1
ME 525
SP2011
Homework #3
Questions and Answers
Due Thursday, February 17, 2011
Prof. Lucht (Email address:
[email protected])
Question:
can you put a .xls version of the spreadsheet on the website. i can not read .xlsx files
from my computer
thank you
Answer:
I have put .xls and .csv versions on the website.
Question:
On part a.
I invoke the partial equilibrium assumption to get
k_1f[H][O_2]=k_1r[O][OH] and k_2f[OH][H_2]=k_2r[H][H2O]. I calculate the rate
coefficients using A,b,E_a(converting E_a to J/gmol) and calculate the unknown rate coefficients
using K_eq via the given curve fit for gibbs free energies.
When I use the given concentrations
from the CHEMKIN results at the different times, I am unable to prove that the partial
equilibrium holds for either time.
Below are my values for the rate coefficients:
k_1f=1.992e10 cm^3/gmols
k_1r=1.203e13
k_2f=1.066e12
k_2r=1.115e9
Am I trying to prove the wrong assumption from above?
Thanks again.
Answer:
Your rate coefficients look close but they do not match mine at either the 1.90 ms
condition (T=935.1 K) or the 1.98 ms condition (T=1818.7 K).
What temperature did you use in
your calculations?
Question:
I have a problem understanding the meaning of the term "evaluate the applicability
of the partial equilibrium assumption ". I found K1 and K2, I took down all the mole fractions at
time 1.95ms, for 1.98ms I should interpolate right? How do I check the applicability of the
partial eq. assumption?
Answer:
The partial equilibrium assumption holds when the forward reaction rate equals the
backwards reactions rate.
For reaction 1, partial equilibrium holds when
k
1f
[H][O2] = k
1r
[O][OH]
There is no need to interpolate, t = 1.98 ms = 1.98e03 s is a table entry.
Question:
When I calculate [N]_ss via k_1[O][N_2]k_2[N][O_2]k_3[OH][N]=0, do I use the
initial concentrations and conditions to calcualte [N]_ss.
It seems like [N]_ss would be a
constant and wouldn't change with time so it would be the same for t=0, 1.98 or 3ms.
Thanks for
any help.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document2
Answer:
The term steady state is always confusing.
The steady state concentration [N]_ss
varies drastically with time in this flame.
Steady state for a radical means that its concentration
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '11
 Lucth
 Thermodynamics, Chemical reaction, Equals sign, omega dot

Click to edit the document details