Problemas Kern

PROBLEMA 2,1 PROBLEMA 2,2

1130.6832 149.716636 (Q/A)1

403.775203 143.549454 (Q/A)2

149.716636 (Q/A)3


q 184.124353 184.124353 184.124353T ºf 500 294.866258 111.770445 100

PROBLEMA 2,8 PROBLEMA H

Q 301439.745 COSTO 6303.4289

T1 q1

T2 q2

q3

T1

PROBLEMA 2,3Resultado

100 29.0472311 L 0.83 1.04 0.05

100

99.9999991


80,0ºF q 0 118.966632TºF 450 101.872606

T1

PROBLEMA T

CONDENSADO 0.25144139

118.96663299.6763951

TEXT

DATOS: INTERIOR T1 T2 TEXTERIORTemperaturas 2200 200Material PLG " Material PLG " MaterialLRC 8 LCA 6 LARTemperatura k Temperatura k Temperatura

392 0.05 932 0.15 3921400 0.113 2102 0.26 1112

y = 6E-05x + 0.0255 y = 9E-05x + 0.0624 1832R² = 1 R² = 1 2552

T SUPUESTA VALOR K T SUPUESTA VALOR K1130.6832 0.09334099 403.775203 0.09873977 R² = 0.9912

1130.68 218.68 VALOR DE T1053.34 200

659.49509

ECUACIONESFORMULA GENERAL RESULTADOS:

T1T2

ECUACIONES PARA CADA PARED

149.716636

143.549454

149.716636

y = -1E-07x2 + 0.0005x + 0.4132

q1

q2

q3

𝑄=𝐾𝐴(∆𝑇/𝐿)

𝑞_(1= ) 𝑘_𝐿𝑅𝐶 ((𝑇_𝐼𝑁𝑇𝐸𝑅𝐼𝑂𝑅−𝑇_1)/(𝐸𝑠𝑝𝑒𝑠𝑜𝑟 𝐿𝑅𝐶/12))𝑄/𝐴=𝑞

𝑞_(2= ) 𝑘_𝐿𝐶𝐴 ((𝑇_1−𝑇_2)/(𝐸𝑠𝑝𝑒𝑠𝑜𝑟 𝐿𝐶𝐴/12))𝑞_(3= ) 𝑘_𝐿𝐴𝑅 ((𝑇_2−𝑇_𝐸𝑋𝑇𝐸𝑅𝐼𝑂𝑅)/(𝐸𝑠𝑝𝑒𝑠𝑜𝑟 𝐿𝐴𝑅/12))

PLG "7

k0.580.850.951.02 Pearson LRC Pearson LCA Pearson LAR

1 1 0.94962567

VALOR K 0.4286

SOLVER Q1=Q2CELDA OBJETIVO Q2=Q3

RESULTADOS:1130.6832

403.775203

y = -1E-07x2 + 0.0005x + 0.4132

200 400 600 800 1000 1200 1400 16000

0.02

0.04

0.06

0.08

0.1

0.12

f(x) = 0.0000625 x + 0.0255R² = 1

f(x) = 0.0000625 x + 0.0255R² = 1

k/T (LRC)

k/T LRC Linear (k/T LRC) Linear (k/T LRC)

T

k

800 1000 1200 1400 1600 1800 2000 22000

0.05

0.1

0.15

0.2

0.25

0.3

f(x) = 9.4017094017094E-05 x + 0.0623760683760684R² = 1

k/T (LCA)

k/T LCA Linear (k/T LCA)

T

k

0 500 1000 1500 2000 2500 30000

0.2

0.4

0.6

0.8

1

1.2

f(x) = 0.235644785742051 ln(x) − 0.819718479678335R² = 0.996248839190914

k/T (LAR)

k/T LAR Logarithmic (k/T LAR)

T

k

0 500 1000 1500 2000 2500 30000

0.2

0.4

0.6

0.8

1

1.2

f(x) = 0.235644785742051 ln(x) − 0.819718479678335R² = 0.996248839190914

k/T (LAR)


