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A piston-cylinder assembly fitted with a slowly rotating paddle wheel contains \(0.13 \mathrm{~kg}\) of air, initially at \(300 \mathrm{~K}\). The air undergoes a constant-pressure process to a final temperature of \(400 \mathrm{~K}\). During the process, energy is gradually transferred to the air by heat transfer in the amount \(12 \mathrm{~kJ}\). Assuming the ideal gas model with \(k=1.4\) and negligible changes in kinetic and potential energy for the air, determine the work done (a) by the paddle wheel on the air and (b) by the air to displace the piston, each in kJ.

Short Answer

Expert verified
Work done by the paddle wheel: 2.6725 kJ.Work done by the air to displace the piston: 3.731 kJ.

Step by step solution

01

Identify given data and assumptions

Given: - Mass of air, m = 0.13 kg- Initial temperature, T1 = 300 K- Final temperature, T2 = 400 K- Heat transfer to the air, Q = 12 kJ- Specific heat ratio for air, k = 1.4It is also given that the ideal gas model is assumed, and changes in kinetic and potential energy are negligible.
02

Write down relevant equations

For an ideal gas undergoing a constant pressure process, apply the first law of thermodynamics, expressed as:\[ Q = \text{Δ}U + W \ \text{where,} \ Q \text{ is heat transfer, } \ \text{Δ}U \text{ is change in internal energy, } \ W \text{ is work done by the gas.} \]The change in internal energy can be written as:\[ \text{Δ}U = m c_v (T_2 - T_1) \]where \(c_v\) is the specific heat at constant volume. For air and assuming ideal gas, \( c_v = \frac{R}{k-1} \), where \( R = 0.287 \text{ kJ/kg.K} \). For constant-pressure process work (\( W_{piston} \)) is given by:\[ W_{piston} = m R (T_2 - T_1) \]
03

Calculate specific heat at constant volume, \( c_v \)

Using the relation \( c_v = \frac{R}{k-1} \) and given \( R = 0.287 \text{ kJ/kg.K} \) and \( k = 1.4 \):\[ c_v = \frac{0.287}{1.4 - 1} = \frac{0.287}{0.4} = 0.7175 \text{ kJ/kg.K} \]
04

Calculate change in internal energy, \( \text{Δ}U \)

Using \( \text{Δ}U = m c_v (T_2 - T_1) \):\[ \text{Δ}U = 0.13 \text{ kg} × 0.7175 \text{ kJ/kg.K} × (400 \text{ K} - 300 \text{ K}) = 0.13 × 0.7175 × 100 = 9.3275 \text{ kJ} \]
05

Calculate work done by the air to displace the piston

Using the expression for work done by the air in a constant-pressure process:\[ W_{piston} = m R (T_2 - T_1) \]Substitute given values:\[ W_{piston} = 0.13 \text{ kg} × 0.287 \text{ kJ/kg.K} × (400 \text{ K} - 300 \text{ K}) = 0.13 × 0.287 × 100 = 3.731 \text{ kJ} \]
06

Apply energy balance to find work done by the paddle wheel

Using the first law of thermodynamics described in step 2:\[ Q = \text{Δ}U + W \rightarrow 12 \text{ kJ} = 9.3275 \text{ kJ} + W_{paddle} \]Solve for \( W_{paddle} \):\[ W_{paddle} = 12 \text{ kJ} - 9.3275 \text{ kJ} = 2.6725 \text{ kJ} \]

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Key Concepts

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

First Law of Thermodynamics
The First Law of Thermodynamics is crucial for understanding energy interactions in thermodynamic systems. It establishes the conservation of energy principle. For a closed system, the law can be expressed as: \[ Q = \text{Δ}U + W \] Here, \( Q \) represents the heat added to the system, \( \text{Δ}U \) is the change in internal energy, and \( W \) denotes the work done by the system. This principle ensures that energy input into a system (\( Q \)) is balanced by the change in the system's internal energy (\( \text{Δ}U \)) and the work it performs (\( W \)). Understanding this balance is key to solving thermodynamics problems.

Constant-Pressure Process
A constant-pressure process occurs when the pressure within the system stays the same throughout the process. In such scenarios, evaluating work becomes straightforward. The work done by a gas during a constant-pressure process can be determined using the formula: \[ W_{piston} = mR (T_2 - T_1) \] Here, \( W_{piston} \) is the work done by the gas to move the piston, \( m \) is the mass of the gas, \( R \) is the specific gas constant, and \( T_1 \) and \( T_2 \) are the initial and final temperatures, respectively. This relationship simplifies calculations by directly linking temperature change to the work done in maintaining constant pressure.

