Question

The Chalk Point, Maryland, generating station supplies electrical power to the Washington, D.C., area. Units 1 and 2 have a gross generating capacity of \(710 .\) MW (megawatt). The steam pressure is \(25 \times 10^{6} \mathrm{Pa}\), and the superheater outlet temperature \(\left(T_{h}\right)\) is \(540 .^{\circ} \mathrm{C} .\) The condensate temperature \(\left(T_{c}\right)\) is \(30.0^{\circ} \mathrm{C}\) a. What is the efficiency of a reversible Carnot engine operating under these conditions? b. If the efficiency of the boiler is \(91.2 \%\), the overall efficiency of the turbine, which includes the Carnot efficiency and its mechanical efficiency, is \(46.7 \%,\) and the efficiency of the generator is \(98.4 \%,\) what is the efficiency of the total generating unit? (Another \(5.0 \%\) needs to be subtracted for other plant losses. c. One of the coal-burning units produces \(355 \mathrm{MW}\). How many metric tons (1 metric ton \(=1 \times 10^{6} \mathrm{g}\) ) of coal per hour are required to operate this unit at its peak output if the enthalpy of combustion of coal is \(29.0 \times 10^{3} \mathrm{kJ} \mathrm{kg}^{-1} ?\)

Short answer

The efficiency of a reversible Carnot engine operating under the given conditions is 62.71%. The overall efficiency of the total generating unit is 43.93%. To operate one of the coal-burning units at its peak output, approximately 100.3 metric tons of coal are required per hour.

Step-by-step solution

a. Calculating the Carnot efficiency

Recall that the efficiency of a reversible Carnot engine can be calculated using the formula: $$ \eta_\text{Carnot} = 1 - \frac{T_c}{T_h} $$ where \(\eta_\text{Carnot}\) is the efficiency, \(T_h\) is the hot temperature (in Kelvin) and \(T_c\) is the cold temperature (in Kelvin). To use this formula, we first convert the given temperatures from Celsius to Kelvin: $$ T_h = 540 + 273.15 = 813.15 \, \text{K} $$ $$ T_c = 30 + 273.15 = 303.15 \, \text{K} $$ Now, we can calculate the Carnot efficiency: $$ \eta_\text{Carnot} = 1 - \frac{303.15}{813.15} = 1 - 0.3729 = 0.6271 $$ The efficiency of a reversible Carnot engine operating under the given conditions is 62.71%.

b. Calculating the total generating unit efficiency

We are given the efficiencies of the boiler, turbine (which includes the Carnot efficiency), and generator as 91.2%, 46.7%, and 98.4%, respectively. Additionally, we need to subtract 5.0% for other plant losses. To find the overall efficiency of the total generating unit, we multiply the individual efficiencies together and subtract the loss percentage: $$ \eta_\text{total} = \eta_\text{boiler} \times \eta_\text{turbine} \times \eta_\text{generator} \times (1 - \text{loss}) $$ So, we have: $$ \eta_\text{total} = (0.912) \times (0.467) \times (0.984) \times (1 - 0.050) = 0.4393 $$ The overall efficiency of the total generating unit is 43.93%.

