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Air within a piston-cylinder assembly, initially at \(30 \mathrm{lbf} /\) in. \({ }^{2}, 510^{\circ} \mathrm{R}\), and a volume of \(6 \mathrm{ft}^{3}\), is compressed isentropically to a final volume of \(1.2 \mathrm{ft}^{3}\). Assuming the ideal gas model with \(k=1.4\) for the air, determine the (a) mass, in lb, (b) final pressure, in lbf/in. \({ }^{2}\), (c) final temperature, in \({ }^{\circ} \mathrm{R}\), and (d) work, in Btu.

Short Answer

Expert verified
Mass: 0.48 lb, Final Pressure: 240 lbf/in\textsuperscript{2}, Final Temperature: 850\textsuperscript{0}R, Work Done: 82.39 Btu.

Step by step solution

01

- Find the mass of the air

Use the ideal gas equation to find the mass of the air. The ideal gas equation is given by \[ PV = nRT \].Rearrange to find the mass: \[ m = \frac{PV}{RT} \],where\(P = 30 \mathrm{lbf/in^2} \), \(V = 6 \mathrm{ft^3} \), \(T = 510 ^{\circ} \mathrm{R} \), and \(R = 53.3 \mathrm{ft \cdot lbf/( lb \cdot R) }\).Convert the pressure to lbf/ft\textsuperscript{2} first by multiplying by 144 (since 1 in\textsuperscript{2} = 144 ft\textsuperscript{2}).So, the equation becomes:\[ m = \frac{30 \times 144 \times 6}{53.3 \times 510} \approx 0.48 \text{ lb} \]
02

- Find the final pressure

Since the process is isentropic for an ideal gas, use the relation for isentropic processes: \[ P_2 V_2^k = P_1 V_1^k \], where \(k = 1.4\).Rearrange the equation to solve for \(P_2\): \[ P_2 = P_1 \left( \frac{V_1}{V_2} \right)^k \]Given \(P_1 = 30 \mathrm{lbf/in^2} \), \(V_1 = 6 \mathrm{ft^3} \), \(V_2 = 1.2 \mathrm{ft^3} \), so\[ P_2 = 30 \left( \frac{6}{1.2} \right)^{1.4} \approx 240 \mathrm{lbf/in^2} \]
03

- Find the final temperature

Using the isentropic relation for ideal gases:\[ \frac{T_2}{T_1} = \left( \frac{V_1}{V_2} \right)^{k-1} \],Rearrange to solve for \(T_2\): \[ T_2 = T_1 \left( \frac{V_1}{V_2} \right)^{k-1} \]Given \( T_1 = 510^{\circ} \mathrm{R} \), so\[ T_2 = 510 \left( \frac{6}{1.2} \right)^{0.4} \approx 850^{\circ} \mathrm{R} \]
04

- Find the work done

For an isentropic process, the work done can be found using the relation:\[ W = \frac{P_1 V_1 - P_2 V_2}{k-1} \]Given \( P_1 = 30 \mathrm{lbf/in^2} \), \( P_2 = 240 \mathrm{lbf/in^2} \), \( V_1 = 6 \mathrm{ft^3} \), \( V_2 = 1.2 \mathrm{ft^3} \), and \( k = 1.4 \):\[ W = \frac{30 \times 144 \times 6 - 240 \times 1.2}{1.4 - 1} \]Convert the units to Btu (1 Btu = 778.17 lbf-ft):\[ W \approx \frac{25920 - 288}{0.4} \div 778.17 \approx 82.39 \text{ Btu} \]

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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 common apparatus used in thermodynamics to study the behavior of gases. It consists of a cylindrical chamber in which a piston can move freely.
This setup allows us to control the volume of the gas by moving the piston, making it possible to observe changes in pressure and temperature.
The initial conditions given in the problem include a pressure of 30 lbf/in², a temperature of 510°R, and a volume of 6 ft³. This information is crucial for applying thermodynamic principles to solve various properties of the gas.
Understanding this setup helps us analyze processes where the volume changes, such as compression, and how it affects the gas inside.
Ideal Gas Law
The Ideal Gas Law is a fundamental equation in thermodynamics which relates pressure, volume, and temperature of an ideal gas. It is given by:
ewline
Isentropic Relations
Isentropic processes are thermodynamic processes that occur at constant entropy. For an ideal gas, this means no heat is transferred in or out of the system. The isentropic relations help us link different state variables through certain constants:
ewline
Work Done in Thermodynamic Processes
In thermodynamic processes, work done refers to the energy transfer due to volume changes against external pressure. For isentropic processes, the work done can be calculated using:
ewline
Specific Heat Ratio (k)
The specific heat ratio, often represented as k, is a crucial property in thermodynamics. It is defined as the ratio of specific heat at constant pressure (C_p) to specific heat at constant volume (C_v):ewline

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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, \(27^{\circ} \mathrm{C}\).

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