q5

• Chris Nkinthorn, 20190625

Prompt: Estimate the centerline distance from the nozzle exit to location where stoichiometric condition is reached in a simple combustor. Find the associated speed and nominal jet width at that location.

Given: Pure methane $\ce{ CH4 }$, combusting with oxygen $\ce{O2}$. Round nozzle geometry, $d= 1\text{ cm}$, with outlet speed $U=20\text{ m/s}$, volume fraction $M_\ce{O2} $ of $21\%$.

Solution

Begin by checking the flow range, so that the form of each equation can be determined: laminar flow has analytic expressions and turbulent flow uses correlations.

$$\sqrt{\frac{J_{s}}{\rho_{a i r}} \frac{\rho_{a i r}}{\rho_{C H_{4}}}} \sim \frac{U_{o} d}{v} \rightarrow \sqrt{\frac{\rho_{C H_{4}}}{\rho_{a i r}}} \frac{U_{o} d}{v}=\sqrt{\frac{16.04}{28.94}} \frac{(20)(1 / 100)}{(1.5E-5)} \sim 10,000 \rightarrow \text { Turbulent Regime}$$

Procedure

1. Get fuel molar fraction from reaction. This expression is held at the location where stoichiometric condition is reached.

2. Calculate fuel mass fraction, Y.

3. Obtain axial location for complete combustion from step 2.

4. Obtain center jet velocity using similarity results.

5. Obtain nominal jet width using half jet spray width.

Stoichiometric Point Relation

The initial equation is molar fraction distribution, which is a restatement of the conservation of mass. This in addition to the chemical reaction, gives the molar proportion of the two reactants. At the stoichiometric point, complete reaction occurs so that the methane burns into carbon dioxide and water only.

\chi_{C H_{4}}+\chi_{O_{2}}=1,\quad \ce{ CH4 + 2O2 -> CO2 + 2H2O }

Another expression for the rate of reaction is the reaction progress equation, linking the molar fractions of the reactants methane and oxygen, together. This is then converted into a mass fraction.

$$\begin{equation} \begin{split} \chi_{C H_{4}}&=\frac{n_{C H_{4}}}{n_{C H_{4}}+n_{O_{2}}\left(1+\frac{79}{21}\right)}=\frac{1}{1+2\left(1+\frac{79}{21}\right)}=0.095\\ &= \chi_{C H_{4}}\left(\frac{M_{C H 4}}{M_{\text {air}+C H_{4}}}\right)=0.095 \frac{16.04}{\left[16.04 * 0.095+28.96^{*}(1-.095)\right]}\\ &=0.0549 \quad\underline{(5.5 \%)} \end{split} \end{equation}$$

This is then set equal to the constraint equation for species mass flux to determine axial location for complete combustion of the reactant.

\begin{equation} \begin{aligned} \overline{Y}(x, r=0) &=C_{8} Y_{o}\left(\frac{\rho_{s}}{\rho}\right)^{1 / 2}\left(\frac{r}{d}\right)^{-1}\cancelto{1}{ g\left(\frac{r}{x}\right)}=0.0549 \\ &=5 Y_{o}\left(\frac{\rho_{C H_{4}}}{\rho_{\text {air}}}\right)\left(\frac{x}{d}\right)^{-1} \\ &=5(1)\left(\frac{16.04}{28.96}\right)\left(\frac{x}{0.01}\right)^{-1} \quad \rightarrow \quad \boxed{x=0.68 \mathrm{m}} \end{aligned} \end{equation}

Jet Velocity at Stoichiometric Point

Use result of similarity analysis to calculate CL jet velocity and jet width:

\begin{equation}\begin{aligned} U_{C L} &=U(x, r=0)=C_{6} U_{o}\left(\frac{\rho_{s}}{\rho}\right)^{1 / 2}\left(\frac{x}{d}\right)^{-1} \cancelto{1}{f\left(\frac{y}{x}\right)^{\prime} }\\ &=6(20)\left(\frac{16.04}{28.96}\right)^{1 / 2}\left(\frac{0.68}{0.01}\right)^{-1} \quad \rightarrow \quad U_{C L}=\boxed{1.3 \text{ m/s}}\end{aligned}\end{equation}

Jet Width at Stoichiometric Point

Get $F(\frac{y}{x})$ profile from gaussian profile plot. Read off where 0.5 is on the y axis then find that the jet half width. Then, scale the approximate value and solve for the nominal width. From either table or plot, $1 / 2 \text { Jet width: }\left(\xi{1 / 2}\right){Y}=0.11$.

$$\frac{R_{\text { nominal }}}{x}=4\left(\xi_{1 / 2}\right)_{Y}\rightarrow R_{\text { nominal }}=4\left[\left(\xi_{1 / 2}\right)_{Y} \right]x_{\text{stoich}}=4(0.11)(0.68) \approx \boxed{0.3 \text{ m}}$$