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One kg of air contained in a piston-cylinder assembly undergoes a process from an initial state where \(T_{1}=300 \mathrm{~K}\), \(v_{1}=0.8 \mathrm{~m}^{3} / \mathrm{kg}\) to a final state where \(T_{2}=420 \mathrm{~K}, v_{2}=\) \(0.2 \mathrm{~m}^{3} / \mathrm{kg}\). Can this process occur adiabatically? If yes, determine the work, in \(\mathrm{kJ}\), for an adiabatic process between these states. If no, determine the direction of the heat transfer. Assume the ideal gas model for air.

Short Answer

Expert verified
The process is not adiabatic; hence, heat is transferred into the system.

Step by step solution

01

Identify the Known Values

Given values are: - Initial temperature, \( T_1 = 300 \text{ K} \)- Initial specific volume, \( v_1 = 0.8 \text{ m}^3/\text{kg} \)- Final temperature, \( T_2 = 420 \text{ K} \)- Final specific volume, \( v_2 = 0.2 \text{ m}^3/\text{kg} \)- Mass of air, \( m = 1 \text{ kg} \)
02

Use the Ideal Gas Law

The ideal gas law is given by \( PV = nRT \). For a specific state, we can express it as \( P_1 v_1 = R T_1 \) and \( P_2 v_2 = R T_2 \). Rearranged, we get:\[ P_1 = \frac{R T_1}{v_1} \]\[ P_2 = \frac{R T_2}{v_2} \]
03

Calculate Initial and Final Pressures

Using the ideal gas law equations derived in Step 2:\[ P_1 = \frac{R \cdot 300}{0.8} = \frac{300R}{0.8} \]\[ P_2 = \frac{R \cdot 420}{0.2} = \frac{420R}{0.2} \]
04

Verify if Process Can be Adiabatic

If the process is adiabatic for an ideal gas, it follows \(P V^{\gamma} = \text{const} \). Therefore, we need to check if:\[ P_1 v_1^{\gamma} = P_2 v_2^{\gamma} \]Use the values of \( P_1 \) and \( P_2 \) from Step 3, and the adiabatic index for air, \( \gamma = 1.4 \) to check this.
05

Evaluate if the Process is Adiabatic

Substitute the values into the adiabatic relationship:\[ \frac{300R}{0.8} \cdot (0.8)^{1.4} \stackrel{?}{=} \frac{420R}{0.2} \cdot (0.2)^{1.4} \]Simplify and solve to verify equality:\[ (300) \cdot (0.8)^{0.4} \stackrel{?}{=} (420) \cdot (0.2)^{0.4} \]Solve each side separately to confirm if they are equal.
06

Conclusion on Adiabatic Process

If the values are equal, the process is adiabatic, and we can use the work equation for an adiabatic process. If they are not equal, the process is not adiabatic, and we determine the direction of heat transfer based on temperatures.
07

Calculate Work for Adiabatic Process

If the process is adiabatic, the work done (W) can be calculated using the following equation:\[ W = C_v (T_1 - T_2) \]Where \( C_v \) for air is approximately 0.718 kJ/(kg·K).
08

Solve for Work

Substitute the given values:\[ W = 0.718 \cdot (300 - 420) \]\[ W = -86.16 \text{kJ} \]
09

Conclusion if Process is not Adiabatic

If the earlier calculations show the process is not adiabatic, heat must be transferred into the system since temperature increases from T1 to T2.

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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 that describes the behavior of an ideal gas. It is usually written as \( PV = nRT \) where:
  • \( P \) is the pressure of the gas
  • \( V \) is the volume it occupies
  • \( n \) is the number of moles of the gas
  • \( R \) is the universal gas constant
  • \( T \) is the temperature in Kelvin
In contexts involving specific volume \( (v) \) and mass \( (m) \), the Ideal Gas Law can be rearranged to: \( P = \frac{R T}{v} \) where:
  • \( P \) is the pressure of the gas
  • \( R \) is the specific gas constant for the gas
  • \( T \) is the temperature in Kelvin
  • \( v \) is the specific volume, which is volume per unit mass
For air, you can use this equation to calculate pressure at different states, assuming the gas behaves ideally.
specific volume
Specific volume \( v \) is an important thermodynamic property that represents the volume occupied by a unit mass of a substance, in this case, air. It is given by the ratio: \( v = \frac{V}{m} \) where:
  • \( V \) is the total volume
  • \( m \) is the mass
In the provided problem, the specific volumes are given at the initial and final states:
  • Initial specific volume, \( v_1 = 0.8 \text{ m}^3/\text{kg} \)
  • Final specific volume, \( v_2 = 0.2 \text{ m}^3/\text{kg} \)
Specific volume helps us understand how 'spread out' the air molecules are at different states, and it's crucial for calculations involving the Ideal Gas Law and determining if a process is adiabatic.
pressure calculation
Calculating pressure accurately in thermodynamic processes is crucial for further analysis. Using the Ideal Gas Law in its form \( P = \frac{RT}{v} \), we can determine the pressure at different states by substituting specific volume and temperature values given in the problem: For the initial state: \( P_1 = \frac{RT_1}{v_1} = \frac{R \times 300}{0.8} \) For the final state: \( P_2 = \frac{RT_2}{v_2} = \frac{R \times 420}{0.2} \) We have:
  • \( T_1 = 300 \text{ K} \)
  • \( v_1 = 0.8 \text{ m}^3/\text{kg} \)
**Result**: \( P_1 = \frac{300R}{0.8} \) Similarly, for the final state:
  • \( T_2 = 420 \text{ K} \)
  • \( v_2 = 0.2 \text{ m}^3/\text{kg} \)
**Result**: \( P_2 = \frac{420R}{0.2} \) These calculations are used to check if the process abides by the adiabatic condition.
adiabatic index
The adiabatic index (\( \gamma \) ) is a specific ratio for a gas, defined as the ratio of the specific heats: \( \gamma = \frac{C_p}{C_v} \) where:
  • \( C_p \) is the specific heat at constant pressure
  • \( C_v \) is the specific heat at constant volume
For air, the adiabatic index is usually taken as \( 1.4 \). In an adiabatic process, where no heat is transferred into or out of the system, the relationship between pressure and volume is given by: \( P_1 v_1^{\gamma} = P_2 v_2^{\gamma} \) To verify if the process is adiabatic, you substitute the initial and final pressures and specific volumes: \( \frac{300R}{0.8} \cdot (0.8)^{1.4} \stackrel{?}{=} \frac{420R}{0.2} \cdot (0.2)^{1.4} \) Solving each side separately determines if they are equal, confirming whether the process can be considered adiabatic. If they are equal, the process is adiabatic, and work done can be calculated using: \( W = C_v (T_1 - T_2) \). If not, the process involves heat transfer.

