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Chapter 7: Problem 3

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.

Short Answer

Expert verified
(a) \( \text{e}\textsubscript{x} = -T_{\text{0}} R \text{ln}\frac{p}{p_{\text{0}}} \). (b) \( \text{e}\textsubscript{x} = c_{p} (T - T_{\text{0}}) - T_{\text{0}} \text{ln}\frac{T}{T_{\text{0}}} \)

Step by step solution

01

- Define Exergy

Exergy is the measure of the maximum useful work possible during a process that brings the system into equilibrium with a heat reservoir. For an ideal gas, the specific exergy, denoted as \(\text{e}\textsubscript{x}\), can be derived from the first and second laws of thermodynamics.
02

Part (a): Given that Temperature \(T = T\textsubscript{0}\)

For an ideal gas at constant temperature \(T = T_{\text{0}}\), the change in exergy \(d \text{e}\textsubscript{x}\) in terms of pressure and temperature is defined as: \[\text{e}\textsubscript{x} = c_{\text{v}} (T - T_{\text{0}}) - T_{\text{0}} R \text{ln}\frac{p}{p_{\text{0}}} \] Since \(T = T_{\text{0}}\) for this part, \(c_{\text{v}} (T - T_{\text{0}})\) becomes zero.
03

Simplify Exergy for Part (a)

The exergy expression simplifies to: \[\text{e}\textsubscript{x} = -T_{\text{0}} R \text{ln}\frac{p}{p_{\text{0}}}\textsubscript{0} \]
04

Part (b): Given that Pressure \(p = p_{0}\)

For an ideal gas at constant pressure \(p = p_{\text{0}}\), the change in exergy \(d \text{e}\textsubscript{x}\) is given by: \[\text{e}\textsubscript{x} = c_{\text{p}} (T - T_{\text{0}}) - T_{\text{0}} (s - s_{\text{0}}) \], where \(s\) and \(s_{\text{0}}\) are the specific entropy values.
05

Relation for Entropy Change

The specific entropy change \(s - s_{\text{0}}\) can be expressed as: \[\frac{R}{c_{\text{p}}} \text{ln}\frac{T}{T_{\text{0}}} \] for constant pressure.
06

Substitute and Simplify for Part (b)

Replacing the entropy change, the expression becomes: \[\text{e}\textsubscript{x} = c_{\text{p}} (T - T_{\text{0}}) - T_{\text{0}} \text{ln}\frac{T}{T_{\text{0}}}\textsubscript{0} \]

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Key Concepts

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

ideal_gas
An ideal gas is a theoretical gas composed of a set of randomly moving, non-interacting point particles. The behavior of an ideal gas is well described by the Ideal Gas Law: \[ PV = nRT \], where:
  • \( P \) is the pressure of the gas
  • \( V \) is the volume
  • \( n \) is the amount of gas in moles
  • \( R \) is the gas constant
  • \( T \) is the absolute temperature
For an ideal gas, this law relates pressure, volume, and temperature in a linear fashion, allowing us to predict how changes in one property will affect the others. This idealization is useful in thermodynamics for simplifying calculations and understanding fundamental principles.
first_law_of_thermodynamics
The first law of thermodynamics, also known as the Law of Energy Conservation, states that energy cannot be created or destroyed in an isolated system. The energy of a system can only be transferred or transformed. This law can be expressed as: \[ \frac{dQ}{dt} = \frac{dU}{dt} + \frac{dW}{dt} \], where:
  • \( dQ \) is the heat added to the system
  • \( dU \) is the change in internal energy
  • \( dW \) is the work done by the system
For an ideal gas, internal energy is a function of temperature alone. Therefore, any change in internal energy directly corresponds to a change in temperature. This concept is crucial for deriving the specific exergy in the original exercise.
second_law_of_thermodynamics
The second law of thermodynamics states that the entropy of an isolated system always increases over time. Entropy is a measure of the disorder or randomness in a system. This law introduces the concept of exergy, which is the maximum useful work obtainable from a system as it reaches equilibrium with its surroundings. For any reversible process, the change in entropy can be determined by the equation: \[ dS = \frac{dQ_{rev}}{T} \], where \( dQ_{rev} \) is the reversible heat exchange and \( T \) is the absolute temperature. This law is crucial for calculating the specific exergy, as it enables us to quantify the irreversibilities and the available energy for work.
specific_heat
Specific heat is the amount of heat per unit mass required to raise the temperature by one degree Celsius. For an ideal gas, there are two specific heats:
  • Specific heat at constant volume \( c_v \)
  • Specific heat at constant pressure \( c_p \)
The relationship between these is given by: \[ c_p - c_v = R \], where \( R \) is the gas constant. In the context of the original exercise, the specific heat at constant pressure \( c_p \) is used to derive the specific exergy when the pressure is constant. Knowing the specific heat values helps us understand how much energy is needed to change the temperature of the gas under different conditions.
entropy_change
Entropy change is a measure of the amount of energy that is not available for work during a thermodynamic process. For an ideal gas, the change in entropy can be expressed as:
  • For constant volume: \( \bigtriangleup S = c_v \text{ln}\frac{T_2}{T_1} \)
  • For constant pressure: \( \bigtriangleup S = c_p \text{ln}\frac{T_2}{T_1} \)
The entropy change is crucial in calculating specific exergy, as seen in Part (b) of the given exercise. A higher entropy change indicates more irreversibility and less available energy for work. Understanding how to compute entropy change is essential for determining the efficiency of thermodynamic processes.

