Answer:
[tex]\eta = 40\%[/tex]
Explanation:
Total quantity of coal per year required is 2.0 million
capacity of electric plant = 1000 MWe
capacity factor 60%
energy content = 22 MBtu per ton
[tex]W_{max} = 1000 MWe[/tex]
[tex]60\% = \frac{w}{w_{max}}[/tex]
where w is actual power
we knwo that 1 Btu = 0.000293 KWH
1MBtu = 293 KWH
Thermal Efficiency is given as
[tex]\eta = \frac{w_{output}}{w_{input}}[/tex]
[tex]w_{output} = 1000\times 0.60 = 600 MW[/tex]
[tex]= 600 \times 10^3 KW[/tex]
[tex]w_{input} = 2\times 10^6 ton/ year \times 22 MBtu/ ton[/tex]
[tex] = \frac{2\times 10^6}{365\times 24} ton/ hr \times 22\times 293 KWH/ ton[/tex]
[tex]= 1.5 \times 10^6 KW[/tex]
[tex]\eta = \frac{600\times 1000}{1.5\times 10^6} = 40\%[/tex]
