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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

An open container of pure ethanol (ethyl alcohol) liquid is placed on a table in a room. Evaporation occurs until all of the ethanol is gone. Where did the ethanol go? If the ethanol and the room air are taken to be a closed system, can the system be regarded as a pure substance during the process? How many phases are present initially and finally? Explain. Using \(p-v-T\) Data

A piston-cylinder assembly contains propane, initially at \(27^{\circ} \mathrm{C}, 1\) bar, and a volume of \(0.2 \mathrm{~m}^{3}\). The propane undergoes a process to a final pressure of 4 bar, during which the pressure- volume relationship is \(p V^{\mathrm{L} .1}=\) constant. For the propane, evaluate the work and heat transfer, each in kJ. Kinetic and potential energy effects can be ignored.

A closed, rigid tank fitted with a paddle wheel contains \(0.1 \mathrm{~kg}\) of air, initially at \(300 \mathrm{~K}, 0.1 \mathrm{MPa}\). The paddle wheel stirs the air for 20 minutes, with the power input varying with time according to \(\dot{W}=-10 t\), where \(\dot{W}\) is in watts and \(t\) is time, in minutes. The final temperature of the air is \(1060 \mathrm{~K}\). Assuming ideal gas behavior and no change in kinetic or potential energy, determine for the air (a) the final pressure, in MPa, (b) the work, in kJ, and (c) the heat transfer, in kJ.

A piston-cylinder assembly contains water, initially saturated liquid at \(150^{\circ} \mathrm{C}\). The water is heated at constant temperature to saturated vapor. (a) If the rate of heat transfer to the water is \(2.28 \mathrm{~kW}\), determine the rate at which work is done by the water on the piston, in \(\mathrm{kW}\). (b) If in addition to the heat transfer rate given in part (a) the total mass of water is \(0.1 \mathrm{~kg}\), determine the time, in \(\mathrm{s}\), required to execute the process.

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