ejercicio practico cte he1
DESCRIPTION
TEMA 4.- APLICACIÓN DEL C.T.E., A UN CASO PRÁCTICO OPCIÓN SIMPLIFICADA. Instalaciones II, ETSAG.TRANSCRIPT
Diapositiva 1
INSTALACIONES-2DEPARTAMENTO DE CONSTRUCCIONES ARQUITECTÓNICAS
CURSO
08-09
Profesor: Dr. Julián Domene García
4º C
TEMA 4.-
APLICACIÓN DEL C.T.E., A UN CASO PRÁCTICO
OPCIÓN SIMPLIFICADA
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Diapositiva 2
EJEMPLO DE C�LCULO
2
El techo est� compuesto de:
� En la parte inferior, enlucido de yeso de 800 kg/m�
� Bovedilla cer�mica de 16 cm con un coeficiente de conductividadt�rmica de 0,23
� Hormig�n en masa con �ridos ligeros de 1600 kg/m�� Encima se hallan dispuestas unas placas de poliestirenoexpandido, con � = 0,034 W/m �K
� Sobre el aislante hay un desv�n seguido de una cubierta conensamblamiento cer�mico
La cubierta es inclinada a dos aguas, con ensamblamiento cer�micoy una superficie de 225,4 m�.
Se dispone de una vivienda unifamiliar de 129 m� de superficie �til,ubicada en Bilbao.
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Diapositiva 3
EJEMPLO DE CÁLCULO
3
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Diapositiva 4
EJEMPLO DE CÁLCULO
4
El muro está constituido de la siguiente manera:� Enlucido exterior de cemento, cuya densidad aparente es2.000 kg/m³� Ladrillo hueco normalizado de 15 cm de espesor y 1.200Kg/m³�Aislante de 5 cm de espesor (poliestireno expandido conhidrofluorcarbones-HFC)� Ladrillo hueco normalizado de 10 cm de espesor�Un enlucido de yeso de densidad aparente 800 kg/m³
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Diapositiva 5
EJEMPLO DE CÁLCULO
5
El suelo de la vivienda está constituido por cuatromateriales diferentes:
� Mosaico, con Us = 0,41 W/m ºK
� Mortero de cemento de densidad aparente 2.000 kg/m³
� 20 cm de hormigón de áridos ligeros de la mismadensidad
� Grava rodada de densidad 1.700 Kg/m³
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Diapositiva 6
EJEMPLO DE CÁLCULO
6
Las ventanas y puertas acristaladas (exteriores), son de carpinteríade madera y doble cristal con cámara de aire de 6 mm, cuyatransmitancia térmica es 3,26 W/m² ºC.
La superficie total acristalada es de 25,2 m² para las ventanasexteriores y de 6,8 para las puertas acristaladas que lindan con elexterior.
Existen dos puertas de madera de pino opaca. Una linda con unlocal no calefactado y la otra con el exterior. Sus transmitancias son1,98 y 3,49 respectivamente.
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Diapositiva 7
EJEMPLO DE CÁLCULO
7
La tabiquería interior está formada por tabicón de ladrillohueco, guarnecidas y enlucidas de yeso por ambas caras.Las puertas interiores son de madera de pino chapadas porlas dos caras y opacas.
¿Calcular las transmitancias térmicas de loscerramientos y su validación, según el CTE?
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Diapositiva 8
8
Superficie fachada = 152,6 m²Superficie huecos en la fachada=25,2+6,8=32 m²Cumple la condición a)
También cumple la segunda condición
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Diapositiva 9
9
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Diapositiva 10
1010
Procedimiento de aplicación de la opción simplificada
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Diapositiva 11
1111
Cálculo según opción simplificada
En nuestro caso son todos de:BAJA CARGA INTERNA
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Diapositiva 12
.
A efectos de comprobación de limitación de condensaciones:clase de higrometría.
Clase de higrometría 3: Todos los espacios designados con tipode uso Residencial y Oficina.
Clase de higrometría 4: Todos los espacios designados con tipode uso Restaurante, etc.
Clase de higrometría 5: Todos los espacios designados con tipo de uso Lavandería, etc.
