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Chapter 2: Problem 77

A gas within a piston-cylinder assembly undergoes a thermodynamic cycle consisting of three processes in series, beginning at state 1 where \(p_{1}=1\) bar, \(V_{1}=1.5 \mathrm{~m}^{3}\), as follows: Process 1-2: Compression with \(p V=\) constant, \(W_{12}=-104 \mathrm{~kJ}\), \(U_{1}=512 \mathrm{~kJ}, U_{2}=690 \mathrm{~kJ} .\) Process 2-3: \(W_{23}=0, Q_{23}=-150 \mathrm{kJJ}\). Process 3-1: \(W_{31}=+50 \mathrm{~kJ}\). There are no changes in kinetic or potential energy. (a) Determine \(Q_{12}, Q_{31}\), and \(U_{3}\), each in kJ. (b) Can this cycle be a power cycle? Explain.

Short Answer

Expert verified
Solve for each step using the first law: \(Q_{12} = 282\) kJ, \(U_{3} = 540\) kJ, \(Q_{31} = -78\) kJ. The cycle cannot be a power cycle because the net work is negative.

Step by step solution

01

Understand the first law of thermodynamics

The first law of thermodynamics states that energy is conserved. For a closed system, this can be written as: \[ \text{螖U} = Q + W \]where \( \text{螖U} \) is the change in internal energy, \( Q \) is the heat added to the system, and \( W \) is the work done by the system.
02

Calculate \(Q_{12}\) for Process 1-2

Using the first law of thermodynamics applied to Process 1-2:\[ \text{螖U}_{12} = Q_{12} + W_{12} \]Given:\(U_{1} = 512 \, \text{kJ}\), \(U_{2} = 690 \, \text{kJ}\),\(W_{12} = -104 \, \text{kJ}\)Calculate the change in internal energy, \( \text{螖U}_{12} = U_{2} - U_{1} = 690 \, \text{kJ} - 512 \, \text{kJ} = 178 \, \text{kJ} \)Substitute into the first law equation:\[ 178 \, \text{kJ} = Q_{12} + (-104 \, \text{kJ}) \]\[ Q_{12} = 178 \, \text{kJ} + 104 \, \text{kJ} = 282 \, \text{kJ} \]
03

Determine \(U_{3}\) using Process 2-3

For Process 2-3, we know from the problem statement:\( W_{23} = 0 \, \text{kJ} \), \(Q_{23} = -150 \, \text{kJ} \)Using the first law of thermodynamics:\[ \text{螖U}_{23} = Q_{23} + W_{23} \]\[ U_{3} - U_{2} = -150 \, \text{kJ} + 0 \, \text{kJ} \]\[ \text{螖U}_{23} = -150 \, \text{kJ} \]Calculate \( U_{3} \):\[ U_{3} = U_{2} + \text{螖U}_{23} = 690 \, \text{kJ} - 150 \, \text{kJ} = 540 \, \text{kJ} \]
04

Calculate \(Q_{31}\) for Process 3-1

Using the first law of thermodynamics for Process 3-1:\[ \text{螖U}_{31} = Q_{31} + W_{31} \]Given:\(U_{1} = 512 \, \text{kJ}\), \(U_{3} = 540 \, \text{kJ}\),\(W_{31} = 50 \, \text{kJ}\)Calculate the change in internal energy, \( \text{螖U}_{31} = U_{1} - U_{3} = 512 \, \text{kJ} - 540 \, \text{kJ} = -28 \, \text{kJ} \)Substitute into the first law equation:\[ -28 \, \text{kJ} = Q_{31} + 50 \, \text{kJ} \]\[ Q_{31} = -28 \, \text{kJ} - 50 \, \text{kJ} = -78 \, \text{kJ} \]
05

Determine if this cycle can be a power cycle

A power cycle produces net work during a cycle. Calculate the net work for the entire cycle:\[ W_{\text{net}} = W_{12} + W_{23} + W_{31} \]Given:\(W_{12} = -104 \, \text{kJ}\), \(W_{23} = 0 \, \text{kJ}\), \(W_{31} = 50 \, \text{kJ}\)\[ W_{\text{net}} = -104 \, \text{kJ} + 0 \, \text{kJ} + 50 \, \text{kJ} = -54 \, \text{kJ} \]A negative net work means the cycle cannot be a power cycle as it does not produce net work output.

