Question

1. Make a way to calculate the energy of 5000gr of coal Calorific Value 475 k cal / kg into electrical energy with the combustion requirements and the conversion factor by yourself.

Answer 1

The given data is as follows:5000gr of coal whose Calorific Value 475 k cal / kg

as there is no mentioning about efficiency of boiler and turbine system we need to assume it as 0.85 and 0.4 respectively

coal available= 5000grams =5kg => the heat energy produced by burning this much coal we get 475*5=2375kcal

by considering efficiency of boiler and turbine as mentioned above the amount of input energy avalable at the input of generator =2375*0.85*0.4=807.5Kcal

we know that 1kcal=4.184Kjouls in general the joule is defined as watt-second so we get

1joule= 1watt-sec=1*10^-3Kwatt-hr/(3600)=2.77*10^-7Kwhr

total electrical energy developed =807.5*4.184*1000*2.777*10^-7Kwhr=0.938Whr here we have considered the efficiency of generator as 100% i.e no losess inthe generator but in practical the efficiency of generator will be arounf 95% hence the electrical ouput genrated is =0.95*0.938Whr =0.891Whr

that means if i use 0.891watts of electrical equipment then i can use it for continuously for 1hr by burning 5000grams of coal .

Note : due to poor efficiency of turbine the available energy at input of generator is reduced to grate extent .