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A vessel contains \(1 \mathrm{lb}\) of air at pressure \(p\) and \(200^{\circ} \mathrm{F}\). Using the ideal gas model, plot the specific exergy of the air, in Btu/lb, for \(p\) ranging from \(0.5\) to 2 atm. The air is at rest and negligible elevation relative to an exergy reference environment for which \(T_{0}=60^{\circ} \mathrm{F}, p_{0}=1 \mathrm{~atm}\).

Short Answer

Expert verified
Convert temperatures to Rankine, calculate internal energy change, entropy change, specific volume change, and insert into exergy formula. Finally, plot specific exergy vs. pressure.

Step by step solution

01

Convert Temperatures to Rankine

Convert the air temperature and the reference temperature from Fahrenheit to Rankine. The formula to convert Fahrenheit to Rankine is \[ T_R = T_F + 459.67 \]Given temperatures:\[ T = 200^{\bullet}F \]\[ T_0 = 60^{\bullet}F \]So,\[ T_R = 200 + 459.67 = 659.67 \text{ R} \]\[ T_{0_R} = 60 + 459.67 = 519.67 \text{ R}\]
02

Define Specific Exergy Formula

The specific exergy of an ideal gas is given by the formula:\[ e = (u - u_0) + p_0 (v - v_0) - T_0 (s - s_0) \]Since air is ideal and specific heats (\( c_p \text{ and } c_v \)) are constant, we use the definition for the specific exergy per unit mass for calculations.
03

Calculate Internal Energy Change

Use the change in internal energy formula: \[ u - u_0 = c_v (T - T_0) \]We assume air, and typically, \[ c_v = 0.171 \frac{Btu}{lb \text{ R}} \]\[ u - u_0 = 0.171 (659.67 - 519.67) = 0.171 \times 140 = 23.94 \frac{Btu}{lb} \]
04

Calculate Entropy Change

Use the entropy change formula: \[ s - s_0 = c_p \text{ln} \left( \frac{T}{T_0} \right) - R \text{ln} \left( \frac{p}{p_0} \right) \]Where:\[ c_p = 0.240 \frac{Btu}{lb \text{ R}} \]\[ R = 0.06855 \frac{Btu}{lb \text{ R}} \]\[ \text{ln} \equiv \text{natural logarithm} \]This step generally involves computing specific values for usually more than one pressure value.
05

Calculate Specific Volume Change

Using the ideal gas relation: \[ v = \frac{RT}{p} \]\[ v_0 = \frac{RT_0}{p_0} \]Where the specific gas constant: \[ R = 0.06855 \frac{Btu}{lb \text{ R}} \]
06

Insert Values into Exergy Formula

Combine all computed expressions into the specific exergy formula:\[ e = (u - u_0) + p_0 (v - v_0) - T_0 (s - s_0) \]We assume calculations for various pressures would provide different exergy values to plot for range of \(0.5 \text{ atm through to } 2 \text{ atm} \)
07

Plot the Graph

Plot the specific exergy values against corresponding pressures to visualize the relationship. Use a software tool or graph paper for plotting.

