generated/annum ÷ hours in a year , 

» hours in a year = 365 × 24 = 8760 hours , 

» unit generated/annum = η overall × heat of combustion , 

»  η overall = thermal efficiency × electrical efficiency ,  = 0.33 ×0.9 = 0.297 

» heat of combustion = coal use/annum × calorific value , 

» So , coal use / annum = 300 rs –––> 1ton ,  1ton––> 1/300 ,  30×10⁵ –––( 30 ×10⁵ ) ÷ 300 = 10⁴ tons , या 10⁷ kgs ,. 

» calorific value of coal = 5000 kcal/kg ,  So , heat of combustion = 10⁷ × 5000 , = 5 × 10¹⁰ kcal , 

» unit generated/annum या heat output =  0.297 × 5× 10¹⁰ = 1487 × 10⁷ kcal , 

» 860 kcal –– 1 kwh ,  » 1 kcal–– 1/860 kWh , » 1485 × 10⁷ kcal ·–– (1485 ×10⁷ × 1) ÷ 860 kcal , kWh , 

» average load on station = (1485 ×10⁷) ÷ 860 ×8760 , = 1971 kW , answer ✓