Generating stations 27

 Question = A diesel engine power plant has 700 kilowatt and two 500 kilowatt generating units .  the fuel consumption is 0.28 kg/kwh and calorific value of fuel oil is 10,200 kcal/kg .  Estimate (1) the fuel oil required for a month of 30 days ? and (2) overall η .? plant capacity factor 40% . 

Solution = »  first generating unit =  700 kW ,  and , 

»  second two generating units = 500 kw × 2 ,

» per unit fuel consumption = 0.28 kg/ kWh , 

» per kg calorific value of fuel oil = 10,200 kcal/kg , 

» plant capacity factor =40% , 

» 1 kWh fuel consumption = 0.28 kg , 

» x kwh fuel consumption = x × 0.28 kg , 

» x or , actual energy produced in a month , kWh = ? 

Formula , 

» plant capacity factor = actual energy produced/month ÷ max . energy that could have been produced/month . 

» maximum energy that could have been produce in a month = plant capacity × hour in a month , = (700 kw + 2 × 500 kw ) × (30days × 24 hours ) , = 1700 × 720 kWh , 

» actual energy produced in a month = 0.4 ×1700 × 720 = 4,89,600 kWh , 

» fuel oil consumption or , x kwh fuel consumption =  4,89,600 × 0.28 , = 1,37,088 kg , answer ✓ 

(2) » efficiency ( η ) overall = output/input , 

» 1 kwh = 860 kcal , 

» output = 4,89,600 kWh × 860 kcal , 

» input = 137088 × 10200 kcal , 

» η overall = ( 489600 kWh ×860 kcal ) ÷ ( 137088 kWh × 10200 kcal ) = 0.3 or , ×100 = 30% , answer ✓