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Key Concepts

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

thermodynamic cycles
A thermodynamic cycle is a series of processes that a system undergoes, returning to its initial state at the end. These cycles are crucial for understanding how energy transfer and work interaction occur in systems like engines and refrigerators.
The cycle discussed in the exercise involves a gas within a piston-cylinder assembly undergoing three processes in series. By the end of these processes, the gas returns to its starting state, fulfilling the definition of a cyclic process.
An essential characteristic of thermodynamic cycles is that the system's internal energy change over one complete cycle is zero since it returns to its initial state.
internal energy
Internal energy (U) is the total energy contained within a system, stemming from the kinetic and potential energies of its molecules. It plays a vital role in understanding energy changes during different thermodynamic processes.
In the provided exercise, internal energy changes (螖U) are calculated for each process to understand the work and heat interactions. For example, during Process 1-2, the internal energy change (螖U_{12}) was found to be 178 kJ. Similarly, for Process 2-3, 螖U_{23} was calculated to be -150 kJ, and for Process 3-1, 螖U_{31} was -28 kJ.
heat transfer
Heat transfer (Q) is the energy transfer between a system and its surroundings due to a temperature difference. In thermodynamics, it is a critical concept as it directly affects the system's internal energy and work done.
Each process in the cycle has heat transfer associated with it. Using the first law of thermodynamics (螖U = Q + W), we calculated the heat transfer for each process. For Process 1-2, Q_{12} was found to be 282 kJ. For Process 2-3, with no work done (W_{23} = 0 kJ), Q_{23} was -150 kJ. For the final process 3-1, Q_{31} was -78 kJ.
work done
Work done (W) in a thermodynamic process is the energy transferred when a force is applied over a distance. It is another key factor, along with heat, that affects the internal energy of the system.
In this problem, we have work done values for each process: W_{12} = -104 kJ for Process 1-2, implying work was done on the system (compression). For Process 2-3, W_{23} = 0 kJ indicates that no work was done. For Process 3-1, W_{31} = 50 kJ shows work done by the system (expansion).
By adding all the work done values, we get the net work for the cycle: W_{net} = -54 kJ.
power cycles
Power cycles are thermodynamic cycles that are designed to produce net work output. This output can be used to generate electricity or perform mechanical work.
To determine whether the cycle from the exercise can be considered a power cycle, we calculated the net work output of the entire cycle: W_{net} = -54 kJ. Since the net work is negative, it indicates that the cycle does not produce net work output but consumes work instead. As a result, this cycle cannot be classified as a power cycle.

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Most popular questions from this chapter

An automobile having a mass of \(900 \mathrm{~kg}\) initially moves along a level highway at \(100 \mathrm{~km} / \mathrm{h}\) relative to the highway. It then climbs a hill whose crest is \(50 \mathrm{~m}\) above the level highway and parks at a rest area located there. For the automobile, determine its changes in kinetic and potential energy, each in kJ. For each quantity, kinetic energy and potential energy, specify your choice of datum and reference value at that datum. Let \(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\).

A gas is contained in a vertical piston-cylinder assembly by a piston with a face area of 40 in. \(^{2}\) and weight of \(100 \mathrm{lbf}\). The atmosphere exerts a pressure of \(14.7 \mathrm{lbf} / \mathrm{in}^{2}\) on top of the piston. A paddle wheel transfers 3 Btu of energy to the gas during a process in which the elevation of the piston increases by \(1 \mathrm{ft}\). The piston and cylinder are poor thermal conductors, and friction between them can be neglected. Determine the change in internal energy of the gas, in Btu.

A construction crane weighing \(12,000 \mathrm{lbf}\) fell from a height of \(400 \mathrm{ft}\) to the street below during a severe storm. For \(g=\) \(32.05 \mathrm{ft} / \mathrm{s}^{2}\), determine the mass, in \(\mathrm{lb}\), and the change in gravitational potential energy of the crane, in \(\mathrm{ft} \cdot \mathrm{lbf}\).

A car magazine article states that the power \(\dot{W}\) delivered by an automobile engine, in hp, is calculated by multiplying the torque \(\mathscr{T}\), in \(\mathrm{ft} \cdot \mathrm{lbf}\), by the rotational speed of the driveshaft \(\omega\), in RPM, and dividing by a constant: $$ \dot{W}=\frac{\mathscr{T}_{\omega}}{\mathrm{C}} $$ What is the value and units of the constant \(C\) ?

A gas contained within a piston-cylinder assembly undergoes two processes, \(A\) and \(B\), between the same end states, 1 and 2 , where \(p_{1}=10\) bar, \(V_{1}=0.1 \mathrm{~m}^{3}, U_{1}=400 \mathrm{~kJ}\) and \(p_{2}=1\) bar, \(V_{2}=1.0 \mathrm{~m}^{3}, U_{2}=200 \mathrm{~kJ}\) : Process A: Process from 1 to 2 during which the pressurevolume relation is \(p V=\) constant. Process B: Constant-volume process from state 1 to a pressure of 2 bar, followed by a linear pressure-volume process to state 2 . Kinetic and potential energy effects can be ignored. For each of the processes \(A\) and \(B\), (a) sketch the process on \(p-V\) coordinates, (b) evaluate the work, in kJ, and (c) evaluate the heat transfer, in kJ.
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