Question 15.3: A coal-fired power station is a huge heat engine. It uses he...

A coal-fired power station is a huge heat engine. It uses heat transfer from burning coal to do work to turn turbines, which are used to generate electricity. In a single day, a large coal power station has 2.50 \times 10^{14} J of heat transfer from coal and 1.48 \times 10^{14} J of heat transfer into the environment. (a) What is the work done by the power station? (b) What is the efficiency of the power station? (c) In the combustion process, the following chemical reaction occurs: C + O _{2} \rightarrow CO _{2}. This implies that every 12 kg of coal puts 12 kg + 16 kg + 16 kg = 44 kg of carbon dioxide into the atmosphere. Assuming that 1 kg of coal can provide 2.5 \times 10^{6} J of heat transfer upon combustion, how much CO _{2} is emitted per day by this power plant?

Strategy for (a)
We can use W=Q_{ h }-Q_{ c } to find the work output W , assuming a cyclical process is used in the power station. In this process, water is boiled under pressure to form high-temperature steam, which is used to run steam turbine-generators, and then condensed back to water to start the cycle again.

Strategy for (b)
The efficiency can be calculated with E f f=\frac{W}{Q_{ h }} since Q_{ h } is given and work W was found in the first part of this example.

Strategy for (c)
The daily consumption of coal is calculated using the information that each day there is 2.50 \times 10^{14} J of heat transfer from coal. In the combustion process, we have C + O _{2} \rightarrow CO _{2}. So every 12 kg of coal puts 12 kg + 16 kg + 16 kg = 44 kg of CO _{2} into the atmosphere.

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Solution for (a)
Work output is given by:

W=Q_{ h }-Q_{ c }.                      (15.28)

Substituting the given values:

W=2.50 \times 10^{14} J -1.48 \times 10^{14} J                   (15.29)

=1.02 \times 10^{14} J.

Solution for (b)
Efficiency is given by: E f f=\frac{W}{Q_{ h }}. The work W was just found to be 1.02 \times 10^{14} J, and Q_{ h } is given, so the efficiency is

E f f=\frac{1.02 \times 10^{14} J }{2.50 \times 10^{14} J }                (15.30)

= 0.408, or 40.8%

Solution for (c)
The daily coal consumption is

\frac{2.50 \times 10^{14} J }{2.50 \times 10^{6} J / kg }=1.0 \times 10^{8} kg.                        (15.31)

Assuming that the coal is pure and that all the coal goes toward producing carbon dioxide, the carbon dioxide produced per day is

1.0 \times 10^{8} kg  coal \times \frac{44 kg CO _{2}}{12 kg coal }=3.7 \times 10^{8} kg CO _{2}.                  (15.32)

This is 370,000 metric tons of CO _{2} produced every day.

Discussion
If all the work output is converted to electricity in a period of one day, the average power output is 1180 MW (this is left to you as an end of chapter problem). This value is about the size of a large-scale conventional power plant. The efficiency found is acceptably close to the value of 42% given for coal power stations. It means that fully 59.2% of the energy is heat transfer to the environment, which usually results in warming lakes, rivers, or the ocean near the power station, and is implicated in a warming planet generally. While the laws of thermodynamics limit the efficiency of such plants—including plants fired by nuclear fuel, oil, and natural gas—the heat transfer to the environment could be, and sometimes is, used for heating homes or for industrial processes. The generally low cost of energy has not made it economical to make better use of the waste heat transfer from most heat engines. Coal-fired power plants produce the greatest amount of CO _{2} per unit energy output (compared to natural gas or oil), making coal the least efficient fossil fuel.

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