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Chapter 3: Problem 81

Ten \(\mathrm{kg}\) of Refrigerant 22 contained in a piston-cylinder assembly undergoes a process for which the pressure-specific volume relationship is \(p v^{n}=\) constant. The initial and final states of the refrigerant are fixed by \(p_{1}=400 \mathrm{kPa}, T_{1}=\) \(-5^{\circ} \mathrm{C}\), and \(p_{2}=2000 \mathrm{kPa}, T_{2}=70^{\circ} \mathrm{C}\), respectively. Determine the work and heat transfer for the process, each in kJ.

Short Answer

Expert verified
Work: 681 kJ, Heat transfer: 1153 kJ.

Step by step solution

01

- Identify the given data

Given: - Mass of Refrigerant 22, \( m = 10 \text{ kg} \) - Initial pressure, \( p_1 = 400 \text{ kPa} \) - Initial temperature, \( T_1 = -5^\text{ o} \text{C} \) - Final pressure, \( p_2 = 2000 \text{ kPa} \) - Final temperature, \( T_2 = 70^\text{ o} \text{C} \)
02

- Determine initial and final specific volumes

Using refrigerant tables for Refrigerant 22: At \(p_1 = 400 \text{ kPa}\) and \(T_1 = -5^\text{ o} \text{C}\), find the specific volume \(v_1\). At \(p_2 = 2000 \text{ kPa}\) and \(T_2 = 70^\text{ o} \text{C}\), find the specific volume \(v_2\). Use values: \(v_1 = 0.1807 \text{ m}^3/\text{kg} \) \(v_2 = 0.0451 \text{ m}^3/\text{kg} \)
03

- Apply the relationship \(p v^n = \text{constant}\)

Using \(p_1 v_1^n = p_2 v_2^n\), solve for \(n\). Substitute values: \( 400 (0.1807)^n = 2000 (0.0451)^n \) Solve for \(n\): \(n = \frac{\text{ln}(400) - \text{ln}(2000)}{\text{ln}(0.0451) - \text{ln}(0.1807)} = 1.263\)
04

- Calculate work done (W)

For polytropic process, work done: \( W = \frac{p_2 v_2 - p_1 v_1}{1 - n} \) Substitute values: \( W = \frac{2000 \times 0.0451 - 400 \times 0.1807}{1 - 1.263} \text{ kPa m}^3/\text{kg} \) Convert kPa m^3 to kJ: \( W = \frac{90.2 - 72.28}{-0.263} = 68.1 \text{ kJ/kg} \) Total work for 10 kg: \( W_{\text{total}} = 68.1 \times 10 = 681 \text{ kJ} \)
05

- Apply the First Law of Thermodynamics

First Law of Thermodynamics: \( \text{Q} = \text{螖U} + \text{W} \) Using refrigerant tables, find \(u_1\) and \(u_2\): \( u_1 = 264.9 \text{ kJ/kg} \) \( u_2 = 312.1 \text{ kJ/kg} \) \( \text{螖U} = m(u_2 - u_1) \) Substitute values: \( \text{螖U} = 10(312.1 - 264.9) = 472 \text{ kJ} \) Calculate Q: \( Q = 472 \text{ kJ} + 681 \text{ kJ} = 1153 \text{ kJ} \)

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Key Concepts

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

Polytropic Process
A polytropic process is a thermodynamic process that follows the equation \(\text{p} \text{v}^n = \text{constant}\). In this type of process, the pressure and specific volume are related in a specific way.

The parameter \(n\) determines the behavior of the process:
  • When \(n = 0\), it represents an isobaric process (constant pressure).
  • When \(n = 1\), it is an isothermal process (constant temperature).
  • When \(n = \infty\), it indicates an isochoric process (constant volume).
For our exercise, the refrigerant undergoes a polytropic process, and we derived \(n = 1.263\) using initial and final states of pressure and specific volume.
First Law of Thermodynamics
The First Law of Thermodynamics states that energy cannot be created or destroyed, only transferred or converted from one form to another.

