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 throttled from the saturated liquid state at \(700 \mathrm{kPa}\) to a pressure of \(160 \mathrm{kPa}\). Determine the temperature drop during this process and the final specific volume of the refrigerant.

Short Answer

Expert verified
Question: Determine the temperature drop and final specific volume of Refrigerant-134a when it undergoes a throttling process from a saturated liquid state at 700 kPa to a pressure of 160 kPa. Answer: To find the temperature drop and final specific volume of Refrigerant-134a, first determine the initial state by looking up the saturation temperature and specific enthalpy at 700 kPa. Apply the isenthalpic process condition to find the final specific enthalpy. Determine the final state by looking up the saturation temperatures, pressures, and specific enthalpies in the property table at a pressure of 160 kPa. Calculate the temperature drop by taking the difference between the initial and final temperatures and find the final specific volume by looking up the final state in the property table.

Step by step solution

01

Determine the initial state

The problem states that the initial state of the refrigerant is a saturated liquid at a pressure of 700 kPa. We can look up the saturation temperature and specific enthalpy at this pressure in a Refrigerant-134a property table. Let's denote the initial temperature as \(T_1\) and the initial specific enthalpy as \(h_1\).
02

Apply the isenthalpic process condition

Since the throttling process is isenthalpic, the specific enthalpy at the final state (\(h_2\)) is equal to the specific enthalpy at the initial state (\(h_1\)). Mathematically, this can be represented as: \(h_2 = h_1\) We already found \(h_1\) in Step 1, so we have the specific enthalpy for the final state directly from the initial state.
03

Determine the final state

Now that we have the final pressure \(P_2=160 \mathrm{kPa}\) and the final specific enthalpy \(h_2\), we can find the final state by looking up the saturation temperatures, pressures, and specific enthalpies in the Refrigerant-134a property table. We need to find the final temperature \(T_2\) and specific volume \(v_2\).
04

Calculate the temperature drop

Now that we have the initial temperature \(T_1\) and final temperature \(T_2\), we can calculate the temperature drop during the throttling process by simply taking the difference between the initial and final temperatures: Temperature drop = \(T_1 - T_2\)
05

Find the final specific volume

We have already found the final specific volume \(v_2\) in Step 3 while determining the final state. The temperature drop and final specific volume are the answers to this problem.

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) used widely in air conditioning and refrigeration systems. Understanding its properties is crucial for designing and analyzing such systems. R-134a is non-flammable, non-toxic, and has zero ozone depletion potential, making it an environmentally safer alternative to older refrigerants like R-12.

Key thermophysical properties of R-134a include its boiling point, specific heat, thermal conductivity, viscosity, and more importantly for this exercise, its temperature-enthalpy relationship. These properties vary with pressure and temperature, influencing the refrigerant's performance during phase change processes. For instance, knowing the saturation temperature at a given pressure can help determine the refrigerant's state — whether it's in a liquid, vapor, or mixed phase — which is essential in thermodynamic calculations.
Isenthalpic Process
The isenthalpic process is a vital concept in thermodynamics, particularly relevant in the analysis of throttling devices such as expansion valves and capillary tubes. During an isenthalpic process, the specific enthalpy (\(h\)) remains constant even though other properties like temperature and pressure may change. This is because throttling is an adiabatic process with no work done on or by the system and negligible heat transfer.

In the context of the example problem where Refrigerant-134a is throttled, the process does not change the enthalpy. By assuming that we undergo an ideal, isenthalpic process, we can track and predict the state of the refrigerant after throttling using the initial enthalpy as an anchor point. Recognizing that the final enthalpy equals the initial enthalpy (\(h_2 = h_1\) simplifies calculations in the absence of external heat transfer or work.
Specific Volume
Specific volume (\(v\) is a fundamental property in thermodynamics, defined as the volume occupied by a unit mass of a substance. It is the inverse of density and provides engineers with insights into the space requirements for a system's design and operation. For example, the specific volume of a refrigerant like R-134a can dictate the size of the components, such as pipes and compressors, in a refrigeration cycle.

