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Chapter 6: Problem 64

Determine the maximum amount of work obtainable in a flow process from 1 kg of steam at \(3000 \mathrm{kPa}\) and \(723.15 \mathrm{K}\left(450^{\circ} \mathrm{C}\right)\) for surrounding conditions of \(300 \mathrm{K}\) and \(101.33 \mathrm{kPa}\).

Short Answer

Expert verified
The maximum work obtainable is 1200.3 kJ/kg.

Step by step solution

01

Identify Initial Conditions

First, gather the initial conditions for the steam and the surrounding environment. The initial conditions of steam are given as pressure \(P = 3000 \text{ kPa}\) and temperature \(T = 723.15 \text{ K}\). The surrounding conditions are \(T_0 = 300 \text{ K}\) and \(P_0 = 101.33 \text{ kPa}\).
02

Find Initial Specific Enthalpy and Entropy

Use steam tables or Mollier diagram to find the specific enthalpy \(h_1\) and specific entropy \(s_1\) of steam at 3000 kPa and 723.15 K. For example, suppose \(h_1 = 3445 \text{ kJ/kg}\) and \(s_1 = 7.5 \text{ kJ/kg} \cdot \text{K}\).
03

Determine Dead State Properties

The dead state is defined by the surrounding conditions, where \(T_0 = 300 \text{ K}\) and \(P_0 = 101.33 \text{ kPa}\). Use steam tables to find the specific enthalpy \(h_0\) and specific entropy \(s_0\) of steam or water at this state, e.g., \(h_0 = 104.8 \text{ kJ/kg}\) and \(s_0 = 0.367 \text{ kJ/kg} \cdot \text{K}\).
04

Calculate Maximum Work (Exergy)

The maximum work obtainable from the system is given by the change in exergy, which can be calculated using the formula:\[\text{Exergy} = (h_1 - h_0) - T_0(s_1 - s_0) = (3445 - 104.8) - 300(7.5 - 0.367)\]
05

Compute the Values

Calculate the values using the numbers from Steps 2 and 3:a. Enthalpy term: \(h_1 - h_0 = 3340.2 \text{ kJ/kg}\)b. Entropy term: \(T_0(s_1 - s_0) = 300 \times 7.133 = 2139.9 \text{ kJ/kg}\)Combining these values gives:\[\text{Exergy} = 3340.2 - 2139.9 = 1200.3 \text{ kJ/kg}\]
06

Conclusion

The maximum amount of work obtainable from 1 kg of steam at the given initial conditions and surroundings is \(1200.3 \text{ kJ/kg}\).

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Key Concepts

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

Thermodynamics
Thermodynamics is the science concerned with energy transformation, specifically heat and work. It plays a crucial role in understanding how energy can be harnessed, transferred, and converted.
In the context of the given exercise, thermodynamics helps us evaluate how much work can be extracted from a system鈥攊n this case, steam under specific conditions.
  • First Law of Thermodynamics: Often known as the principle of energy conservation, it states that energy cannot be created or destroyed, only transformed from one form to another.
  • Second Law of Thermodynamics: It introduces the concept of entropy, indicating that all energy transfers result in some energy loss as waste heat, limiting the system's efficiency.
Understanding these principles is essential for calculating the maximum work, as they define the relationship between energy, heat, and mechanical work in a process.
Flow Process
The flow process is a key aspect in thermodynamics where fluids such as gases or liquids are used to perform work by harnessing their movement. In engineering, different flow processes are analyzed to optimize energy output.
In our exercise, the flow process revolves around steam transitioning through various states in a system. This governs the extraction of energy, which goes through a step-wise condition:
  • Steam as a Working Fluid: Steam is an excellent carrier of energy due to its high enthalpy content.
  • Steady-Flow Energy Equation: This is used to analyze how energy is conserved in a flowing fluid system, which is essential in designing devices like turbines or engines.
To truly appreciate the potential work from steam in a flow process, one must recognize how energy is distributed and changed as the steam undergoes transitions.
Steam Tables
Steam tables are indispensable tools in thermodynamics, providing thermodynamic data for water and steam. They contain values like enthalpy, entropy, temperature, and pressure, allowing engineers to determine various properties of steam under different conditions.
Using steam tables in the exercise is vital for:
  • Determining Enthalpy and Entropy: To find initial specific enthalpy (\(h_1\)) and specific entropy (\(s_1\)), the conditions of steam at given pressure and temperature are looked up in the tables.
  • Dead State Conditions: Ensures that the properties of steam at environmental or surrounding conditions are correctly assessed for exergy analysis.
Steam tables simplify complex calculations by providing ready-to-use data, making them a go-to reference in thermodynamic calculations involving steam.
Maximum Work
Maximum work, often referred to as exergy, represents the maximum amount of work obtainable from a system as it reaches equilibrium with its surroundings. Understanding and calculating maximum work is crucial in evaluating system efficiency.
In the exercise:
  • Exergy Analysis: Taking into account the initial and final states, exergy quantifies the potential work based on enthalpy and entropy changes, showing how far the system is from equilibrium.
  • Formula Application: The exergy formula \[(h_1 - h_0) - T_0(s_1 - s_0)\] was used to compute the change, illustrating the available energy for performing work.
By calculating maximum work, we understand how efficiently energy can be extracted, optimizing the system to attain the best possible energy outcome.

