Calculo de Instalacion de Calefaccion Bitubular

7/23/2019 Calculo de Instalacion de Calefaccion Bitubular

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Diapositiva 1

INSTALACIONES-2DEPARTAMENTO DE CONSTRUCCIONES ARQUITECTNICAS

CURSO

08-09

Profesor: Dr. D. Julin Domene Garca

4 C

DISEO Y CLCULO DE LAINSTALACIN DE CALEFACCINBITUBULAR EN UNA VIVIENDA

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Diapositiva 2

EJEMPLO DE CLCULO

2

Se dispone de una vivienda unifamiliar de129 m de superficie til, ubicada en Bilbao.Cuyas caractersticas geometricas sepueden observar:

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Diapositiva 3

EJEMPLO DE CLCULO

3

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Diapositiva 4

4

Resumen de las prdidas totales en la vivienda:

SUMATOTAL 10.115 W

Local QSI Infiltraciones QT

Saln 3.757 472 4.229

Bao 239 22

261

Cocina 678 56

734

Dormitorio 1 735 127

862

Dormitorio 2 993 213

1.206

Dormitorio 3 604 53

657

Dormitorio 4 991 278 1.269

Vestbulo 787 110

897

Prdidas totales= prdidas por transmisin + prdidas por infiltraciones:

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Diapositiva 5

5

Con estos valores, se tendr que hallar elmodelo de radiador idneo para suministrarsta potencia en las condiciones establecidas.

Los radiadores se montan en la viviendacompletamente exentos, debajo de lasventanas, con lo cual el rendimiento de los

mismos ser el mximo, es decir el 100 %.

SELECCIN DE LOS RADIADORES

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Diapositiva 6

6

SELECCIN DE LOS RADIADORES

Los valores expuestos en la tabla siguiente, estncalculados para salto trmico de 60 C,correspondientes a la antigua norma UNE 9015,sustituida por la norma europea UNE-EN 442,despus de la entrada en vigor del RITE, en la cualsu valor es de 50 C. Encontrando la diferenciaposteriormente.

Los radiadores se han elegido delcatlogo general de la marca comercialRoca.

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Diapositiva 7

7

TEMA 7

1. Saln comedor:Prdidas 4.229 W = 3.636 Kcal/h

1 radiador

2 radiadores

Clculo de los radiadores

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Diapositiva 8

8

TEMA 7 Prdidas Saln = 4.229 W = 3.636 kcal/h

2 Radiadores Dubal 70, Emisin = 1.196,3x2 = 2.392,6 kcal/h

Con 7 elementos cada uno

1 Radiador Dubal 30, de 1.273,5 Kcal/h

Con 15 elementos

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Diapositiva 9

9

TEMA 7

1. Saln comedor:Prdidas 4.229 W = 3.636 Kcal/h

1 radiador 1.273 Kcal/h

2 radiadores 2.393Kcal/h

SUMA 3.666 Kcal/h

Clculo de los radiadores

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Diapositiva 10

EJEMPLO DE CLCULO

10

1

2 34

5

67

8 9

10

11

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Diapositiva 11

11

Local W Kcal/h N Radiador Elementos Emisincalorfica

Saln 4 .229 3.6361

2 y 3Dubal 30

2xDubal 7015

2x71.273

2x1.196

Bao 261 224 4 Dubal 30 3 255

Cocina 734 631 5 Dubal 45 6 677

Dormitorio 1 862 741 6 Dubal 60 6 886

Dormitorio 2 1.206 1.037 7 Dubal 70 7 1.196

Dormitorio 3 657 565 8 Dubal 45 6 677

Dormitorio 4 1.269 1.091 9 Dubal 70 7 1.196

Vestbulo 897 771 10 y 11 Dubal 45

Dubal 3044

451340

10.115 8.697 72 9.343

De la misma manera y resumiendo:

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Diapositiva 12

12

Queda definido el caudal de agua que circula por el dispositivoemisor, como:

C = Q/cxt

En la que:

C = caudal msico de agua que circula por el emisor en kg/s.

