Question 2.12: Using data from the steam tables in Appendix A, estimate the...

Using data from the steam tables in Appendix A, estimate the constant pressure heat capacity of superheated steam at 350 kPa, 200°C and at 700 kPa, 200°C. Are the answers very different from each other? (Note: You may need to use the “limit” definition of the derivative to help you get started.)

The Blue Check Mark means that this solution has been answered and checked by an expert. This guarantees that the final answer is accurate.
Learn more on how we answer questions.

There is no entry in the steam tables at 350 kPa (3.5 bar). Interpolation between the data at 3 and 4 bar, and 200 ˚C, can be used to find the specific enthalpy at 3.5 bar and 200 ˚C. But what we actually seek is the change in specific enthalpy, so we will instead look at the temperatures above and below 200 ˚C.

Steam at 350 kPa and 150°C →Interpolation Needed Interpolate

\begin{array}{|c|c|c|c|} \hline & 1 & & 2\\\hline \text{Enthalpy (y) }&2761.2 & ??? & 2752.8\\\hline \text{Pressure (x) }& 3 & 3.5 & 4\\\hline\end{array}

\begin{gathered}y=\frac{\left(y_{2}-y_{1}\right)\left(x-x_{1}\right)}{\left(x_{2}-x_{1}\right)}+y_{1}\\\mathrm{y}=\frac{(2752.8-2761.2)(3.5-3)}{(4-3)}+2761.2=2757.0 \frac{\mathrm{kJ}}{\mathrm{kg}}\end{gathered}

Steam at 350 kPa and 250°C → Interpolation Needed Interpolate

\begin{array}{|c|c|c|c|}\hline & 1 & & 2\\\hline \text{Enthalpy (y) }&2967.9 & ??? & 2964.5\\\hline \text{Pressure (x) }& 3 & 3.5 & 4\\\hline\end{array}

\begin{gathered}y=\frac{\left(y_{2}-y_{1}\right)\left(x-x_{1}\right)}{\left(x_{2}-x_{1}\right)}+y_{1}\\\mathrm{y}=\frac{(2964.5-2967.9)(3.5-3)}{(4-3)}+2967.9=2966.2 \frac{\mathrm{kJ}}{\mathrm{kg}}\end{gathered}

Constant pressure heat capacity \left(\mathrm{C}_{\mathrm{P}}\right) can be estimated as

\mathrm{C}_{\mathrm{P}} \approx \frac{\Delta \mathrm{H}}{\Delta \mathrm{T}}=\frac{2966.2 \frac{\mathrm{kJ}}{\mathrm{kg}}-2757.0 \frac{\mathrm{kJ}}{\mathrm{kg}}}{250^{\circ} \mathrm{C}-150^{\circ} \mathrm{C}}=2.092 \frac{\mathrm{kJ}}{\mathrm{kg}^{\circ} \mathrm{C}}

If we attempt to apply a directly analogous procedure at 700 kPa (7 bar) there is a complication:

Water at 700 kPa and 150°C →Liquid
Since these two intensive properties do not yield a vapor, we must use the enthalpy of saturated vapor at 700 kPa. This temperature is 164.9°C

Steam at 700 kPa and 164.9°C → 2762.8 \frac{\mathrm{kJ}}{\mathrm{kg}}

Steam at 700 kPa and 250°C → Interpolation Needed
Interpolate

\begin{array}{|c|c|c|c|} \hline & 1 & & 2\\\hline \text{Enthalpy (y) }&2957.6 & ??? & 2950.4\\\hline \text{Pressure (x) }& 6 & 7 & 8\\\hline\end{array}

\begin{gathered}y=\frac{\left(y_{2}-y_{1}\right)\left(x-x_{1}\right)}{\left(x_{2}-x_{1}\right)}+y_{1}\\\mathrm{y}=\frac{(2950.4-2957.6)(7-6)}{(8-6)}+2957.6=2954.0 \frac{\mathrm{kJ}}{\mathrm{kg}}\\\mathrm{C}_{\mathrm{P}} \approx \frac{\Delta \mathrm{H}}{\Delta \mathrm{T}}=\frac{2954.0 \frac{\mathrm{kJ}}{\mathrm{kg}}-2762.8 \frac{\mathrm{kJ}}{\mathrm{kg}}}{250^{\circ} \mathrm{C}-164.9^{\circ} \mathrm{C}}=2.247 \frac{\mathrm{kJ}}{\mathrm{kg}^{\circ} \mathrm{C}} \end{gathered}

These answers are fairly similar, but different enough to demonstrate that \mathrm{C}_{\mathrm{P}} can in reality be a function of pressure as well as temperature. This is why the distinction between “ideal gas heat capacity” and simply “heat capacity” is important; ideal gas heat capacity is only a function of temperature.

Conceptually, we could get more accurate estimates of C_{P} by using values of \widehat{H} that were closer to the actual temperature of interest; e.g., 190 and 210°C instead of 150 and 250°C. But that data is not included in the available version of the steam table.

Related Answered Questions