T

k

200 400 600 800 1000 1200 1400 16000

0.02

0.04

0.06

0.08

0.1

0.12

f(x) = 0.0000625 x + 0.0255R² = 1

f(x) = 0.0000625 x + 0.0255R² = 1

k/T (LRC)

k/T LRC Linear (k/T LRC) Linear (k/T LRC)

T

k

800 1000 1200 1400 1600 1800 2000 22000

0.05

0.1

0.15

0.2

0.25

0.3

f(x) = 9.4017094017094E-05 x + 0.0623760683760684R² = 1

k/T (LCA)

k/T LCA Linear (k/T LCA)

T

k

0 500 1000 1500 2000 2500 30000

0.2

0.4

0.6

0.8

1

1.2

f(x) = 0.235644785742051 ln(x) − 0.819718479678335R² = 0.996248839190914

k/T (LAR)


T

k

0 500 1000 1500 2000 2500 30000

0.2

0.4

0.6

0.8

1

1.2

f(x) = 0.235644785742051 ln(x) − 0.819718479678335R² = 0.996248839190914

k/T (LAR)


T

k

T ( C ) K (BTU/hra ft^2) PULGADAS T supuesta Temp mediasT1 902.09478

LRC 392 0.05 7 T2 504.2808581400 0.113

LCA 932 0.15 6 Tinterior 15002102 0.26 Texterior 100

LAR 392 0.58 ?1112 0.851832 0.952552 1.02

K1 k2 k30.09756284 0.1256869 0.5987428085838

ECUACIONES DEL SISTEMA

(Q/A)1(Q/A)2(Q/A)3

𝑞_(1= ) 𝑘_𝐿𝑅𝐶 ((𝑇_𝐼𝑁𝑇𝐸𝑅𝐼𝑂𝑅−𝑇_1)/(𝐸𝑠𝑝𝑒𝑠𝑜𝑟 𝐿𝑅𝐶/12))𝑞_(2= ) 𝑘_𝐿𝐶𝐴 ((𝑇_1−𝑇_2)/(𝐸𝑠𝑝𝑒𝑠𝑜𝑟 𝐿𝐶𝐴/12))

𝑞_(1=)𝑞_(2=)𝑞_(3=)

𝑞_(3= ) 𝑘_𝐿𝐴𝑅 ((𝑇_𝐸𝑋𝑇𝐸𝑅𝐼𝑂𝑅−𝑇_1)/(𝐸𝑠𝑝𝑒𝑠𝑜𝑟 𝐿𝐴𝑅/12))𝑞_(3= ) 𝑘_𝐿𝐴𝑅 ((𝑇_2−𝑇_𝐸𝑋𝑇𝐸𝑅𝐼𝑂𝑅)/?)

Temp mediasTm 1 1201.04739Tm 2 703.187819Tm 3 302.140429

Resultado29.0472311

100100

99.9999991

200 400 600 800 1000 1200 1400 16000

0.02

0.04

0.06

0.08

0.1

0.12

f(x) = 0.0000625 x + 0.0255

LRC

800 1000 1200 1400 1600 1800 2000 22000

0.05

0.1

0.15

0.2

0.25

0.3

f(x) = 9.4017094017094E-05 x + 0.0623760683760684

LCA

0 500 1000 1500 2000 2500 30000

0.2

0.4

0.6

0.8

1

1.2

f(x) = 0.000197222222222222 x + 0.559688888888889

LAR

200 400 600 800 1000 1200 1400 16000

0.02

0.04

0.06

0.08

0.1

0.12

f(x) = 0.0000625 x + 0.0255

LRC

800 1000 1200 1400 1600 1800 2000 22000

0.05

0.1

0.15

0.2

0.25

0.3

f(x) = 9.4017094017094E-05 x + 0.0623760683760684

LCA

0 500 1000 1500 2000 2500 30000

0.2

0.4

0.6

0.8

1

1.2

f(x) = 0.000197222222222222 x + 0.559688888888889

LAR

LC LM LRCOMUNT K T K T

392 0.67 399 2.2 681202 0.85 1202 1.62399 1 2192 1.1

1500 1500 1500

T °F 2500 1400 400 200

L 0.83 1.04 0.05

LC LM LRC

k 1.1251 1.5638 0.4

Se fijo el valor minimo de "q" respecto de las temperaturas -100 del valor maximo

q(BTU/hrpie^2)

