Question

You want a company to build a thermal power plant in your neighborhood in order to supply electrical power. In considerations for the environment, it is required that the efficiency of the plant’s system would be at least 70% . You got a call from a company that said they could meet this efficiency. They claim that they could burn their coal in a way such that the hot reservoir for their heat engine is maintained at 626 degrees Celsius. You have also learned from local environmental scientists that the river running next to the power plant has an average temperature of 21 degrees Celsius, which would act as the cold reservoir for the power plant. The hot and cold reservoirs are respectively the heat supplier and the heat dump for a heat engine that performs mechanical work. The power plant’s system then converts this mechanical work into electric energy.

Draw the PV diagram for a singular heat cycle of the mechanical heat engine. Assume that it is a Carnot Engine that starts off from an initial state 1; next, it expands isothermally to another state 2, whereafter it cools adiabatically to state 3; it then compresses isothermally to state 4; and lastly, its temperature rises adiabatically such that its state returns to the original state 1. Clearly label each state, and the parts of the cycle as adiabatic/isothermal. Once again assuming that the heat engine is a Carnot Engine, determine whether it is possible that the company’s power plant could meet the required efficiency (Assume that all of the mechanical work is turned into electrical energy).

Answer 1

In PV diagram the area under the complete cycle path represents the total work that is done during one cycle.

A single complete cycle can be represented by the 4 states as follows,

Now in the above mentioned question we are provided with the following details,

T_H = 626 + 273.15 = 899.15 Kelvin

T_C = 21 + 273.15 = 294.15 Kelvin

We are asked to find the efficiency of the heat engine, to obtain that we will use the following equation,

i.e,

Efficiency = (1 - T_L / T_H) x 100

Substituting values we get,

Efficiency = (1 - 0.0335) × 100

Efficiency = 96.6 %

That is greater than 70 %, hence the thermal power plant can be successfully built.