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 at \(800 \mathrm{kPa}\) and \(25^{\circ} \mathrm{C}\) is throttled to a temperature of \(-20^{\circ} \mathrm{C}\). Determine the pressure and the internal energy of the refrigerant at the final state.

Short Answer

Expert verified
Question: A throttling process of refrigerant-134a starts from an initial pressure of 800 kPa and an initial temperature of 25°C and ends at a final temperature of -20°C. Determine the final pressure and internal energy of the refrigerant at the final state. Answer: Using the R134a Tables and the isenthalpic process, the final pressure and internal energy of the refrigerant-134a at -20°C are __(P_2)__ and __(u_2)__, respectively.

Step by step solution

01

Find Initial Properties

Using the R134a Tables, look for the properties of the refrigerant at the given initial conditions: \(800\,\text{kPa}\) and \(25^{\circ}\text{C}\). Note down the specific enthalpy (\(h_1\)) and specific internal energy (\(u_1\)) at this initial state.
02

Throttling Process

A throttling process is an isenthalpic process, which means the specific enthalpy remains constant. Thus, the specific enthalpy at the final state is the same as the initial state: \(h_2=h_1\).
03

Find Final Properties

Using the R134a Tables, find the properties of the refrigerant at the final temperature, \(-20^{\circ}\text{C}\), and with the specific enthalpy found in Step 2 (\(h_2=h_1\)). This will give the final pressure (\(P_2\)) and specific internal energy (\(u_2\)).
04

Determine the Final Pressure and Internal Energy

Using the information obtained in Step 3, state the final pressure (\(P_2\)) and internal energy (\(u_2\)) of the refrigerant at the final state.

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!

Key Concepts

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

Refrigerant-134a Properties
Refrigerant-134a, also known as R-134a, is a hydrofluorocarbon (HFC) commonly used in air conditioning and refrigeration systems. It has gained popularity as a replacement for CFCs and HCFCs which are known to deplete the ozone layer.

Understanding its properties is essential for engineers and technicians working with refrigeration systems. R-134a operates efficiently over a broad range of temperatures and pressures, making it versatile in different climates and applications. Its key thermodynamic properties include specific enthalpy and specific internal energy, which determine the energy content at various states. By utilizing pressure-enthalpy charts or thermodynamic tables for R-134a, professionals can predict how it will behave under changing conditions in a system.
Isenthalpic Process
An isenthalpic process is a thermodynamic process during which the enthalpy of the fluid remains constant. This is a typical characteristic of a throttling process, not common in everyday life, but crucial in the design of refrigeration and air conditioning systems, as well as in other engineering applications.

In the context of the exercise, when Refrigerant-134a is throttled, its enthalpy before and after the process remains the same (\(h_1 = h_2\)). Despite a change in pressure and possibly other properties like temperature, no heat is transferred to or from the refrigerant, and no work is done by or on the refrigerant during throttling.
Internal Energy
The concept of internal energy in thermodynamics refers to the total energy contained within a system, accounting for both the kinetic and potential energy of its molecules. In the case of Refrigerant-134a, internal energy plays a significant role in determining how the refrigerant will act when it is compressed, expanded, or when it undergoes a phase change.

Knowing the internal energy of a refrigerant before and after a process, such as throttling, can provide insights into the state of the refrigerant, which is not always apparent from other properties. The relationship between internal energy and temperature is particularly important, as it can indicate phase changes and the refrigeration capacity of the substance.
Pressure-Temperature Relationship
The pressure-temperature relationship is a fundamental concept in thermodynamics that describes how the pressure of a substance correlates with its temperature, particularly when dealing with phase changes between the liquid and vapor states. For refrigerants like R-134a, this relationship is predictable and is characterized by its saturation curve.

In the given exercise, the throttling process results in a decrease in temperature without a change in enthalpy. By consulting the R-134a tables or diagrams, it's possible to determine the new saturation pressure at the final lower temperature, effectively solving for the refrigerant's final pressure which is crucial in designing refrigeration systems and for troubleshooting them when performance issues arise.

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 \(110-V\) electric hot-water heater warms \(0.1 \mathrm{L} / \mathrm{s}\) of water from 18 to \(30^{\circ} \mathrm{C}\). Calculate the current in amperes that must be supplied to this heater.

A \(0.3-\mathrm{m}^{3}\) rigid tank is filled with saturated liquid water at \(200^{\circ} \mathrm{C}\). A valve at the bottom of the tank is opened, and liquid is withdrawn from the tank. Heat is transferred to the water such that the temperature in the tank remains constant. Determine the amount of heat that must be transferred by the time one-half of the total mass has been withdrawn.

Cold water enters a steam generator at \(20^{\circ} \mathrm{C}\) and leaves as saturated vapor at \(200^{\circ} \mathrm{C}\). Determine the fraction of heat used in the steam generator to preheat the liquid water from \(20^{\circ} \mathrm{C}\) to the saturation temperature of \(200^{\circ} \mathrm{C}\).

In a shower, cold water at \(10^{\circ} \mathrm{C}\) flowing at a rate of \(5 \mathrm{kg} / \mathrm{min}\) is mixed with hot water at \(60^{\circ} \mathrm{C}\) flowing at a rate of \(2 \mathrm{kg} / \mathrm{min} .\) The exit temperature of the mixture is \((a) 24.3^{\circ} \mathrm{C}\) (b) \(35.0^{\circ} \mathrm{C}\) \((c) 40.0^{\circ} \mathrm{C}\) \((d) 44.3^{\circ} \mathrm{C}\) \((e) 55.2^{\circ} \mathrm{C}\)

A constant-pressure \(R-134\) a vapor separation unit separates the liquid and vapor portions of a saturated mixture into two separate outlet streams. Determine the flow power needed to pass \(6 \mathrm{L} / \mathrm{s}\) of \(\mathrm{R}-134 \mathrm{a}\) at \(320 \mathrm{kPa}\) and 55 percent quality through this unit. What is the mass flow rate, in \(\mathrm{kg} / \mathrm{s}\), of the two outlet streams?

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