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In today's lecture we willbe talking about what are the compliance options. So, when we are talking about the differentproperties of the materials, the different factors, the different parameters related to buildingenergy efficiency how do we go about them to comply as per the rating systems, as per thecodes.So, in this case when we are talking about the Indian scenario, we have our code which iscalled Energy Conservation Building Code. And, the requirements for energy efficientbuildings come from this code which is ECBC. When we are talking about ECBC and whenwe are talking about the building envelope at large, there are few components distinctlywhich we have also seen in the previous lecture. So, we have opaque walls and roofs and thenwe have fenestrations and skylights, all of these together form the building envelope; otherfactors include lighting, what kind of lighting, what is the wattage, what is the efficacy.So, lighting power density is largely what we are talking about and then we also have HVAC,pumps, electrical power and all the electrical systems, electrical and mechanical systems. So,we will start dealing with each of these components one by one and see what ECBCprescribes.(Refer Slide Time: 03:19)So, first of all we take the case of opaque walls. Now, when we are talking about opaquewalls ECBC prescribes the maximum U value or U factor for a complete wall assembly. So,when we say a complete was wall assembly, it implies there will be a plaster on one side of itand then there will be a brick wall. There might be an insulation in between, then anothercladding may be a wall or another hard material and then another layer of plaster or finishingmaterial.So, this entire set of materials is the wall assembly, it is the complete wall assembly and whenwe are talking about the properties of this wall; we are talking about the properties of thiscomplete wall assembly and not an individual material. So, we can look at the complete Uvalue of this wall assembly or alternatively we can also look at the minimum R value which isthe resistance value which is prescribed for insulation alone.So, only the R value of insulation can be can also be met with, this is the prescriptiveapproach; so, where these values have to be met. So, if you look at these values; so, the Uvalues have been distinctly defined for buildings depending upon their purpose, dependingupon their function.So, this offer all building types except hotels, business purpose buildings and schools all otherbuilding types have these specified U values for walls as this. So, we see for composite, hotdry and warm humid climates, the U values are 0.4. For temperate climates the U value isprescribed to be higher. Now, if you remember the concept of U value, the U value which isthe thermal transmittance property of any material or an assembly of materials is the propertyby virtue of it.The heat is transferred from one side to the other side based upon the temperature differential.So, if we look at the composite climate, hot dry climate and warm humid climate thetemperature differential is quite high. At times it is in the range of the in peak summers whichmay be in the range of 20 degree centigrade, but this is only during the peak summers. Inextreme winters it may be around 15 degrees and rest of the time of the year, it may besomewhere in between them, where it may be as less as 5 degrees or it may be as high as 20degrees.When we are looking at temperate climate, now temperate climate if you remember theclimatic that our discussion on climate and weather; the temperate climate has very lessdiagonal range. So, the difference between indoors and outdoors will often be less, very less.So, we are looking at a time temperature range, the difference between indoor and outdoor inthe range of a 0 to 10 degrees only; so, maximum is 10 degrees. So, having very low U valueswill not affect much because anyways the temperature differential is going to be much lesser.While, when we look at the cold climates the extremely cold season in cold climates has ahigher temperature differential between indoors and out, the outdoors may go sub-zero. So,we may be having temperatures which are like minus 10, minus 5 degree centigrade whileindoors we may be wanting to maintain a temperature of around 20 degree centigrade.Thereby, the temperature difference increases to around 30 degree and therefore, we wouldsee that the for all the buildings the maximum, if emphasis is on U values for cold climatesand they have been prescribed to be the lowest.With this we can see that different buildings have different, different building types havedifferent U values where for school buildings which are going to be occupied largely duringthe day; they have been provided with most lenient values, but the difference from climate toclimate that variation remains the same.(Refer Slide Time: 07:57)Another opaque component is roof and the for roofs also which is also opaque maximum Ufactor and my minimum R value is prescribed exactly same as that for the walls. And, we cansee the same trend following here for the climate variation. Now, here we would see thatthere is no difference between these three and temperate as well because, the roof receives theheat is largely because of the solar radiation falling onto the surface.And, in temperate climates the heat received because of solar radiation is same as that ofthese 3 climates and hence no variation in U value for these 4 climates. While, in coldclimates again the temperature differential is quite high. Therefore, the U value assembly Ufactor is proposed to be low as compared to all other. Again we would see that the schoolbuilding since there are daytime use buildings they are proposed to have relatively higher Uvalues. So, more lenient properties are allowed for roofs of school buildings, opaquecomponents of school buildings.Now, how do we calculate this U value? So, often the U values the resistance or conductivityvalues for individual materials are available, but for the assemblies it is not available.