Generating stations 25

 Question= A diesel power station has fuel consumption of 0.28 kg/kwh the calorific value of fuel begin 10,000 kcal/kg . determine (1) the overall efficiency , and (2) efficiency of the engine if alternator efficiency is 95% . 

Solution =•  Fuel consumption = 0.28 kg/kWh , • calorific value of fuel = 10,000 kcal/kg , 

(1) efficiency ( η ) overall = electrical output in heat unit ÷ heat of combustion , 

• So first , 

» electrical output in heat units or , heat equivalent of 1kwh =  860 kcal , 

» heat of combustion or , heat produced by 0.28 kg of oil = 10,000 × 0.28 = 2800  kcal , 

» η overall = 860 ÷ 2800 = 0.307 = 30.7 % , answer ✓

(2) engine η =  η overall ÷ η alternator , = 30.7 ÷ 0.95 = 32.3 % , answer ✓