Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Refrigerant-134a is expanded adiabatically from 100 psia and \(100^{\circ} \mathrm{F}\) to a saturated vapor at 10 psia. Determine the entropy generation for this process, in Btu/lbm-R.

Short Answer

Expert verified
Question: Determine the entropy generation during an adiabatic process involving the expansion of Refrigerant-134a from an initial state of 100 psia and 100°F to a final state of saturated vapor at 10 psia. Answer: The entropy generation for this adiabatic process of Refrigerant-134a is equal to the change in specific entropy (\(\Delta s\)), which can be calculated using the property data for Refrigerant-134a and the formula \(S_{gen} = \Delta s = s_2 - s_1\). The final value of entropy generation will be in Btu/lbm-R.

Step by step solution

01

Identify the process and important parameters

Since this is an adiabatic process, there is no heat transfer between the system and its surroundings. Our goal is to calculate the entropy generation during this process. To do so, we need the property data for Refrigerant-134a, which we can find in the tables provided in standard textbooks or online resources.
02

Determine the initial and final entropy values

Using the Refrigerant-134a tables, we can determine the entropy at the initial and final states. For the initial state (100 psia, 100°F), look up the corresponding specific entropy value (denoted as \(s_1\)) . For the final state (saturated vapor at 10 psia), look up the specific entropy of vapor at 10 psia (denoted as \(s_2\)).
03

Calculate the entropy change during the process

Now that we have the individual entropy values for both initial and final states, we can calculate the change in specific entropy during the process using the following formula: \(\Delta s = s_2 - s_1\)
04

Use the adiabatic process relation to find the entropy generation

An adiabatic process implies that the total change in specific entropy should be equal to the specific entropy generation (\(S_{gen}\)) during the process. Thus, we can directly use the computed entropy change (\(\Delta s\)) as the entropy generation: \(S_{gen} = \Delta s\) Now, we have the entropy generation for this adiabatic process of Refrigerant-134a in Btu/lbm-R.

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 Vaia!

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 hot-water pipe at \(80^{\circ} \mathrm{C}\) is losing heat to the surrounding air at \(5^{\circ} \mathrm{C}\) at a rate of \(2200 \mathrm{W}\). Determine the rate of entropy generation in the surrounding air, in \(\mathrm{W} / \mathrm{K}\).

Steam enters an adiabatic turbine at \(8 \mathrm{MPa}\) and \(500^{\circ} \mathrm{C}\) at a rate of \(18 \mathrm{kg} / \mathrm{s}\), and exits at \(0.2 \mathrm{MPa}\) and \(300^{\circ} \mathrm{C}\) The rate of entropy generation in the turbine is \((a) 0 \mathrm{kW} / \mathrm{K}\) \((b) 7.2 \mathrm{kW} / \mathrm{K}\) \((c) 21 \mathrm{kW} / \mathrm{K}\) \((d) 15 \mathrm{kW} / \mathrm{K}\) \((e) 17 \mathrm{kW} / \mathrm{K}\)

\(3-\mathrm{kg}\) of helium gas at \(100 \mathrm{kPa}\) and \(27^{\circ} \mathrm{C}\) are adiabati cally compressed to 900 kPa. If the isentropic compression efficiency is 80 percent, determine the required work input and the final temperature of helium.

Steam enters a diffuser at 20 psia and \(240^{\circ} \mathrm{F}\) with a velocity of \(900 \mathrm{ft} / \mathrm{s}\) and exits as saturated vapor at \(240^{\circ} \mathrm{F}\) and \(100 \mathrm{ft} / \mathrm{s}\). The exit area of the diffuser is \(1 \mathrm{ft}^{2}\). Determine (a) the mass flow rate of the steam and ( \(b\) ) the rate of entropy generation during this process. Assume an ambient temperature of \(77^{\circ} \mathrm{F}\).

A refrigerator with a coefficient of performance of 4 transfers heat from a cold region at \(-20^{\circ} \mathrm{C}\) to a hot region at \(30^{\circ} \mathrm{C}\). Calculate the total entropy change of the regions when \(1 \mathrm{kJ}\) of heat is transferred from the cold region. Is the second law satisfied? Will this refrigerator still satisfy the second law if its coefficient of performance is \(6 ?\)

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.

Sign-up for free