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Chapter 7: Problem 13

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 \(T_{0}=537^{\circ} \mathrm{R}\), \(p_{0}=1 \mathrm{~atm} .\)

Short Answer

Expert verified
Exergy at state (a): \( e_a = T_0 R (2.5 - 1.5 \, \text{ln}(2)) \), and at state (b): \( e_b = T_0 R (-1.25 + 1.5 \, \text{ln}(2)) \).

Step by step solution

01

Understand the Exergy Equation

Exergy, or available energy, for an ideal gas can be determined using the specific exergy formula in relation to a dead state (ambient condition) which is given as: \[ e = (h-h_0) - T_0 (s-s_0) \]Here, \( h \) and \( s \) denote specific enthalpy and specific entropy respectively, and the subscript 0 denotes properties at the dead state.
02

Enthalpy and Entropy for Ideal Gas

For an ideal gas, the specific enthalpy and specific entropy differences can be defined as:\[ h - h_0 = c_p(T - T_0) \]\[ s - s_0 = c_p \, \text{ln} \left(\frac{T}{T_0}\right) - R \, \text{ln} \left(\frac{p}{p_0}\right) \]Where \( c_p \) is specific heat at constant pressure and \( R \) is the specific gas constant for argon. For argon, given \( k = 1.67 \), we have \( c_p = \frac{k R}{k-1} = 2.5R \).
03

State (a) Calculation

For (a) where \(p = 2p_0\) and \(T=2T_0\):First calculate \(h - h_0\):\[ h - h_0 = c_p(T - T_0) = 2.5R(2T_0 - T_0) = 2.5R(T_0) \]Next, calculate \(s - s_0\):\[ s - s_0 = 2.5R \, \text{ln} \left(2\right) - R \, \text{ln} \left(2\right) = 2.5R \, \text{ln}(2) - R \, \text{ln}(2) = 1.5R \, \text{ln}(2) \]Now plug these into the exergy equation:\[ e = 2.5R(T_0) - T_0 \left(1.5R \, \text{ln}(2)\right) = T_0 R (2.5 - 1.5 \, \text{ln}(2)) \]
04

State (b) Calculation

For (b) where \(p = p_0 / 2\) and \(T = T_0 / 2\):First calculate \(h - h_0\):\[ h - h_0 = c_p(T - T_0) = 2.5R(T_0/2 - T_0) = -1.25R(T_0) \]Next, calculate \(s - s_0\):\[ s - s_0 = 2.5R \, \text{ln} \left(0.5\right) - R \, \text{ln} \left(0.5\right) = 2.5R \, \text{ln}(0.5) - R \, \text{ln}(0.5) = 1.5R \, \text{ln}(0.5) \]Now plug these into the exergy equation:\[ e = -1.25R(T_0) - T_0 \left(1.5R \, \text{ln}(0.5)\right) = T_0 R (-1.25 + 1.5 \, \text{ln}(2)) \]
05

Summarize the Results

The specific exergy for state (a) is:\[ e_a = T_0 R (2.5 - 1.5 \, \text{ln}(2)) \]The specific exergy for state (b) is:\[ e_b = T_0 R (-1.25 + 1.5 \, \text{ln}(2)) \]

