# EXAM MIE516

Question 2

Approach: Integrate Equations (6.10) & (6.15) of the text book using Euler method to find Fuel concentration and reactor temperature as a function of time. Euler method is an explicit Numerical scheme, meaning that the unknown values in each time‐step (temperature and species) are related to the known parameters in the previous time‐step. First let’s find the conservation equations. Stoichiometric methane combustion reaction is: CH4+2(O2+3.76N2) → CO2+2H2O+7.52N2

&

&

ω O 2 = 2ω F

&

ωN 2 = 0

&

&

ω CO 2 = −ω F

&

&

ω H 2O = −2ω F

∴

&

∑ω

=0

Re ac tan ts & Pr oducts

We will have one species conservation equation (IC.6.15) for each of CH4, O2, N2, CO2 and H2O concentrations and one equation for temperature. All of these equations use fuel consumption rate which is in the following form (from table 5.1): ⎛ mole ⎞

&

{ω F }n = { .3E8 × exp(−24358 / T ) × [CH 4]−0.3 [O 2]1.3 }n ⎜ 1

⎟

⎝ cc. sec ⎠

Where “n” indicates the time‐step number. Assuming zero heat transfer to the reactor, the temperature equation is:

&

⎧ − ∑ (h f ,i + c p ,i (T − Tref ))ω i ⎫

⎪

⎪ Re ac.& Pr od .

⎧ dT ⎫

⎨ ⎬ =⎨

⎬

X

⎩ dt ⎭ n ⎪

∑Pr[od . i ]c p,i

⎪

Re ac .&

⎭n

⎩

{T }n+1 = {T }n + ⎧ dT ⎫ × dt

⎨ ⎬

⎩ dt ⎭ n

Where Tref=298K. Species equations are in the following form if we utilize Euler’s method: ⎧

⎧ d[ X i ] ⎫

⎛ 1 dT ⎞⎫

&

⎟⎬

⎨

⎬ = ⎨ωi − [ X i ]⎜

⎝ T dt ⎠⎭ n

⎩ dt ⎭ n ⎩

⎧ d[ X i ] ⎫

{[ X i ]}n+1 = {[ X i ]}n + ⎨

⎬ × dt

⎩ dt ⎭ n

Initial condition:

T=1000K,

X_H2O=X_CO2=0,

X_FUEL=0.095, X_O2=0.19, X_N2=0.715

Table of constants:

parameter UNITS

CH4

Cp

J/mole/K 71.6

at 1000K

O2

34.936

N2

32.762

CO2

54.36

H2O

41.315

hf

J/mole ‐74831 0

0

‐393546 ‐241845

MW

kg/mole 0.016

0.032

0.028

0.044

0.018

Here, 1ms was chosen as the time‐step, but any value that shows the same trend is OK. First 5 time‐steps calculated with Excel:

t

wF

N/V

[F]

s

mole/m3/s mole/m3 mole/m3

0.0E+00

‐0.54

670 64

1.0E‐03

‐0.54

670 64

2.0E‐03

‐0.54

670 64

3.0E‐03

‐0.54

670 64

4.0E‐03

‐0.54

670 64

[O2]

mole/m3

1.27E+02

1.27E+02

1.27E+02

1.27E+02

1.27E+02

[N2]

mole/m3

4.79E+02

4.79E+02

4.79E+02

4.79E+02

4.79E+02

[CO2]

mole/m3

0.00E+00

5.38E‐04

1.08E‐03

1.62E‐03

2.15E‐03

[H2O]

mole/m3

0.00E+00

1.08E‐03

2.15E‐03

3.23E‐03

4.31E‐03

DT/dt

K/s

17.5458934

17.5532489

17.5606102

17.5679774

17.5753506

T

K

1000

1000

1000

1000

1000

Here is the result:

Note: Auto‐ignition time for methane is around 2.5 seconds which is 4 orders of magnitude higher than Octane. It means that Octane is easier and faster to ignite. This is consistent with the octane rating as octane number is higher for methane.

Question 3

According to this graph (Gauthier, Davidson and Hanson, 2004) from the lecture notes:

This graph is suggesting that for a fixed T, all ignition delay times for different pressures, must fall roughly into a single point on the graph if scaled for the pressure. So for two different conditions with the same T and different P, scaled ignition delay times should be the same: ⎛ P ⎞

τ 1 /⎜ 1 ⎟

⎝ 55atm ⎠

τ 2 ⎛ P1 ⎞

=⎜ ⎟

τ 1 ⎜ P2 ⎟

⎝ ⎠

−1.05

1.05

⎛ P ⎞

= τ 2 /⎜ 2 ⎟

⎝ 55atm ⎠

⎛1⎞

=⎜ ⎟

⎝2⎠

1.05

−1.05

= 0.48

So the new ignition delay time is 0.48 times the original one.

Question4

Question 5

From Isothermal jet theory we have

⎛ Qf ⎞

Lf ∝ ⎜

⎟

⎜ D ⎟

⎠

⎝

From ideal gas law, for a fixed temperature, pressure is proportional to the density. In addition, for a fixed mass flow rate and cross section, volumetric flow rate is inversely proportional to the ...

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