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A coal-burning power plant produces \(3000 .\) MW of thermal energy, which is used to boil water and produce supersaturated steam at \(300 .{ }^{\circ} \mathrm{C}\). This high-pressure steam turns a turbine producing \(1000 .\) MW of electrical power. At the end of the process, the steam is cooled to \(30.0^{\circ} \mathrm{C}\) and recycled. a) What is the maximum possible efficiency of the plant? b) What is the actual efficiency of the plant? c) To cool the steam, river water runs through a condenser at a rate of \(4.00 \cdot 10^{7} \mathrm{gal} / \mathrm{h} .\) If the water enters the condenser at \(20.0^{\circ} \mathrm{C},\) what is its exit temperature?

Short Answer

Expert verified
Based on the given information, we found that the maximum possible efficiency of the coal-burning power plant is 47.1%. The actual efficiency of the plant is 33.3%. The heat removed by river water from the plant is 2000 MW, and the exit temperature of the river water is approximately 31.4°C.

Step by step solution

01

Determine the maximum possible efficiency of the plant

Using the Carnot efficiency, we can calculate the maximum possible efficiency of the plant. The formula for Carnot efficiency is given by: Carnot Efficiency = \(1 - \frac{T_{cold}}{T_{hot}}\) Here, \(T_{cold}\) is the temperature of the cold reservoir, and \(T_{hot}\) is the temperature of the hot reservoir. We need to convert the temperatures to Kelvin before we plug them into the formula. \(T_{cold} = 30.0^{\circ} \mathrm{C} + 273.15 = 303.15\) K \(T_{hot} = 300^{\circ} \mathrm{C} + 273.15 = 573.15\) K Now, we can plug these into the Carnot Efficiency formula: Carnot Efficiency = \(1 - \frac{303.15}{573.15} \approx 0.471\) or 47.1%
02

Determine the actual efficiency of the plant

The actual efficiency of the plant can be calculated using the formula: Actual Efficiency = \(\frac{Power \: Output}{Power \: Input}\) The power output is given as 1000 MW and the power input as 3000 MW. So we have: Actual Efficiency = \(\frac{1000}{3000} = 0.333\) or 33.3%
03

Calculate the heat removed by river water

We can determine the amount of heat removed by the river water by using the conservation of energy as follows: Heat input - Heat output = Heat removed by river water Heat removed by river water = Power input - Power output Heat removed by river water = 3000 MW - 1000 MW = 2000 MW
04

Find the exit temperature of river water

We need to determine the mass flow rate of the river water before we can calculate the exit temperature. The flow rate of river water is given as \(4.00 \cdot 10^7\) gal/h. We need to convert this to kg/s as follows: Flow rate = \(4.00 \cdot 10^7 \frac{gal}{h} \times \frac{3.78541\: liters}{gal} \times \frac{1\: kg}{liter} \times \frac{1\: h}{3600\: s} \approx 4.2 \times 10^4 \frac{kg}{s}\) Now, we can use the formula: \(q = mc∆T\) to find the change in temperature of the river water. Here, \(q\) is the heat removed by the river water, \(m\) is the mass flow rate, \(c\) is the specific heat capacity of water, and \(∆T\) is the change in temperature. Rearranging the formula to find \(∆T\): \(∆T = \frac{q}{mc}\) We are given \(q = 2000\) MW, which we convert to Watts (1 MW = \(10^6\) W): \(q = 2000 \times 10^6\: W\) The specific heat capacity of water is approximately \(4.18 \times 10^3\) J/kg K. We can now plug in the values to find the change in temperature: \(∆T = \frac{2000 \times 10^6}{4.2 \times 10^4 \times 4.18 \times 10^3} \approx 11.4\)°C Finally, we can find the exit temperature by adding the change in temperature to the initial temperature of the river water: Exit temperature = \(20.0^{\circ}\mathrm{C} + 11.4^{\circ}\mathrm{C} \approx 31.4^{\circ}\mathrm{C}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Carnot Efficiency
Understanding Carnot efficiency is essential for grasping the limits of energy conversion in heat engines, such as power plants. It represents the highest possible efficiency that any engine can achieve, based on the temperature difference between the heat source and the sink. The formula for Carnot efficiency is \(1 - \frac{T_{cold}}{T_{hot}}\), where \({T_{hot}}\) and \({T_{cold}}\) must be in absolute temperature units (Kelvin, K).

