mechanical engineering Gaziantep University combustion

1. UNIVERSITY OF GAZIANTEP
ENGINEERING FACULTY
MECHANICAL ENGINEERING DEPARTMENT
ME 438 THEORY OF COMBUSTION
TAKE HOME MIDTERM
SUBMITTED TO: Dr. AYŞEGÜL ABUŞOGLU
SUBMITTED BY:
ÖZGECAN ARACI
BURAK BARLAS
YÜKSEL ERASLAN
İBRAHİM İLKER ERDEM
YAŞAR GÜRBÜZ
ZÜHEYİR NURAL
ERDİ KARAÇAL
DECEMBER 2013

2. QUESTION 1
An inventor has devised on atmospheric-pressure process to manufacture methanol.The
inventor claims he has developed a catalyst that promotes the economical reaction of CO and H2 to
yield methanol; however a cheap supply of CO and H2 is needed.The inventor proposes burning
natural gas (CH4) in oxygen under fuel-rich conditions to yield a gas mixture of CO,CO2,H2O and H2.
A)If methane burns in oxygen at an equivalence ratio I=1.5 and the combustion reactions go to
equilibrium, What will be the resulting gas composition? Assume the combustion temperature is
controlled to 1500 K.
B)What would be the composition if the temperature were controlled to 2500 K ?

3. For the first question which we have given for take home midterm,first we must write the
equation.
 Our aim is evaluating the compositions of exhaust gas.
 Hidrogen and Carbon monoxide is necessary to produce etylene.
We have read lots of thing about how it is made etylene and finally we found that our course
book gave us some formulations about how we can burn methane to produce raw materials of etylene
and its compositions.After this point all the calculations were done by using the Course book which
was written by R Turns.
 CHସ + ଶ ைమ
ః → bCOଶ + cCO + ݀ܪଶܱ + eHଶ
 #஼ୟ୲୭୫ୱ
#ுୟ୲୭୫ୱ = ଵ
ସ = ௕ା௖
ଶௗାଶ௘ (***)
#ுୟ୲୭୫ୱ = ସ⁄ః
ସ = ଵ
ః = ଵ
ଵ.ହ = ଶ
ଷ = ௖ାଶ௕ାௗ
ଶௗାଶ௘
 #ைୟ୲୭୫ୱ
1 = ܲେ୓ + ܲେ୓మ + ܲுమை + ܲுమ
ܲେ୓ means that partial pressure of CO gas
ܲେ୓మ means that partial pressure of COଶ gas
 Equilibrium; CO + ܪଶܱ ⇌ COଶ + ܪଶ (***)
Simultaneously Solving the equations 2 star (***) equations
ܭ௉ (ܶ) =
ܲେ୓మ ∗ ܲுమ
ܲେ୓ ∗ ܲுమ଴
ܥ௫ܪ௬ + aOଶ → bCOଶ + cCO + dHଶܱ + eHଶ
We must find the coefficients of this equation a,b,c,d and e.
At course book Eqns;2.72 , 2.73, 2.74 , 2.75 , 2.76 can be used to find a,b,c,d,e

4. We have used our course book and coefficient of ‘’b’’ has the most complicated formulation.
ܾ =
2ܽ(ܭ௣ − 1) + ݔ + ݕ⁄2
2(ܭ௣ − 1)
−
1
2(ܭ௣ − 1)
∗ [(2ܽ(ܭ௣ − 1) + ݔ + ݕ⁄2)ଶ − 4ܭ௣(ܭ௣ − 1)(2ax − ݔଶ)]ଵ⁄ଶ
ܿ = ݔ − ܾ eqn. 2.73ܽ
݀ = 2ܽ − ܾ − ݔ eqn. 2.73ܾ
݁ = −2ܽ + ܾ + ݔ + ݕ/2eqn. 2.73ܿ
When we look at the formulation of ‘’b’’ ,We can see a value of Kp.
Kp is function of temperature and we found table 2.3 at our book
Table 1 has a values of Kp at different temperatures.Our Excel program can calculate the compositions
of exhaust gas at different temperatures.
T(K) Kp
298 105000
500 138,3
1000 1,443
1500 0,3887
2000 0,22
2500 0,1635
3000 0,1378
3500 0,1241
Table 1 : Kp Values
Calculations have been done at Excel program and Wolfram Mathematica Program.
At the Table 2 we can see the compositions of exhaust gas at different temperatures.
As a result we can say that changing the temperature 1500 to 2500 Kelvin don’t change the
compositions too much.
We have asked at question for temperatures of 1500 and 2500 K
TABLE 2
Compositions of exhaust gas (%)
Temp (K) CO2 CO H20 H2
298 33,33 0 22 44,67
500 32,87 0,47 22,47 44,2
1000 20,44 12,9 34,9 31,77
1500 13,29 20,04 42,04 24,63
2000 10,41 22,92 44,92 21,75
2500 9,03 24,3 46,3 20,37
3000 8,28 25,05 47,05 19,61
3500 7,84 25,5 47,5 19,17
Table 2:Compositions of exhaust gas