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Key Concepts

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

Ideal Gas Law
The ideal gas law is a fundamental equation in thermodynamics. It relates the pressure, volume, and temperature of an ideal gas using the formula:
\[ PV = nRT \]
In this equation:
* \(P\) represents the pressure of the gas.
* \(V\) represents the volume of the gas.
* \(n\) is the number of moles of the gas.
* \(R\) is the universal gas constant.
* \(T\) is the absolute temperature of the gas.
For specific exergy calculations, we often deal with the specific volume (\(v\)). The specific volume is the volume per unit mass (\text{lb}). The ideal gas law can be adapted to include specific volume in the following way: \( v = \frac{RT}{P} \). For constant specific gas constant \(R = 0.06855 \frac{Btu}{lb \cdot R} \).
Understanding the ideal gas law allows us to interrelate changes in state properties such as pressure, volume, and temperature, key knowledge for calculating exergy.
Internal Energy Change
Internal energy change is a crucial concept in thermodynamics, particularly for specific exergy calculations. For an ideal gas, the change in internal energy \((u - u_0)\) is directly related to the change in temperature, represented by:
\[ u - u_0 = c_v (T - T_0) \]
Here:
* \(c_v\) is the specific heat at constant volume.
* \(T\) is the initial temperature and \(T_0\) is the reference temperature, both in absolute units (Rankine).
For air, the specific heat at constant volume \(c_v\) is typically \(0.171 \frac{Btu}{lb \cdot R} \). This direct relationship simplifies exergy calculations by allowing internal energy changes to be derived solely from temperature differences.
Entropy Change
The change in entropy \((s - s_0)\) in thermodynamic processes is significant for evaluating energy efficiency and exergy. For an ideal gas, the change in entropy comprises both temperature and pressure changes and is given by:
\[ s - s_0 = c_p \ln\left( \frac{T}{T_0} \right) - R \ln\left( \frac{P}{P_0} \right) \]
Where:
* \(c_p\) is the specific heat at constant pressure.
* \(R\) is the specific gas constant.
* \(ln\) denotes the natural logarithm.
In our example, \(c_p\) is \(0.240 \frac{Btu}{lb \cdot R} \) and \(R\) is \(0.06855 \frac{Btu}{lb \cdot R} \).
By calculating the entropy change for various pressures, we can better understand how pressure and temperature variations affect exergy.
Specific Volume
Specific volume \((v)\) is defined as the volume occupied by a unit mass of a substance, and it plays a significant role in the ideal gas law and specific exergy calculations. It can be derived from the ideal gas law using:
\[ v = \frac{RT}{P} \]
For consistency, the reference specific volume \(v_0\) at the reference state is:
\[ v_0 = \frac{RT_0}{P_0} \]
Where:
* \(T_0\) and \(P_0\) are the reference temperature and pressure respectively.
* \(R\) is the specific gas constant, and \(T\) and \(P\) are the temperature and pressure of the gas.
The specific volume helps to complete the exergy balance as it translates pressure, temperature, and volume changes into energy terms.

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Most popular questions from this chapter

Refrigerant \(134 a\) as saturated vapor at \(-10^{\circ} \mathrm{C}\) enters a compressor operating at steady state with a mass flow rate of \(0.3 \mathrm{~kg} / \mathrm{s}\). At the compressor exit the pressure of the refrigerant is 5 bar. Stray heat transfer and the effects of motion and gravity can be ignored. If the rate of exergy destruction within the compressor must be kept less than \(2.4 \mathrm{~kW}\), determine the allowed ranges for (a) the power required by the compressor, in \(k W\), and (b) the exergetic compressor efficiency. Let \(T_{0}=298 \mathrm{~K}, p_{0}=1\) bar.

An ideal gas is stored in a closed vessel at pressure \(p\) and temperature \(T\). (a) If \(T=T_{0}\), derive an expression for the specific exergy in terms of \(p, p_{0}, T_{0}\), and the gas constant \(R\). (b) If \(p=p_{0}\), derive an expression for the specific exergy in terms of \(T, T_{0}\), and the specific heat \(c_{p}\), which can be taken as constant. Ignore the effects of motion and gravity.

Consider \(100 \mathrm{~kg}\) of steam initially at 20 bar and \(240^{\circ} \mathrm{C}\) as the system. Determine the change in exergy, in \(\mathrm{kJ}\), for each of the following processes: (a) The system is heated at constant pressure until its volume doubles. (b) The system expands isothermally until its volume doubles. Let \(T_{0}=20^{\circ} \mathrm{C}, p_{0}=1\) bar and ignore the effects of motion and gravity.

One lbmol of carbon monoxide gas is contained in a \(90-\mathrm{ft}^{3}\) rigid, insulated vessel initially at 5 atm. An electric resistor of negligible mass transfers energy to the gas at a constant rate of \(10 \mathrm{Btu} / \mathrm{s}\) for \(2 \mathrm{~min}\). Employing the ideal gas model and ignoring the effects of motion and gravity, determine (a) the change in exergy of the gas, (b) the electrical work, and (c) the exergy destruction, each in Btu. Let \(T_{0}=70^{\circ} \mathrm{F}\), \(p_{0}=1 \mathrm{~atm}\).

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

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