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Chapter 3: Problem 7

The following table lists temperatures and specific volumes of water vapor at two pressures: $$ \begin{array}{lccc} {}{}{p=1.0 \mathrm{MPa}} &{}{c}{p=1.5 \mathrm{MPa}} \\ \hline T\left({ }^{\circ} \mathrm{C}\right) & v\left(\mathrm{~m}^{3} / \mathrm{kg}\right) & T(\mathrm{C}) & v\left(\mathrm{~m}^{3} / \mathrm{kg}\right) \\ \hline 200 & 0.2060 & 200 & 0.1325 \\ 240 & 0.2275 & 240 & 0.1483 \\ 280 & 0.2480 & 280 & 0.1627 \end{array} $$ Data encountered in solving problems often do not fall exactly on the grid of values provided by property tables, and linear interpolation between adjacent table entries becomes necessary. Using the data provided here, estimate (a) the specific volume at \(T=240^{\circ} \mathrm{C}, p=1.25 \mathrm{MPa}\), in \(\mathrm{m}^{3} / \mathrm{kg}\). (b) the temperature at \(p=1.5 \mathrm{MPa}, v=0.1555 \mathrm{~m}^{3} / \mathrm{kg}\), in \({ }^{\circ} \mathrm{C}\). (c) the specific volume at \(T=220^{\circ} \mathrm{C}, p=1.4 \mathrm{MPa}\), in \(\mathrm{m}^{3} / \mathrm{kg}\).

Short Answer

Expert verified
For (a): 0.1879 \mathrm{~m}^{3}/\mathrm{kg}, For (b): 265^{\circ} \mathrm{C}, For (c): 0.1665 \mathrm{~m}^{3}/\mathrm{kg}

Step by step solution

01

- Identify Known Data Points for (a)

To estimate the specific volume at \(T=240^{\circ} \mathrm{C}, p=1.25 \mathrm{MPa}\), note that this pressure lies between \(1.0 \mathrm{MPa}\) and \(1.5 \mathrm{MPa}\). Therefore, use linear interpolation between these pressures at \(T=240^{\circ} \mathrm{C}\).
02

- Linear Interpolation for Part (a)

Use the formula for linear interpolation: \[ v = v_1 + \frac{(v_2 - v_1)}{(p_2 - p_1)}(p - p_1) \] Here, \(v_1 = 0.2275 \mathrm{~m}^{3}/\mathrm{kg}\), \(v_2 = 0.1483 \mathrm{~m}^{3}/\mathrm{kg}\), \(p_1 = 1.0 \mathrm{MPa}\), \(p_2 = 1.5 \mathrm{MPa}\), and \(p = 1.25 \mathrm{MPa}\).
03

- Calculate Specific Volume for Part (a)

Substitute the values into the interpolation formula: \[ v = 0.2275 + \frac{(0.1483 - 0.2275)}{(1.5 - 1.0)}(1.25 - 1.0) \] \[ v \approx 0.1879 \mathrm{~m}^{3} / \mathrm{kg} \] So, the specific volume at \(T=240^{\circ} \mathrm{C}, p=1.25 \mathrm{MPa}\) is approximately \(0.1879 \mathrm{~m}^{3} / \mathrm{kg}\).
04

- Identify Known Data Points for (b)

To estimate the temperature at \(p=1.5 \mathrm{MPa}, v=0.1555 \mathrm{~m}^{3}/\mathrm{kg}\), note that this specific volume lies between the volumes at \(T=240^{\circ} \mathrm{C}\) and \(T=280^{\circ} \mathrm{C}\), both at \(p=1.5 \mathrm{MPa}\).
05

- Linear Interpolation for Part (b)

Use the formula for linear interpolation: \[ T = T_1 + \frac{(T_2 - T_1)}{(v_2 - v_1)}(v - v_1) \] Here, \(T_1 = 240^{\circ} \mathrm{C}\), \(T_2 = 280^{\circ} \mathrm{C}\), \(v_1 = 0.1483 \mathrm{~m}^{3}/\mathrm{kg}\), \(v_2 = 0.1627 \mathrm{~m}^{3} / \mathrm{kg}\), and \(v = 0.1555 \mathrm{~m}^{3} / \mathrm{kg}\).
06

