A piston-cylinder device initially contains 2 L of air at 100 kPa and 25^∘ C. Air is now compre

step 1 Let us solve part A of this problem. So in part A, calculate the energy of air at initial stage.

A piston-cylinder device initially contains 2 L of air at...

A piston-cylinder device initially contains $2 \mathrm{~L}$ of air at $100 \mathrm{kPa}$ and $25^{\circ} \mathrm{C}$. Air is now compressed to a final state of $600 \mathrm{kPa}$ and $150^{\circ} \mathrm{C}$. The useful work input is $1.2 \mathrm{~kJ}$. Assuming the surroundings are at $100 \mathrm{kPa}$ and $25^{\circ} \mathrm{C}$, determine ( $a$ ) the exergy of the air at the initial and the final states, (b) the minimum work that must be supplied to accomplish this compression process, and (c) the second-law efficiency of this process.

A piston-cylinder device initially contains $2 \mathrm{~L}$ of air at $100 \mathrm{kPa}$ and $25^{\circ} \mathrm{C}$. Air is now compressed to a final state of $600 \mathrm{kPa}$ and $150^{\circ} \mathrm{C}$. The useful work input is $1.2 \mathrm{~kJ}$. Assuming the surroundings are at $100 \mathrm{kPa}$ and $25^{\circ} \mathrm{C}$, determine
( $a$ ) the exergy of the air at the initial and the final states,
(b) the minimum work that must be supplied to accomplish this compression process, and (c) the second-law efficiency of this process.

Key Concepts

Second-Law Efficiency

Second-law efficiency is a performance metric that compares the useful output or required input of a real process to that of an ideal, reversible process. It indicates how effectively a process uses available exergy by accounting for losses due to irreversibility, providing insight into the thermodynamic quality of the energy conversion or work process.

Minimum Work and Reversible Processes

The minimum work required to achieve a process is associated with a reversible path between the initial and final states. In thermodynamics, a reversible process is one with no irreversibilities, meaning no exergy is destroyed. This concept is used to determine the lower limit on work input or output, against which real processes can be compared.

Exergy

Exergy is a measure of the maximum useful work obtainable from a system as it comes to equilibrium with a given reference environment. It quantifies the quality or usefulness of energy and accounts for the irreversibility of real processes. Exergy analysis helps identify where and how energy is degraded in thermodynamic processes.

Reference Environment (Dead State)

The reference environment, or dead state, is the state of the surroundings (typically defined by a specified pressure and temperature) with which the system eventually equilibrates. This concept is crucial in exergy analysis since it defines the baseline for determining the available energy in the system.

Transcript

00:01 Let us solve part a of this problem.

00:04 So in part a, calculate the energy of air at initial stage.

00:10 So i will use the formula x1 is equal to u minus u0 plus p multiplication v minus v0 v0 minus t not, minus t not multiplication s minus s not, plus m v squared by 2 plus m g z assuming air as an ideal gas kinetic energy and potential energy are negligible at initial state the property of air is similar to that state so here i can write the value of x1 is equal to 0 this is the energy of initial state now calculate the energy of air at the final state so i can write the next expression at x2 is equal to m multiplication cv multiplication t2 minus t not minus t not multiplication h2 minus s not plus p not multiplication v2 minus v1 let this be equation number one now for ideal gas for ideal gas i can write a expression as p1 v1 is equal to p1 v1 is equal to m r t1 or m is equal to p1 v1 by r t1 that will be equal to hundred multiplication 10 to the power 3 multiplication 2 multiplication 10 to the bar minus 3 by 287 multiplication 298 which is equal to 0 .00234 kg as the value.

02:06 Now i will use the formula p1 v1 by t1 is equal to p2 v2 by t2 so the next expression is v2 is equal to p1 v1 t2 by t1 p2 now simplifying it further it will be equal to 100 multiplication 10 to the power 3 multiplication 2 multiplication 10 to the power minus 3 multiplication 4 to 3 by 298 multiplication 600 multiplication 10 to the power 3 so you will get the value of v2 edge v2 is equal to 0 .00 473 meter cube now i will use the formula as 2 minus s not is equal to cp lawn cp lawn t2 by t knot minus r lawn p2 by p not that will be equal to 1 .009 lawn 423 by 29 8 minus 0 .287 lawn 600 by 100 so finally you will get this value as minus 0 .1608 kilojoule per k g kelvin...

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