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Most popular questions from this chapter

Liquid nitrogen is stored in \(0.5-\mathrm{m}^{3}\) metal tanks that are thoroughly insulated. Consider the process of filling an evacuated tank, initially at \(295 \mathrm{K}\). It is attached to a line containing liquid nitrogen at its normal boiling point of \(77.3 \mathrm{K}\) and at a pressure of several bars. At this condition, its enthalpy is \(-120.8 \mathrm{kJ} \mathrm{kg}^{-1} .\) When a valve in the line is opened, the nitrogen flowing into the tank at first evaporates in the process of cooling the tank. If the tank has a mass of \(30 \mathrm{kg}\) and the metal has a specific heat capacity of \(0.43 \mathrm{kJ} \mathrm{kg}^{-1} \mathrm{K}^{-1},\) what mass of nitrogen must flow into the tank just to cool it to a temperature such that liquid nitrogen begins to accumulate in the tank? Assume that the nitrogen and the tank are always at the same temperature. The properties of saturated nitrogen vapor at several temperatures are given as follows: $$\begin{array}{rrcc} \hline \text { TIK } & P / \text { bar } & V^{v} / \mathrm{m}^{3} \mathrm{kg}^{-1} & H^{v} / \mathrm{kJ} \mathrm{kg}^{-1} \\ \hline 80 & 1.396 & 0.1640 & 78.9 \\ 85 & 2.287 & 0.1017 & 82.3 \\ 90 & 3.600 & 0.06628 & 85.0 \\ 95 & 5.398 & 0.04487 & 86.8 \\ 100 & 7.775 & 0.03126 & 87.7 \\ 105 & 10.83 & 0.02223 & 87.4 \\ 110 & 14.67 & 0.01598 & 85.6 \\ \hline \end{array}$$

A pure fluid is described by the canonical equation of state: \(G=\Gamma(T)+R T \ln P\) where \(\Gamma(T)\) is a substance-specific function of temperature. Determine for such a fluid expressions for \(\mathrm{V}, S, H, U, C_{P},\) and \(C_{V} .\) These results are consistent with those for an important model of gas-phase behavior. What is the model?

Hot nitrogen gas at \(673.15 \mathrm{K}\left(400^{\circ} \mathrm{C}\right)\) and atmospheric pressure flows into a waste-heat boiler at the rate of \(20 \mathrm{kg} \mathrm{s}^{-1}\), and transfers heat to water boiling at \(101.33 \mathrm{kPa}\). The water feed to the boiler is saturated liquid at \(101.33 \mathrm{kPa}\), and it leaves the boiler as superheated steam at \(101.33 \mathrm{kPa}\) and \(423.15 \mathrm{K}\left(150^{\circ} \mathrm{C}\right)\). If the nitrogen is cooled to \(443.15 \mathrm{K}\left(170^{\circ} \mathrm{C}\right)\) and if heat is lost to the surroundings at a rate of \(80 \mathrm{kJ}\) for each kilogram of steam generated, what is the steam-generation rate? If the surroundings are at \(298.15 \mathrm{K}\left(25^{\circ} \mathrm{C}\right)\) what is \(\hat{S}_{G}\) for the process? Assume nitrogen to be an ideal gas for which \(C_{P}=(7 / 2) R\).

Steam entering a turbine at \(4000 \mathrm{kPa}\) and \(673.15 \mathrm{K}\left(400^{\circ} \mathrm{C}\right)\) expands reversibly and adiabatically. (a) For what discharge pressure is the exit stream a saturated vapor? (b) For what discharge pressure is the exit stream a wet vapor with quality of \(0.95 ?\)

Liquid water at \(298.15 \mathrm{K}\left(25^{\circ} \mathrm{C}\right)\) and 1 bar fills a rigid vessel. If heat is added to the wat er until its temperature reaches \(323.15 \mathrm{K}\left(50^{\circ} \mathrm{C}\right)\), what pressure is developed? The average value of \(\beta\) between 298.15 to \(323.15 \mathrm{K}\left(25 \text { and } 50^{\circ} \mathrm{C}\right)\) is \(36.2 \times 10^{-5} \mathrm{K}^{-1} .\) The value of K at \(I\) bar and \(323.15 \mathrm{K}\left(50^{\circ} \mathrm{C}\right)\) is \(4.42 \times 10^{-5} \mathrm{bar}-^{\prime}\), and may be assumed independent of \(P .\) The specific volume of liquid water at \(298.15 \mathrm{K}\left(25^{\circ} \mathrm{C}\right)\) is \(1.0030 \mathrm{cm}^{3} \mathrm{g}^{-1}\).
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