LRCOMUNK

0.4

0 500 1000 1500 2000 2500 30000

0.2

0.4

0.6

0.8

1

1.2

f(x) = 0.000161462609355827 x + 0.625093266947394R² = 0.97356630498398

LCLC Linear (LC)

0 500 1000 1500 2000 25000

0.5

1

1.5

2

2.5

f(x) = − 0.000609348649250645 x + 2.40375314220257R² = 0.98737917665114

LMLM Linear (LM)

184.12435288 184.124353 184.12435288T - °F 500 294.866258 111.77044509 100

L 0.0416666667 0.08333333 0.0416666667 ?K LM Mn

K 0.02 0.0225 0.35

q 0

KAPOK LANA MINERAL T K T K

68 0.02 86 0.0225

q (BTU/hrpie lin)

MAGNESITA MOLIDA T K

117 0.35

25% NaCl ----->Concent. 0.25 Q = 30000 lb/hK = 0.020 TEXT = 68°F

1.- 𝒒=2𝜋𝑘/(𝑙𝑛 𝐷_𝑜/𝐷_𝑖 )(𝑡_2−𝑡_1)

SUSTITUYENDO VALORES DADOS :𝑞=2𝜋𝑘/(𝑙𝑛 3.38/2.38) (90−0)=32.242 𝐵𝑡𝑢/(ℎ𝑟𝑝𝑖𝑒𝑙𝑖𝑛.) CALCULANDO Q Y POSTERIORMENTE MULTIPLICANDO POR L =60 ft𝑄=32.242 𝐵𝑡𝑢/(ℎ𝑟𝑝𝑖𝑒𝑙𝑖𝑛.) (60𝑝𝑖𝑒𝑠)=1934.526 USANDO LA ECUACION 2 :

2.- 𝑄=𝑚𝐶𝑝∆𝑇 OBTENIENDO EL Cp DE LA SOLUCION DE SODIO :𝐶𝑝=𝐶𝑝 𝑠𝑜𝑙. −𝐶𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑐𝑖𝑜𝑛 𝑑𝑒𝑙 𝑠𝑜𝑙𝑖𝑑𝑜=1−0.25 DESPEJANDO LA T1 DE LA ECUACION 2𝑄/𝑚𝐶𝑝+ 𝑇_2=𝑇_1 1934.526/(32.242(.75))+ 0=𝑇_1 𝑻_𝟏=𝟖𝟎.𝟎 °𝑭

q(BTU/hrpie lin) 0 118.966632 118.966632T°F 450 101.872606 99.6763951 70

T1 T extL 0.08333333 0.08333333

M K MK 0 2.31277644

ha 1.80146766hr 1.11799986hc 0.6834678T°R 909.67 561.542606 559.346395 529.67

q -118.966632

KAPOK Magnesita T K T K

68 0.02 339 2.21220 1.62192 1.1

301.439745 301.439745T°F 429.694357 85.2629649 70

Tprom 257.478661 TambL

Material K AireK 0.02 0.01834957

ha 1.7961545

hr 1.07361313

hc 0.72254137T°R 889.364357 544.932965 529.67

q 7.19107E-07

L (pies) 1000

Q 301439.745

q (BTU/hrpie lin)