(Refer Slide Time: 09:30)So, in ECBC the method for calculating the U value for an assembly is also given, let usquickly look at this calculation here. So, what we have to do is we have to calculate the Ufactor for a roof assembly which is made up of mainly RCC slab which is 150 mm thick. Ithas an insulation of 5 centimeter 50 mm on top of it and which is the expanded polystyreneXPS. And, then on top of it we have tiles which is 40 mm thick and in the bottom on theinside it has a plaster which is 10 mm thick. So, we have to calculate the collective U valuefor this roof assembly.(Refer Slide Time: 10:14)Looking at ECBC, ECBC also has annexure where individual properties for these materialsseparately are given. So, if we look at this for each of these 4 layers; 1 2 3 4 for each of theselayers we first write down the thickness of this material in meters. And, from ECBC annexurewe note down the conductivity value for the given materials. So, this is for tile, this is forinsulation, this is for the RCC slab and for plaster. First of all we calculate the resistancevalues for each of this by dividing the thickness of the material by its conductivity value.So, for each material its resistance values will be calculated, once we have calculated theresistance values we calculate the total resistance offered by the roof which is the sum total ofall these different resistances. Now, here for simplicity we are taking the resistances offeredby these 4 layers only, 4 layers of materials. But, when we look at it scientifically and as perthe definitions there is a thin layer of air above the style and also beneath this. On both sidesthere is a thin layer of air which also offers resistance to heat transfer.Now, whenever we are calculating the overall resistance offered we would also calculate R 0and R i. So, this is R outside resistance offered by air outside and resistance offered by airinside and that is dependent upon the speed of the air because the air. So, small eddies areformed on the surface of this layer, both the sides. This is dependent upon the air speed andbased upon that the resistance offered by the air is calculated.These values are also given in ECBC and we can take it from there. For the ease of acalculation for simplicity we have not taken these into account, but ideally we will be takingthese two resistances offered also into account and add them here. So, R o and R i will also bethere, the total resistance of the material then calculated the inverse of it 1 by R will give usthe overall U factor of the opaque assembly, here it was for the roof. So, this is how the Uvalue will be calculated.Now, when we are looking at the compliance using prescriptive method; so, we have toachieve the U value less than the U value prescribed for the wall assembly or the roofassembly as per ECBC. So, if they prescribed U value for the roof assembly has been given tobe 0.3 we have to achieve a U value which is less than or equal to 0.3; because that is themaximum U value which is prescribed. If you look at the code, this is much higher than whatis prescribed in the code which implies that either the insulation needs to be increased.So, if we increase the thickness of this insulation from 0.05 meter to say around 0.1 meters.So, instead of this if we increase it to 0.1, this will be increased by 2 and we will get aresistance of 1.4 Kelvin meter square per Watt. And, when we add it up here, it is furtheradded to make it approximately 1.575. And, if we calculate the inverse of it, it comes out tobe 1 by 1.575.So, we get a U value of 0.634 which is still higher than the U value which is prescribed by thecode, but we can see that just by increasing the thickness of insulation the U value hassubstantially reduced. So, we have to do those permutation combination to achieve the Uvalue of the root within the prescribed limit.(Refer Slide Time: 14:55)The next prescription is about the cool roofs, we have also seen how cool roofs work. So, forIndia the prescription is that the initial solar reflectance should not be less than 0.70 and theinitial emittance should not be less than 0.75. This is for the roofs with slope less than 20degree, for roofs which have slope higher than 20 degree the values would vary. Thereflectance values especially would be much lower because, they are more the high slope ismore likely to cause the glare on surrounding buildings.So, this is one of the prescriptions as per the US DOE, where they have prescribed the initialsolar reflectance for different slopes, where the initial solar reflectance has been prescribed tobe higher than 65 percent for the low sloped roofs initially. And, the best available solarreflectance in the US market is around 87 percent. So, when we are talking about cool roofs,we are actually looking at the solar reflectance and the emittance of the material.