Calculate the efficiency of a fossil-fuel power plant that consumes 380metric tons of coal each hour to produce useful work at the rate of 750 MW. The heat of combustion of coal (the heat due to burning it) is 28 MJ/kg.

Amount of coal that is used each hour,$\mathrm{m}=380\mathrm{metric}\mathrm{ton}=380000\mathrm{kg}$

The rate of work produced in one hour,$\left|\mathrm{W}\right|=750\mathrm{MW}$

The heat of combustion of coal,$\left|{\mathrm{Q}}_{\mathrm{s}}\right|=28\mathrm{MJ}/\mathrm{kg}$

An engine is a device that, operating in a cycle, extracts energy as heat from a high-temperature reservoir and does a certain amount of work. The efficiency is given by the ratio of the energy that we get to that of energy that we pay for. Again we know that, the value of one metric ton = 1000 kg.

Formula:

The efficiency of a Carnot cycle,

$\mathrm{\epsilon }=\frac{\mathrm{energy}\mathrm{we}\mathrm{get}}{\mathrm{energy}\mathrm{wepay}\mathrm{for}}=\frac{\left|\mathrm{W}\right|}{\left|\mathrm{Q}\right|}$ (1)

The total energy generated as heat by burning 380000kg in one hour is given as:

$$\begin{gathered} \mathrm{Q}=380000\mathrm{kg}\times 28\mathrm{MJ}/\mathrm{kg} \\ =10.6\times {10}^6\mathrm{MJ} \end{gathered}$$

The work done in one hour is given using the rate of work produced in one hour as:

$$\begin{gathered} \mathrm{W}=750\mathrm{MW}\times 3600\mathrm{s} \\ =2.7\times {10}^6\mathrm{MJ} \end{gathered}$$

Efficiency by using the above values in equation (1) as given:

$$\begin{gathered} \mathrm{\epsilon }=\frac{2.7\times {10}^6\mathrm{J}}{10.6\times {10}^6\mathrm{J}} \\ \mathrm{\epsilon }=0.25 \\ \mathrm{\epsilon }\%=25\% \end{gathered}$$

Hence, the value of the efficiency is 25%