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Chapter 2: Problem 35

Air contained within a piston-cylinder assembly undergoes three processes in series: Process 1-2: Compression during which the pressure-volume relationship is \(p V=\) constant from \(p_{1}=10 \mathrm{lbf} / \mathrm{in}^{2}, V_{1}=4 \mathrm{ft} 3\) to \(p_{2}=50 \mathrm{lbf} / \mathrm{in}^{2}{ }^{2}\) Process 2-3: Constant volume from state 2 to state 3 where \(p=10 \mathrm{lbf} / \mathrm{in}^{2}\) Process 3-1: Constant pressure expansion to the initial state. Sketch the processes in series on \(p-V\) coordinates. Evaluate (a) the volume at state 2 , in \(\mathrm{ft}^{3}\), and (b) the work for each process, in Btu.

Short Answer

Expert verified
Volume at state 2 is 0.8 ft鲁. Work is -2.58 Btu (process 1-2), 0 Btu (process 2-3), and 5.94 Btu (process 3-1).

Step by step solution

01

Understand the given data

Identify the states and the processes:- Initial state (1): \(p_1=10 \text{ lbf/in}^2, V_1=4 \text{ ft}^3\)- State (2): Given that \(pV = \text{constant} \implies p_1V_1 = p_2V_2\)- State (3): Given that \(V_3= V_2\) and \(p_3 = 10 \text{ lbf/in}^2\).
02

Solve for Volume at State 2

Using the principle that pressure-volume relationship is constant during process 1-2 and given \(p_1V_1 = p_2V_2\), we can solve for \(V_2\): Substitute values:\[10 \text{ lbf/in}^2 \times 4 \text{ ft}^3 = 50 \text{ lbf/in}^2 \times V_2\]Solving for \(V_2\) yields 0.8 \( \text{ft}^3\).
03

Sketch the processes on p-V coordinates

Plot the processes on a pressure-volume diagram:- From \( (10, 4)\) to \( (50, 0.8)\) for process 1-2 (compression), it will be a hyperbolic curve.- Vertical line from \( (50, 0.8)\) to \( (10, 0.8)\) for process 2-3 (constant volume).- Horizontal line from \( (10, 0.8)\) to \( (10, 4)\) for process 3-1 (constant pressure).
04

Calculate work for each process

For process 1-2: \(W_{1-2} = \int_{V_1}^{V_2} p dV\)Since \(pV = \text{constant}\), we use: \(p = \frac{p_1V_1}{V}\)\[W_{1-2} = p_1V_1 \int_{V_1}^{V_2} \frac{1}{V} dV\] =\[ p_1V_1 \ln\left(\frac{V_2}{V_1}\right) \] =\[ 10 \text{ lbf/in}^2 \times 144 \text{ in}^2/\text{ft}^2 \times 4 \text{ ft}^3 \times \ln\left(\frac{0.8}{4}\right) = -2000 \text{ lbf-ft} = -2.58 \text{ Btu}\]\For process 2-3:\(W_{2-3}=0 \text{ Btu}\) (constant volume process)\For process 3-1:\(W_{3-1} = p_{3}(V_{1} - V_2)\)=\[ 10 \text{ lbf/in}^2 \times 144 \text{ in}^2/\text{ft}^2 \times (4 - 0.8) \text{ ft}^3 = 4608 \text{ lbf-ft} = 5.94 \text{ Btu}\]

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Key Concepts

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

piston-cylinder assembly
A piston-cylinder assembly is a fundamental device in thermodynamics used to understand various processes involving gases. Imagine a hollow cylinder with a movable piston at one end. This setup can compress or expand the gas inside. Pistons can move up and down to change the volume of the gas. It makes the cylinder either smaller or larger, respectively. This movement is vital for studying how pressure and volume interact in different thermodynamic processes. In our exercise, the air inside the piston-cylinder undergoes a series of processes, where we analyze how pressure and volume change.
pressure-volume relationship
Understanding the pressure-volume relationship is crucial in thermodynamic processes. In our exercise, the initial compression process (1-2) follows a specific relationship where pressure multiplied by volume remains constant (pV = constant). This means if the volume decreases, the pressure must increase to maintain the same product. Process 2-3 has a constant volume, meaning no change in volume but a change in pressure. Finally, the expansion process (3-1) keeps the pressure constant while increasing the volume. These changes are typically represented on a p-V diagram, showcasing how both properties interact during different stages.
work calculation
Work calculation in thermodynamic processes often involves understanding how pressure and volume changes affect work done on or by the system. For our piston-cylinder assembly:

