1、CGSB CAN/.CGSB- 12.4-ML tt 1874650 0001193 T71 CANICGSB-12.4-M91 SupersededRemplace NATIONAL NORME STANDARD N AT1 O NA LE CAN/CGSB-12.4-M76 OF CANADA DU CANADA I 1 Heat Verre Absorbing athermane Glass * Canadian General Standards Board ss Office des normes gnrales du Canada O this surface is not ill
2、uminated by direct sunlight so its brightness is very low. Thus very little radiation enters the radiometer by reflection compared with the solar radiation that comes through the glass. The long-wave radiation emitted by the glass is excluded from the radiometer by a quariz window at the entrance to
3、 the radiometer tube. rflchis proviennent de la face intrieure peinte en noir de cet abat-jour; cette face prsente une luminosit trs faible du fait quelle nest pas directement claire par la lumire du soleil. Ainsi, une trs faible quantit de rayons rflchis pntrent dans le radiomtre comparativement au
4、x rayons solaires qui traversent le verre. Les rayons grandes ondes mis par le verre sont carts du radiomtre au moyen dune fentre de quae place lentre du tube du radiomtre. Radiometer in position for transmission test Radiomtre install pour lessai de transmission Radiometer in position for reflectio
5、n test Radiomtre install FIGURE Al Arrangement for Transmission and Reflection Test on Glazing Unit Montage pour lessai de transmission et de rflexion duo panneau de verre A2.4 The procedure for determining the reflection factor is similar: La mthode de calcul du facteur de rflexion est semblable: A
6、2.4.1. The radiometer is aimed at the sun and the voltage, el, generated by the thermopile is measured. A2.4.2 The glazing unit is placed behind the radiometer and positioned so that the direct rays from the sun have the desired angle of incidence at the glass surface. The radiometer is then aimed a
7、t the image of the sun produced by the glazing unit. The thermopile output voltage, e3, is measured. A2.4.3 The solar reflection factor for the glazing unit is simply p = edel. Pointer le radiomtre vers le soleil et mesurer la tension, el, produite par la thermopile. Placer le panneau derrire le rad
8、iomtre en lorientant de faon que les rayons mis directement par le soleil aient langle dincidence voulu sur la surface du verre. Pointer ensuite le radiomtre vers le reflet du soleil mis par le panneau de verre. Mesurer la tension de sortie, e3, de la thermopile. Le facteur de rflexion de la lumire
9、du soleil pour ce panneau est simplement le suivant: p = edel. O A2.4.4 In this case it is important that the brightness of the background seen through the window by the radiometer Dans ce casci, il est important que la luminosit du fond, perue travers la fentre par le radiomtre, soit trs faible par
10、 A2 CAN/CGSB-I 2.4-M91 CGSB CANICGSB- 12.4-Pl91 Xf 1874b50 OOOLZOb 45T be very low compared with the brightness of the suns image. The black-painted shade behind the unit shown in Figure Al ensures that this is the case. rapport celle du reflet du soleil. Voil ce quassure labat-jour peint en noir il
11、lustr la figure Al. A3. THERMAL CONDUCTANCE OF ME INTERPANE CONDUCTIVIT MERMIQUE DE LA LAME DAIR ENTRE AIRSPACE DEUX VITRES A3.1 Heat is transferred across an airspace in a window by two quite separate processes: by thermal radiation and by conduction, or a combination of conduction and convection.
12、The rate of heat transfer by each process is dependent on the difference between the temperatures of surfaces that bound the space. The ratio of the heat flux to the temperature difference is the thermal conductance of the airspace. A3.2 The conductance associated with the radiant heat transfer depe
13、nds on the emissivity of the enclosing surfaces as well as on the mean temperature. The thin film of metal that is applied to the surface of the glass in a reflective glazing unit generally reduces the emissivity of the surface and hence reduces the conductance of the airspace. The conductance of th
14、e airspace in this type of unit can be determined experimentally using a guarded hot plate apparatus of the type normally used to determine the thermal conductivity of materials. A sample is made by taking two sheets of glass of the same types that are used in the glazing unit, and separating them b
15、y about 6 mm by using small pieces of flat glass ai each corner. The thickness of the airspace is then the same as the thickness of the glass spacer pieces, which can be measured accurately with a micrometer. The two sheets of glass can be held together by a strip of adhesive tape applied all around
16、 the perimeter. This also completes the enclosure of the airspace. A pair of identical samples are placed in a guarded hot plate and their overall thermal resistance is measured in the normal way. The resistance of the airspace is obtained by subtracting the thermal resistance of the two sheets of g
