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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 steam turbine operating at steady state develops \(9750 \mathrm{hp}\). The turbine receives 100,000 pounds of steam per hour at \(400 \mathrm{lbf} / \mathrm{in} .^{2}\) and \(600^{\circ} \mathrm{F}\). At a point in the turbine where the pressure is \(60 \mathrm{lbf} / \mathrm{in}^{2}\). and the temperature is \(300^{\circ} \mathrm{F}\), steam is bled off at the rate of \(25,000 \mathrm{lb} / \mathrm{h}\). The remaining steam continues to expand through the turbine, exiting at \(2 \mathrm{lbf} / \mathrm{in}^{2}\) and \(90 \%\) quality. (a) Determine the rate of heat transfer between the turbine and its surroundings, in Btu/h. (b) Devise and evaluate an exergetic efficiency for the turbine. Kinetic and potential energy effects can be ignored. Let \(T_{0}=\) \(77^{\circ} \mathrm{F}, p_{0}=1 \mathrm{~atm}\).

A vessel contains carbon dioxide. Using the ideal gas model (a) determine the specific exergy of the gas, in Btu/lb, at \(p=80 \mathrm{lbf}^{2} \mathrm{in}^{2}\) and \(T=180^{\circ} \mathrm{F}\). (b) plot the specific exergy of the gas, in Btu/b, versus pressure ranging from 15 to \(80 \mathrm{lbf} / \mathrm{in}^{2}\), for \(T=80^{\circ} \mathrm{F}\). (c) plot the specific exergy of the gas, in Btu/lb, versus temperature ranging from 80 to \(180^{\circ} \mathrm{F}\), for \(p=15 \mathrm{lbf} / \mathrm{in}^{2}{ }^{2}\) The gas is at rest and zero elevation relative to an exergy reference environment for which \(T_{0}=80^{\circ} \mathrm{F}, p_{0}=15 \mathrm{lbf} / \mathrm{in} .^{2}\)

An electric water heater having a \(200-L\) capacity heats water from 23 to \(55^{\circ} \mathrm{C}\). Heat transfer from the outside of the water heater is negligible, and the states of the electrical heating element and the tank holding the water do not change significantly. Perform a full exergy accounting, in kJ, of the electricity supplied to the water heater. Model the water as incompressible with a specific heat \(c=4.18 \mathrm{~kJ} / \mathrm{kg}+\mathrm{K}\). Let \(T_{0}=23^{\circ} \mathrm{C}\).

A counterflow heat exchanger operating at steady state has water entering as saturated vapor at 5 bar with a mass flow rate of \(4 \mathrm{~kg} / \mathrm{s}\) and exiting as saturated liquid at 5 bar. Air enters in a separate stream at \(320 \mathrm{~K}, 2\) bar and exits at \(350 \mathrm{~K}\) with a negligible change in pressure. Heat transfer between the heat exchanger and its surroundings is negligible. Determine (a) the change in the flow exergy rate of each stream, in \(\mathrm{kW}\). (b) the rate of exergy destruction in the heat exchanger, in \(\mathrm{kW}\). Ignore the effects of motion and gravity. Let \(T_{0}=300 \mathrm{~K}\), \(p_{0}=1 \mathrm{bar}\).

Determine the specific exergy of argon at (a) \(p=2 p_{0}\), \(T=2 T_{0}\), (b) \(p=p_{0} / 2, T=T_{0} / 2\). Locate cach state relative to the dead state on temperature-pressure coordinates. Assume ideal gas behavior with \(k=1.67\). Let \(T_{0}=537^{\circ} \mathrm{R}\), \(p_{0}=1 \mathrm{~atm} .\)

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