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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

One-half pound of air is contained in a closed, rigid, insulated tank. Initially the temperature is \(520^{\circ} \mathrm{R}\) and the pressure is \(14.7\) psia. The air is stirred by a paddle wheel until its temperature is \(600^{\circ} \mathrm{R}\). Using the ideal gas model, determine for the air the change in exergy, the transfer of exergy accompanying work, and the exergy destruction, all in Btu. Ignore the effects of motion and gravity and let \(T_{0}=537^{\circ} \mathrm{R}, p_{0}=14.7\) psia.

Two kilograms of carbon monoxide in a piston-cylinder assembly, initially at 1 bar and \(27^{\circ} \mathrm{C}\), is heated at constant pressure with no internal irreversibilities to a final temperature of \(227^{\circ} \mathrm{C}\). Employing the ideal gas model, determine the work, the heat transfer, and the amounts of exergy transfer accompanying work and heat transfer, each in kJ. Let \(T_{0}=300 \mathrm{~K}, p_{0}=1\) bar and ignore the effects of motion and gravity.

Nitrogen \(\left(\mathrm{N}_{2}\right)\) at 25 bar, \(450 \mathrm{~K}\) enters a turbine and expands to \(2 \mathrm{bar}, 250 \mathrm{~K}\) with a mass flow rate of \(0.2 \mathrm{~kg} / \mathrm{s}\) The turbine operates at steady state with negligible heat transfer with its surroundings. Assuming the ideal gas model with \(k=\) \(1.399\) and ignoring the effects of motion and gravity, determine (a) the isentropic turbine efficiency. (b) the exergetic turbine efficiency. Let \(T_{0}=25^{\circ} \mathrm{C}, p_{0}=1 \mathrm{~atm}\).

A rigid, well-insulated tank consists of two compartments, each having the same volume, separated by a valve. Initially, one of the compartments is evacuated and the other contains \(0.25 \mathrm{lbmol}\) of a gas at \(50 \mathrm{lbf} / \mathrm{in} .^{2}\) and \(100^{\circ} \mathrm{F}\). The valve is opened and the gas expands to fill the total volume, eventually achieving an equilibrium state. Using the ideal gas model (a) determine the final temperature, in \({ }^{\circ} \mathrm{F}\), and final pressure, in lbf/in. \({ }^{2}\) (b) evaluate the exergy destruction, in Btu. (c) What is the cause of exergy destruction in this case? Let \(T_{0}=70^{\circ} \mathrm{F}, p_{0}=1 \mathrm{~atm}\).

Saturated liquid water at 0.01 MPa enters a power plant pump operating at a steady state. Liquid water exits the pump at \(10 \mathrm{MPa}\). The isentropic pump efficiency is \(90 \%\). Property data are provided in the accompanying table. Stray heat transfer and the effects of motion and gravity are negligible. Let \(T_{0}=300 \mathrm{~K}, p_{0}=100 \mathrm{kPa}\). Determine (a) the power required by the pump and the rate of exergy destruction, each in \(\mathrm{kJ}\) per kg of water flowing, and (b) the exergetic pump efficiency. \begin{tabular}{lccc} \multicolumn{1}{c}{ State } & \(p(\mathrm{MPa})\) & \(h(\mathrm{~kJ} / \mathrm{kg})\) & \(s(\mathrm{k}] / \mathrm{kg} \cdot \mathrm{K})\) \\ \hline Pump inlet & \(0.01\) & \(191.8\) & \(0.6493\) \\ Pump exit & 10 & \(204.5\) & \(0.6531\) \end{tabular}
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