(Refer Slide Time: 16:15)The next component which is also very important and lot of work is required here, because alot of heat transfer is possible through the vertical fenestration. We are talking about largelyglazed windows here. So, when we are talking about vertical fenestration, we have tworequirements. One is the assembly U factor which is the U factor taking into account bothglass and frame and we also have the SHGC requirements for the same fenestration.The as per the ECBC the U factor and SHGC requirement are given for the fenestration basedupon the orientation and latitude. So, we have U value irrespective of the latitude andorientation and for SHGC we have SHGC maximum values of SHGC prescribed for thenon-north windows. And, maximum SHGC prescribed for the north facing windows basedupon their latitudes greater than equal to 15 degree north and less than 15 degree north. Thisis based upon the amount of solar radiation, the direct light which is received on the northernside.So, based upon this the U values and SHGCs are given, this is the maximum values of it.Again if we see the U value for all the windows in all the climates they have been prescribedto be same, because we have also seen that the larger percentage of the total heat gainedthrough a fenestration is because of SHGC. So, greater emphasis has been placed here forSHGC. If we see for all the non north windows the SHGC in 4 climates has been prescribedto be quite low 0.27, implies only 27 percent of the direct solar radiation, the heat falling onthe window is transmitted inside permitted inside. While, for cold climate the SHGC has beenprescribed to be much higher.So, the so that the window is able to allow maximum amount of heat indoors in case of coldclimate. If we look at the SHGC for the north facing windows for latitude greater than equalto north 15 degree north, we see that higher SHGC values have been prescribed. Because, allthe places with latitude greater than equal to 15 degree north will not receive any direct solarradiation from the north and hence there is no requirement to keep the SHGC low. However,for cold we still are keeping it very high.In all other windows for places which are having their latitudes less than 15 degree north, theSHGC has again been prescribed the same as the non-north windows which is 0.27. So, whenwe are selecting the fenestration we have to meet the prescriptive requirement. Now, when wewere having this discussion about the SHGC we talked about the availability of direct solarradiation onto the window. Now, if the shading is provided on top of this window outside, thedirect solar radiation which is incident on the window, the fenestration will reduce.And, there the requirement for this SHGC may not remain that stringent, it will becomelenient which we will see in subsequent slides.(Refer Slide Time: 20:25)The next important factor here is daylighting, here we are talking about the amount of naturallight available in a habitable space. So, we are talking about the amount of light daylightwhich is available on a work plane and useful daylight which is available on a work planethroughout the year. So, for 90 percent of the potential daylit time in a year.So, when we are talking about compliance, we are talking about minimum 40 percent of thearea, total floor area shall remain potentially lit, shall remain daylit for at least 90 percent timeof the daylit time in an year. Now, when we are talking about the compliance here, what weare looking at is we are looking at the daylit areas.(Refer Slide Time: 21:21)So, these are the windows if you can see. So, from these windows this particular plant isusing a simulation software, where the daylit areas have been calculated. Now, these daylitareas have been calculated taking into account that this area remains delayed for 90 percentdaylit time in the year and the minimum that is required is 40 percent. So, if the total floorarea was 1254 meter square at least 502 square meter of this total floor area is supposed to belit.(Refer Slide Time: 22:04)Now, how do we prove the compliance? So, one method is by using simulation tools whichwe will subsequently see in other portions of the ongoing discussion. When we are looking atthe manual calculation method, we have to calculate the Daylight Extent Factors or DEF.Now, the values for these are to be calculated based upon the tables which are given in ECBCand the compliance criteria remains the same.(Refer Slide Time: 22:33)So, when we have to calculate the DEF, first of all we need to locate the latitude of the place.For example, in this case where we are discussing about the DEF, this is in Delhi which has alatitude greater than 15 degrees, next we have to know the VLT of the glass. So, the VLT ofthe glazing for this window is 0.39 here, this window also has a projection an overhang andthe projection factor of it has been given to be 0.4. So, these are the basic components whichwe require, the latitude, the VLT, the projection factor and then we calculate the DEF basedupon the table from the ECBC.So, when we know that the VLT is greater than 0.3 and the latitude is greater than 15 degreenorth, here we are talking about window with a shading with a projection factor of zero pointgreater than equal to 0.4. Based upon this we would calculate the DEFs for all the windowson different orientations with and without light shelf, here it is with light shelf. So, we can seethat this is the relevant numbers of DEF which is to be used in calculations here. Based uponthis we get these numbers: north 3.5, south 3.0, east 2.1 and west 1.8.And, the head height is also available as per the design, now based upon all this we have tocalculate the area which