Question

A typical coal-fired power plant burns 350 metric tons of coal every hour to generate 2.6×106 MJ of electricity. 1 metric ton = 1000 kg; 1 metric ton of coal has a volume of 1.5 m3. The heat of combustion is 28 MJ/kg. Assume that all heat is transferred from the fuel to the boiler and that all the work done in spinning the turbine is transformed into electrical energy. What is the power plant's thermal efficiency?

Answer 1

0.00095%

Step-by-step explanation:

Mass of coal needed every day = Mass needed every hour × 24

`350* 1000* 24=8400000\ kg`

Thermal energy released per hour = Mass × Heat of combustion

`350* 1000* 28* 10^6=9800000000000\ J`

Power released per second = Energy / 3600 seconds

`P=(E)/(t)\\\Rightarrow P=(9800000000000)/(3600)\\\Rightarrow P=2722222222.22\ W`

Thermal efficiency = (Electrical power / Mechanical power) × 100

`(2.6* 10^6)/(2722222222.22)* 100=0.00095\ \%`

Thermal efficiency of the plant is 0.00095%