Question

A cryogenic liquid rocket engine (expander cycle) burns liquid hydrogen and liquid oxygen at stoichiometry. The hydrogen mass-flow rate is $\dot m_{H_2}=32\,\text{kg/s}$, and the oxygen mass-flow rate satisfies $\dot m_{O_2}/\dot m_{H_2}=8$. Assuming the forward reaction dominates, find the rate of formation of $\mathrm{H_2O}$ (kmol/s). (round off to the nearest integer)

Hint:

Convert mass flow to molar flow using molar masses, check the limiting reactant via stoichiometry, and map directly to product molar rate.

Answer 1

Step 1: Compute molar feed rates.
\[ \dot m_{O_2}=8\times 32=256\ \text{kg/s}. \] Molar masses: $M_{H_2}=2\,\text{kg/kmol}$, $M_{O_2}=32\,\text{kg/kmol}$. \[ \dot n_{H_2}=\frac{32}{2}=16\ \text{kmol/s}, \dot n_{O_2}=\frac{256}{32}=8\ \text{kmol/s}. \]

Step 2: Stoichiometry.
\[ 2\,\mathrm{H_2}+ \mathrm{O_2}\ \rightarrow\ 2\,\mathrm{H_2O}. \] The given feeds satisfy \(2:\!1\) exactly (since $16:8=2:1$), so both react completely.

Step 3: Product formation rate.
Per 1 kmol of $\mathrm{O_2}$, $2$ kmol of $\mathrm{H_2O}$ form. With $8$ kmol/s of $\mathrm{O_2}$, \[ \dot n_{\mathrm{H_2O}}=2\times 8=16\ \text{kmol/s}. \] \[ \boxed{16\ \text{kmol/s}} \]