Q = potencia calorficaen W

t = diferencia temperat ura de entrada y s alida del emisor

C = calor especfico del agua 4.185 J (kg C)

Se toma como base para el clculo una diferencia de 20C

Con lo que en el saln-comedor:

C = 3.636/4.185x20 = 0,043kg/s

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Diapositiva 13

13

El restode valores:

Local W Kc al /h N Ra dia do r Ele me nt os Emisincalorfica Caudal

Saln 4 .229 3.6361

2 y 3Dubal 30

2xDubal 7015

2x71.273

2x1.196 0,043

Bao 261 224 4 Dubal 30 3 255 0,003

Cocina 734 631 5 Dubal 45 6 677 0,009

Dormitorio 1 862 741 6 Dubal 60 6 886 0,01

Dormitorio 2 1.206 1.037 7 Dubal 70 7 1.196 0,014

Dormitorio 3 657 565 8 Dubal 45 6 677 0,008

Dormitorio 4 1.269 1.091 9 Dubal 70 7 1.196 0,015

Vestbulo 897 771 10 y 11 Dubal 45

Dubal 3044

451340

0,011

10.115 8.697 72 9.343 0,113

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Diapositiva 14

14

sta es sustituida por la norma europea UNE-EN442, despus de la entrada en vigor del RITE, enla cual su valor es de 50 C. Encontrando la

diferencia posteriormente.

Los valores expuestos en la tabla siguiente, estncalculados para salto trmico de 60 C,correspondientes a la antigua norma UNE 9015.

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Diapositiva 15

15

TEMA 7 Prdidas Saln = 4.229 W = 3.636 kcal/h

2 Radiadores Dubal 70, Emisin = 1.191,3x2 = 2.382,6 kcal/h

Con 10 elementos cada uno

1 Radiador Dubal 30, de 1.283,4 Kcal/h

Con 18 elementos

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Diapositiva 16

16

Comparacin entre las normas UNE 9015 y UNE-EN 442-Radiadores

Norma UNE9015 (60C)

Norma UNE 442(50C)

Local W Kcal/h N Radiador Elementos Emisin

calorfica Radiador Elementos Emisin

calorfica

Saln 4.229 3.6361

2 y 3Dubal 30

2xDubal 7015

2x71.273

2x1.196Dubal 30

2xDubal 70

182x10

1.2832x1.191

Bao 261 224 4 Dubal 30 3 255 Dubal 30 4 285

Cocina 734 631 5 Dubal 45 6 677 Dubal 45 8 635

Dormitorio 1 862 741 6 Dubal 60 6 886 Dubal 60 8 831

Dormitorio 2 1.206 1.037 7 Dubal 70 7 1.196 Dubal 70 10 1.191

Dormitorio 3 657 565 8 Dubal 45 6 677 Dubal 45 8 635

Dormitorio 4 1.269 1.091 9 Dubal 70 7 1.196 Dubal 70 10 1.191

Vestbulo 8 97 771 10 y 1 1 Dubal 45

Dubal 3044

451340

Dubal 45Dubal 30

65

477356

10.115 8.697 72 9.343 87 9.266

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Diapositiva 17

17

Como se aprecia la aplicacin del salto trmicode la UNE-EN 442 (10) supone un aumento deldoble en lo referente al gasto, y por lo tantoconlleva un sobredimensionamiento de lastuberas.

En cuanto a los caudales:

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Diapositiva 18

18

Comparacin entre las normas UNE 9015 y UNE-EN 442CAUDALES

Norma UNE9015 (60C)

Norma UNE 442(50C)

