Question

The following table lists temperatures and specific volumes of water vapor at two pressures: \begin{tabular}{ccccc} \hline\(p=1.0 \mathrm{MPa}\) & & \multicolumn{2}{c}{\(p=1.5 \mathrm{Mpa}\)} \\ \cline { 1 - 2 } \hline\(\left({ }^{\circ} \mathrm{C}\right)\) & \(v\left(\mathrm{~m}^{3} / \mathrm{kg}\right)\) & & \(T\left({ }^{\circ} \mathrm{C}\right)\) & \(v\left(\mathrm{~m}^{3} / \mathrm{kg}\right)\) \\ \hline 200 & \(0.2060\) & 200 & \(0.1325\) \\ 240 & \(0.2275\) & 240 & \(0.1483\) \\ 280 & \(0.2480\) & 280 & \(0.1627\) \\ \hline \end{tabular} 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

a) 0.188 \ \mathrm{m}^3/\mathrm{kg}\, b) 257^{\circ} C\, c) 0.1676 \ \mathrm{m}^3/\mathrm{kg} \

Step-by-step solution

Linear Interpolation Formula

First, recall the linear interpolation formula to find a value between two known values: \(y = y_1 + \frac{(x - x_1)(y_2 - y_1)}{x_2 - x_1}\). The known values \((x_1, y_1)\) and \((x_2, y_2)\) are used to calculate the unknown value \(y\) at a point \(x\) between \(x_1\) and \(x_2\).

Specific Volume at T=240°C, p=1.25 MPa

Using interpolation between pressures at 1.0 MPa and 1.5 MPa for \(T = 240^{\circ} C\):\(v_{1} = 0.2275 \ m^3/kg\) at \(p_1 = 1.0\ MPa\),\(v_{2} = 0.1483 \ m^3/kg\) at \(p_2 = 1.5\ MPa\),\(p = 1.25\ MPa\).Using interpolation:\[v = v_1 + \frac{(p - p_1)(v_2 - v_1)}{p_2 - p_1}\ = 0.2275 + \frac{(1.25 - 1.0)(0.1483 - 0.2275)}{1.5 - 1.0}\ = 0.188\ \mathrm{m}^3/\mathrm{kg}\]

Temperature at p=1.5 MPa, v=0.1555 m³/kg

Interpolating between volumes at 240°C and 280°C for \(p = 1.5\ MPa\):\(T_{1} = 240^{\circ} C\), \(v_1 = 0.1483 \mathrm{m}^3/\mathrm{kg}\),\(T_{2} = 280^{\circ} C\), \(v_2 = 0.1627 \mathrm{m}^3/\mathrm{kg}\),\(v = 0.1555 \mathrm{m}^3/\mathrm{kg}\).Using interpolation:\[T = T_1 + \frac{(v - v_1)(T_2 - T_1)}{v_2 - v_1}\ = 240 + \frac{(0.1555 - 0.1483)(280 - 240)}{0.1627 - 0.1483}\ = 257^{\circ} C\]

Specific Volume at T=220°C, p=1.4 MPa

Since 220°C is between 200°C and 240°C, interpolate at 220°C for \(p = 1.4\ MPa\):First, interpolate for \(v\) at 220°C and 1.0 MPa and 1.5 MPa:For 1.0 MPa:\(T_{1} = 200^{\circ} C\), \(v_1 = 0.2060 \mathrm{m}^3/\mathrm{kg}\),\(T_{2} = 240^{\circ} C\), \(v_2 = 0.2275 \mathrm{m}^3/\mathrm{kg}\),\[v_{220,1.0} = 0.2060 + \frac{(220-200)(0.2275-0.2060)}{240-200}\ = 0.21675\mathrm{m}^3/\mathrm{kg}\]For 1.5 MPa:\(T_{1} = 200^{\circ} C\), \(v_1 = 0.1325 \mathrm{m}^3/\mathrm{kg}\),\(T_{2} = 240^{\circ} C\), \(v_2 = 0.1483 \mathrm{m}^3/\mathrm{kg}\),\[v_{220,1.5} = 0.1325 + \frac{(220-200)(0.1483-0.1325)}{240-200}\ = 0.1404\mathrm{m}^3/\mathrm{kg}\]Finally Interpolate for 1.4 MPa:\(v_{220, 1.0} = 0.21675\ \mathrm{m}^3/\mathrm{kg}\),\(v_{220, 1.5} = 0.1404 \mathrm{m}^3/\mathrm{kg}\),\[v = 0.21675 + \frac{(1.4-1.0)(0.1404-0.21675)}{1.5-1.0}\ = 0.1676 \ \mathrm{m}^3/ \mathrm{kg}\]

Key concepts

Specific Volume

Specific volume refers to the volume occupied by a unit mass of a substance. It is a crucial property in thermodynamics as it helps define the state of a material. In the SI system, the specific volume is usually measured in \(\text{m}^3/\text{kg}\). It is the inverse of density. For instance, in the provided exercise, we look at how the specific volume of water vapor changes with temperature and pressure.
By using a concept called linear interpolation, we can estimate specific volumes at temperatures and pressures not directly listed in property tables. Linear interpolation assumes that the change between two known points is linear, which simplifies calculations. This method is particularly useful in thermodynamics for estimating properties between tabulated data points.
For example, to find the specific volume at a temperature of 240°C and a pressure of 1.25 MPa, we use the formula: \[ v = v_1 + \frac{(p - p_1)(v_2 - v_1)}{p_2 - p_1} \] Here, \(v_1\) and \(v_2\) are the known specific volumes at pressures \(p_1\) and \(p_2\), respectively.

Thermodynamic Properties

Thermodynamic properties are characteristics of a system that are used to describe its state and predict its behavior. Common thermodynamic properties include temperature, pressure, and specific volume. These properties are interrelated and depend on the system’s state.
Understanding these properties allows us to predict how a system will react to changes. For instance, when solving the given exercise, knowing the specific volumes at various temperatures and pressures helps determine the missing values through interpolation.
Thermodynamic tables summarize these properties and provide a convenient reference for engineers and scientists. However, since exact values may not always be available, interpolation is a valuable tool. This practice is common in solving many thermodynamics problems, ensuring that calculated values are within a realistic and reliable range.

Pressure-Temperature Relationship

The pressure-temperature (P-T) relationship is critical in thermodynamics. It describes how pressure and temperature affect each other and the material's properties. When the temperature increases at constant pressure, the specific volume often increases, indicating a direct relationship.
The exercise examines this relationship by providing specific volume data at different pressures and temperatures. By analyzing this data, students can use interpolation to estimate values at intermediate conditions, which is crucial for real-world applications.
For example, consider estimating the temperature at a specific volume of 0.1555 \(\text{m}^3/\text{kg}\) when the pressure is 1.5 MPa. Here, the properties at known temperatures help us interpolate, giving us a temperature where the specific volume meets the given value under the specified pressure. The formula used will be:
\[ T = T_1 + \frac{(v - v_1)(T_2 - T_1)}{v_2 - v_1} \] Understanding these kinds of relationships and the ability to estimate properties accurately is foundational in thermodynamics and engineering.