Generating stations 8

 Question = A 75 MW steam power station used of calorific value of 6400 kcal/kg . Thermal efficiency of the station is 30% while electrical efficiency is 80% . Calculate the coal consumption per hour when the station is delivering it's full output ? 

Solution =formula , 

 coal consumption/hour = H/calorific value ,( H = heat produced/hour ) ,  steam power station = 75 MW , coal calorific value = 6400 kcal/kg , thermal efficiency= 30% , electrical efficiency = 80 % ,  

» coal consumption/hour = ? . 

» heat produced/hour , H = electrical output in heat unit ÷ η overall , 

» η overall =  η thermal ×  η electrical , = 0.30 × 0.80 = 0.24 . 

» now , electrical output in heat unit or unit generated/hour = 75 MW × 1 = 75 MWH or , 75 × 10⁶ wh ÷ 1000 = 75 × 10³ kWh , 

» 1 kWh = 860 kcal , 

» 75 × 10³ kWh = 75 × 10³ × 860 , 

» H = 75 × 10³ × 860 / 0.24 , = 268750000 kcal ,or , 26875 × 10⁴ kcal ,  

» So , coal consumption/ hour = 26875 × 10⁴ kcal ÷ 6400 kcal/kg = 41992.18 kg or, 42 tons , answer ✓