Local W Kcal/h N Elemento

sEmisincalorfica Caudal

Elementos Emisincalorfica Caudal

Saln 4.229 3.6361

2 y 315

2x71.273

2x1.196 0,043 18

2x10

1.2832x1.191 0,086

Bao 261 224 4 3 255 0,003 4 285 0,006

Cocina 734 631 5 6 677 0,009 8 635 0,018

Dormitorio 1 862 741 6 6 886 0,01 8 831 0,02

Dormitorio 2 1.206 1.037 7 7 1.196 0,014 10 1.191 0,028

Dormitorio 3 657 565 8 6 677 0,008 8 635 0,016

Dormitorio 4 1.269 1.091 9 7 1.196 0,015 10 1.191 0,030

Vestbulo 8 97 771 10 y 1 1 4

4451340

0,011 6

5

477356 0,022

10.115 8.697 72 9.343 0,113 87 9.266 0,226

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Diapositiva 19

Distribucin instalacin tuberas

19A

B

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Diapositiva 20

20

RED HIDRULICA DE CALEFACCINDISTRIBUCIN

RAMAL 1

RAMAL 2

AB

C D E F

G H I J K

RAD 1Comedor-Estar

RAD 2

Comedor-Estar

RAD 3

Comedor-Estar

RAD 4

Bao

RAD 5

Cocina

RAD 6Dormitorio 1

RAD 7Dormitorio 2

RAD 8Dormitorio 3

RAD 9Dormitorio 4

RAD 10Pasillo

RAD 11Vestbulo

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Diapositiva 21

21

DISEO DE LA RED DE TUBERIAS

Norma UNE9015 (60C)

Norma UNE 442(50C)

N Elementos Caudal Elementos Caudal (kg/s)

12 y 3

152x7 0,043

182x10

0,0,86

4 3 0,003 4 0,006

5 6 0,009 8 0,018

6 6 0,01 8 0,02

7 7 0,014 10 0,028

8 6 0,008 8 0,016

9 7 0,015 10 0,030

10 y 11 4

4 0,011

65

0,022

72 0,113 87 0,226

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Diapositiva 22

22


Norma UNE 442(50C)

N Elementos Caudal (kg/s) Kg/h l/h

12 y 3

182x10

0,086 310

4 4 0,006 21,6

5 8 0,018 64,8

6 8 0,02 72,0

7 10 0,028 100,8

8 8 0,016 54,1

9 10 0,030 108,0

10 y 11 6

5 0,022 79,2

87 0,226 810,5 810

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Diapositiva 23

23


Caudal total: 426,9 + 383,6 = 810,5

Ramal 1 Kg/h l/h Ramal 2 l/h l/h

2 100 1 110 110

3 100 200 11 36,2 146,2

5 64,8 264,8 6 72 218,2

8 54,1 318,9 10 43 261,2

9 108 426,9 4 21,6 282,8

7 100,8 383,6

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Diapositiva 24

24

RED HIDRULICA DE CALEFACCINCAUDALES

RAMAL 1

RAMAL 2

AB

C D E F

G H I J K

RAD 1Comedor-Estar

RAD 2

Comedor-Estar

RAD 3

Comedor-Estar

RAD 4

Bao

RAD 5

Cocina

RAD 6Dormitorio 1

RAD 7Dormitorio 2

RAD 8Dormitorio 3

RAD 9Dormitorio 4

RAD 10Pasillo

RAD 11Vestbulo

810,5

426,9

383,6

108

318,9

54,1

264,8

64,8

200

100

100

100,8

282,8

21,6

261,2

43

218,272

146,2

36,2

110

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Diapositiva 25

25

CLCULO DE LA RED HIDRULICA DECALEFACCIN

Se utilizar tubera de cobre para la ejecucin del circuito detuberas.

El clculo se efectuartratandode minimizar las prdidasde carga

Aunque la normativa prev que stas no superen los 40mmcda/m, usualmente se trabaja con prdidas entre 12 y16 mmcda/m. Sin sobrepasar los 20.

Por lo tanto, en nuestro caso se elegirn dimetrossuficientementegrandes para que estoocurra.

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Diapositiva 26

26

Podemos encontrar los dimetros en el bacocorrespondiente a las tuberas de cobre, teniendo en cuentaas mismo la velocidad del agua, que segn el RITE, nodebe pasar de 0,5 m/s en los montantes y 1,5 en losdistribuidores.

NOTA:

El anterior reglamento, recomendaba que la prdida de carga fuera inferior a 40 mmcdapor metro, y la velocidad de 2 m/s en locales habitados y 3 en tuberas enterradas.

