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Nitrogen at \(200^{\circ} \mathrm{C}, 300 \mathrm{kPa}\) is in a piston/cylinder, volume \(5 \mathrm{~L}\), with the piston locked with a pin. The forces on the piston require a pressure inside of \(200 \mathrm{kPa}\) to balance it without the pin. The pin is removed, and the piston quickly comes to its equilibrium position without any heat transfer. Find the final \(P, T\) and the entropy generation due to this partly unrestrained expansion.

Short Answer

Expert verified
Final pressure is 200 kPa, temperature can be found from the adiabatic relation, and entropy generation is nonzero due to irreversibility.

Step by step solution

01

Understand the Initial Condition

The initial condition of the nitrogen gas is given by the following parameters:- Temperature: \(200^{\circ} C\),- Pressure: \(300 \text{ kPa}\),- Volume: \(5 \text{ L}\).With these values, we will use the ideal gas law and tables if needed to obtain properties such as specific volume and specific internal energy.
02

Final Condition After Expansion

When the pin is removed, and the piston reaches a new equilibrium without heat transfer (an adiabatic process), the final pressure \(P_f\) is given as \(200 \text{ kPa}\). For an adiabatic process, the relationship \(P V^\gamma = \text{constant}\) holds, where \(\gamma\) is the ratio of specific heats. Find the new volume if possible using ideal gas tables or equations.
03

Calculate Final Temperature

Assuming that the specific heats do not change significantly, we can use the relation for adiabatic processes to find the final temperature. For ideal gases:\[ T_f = T_i \left( \frac{P_f}{P_i} \right)^{\frac{\gamma-1}{\gamma}} \]Calculate \(T_f\) using the given \(P_f = 200 \text{ kPa}\) and \(P_i = 300 \text{ kPa}\). Set \(\gamma \approx 1.4\) for nitrogen.
04

Calculate Entropy Generation

Entropy change for an adiabatic process (without heat transfer) should ideally be zero. However, due to irreversibilities, there will be entropy generation. Using the relation:\[ \Delta S_{gen} = \Delta S - \frac{Q_{rev}}{T} = \Delta S \]Since \(\Delta S\ge 0\), calculate using the initial and final states of nitrogen.

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Key Concepts

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

Adiabatic Process
An adiabatic process is a thermodynamic process where no heat is transferred into or out of the system. Because there is no exchange of heat, any change in the internal energy of the system must come from work done on or by the system. When a gas undergoes an adiabatic process in a piston-cylinder setup, such as the situation in the problem, the process is characterized by changes in pressure, volume, and temperature without heat transfer. Some key attributes of adiabatic processes include:
  • The total heat exchange (\( Q \)) is zero.
  • Work done (\( W \)) is solely due to changes in the system.
  • They can be further categorized as reversible or irreversible, impacting the entropy.
In the context of the exercise, when the pin in the cylinder is removed, the gas expands without heat exchange, making the process adiabatic. Understanding this helps us determine that any observed changes are due to volume and pressure alterations, not external factors like heat transfer.
Ideal Gas Law
The Ideal Gas Law is a fundamental principle connecting the pressure, volume, temperature, and amount of an ideal gas. The law is expressed as \( PV = nRT \), where:
  • \( P \) represents the pressure of the gas.
  • \( V \) is the volume of the gas.
  • \( n \) is the number of moles of the gas.
  • \( R \) is the universal gas constant.
  • \( T \) is the temperature of the gas in Kelvin.
In practice, the Ideal Gas Law is used when dealing with gases under conditions where they behave ideally, meaning intermolecular forces and molecular sizes are negligible. In our exercise, applying the Ideal Gas Law initially helps to establish baseline conditions before the gas undergoes expansion. By knowing the initial state parameters such as temperature, volume, and pressure, we can predict how other variables will shift during operations like the removal of the locking pin.
Specific Heat Ratio
The Specific Heat Ratio, denoted as \( \gamma \), is the ratio of specific heats at constant pressure (\( C_p \)) to that at constant volume (\( C_v \)):\[ \gamma = \frac{C_p}{C_v} \]This ratio is important in analyzing and understanding adiabatic processes because it defines how gases behave when they are compressed or expanded without heat exchange.
  • \( \gamma > 1 \) for all gases.
  • Typical values are 1.4 for diatomic gases like air and nitrogen.
  • It influences the relationship between pressure and volume in adiabatic processes \( (P V^\gamma = \text{constant}) \).
In the given exercise, the specific heat ratio is significant because it aids in calculating the final temperature following the adiabatic expansion. Knowing \( \gamma \) allows us to apply formulas that describe the behavior of gases as they undergo such expansions. This factor ultimately helps in determining the change in the state of the gas post-expansion.
Unrestrained Expansion
Unrestrained expansion involves a sudden change in volume without any initial constraint, often leading to irreversible processes. When considering unrestrained expansion, entropy generation is a key aspect to examine, as it provides insight into irreversible losses during the process. This type of expansion typically implies:
  • Rapid changes where the system doesn’t stay in equilibrium.
  • Significant entropy generation due to non-reversible work.
In the context of the exercise, when the pin is removed from the piston, the gas undergoes unrestrained expansion. As a result, there is likely an increase in entropy, due to the irreversible nature of the expansion. This understanding allows for calculations of entropy generation, which can be crucial for evaluating system efficiency and identifying areas to reduce energy losses.