Clasificación de los espacios (según el apartado 3.1.2)
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Diapositiva 13
La permeabilidad de las carpinterías de los huecos y lucernarios de loscerramientos, se limita en función de:
El clima de la localidadLa zonificaciónclimática
El edificio se encuentra en zona climática C, en este caso, la permeabilidadal aire de las carpinterías, medida con una sobrepresión de 100 Pa, tendráunos valores inferiores a 27 m³/h m². (apartado 2.3.3)
Por lo tanto, para que se CUMPLA la limitación de permeabilidad de loshuecos, estos deben ser de clase 2, clase 3 o clase 4. (apartado 3.2.4).
Ensayo: UNE EN 1026:2000Clasificación:UNE EN 12207:2000
Cumplimiento de las limitaciones de permeabilidad
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Diapositiva 14
1414
Cálculo según opción simplificada
M1
UM1
H
UH
S1
US1UC1 C1
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Diapositiva 15
1515
Cálculo según opción simplificada
H
NH
M2
UM2
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Diapositiva 16
16
Cálculo según opción simplificada
Se calcula la transmitancia térmica media del cerramiento, una vezrealizadas las de los elementos del mismo, mediante las expresionessiguientes, que se obtienen en la Tabla 3.1, del DB HE1:
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Diapositiva 17
17
Cálculo según opción simplificada
Se calcula la transmitancia térmica media del cerramiento, una vezrealizadas las de los elementos del mismo, mediante la expresión:
� �� �
�
����
PFM
PFPFMMMm AA
UAUAU
Siendo:
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Diapositiva 18
1818
E.1.1.- Cerramientos en contacto con el aire exterior
15 150 50
Enlucido de cemento
Ambiente
exterior
Ladrillo doble hueco 15 cm
Ladrillo hueco 10 cm
Enlucido de yeso
Aislante
Ambiente
interior
100 15
Resistencia superficial interiorResistencia superficial exterior
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Diapositiva 19
19
Las resistencias térmicas superficiales se obtienen de la Tabla E-1
E.1.1.- Cerramientos en contacto con el aire exterior
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Diapositiva 20
2020
Muro de cerramiento opaco con contacto al exterior
Material Capa Espesor (m)
Rs � (*) Rj=ej/�jm��K/W
Exterior 0 -Enlucido de
cemento1 0,015
Ladrillo doble hueco
2 0,15
C�mara aire 3 0,050 0,029
Ladrillo hueco 4 0,100Enlucido de
yeso5 0,015
Ambiente interior
6 -
RT
(*) Valores obtenidos según la norma UNE EN ISO 10456:2001
0,04
0,13
1,40
0,49
0,49
0,30
0,04
0,13
0,0107
0,33
0,204
0,05
1,72
2,485
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Diapositiva 21
21
Por lo que la transmitancia será:
UM1 = 1/2,485 = 0,40
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Diapositiva 22
22
Cálculo de superficies de los cerramientos:
1. Cerramiento exterior
Muro exterior: 56,50 x 2,7 = 152,60 m²Huecos: 25,2 m²Puertas exteriores = 6,8 m²
Superficie = 152,6 – 25,2 – 6,8 = 120,6 m²
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Diapositiva 23
23
Se dispone de una vivienda unifamiliar de 129 m² de superficie útil, ubicadaen la comarca del Vallés Occidental, provincia de Barcelona.
N
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Diapositiva 24
24
Cálculo de superficies de los cerramientos segúnorientación :
Área de la fachada
Área ventanas
Área muro cálculo
N 34,44 5,8 28,64
Este 38 7,6 30,4
Oeste 38,7 6,3 32,4
Sur 41,46 12,3 29,16
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Diapositiva 25
25
Puentes térmicos: UPF
En nuestro caso, se tendrán los puentes térmicos:
� Contorno de huecos
� Caja de persianas� Pilares en fachada
Contorno de huecos, y pilares en fachada, en nuestro caso no se deberántener en cuenta, al ser < 0,5 m²
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Diapositiva 26
26
Puentes térmicos: UPF
Caja de persianas
Se tratarán como cámaras de aire medianamente ventiladas; para ello, laresistencia térmica será la mitad de la encontrada en la tabla E-2, apartadoE-1.1.- 6)
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Diapositiva 27
27
Puentes térmicos: UPF
Caja de persianas:
Se tomará por lo tanto, el valor:
R = 0,18/2 = 0,09 m²K/W
Y su transmitancia térmica:
11,1109,011
3 ���R
UPF
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Diapositiva 28
2828
Cálculo según opción simplificada
Para calcular la transmitancia térmicamedia del cerramiento:
� �� �
�
����
PFM
PFPFMMMm AA
UAUAU
