ME 475/675 Introduction to

CombustionLecture 19

Chemical time scales, Example, Partial Equilibrium, Example

Announcements• HW 7, Due Monday, 10/12/2015• This is a long homework.• We are finishing the material this lecture so start it now.

• Midterm Solutions posted outside PE 215

Chemical Time Scales• Which reactions are “fast” (reach steady state or partial equilibrium) compared to

other?• How long does it take for a reactant with a smaller initial amount to significantly

decrease?• Uni-molecular Reaction• ; (1st order separable differential equation)

• Units:

• Assume temperature T changes slowly, so that • ; • When (chemical time scale)

• Only 37% of the initial amount of A remains

Bi-molecular Reaction• ;

• Assume A has smaller initial amount: at • But and are related (decrease from initial concentrations by the same amount), so

• and

• Where is a positive constant (initially more B than A)

• (1st order separable)• (partial fractions) • •

Bi-molecular Reaction Chemical Time

• At (find this) • and

•

• For , , and

Ter-molecular• ; note is constant

• For , , and

• Time scale increases as increases, but only by 67%

• To find , need , , ,

0 0.2 0.4 0.6 0.81

1.2

1.4

1.6

1.717

1.0

Tr r( )

10 r[ 𝐴 ]0[𝐵 ]0

𝜏 h𝑐 𝑒𝑚 ,𝑡

𝜏 h𝑐 𝑒𝑚 ,𝑡 ,0

Example 4.5page 131

Example 4.5page 131

• Work in Excel, conclude:• Time scale t decreases at temperature T increases• Ter-molecular reaction is slow

T [K] 1344.3 2199.2P [atm] 1 1

Xi Xi [i] [i]CH4 2.01E-04 3.77E-06 1.82E-09 2.09E-11N2 0.7125 0.7077 6.46E-06 3.92E-06CO 4.08E-03 1.11E-02 3.70E-08 6.14E-08OH 1.82E-04 3.68E-03 1.65E-09 2.04E-08H 1.42E-08 6.63E-04 1.29E-13 3.68E-09CH 2.08E-09 9.15E-09 1.89E-14 5.07E-14H2O 1.86E-01 1.82E-01 1.69E-06 1.01E-06M 8.99E-01 8.89E-01 8.15E-06 4.93E-06

k k t [sec] t [sec] t [ms] t [ms]i 3.15053E+12 1.09039E+13 2.77E-04 4.50E-06 0.2765 0.0045 0.045183 0.001917ii 3.26803E+11 6.04841E+11 8.41E-05 3.08E-05 0.0841 0.0308 0.013738 0.013109iii 380905484.3 12745197557 4.07E-04 2.00E-05 0.4066 0.0200 0.066431 0.00852iv 1.21739E+16 4.54876E+15 6.12E-03 2.35E-03 6.1203 2.3489 1 1

Partial Equilibrium

• Some reactions of a mechanism chain are much faster in both directions (forward and reverse) than others (as described in Example 4.3)•Usually chain propagating (or branching) reactions are bi-

molecular and faster than ter-molecular recombination reactions • Treat fast reactions as if they are equilibrated• This allows them to be treated using algebraic equations and

reduces the number of differential equations that must be solved.

Example: Shuffle Reaction (Stable , , ) • 1f• 1r

• 2f• 2r

• 3f• 3r

• 4f (ter-molecular recombination) • , need and = fn(, , )

• Assume reactions 1, 2 and 3 are all in partial equilibrium• Species , , , N=7

• Stable: , , • Intermediaries:

Find intermediary concentrations ()

• ; 1

• ; 2

• ; 3

• Goal: find , , and = fn(, )• From 3: 4

• Plug into 1: ; 5

• Plug 4 and 5 into 2: • 6

• Plug 6 into 4:

Problem 4.20 page 148• In the combustion of hydrogen, the following reactions involving radicals are fast

in both the forward and reverse directions:

• Use the assumption of partial equilibrium to derive algebraic expression for the molar concentrations of the three radical species, , and , in terms of the kinetic rate coefficients and the molar concentrations of reactant and product species , , and .