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Most popular questions from this chapter

A farmer runs a heat pump using \(2.5 \mathrm{hp}\) of power input. It keeps a chicken hatchery at a constant \(86 \mathrm{~F}\) while the room loses \(20 \mathrm{Btu} / \mathrm{s}\) to the colder outside ambient air at \(50 \mathrm{~F}\). What is the rate of entropy generated in the heat pump? What is the rate of entropy generated in the heat loss process?

One kilogram of ammonia \(\left(\mathrm{NH}_{3}\right)\) is contained in a \(1 \mathrm{~kg}\) steel spring-loaded piston/cylinder, similar to Fig. P6.82, as saturated liquid at \(-20^{\circ} \mathrm{C}\). Heat is added from a reservoir at \(100^{\circ} \mathrm{C}\) until a final condition of \(800 \mathrm{kPa}, 70^{\circ} \mathrm{C}\) is reached. Find the work, heat transfer, and entropy generation, assuming the process is internally reversible.

A piston/cylinder contains air at \(300 \mathrm{~K}, 100 \mathrm{kPa}\). A reversible polytropic process with \(n=1.3\) brings the air to \(500 \mathrm{~K}\). Any heat transfer if it comes in is from a \(325^{\circ} \mathrm{C}\) reservoir, and if it goes out it is to the ambient at \(300 \mathrm{~K}\). Sketch the process in a \(P-v\) and a \(T-s\) diagram. Find the specific work and specific heat transfer in the process. Find the specific entropy generation (external to the air) in the process

In a sink, \(5 \mathrm{~L}\) of water at \(70^{\circ} \mathrm{C}\) is combined with \(1 \mathrm{~kg}\) of aluminum pots, \(1 \mathrm{~kg}\) of flatware (steel), and \(1 \mathrm{~kg}\) of glass, all put in at \(20^{\circ} \mathrm{C}\). What is the final uniform temperature and change in stored entropy, neglecting any heat loss and work?

A piston/cylinder contains \(1 \mathrm{lbm}\) water at 40 psia, \(600 \mathrm{~F}\), and it now cools to \(280 \mathrm{~F}\) in an isobaric process. The heat goes into a heat engine that rejects heat to the ambient at \(77 \mathrm{~F}\), shown in Fig. P6.26, and the whole process is assumed reversible. Find the heat transfer out of the water and the work given out by the heat engine.

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