will be daylit because of this window here. So, we have to knowthese dimensions A and B, A is the depth up to which the daylight will be available and B isthe horizontal dimension up to which the daylight will be available; considering that there isno opaque partition in this given area. In case there is an opaque partition higher than the headheight of the fenestration the extent will reduce.So, once we have calculated; so, A which is the depth, it is in the direction which isperpendicular to the fenestration. And, the day light area extends to the head height of thefenestration multiplied by the DEF which we have calculated or the distance till an opaquepartition higher than the head height of the fenestration is appears erected. So, if we look atthis particular case, we have a DEF of say 3.5 in north.So, if we are calculating it for the north fenestration we would say we would see that this hasa DEF of 3.5 and the head height is 3. So, we have A available as 3.5 multiply it by 3, that is9.5. And, in horizontal direction it is equal to the width of the fenestration plus either 1 meteron each side of the aperture or the distance to an opaque partition.(Refer Slide Time: 26:15)So, for B if this is W we are talking about W plus 2 meters. On the basis of this we willcalculate the daylit area which is available for each of these windows on different directions,orientations. So, we will calculate it based upon the DEF for north and south east and westand together we will calculate the total area which is which will meet the UDI requirement,the usable day light requirements for the as per the prescriptions of ECBC.And, as per these calculations this comes out to be 49.2 percent which implies that thisparticular plan, this particular design is compliant as far as day lighting requirements areconcerned. (Refer Slide Time: 27:02)Next in continuation of the fenestration we are talking about the shading devices. So, we havevery quickly in passing discuss about the projection factor which is PF. When we are talkingabout projection factor we are taking the shading which is offered by the by the projection.So, in this case the horizontal projection is 2 feet and the vertical and the height of thewindow is this height, where effectively the point at which the projection has been installedthat, but a bottom most point we calculate the total height.So, this is the W by H, where W is the width of the projection and H is the height of thewindow from its shelf to the bottom most point of the shading device. So, W by H is theprojection factor and we calculate the projection factor with this given data. Now, wheneverwe have projection factor as we have seen in the previous lecture as well; it will limit theamount of sun which is falling directly onto the window, the glazing fenestration. So, thisportion of the sunlight has already been cut off.(Refer Slide Time: 28:27)To account for this projection and this shading which is provided by this shading device, wetake into account the shading equivalent factor. So, if we have a window which remainsshaded because of the presence of an overhang throughout the year, there is no direct lightfalling onto it. And, practically the amount of heat which is transferred inside because of thedirect radiation falling onto it will become very close to 0. There is not practically 0 becausethere is the diffused component of the solar radiation which is also available.So, there is one direct component of the radiation which will be reduced to 0 close to 0, butthere will always be a diffused component of the solar radiation which will always be incidenton the glazing, on the fenestration. So, to calculate the shading equivalent factor we need toknow the projection factor of the shading device. Once we know the projection factor and wealso know what type of shading device it is, it could be an overhang or fin, it could be only anoverhang or it could just be fin. So, depending upon the design of the shading device.Then for different latitudes these tables are available in ECBC, based upon the latitude theprojection factor, the type of shading device and the orientation, 8 orientations have beentaken; SEF can be picked up from the tables given in ECBC.(Refer Slide Time: 30:04)Once we have picked up from the table, here we are taking a quick example again. So, this isa two storied office building in Delhi again and we have to achieve an ECBC compliance. So,this has a rectangular layout of 90 meter by 30 meter and the windows are all 1.8 meter inlength and 2.165 in height. So, we see that this is zero point; so, this is 2.165 and theprojection of this roof slab is further 0.3 meter above the window. So, the total height of theprojection, the shading device from the shell of the window is 2.165 plus 0.3 meter which isthe vertical dimension and horizontal projection of this roof is given to be 0.85 meter.So, we calculate the projection factor as H by V and this comes out to be 0.345. If we go backto the table which is given here; so, we have a projection factor of 0.345. We have so basedupon this projection factor which we calculate here, based upon the latitude for Delhi whichis greater than 15 degree north. And, the orientation of different windows; 1: we will have theSHGC which is prescribed as per the initial tables which we have seen and then we willcalculate the SEF.