Combustion stoichiometry¶

• Combustion occurs between Fuel and Air (or oxidizer)
• Need to get the mixing ratio / stoichiometry right.

Air properties¶

• Use $x$ for mole fraction, and $y$ for mass fraction.

species	x	y
N$_2$	0.7809	0.75532
O$_2$	0.2095	0.23144
Ar	0.0096	0.01324

• Lump Ar with N$_2$:

species	x	y
N$_2$	0.79	0.77
O$_2$	0.21	0.23

• n$_{N2}$/n$_{O2}$ = 3.76
• n$_{air}$/n$_{O2}$ = 4.76

Mean molecular weight¶

• M = 28.96 kg/kmol
• M = 29 kg/kmol

$$ (M) = \sum_k x_k (M_k),$$ $$ \left(\frac{1}{M}\right) = \sum_ky_k\left(\frac{1}{M_k}\right).$$ $$ \phantom{xxxxxx}\rightarrow M = \frac{1}{\sum_k \frac{y_k}{M_k}}.$$

• Intuitively, for $M$ in terms of $x_k$
  • $(M)$ has units of kg/kmol.
  • For a basis of 1 total kmol, the $x_k$ are the kmol of each species.
  • Then $x_kM_k$ is the kg of each species.
  • Sum these up and we have the total kg, with 1 kmol assumed, so the sum is M.
• Similarly, for $M$ in terms of $y_k$
  • Work with (1/M) with units of kmol/kg
  • For a basis of 1 total kg, the $y_k$ are the kg of each species.
  • Then $y_k(1/M_k)$ is the kmol of each species.
  • Sum these up and we have the total kmol, with 1 kg assumed, so the sum is (1/M).

Convert $x_k$ to $y_k$¶

$$ x_kM_k = y_kM.$$

To remember this, we have a form that looks like $xM=yM$. $x$ comes before $y$ in the alphabet. Then group the $k$ subscripts together on the first three terms.

Solve for $x_k$ and $y_k$ as needed: $$ y_k = \frac{x_kM_k}{M},$$ $$ x_k = \frac{y_kM}{M_k}.$$

If you want to derive the equation. Start with $y_k$, say, and get $x_k$:

• $y_k$ is $m_k/m_t$, ($t$ for total).
• Then convert this to $x_k$, which is $n_k/n_t$.
• Change the $m_k$ to $n_k$ by dividing by $M_k=m_k/n_k$:
  • $y_k/M_k = \frac{m_k}{m_t}\frac{n_k}{m_k} = \frac{n_k}{m_t}$
• Similarly, change the $m_t$ to $n_t$ by multiplying by $M=m_t/n_t$.
• The result is $x_k = y_k\frac{M}{M_k}$.
• Then can solve for $y_k$ to get $y_k = x_k\frac{M_k}{M}$.

Stoichiometry¶

• Rich (too much fuel)
• Lean (too much air)
• Stoichiometric (just right)

Air-to-fuel ratio¶

$$ \frac{A}{F} = \frac{m_{air}}{m_{fuel}}.$$

• Mass Basis
• $0 \le A/F \le \infty$

Equivalence ratio¶

$$\Phi = \frac{F/A}{(F/A)_{stoic}}$$

• Mass or mole basis (same).
• $0 \le \Phi \le \infty$
• $\Phi=0$ is lean, $\Phi>1$ is rich.
• $\Phi$ is common in practical applications.

Mixture fraction¶

$$\xi = \frac{m_f}{m_f + m_a}$$

• Notation, $\xi$, or $f$, or $Z$ is common.
• $0\le \xi\le 1$
• $\xi=0$ is pure air, $\xi=1$ is pure fuel.
• $\xi$ is common in modeling.
• $\xi$ can be defined in terms of any two distinct streams, not just fuel and air.

Ideal gas law¶

• $PV = nRT$
• Solve for $n$ and divide by $V$, with $n/V = c$, that is molar concentration:
  • $c = P/RT$
• Multiply through by $M$, with $\rho=cM$, that is, mass density:
  • $\rho = MP/RT$
• Here, $R$ is the gas constant, $R=8314.46$ J/kmol$\cdot$K.

