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Four kilograms of a two-phase liquid-vapor mixture of water initially at \(300^{\circ} \mathrm{C}\) and \(x_{1}=0.5\) undergo the two different processes described below. In each case, the mixture is brought from the initial state to a saturated vapor state, while the volume remains constant. For each process, determine the change in exergy of the water, the net amounts of exergy transfer by work and heat, and the amount of exergy destruction, each in kJ. Let \(T_{0}=300 \mathrm{~K}, p_{0}=1\) bar, and ignore the effects of motion and gravity. Comment on the difference between the exergy destruction values. (a) The process is brought about adiabatically by stirring the mixture with a paddle wheel. (b) The process is brought about by heat transfer from a thermal reservoir at \(610 \mathrm{~K}\). The temperature of the water at the location where the heat transfer occurs is \(610 \mathrm{~K}\).

Short Answer

Expert verified
螖EX calculated using steam tables and given formulas. Adiabatic process typically shows higher exergy destruction compared to heat transfer process.

Step by step solution

01

Understand Initial and Final States

The initial state of the water is a two-phase mixture at 300掳C with a quality (x鈧) of 0.5. The final state is a saturated vapor. The initial and final masses are constant at 4 kg.
02

Identify Required Equations and Given Data

Given data include the ambient temperature (T鈧 = 300 K) and pressure (p鈧 = 1 bar). The process is either adiabatic (no heat transfer) or involves heat transfer from a thermal reservoir at 610 K.
03

Determine Specific Properties of Water

Use steam tables to find the specific enthalpy (h) and specific entropy (s) for the initial and final states. For the initial state at 300掳C and x鈧 = 0.5:h鈧 = h鈧 + x鈧 (h岬 - h鈧)s鈧 = s鈧 + x鈧 (s岬 - s鈧)For the final state (saturated vapor):h鈧 = h岬モ瑲s鈧 = s岬モ瑲
04

Calculate exergy change

The change in exergy (螖EX) is given by:螖EX = mdef [(h鈧-h鈧) - T鈧 (s鈧-s鈧)]where m = 4 kg T鈧 is the ambient temperature.Calculate 螖EX for both processes.
05

Calculate Exergy Transfer by Work and Heat 鈥 Adiabatic Process

For the adiabatic process, there is no heat transfer. The work done W such as the work done by the paddle wheel is equal to the change in exergy EX_work = 螖EX since no heat is transferred.
06

Calculate Exergy Transfer by Work and Heat 鈥 Heat Transfer Process

The heat transfer process involves heat transfer at 610 K. Use:EX_heat = Q (1 - T鈧/T) = Q (1 - 300/610)where Q is the heat transferred.
07

Calculate Exergy Destruction for Both Processes

Exergy destruction (EX_dest) can be found using the balance:EX_dest = EX_in - EX_outFor the adiabatic process, all exergy change is due to work.For the heat transfer process, the difference is due to heat transfer.
08

Comment on Exergy Destruction Values

Compare exergy destruction for both processes. Adiabatic process often involves more exergy destruction due to the inefficiencies related to mechanical work.

