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principle of combustion, Essays (university) of Mechanical Engineering

homework principle of combustion

Typology: Essays (university)

2019/2020

Uploaded on 07/27/2020

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1. Equivalence ratio (ϕ) is frequently more meaningful than air-fuel (AFR) ratio when comparing different fuels,
because ϕ calculates the actual ratio of air-fuel (AFR)actual over the stoichiometric ratio of air-fuel (AFR)st. Thus if
(AFR)actual is higher than (AFR)st or ϕ>1 , the mixture is fuel rich. On the contrary, if (AFR) actual is smaller than (AFR)st or
ϕ<1 , the mixture is fuel lean. This way, we can choose which fuels is more efficient by comparing the ϕ results, the
closest result to 1 (stoichiometric) is the better/more efficient.
2. The chemical structure of 3-ethyl-3-methylpentane C8H18:
3. -skipped
4. Assumption: Tref 298o K, Pref 1 atm = 101325 Pa, Tfinal 1902o K, Tavg 1100o K
a. flow rates of propane and air
100 KW = 100 KJ/s
Assume a complete combustion, the moles fractions are 16.67% for C3H8 and 83.33% for air or O2 & N2
b. equivalence ratio ϕ and excess air percentage
stoichiometric condition:
C3H8 + 5 (O2+3.76 N2) 3 CO2 + 4 H2O + 18.8 N2
with the help of excel, the moles are found w.r.t. the oxygen contentration of 3% in the dry-basis product:
C3H8 + 5.355 (O2+3.76 N2) 3 CO2 + 4 H2O + 20.135 N2 + 0.71 O2
ϕ=
(F
A)
(
F
A
)
st
=
(1
5.355 )
(1
5)
=0.934
%of excess air =100% ×1ϕ
ϕ=100 % ×10.934
0.934 =7.098 %
c. adiabatic flame temp (Tad) by manual calculation
Result = 2280 K
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1. Equivalence ratio (ϕ) is frequently more meaningful than air-fuel (AFR) ratio when comparing different fuels,

because ϕ calculates the actual ratio of air-fuel (AFR)actual over the stoichiometric ratio of air-fuel (AFR)st. Thus if

(AFR)actual is higher than (AFR)st or ϕ>1 , the mixture is fuel rich. On the contrary, if (AFR)actual is smaller than (AFR)st or

ϕ<1 , the mixture is fuel lean. This way, we can choose which fuels is more efficient by comparing the ϕ results, the

closest result to 1 (stoichiometric) is the better/more efficient.

2. The chemical structure of 3-ethyl-3-methylpentane C 8 H 18 :

3. -skipped

4. Assumption: Tref 298o^ K, Pref 1 atm = 101325 Pa, Tfinal 1902o^ K, Tavg 1100o^ K

a. flow rates of propane and air

100 KW = 100 KJ/s

Assume a complete combustion, the moles fractions are 16.67% for C 3 H 8 and 83.33% for air or O 2 & N 2

b. equivalence ratio ϕ and excess air percentage

stoichiometric condition:

C 3 H 8 + 5 (O 2 +3.76 N 2 )  3 CO 2 + 4 H 2 O + 18.8 N 2

with the help of excel, the moles are found w.r.t. the oxygen contentration of 3% in the dry-basis product:

C 3 H 8 + 5.355 (O 2 +3.76 N 2 )  3 CO 2 + 4 H 2 O + 20.135 N 2 + 0.71 O 2

ϕ = ( F A )

F

A )

st

( 1

) ( 1 5 ) =0. % of excess air = 100 % × 1 − ϕ ϕ = 100 % × 1 −0.

=7.098 %

c. adiabatic flame temp (Tad) by manual calculation

Result = 2280 K

d. adiabatic flame temp (Tad) by NASA CEA

NASA-GLENN CHEMICAL EQUILIBRIUM PROGRAM CEA2, FEBRUARY 5, 2004

BY BONNIE MCBRIDE AND SANFORD GORDON

REFS: NASA RP-1311, PART I, 1994 AND NASA RP-1311, PART II, 1996

CEA analysis performed on Tue 24-Mar-2020 17:02:

Problem Type: "Combustion at Assigned Volume"

prob case=1______________5832 uv

Density (1 value):

rho,m**3/kg= 500

Equivalence based on Fuel/Oxid. wt ratio (Eq 9.19*) (1 value):

phi= 0.

You selected the following fuels and oxidizers:

reac fuel C3H8(L) mole=100.0000 t,k= 298.000 rho,g/cc= 500.000 h,kj/mole=-

oxid Air mole=100.0000 t,k= 298.

You selected these options for output:

short version of output

output short

Proportions of any products will be expressed as Mole Fractions.

Heat will be expressed as siunits

output siunits

Input prepared by this script:prepareInputFile.cgi

IMPORTANT: The following line is the end of your CEA input file!

end REACTANT C3H8(L) HAS BEEN DEFINED FOR THE TEMPERATURE 231.08K ONLY. YOUR TEMPERATURE ASSIGNMENT 298.00 IS MORE THAN 10 K FROM THIS VALUE. (REACT) ERROR IN REACTANTS DATASET (INPUT)