---### Sample Problem: Thermodynamics - Refrigerant R-134a in a Rigid Container**Problem Statement:**10 kg of R-134a at 300 kPA fills a rigid container whose volume is 14 L. Determine the temperature and total enthalpy in the container. The container is now heated until the pressure is 600 kPa. Determine the temperature and total enthalpy when the heating is completed.**Initial Conditions:**- Refrigerant: R-134a- Initial Pressure: 300 kPa- Mass: 10 kg- Volume: 14 L**Heating Process:**- Final Pressure: 600 kPa**Diagram Overview:**A rectangular box represents the rigid container filled with R-134a. The following details are written inside the box:- **Substance:** R134a- **Initial Pressure:** 300 kPa- **Mass:** 10 kg- **Volume:** 14 LAn arrow pointing to the container is labeled "Q," indicating that heat is added to the system.**Step-by-Step Solution:**1. **Determine the Initial State:** a. Given values: - P1 = 300 kPa - V = 14 L (convert to m³: 14 L = 0.014 m³) - m = 10 kg b. Calculate specific volume (v): \[ v = \frac{V}{m} = \frac{0.014 \, \text{m}^3}{10 \, \text{kg}} = 0.0014 \, \text{m}^3/\text{kg} \] c. From the refrigeration tables, locate the temperature (T1) and specific enthalpy (h1) for R-134a at P1 = 300 kPa and specific volume v = 0.0014 m³/kg.2. **Determine the Final State:** a. Given values after heating: - P2 = 600 kPa b. Since the volume is rigid, specific volume remains constant: \[ v = 0.0014 \, \text{m}^3/\text{kg} \] c. From the refrigeration tables, locate the temperature (T2) and specific enthalpy (h2)

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5. 10 kg of R-134a at 300KPA fills a rigid container whose volume is 14 L. Determine the temperature and total enthalpy in the container. The container is now heated until the pressure is 600 kPa. Determine the temperature and total enthalpy when the heating is completed.

5. 10 kg of R-134a at 300KPA fills a rigid container whose volume is 14 L. Determine the temperature and total enthalpy in the container. The container is now heated until the pressure is 600 kPa. Determine the temperature and total enthalpy when the heating is completed.

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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Concept explainers

Chemical and Phase Equilibrium

Phase diagrams or equilibrium phase diagrams are the graphical representations that show the phase transformations of the constituting elements with respect to the pressure, temperature, and compositions. The diagrams show the conditions that limit the different phases, namely solid, liquid, and gas. The diagram can also be utilized to verify the relations between the state variables. Most of the phase diagrams are temperature-pressure phase diagrams, with the variable parameter as the composition. Usually other variables such as solubility and pressure parameters are not indicated in the phase diagram. For a one-component system (unary phase diagram), the phase diagram is generally a function of temperature and pressure as the state variables, or temperature and volume. However, for a binary system (two-phase) having two components, the phase diagram is a function of temperature and composition as the state variables only. Coexisting of the different phases of the constituting elements is indicated in the phase diagram. A ternary phase diagram can be characterized by consisting of three phases in equilibrium.

Question

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### Sample Problem: Thermodynamics - Refrigerant R-134a in a Rigid Container

**Problem Statement:**

10 kg of R-134a at 300 kPA fills a rigid container whose volume is 14 L. Determine the temperature and total enthalpy in the container. The container is now heated until the pressure is 600 kPa. Determine the temperature and total enthalpy when the heating is completed.

**Initial Conditions:**
- Refrigerant: R-134a
- Initial Pressure: 300 kPa
- Mass: 10 kg
- Volume: 14 L

**Heating Process:**
- Final Pressure: 600 kPa

**Diagram Overview:**

A rectangular box represents the rigid container filled with R-134a. The following details are written inside the box:

- **Substance:** R134a
- **Initial Pressure:** 300 kPa
- **Mass:** 10 kg
- **Volume:** 14 L

An arrow pointing to the container is labeled "Q," indicating that heat is added to the system.

**Step-by-Step Solution:**

1. **Determine the Initial State:**

a. Given values:
- P1 = 300 kPa
- V = 14 L (convert to m³: 14 L = 0.014 m³)
- m = 10 kg

b. Calculate specific volume (v):
\[
v = \frac{V}{m} = \frac{0.014 \, \text{m}^3}{10 \, \text{kg}} = 0.0014 \, \text{m}^3/\text{kg}
\]

c. From the refrigeration tables, locate the temperature (T1) and specific enthalpy (h1) for R-134a at P1 = 300 kPa and specific volume v = 0.0014 m³/kg.

2. **Determine the Final State:**

a. Given values after heating:
- P2 = 600 kPa

b. Since the volume is rigid, specific volume remains constant:
\[
v = 0.0014 \, \text{m}^3/\text{kg}
\]

c. From the refrigeration tables, locate the temperature (T2) and specific enthalpy (h2)

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