termo terminado.docx
TRANSCRIPT
![Page 1: TERMO TERMINADO.docx](https://reader036.vdocuments.co/reader036/viewer/2022081000/55cf92da550346f57b9a1268/html5/thumbnails/1.jpg)
2-) En un proceso de combustión se quema metano CH 4 con 20% de exceso de aire.
Determinar:
1)ra / c (t,r) ; rc /a (t;r)
2) % Volumétrico de productos en base seca.
SOLUCION
A=1+exc=1+0,2=1,2
ECUACION REAL
CH 4+1,2b (O2+3.76N2 )→α CO2+β H 2O+δO2+γ N2
C :1=α ;H :4=2∗β→ β=2
O :2 (1,2 )b=2∗α+β+2δ→δ=1,2b−2
N :2 (1,2 )b (3,76 )=2 γ→γ=4,512b
ECUACION ESTEQUIOMETRICA
CH 4+b (O2+3.76N2 )→CO2+2H 2O+3,76b N 2
O :2b=2+2→b=2
→δ=0,4 ; γ=9,024
ECUACION REAL
CH 4+2,4 (O2+3.76N 2)→CO2+2H 2O+0,4O2+9,024N2
ECUACION ESTEQUIOMETRICACH 4+2 (O2+3.76N2 )→CO2+2H 2O+7,52N 2
r( ac )t
=2 (32+3.76∗28 )
12+4=17,16 KGaire
KGcomb→r
( ac )t=0,0583
r( ac )r
=2,4 (32+3,76∗28)
12+4=20,592 KGaire
KGcomb→r
( ac )t=0,0486
N t=1+0.4+9,024=10,424 (base seca )
![Page 2: TERMO TERMINADO.docx](https://reader036.vdocuments.co/reader036/viewer/2022081000/55cf92da550346f57b9a1268/html5/thumbnails/2.jpg)
%CO2=1
10,424∗100=9,59%
%O 2=0,410,424
∗100=3,837%
%N 2=9.02410,424
∗100%=86,569%
3) En un proceso de combustión se quema metano CH 4con 20% de defecto de aire.Determinar:
1) r ac
(t , r ) ;r ac
(t , r )
2) % volumétrico de productos en base seca.
SOLUCION
A=1−¿¿1−0,2=0,8
ECUACION REAL
CH 4+0,8b (O2+3.76N2 )→αCO2+ βCO+γ H 2O+θO2+φ N2
C :1=α+β
H :4=2 γ→γ=2
O :2 (0,8b )=2α+β+γ+2θ=α+1+2+θ→θ=1,6b−α−32
N :2 (0,8 )b∗3,76=2φ→φ=3,008b
ECUACION ESTEQUIOMETRICA
CH 4+b (O2+3.76N2 )→(1−β)CO2+2H 2O+3,76b N2
C :1=1−β→ β=0
O :2b=2 (1−β )+2→b=2 , φ=6.016 , θ=0.2−α2
ECUACION REAL
CH 4+1,6 (O2+3.76N2 )→αCO2+βCO+2H 2O+¿
C :1=α+β
O :2 (1,6 )=2α+ β+2+0,1−α2
![Page 3: TERMO TERMINADO.docx](https://reader036.vdocuments.co/reader036/viewer/2022081000/55cf92da550346f57b9a1268/html5/thumbnails/3.jpg)
3,2=α+1+2+0,1−α2→α=0,2 ; β=0,8
CH 4+1,6 (O2+3.76N2 )→0,2CO2+0,8CO+2H 2O+6,016N 2
r( ac )r
=1,6 (32+3,76∗28)
12+4=13,728 KGaire
KGcomb→r
( ac )r=0,073 KG comb
KG aire
r( ac )t
=2(32+3,76∗28)
12+4=17,16 KGaire
KGcomb→r
( ac )t=0,058 KG comb
KG aire
N t=0,2+0,8+6,016=7,016
%CO2=0,27,016
∗100%=2,851%
%CO= 0,87,016
∗100%=11,403%
%N 2=6,0167,016
∗100%=85,747%
![Page 4: TERMO TERMINADO.docx](https://reader036.vdocuments.co/reader036/viewer/2022081000/55cf92da550346f57b9a1268/html5/thumbnails/4.jpg)