CLCULO DE LAS TUBERAS

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Diapositiva 27

27

DISEO DE LA RED DE TUBERIAS- RAMAL - 1

Norma UNE 442(50C)

Tramo l/h l/m

A-B 810 13,5

B-C 426 7,1

C-D 318 5,3

D-E 265 4,4

E-F 200 3,3

F-1 100 1,6

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Diapositiva 28

28

13,5 l/m

V=1 m/s

TRAMO A-B

Dimetro interior 20 mm (Tubo de cobre de 22x1)

Prdidas de carga 20 mmcda/m

V=0,65 m/s

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Diapositiva 29

29

En el punto B, la tubera se bifurca en dos ramales, uno para lateral de lavivienda, siendo:

Tramo Q D VPrdida de

carga

unitaria

L

A-B l/m mm m/s Mmcda/m m

13,5 20 0,65 20 7,1

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Diapositiva 30

30

Tramo Q D VPrdida de

carga

unitaria

L

A-B 13,5 20 0,65 20 7,1

B-C 7,1 16,5 0,38 14 0,65

C-D 5,3 14,5 0 ,37 14 4,15

D-E 4,4 14,5 0 ,33 12 7,95

E-F 3,3 14,5 0,25 7 1,25

F-2 1,6 10,5 0 ,23 9,5 6,3

C-9 10,5 0,22 8 2,4

D-8 10,5 0,20 9 2,66

E-5 8,5 0,20 9 1,77

F-3 10,5 0,23 9,5 1,77

RAMAL 1

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Diapositiva 31

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Tramo L D V

Prdida

de carga

unitaria

B-G 4,65 16,5 0,33 10

G-H 3,05 14,5 0,33 12

H-I 1 14,5 0,28 9

I-J 0,95 14,5 0,25 7

J-K 4,85 12,5 0,23 7

G-7 2,96 10,5 0,2 7

H-4 3,61 8,5 0,1 2,5

I-10 1,92 8,5 0,11 3

J-6 5,96 10,5 0,16 4,3

K-11 2,66 8,5 0,14 4,5

K-1 13,70 10,50 0,24 10

RAMAL 2

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Diapositiva 32

32

RED HIDRULICA DE CALEFACCINDIAMETRO TUBERIAS

RAMAL 1

RAMAL 1

AB

C D E F

G H I J K

RAD 1Comedor-Estar

RAD 2

Comedor-Estar

RAD 3

Comedor-Estar

RAD 4

Bao

RAD 5

Cocina

RAD 6Dormitorio 1

RAD 7Dormitorio 2

RAD 8Dormitorio 3

RAD 9Dormitorio 4

RAD 10Pasillo

RAD 11Vestbulo

20 mm

16,5

16,5

10,5

14,5

8,5

14,5

8,5

14,5

10,5

10,5

10,5

14,5

8,5

14,5

8,5

10,5

12,5

8,5

10,5

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Diapositiva 33

33

El paso siguiente consistir en mostrar de que forma puede estimarsela prdida de carga principal (nicamente de los tramos rectos de lastuberas)de la lnea de alimentacinde los radiadores.

Prdidas de carga principales

Se tendr que determinar qu circuito es el que genera una mayorresistenciaal paso del lquido.

Parece lgico que ser el de mayor longitud, en nuestro caso setratardel ramal II.

El circuito hidrulico completo, con la lnea de retorno, se puede veren el siguiente esquema:

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Diapositiva 34

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RAMAL 1

RAMAL 2

RadiadorDormitorio 2

RadiadorBao

RadiadorPasillo

RadiadorDormitorio 1

RadiadorVestbulo

RadiadorComedor-Estar

CALDERA

A A

B

B

GHIJK

111 6 10 4 7

1.480 555 966

415 332

1.385

5.133 W

2.035 3.0013.416

3.748

Potencia calorfica 10.7 76 W

Para estimar las prdidas de carga

principales se confecciona la tablasiguiente

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Diapositiva 35

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Datos de los planos de tuberas Clculos hidrulicos

Capacidad delramal

Caudal deagua

Longitud deltramo Dimetro

Velocidaddel fluido

Prdida decarga unitaria

Prdida decarga del tramo

C L D V Hf/L hf

Tramo W Kg/h M mm m/s mmcda mmcda

AB 10.776 810 7,1 20 0,47 15 107,7

BG 5.133 383,6 4,65 16,5 0,33 10 46,5

GH 3.748 282,8 3,05 14,5 0,33 12 36,6

HI 3.416 261,2 1 14,5 0,28 9 9

IJ 3 .001 218,2 0,95 14,5 0,25 7 6,7

JK 2.035 146,2 4,85 12,5 0,23 7 34

K1 1.480 110 13,7 10,5 0,24 10 137

Prdidas de carga debidas a la longitud de la tuberia de ida ( sin accesorios) 337,5

