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Two \(\mathrm{kg}\) of a two-phase, liquid-vapor mixture of carbon dioxide \(\left(\mathrm{CO}_{2}\right)\) exists at \(-40^{\circ} \mathrm{C}\) in a \(0.05 \mathrm{~m}^{3}\) tank. Determine the quality of the mixture, if the values of specific volume for saturated liquid and saturated vapor \(\mathrm{CO}_{2}\) at \(-40^{\circ} \mathrm{C}\) are \(v_{\mathrm{f}}=0.896 \times 10^{-3} \mathrm{~m}^{3} / \mathrm{kg}\) and \(v_{\mathrm{g}}=3.824 \times 10^{-2} \mathrm{~m}^{3} / \mathrm{kg}\) respectively.

Short Answer

Expert verified
The quality of the mixture is approximately 0.645.

Step by step solution

01

Write Down Given Data

Identify and note all the given data from the problem:\( m = 2 \, \text{kg} \) (mass of CO2)Volume of tank, \( V = 0.05 \, \text{m}^3 \)Specific volume of saturated liquid CO2 at -40°C, \( v_f = 0.896 \times 10^{-3} \, \text{m}^3 / \text{kg} \)Specific volume of saturated vapor CO2 at -40°C, \( v_g = 3.824 \times 10^{-2} \, \text{m}^3 / \text{kg} \)
02

Calculate Specific Volume of Mixture

The specific volume of the mixture, \( v \), can be found using the formula: \( v = \frac{V}{m} \).\[ v = \frac{0.05 \, \text{m}^3}{2 \, \text{kg}} = 0.025 \, \text{m}^3 / \text{kg} \]
03

Use Quality of Mixture Formula

The quality of the mixture, \( x \), is given by the formula: \[ v = v_f + x (v_g - v_f) \] Substituting the values: \[ 0.025 = 0.896 \times 10^{-3} + x (3.824 \times 10^{-2} - 0.896 \times 10^{-3}) \]
04

Solve for Quality, x

First, simplify the equation: \[ 0.025 = 0.000896 + x (0.03824 - 0.000896) \] \[ 0.025 = 0.000896 + x (0.037344) \] Now, isolate \( x \) on one side of the equation: \[ 0.025 - 0.000896 = x (0.037344) \] \[ 0.024104 = x (0.037344) \] \[ x = \frac{0.024104}{0.037344} \] \[ x \approx 0.645 \]

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Key Concepts

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

specific volume
In thermodynamics, specific volume is a fundamental property. It represents the volume occupied by a unit mass of a substance. The formula for specific volume is: \( v = \frac{V}{m} \). Here, \( V \) is the total volume and \( m \) is the mass.

Specific volume is particularly important in analyzing phases of a substance such as liquids and vapors. It helps determine the behavior of substances when they change phases.
In the given exercise, specific volume of the CO2 mixture is found to be 0.025 \( \text{m}^3/\text{kg} \). This is calculated by dividing the tank volume (0.05 \( \text{m}^3 \)) by the mass (2 kg).
saturated liquid
A saturated liquid is a liquid that is about to evaporate. At this state, the liquid has absorbed as much heat as possible without changing phase. This is known as the saturation point.

For the exercise, at -40°C, the specific volume for saturated liquid CO2 is given as \( v_f = 0.896 \times 10^{-3} \text{m}^3 / \text{kg} \). It is crucial in determining other properties of the liquid-vapor mixture.
The heat required to change the phase from liquid to vapor at this temperature causes the specific volume to increase.
saturated vapor
A saturated vapor is a vapor that is about to condense. At this stage, the vapor is in equilibrium with its liquid phase.

In the exercise, the specific volume for saturated vapor CO2 at -40°C is \( v_g = 3.824 \times 10^{-2} \text{m}^3 / \text{kg} \). This value is larger than that of the saturated liquid because vapor occupies more volume than liquid.
Knowing this property is essential for calculating the quality of the mixture and other thermodynamic properties.
two-phase mixture
A two-phase mixture contains both liquid and vapor phases in equilibrium. Understanding the quality of the mixture helps to quantify the proportion of liquid and vapor present.

