A typical coal-fired power plant burns 310 metric tons of coal every hour to generate 790 MW of electricity. 1 metric ton= 1000 kg. The density of coal is 1500 {\rm kg}/{\rm m}^{3} and its heat of combustion is 28 \;\rm 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.

Question

A typical coal-fired power plant burns 310 metric tons of coal every hour to generate 790 MW of electricity. 1 metric ton= 1000 kg. The density of coal is 1500 {\rm kg}/{\rm m}^{3} and its heat of combustion is 28 \;\rm 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

heat in fuel:

massflowrate*heat/kg
310tons/hr*1000kg/ton*1hr/3600sec*28MJoule/kg

efficiencty= 790MW/above

check my thinking. That is about typical for a thermal plant. In very high pressure, high temp (3000F, 3000PSI), efficencies are boosted some, but the elec generators hold the eff lower.

Answer 2

To calculate the power plant's thermal efficiency, we need to determine the electricity generated by the combustion of coal and compare it to the heat input.

First, let's calculate the heat input. We know that the power plant burns 310 metric tons of coal every hour, and each metric ton is equal to 1000 kg. So, the mass of coal burned every hour is:

Mass of coal burned per hour = 310 metric tons/hour × 1000 kg/metric ton = 310,000 kg/hour

Next, we can calculate the energy released by the combustion of this amount of coal. The density of coal is given as 1500 kg/m³, and the heat of combustion is 28 MJ/kg. Therefore, the energy released per hour is:

Energy released per hour = Mass of coal burned per hour × Heat of combustion
= 310,000 kg/hour × 28 MJ/kg
= 8,680,000 MJ/hour

Now, let's calculate the electricity generated. The power generated by the power plant is given as 790 MW (megawatts). Since power is the rate of energy transfer, we need to convert the energy released from MJ (megajoules) to megawatts. Since 1 MW is equal to 1 million joules per second, we have:

Power generated = Energy released per hour / time in hours
= (8,680,000 MJ/hour) / 1 hour
= 8,680,000 MJ/hour × (1 MW / 1 million J)
= 8,680 MW

Finally, we can calculate the thermal efficiency of the power plant by dividing the electrical power generated by the heat input:

Thermal efficiency = (Power generated / Heat input) × 100%

Given that all work done in spinning the turbine is transformed into electrical energy, the power generated is equal to the power input from the heat source:

Thermal efficiency = (Power generated / Power input) × 100%
= (8,680 MW / 8,680 MW) × 100%
= 100%

Therefore, the power plant's thermal efficiency is 100%.