/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 27 Methane gas \(\left(\mathrm{CH}_... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Methane gas \(\left(\mathrm{CH}_{4}\right)\) enters a compressor at \(298 \mathrm{~K}, 1\) bar and exits at 2 bar and temperature \(T\). Employing the ideal gas model, determine \(T\), in \(\mathrm{K}\), if there is no change in specific entropy from inlet to exit.

Short Answer

Expert verified
366.54 K

Step by step solution

01

- Understand Given Data

Methane (\(\mathrm{CH}_4\)) enters a compressor at 298 K and 1 bar, and exits at 2 bar with an unknown temperature \(T\). There is no change in specific entropy.
02

- Ideal Gas Model

Since we are applying the ideal gas model, recall that for an isentropic process involving an ideal gas, the relation between temperature and pressure is given by: \[ \left( \frac{T_2}{T_1} \right) = \left( \frac{P_2}{P_1} \right)^{(k-1)/k} \]
03

- Specific Heat Ratio

For methane (\(\mathrm{CH}_4\)), the ratio of specific heats (\(k\)) is approximately 1.31.
04

- Apply Given Values

Let \(T_1 = 298\) K, \(P_1 = 1\) bar, and \(P_2 = 2\) bar. Substitute these values into the isentropic relation: \[ \left( \frac{T_2}{298} \right) = \left( \frac{2}{1} \right)^{(1.31-1)/1.31} \]
05

- Simplify and Solve

Simplify the exponent first: \[ \left( \frac{2}{1} \right)^{0.31/1.31} \approx 1.23 \]Then solve for \(T_2\): \[ T_2 = 298 \times 1.23 \approx 366.54 ~ \text{K} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Isentropic Process
An isentropic process is a thermodynamic process in which the entropy of the system remains constant. This means there is no heat transfer into or out of the system, and any changes in pressure or volume are adiabatic. A perfect example is the compression of gases in isolated conditions. For an ideal gas, the isentropic process can be expressed mathematically using the relationship between temperature and pressure. Understanding this no-change-in-entropy principle is crucial when solving problems involving compressors and turbines, where efficiency maximization is often desired.
Specific Heat Ratio
The specific heat ratio, often denoted by the symbol \(k\) or sometimes \(\gamma \), is the ratio of the specific heat at constant pressure (\(C_p\)) to the specific heat at constant volume (\(C_v\)). It is a critical value in thermodynamics, particularly when dealing with ideal gases. For methane (\(\mathrm{CH}_4\)), this ratio is approximately 1.31. This ratio helps in determining the relationship between different thermodynamic properties of the gas. Knowing the specific heat ratio is essential for calculating the final temperature in processes like those involving compressors, where temperature and pressure changes occur without heat exchange.
Temperature-Pressure Relationship
In thermodynamics, the relationship between temperature and pressure of an ideal gas undergoing an isentropic process is given by the formula: \[ \left( \frac{T_2}{T_1} \right) = \left( \frac{P_2}{P_1} \right)^{(k-1)/k} \] This equation states that the ratio of the final and initial temperatures is directly related to the ratio of the final and initial pressures, raised to the power of \(\frac{k-1}{k}\), where \(k\) is the specific heat ratio. Using this formula, you can determine unknown variables such as the final temperature (\(T_2\)) given initial conditions and specific heat ratio. In the given exercise, the initial temperature and pressures were substituted into this formula to find the exit temperature after compression. Such relationships are pivotal for engineers to understand and predict outcomes in practical applications like compressors and turbines.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A 5-kilowatt pump operating at steady state draws in liquid water at 1 bar, \(15^{\circ} \mathrm{C}\) and delivers it at 5 bar at an elevation \(6 \mathrm{~m}\) above the inlet. There is no significant change in velocity between the inlet and exit, and the local acceleration of gravity is \(9.8 \mathrm{~m} / \mathrm{s}^{2} .\) Would it be possible to pump \(7.5 \mathrm{~m}^{3}\) in \(10 \mathrm{~min}\) or less? Explain.

Of increasing interest today are turbines, pumps, and heat exchangers that weigh less than 1 gram and have volumes of 1 cubic centimeter or less. Although many of the same design considerations apply to such micromachines as to corresponding full-scale devices, others do not. Of particular interest to designers is the impact of irreversibilities on the performance of such tiny devices. Write a report discussing the influence of irreversibilities related to heat transfer and friction on the design and operation of micromachines.

An electric motor operating at steady state draws a current of 10 amp with a voltage of \(220 \mathrm{~V}\). The output shaft rotates at 1000 RPM with a torque of \(16 \mathrm{~N} \cdot \mathrm{m}\) applied to an external load. The rate of heat transfer from the motor to its surroundings is related to the surface temperature \(T_{\mathrm{b}}\) and the ambient temperature \(T_{0}\) by \(\mathrm{hA}\left(T_{\mathrm{b}}-T_{0}\right)\), where \(\mathrm{h}=100 \mathrm{~W} / \mathrm{m}^{2}\). \(\mathrm{K}, \mathrm{A}=0.195 \mathrm{~m}^{2}\), and \(T_{0}=293 \mathrm{~K}\). Energy transfers are considered positive in the directions indicated by the arrows on Fig. P6.51. (a) Determine the temperature \(T_{\mathrm{b}}\), in \(\mathrm{K}\). (b) For the motor as the system, determine the rate of entropy production, in \(\mathrm{kW} / \mathrm{K}\). (c) If the system boundary is located to take in enough of the nearby surroundings for heat transfer to take place at temperature \(T_{0}\), determine the rate of entropy production, in \(\mathrm{kW} / \mathrm{K}\), for the enlarged system.

How might you explain the entropy production concept in terms a child would understand?

A system consists of \(2 \mathrm{~m}^{3}\) of hydrogen gas \(\left(\mathrm{H}_{2}\right)\), initially at \(35^{\circ} \mathrm{C}, 215 \mathrm{kPa}\), contained in a closed rigid tank. Energy is transferred to the system from a reservoir at \(300^{\circ} \mathrm{C}\) until the temperature of the hydrogen is \(160^{\circ} \mathrm{C}\). The temperature at the system boundary where heat transfer occurs is \(300^{\circ} \mathrm{C}\). Modeling the hydrogen as an ideal gas, determine the heat transfer, in \(\mathrm{kJ}\), the change in entropy, in \(\mathrm{kJ} / \mathrm{K}\), and the amount of entropy produced, in \(\mathrm{kJ} / \mathrm{K}\). For the reservoir, determine the change in entropy, in \(\mathrm{kJ} / \mathrm{K}\). Why do these two entropy changes differ?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.