Balanced reactions¶

Methane combustion¶

• Normally write 1 mole of fuel for convenience.
• 1 mole O$_2$ gives 4.75 moles Air = O$_2$ + 3.76 N$_2$.
• Mole basis
• Products
  • Carbon $\rightarrow$ CO$_2$
  • Hydrogen $\rightarrow$ H$_2$O

$$(CH_4) + \alpha (air) \rightarrow b(products)$$ $$CH_4 + \alpha O_2 + 3.76\alpha N_2 \rightarrow \beta CO_2 + \gamma H_2O + 3.76\alpha N_2$$

• Solve for $\alpha$, $\beta$, $\gamma$, using elemental balnces on $C$, $H$, and $O$.
  • C $\rightarrow$ $1 = \beta$ $\rightarrow$ $\beta = 1$
  • H $\rightarrow$ $4 = 2\gamma$ $\rightarrow$ $\gamma = 2$
  • O $\rightarrow$ $2\alpha = 2\beta + \gamma$ $\rightarrow$ $\alpha = 2$.

$$CH_4 + 2O_2 + 7.52N_2 \rightarrow CO_2 + 2H_2O + 7.52N_2$$

• Note, no change in moles for methane combustion.
• Note the high $m_f/m_a = M_{CH4}/(9.52M_{air}) = 16/(9.52*29) = 0.058$
• Note: A/F = 17.2.

Generalize the fuel¶

$$C_xH_y + \alpha O_2 + 3.76\alpha N_2 \rightarrow \beta CO_2 + \gamma H_2O + 3.76\alpha N_2$$

• (x and y are number of carbons and hydrogens here, not mole and mass fractions).
• C, H balances $\rightarrow$ $\beta = x$, $\gamma = y/2$.
• O balance $\rightarrow$ $2\alpha = 2\beta + \gamma$ $\rightarrow$ $\alpha = x + y/4$

$$C_xH_y + \left(x+\frac{y}{4}\right)O_2 + 3.76\left(x+\frac{y}{4}\right)N_2 \rightarrow xCO_2 + \frac{y}{2}H_2O + 3.76\left(x+\frac{y}{4}\right)N_2$$

• When is there no change in moles?

  • Reactants: $1 + \alpha + 3.76\alpha$
  • Products: $x + y/2 + 3.76\alpha$.
  • $\rightarrow$ (1+x + y/4 = x + y/2).
  • $\rightarrow$ $y=4$
  • No change in moles for species with 4 hydrogens: methane, ethylene, etc.

• Ethylene is special

  • $y=$ so no change in moles.
  • $M_{C2H4} = 28\approx M_{air}$, so $\rho_{air}\approx\rho_{fuel}$
    • $\rho = MP/RT$

Lean combustion¶

• Too much air (the usual case).

• Assume a fixed quantity of fuel, $F$.

• Fraction excess air, use subscript $s$ for stoichiometric: $$E = \frac{A - A_s}{A_s},$$

• Divide the top and bottom by $F$: $$E = \frac{A/F - A_s/F}{A_s/F} = \frac{A/F}{A_s/F} - 1.$$

• At stoichiometric conditions, $A_s$ is stoichiomtric amount of air for the given amount of fuel, and we have $A_s/F = (A/F)_s$

$$E = \frac{A/F}{(A/F)_s} - 1 = \frac{1}{\Phi} - 1.$$

Write the balanced reaction with excess air:

$$C_xH_y + (1+E)\left(x+\frac{y}{4}\right)(O_2 + 3.76 N_2) \rightarrow xCO_2 + \frac{y}{2}H_2O + E\left(x+\frac{y}{4}\right)O_2 + (1+E)3.76\left(x+\frac{y}{4}\right)N_2$$

Rich combustion¶

• Too much fuel.
• This case is easy.
  • If the stoichiometric air is taken as an air basis, then $\Phi=F/F_s$.
  • Or, $F = \Phi F_s = \Phi$ when we use $F_s=1$ as in our generic balance equation for $C_xH_y$. $$\Phi C_xH_y + \left(x+\frac{y}{4}\right)O_2 + 3.76\left(x+\frac{y}{4}\right)N_2 \rightarrow xCO_2 + \frac{y}{2}H_2O + 3.76\left(x+\frac{y}{4}\right)N_2 + (\Phi-1)C_xH_y$$

Example¶

• Natural gas boiler fired with 20% excess air.

$$CH_4 + (1.2)\cdot(2 O_2 + 7.52 N_2) \rightarrow CO_2 + 2H_2O + (1.2)\cdot 7.52 N_2 + (0.2)\cdot 2 O_2$$

• Typical boilers run with excess air to compensate for imperfect mixing of fuel and air to prevent emissions of fuel or partially burnt species, such as CO.

• With 20% excess air, we get 3% O$_2$ (by mole) in the flue gas.

Typical systems

• Natural Gas: 10-20% Excess air

• Oil: 10-20

• Coal: 20-25

• Stoker: 35-40

• Often, the O$_2$ in the flue gas is given on a dry basis. See Turns Chapter 15.