17、lass from the measured total resistance. The conductance of the airspace is just the reciprocal of its thermal resistance. The conductance obtained in this way is for the particular thickness and mean temperature used for the test. Values for other thicknesses and mean temperatures can be deduced fr
18、om this value. A3.3 O A3.4 La chaleur traverse la lame dair qui spare deux vitres dune fentre de deux faons tout fait diffrentes: par rayonnement thermique et par conduction, ou par une combinaison de conduction et de convection. Le taux de transfert de la chaleur selon chaque mthode est fonction de
19、 la diffrence de temprature entre les deux surfaces qui limitent la lame. Le rapport du flux de chaleur la diffrence de temprature reprsente la conductivit thermique de la lame dair. La conductivit associe au transfert de la chaleur rayonnante est fonction de Imissivit des surfaces qui retiennent la
20、ir, ainsi que de la temprature moyenne. La mince couche de mtal qui est applique la surface du verre dune vitre rflchissante rduit en gnral Imissivit de la surface et, par consquent, rduit la conductivit de la lame dair. La conductivit de la lame dair de ce genre de vitrage peut tre dtermine exprime
21、ntalement laide dune plaque chauffante grillage du modle habituellement utilis pour dterminer la conductivit thermique des matriaux. Lchantillon se compose de deux vitres du mme type que celui des vitres constituant le panneau, spares denviron 6 mm par de petits fragments de verre plat placs chaque
22、coin. Lpaisseur de la lame dair est alors la mme que celle des fragments de verre qui servent de cales despacement et peut tre mesure avec prcision au moyen dun micromtre. Les deux vitres peuvent tre retenues ensemble par un ruban adhsif appliqu sur leur pourtour. Ce ruban sert galement enfermer la
23、lame dair. Placer deux chantillons identiques sur une plaque chauffante grillage et mesurer leur rsistance thermique totale de la manire habituelle. Calculer la rsistance de la lame dair en soustrayant la rsistance thermique des deux vitres de la rsistance totale. La conductivit de la lame dair est
24、inversement proportionnelle sa rsistance thermique. La conductivit ainsi obtenue vaut pour lpaisseur et la temprature moyenne utilises pour le prsent essai. Les valeurs applicables dautres paisseurs et tempratures moyennes peuvent tre tablies partir de cette valeur. A3.5 The conductance due to heat
25、conduction across an La conductivit due la conduction de la chaleur travers une lame dair dune paisseur infrieure 12.5 mm est calcule comme suit: airspace of less than 12.5 mm thickness is he = WW he = KJW where: K = thermal conductivity of the air and, W = width of the airspace. o: K = conductivit
26、thermique de lair et W = largeur de la lame dair. La conductivit thermique de lair une temprature atm est donne par la formule suivante: K = 0.02423 (1 + 0.00324t)W/(m.K) The thermal conductivity of air at temperature “t“ is given K .I 0.02423 (1 + 0.00324t) W/(m.K) where: t = degrees Celsius. o: t
27、degrs Celsius. by the expression: CANICGSB-12.4-WI A3 CGSB CAN/CGSB- L2-4-M91 * 1874650 0001207 396 Thus the experimentally determined total conductance can be separated into its two components by calculating he and subtracting it from the total to get h. This is illustrated in the following example
28、: The thermal resistance of a double glazing unit composed of two sheets of 6 mm thick clear glass and having an airspace thickness of 6 mm was determined by test to be 0.2126 m2.W at a mean temperature of 23.9OC. The resistance of each sheet of glass was known to be 0.0092 m2-K/W. The thermal resis
29、tance of the airspace is, therefore R = 0.2126 - 2 x 0.0092 = 0.1942 m2WW and the thermal conductance is 1 0.1942 h = - = 5.149 W/(MZ.K) The conductance due to heat conduction is 0.02423 (1 + 0.00324 x 23.9) 0.006147 4.247 W/(m2.K) he Hence the conductance due to radiant energy transfer is h, = 5.14
30、9 - 4.247 = 0.902 W/(rnz.K). If both surfaces of the airspace had had an emissivity of unity, the value of the radiant conductance would have been as indicated in figure A2; .e. h, = 5.945 W/(m2.K) for a mean temperature of 23.9“C. The ratio of h, to h, is independent of the temperature and hence ca
31、n be used in conjunction with figure A2 to determine hr for any mean temperature. In this case, h, 0.902 -I - = 0.152 h, 5.945 The radiant conductance of a similar airspace at a mean temperature of 35C is h, I 0.152 x 6.644 = 1.010 Wl(m2.K) where: h: = 6.644 is taken from figure A2. The value of h.