(Refer Slide Time: 31:54)So, for calculating the SEF we will get the coefficients from the ECBC based upon this data.So, if we look at the projection factor and the latitudes following the table 4 dash 12 and 4dash 13 we will calculate the coefficients C 3 C 2 and C 1. And, put it in this equation wherewe use the projection factor which is calculated and substitute the coefficients, 4 coefficientsto calculate the SEF for each direction. So, SEF of east comes out to be 1.296. Now, this SEFof east will be multiplied by the SHGC which is prescribed in the prescriptive tables and wecalculate the effective SHGC.Now, this is the effective SHGC. So, once we have calculated the SEF we divide the SHGCwhich is prescribed for the given orientation as per the prescriptive tables and divide theSHGC by the SEF to obtain the effective SHGC of that fenestration. So, instead of using afenestration which is having lower SHGC we can actually use an SHGC which is higher.And, still obtain an SHGC which is well within the prescriptive limits by dividing it with theequivalent shading factor the SEF and we can still comply with the requirements the ECBC.So, once we have provided the shading compliance can be shown based upon thesecalculations as per ECBC. Now, so far this is all prescriptive requirement whatever is givenin the code will be met up met with.(Refer Slide Time: 34:09)Again, we also have the prescriptive requirements for the skylights. So, for skylights for allclimatic zones the maximum U factor and maximum SHGC has been prescribed. So,whenever we are going in for skylights we will use these U factors and SHGCs.(Refer Slide Time: 34:27)Now, this was how the compliance can be shown when we are talking about the prescriptiveapproach, where whatever value has been given in the code will be met. So, if we have to goahead with the prescriptive approach and if you want to comply using it, a glass which has theU value as prescribed and an SHGCs as prescribed will be used. In case there is a shadingdevice using SEF the compliance will be shown; all this is prescriptive. Now, sometimes wemay decide that we will invest more in insulating the roof, but by virtue of the design there isnot enough insulation which is required on say the northern side of the building.Or, because there is another building coming up so, certain portions of the building will noteven receive the direct solar radiation. So, the SHGC of the glass; so, the glass will beselected such that SHGC is much higher and we are not investing high on glass. So, how willthe compliance be shown? In such a case we use the envelope trade off method, in this theenergy performance factor of the proposed building will be compared with that of a basebuilding. And, this overall envelope performance factor is a sum total of EPF of roof, EPF ofwall and EPF of fenestration.So, what we are doing here is fundamentally we are compensating for a reduced performanceof one component with an increased performance of say another component. This is the tradeoff, but it is only between the components of the envelope. So, we are looking only at roofwall and fenestration here and using their performances and efficiencies interchangeably. So,when we are talking about an envelope trade off method, we will calculate the EPF of each ofthese components separately using these equations.Here the coefficients for the roof or wall or the fenestration for SHGC and U values, they aretaken from the tables given in ECBC. For the calculation of EPF of roof and wall, wemultiply the U values with the areas and we calculate the sum total of it for different types ofmaterials which has prescribed on the roof and wall. So, there may be combination of thematerials which is used and then for EPF fenestration we calculate it for all the differentdirections.So, we calculate it for the north, south, east and west separately; for both the properties that isU value and the SHGC with relevant SEFs. Using this the overall EPF of the building will becalculated and EPF of proposed case, proposed building has to be lesser than that of the basecase. Now, this base case has the EPF calculated as per the U value given in the prescriptiveapproach method. So, the U value will be taken as what is given in the prescriptive tables andthe area will be the total area. So, once we calculate the base case PEPF, we compare it theEPF of the proposed case.(Refer Slide Time: 38:17)Let us quickly take an example here. So, this one is a 600 meter square single storey daytimeuse office building in Roorkee and here we are trying to achieve the ECBC level compliance.So, it has a band of windows; so, the height of the window is 1.2 meter, the total height of thebuilding is 3 meters and there is no shading which is proposed.And, the materials of the for the building are given; so, for roof, for external wall assembly,for glazing all the properties are given here. We also have the VLT given and overall thebuilding is a oriented in such a manner that the shorter dimension of the building which is the20 meter side is oriented towards north. (Refer Slide Time: 39:03)So, for the proposed case the shading device has been proposed here, in base case there wasno shading device, while in proposed case there is an overhang of 0.6 meter proposed that is 2feet. And, this is right above the window which is 1.2 meter high.So, we calculate the projection factor as 0.5, based upon the tables we will calculate the wewill obtain the coefficients substituting this projection factor here. This is the table whichshows how the SEFs from the tables have been obtained and we calculate the effective SEF,the total SEF.(Refer Slide Time: 39:49)Now, here this very quickly shows how the difference between the base case and proposedcase is.