- Calculate Temperature for Part (b)

Substitute the values into the interpolation formula: \[ T = 240 + \frac{(280 - 240)}{(0.1627 - 0.1483)}(0.1555 - 0.1483) \] \[ T \approx 265^{\circ} \mathrm{C} \] So, the temperature at \(p=1.5 \mathrm{MPa}, v=0.1555 \mathrm{~m}^{3} / \mathrm{kg}\) is approximately \(265^{\circ} \mathrm{C}\).
07

- Identify Known Data Points for (c)

To estimate the specific volume at \(T=220^{\circ} \mathrm{C}, p=1.4 \mathrm{MPa}\), note that \(T=220^{\circ} \mathrm{C}\) lies between \(T=200^{\circ} \mathrm{C}\) and \(T=240^{\circ} \mathrm{C}\). Linear interpolation can be used between these temperatures at \(p=1.4 \mathrm{MPa}\).
08

- Linear Interpolation in 2 Steps for Part (c)

First, perform interpolation for \(T=220^{\circ} \mathrm{C} \) at both \(p=1.0 \mathrm{MPa}\) and \(p=1.5 \mathrm{MPa}\) to find the intermediate volumes: \[ v_{1.0} = 0.2060 + \frac{(0.2275-0.2060)}{(240-200)}(220-200) \] \[ v_{1.0} \approx 0.21675 \mathrm{~m}^{3} / \mathrm{kg} \] Similarly, for \(p=1.5 \mathrm{MPa}\), \[ v_{1.5} = 0.1325 + \frac{(0.1483-0.1325)}{(240-200)}(220-200) \] \[ v_{1.5} \approx 0.1404 \mathrm{~m}^{3} / \mathrm{kg} \]
09

- Final Interpolation for Part (c)

Now perform interpolation at \(T=220^{\circ} \mathrm{C}\) for intermediate pressures: \[ v = v_{1.0} + \frac{(v_{1.5} - v_{1.0})}{(1.5 - 1.0)}(1.4 - 1.0) \] \[v = 0.21675 + \frac{(0.1404-0.21675)}{(1.5-1.0)}(1.4-1.0)\] \[ v \approx 0.1665 \mathrm{~m}^{3}/ \mathrm{kg} \] So, the specific volume at \(T=220^{\circ} \mathrm{C}, p=1.4 \mathrm{MPa}\) is approximately \(0.1665 \mathrm{~m}^{3} / \mathrm{kg}\).

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Key Concepts

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

Thermodynamics
Thermodynamics is the branch of physics that deals with the relationships between heat and other forms of energy. It studies how thermal energy is converted to and from other forms of energy and how it affects matter. In thermodynamics, we often use terms like pressure, temperature, and volume to describe the state of a system.
Specific Volume
Specific volume is a thermodynamic property that represents the volume occupied by a unit mass of a substance. It is defined as the inverse of density and is usually given in units of \( \text{m}^3/\text{kg} \). In the context of the provided exercise, specific volume helps us understand how much space the water vapor occupies under different pressure and temperature conditions.
Pressure-Temperature Relationship
The pressure-temperature relationship in thermodynamics is vital for understanding the behavior of substances under different conditions. For gases and vapors, changes in temperature and pressure can significantly affect their specific volumes. This relationship is often visualized using a PT-diagram, which plots pressure against temperature to show the phase changes and properties of substances like water vapor.
Property Tables
Property tables, such as steam tables, provide essential thermodynamic data for substances at various conditions of temperature and pressure. They typically include values like specific volume, enthalpy, and entropy. Engineers and scientists use these tables to find accurate data for calculations, especially when values do not fall exactly on the provided grid. Linear interpolation is a common method used to estimate values between the given points in property tables.

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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.

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A piston-cylinder assembly contains air, initially at 2 bar, \(300 \mathrm{~K}\), and a volume of \(2 \mathrm{~m}^{3}\). The air undergoes a process to a state where the pressure is 1 bar, during which the pressure-volume relationship is \(p V=\) constant. Assuming ideal gas behavior for the air, determine the mass of the air, in \(\mathrm{kg}\), and the work and heat transfer, each in \(\mathrm{kJ}\).

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