200 250 300 350 400 4500.1040.1060.108

0.110.1120.1140.1160.118

0.120.1220.124

f(x) = 6.66666666666667E-05 x + 0.0968666666666667R² = 1

Asbesto

Diam. Int Diam.ext KAPOK Asbesto3.5 3.068 T K T K

68 0.02 140 0.114-328 0.04332 0.0932 0.087

212 0.111

392 0.123

752 0.129-328 0.0932 0.135

68 0.43

200 250 300 350 400 4500.1040.1060.108

0.110.1120.1140.1160.118

0.120.1220.124

f(x) = 6.66666666666667E-05 x + 0.0968666666666667R² = 1

Asbesto

-200 -100 0 100 200 300 400 500 600 7000

0.005

0.01

0.015

0.02

0.025

0.03

f(x) = 2.36666666666667E-05 x + 0.0131626666666667R² = 0.999383219155451

Aire

AIRET °F K

-148 0.009532 0.014

212 0.0183392 0.0226

572 0.0265

-200 -100 0 100 200 300 400 500 600 7000

0.005

0.01

0.015

0.02

0.025

0.03

f(x) = 2.36666666666667E-05 x + 0.0131626666666667R² = 0.999383219155451

Aire

Tamb 60ºF

0.65616797900261000 lb/hr Pipe

Combustible P=85.3 lb/in^2 m 2" ced 40T ambiente

HornoComo se analizo en clase suponemos que esta en el piso y unicamente se calcula la cara superior

Placas verticales Cara caliente superiorq (BTU/hrpie lin) 199333.013 199333.078 q

T°F 2200 475.608756 60 T°FT ext Tamb

T°R 2659.67 935.278756 519.67 T°R

L 0.08 L

M LR MK 0.67120219 K

ha 2.78486806 hahr 1.43032622 hrhc 1.35454185 hc

t prom 1337.80438 t promCleda objetivo -0.06434984 Celda objetiv

797332.31 Q

ESPESOR LR=1 ft

PC=1550 Btu/lbTFA=2200°FSqr=1.2$=12$/litro

Q4

Q 1=(𝐴 𝑘_𝑎∙(2200−𝑇_𝑒𝑥𝑡 ))/𝐿 𝑄2=𝐴 ℎ_𝑎∙(𝑇_𝑒𝑥𝑡−𝑇_𝑎𝑚𝑏 )

PARAMETRO VALOR Ladrillo Refractario Aislante$ 12 Horno TuberiaT flama (°F) 2200 T°F K T°FP calori (BTU/1550 399 0.2018 20Pman. (lb/in285.3 1600 0.8023 300Patm. (lb/in214.7 700

Aislante Pabs. (lb/in2)1001/2" espesor T amb (°F) 60 λ 888.81

62.43 P lb/in^2 1000Sqr 1.2 T°F 327.81Flujo (lb/h) 1000 T amb (°F) 60Cp (BTU/lb °F1 74.916

Espesor (ft) 1 Q latente (BTU/h) 888810

L (ft) 13.1233596 Q sensible(BTU/h) 267810Area (ft^2) 172.222567 PQ 997439.615

Como se analizo en clase suponemos que esta en el piso y unicamente se calcula la cara superior 2154059.62

Cara caliente superior200107.305 200107.283 W (lb/h) 1389.71588

2200 456.504042 60 Q (Ft^3/h) 18.5503214T ext Tamb Q (L/h) 525.285742

2659.67 916.174042 519.67

0.08 $ = 6303.4LR0.66642601

2.93038861

1.23470235

1.695686261328.25202

0.0215422

200107.305

ᵨH2O (lb/ft3)

ᵨ(lb/ft3)

QT (BTU/h)

𝑄2=𝐴 ℎ_𝑎∙(𝑇_𝑒𝑥𝑡−𝑇_𝑎𝑚𝑏 )

AislanteTuberia

K0.01360.06680.1428

QT = W* P.C

W = QT / P.C

,Tamb 60ºF

tablasP=85.3 lb/in^2 m Pabs. (lb/in2) 100 327.81°F Tuberia

L 1000 2" cd 40

Flujo (lb/h)λ

q 223.483618 223.48361762258 Calor latenteT°F 327.81 162.25499416091 60 qT=

T ext T amb CondensadoT°R 787.48 621.92499416091 519.67 %L 0.04166667M AK 0.075362ha 2.46987991hr 1.29724732hc 1.17263259

q 5.11591E-13

Aislante1/2" espesor

1000888.81

888810223483.617620.251441385225.144138525

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