Generating stations 6

 Question = A 100 MW  steam station uses coal of calorific value 6400 kcal/kg . Thermal efficiency of station is 30% and electrical efficiency is 92% . Calculate the coal consumption per hour when the station is delivering it's full output . 

Solution = calorific value =  6400 kcal/kg , 

» heat produced/hours = H = electrical output in heat units /  η overall , 

» η overall =  η thermal × η electrical ,  = 0.30 × 0.82 = 0.276 , 

» 100 MW or , 100 MW × 1hour , = 100 MWH , 

Or , 100 × 10⁶ wh ÷ 1000 , = 100 × 10³ kWh or , 10⁵ kWh , 

» now. electrical output in heat unit or , unit generated/hour = ( 100 × 10³ ) = 10⁵ kWh or , 10⁵ × 860 kcal , 

» details of 860 kcal  ,

 ( 1 kwh = 1000 watts × 3600 seconds = 36 ×10⁵ joules , 

» 4.18 joules = 1 calories ,  

» 1 joules = 1/4.18 calories ,  

» 36 ×10⁵ joules =  ( 36 × 10⁵ j × 1c )  ÷ 4.18 j , =  860 ×10³ calories , or 860 kcal , )

» heat produced/hours = H = 10⁵ ×860 / 0.276 ,= 311.6 × 10⁶  kcal , 

And, 

» coal consumption/hour = 311.6 ×10⁶ kcal / 6400 kcal/kg , = 48687 kg answer ✓