**Title: Analysis of an Isothermal Process for Refrigerant-134a****Introduction to the Problem:**Refrigerant-134a at an initial pressure of 320 kPa and a temperature of 40°C undergoes an isothermal process within a closed system. During this process, the quality of the refrigerant reaches 65 percent. The task is to determine the magnitude of work and heat transfer required per unit mass of the refrigerant. For calculations, refer to the thermodynamic tables specific to R-134a.**System Description:**- **Refrigerant**: R-134a- **Initial Conditions**: - Pressure: 320 kPa - Temperature: 40°C**Process Details:**The process is isothermal, which means that the temperature remains constant throughout the process at 40°C.**Results:**- **Work Required**: 41.823 kJ/kg- **Heat Transfer Required**: 102.783 kJ/kg**Diagram Explanation:**The diagram displays a closed system with R-134a. It visually shows the initial conditions of pressure and temperature. The system remains at 320 kPa and 40°C throughout the process. The quality of the refrigerant changes as indicated in the problem, affecting the energy calculations.**Conclusion:**In this example, the tools of thermodynamics are used to calculate the work and heat transfer necessary for the isothermal transition of refrigerant-134a within specified constraints. Such analysis is crucial for understanding energy requirements in HVAC systems and other industrial applications involving refrigerants.

Answered: Refrigerant-134a at 320 kPa and 40°C…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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**Title: Analysis of an Isothermal Process for Refrigerant-134a**

**Introduction to the Problem:**

Refrigerant-134a at an initial pressure of 320 kPa and a temperature of 40°C undergoes an isothermal process within a closed system. During this process, the quality of the refrigerant reaches 65 percent. The task is to determine the magnitude of work and heat transfer required per unit mass of the refrigerant. For calculations, refer to the thermodynamic tables specific to R-134a.

**System Description:**

- **Refrigerant**: R-134a
- **Initial Conditions**:
- Pressure: 320 kPa
- Temperature: 40°C

**Process Details:**

The process is isothermal, which means that the temperature remains constant throughout the process at 40°C.

**Results:**

- **Work Required**: 41.823 kJ/kg
- **Heat Transfer Required**: 102.783 kJ/kg

**Diagram Explanation:**

The diagram displays a closed system with R-134a. It visually shows the initial conditions of pressure and temperature. The system remains at 320 kPa and 40°C throughout the process. The quality of the refrigerant changes as indicated in the problem, affecting the energy calculations.

**Conclusion:**

In this example, the tools of thermodynamics are used to calculate the work and heat transfer necessary for the isothermal transition of refrigerant-134a within specified constraints. Such analysis is crucial for understanding energy requirements in HVAC systems and other industrial applications involving refrigerants.

Expert Solution

Step 1

To find :

The required magnitude of work

Given :

The pressure is $P_{1} = 320 kPa$ ,

The Temperature is $T_{1} = 40 ° C$ ,

For isothermal process

The Temperature is $T_{2} = 40 ° C$ ,

The quality is $x_{2} = 0.65$

The properties are taken From the table $134 ‐ a$ at pressure $P_{1} = 320 kPa$ and $T_{1} = 40 ° C$ are,

$ν_{1} = 0.0753 \frac{m^{3}}{kg}, h_{1} = 261.6 \frac{kJ}{kg}$

The properties are taken From the table $134 ‐ a$ at $T_{2} = T_{1} = 40 ° C$ are,

$\begin{array}{rcl} U_{f 2} & = & 107.3 \frac{kJ}{kg}, U_{f g 2} = 143.6 \frac{kJ}{kg} \\ ν_{f 2} & = & 0.000872 \frac{m^{3}}{kg}, v_{g_{2}} = 0.01995 \frac{m^{3}}{kg} \end{array}$

Formula used :

The work done can be obtained as,

$W = P_{1} ν_{1} \ln (\frac{v_{2}}{v_{1}})$

where $P_{1}, v$ is pressure and volume.

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