Question 4 of 8: Adiabatic Mixing of Steam and Syngas

Question 4: Adiabatic Mixing of Steam and Syngas (Part B — 30%)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A steam stream (\(\dot n_1 = 8.3\times10^{6}\) mol/h, \(H_1=63.501\) kJ/mol) mixes adiabatically with a syngas stream (\(\dot n_2\), \(H_2=18.344\) kJ/mol); the combined stream leaves at \(\dot n_3\) with \(H_3=41.698\) kJ/mol. The mixer is adiabatic with no shaft work.

Find. \(\dot n_2\) (syngas) and \(\dot n_3\) (mixture).

Approach

Two unknowns (\(\dot n_2,\dot n_3\)) require two equations: a mole balance and an energy balance. The adiabatic, work-free mixer conserves total enthalpy, giving a single equation in \(\dot n_2\) once the mole balance eliminates \(\dot n_3\).

1. Mole and energy balances. $$\dot n_3 = \dot n_1 + \dot n_2,\qquad \dot n_1 H_1 + \dot n_2 H_2 = \dot n_3 H_3.$$
2. Eliminate \(\dot n_3\). Substituting the mole balance, $$\dot n_1 H_1 + \dot n_2 H_2 = (\dot n_1+\dot n_2)H_3 \;\Rightarrow\; \dot n_1(H_1-H_3) = \dot n_2(H_3-H_2).$$
3. Solve for the syngas flow. $$\dot n_2 = \dot n_1\,\frac{H_1-H_3}{H_3-H_2} = 8.3\times10^{6}\,\frac{63.501-41.698}{41.698-18.344} = 8.3\times10^{6}(0.9336).$$

   \(\dot n_2 \approx 7.75\times10^{6}\) mol syngas/h.

4. Total mixture flow. $$\dot n_3 = 8.3\times10^{6} + 7.749\times10^{6} = 1.605\times10^{7}\ \mathrm{mol/h}.$$

   \(\dot n_3 \approx 1.605\times10^{7}\) mol/h.