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Most popular questions from this chapter

Determine the specific exergy, in Btu, of one pound mass of (a) saturated liquid Refrigerant \(134 a\) at \(-5^{\circ} \mathrm{F}\) - (b) saturated vapor Refrigerant \(134 a\) at \(140^{\circ} \mathrm{F}\). (c) Refrigerant \(134 \mathrm{a}\) at \(60^{\circ} \mathrm{F}, 20 \mathrm{lbf} / \mathrm{in} .^{2}\) (d) Refrigerant \(134 \mathrm{a}\) at \(60^{\circ} \mathrm{F}, 10 \mathrm{lbf} / \mathrm{in}^{2}{ }^{2}\) In each case, consider a fixed mass at rest and zero elevation relative to an exergy reference environment for which \(T_{0}=\) \(60^{\circ} \mathrm{F}, p_{0}=15 \mathrm{lbt} / \mathrm{in}^{2}\)

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 \(T_{0}=70^{\circ} \mathrm{F}\).

A stream of hot water at \(300^{\circ} \mathrm{F}, 500 \mathrm{lbf} / \mathrm{in}^{2}\), and a velocity of \(20 \mathrm{ft} / \mathrm{s}\) is obtained from a geothermal supply. Determine the specific flow exergy, in Btu/lb. The velocity is relative to the exergy reference environment for which \(T_{0}=77^{\circ} \mathrm{F}, p_{0}=\) \(1 \mathrm{~atm}\). Neglect the effect of gravity.

Steam enters an insulated turbine operating at steady state at \(120 \mathrm{lbf} / \mathrm{in}^{2}, 600^{\circ} \mathrm{F}\), with a mass flow rate of \(3 \times 10^{5}\) \(\mathrm{lb} / \mathrm{h}\) and expands to a pressure of \(10 \mathrm{lbf} / \mathrm{in}^{2}\). The isentropic turbine efficiency is \(80 \%\). If exergy is valued at 8 cents per \(\mathrm{kW} \cdot \mathrm{h}\), determine (a) the value of the power produced, in \(\$ / \mathrm{h}\). (b) the cost of the exergy destroyed, in \(\$ / h\). (c) Plot the values of the power produced and the exergy destroyed, each in \(\$ / h\), versus isentropic efficiency ranging from 80 to \(100 \%\). Ignore the effects of motion and gravity. Let \(T_{0}=70^{\circ} \mathrm{F}\), \(p_{0}=1 \mathrm{~atm} .\)

Air enters a counterflow heat exchanger operating at steady state at \(27^{\circ} \mathrm{C}, 0.3 \mathrm{MPa}\) and exits at \(12^{\circ} \mathrm{C}\). Refrigerant 134 a enters at \(0.4 \mathrm{MPa}\), a quality of \(0.3\), and a mass flow rate of \(35 \mathrm{~kg} / \mathrm{h}\). Refrigerant exits at \(10^{\circ} \mathrm{C}\). Stray heat transfer is negligible and there is no significant change in pressure for either stream. (a) For the Refrigerant 134a stream, determine the rate of heat transfer, in \(\mathrm{kJ} / \mathrm{h}\). (b) For each of the streams, evaluate the change in flow exergy rate, in \(\mathrm{kJ} / \mathrm{h}\), and interpret its value and sign. Let \(T_{0}=22^{\circ} \mathrm{C}, p_{0}=0.1 \mathrm{MPa}\), and ignore the effects of motion and gravity.
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