Siendo:
31
3311
PFM
PFPFMMMm AA
UAUAU�
����
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Diapositiva 29
2929
Cálculo según opción simplificada
Sustituyendo valores:
90,36,12011,1190,340,06,120
�����MmU
73,0�MmU W/m²K
limMMm UU �Ahora comprobaremosque:
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Diapositiva 30
3030
Comprobación del valor obtenido con la tabla 2.2
=0,73
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Diapositiva 31
31
E.1.4.1.- Transmitancia térmica de huecos
E.1.4.- Huecos y lucernarios
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Diapositiva 32
32
Transmitancia térmica de huecos
Superficie acristalada = 25,2 + 6,8 = 32 m²
FM = 0,151
UHV = 3,0 W/m²K, para acristalamiento doble con cámara de 6 mm deespesor y carpintería de madera
UHm = 3,3 W/m²K, para carpinteríade madera
UH = (1-0,15)3,0 + 0,15 x 3,3 = 3,04 W/m²K
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Diapositiva 33
3333
Cálculo según opción simplificada
Para calcular la transmitancia térmicamedia del cerramiento:
HH
HHHm U
AUA
U ��
��
�
Siendo:
04,3�HmU W/m²K
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Diapositiva 34
3434
Comprobación del valor obtenido con la tabla 2.2
3,4>3,04
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Diapositiva 35
35
E.1.2.- Cerramientos en contacto con el terreno
E.1.2.1.- Suelos en contacto con el terreno
Para el cálculo de la transmitancia US, utilizaremos el Caso 1:Soleras o losas apoyadas sobre el nivel del terreno o como máximo0,50 m por debajo de éste, por ser el que nos ocupa.
El valor de la transmitancia se obtendrá de la tabla E-3:
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Diapositiva 36
36
E.1.2.- Cerramientos en contacto con el terreno
Ra = Resistencia térmica del aislante = 0,050/0,029 = 1,72(Poliestireno expandido con HFC); y como el aislamientoes continuo, se utiliza la columna D>1,5
B’ =Longitud característica de la solera
56,45,565,0
1295,0
' ��
��
�P
AB
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Diapositiva 37
37
E.1.2.- Cerramientos en contacto con el terreno
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Diapositiva 38
3838
Cálculo según opción simplificada
Para calcular la transmitancia térmica media de lasolera:
SS
SSSm U
AUA
U ��
���
Siendo:
48,0�SmU W/m²K
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Diapositiva 39
3939
Comprobación del valor obtenido con la tabla 2.2
48,0�
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Diapositiva 40
40
La transmitancia térmica U (W/m²K) viene dada por la siguiente expresión:
U = UP x bsiendo
UP la transmitancia térmica de la partición interior en contacto con el espacio no habitable, calculada según el apartado E.1.1,
Transmitancia de la cubierta
El coeficiente de reducción de temperatura b para espaciosadyacentes no habitables (trasteros, despensas, garajesadyacentes...) y espacios no acondicionados bajo cubierta inclinadase podrá obtener de la tabla E.7 en función de:� La situación del aislamiento térmico� Del grado de ventilación del espacio� De la relación de áreas entre la partición interior y el cerramiento(Aiu/ Aue).
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Diapositiva 41
4141
UP
Habitable
No habitable
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Diapositiva 42
42
Transmitancia UP
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Diapositiva 43
4343
Material Capa
Espesor (m)
Rs � (*) Rj=ej/�jm��K/W
No habitable 0 -
poliestireno expandido tipo IV
1 0,05
Hormigón en masa con áridos ligeros
2 0,03
bovedilla cerámica 3 0,160
Enlucido de yeso 4 0,015Ambiente interior
habitable5 -
RT
Valores obtenidos según la norma UNE EN ISO 10456:2001
0,10
0,034
0,73
0,30
0,10
1,66
0,04
0,05
0,10
2,65
CÁLCULO UP
0,10
0,700,23
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Diapositiva 44
44
Cubiertas en contacto con el aire exterior
La transmitancia del cerramiento en contacto con el local no habitable,será:
Siendo el valor de la resistencia térmica total de 2,65 m²K/W
37,065,211
���T
P RU W/m²K
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Diapositiva 45
4545
Se distinguen dos grados de ventilación en función del nivel de estanqueidad del espaciodefinido en la tabla E.8:
CASO 1: espacio ligeramente ventilado, que comprende aquellos espacios con un nivel de
estanqueidad 1, 2 o 3;
CASO 2: espacio muy ventilado, que comprende aquellos espacios con un nivel deestanqueidad 4 o 5.
