Question 14.12: Use of van’t Hoff Equation (a) Use van’t Hoff equation to es...

Use of van’t Hoff equation to obtain enthalpy of combustion. Use the open-steady combustion TESTcalc to verify.

Assumptions
The IG mixture model is applicable.

Analysis
In Table G-3, K, for the oxidation of CO, is listed at 1800 K and 2200 K. We use these values to estimate the heat of combustion at T = 2000 K from Eq. (14.57):

\ln \frac{K_{2}}{K_{1}} \cong \frac{-\Delta \bar{h}_{C}^{ o }}{\bar{R}}\left(\frac{1}{T_{2}}-\frac{1}{T_{1}}\right)                         (14.57)

\begin{aligned}\ln \frac{K_{2}}{K_{1}} & \cong \frac{\Delta \bar{h}_{T}^{ o }(T)}{\bar{R}}\left(\frac{1}{T_{2}}-\frac{1}{T_{1}}\right) \\\Rightarrow \quad \ln \frac{e^{-5.12}}{e^{-8.497}} & \cong \frac{\Delta \bar{h}_{T}^{ o }(T)}{8.314}\left(\frac{1}{2200}-\frac{1}{1800}\right) \\\Rightarrow \quad \Delta \bar{h}_{T}^{ o }(T) & \cong-277,956 kJ / kmol\end{aligned}

Launch the IG open-steady, non-premixed combustion TESTcalc and leave the Mole radio button selected. To set up the reaction, select CO in the fuel block and choose Pure Oxygen from the action menu. Evaluate the fuel, oxidizer, and product states, each at 2000 K. In the device panel, load the states, enter Wdot_ext=0 and calculate Qdot as -277,820 kW; therefore, the heat of combustion is -277,820 kJ/kmol.

Discussion
High accuracy from the van’t Hoff equation is possible when the heat of combustion does not vary significantly over the temperature range. Using the Super-Calculate button, it is a simple task to show that the enthalpy of combustion changes from -278,852 kJ/kmol at 1800 K to -276,785 kJ/kmol at 2200 K (after changing T1, click Calculate and then Super-Calculate). In percentage terms, this change is less than 0.75%.