Question

It takes 2.0 million tons of coal per year to feed a 1000-MWe steam-electric plant. Assuming that the plant has a 60% capacity factor, what is its efficiency? Coal has an energy content of 22 MBtu per ton. Assume two significant digits.

Answer 1

`\eta = 40\%`

Step-by-step 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

`W_(max) = 1000 MWe`

`60\% = (w)/(w_(max))`

where w is actual power

we knwo that 1 Btu = 0.000293 KWH

1MBtu = 293 KWH

Thermal Efficiency is given as

`= 600 * 10^3 KW`

`= (2* 10^6)/(365* 24) ton/ hr * 22* 293 KWH/ ton`

`= 1.5 * 10^6 KW`

`\eta = (600* 1000)/(1.5* 10^6) = 40\%`