Las perdidas de carga debidas a la longitud de la tubera de ida ms larga (ramal 2) sin tener encuenta la existencia de resistencias aisladas (accesorios) son de 337,5 mmca

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Diapositiva 36

36

Sabiendo que las prdidas de carga originadas por codos, llaves,radiadores, vlvulas, etc., suponen aproximadamente del 30 al 50 % delas prdidas totales, que son las que deber vencer la bomba,sirvindonos para su clculo.

En nuestro caso sern:

337,5 x 0,50 = 169 mmcda

Prdidas de carga secundarias

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37

Sern la suma de las calculadas, principales y secundarias, de talforma que:

337,5 + 169 = 506,5 mmcda

Prdidas de carga totales

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Diapositiva 38

Segn el caudal calculado de 6,4 l/m,equivalente a 0,38 m/h y una prdida decarga total de 5,06 mcda, segn las curvasde caudal presin de Roca:

38

Clculo de la bomba aceleradora

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39

Clculo de la bomba aceleradora

0,38 m/h

5,06 mcda

PC-1055

7,6

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40

Actualmente la bomba de circulacin se encuentra, lamayor parte de veces instalada en el conjunto de lacaldera. Por esto el fabricante, que conoce sta, decide sutipo y caractersticas.

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41

La potencia de la caldera o calderas, debe ser capaz desuministrar la debida a la calefaccin y la necesaria para ACS,siempreque el grupo trmico mixto sea inferior a 70 KW.

l cul o de l a cal der a y el quemador

Ser la necesaria para contrarrestar las prdidas de calor atravs de los elementos constructivos, el aire de renovacindebido a la infiltracin e incluso la prdida de calor disipada enlas tuberas.

La Directiva Europea 92/42/CEE, exige que los generadorestengan un rendimiento mnimo en funcin de su potencia til ysegn el tipo de combustible. RD 275/1995.

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42


El clculo de la caldera se realiza:

Potencia total demandada = QT=(SQ1+SQ2)xA

Q1=potenciacalculada en radiadores (kcal/h)

Q2=potenciadisipada en tuberas

A=aumentode inerciade valor 1,2

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43

En nuestro caso:

Potenciacalculadaradiadores= 9.266 w = 9,26 kw

Potenciatotal = (9,26/0,95) 1,2 = 11,64 kW.

Se adoptar una caldera tipo Lidia,modeloLidia20 GT, de Roca.

Grupo trmico de fundicin, para instalaciones de calefaccinpor agua caliente hasta 4 bar y 100 C.

De 15000 kcal/h (17,4 kW), y resto de caractersticas que se

pueden observar en:


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45

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En cuanto al quemador, igual que la bomba, el fabricante, porconocer las caractersticas de la caldera, procede a la seleccindel mismo. En nuestro caso un quemador de gasoil Neotronic ykadet-tronic,modelo , 2RS-L, integradoen el conjunto.

l cul o del quemador

Cumplir lo establecido en la IT correspondiente, en cuanto a suregulacin. Bsicamente se elige en funcin de la longitud yanchurade la llama.

El consumo horario de combustible viene expresado por la

expresin:Q = Qt/Pci x

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47

Se deben cumplir una serie de consideraciones con estoselementos:

Es recomendable un vaso de expansinpor caldera

La alimentacin del agua al sistema no puede realizarse conuna conexin directa a la red de distribucin urbana. Se debehacer preferiblemente por el vaso.

No debe existir ningn elemento de corte entre la caldera y elvaso de expansin.

l cul o del vaso de expansin

Debe calcularse el contenido de agua de la instalacin llena,contando todos los elementos que la componen, radiadores,

tuberas y caldera, que junto al factor de dilatacin del agua,nos proporcionar la capacidad del vaso

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48


Si es abierto,se debe calcular segn la norma UNE100155-88.