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Key Concepts

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

Ideal Gas Behavior
The calculation of specific exergy assumes that argon behaves as an ideal gas. An ideal gas is a theoretical gas composed of a set of randomly moving, non-interacting point particles. Real gases approximate this behavior at high temperatures and low pressures. The ideal gas law is expressed as \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the gas constant, and \( T \) is temperature. For argon, it's important to note that argon closely follows the ideal gas law under common laboratory conditions. This simplification helps in analyzing other properties without considering complex interactions.
Specific Enthalpy
Specific enthalpy (\( h \)) is the enthalpy per unit mass. For an ideal gas, the change in specific enthalpy is directly proportional to the change in temperature. The relation is given by \( h - h_0 = c_p (T - T_0) \), where \( c_p \) is the specific heat at constant pressure. Specific enthalpy represents the total energy (internal energy plus the product of pressure and volume) of the gas. This relationship becomes simpler under the assumption of ideal gas behavior, allowing us to tie enthalpy changes to temperature changes only, without worrying about pressure variations.
Specific Entropy
Specific entropy (\( s \)) measures the disorder or randomness in the system. For an ideal gas, the difference in specific entropy is given by \( s - s_0 = c_p \, \text{ln}(\frac{T}{T_0}) - R \, \text{ln}(\frac{p}{p_0}) \). Here, \( c_p \) is the specific heat at constant pressure, and \( R \) is the specific gas constant for argon. Specific entropy change takes into account both temperature and pressure variations. The logarithmic nature of the formula indicates how entropy responds more sensitively to relative changes rather than absolute values, reflecting the physical concept of entropy as a measure of energy dispersion in the system.
Thermodynamic Dead State
The thermodynamic dead state is a reference state at which a system is in equilibrium with its environment and can no longer do any useful work. In this problem, the dead state conditions are given as \( T_0 = 537^{\circ} \mathrm{R} \) and \( p_0 = 1 \mathrm{~atm} \). At this state, the specific exergy is zero. Understanding the dead state is crucial to calculate exergy because exergy is defined relative to this equilibrium state. Essentially, it mirrors how far the actual state is from complete equilibrium and hence, how much useful work it can potentially deliver before reaching equilibrium.
Specific Heat at Constant Pressure
Specific heat at constant pressure (\( c_p \)) for a gas is the amount of heat required to raise the temperature of a unit mass of the gas by one degree while keeping the pressure constant. For argon, given the ratio of specific heats (\( k = 1.67 \)), the specific heat at constant pressure can be calculated using \( c_p = \frac{k R}{k - 1} = 2.5R \). The value of \( c_p \) is essential for determining changes in enthalpy and entropy. Knowing \( c_p \) allows us to link energy changes directly to temperature changes, thus simplifying the computation of specific enthalpy and specific entropy for ideal gases.

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Most popular questions from this chapter

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 \(610 \mathrm{~K}\).

A pump operating at steady state takes in saturated liquid water at \(65 \mathrm{lbf} / \mathrm{in}^{2}\) at a rate of \(10 \mathrm{lb} / \mathrm{s}\) and discharges water at \(1000 \mathrm{lbf} / \mathrm{in}^{2}\). The isentropic pump efficiency is \(80.22 \%\). Heat transfer with the surroundings and the effects of motion and gravity can be neglected. If \(T_{0}=75^{\circ} \mathrm{F}\), determine for the pump (a) the exergy destruction rate, in Btu/s (b) the exergetic efficiency.

For an ideal gas with constant specific heat ratio \(k\), show that in the absence of significant effects of motion and gravity the specific flow exergy can be expressed as $$ \frac{\mathrm{e}_{i}}{c_{p} T_{0}}=\frac{T}{T_{0}}-1-\ln \frac{T}{T_{0}}+\ln \left(\frac{p}{p_{0}}\right)^{(k-1) k k} $$ (a) For \(k=1.2\) develop plots of \(e_{9} / c_{p} T_{0}\) versus for \(T / T_{0}\) for \(p / p_{0}=0.25,0.5,1,2,4\). Repeat for \(k=1.3\) and 1.4. (b) The specific flow exergy can take on negative values when \(p / p_{0}<1\). What does a negative value mean physically?

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 \(T_{0}=80^{\circ} \mathrm{F}, p_{0}=15 \mathrm{lbf} / \mathrm{in} .^{2}\)

Argon enters an insulated turbine operating at steady state at \(1000^{\circ} \mathrm{C}\) and \(2 \mathrm{MPa}\) and exhausts at \(350 \mathrm{kPa}\). The mass flow rate is \(0.5 \mathrm{~kg} / \mathrm{s}\) and the turbine develops power at the rate of \(120 \mathrm{~kW}\). Determine (a) the temperature of the argon at the turbine exit, in \({ }^{\circ} \mathrm{C}\). (b) the exergy destruction rate of the turbine, in \(k W\). (c) the turbine exergetic efficiency. Neglect kinetic and potential energy effects. Let \(T_{0}=20^{\circ} \mathrm{C}\), \(p_{0}=1\) bar.
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