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Diapositiva 46
4646
Se distinguen dos grados de ventilación en función del nivel de estanqueidad del espaciodefinido en la tabla E.8:
CASO 1: espacio ligeramente ventilado, que comprende aquellos espacios con un nivel de
estanqueidad 1, 2 o 3;
CASO 2: espacio muy ventilado, que comprende aquellos espacios con un nivel deestanqueidad 4 o 5.
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Diapositiva 47
4747
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Diapositiva 48
4848
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Diapositiva 49
4949
Cálculo según opción simplificada
Para calcular la relación entre las superficies del forjado y la cubierta, setendrá:
57,04,225
129��
ue
iu
AA
Con lo cual en la tabla siguiente:
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Diapositiva 50
5050
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Diapositiva 51
5151
Cálculo según opción simplificada
Con el coeficiente b = 0,96, la transmitancia de lacubierta, será:
35,096,037,0 ����� bUU PW/m²K
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Diapositiva 52
5252
Comprobación del valor obtenido con la tabla 2.2
> 0,35
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Diapositiva 53
5353
Fichas justificativas de la opción simplificadaC-1 X
M1 28,64 0,73 20,90
0,94Caja persianas 0,6 11,11
11,11
11,11
11,11
6,6630,4 0,73
0,73
0,73
M1
M1
M1
Caja persianas
Caja persianas
Caja persianas
0,55
0,57
0,57
32,4
29,16
22,19
6,1123,65
6,3321,28
6,33
29,2427,56
0,91
30,9528,3
0,9
32,9729,98
0,92
29,7327,61
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Diapositiva 54
5454
Fichas justificativas de la opción simplificada
S1 129 0,5
0,5
129,064,564,5
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Diapositiva 55
5555
Fichas justificativas de la opción simplificada
C1 129 0,35
0,35
45,15 12945,15
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Diapositiva 56
56
TEMA 4
56
TEMA 4 Fichas justificativas de la opción simplificadaH 5,8 3,26
3,26
18,918,95,8
H 7,6 3,26 24,77
H 6,3 3,26 20,53
H 12,3 3,26 40,09
7,624,77
3,26
6,3
12,3
20,53
40,09
3,26
3,26
Factor solar
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Diapositiva 57
57
Factor solar modificado de huecos y lucernariosEl factor solar modificado en el hueco FH o en el lucernario FL se determinará utilizando la siguiente expresión:
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Diapositiva 58
58
Factor solar modificado de huecos y lucernarios
Valor de FS:Las persianas dispondrán de lamas verticales, con un ángulode inclinación de 60º.
Se utilizará la tabla E-13 del apéndice correspondiente
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Diapositiva 59
59
Factor solar modificado de huecos y lucernarios
Valor de = 0,75; obtenido de la base de datos Lider
FM = 0,15, obtenido anteriormente
Según la tabla E-10, el valor de ,será:�
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Diapositiva 60
60
Factor solar modificado de huecos y lucernarios
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Diapositiva 61
61
Factor solar modificado de huecos y lucernarios
Sustituyendo estos valores, para el sur y el este, tendremos:
Para el oeste:
� �� � 20,075,003,304,0151,075,0151,0132,0 ��������SF
� �� � 19,075,003,304,0151,075,0151,0129,0 ��������OF
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Diapositiva 62
62
TEMA 4
62
TEMA 4 Fichas justificativas de la opción simplificadaH 5,8 3,26
3,26
18,918,95,8
H 7,6 3,26 24,77
H 6,3 3,26 20,53
H 12,3 3,26 40,09
7,624,77
3,26
6,3
12,3
20,53
40,09
3,26
3,26
Factor solar
0,20
0,20
0,19
1,52
1,52
0,201,19
1,19
0,192,46
2,46
0,20
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Diapositiva 63
63
TEMA 4
63
TEMA 4 Fichas justificativas de la opción simplificadaC-1 X
0,73------
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0,500,3
3,26T-2.1
0,95
0,650,53
4,40
0,940,910,9
0,92
T-2.20,73X
3,043,043,043,04
3,43,9
4,4
0,50,5 0,35 0,41
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Diapositiva 64
64
TEMA 4
64
TEMA 4 Fichas justificativas de la opción simplificada
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