Calderos

ANALISIS DE COMBUSTION.
Datos del análisis de gases.

Anterior Nuevo

Valores para el cálculo:
AT=15.6 ℃
CO2=15.6%
AT=258.9 ℃
EF=86%
O2=1.4%
CO=4482mg/m3
ppm=3913.7
CO=0.392

Composición del Bunker.

Formula química.
7.3 C 9 H 0.0625 S 0.047 O2 0.0125 N2
Ecuación estequiometria conel análisis de gases.

x 7.3 C 9 H 0.0625 S 0.047 O2+0.0125 N2+1.058*a O2+3.76N2
15.6CO2+1.4*0.058O2+0.392CO+0.0625 SO2+ b H2O+c*1.058 N2

C 7.3x=15.6+0.392
x=2.1907

H 9x=2bb=9.936

O2 0.047*2.1907+1.058*a=15.6+1.4*0.058+
0.3922+0.0625+ 9.936/2
a=19.664

N2 0.0125*2.1907+3.76*19.664*1.058=1.058*c
c=73.963

2.1907 7.3 C 9 H 0.0625 S 0.047 O2+0.0125 N2+1.058*19.664 O2+3.76N2
15.6CO2+0.0812O2+0.392CO+0.0625 SO2+ 9.936 H2O+78.253 N2

Ecuaciónbalanceada para 1 mol de combustible.
7.3 C 9 H 0.0625 S 0.047 O2+0.0125 N2+9.4967 O2+3.76N2

7.121CO2+0.03707O2+0.178CO+0.2831 SO2+ 4.5 H2O+35.721N2

AC=mairemcomb
AC=9.5054.762912*7.3+4.5*2+32*0.0625+0.047*32+0.0125*28
=15.913kg airekg comb
CALCULO DE FLUJO MASICO DEL COMBUSTIBLE.

Caudal anterior = 3.5 gal/min
Caudal = 2.9 gal/min
m=Q*ρ
mcomb=0.01098m3min*965kgm3mcomb=10.592kg/min

Calculo de calor de salida

SUSTANCIA | hfº KJ/Kmol | Hf288KJ/Kmol | Hf 380KJ/Kmol | Hf532 KJ/Kmol |
7.3 C 9 H 0.0625 S 0.047 O2 0.0125 N2 | -342587.5 | …… | …… | …… |
O2 | 0 | 8384.4 | 11109 | 15771 |
N2 | 0 | 8373.8 | 11055 | 15528 |
H2O | -285830 | ……. | 10843 | 15287 |
CO2 | -393520 | ……. | 12552 | 19120 |
CO | -110530 | ……. | 11058 | 15557 |Q=3217515.3529kJkmol
Tasa de transferencia de calor.

Qsal=3217515.3529kJkmol*1100.45kg*10.592kgmin=5654.5kw

Temperatura de flama adiabatica.

SUSTANCIA | hfº KJ/Kmol | hf288KJ/Kmol |
7.3 C 9 H 0.0625 S 0.047 O2 0.0125 N2 | | ****** |
O2 | 0 | 8634.2 |
N2 | 0 | 8621.6 |
H2O | -285830 | 8175.4 |
CO2 | -393520 | 8989.8 |
CO | -110530 | 8373.6 |

7.121 kmolCO2-393520+hCO2-8989.8
+4.5 kmol H2O-241820+hH2O-8175.4
+0.03707 kmol O20+hO2-8934.2+35.72 kmol N20+hN2-8621.6
+0.178 kmol CO-110530+hCO-8373.6
=1kmol7.3 C 9 H 0.0625 S 0.047 O2 0.0125 N2(-342587.5 KJKmol)
7.121hCO2+4.5hH2O+0.03707hO2+35.72hN2+0.178hCO=3978123.6

3978123.647.556=83651

COMPONENTE | TEMPERATURA |
O2 | 2450 |
N2 | 2550 |
H2O | 2050 |
CO2 | 1720 |
CO | 2550 |T=2550K
COMPONENTE | ENTALPIAS |
O2 | 89004*0.03707 |
N2 | 84814*35.72 |
H2O | 111565*4.5 |
CO2 | 134368*7.121 |
CO | 85537*0.178 |
TOTAL | 4506958.072 |

T=2200K
COMPONENTE | ENTALPIAS |
O2 | 75484*0.03707 |
N2 | 72040*35.72 |
H2O | 92940*4.5 |
CO2 | 112939*7.121 |
CO | 72688*0.178 |
TOTAL | 3811474.074 |

INTERPOLACION.
T=2550K 4506958.072
T= X3978123.6
T=2200K 3811474.074
Temperatura de flama adiabatica
T=2283.9 K

Transferencia de calor en el paso 1
Distribución de tubos en el espejo

DESCRIPCION DE TUBOS EN EL CALDERO |
PASO | CANTIDAD | LONGITUD |
| | |
2 | 139 | 6.578m =2 59” |
3 | 190 | 7.340m = 289” |
TOTAL | 329 | |

Diámetro interior tubo: 3”= 76.2 mm
Espesor tubo0.95”=2.41mm

Dext=1438 mm=1.438m
Dint=1400 mm=1.400m
L=6.578m
Tflama=2010.75 °C
2010.75∙86.0100=1729.245

Se considera que la temperatura es constante a lo largo del hogar.
R=1A1h1+RfiA1+lnDextDint2πKl+Rf0Ao+1Aoho
mcomb=10.592kg/min=0.176533kg/s
AC=15.913kg airekg comb
m=mcomb+maire
m=mcomb+AC∙mcomb
m=mcomb(1+AC)
m=0.176533 Kgs(1+15.913)

m=2.9857Kgs

Ao=πDext∙L=π*1.438*6.578=29.71m2...