Using the Steam Tables to Show That the Stability Conditions Are Satisfied for Steam
Show that Eqs. 7.2-12 and 7.2-13 are satisfied by superheated steam.
C_{\mathrm{V}}>0 (7.2.12)
\left(\frac{\partial P}{\partial \underline{V}}\right)_{T}<0 \quad \text { or } \quad \kappa_{T}>0 (7.2.13)
It is easiest to use Eq. 7.2-13 in the form (\partial P / \partial \underline{V})_{T}<0, which requires that the volume decrease as the pressure increases at constant temperature. This is seen to be true by using the superheated steam table and observing that \hat{V} decreases as P increases at fixed temperature. For example, at 1000°C
| P (MPa) | 0.50 | 0.8 | 1 | 1.4 | 1.8 | 2.5 |
| \hat{V}\left(\mathrm{~m}^{3} / \mathrm{kg}\right) | 1.1747 | 0.7340 | 0.5871 | 0.4192 | 0.3260 | 0.2346 |
Proving that C_{\mathrm{V}}>0 or \hat{C}_{\mathrm{V}}>0 is a bit more difficult since
\hat{C}_{V}=\left(\frac{\partial \hat{U}}{\partial T}\right)_{V}
and the internal energy is not given at constant volume. Therefore, interpolation methods must be used. As an example of how the calculation is done, we start with the following data from the superheated vapor portion of the steam tables.
| P = 1.80 Mpa | P = 2.00 Mpa | |||
| T (°C) | \hat{V}\left(\mathrm{~m}^{3} / \mathrm{kg}\right) | \hat{U}(\mathrm{~kJ} / \mathrm{kg}) | \hat{V}\left(\mathrm{~m}^{3} / \mathrm{kg}\right) | \hat{U}(\mathrm{~kJ} / \mathrm{kg}) |
| 800 | 0.2742 | 3657.6 | 0.2467 | 3657 |
| 900 | 0.3001 | 3849.9 | 0.2700 | 3849.3 |
| 1000 | 0.3260 | 4048.5 | 0.2933 | 4048.0 |
To proceed, we need values of the internal energy at two different temperatures and the same specific volume. We will use P = 2.00 MPa and T = 1000°C as one point; at these conditions \hat{V}=0.2933 \mathrm{~m}^{3} / \mathrm{kg} and \hat{U}=4048.0 \mathrm{~kJ} / \mathrm{kg}. We now need to find the temperature at which \hat{V}=0.2933 \mathrm{~m}^{3} / \mathrm{kg} at P=1.80 \mathrm{MPa}. We use linear interpolation for this:
\frac{T-800}{900-800}=\frac{\hat{V}(T, 1.80 \mathrm{MPa})-\hat{V}\left(800^{\circ} \mathrm{C}, 1.80 \mathrm{MPa}\right)}{\hat{V}\left(900^{\circ} \mathrm{C}, 1.80 \mathrm{MPa}\right)-\hat{V}\left(800^{\circ} \mathrm{C}, 1.80 \mathrm{MPa}\right)}=\frac{0.2933-0.2742}{0.3001-0.2742}
so that T = 873.75°C. Next we need the internal energy \hat{U} at T = 873.75°C and P = 1.80 MPa (since at these conditions \hat{V}=0.2933 \mathrm{~m}^{3} / \mathrm{kg} ).
Again using linear interpolation,
\begin{aligned}\frac{873.75-800}{900-800} &=\frac{\hat{U}\left(873.75^{\circ} \mathrm{C}, 1.80 \mathrm{MPa}\right)-\hat{U}\left(800^{\circ} \mathrm{C}, 1.80 \mathrm{MPa}\right)}{\hat{U}\left(900^{\circ} \mathrm{C}, 1.80 \mathrm{MPa}\right)-\hat{U}\left(800^{\circ} \mathrm{C}, 1.80 \mathrm{MPa}\right)} \\\\&=\frac{\hat{U}\left(873.75^{\circ} \mathrm{C}, 1.80 \mathrm{MPa}\right)-3657.6}{3849.9-3657.6}\end{aligned}
we find that
\hat{U}\left(873.75^{\circ} \mathrm{C}, 1.80 \mathrm{MPa}\right)=\hat{U}\left(873.75^{\circ} \mathrm{C}, 0.2933 \mathrm{~m}^{3} / \mathrm{kg}\right)=3799.4 \mathrm{~kJ} / \mathrm{kg}
Finally, replacing the derivative by a finite difference, and for the average temperature [i.e., T = (1000+873.75) / 2=936.9°C], we have
\begin{aligned}\hat{C}_{\mathrm{V}}\left(T=936.9^{\circ} \mathrm{C}, \hat{V}=0.2933 \mathrm{~m}^{3} / \mathrm{kg}\right) \\\\& \approx \frac{\hat{U}\left(1000^{\circ} \mathrm{C}, 0.2933 \mathrm{~m}^{3} / \mathrm{kg}\right)-\hat{U}\left(873.75^{\circ} \mathrm{C}, 0.2933 \mathrm{~m}^{3} / \mathrm{kg}\right)}{1000^{\circ} \mathrm{C}-873.75^{\circ} \mathrm{C}} \\\\&=\frac{4048.0-3799.4}{1000-873.75}=1.969 \frac{\mathrm{kJ}}{\mathrm{kg} \mathrm{K}}>0\end{aligned}
Similarly, we would find that \hat{C}_{\mathrm{V}}>0 at all other conditions.