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Key Concepts

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

Two-Phase Liquid-Vapor Mixture
In thermodynamics, a two-phase liquid-vapor mixture refers to a system that contains both liquid and vapor phases of a substance. At any given temperature, the mixture will have a particular quality, denoted as x鈧. This quality represents the ratio of the mass of vapor to the total mass. For instance, in the exercise, the initial quality of the water is 0.5. This means 50% of the mixture's mass is in the vapor phase while the remaining 50% is in the liquid phase.
This mixture is particularly significant in exergy analysis because its properties, such as specific enthalpy (h) and specific entropy (s), vary with temperature and pressure. These properties are critical in determining the energy transformations within thermodynamic systems. In the provided exercise, the initial state of the mixture is at 300掳C, which helps determine the specific properties using steam tables.
Adiabatic Process
An adiabatic process is a thermodynamic process where no heat is exchanged with the surroundings. This means that the only energy transfer possible is through work. In the context of the exercise, the adiabatic process involves stirring the water mixture with a paddle wheel.
This essentially means that the change in exergy of the system is only influenced by the work done on the system and not by heat transfer. The equation used for calculating the change in exergy (螖EX) remains the same, but for an adiabatic process, the term involving heat transfer (Q) is zero and thus simplifies the calculation. Typically, adiabatic processes are less efficient compared to processes involving heat transfer because they result in higher exergy destruction due to mechanical inefficiencies.
Exergy Destruction
Exergy destruction is a measure of the irreversibility of a process. It quantifies the amount of exergy that is not converted into useful work or transferred as heat but instead is destroyed due to inefficiencies in the system. In the exercise, exergy destruction is calculated for both the adiabatic process and the process involving heat transfer.
For the adiabatic process, higher exergy destruction is expected owing to the inefficiencies of mechanical work done by the paddle wheel. For the heat transfer process, exergy destruction is connected to the difference between the input and output exergy involving heat climate and how effectively it's being transferred. This analysis is pivotal in optimizing thermal systems and improving their efficiency. The formula for exergy destruction (EX_dest) is:
.EX_dest = EX_in - EX_out
This identifies the losses encountered and highlights areas for process improvement.
Paddle Wheel Work
Paddle wheel work is used in the context of an adiabatic process where mechanical work is done on a system to bring about a change in state without heat exchange. In the exercise, the mixture undergoes an adiabatic process driven by the work done by a paddle wheel.
This paddle-wheel setup stirs the mixture, causing its internal energy to rise solely due to mechanical input. This work directly translates to a change in the system's exergy, which can be calculated using the specific properties of the mixture. The paddle wheel work signifies the practical applications in engines and turbines where such mechanical work is common, although it often results in higher exergy destruction due to mechanical inefficiencies.
Heat Transfer Process
Heat transfer processes involve the exchange of thermal energy between a system and its surroundings. In thermodynamics, these processes are not adiabatic and include a term involving heat transfer (Q) in exergy calculations. In the exercise, one of the processes involves heat transfer from a thermal reservoir at 610 K to the water mixture.
This type of process typically results in lower exergy destruction compared to adiabatic processes, as it allows for more efficient energy exchange. The exergy transfer by heat is calculated using the formula:
.EX_heat = Q(1 - T鈧/T)
where Q denotes the heat transferred, T鈧 is the ambient temperature, and T is the temperature of the thermal reservoir. This equation helps in understanding how efficiently the thermal energy is being utilized, guiding improvements in thermal system design and operation.

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Most popular questions from this chapter

A gearbox operating at steady state receives 4 hp along the input shaft and delivers 3 hp along the output shaft. The outer surface of the gearbox is at \(130^{\circ} \mathrm{F}\). For the gearbox, (a) determine, in Btu/s, the rate of heat transfer and (b) perform a full exergy accounting, in Btu/s, of the input power. Let \(T_{0}=70^{\circ} \mathrm{F}\).

Saturated water vapor at \(400 \mathrm{lbf} / \mathrm{in}^{2}\) enters an insulated turbine operating at steady state. At the turbine exit the pressure is \(0.6\) lbf/in. \({ }^{2}\) The work developed is 306 Btu per pound of steam passing through the turbine. Kinetic and potential energy effects can be neglected. Let \(T_{0}=60^{\circ} \mathrm{F}\), \(p_{0}=1 \mathrm{~atm}\). Determine (a) the exergy destruction rate, in Btu per pound of steam expanding through the turbine. (b) the isentropic turbine efficiency- (c) the exergetic turbine efficiency.

Steam at \(200 \mathrm{lbf} / \mathrm{in}^{2}, 660^{\circ} \mathrm{F}\) enters a turbine operating at steady state with a mass flow rate of \(16.5 \mathrm{lb} / \mathrm{min}\) and exits at 14.7 lbffin. \({ }^{2}, 238^{\circ}\) F. Stray heat transfer and the effects of motion and gravity can be ignored. Let \(T_{0}=537^{\circ} \mathrm{R}, p_{0}=14.7 \mathrm{lbf} / \mathrm{in}^{2}{ }^{2}\) Determine for the turbine (a) the power developed and the rate of exergy destruction, each in Btu/min, and (b) the isentropic and exergetic turbine efficiencies.

Carbon monoxide \((\mathrm{CO})\) enters an insulated compressor operating at steady state at 10 bar, \(227^{\circ} \mathrm{C}\), and a mass flow rate of \(0.1 \mathrm{~kg} / \mathrm{s}\) and exits at 15 bar, \(327^{\circ} \mathrm{C}\). Determine the power required by the compressor and the rate of exergy destruction, each in \(\mathrm{kW}\). Ignore the effects of motion and gravity. Let \(T_{0}=17^{\circ} \mathrm{C}, p_{0}=1\) bar.

Water vapor at \(6 \mathrm{MPa}, 600^{\circ} \mathrm{C}\) enters a turbine operating at steady state and expands adiabatically to \(10 \mathrm{kPa}\). The mass flow rate is \(2 \mathrm{~kg} / \mathrm{s}\) and the isentropic turbine efficiency is \(94.7 \%\). Kinetic and potential energy effects are negligible. Determine (a) the power developed by the turbinc, in \(\mathrm{kW}\). (b) the rate at which exergy is destroyed within the turbine, in \(\mathrm{kW}\). (c) the exergetic turbine efficiency. Let \(T_{0}=298 \mathrm{~K}, p_{0}=1 \mathrm{~atm}\).

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