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A certain electric generating plant produces electricity by using steam that enters its turbine at a temperature of \(320^{\circ} \mathrm{C}\) and leaves it at \(40^{\circ} \mathrm{C}\). Over the course of a year, the plant consumes \(4.4 \times 10^{16} \mathrm{~J}\) of heat and produces an average electric power output of \(600 \mathrm{MW}\). What is its second-law efficiency?

Short Answer

Expert verified
The second-law efficiency of the steam turbine in the power plant is calculated in step 5. All required values are either given in the exercise or calculated in the previous steps. By substituting these values into the second law efficiency formula, you can find the plant's second-law efficiency.

Step by step solution

01

Convert temperatures from Celsius to Kelvin

Before starting the calculations, it is necessary to convert the temperatures provided from Celsius to Kelvin as thermodynamic calculations generally involve temperature in Kelvin. The conversion from Celsius to Kelvin is done by adding 273.15 to the Celsius temperature. Therefore, the input and exhaust temperatures in Kelvin would be \( T1 = 320^{\circ} C + 273.15 = 593.15 K \) and \( T2 = 40^{\circ} C + 273.15 = 313.15 K \) respectively.
02

Calculate Carnot efficiency

The Carnot efficiency is given by the formula \( \eta_{carnot} = 1 - T2 / T1 = 1 - 313.15 K / 593.15 K \). Calculating this quantity yields the theoretical maximum efficiency a heat engine can achieve working between the two temperatures provided.
03

Convert power output to energy

You need to convert the power output to energy output. 1 year contains 365.25 days, each day has 24 hours, each hour has 60 minutes, each minute has 60 seconds. So, the energy output over a year is \( E_{out} = 600 MW \times 365.25 \times 24 \times 60 \times 60 \).
04

Calculate actual efficiency

The actual efficiency, also known as the first-law efficiency, can be calculated using the formula \( \eta_{actual} = E_{out} / E_{in} \). Substitute the given and calculated values to find the actual efficiency. Note, the first-law efficiency always exceeds the Carnot efficiency.
05

Calculate the second-law efficiency

The second law efficiency is defined as the ratio of the first-law efficiency to the Carnot efficiency and can be calculated by using the formula \( \eta_{second-law} = \eta_{actual} / \eta_{carnot} \).

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Key Concepts

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

Second-law efficiency
Second-law efficiency is an important concept in thermodynamics. It provides a measure of how well a real engine performs compared to the idealized Carnot engine. Carnot engines are theoretical constructs that operate at maximum possible efficiency between two temperature reservoirs. While the first-law efficiency measures the ratio of output energy to input energy, second-law efficiency tells us how close a system is operating to the theoretical best. It is calculated by dividing the actual, or first-law, efficiency by the Carnot efficiency:\[ \eta_{\text{second-law}} = \frac{\eta_{\text{actual}}}{\eta_{\text{Carnot}}} \]This results in a value that gives insight into not only how much energy is conserved in transformation but also how effectively the engine uses the input heat energy compared to the perfect scenario.
  • Second-law efficiency is always less than or equal to 1.
  • A higher second-law efficiency indicates a better-performing engine.
  • This efficiency helps in optimizing processes and understanding real-world limitations.
Carnot cycle
The Carnot cycle is a theoretical cycle that serves as an idealized model for heat engines. It consists of four reversible processes: two isothermal (constant temperature) and two adiabatic (no heat exchange). The cycle's main significance lies in its role in establishing the maximum efficiency limit for heat engines operating between two temperatures.The efficiency of a Carnot engine is given by the formula:\[ \eta_{\text{Carnot}} = 1 - \frac{T_{\text{cold}}}{T_{\text{hot}}} \]where \(T_{\text{cold}}\) and \(T_{\text{hot}}\) are the temperatures of the cold and hot reservoirs, respectively, in Kelvin.
  • Carnot efficiency sets the upper limit for any real engine’s efficiency.
  • It emphasizes the importance of operating between larger temperature differences for higher efficiency.
  • Real engines, due to various losses, operate below the Carnot efficiency.
Understanding the Carnot cycle helps engineers focus on reducing irreversibilities and approaching this ideal as closely as possible in practical designs.
Heat engines
A heat engine is a system that converts thermal energy (heat) into mechanical work. This process involves extracting heat from a high-temperature source, converting part of it into work, and then releasing the remaining heat to a cold sink. The key components and concepts of heat engines include:
  • Working substance: the fluid or gas that absorbs and rejects heat while performing work.
  • Heat source and sink: two reservoirs providing and absorbing heat, crucial for the cycle’s function.
  • Efficiency: the ratio of work output to heat input, indicating how well the engine performs.
Efficiency is critical in assessing heat engines. The ideal engine is a Carnot engine which indicates maximum possible efficiency via the Carnot cycle. However, real heat engines are less efficient due to factors like friction and heat losses.
  • Practical engines include internal combustion engines, steam turbines, and refrigeration cycles.
  • Improvements in efficiency can be achieved by increasing temperature differences, enhancing insulation, and using better materials.
Heat engines are a fundamental concept in thermodynamics, forming the basis for various applications like power plants, vehicle engines, and industrial machines.

