Calculo de temperaturas en aislante de calentador cilindrico vertical

CALCULO DE TRANSFERENCIA DE CALOR EN LA PARED CILINDRICA VERTICAL EN ZONA DE RADIACIÓN DEL CALENTADOR “2B” DE LA PLANTA CATALITICA No. 2 “FCC”
Thermal Conductivity Kaowool Blanket 8#/ft3 Temperaturas Cond term POR CURVA T °F T °C W/ m °C W/ m °C 0.0606 500 260 0.06 0.1282 1000 538 0.12 0.1758 1300 704 0.1758 0.2107 1501 816 0.21 0.2672 1800 982 0.26 0.3080 1999 1093 0.3

Thermal ConductivityKaowool Blanket 8#/ft3
0.35 0.3 0.25 0.2 0.15 0.1 0.05 0 0 500 1000 1500 2000 2500 k W/m °C

°F

Thermal Conductivity Kaowool Blanket 6#/ft3 Temperaturas Cond term POR CURVA T °F T °C W/ m °C W/ m °C 0.0708 500 260 0.07 0.1509 1000 538 0.15 0.2084 1300 704 0.2084 0.2510 1501 816 0.25 0.3203 1800 982 0.32

0.4 0.3 0.2 0.1

Thermal Conductivity Kaowool Blanket 6#/ft3

k W/m °C

0 0 5001000 1500 2000

°F

0.5

Thermal Conductivity Kaowool Blanket 4#/ft3

Thermal Conductivity Kaowool Blanket 4#/ft3 Temperaturas Cond term POR CURVA T °F T °C W/ m °C W/ m °C 0.0705 500 260 0.08 0.2061 1000 538 0.19 0.2873 1300 704 0.28732 0.3417 1501 816 0.33 0.4227 1800 982 0.43

0.4 0.3 0.2 0.1 0 0 500 1000 1500 2000 k W/m °C

°Fhttp://www.barteltinsulation.com/pdfs/CERAMICBLANKET.pdf ESPESOR mm 25.4 50.8 50.8 38.1 165.1 DENSIDAD # / ft3 kg/m3 8 128.3 6 96.2 6 96.2 4 64.1 CALCULO DE AREAS CALENTADOR 2B DIAMETRO = 6.5 ALTURA (L) = 16.9 PERIMETRO= 20.3 AREA = 341.72

CAPA 1 2 3 4 SUMA

inch 1 2 2 1.5 6.5

AISLANTE KAOWOOL BLANKET 8# KAOWOOL BLANKET 6# KAOWOOL BLANKET 6# KAOWOOL BLANKET 4#

m m m m2

RADIOS [r]

r1 = r2 =

m 3.0607 3.0861

ESPESOR [E]

E1 =E2 =

m 0.0254 0.0508

COND. TERMICA

k1 = k2 =

W/m °C… a 704 °C 0.1758 0.2084 ALTURA (L) =

16.9

m

4 3

r3 = r4 = r5 =

3.1369 3.1877 3.2258

E3 = E4 = ETOTAL =

0.0508 0.0381 0.1651

k3 = k4 = hconv = Tcal = Tmetal = Tfria =

0.2084 0.28732 4.2355 704 60 20

3 2 W/m2 °C ° C (Cara Caliente) ° C (Cara fria) ° C (Temp. ambiente) r1 1

RESISTENCIAS TERMICAS

R0 =R1 = R2 = R3 = R4 = R5 = RTOTAL =

1/[h1A1] = ln(r2/r1)/[2pLk1] = ln(r3/r2)/[2pLk2] = ln(r4/r3)/[2pLk3] = ln(r5/r4)/[2pLk4] = 1/[haireA5] = R 0 + R 1 + R2 + R3 + R4 + R5

°C / W 0 0.0004 0.0007 0.0007 0.0004 0.0007 0.0030

r2 r3 r4 r5

Transferencia de Calor Estacionaria (Q*) (Q*) = [Tcal - Tfria]/Rtotal (Q*) = 228,590 W T1 T2 T3 T4 k1 k2 k3 k4 W/m2 °C W/m2 °C T5 T6 CALCULO DE h rad PORSOLVER ERROR = 9.389 0.525 hrad = hrad = -1E-06 0.09241 0.52461 (BTU/HR)/ FT2 °F W/m2 °C

PERDIDA DE CALOR (Q HL) = QHL = 668.93 W/m2 CALCULOS DE TEMPERATURAS INTERMEDIAS T1 = T2 = T3 = T4 = T5 = Tfria (T6) = Tcal Tcal - (Q*)*R1 T2 - (Q*)*R2 T3 - (Q*)*R3 T4 - (Q*)*R4 T5 - (Q*)*R5 °C 704 602.6 433.5 267.2 177.9 20.0 DIF °C 101.4 169.1 166.3 89.2 157.9 PROM °C 653.3 518.0 350.3 222.5 99.0 PROM °F1175.9 932.5 630.6 400.6 178.1 hconv = hrad = POR CURVA W/ m °C 0.1554 0.1388 0.0898 0.0436

→ → → → →

RESISTENCIAS TERMICAS

R0 = R1 = R2 = R3 = R4 = R5 = RTOTAL =

1/[h1A1] = ln(r2/r1)/[2pLk1] = ln(r3/r2)/[2pLk2] = ln(r4/r3)/[2pLk3] = ln(r5/r4)/[2pLk4] = 1/[haireA5] = R 0 + R 1 + R2 + R3 + R4 + R5

°C / W 0 0.0005 0.0011 0.0017 0.0026 0.0003 0.0062

Tcal = Tfria =

704 20

° C(Cara Caliente) ° C (Temp. ambiente) DIFERENCIA DE CALOR ESTACIONARIO (Q*) = -117,737 51.5 W

Transferencia de Calor Estacionaria (Q*) (Q*) = [Tcal - Tfria]/Rtotal (Q*) = 110,853 W

% DESV.= PERDIDA DE CALOR (Q HL) = QHL = 324.40 W/m2

CALCULOS DE TEMPERATURAS INTERMEDIAS T1 = T2 = T3 = T4 = T5 = Tfria (T6) = Tcal Tcal - (Q*)*R1 T2 - (Q*)*R2 T3 - (Q*)*R3 T4 - (Q*)*R4 T5 - (Q*)*R5 °C 704 648.4525.3 338.0 52.7 20.0 DIF °C 55.6 123.1 187.3 285.3 32.7 PROM °C 676.2 586.8 431.6 195.4 36.4 PROM °F 1217.1 1056.3 777.0 351.7 65.4 hconv = hrad = POR CURVA W/ m °C 0.1621 0.1611 0.1126 0.0303 5.559 2.301

→ → → → →

k1 k2 k3 k4 W/m2 °C W/m2 °C

CALCULO DE h rad POR SOLVER ERROR = hrad = hrad = 2.41585E-13 0.40541 2.30149 (BTU/HR)/ FT2 °F W/m2 °C

RESISTENCIAS TERMICAS

R0 = R1 = R2 =...