32、for an airspace of 12.5 mm width at a mean temperature of 35C is 0.02423 (1 + 0.00324 x 35) 2.124 W/(,qq 0.01 27 he Thus the total conductance of this type of airspace with a width of 12.5 mm and a mean temperature of 35C would be h = 1.010 + 2,124 = 3.134 W/(m2.K) Ainsi, la conductivit totale dterm
33、ine exprimentalement peut tre spare en ses deux composantes en calculant he et en le soustrayant du total, pour obtenir h, comme lillustre lexemple suivant: Un essai a rvl que la rsistance thermique dun panneau double vitrage compos de deux vitres de verre transparent de 6 mm dpaisseur spares par un
34、e lame dair dune paisseur de 6mm correspond 0.2126 m2.W une temprature moyenne de 23.9“C. La rsistance thermique de chaque vitre gale 0.0092 m2.W. La rsistance thermique de la lame dair gale donc R I 0.21 26 - 2 x 0.0092 et la conductance thermique gale 0.1 942 m2.W h = - 1 I 5.149 W/(MZ.K) O. 1 942
35、 La conductivit due la conduction de la chaleur gale 0.02423 (1 + 0.00324 x 23.9) = 4.247 Wl(m2.K) 0.0061 47 he Ainsi, la conductivit due au transfert de lnergie rayonnante h, I 5.149 - 4.247 = 0.902 W/(m2-K). Si les deux surfaces limitant la lame dair avaient eu une missivit gale un, la valeur de l
36、a conductivit nergtique aurait t celle indique la figure A2; c.-d. h, I 5.945 W/(m.K) pour une temprature moyenne de 23.9“C. Le rapport de h, h, nest pas fonction de la temprature et peut donc tre utilis de concert avec la figure A2 pour dterminer hr pour toute temprature moyenne. Dans ce cas, gale
37、hr 0.902 - = - = 0.152 h: 5.945 La conductivit nergtique dune lame dair semblable une temprature moyenne de 35C gale h, = 0.1 52 x 6.644 = 1 .O10 W/(m2.K) o: h, = 6.644 tir de la figure A2. La valeur de he pour une lame dair de 12.5 mm de largeur une temprature moyenne de 35C gale 0.02423 (1 + 0.003
38、24 x 35) = 2. 24 W/(m2.K) 0.01 27 he 3 Ainsi, la conductivit totale de ce type de lame dair qui a une largeur de 12.5 mm et une temprature moyenne de 35C gale h = 1.010 + 2.124 I 3.134 W/(mZ.K) A4 CAN/CGCB-12.4-M91 When the thickness of an airspace exceeds about Lorsque ipaisseur dune lame dair dpas
39、se environ 12.5 mm, 12.5 mm, convection begins to become significant. The ia convection prend de limportance. La valeur de h, pour les value o hefor wide airspaces can be determined by using lames dair larges peut tre dtermine au moyen des the graphs given in reference 1. graphiques mentionns ia rfr
40、ence 1. CAN/CGSB-12.4-M91 A5 A6 7 6 5 4 3 2 1 FIGURE A2 Degrees Ceisius Degrs Celsius 20 -1 o O 20 30 44 CGSB CANICGSB- L2-4-fl9L * 1874650 0001210 980 A4. A4.1 A4.2 CALCULATION OF SHADING COEFFICIENT Single or Laminated Glass - The calculation of the shading coefficient for single or laminated glas
41、s requires the following data: a. The solar transmission factor, z b. The solar reflection factor, p c. The thermal resistance between outside surface of the glass and outside air, Ra d. The thermal resistance of the glass, RG e. The thermal resistance between the inside surface of The fraction of t
42、he incident solar energy that is the glass and the inside air, Ri. transferred through glass is Ro + 0.5 RG) (1 - T - p) X=Z+l Ra + RG + Ri) The shading coefficient is the ratio of X for the particular glass to X for standard 3 mm thick sheet glass. The value of S.C. depends on the value used for th
43、e factor (Ro + 0.5 RG)/(R + RG + Ri). A value of 0.5 is appropriate for this factor when calculating the shading coefficient for singleglazing units (.e. assuming that R, = Ri). EXAMPLE For an incident angle of 30“ the transmission and reflection for 3 mm thick clear sheet glass are 730 = 0.868 and
44、pm I 0.081 thus: )(.tyldyd 0.868 + 0.5 x 0.051 I 0.894 The values for a 6 mm thick sheet of heat absorbing (HA) glass that transmits 50% at normal incidence are 730 = 0.484 and pJa = 0.056 thus: Xm = 0.484 + 0.5 x 0.460 = 0.714 and the shading coefficient for the heat absorbing glass is Xw 0.714 Xmn
45、bd 0.894 S.C. = - = - 0 0.799 Double Glazing Units -The calculation is slightly more complicated for a double glazing unit. The first step is to calculate the absorption factor for each of the panes. This requires the measurement of the transmission and reflection factors for one of the panes separa
46、tely as well CALCUL DU COEFFICIENT DOMBRAGE Vitrage simple ou feuillet - Les donnes suivantes sont requises pour le calcul du coefficient dombrage des vitrages simples ou feuillets: a. Le facteur de transmission de la lumire du soleil, T b. Le facteur de rflexion de la lumire du soleil, p c. La rsis
47、tance thermique entre la surface extrieure du verre d. La rsistance thermique du verre, RG e. La rsistance thermique entre la surface intrieure du verre La fraction de lnergie solaire incidente qui traverse le verre et lair extrieur, Ra et lair intrieur, R. gale Ro + 0.5 RG) (1 - T - pl X=T+( Ra + R
48、G + Ri) Le coefficient dombrage est le rapport de X pour le verre en question X pour les vitres en verre talons de 3 mm dpaisseur. La valeur du C.O. est fonction de la valeur utilise pour le facteur (Ra + 0.5 RG)/(R + RG + Ri). Une valeur de 0.5 convient aux fins du calcul du coefficient dombrage des panneaux de verre simple vitrage (si on suppose que R, = R,). EXEMPLE A un angle dincidence de 30, les facteurs de transmission et de rflexion du verre vitres transparent de 3 mm dpaisseur galent 730 = 0.868 et pm I 0.