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Most popular questions from this chapter

Steam enters a turbine operating at steady state at \(6 \mathrm{MPa}, 600^{\circ} \mathrm{C}\) with a mass flow rate of \(125 \mathrm{~kg} / \mathrm{min}\) and exits as saturated vapor at \(20 \mathrm{kPa}\), producing power at a rate of 2 MW. Kinetic and potential energy effects can be ignored. Determine (a) the rate of heat transfer, in \(\mathrm{kW}\), for a control volume including the turbine and its contents, and (b) the rate of entropy production, in \(\mathrm{kW} / \mathrm{K}\), for an enlarged control volume that includes the turbine and enough of its surroundings that heat transfer occurs at the ambient temperature, \(27^{\circ} \mathrm{C}\).

Air enters the turbine of a jet engine at \(1190 \mathrm{~K}, 10.8\) bar and expands to \(5.2\) bar. The air then flows through a nozzle and exits at \(0.8\) bar. Operation is at steady state, and the flow is adiabatic. The nozzle operates with no internal irreversibilities, and the isentropic turbine efficiency is \(85 \%\). The air velocities at the turbine inlet and exit are negligible. Assuming the ideal gas model for the air, determine the velocity of the air exiting the nozzle, in \(\mathrm{m} / \mathrm{s}\).

Refrigerant 22 enters the heat exchanger of an airconditioning system at \(80 \mathrm{lbf} / \mathrm{in}^{2}\) with a quality of \(0.2\). The refrigerant stream exits at \(80 \mathrm{lbf} / \mathrm{in}^{2}, 60^{\circ} \mathrm{F}\). Air flows in counterflow through the heat exchanger, entering at \(14.9 \mathrm{lbf}\) in. \(^{2}, 80^{\circ} \mathrm{F}\), with a volumetric flow rate of \(100,000 \mathrm{ft}^{3} / \mathrm{min}\) and exiting at \(14.5 \mathrm{lbf} / \mathrm{in}^{2}, 65^{\circ} \mathrm{F}\). Operation is at steady state, stray heat transfer from the outside of the heat exchanger to the surroundings can be neglected, and kinetic and potential energy effects are negligible. Assuming ideal gas behavior for the air, determine the rate of entropy production in the heat exchanger, in Btu/min \({ }^{\circ}{ }^{\circ} \mathrm{R}\).

Ammonia enters the compressor of an industrial refrigeration plant at 2 bar, \(-10^{\circ} \mathrm{C}\) with a mass flow rate of \(15 \mathrm{~kg} / \mathrm{min}\) and is compressed to 12 bar, \(140^{\circ} \mathrm{C}\). Heat transfer occurs from the compressor to its surroundings at a rate of \(6 \mathrm{~kW}\). For steady-state operation with negligible kinetic and potential energy effects, determine (a) the power input to the compressor, in \(\mathrm{kW}\), and (b) the rate of entropy production, in \(\mathrm{kW} / \mathrm{K}\), for a control volume enclosing the compressor and its immediate surroundings such that the heat transfer occurs at \(300 \mathrm{~K}\).

A cylindrical copper rod of base area A and length \(L\) is insulated on its lateral surface. One end of the rod is in contact with a wall at temperature \(T_{\mathrm{H}}\). The other end is in contact with a wall at a lower temperature \(T_{\mathrm{C}}\). At steady state, the rate at which energy is conducted into the rod from the hot wall is $$ \dot{Q}_{\mathrm{H}}=\frac{\kappa \mathrm{A}\left(T_{\mathrm{H}}-T_{\mathrm{C}}\right)}{L} $$ where \(\kappa\) is the thermal conductivity of the copper rod. (a) For the rod as the system, obtain an expression for the time rate of entropy production in terms of \(\mathrm{A}, L, T_{\mathrm{H}}, T_{\mathrm{C}}\), and \(\kappa\). (b) If \(T_{\mathrm{H}}=327^{\circ} \mathrm{C}, T_{\mathrm{C}}=77^{\circ} \mathrm{C}, \kappa=0.4 \mathrm{~kW} / \mathrm{m} \cdot \mathrm{K}, \mathrm{A}=0.1 \mathrm{~m}^{2}\), plot the heat transfer rate \(\dot{Q}_{\mathrm{H}}\), in \(\mathrm{kW}\), and the time rate of entropy production, in \(\mathrm{kW} / \mathrm{K}\), each versus \(L\) ranging from \(0.01\) to \(1.0 \mathrm{~m}\). Discuss.

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