The quality \( x \) is the ratio of the mass of vapor to the total mass: \( x = \frac{\text{mass of vapor}}{\text{total mass}} \).
This concept is used in the formula \( v = v_f + x (v_g - v_f) \), where the specific volume of the mixture is a function of the specific volumes of the saturated liquid and vapor, and the quality.
In the exercise, solving for quality \( x \) involves inserting values into the equation and simplifying. It shows how specific volumes relate to the amount of liquid and vapor phases in a mixture.

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Most popular questions from this chapter

Natural Refrigerants-Back to the Future (see box Sec. 3.3). Although used for home appliances in Europe, hydrocarbon refrigerants have not taken hold in the United States thus far owing to concerns about liability if there is an accident. Research hydrocarbon refrigerant safety. Write a report including at least three references.

Two kilograms of Refrigerant 22 undergo a process for which the pressure- volume relation is \(p v^{1.05}=\) constant. The initial state of the refrigerant is fixed by \(p_{1}=2\) bar, \(T_{1}=-20^{\circ} \mathrm{C}\), and the final pressure is \(p_{2}=10\) bar. Calculate the work for the process, in \(\mathrm{kJ}\).

Steam is contained in a closed rigid container with a volume of \(1 \mathrm{~m}^{3}\). Initially, the pressure and temperature of the steam are 7 bar and \(500^{\circ} \mathrm{C}\), respectively. The temperature drops as a result of heat transfer to the surroundings. Determine the temperature at which condensation first occurs, in \({ }^{\circ} \mathrm{C}\), and the fraction of the total mass that has condensed when the pressure reaches \(0.5\) bar. What is the volume, in \(\mathrm{m}^{3}\), occupied by saturated liquid at the final state?

Using the tables for water, determine the specified property data at the indicated states. Check the results using \(I T\). In each case, locate the state by hand on sketches of the \(p-v\) and \(T-v\) diagrams. (a) At \(p=3\) bar, \(T=240^{\circ} \mathrm{C}\), find \(v\) in \(\mathrm{m}^{3} / \mathrm{kg}\) and \(u\) in \(\mathrm{kJ} / \mathrm{kg}\). (b) At \(p=3\) bar, \(v=0.5 \mathrm{~m}^{3} / \mathrm{kg}\), find \(T\) in \({ }^{\circ} \mathrm{C}\) and \(u\) in \(\mathrm{kJ} / \mathrm{kg}\). (c) At \(T=400^{\circ} \mathrm{C}, p=10\) bar, find \(v\) in \(\mathrm{m}^{3} / \mathrm{kg}\) and \(h\) in \(\mathrm{kJ} / \mathrm{kg}\). (d) At \(T=320^{\circ} \mathrm{C}, v=0.03 \mathrm{~m}^{3} / \mathrm{kg}\), find \(p\) in \(\mathrm{MPa}\) and \(u\) in \(\mathrm{kJ} / \mathrm{kg}\) (e) At \(p=28 \mathrm{MPa}, T=520^{\circ} \mathrm{C}\), find \(v\) in \(\mathrm{m}^{3} / \mathrm{kg}\) and \(h\) in \(\mathrm{kJ} / \mathrm{kg}\). (f) At \(T=100^{\circ} \mathrm{C}, x=60 \%\), find \(p\) in bar and \(v\) in \(\mathrm{m}^{3} / \mathrm{kg}\). (g) At \(T=10^{\circ} \mathrm{C}, v=100 \mathrm{~m}^{3} / \mathrm{kg}\), find \(p\) in \(\mathrm{kPa}\) and \(h\) in \(\mathrm{kJ} / \mathrm{kg}\). (h) At \(p=4 \mathrm{MPa}, T=160^{\circ} \mathrm{C}\), find \(v\) in \(\mathrm{m}^{3} / \mathrm{kg}\) and \(u\) in \(\mathrm{kJ} / \mathrm{kg}\).

A closed system consists of an ideal gas with mass \(m\) and constant specific heat ratio \(k\). If kinetic and potential energy changes are negligible, (a) show that for any adiabatic process the work is $$ W=\frac{m R\left(T_{2}-T_{1}\right)}{1-k} $$ (b) show that an adiabatic polytropic process in which work is done only at a moving boundary is described by \(p V^{k}=\) constant.

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