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Most popular questions from this chapter

The number of microstates that correspond to any given macrostate is easily illustrated with randomly flipped coins. Each coin has a \(50 / 50\) chance of being a head (H) or a tail (T). In the case of five coins, it is not too difficult to discern the number of ways that one might arrive at five heads, zero tails \((5 \mathrm{H}, 0 \mathrm{~T})\); four heads, one tail \((4 \mathrm{H}, 1 \mathrm{~T})\); three heads, two tails \((3 \mathrm{H}, 2 \mathrm{~T})\); two heads, three tails \((2 \mathrm{H}\), \(3 \mathrm{~T})\); one head, four tails \((1 \mathrm{H}, 4 \mathrm{~T})\); and zero heads, five tails \((0 \mathrm{H}, 5 \mathrm{~T})\). In general terms, the number of microstates can be calculated according to the following formula: \(W_{n}=\frac{\left(n_{\mathrm{H}}+n_{\mathrm{T}}\right) !}{n_{\mathrm{H}} ! n_{\mathrm{T}} !} \quad n\) is the number of heads The probability of any given macrostate is found by taking the ratio of the number of microstates to the sum of the number of microstates: $$ P_{n}=\frac{W_{n}}{\sum_{n} W_{n}} $$ The entropy of any given macrostate is found by multiplying the Boltzmann constant \(\left(k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K}\right)\) by the natural logarithm of the number of microstates: $$ S_{n}=k_{\mathrm{B}} \ln \left(W_{n}\right) $$ (a) Complete a table such as the one below for a system of 5 coins. (b) Without writing out the individual microstates, repeat for 20 coins. (c) Comment on the process for Avogadro's number of coins. Hint: Use the Stirling approximation, \(\ln N !=N \ln N-N\), to rewrite \(\ln W_{n}\) in the entropy function. SSM

Carnot's heat engine employs A. two adiabatic processes and two isothermal processes. B. two adiabatic processes and two isobaric processes. C. two adiabatic processes and two isochoric processes. D. two isothermal processes and two isochoric processes. E. two isothermal processes and two isobaric processes. SSM

An ideal gas trapped inside a thermally isolated cylinder expands slowly by pushing back against a piston. The temperature of the gas A. increases. B. decreases. C. remains the same. D. increases if the process occurs quickly. E. remains the same if the process occurs quickly. SSM

An engine doing work takes in \(10 \mathrm{~kJ}\) and exhausts \(6 \mathrm{~kJ}\). What is the efficiency of the engine? SSM

One mole of an ideal monatomic gas \((\gamma=1.66)\), initially at a temperature of \(0.00^{\circ} \mathrm{C}\), undergoes an adiabatic expansion from a pressure of 10 atm to a pressure of \(2 \mathrm{~atm}\). Find the work done on the gas.

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