It is often expressed as \Q = \Delta U + W\. Here,
  • \(\text{Q}\) is the heat added to the system.
  • \(\text{\u0394 U}\) is the change in internal energy.
  • \(\text{W}\) is the work done by the system.
In our case, we calculated the values of heat transfer (Q), change in internal energy (\(\text{\u0394 U}\)), and work done (W) using refrigerant details and it's tables.
Specific Volume
Specific volume is the volume per unit mass of a substance, often expressed in \(m^3/\text{kg}\). It is an intrinsic property of substances and plays a critical role in thermodynamic calculations.

Using refrigerant tables, we found the specific volumes (\(v_1\) and \(v_2\)) at the initial and final states of the refrigerant.
  • At \(p_1 = 400 \text{kPa}\) and \(T_1 = -5^\text{oC}\), \(v_1 = 0.1807 \text{m}^3/\text{kg}\).
  • At \(p_2 = 2000 \text{kPa}\) and \(T_2 = 70^\text{oC}\), \(v_2 = 0.0451 \text{m}^3/\text{kg}\).
This information is crucial for understanding the polytropic process and solving for unknowns like the parameter \(n\).
Refrigerant Tables
Refrigerant tables provide essential properties such as pressure, temperature, specific volume, internal energy, enthalpy, and entropy for different states of a refrigerant.

These tables help to determine the exact state of a refrigerant under varying conditions. For instance, you can find specific volume or internal energy for a given pressure and temperature.
  • In our problem, we used refrigerant tables to find \(v_1\) and \(v_2\), which are critical for calculating the work done and heat transfer.
  • We also derived internal energy values (\(u_1 \text{ and } u_2\)) to apply the First Law of Thermodynamics.
Using refrigerant tables simplifies these calculations and provides accurate results, helping us to solve complex thermodynamic problems efficiently.

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Most popular questions from this chapter

A piston-cylinder assembly fitted with a slowly rotating paddle wheel contains \(0.13 \mathrm{~kg}\) of air, initially at \(300 \mathrm{~K}\). The air undergoes a constant-pressure process to a final temperature of \(400 \mathrm{~K}\). During the process, energy is gradually transferred to the air by heat transfer in the amount \(12 \mathrm{~kJ}\). Assuming the ideal gas model with \(k=1.4\) and negligible changes in kinetic and potential energy for the air, determine the work done (a) by the paddle wheel on the air and (b) by the air to displace the piston, each in kJ.

Four-tenth \(\mathrm{lb}\) of air, initially at \(540^{\circ} \mathrm{R}\), is contained in a closed, rigid tank fitted with a paddle wheel that stirs the air until its temperature is \(740^{\circ} \mathrm{R}\). The driveshaft of the paddle wheel rotates for \(60 \mathrm{~s}\) at 100 RPM with an applied torque of \(20 \mathrm{ft}\) - lbf. Assuming ideal gas behavior for the air, determine the work and heat transfer, each in Btu. There are no overall changes in kinetic or potential energy.

Water in a piston-cylinder assembly, initially at a temperature of \(99.63^{\circ} \mathrm{C}\) and a quality of \(65 \%\), is heated at constant pressure to a temperature of \(200^{\circ} \mathrm{C}\). If the work during the process is \(+300 \mathrm{~kJ}\), determine (a) the mass of water, in \(\mathrm{kg}\), and (b) the heat transfer, in kJ. Changes in kinetic and potential energy are negligible.

\(3.110\) A balloon filled with helium, initially at \(27^{\circ} \mathrm{C}, 1\) bar, is released and rises in the atmosphere until the helium is at \(17^{\circ} \mathrm{C}, 0.9\) bar. Determine, as a percent, the change in volume of the helium from its initial volume.

A closed, rigid tank whose volume is \(1.5 \mathrm{~m}^{3}\) contains Refrigerant \(134 \mathrm{a}\), initially a two-phase liquid-vapor mixture at \(10^{\circ} \mathrm{C}\). The refrigerant is heated to a final state where temperature is \(50^{\circ} \mathrm{C}\) and quality is \(100 \%\). Locate the initial and final states on a sketch of the \(T-v\) diagram. Determine the mass of vapor present at the initial and final states, each in \(\mathrm{kg}\).
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