In the context of refrigeration and air-conditioning systems, the specific volume of the refrigerant changes significantly during phase transitions from liquid to vapor and vice versa. When dealing with the throttling process, as shown in the exercise, understanding specific volume before and after the process is crucial to determine the effect on the refrigerant's density and to ensure proper function of the related thermal system components.
Pressure-Enthalpy Diagram
The pressure-enthalpy (\(P-h\) diagram is a valuable tool in visualizing the thermodynamic processes of refrigerants like R-134a. It plots pressure on the vertical axis and enthalpy on the horizontal axis, showing all the possible states the refrigerant can exist in. Key regions in this diagram include the saturated liquid line, the saturated vapor line, and the superheated vapor region. It also includes constant entropy lines, which are useful in analyzing isentropic processes.

For the throttling process, the pressure-enthalpy diagram helps identify states before and after the process. Since throttling is an isenthalpic process, the initial and final points would lie on a vertical line that corresponds to constant enthalpy. This makes it easier to illustrate and understand the changes in the refrigerant's state during a throttling process, aiding in efficient system design and problem-solving.

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

The air in an insulated, rigid compressed-air tank whose volume is \(0.5 \mathrm{m}^{3}\) is initially at \(4000 \mathrm{kPa}\) and \(20^{\circ} \mathrm{C}\) Enough air is now released from the tank to reduce the pressure to 2000 kPa. Following this release, what is the temperature of the remaining air in the tank?

In a heating system, cold outdoor air at \(7^{\circ} \mathrm{C}\) flowing at a rate of \(4 \mathrm{kg} / \mathrm{min}\) is mixed adiabatically with heated air at \(70^{\circ} \mathrm{C}\) flowing at a rate of \(3 \mathrm{kg} / \mathrm{min} .\) The exit temperature of the mixture is \((a) 34^{\circ} \mathrm{C}\) (b) \(39^{\circ} \mathrm{C}\) \((c) 45^{\circ} \mathrm{C}\) \((d) 63^{\circ} \mathrm{C}\) \((e) 77^{\circ} \mathrm{C}\)

In large steam power plants, the feedwater is frequently heated in a closed feedwater heater by using steam extracted from the turbine at some stage. Steam enters the feedwater heater at \(1 \mathrm{MPa}\) and \(200^{\circ} \mathrm{C}\) and leaves as saturated liquid at the same pressure. Feedwater enters the heater at \(2.5 \mathrm{MPa}\) and \(50^{\circ} \mathrm{C}\) and leaves at \(10^{\circ} \mathrm{C}\) below the exit temperature of the steam. Determine the ratio of the mass flow rates of the extracted steam and the feedwater.

The heat of hydration of dough, which is \(15 \mathrm{kJ} / \mathrm{kg}\) will raise its temperature to undesirable levels unless some cooling mechanism is utilized. A practical way of absorbing the heat of hydration is to use refrigerated water when kneading the dough. If a recipe calls for mixing \(2 \mathrm{kg}\) of flour with \(1 \mathrm{kg}\) of water, and the temperature of the city water is \(15^{\circ} \mathrm{C}\), determine the temperature to which the city water must be cooled before mixing in order for the water to absorb the entire heat of hydration when the water temperature rises to \(15^{\circ} \mathrm{C}\). Take the specific heats of the flour and the water to be 1.76 and \(4.18 \mathrm{kJ} / \mathrm{kg} \cdot^{\circ} \mathrm{C},\) respectively.

A long roll of 1 -m-wide and 0.5 -cm-thick \(1-\mathrm{Mn}\) manganese steel plate \(\left(\rho=7854 \mathrm{kg} / \mathrm{m}^{3}\right)\) coming off a furnace is to be quenched in an oil bath to a specified temperature. If the metal sheet is moving at a steady velocity of \(10 \mathrm{m} / \mathrm{min}\) determine the mass flow rate of the steel plate through the oil bath.

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