Change in Internal Energy
Internal energy is a measure of the energy contained within a system, resulting from the kinetic and potential energies of its molecules. For ideal gases, the change in internal energy (\( \text{Δ}U \)) depends only on the temperature change and can be computed using: \[ \text{Δ}U = mc_v (T_2 - T_1) \] In this equation, \( m \) is the mass of the gas, \( c_v \) is the specific heat capacity at constant volume, and \( T_1 \) and \( T_2 \) are the temperatures before and after the process. This formula shows that \( \text{Δ}U \) is proportional to the mass of the gas and the temperature change, given a specific heat capacity.

Specific Heat
Specific heat is an important material property indicating the amount of heat needed to change the temperature of a unit mass of a substance by one degree. For gases, there are two specific heats:
  • Specific heat at constant volume (\( c_v \))
  • Specific heat at constant pressure (\( c_p \))
In ideal gas scenarios, these are related to the gas constant \( R \) and the specific heat ratio (\( k \)) through the formulas: \[ c_v = \frac{R}{k-1} \] and \[ c_p = \frac{kR}{k-1} \] For air, with \( R = 0.287 \text{ kJ/kg.K} \) and \( k = 1.4 \), the values are determined accordingly. Specific heat values are essential in thermodynamic calculations to understand how energy transfer affects temperature changes under different conditions.

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Most popular questions from this chapter

One lb of oxygen, \(\mathrm{O}_{2}\), undergoes a power cycle consisting of the following processes: Process 1-2: Constant-volume heating from \(p_{1}=20 \mathrm{lbf} / \mathrm{in}^{2}{ }^{2}\), \(T_{1}=500^{\circ} \mathrm{R}\) to \(T_{2}=820^{\circ} \mathrm{R}\). Process 2-3: Adiabatic expansion to \(v_{3}=1.432 v_{2}\). Process 3-1: Constant-pressure compression to state \(1 .\) Sketch the cycle on a \(p-v\) diagram. Assuming ideal gas behavior, determine (a) the pressure at state 2 , in lbf/in. \({ }^{2}\) (b) the temperature at state 3 , in \({ }^{\circ} \mathrm{R}\). (c) the heat transfer and work, each in Btu, for all processes. (d) the thermal efficiency of the cycle.

A closed, rigid tank fitted with a paddle wheel contains \(0.1 \mathrm{~kg}\) of air, initially at \(300 \mathrm{~K}, 0.1 \mathrm{MPa}\). The paddle wheel stirs the air for 20 minutes, with the power input varying with time according to \(\dot{W}=-10 t\), where \(\dot{W}\) is in watts and \(t\) is time, in minutes. The final temperature of the air is \(1060 \mathrm{~K}\). Assuming ideal gas behavior and no change in kinetic or potential energy, determine for the air (a) the final pressure, in MPa, (b) the work, in kJ, and (c) the heat transfer, in kJ.

Ammonia in a piston-cylinder assembly undergoes two processes in series. Initially, the ammonia is saturated vapor at \(p_{1}=100 \mathrm{lbf} / \mathrm{in} .^{2}\) Process \(1-2\) involves cooling at constant pressure until \(x_{2}=75 \%\). The second process, from state 2 to state 3 , involves heating at constant volume until \(x_{3}=100 \%\). Kinetic and potential energy effects are negligible. For \(1.2 \mathrm{lb}\) of ammonia, determine (a) the heat transfer and work for Process 1-2 and (b) the heat transfer for Process 2-3, all in Btu.

Five \(\mathrm{kg}\) of butane \(\left(\mathrm{C}_{4} \mathrm{H}_{10}\right)\) in a piston-cylinder assembly undergoes a process from \(p_{1}=5 \mathrm{MPa}, T_{1}=500 \mathrm{~K}\) to \(p_{2}=3 \mathrm{MPa}\), during which the relationship between pressure and specific volume is \(p v=\) constant. Determine the work, in kJ.

Check the applicability of the ideal gas model for (a) water at \(600^{\circ} \mathrm{F}\) and pressures of \(900 \mathrm{lbf} / \mathrm{in} .{ }^{2}\) and 100 lbf/in. (b) nitrogen at \(-20^{\circ} \mathrm{C}\) and pressures of 75 bar and 1 bar.

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