Su volumenser:

Vv=0,06xVt

Vt, es el volumen total de agua incluyendo el de la tubera deconexindesde lacaldera al vaso, que tendr un dimetrode:

D = 15+1,5vP

P=potencia instalada en kW, en todo caso, D=25 mm (segn UNE100157-89)

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En nuestro caso, ser un vaso de expansin cerrado,con compresor, con lo cual:

Vu= Vxfd

Vu=volumen til del vaso

V= Volumen del agua de la instalacin

Fd= Factor de dilatacin, que en la EN 442, adopta como valor medio la temperatura de 70 C


Dilatacin del agua

10 C 0,027 %

20 C 0,177 %

30 C 0,435 %

40 C 0,782 %

50 C 1,21 %

60 C 1,71 %70 C 2,27 %

80 C 2,90 %

90 C 3,59 %

100 C 4,34 %

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V, tendr un valor:

Radiadores: 68 elementosx 0,43 litros + 19x0,36 = 36,08 litros

Caldera:13 litros

Tuberas: 71 m x 0,079 l/m = 5,6 litros ( Segn tabla norma UNE-EN1057)

Factor de dilatacin: 0,0228 (temperatura 70 C, EN 442).

El volumen del vaso ser tan pequeo que:

Se dispondr de un modelo vasoflex de Roca de 8 litros decapacidad, que cumplircon lo quese necesita.

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La cantidad de agua a evacuar por la vlvula de seguridad secalcula:

G = Q/500

G = capacidad de descarga

Q = potencia de la instalacin en kcal/h (o en kW)

Vlvul a de seguridad

La tubera de descarga, la encontramos en la tabla 2 de la normaUNE 100157-89, donde la obtenemos directamente en funcin dela potenciade la instalacin.

El dimetro mnimo de sta, ser de 20 mm.

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Se calcula para una reserva de un mes, considerando el de Enero. Por medio del mtodode los grados-da:

Capacidad depsito combustible

Donde: V=Volumen depsito de combustible

Q=Potencia de la caldera (kcal/h)

G=nmero de grados-da mes de Enero (UNE 24026)

V=coeficiente de uso

I=coeficiente de intermitencia

R=rendimiento de la instalacin=0,85P=poder calorfico de la unidad de combustible (10500 kcal/kg, para el gasleo C)

352

1050085,025

85,01243,2571500024

PRtt

IvGQV

ei

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IT 1.3.4.1.3.1 Evacuacin de los productos de lacombustin

Chimenea

La evacuacin de los productos de la combustin enlas instalaciones trmicas se realizar de acuerdocon las normas generales expuestas en el prrafoanterior.

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En los edificios de nueva construccin en losque se prevea una instalacin trmica, laevacuacin de los productos de lacombustin del generador se realizar por unconducto por la cubierta del edificio, en elcaso de instalacin centralizada.

Chimenea

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IT 1.3.4.1.3.2 Diseo y dimensionado de chimeneas

Chimenea

Es vlido el dimensionado de las chimeneas de acuerdo a loindicadoen las normas:

UNE-EN 13384-1

UNE-EN 13384-2

UNE 123001, segn el caso.

La chimenea ser de material resistente a la accin agresiva de losproductos de la combustin y a la temperatura, con la estanquidadadecuadaal tipo de generador empleado.

En el caso de chimeneas metlicas la designacin segn la normaUNE-EN 1856-1 o UNE-EN 1856-2 de la chimenea elegida en cadacaso y para cada aplicacin ser de acuerdo a lo establecido en lanorma UNE 123001.

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Para la evacuacin de los productos de lacombustin de calderas, la seccin de la chimenea,su material y longitud sern los certificados por elfabricante de la caldera.

Chimenea

Pero conociendo la potencia de la instalacin (kcal/ho kW) as como la altura de la chimenea, la obtencindel dimetro es inmediata, por medio del diagrama

comercial. Siendo ste el procedimiento ms seguro.

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Chimenea

La seccin de los conductos de humo, secalcula de acuerdo con el volumen de gasesprevisibles, segn la expresin de Sander,con la cual se hacen los bacoscorrespondientes:

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S = 213 cm

Seccincircular:

213 = 3,14x R con loque

R = 8,25 cm

Con estos datos:

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Calculo de Instalacion de Calefaccion Bitubular

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