  • In process 1-2 (compression), we compute work using the integral of pressure with respect to volume, which involves calculus. Since pV is constant, the formula simplifies to where work is determined by the natural logarithm of volume ratios.
  • In process 2-3 (constant volume), the work is zero since there's no volume change.
  • In process 3-1 (constant pressure), we calculate work by multiplying the constant pressure by the change in volume.
These detailed calculations help understand how much energy is transferred in each thermodynamic process, measured in units like Btu (British thermal units).

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Most popular questions from this chapter

Nitrogen \(\left(\mathrm{N}_{2}\right)\) gas within a piston-cylinder assembly undergoes a compression from \(p_{1}=20\) bar, \(V_{1}=0.5 \mathrm{~m}^{3}\) to a state where \(V_{2}=2.75 \mathrm{~m}^{3}\). The relationship between pressure and volume during the process is \(p V^{1.35}=\) constant. For the \(\mathrm{N}_{2}\), determine (a) the pressure at state 2 , in bar, and (b) the work, in kJ.

A composite plane wall consists of a 3-in.-thick layer of insulation \(\left(\kappa_{\mathrm{s}}=0.029 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{R}\right)\) and a \(0.75\)-in.-thick layer of siding \(\left(\kappa_{\mathrm{s}}=0.058 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{R}\right)\). The inner temperature of the insulation is \(67^{\circ} \mathrm{F}\). The outer temperature of the siding is \(-8^{\circ} \mathrm{F}\). Determine at steady state (a) the temperature at the interface of the two layers, in \({ }^{\circ} \mathrm{F}\), and (b) the rate of heat transfer through the wall in Btu per \(\mathrm{ft}^{2}\) of surface area.

Four kilograms of carbon monoxide \((\mathrm{CO})\) is contained in a rigid tank with a volume of \(1 \mathrm{~m}^{3}\). The tank is fitted with a paddle wheel that transfers energy to the \(\mathrm{CO}\) at a constant rate of \(14 \mathrm{~W}\) for \(1 \mathrm{~h}\). During the process, the specific internal energy of the carbon monoxide increases by \(10 \mathrm{~kJ} / \mathrm{kg}\). If no overall changes in kinetic and potential energy occur, determine (a) the specific volume at the final state, in \(\mathrm{m}^{3} / \mathrm{kg}\). (b) the energy transfer by work, in kJ. (c) the energy transfer by heat transfer, in kJ, and the direction of the heat transfer.

The pistons of a V-6 automobile engine develop 226 hp. If the engine driveshaft rotational speed is \(4700 \mathrm{RPM}\) and the torque is \(248 \mathrm{ft} \cdot \mathrm{lbf}\), what percentage of the developed power is transferred to the driveshaft? What accounts for the difference in power? Does an engine this size meet your transportation needs? Comment.

An object whose weight is 100 lbf experiences a decrease in kinetic energy of \(500 \mathrm{ft}\) - lbf and an increase in potential energy of \(1500 \mathrm{ft}\) - lbf. The initial velocity and elevation of the object, each relative to the surface of the earth, are \(40 \mathrm{ft} / \mathrm{s}\) and \(30 \mathrm{ft}\), respectively. If \(g=32.2 \mathrm{ft} / \mathrm{s}^{2}\), determine (a) the final velocity, in \(\mathrm{ft} / \mathrm{s}\). (b) the final elevation, in \(\mathrm{ft}\).
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