For a coal-burning power plant, the hot source is the steam produced to run the turbines, and the cold sink is the environment into which waste heat is discharged. This ideal efficiency is not achievable in practice due to real-world frictions and other irreversibilities but serves as a benchmark to measure actual engine performance against. The closer a real engine gets to its Carnot efficiency, the better its design and performance.
Energy Conversion
When discussing energy conversion, especially in the context of power plants, it's about transforming one form of energy into another. Our coal-burning power plant takes chemical energy from coal and converts it into thermal energy. The thermal energy then generates high-pressure steam, which carries out mechanical work by turning turbines. This mechanical energy is finally transformed into electrical energy - the product delivered to consumers.

In this energy conversion pathway, efficiency is paramount because not all the initial energy is converted into electricity; some is lost as waste heat. The measure of what fraction of thermal energy is turned into electrical energy is known as the plant’s actual efficiency. By optimizing each conversion step, from burning coal to generating power, the efficiency of the power plant can be improved, leading to less fuel consumption and fewer emissions.
Conservation of Energy
The law of conservation of energy is a fundamental principle in physics that states that energy cannot be created or destroyed, only transformed from one form to another. This law underpins how power plants operate, where the aim is to convert energy from fuel into useful work as efficiently as possible.

In the context of our coal-burning power plant, the conservation of energy principle helps us understand that the thermal energy not converted into electrical energy must be accounted for. It is primarily removed from the system in the form of waste heat. This heat is often transferred to the environment, in this case, river water, which is used as a coolant. The water absorbs the waste heat, thus conserving the energy by changing its temperature. A precise calculation of this heat transfer, adhering to the conservation principle, ensures accurate estimates of environmental impact and helps in designing appropriate cooling systems.

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Most popular questions from this chapter

The temperature at the cloud tops of Saturn is approximately \(150 .\) K. The atmosphere of Saturn produces tremendous winds; wind speeds of \(600 . \mathrm{km} / \mathrm{h}\) have been inferred from spacecraft measurements. Can the wind chill factor on Saturn produce a temperature at (or below) absolute zero? How, or why not?

A Carnot refrigerator is operating between thermal reservoirs with temperatures of \(27.0^{\circ} \mathrm{C}\) and \(0.00^{\circ} \mathrm{C}\) a) How much work will need to be input to extract \(10.0 \mathrm{~J}\) of heat from the colder reservoir? b) How much work will be needed if the colder reservoir is at \(-20.0^{\circ} \mathrm{C}\) ?

A heat engine cycle often used in refrigeration, is the Brayton cycle, which involves an adiabatic compression, followed by an isobaric expansion, an adiabatic expansion, and finally an isobaric compression. The system begins at temperature \(T_{1}\) and transitions to temperatures \(T_{2}, T_{3},\) and \(T_{4}\) after respective parts of the cycle. a) Sketch this cycle on a \(p V\) -diagram. b) Show that the efficiency of the overall cycle is given by \(\epsilon=1-\left(T_{4}-T_{1}\right) /\left(T_{3}-T_{2}\right)\).

A refrigerator has a coefficient of performance of \(5.00 .\) If the refrigerator absorbs 40.0 cal of heat from the low-temperature reservoir in each cycle, what is the amount of heat expelled into the high-temperature reservoir?

A Carnot engine takes an amount of heat \(Q_{H}=100 .\) J from a high- temperature reservoir at temperature \(T_{H}=1000 .{ }^{\circ} \mathrm{C},\) and exhausts the remaining heat into a low-temperature reservoir at \(T_{\mathrm{L}}=10.0^{\circ} \mathrm{C}\). Find the amount of work that is obtained from this process.

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