c. Calculating the mass of coal per hour

We are given that one of the coal-burning units produces 355 MW and the enthalpy of combustion of coal is \(29.0 \times 10^3 \, \text{kJ kg}^{-1}\). We need to find the mass of coal required per hour to operate this unit at its peak output. To do this, we will first convert the unit's power output to kilojoules per hour: $$ P = 355 \, \text{MW} = 355 \times 10^6 \, \text{W} = 355 \times 10^6 \frac{\text{J}}{\text{s}} \times \frac{3600 \, \text{s}}{1 \, \text{h}} = 1.278 \times 10^9 \, \text{kJ h}^{-1} $$ Next, we will divide the power output by the overall efficiency to find the actual energy input per hour: $$ E_\text{input} = \frac{P}{\eta_\text{total}} = \frac{1.278 \times 10^9 \, \text{kJ h}^{-1}}{0.4393} = 2.909 \times 10^9 \, \text{kJ h}^{-1} $$ Now, we can find the mass of coal required per hour by dividing the energy input by the enthalpy of combustion of coal: $$ m_\text{coal} = \frac{E_\text{input}}{H_\text{c}} = \frac{2.909 \times 10^9 \, \text{kJ h}^{-1}}{29.0 \times 10^3 \, \text{kJ kg}^{-1}} = 1.003 \times 10^5 \, \text{kg h}^{-1} $$ To convert this to metric tons, we divide by \(1 \times 10^6 \, \text{g metric ton}^{-1}\): $$ m_\text{coal} \text{(metric tons)} = \frac{1.003 \times 10^5 \, \text{kg h}^{-1}}{1 \times 10^6 \, \text{g metric ton}^{-1}} \times \frac{1000 \, \text{g}}{1 \, \text{kg}} = 1.003 \times 10^2 \, \text{metric tons h}^{-1} $$ To operate this unit at its peak output, approximately 100.3 metric tons of coal are required per hour.

Key concepts

Thermodynamics

Thermodynamics is a branch of physics that deals with the relationships between heat and other forms of energy. It explains how energy is converted from one form to another and how it affects matter. At the heart of thermodynamics are the three laws, which describe the behavior of energy in physical systems.

• The First Law, also known as the Law of Energy Conservation, states that energy cannot be created or destroyed, only transformed from one form to another.
• The Second Law of Thermodynamics introduces the concept of entropy, indicating that energy transfer is never 100% efficient since some energy is always lost as waste heat.
• The Third Law states that as a system approaches absolute zero, the entropy approaches a constant minimum.

Understanding these principles helps us grasp how engines work, predict energy usage, and solve practical problems related to energy resources and technology.

Heat Engine

A heat engine is a system that converts heat or thermal energy—and chemical energy—to mechanical energy, which can then be used to perform work. A classic example of a heat engine is the Carnot engine, which is an idealized version often used as a standard by which real engines are compared.
A heat engine typically works in cycles:

• First, it takes heat from a high-temperature reservoir.
• Then, it converts part of this heat to do work.
• Finally, it releases the remaining heat into a low-temperature reservoir.

The efficiency of a heat engine is determined by the temperature difference between the hot (source) and cold (sink) reservoirs. The greater this difference, the more efficient the engine can be. This is important in engineering to design systems that maximize output while minimizing energy waste.

Coal Combustion

Coal combustion refers to the process of burning coal to produce energy. It is one of the oldest sources of energy, relying on the chemical reactions of carbon in coal with oxygen from the air to produce carbon dioxide, water, and energy in the form of heat.
The energy released from coal combustion can be used to generate electricity or mechanical work. During the process, the enthalpy of combustion provides the total heat energy released per unit mass of coal burned. This is typically measured in kilojoules per kilogram (kJ/kg).
Using coal efficiently, especially in power plants, requires optimizing combustion, reducing pollutants, and capturing as much energy as possible from the combustion process. This is crucial for meeting environmental standards and reducing the negative impacts of coal burning on the environment.

Energy Conversion

Energy conversion is the process of changing energy from one form to another. In the context of power plants, such as those using heat engines, this involves converting thermal energy into mechanical energy and then into electrical energy.
More specifically, in a power plant:

• Heat generated from fuel combustion (like coal) is used to create steam.
• The steam drives a turbine, converting thermal energy into mechanical energy.
• Finally, a generator converts this mechanical energy into electrical energy, which can power homes and industries.

Optimizing energy conversion processes is essential for improving the overall efficiency of power plants. This involves not only the machinery's design but also the operations and resource management systems to minimize energy losses. Efficient energy conversion leads to lower operating costs and reduced environmental impact.