Welcome back, this lecture is a continuation of my earlier lecture. And the specific objectives
of this particular lecture are to introduce solar radiation through fenestration, define
solar heat gain factor and shading coefficient discuss effects of external shading, introduce
ventilation and infiltration and discuss estimation of heating and cooling loads due to ventilation
and infiltration.
So at the end of the lecture you should be able to estimate heat transmitted into a building
due to solar radiation through fenestration using values of solar heat gain factor and
shading coefficient from tables calculate dimensions of the overhangs for external shading,
explain the importance of ventilation and estimate heating and cooling loads due to
ventilation and infiltration. Before that let me just work out an example.
On calculation of solar radiation on a south facing wall the example is like this you have
to calculate the total solar radiation incident on a south facing vertical surface at solar
noon on June twenty first and December twenty first using the data given below the latitude
angle is given to be twenty three degree centigrade and the hour angle is zero degrees because
it is at solar noon okay. This is not given but you have to infer this
from the given information and the declination. Declination is plus twenty three five degrees
on June twenty first and minus twenty three point five degrees on December twenty first
and the tilt angle, tilt angle here is ninety degrees because it is a vertical surface and
the wall azimuth angle zeta is zero degree centigrade. Because the wall is south facing
and finally it is given that the reflectivity of the ground is point six. So this is the
information given and based on this we will have to calculate the total solar radiation.
Okay. So here step wise procedure is shown first what we do is we find out the altitude
angle beta since we are doing the calculations at solar noon beta is nothing but beta max.
Thus we given by pi by two minus absolute value of l minus d where l is the latitude
and d is the declination so you find that the altitude angle works out to be eighty
nine point five three degree degrees. First I am doing the calculation for June twenty
first okay. Here I take the declination as twenty three point five degrees next at solar
noon we have to calculate what is the solar azimuth angle since we are making the calculations
at solar noon the solar azimuth angle is either one eighty degrees or zero degrees. It is
zero degrees if altitude is less then declination. So you have to take solar azimuth angle to
be zero degrees in this case okay. And next we have to calculate what is so wall
solar azimuth angle alpha in the wall solar azimuth angle alpha is given by one eighty
minus gamma plus zeta and here you find that gamma is zero and zeta is also zero. So you
find that the value of alpha is one eighty degrees for this particular case okay. Once
you know the value of alpha and the value of beta calculate the incidents angle since
this is vertical wall the incidents angle is given by theta vertical is cos inverse
cos beta into cos alpha beta is eighty-nine point five three degrees and alpha is one
eighty degrees. So from this you find that the incidents angle is eighty nine point five
three degrees once you know the incident angle they can calculate what is the direct radiation
on the surface the direct radiation is given by Idn cos theta.
So first let us find out Idn from the ASHRAE model Idn is given by A multiplied by exponential
within bracket minus B by sin beta where beta is the altitude angle that is eighty nine
point five three degrees and a is ten for summer. So the take the value of A as ten
eighty and B which is called as the extinction coefficient it takes a value of point two
one for summer this aspects have discussed in the last lecture okay. So you take this
values of ten eighty for A and point two one for B and substitute the value of beta you
find that the direct normal radiation is eight seventy five point four watt per meter square.
So we have to find out what is the direct radiation on the vertical surface. So that
you have to find out by multiplying this with cos theta where theta is the angle of incident
okay. That is the eighty-nine point five three degrees so if you find that then you will
find that Idn cos theta works out to be seven point one eight watt per meter square okay.
Next we have to find out the diffuse radiation.
Diffuse radiation for that we have to find out the view factor view factor is given by
one plus cos epsilon by two and this ninety degrees so you find the view factor is point
five okay. This is for the diffuse radiation once you know the view factor diffuse radiation
given by again based on the ASHRAE model Id is equal to C into I subscript Dn into F subscript
ws where t c takes the value of point one three five for summer. Since I am making the
calculations for June I take a value of point one three five and Idn v what is so to be
eight seventy-five point four as shown in the last slide and Fws is point if substitute
all these values you if substitute all these values you find that the diffuse radiation
is fifty nine point one watt per meter square. Next we have to find out the reflected radiation
reflected radiation also have the view factor between the ground and the wall. So first
let us find the view factor between the ground and the wall that is F subscript wg this one
that is given by one minus cos epsilon by two. So this what so to be point five then
based on the ASHRAE model the reflected radiation from the ground is equal to Idn plus I diffuse
multiplied by rho g into Fwg where rho g is the reflective of the ground which is given
as point six and Fwg is point five. So if you substitute the value of Idn and I diffuse
you find that the reflected radiation works out to be three eighty point three six watt
per meter square. So the total incident radiation is nothing but some total of direct radiation
contribution diffuse radiation and reflected radiation. So the direct radiation total radiation
works out to be three forty six point six four watt per meter square okay. So this is
a fatedly straight forward procedure. This is for a vertical wall if the surface
is inclined then you have to use the tilt angle and if it is facing any other direction
other then south then you have to use proper surface azimuth angle okay as explained in
the last lecture. The same manner you can calculate the all this parameters for December
twenty first okay. For December twenty first the declination angle will be minus twenty
three three point five degrees okay. So now let me show a table where the solar radiation
results for June twenty first and December twenty first are given.
So this table shows the comparison between June twenty first and December twenty first
June twenty first you find that the incidents angle is eighty-nine point five three whereas
for December twenty first it forty three point five three the direction radiation Idn is
eight seventy-five point four watt per meter square. In case of June twenty first and its
thousand three point seven five ion case of December. Similarly the contribution of direct
radiation is nothing but Idn cos theta is seven point one eight watt per meter square
for June seven twenty seven point seven watt per meter square for December okay. And the
diffuse radiation is fifty-nine point one and twenty nine point one and reflected radiation
is two eighty point three six and three nine point nine in case of December. So the total
incident radiation on the vertical surface is three forty six four watt per meter square
on June twenty first and ten sixty six point seven watt per meter square on December twenty
first okay.
This example actually gives us some useful information that information is that this
is a south facing wall you observe here that for south facing wall. If you get the total
radiation incident on the wall is three forty six point six four watt per meter square in
summer. Whereas it is almost eleven hundred watt per meter square in winter okay. And
here for summer this value is coming because of the reflected radiation. Because we have
taken point six as observe reflectivity of the ground if the ground is not so reflective
you find that the total incident radiation on the vertical surface on June will be very
small whereas it is very high on the December twenty first or that is in winter okay.
So this tells us this gives us an important information that is that for south facing
walls the solar radiation incident on a vertical surface is much less in June whereas it is
much higher in winter okay. So since we want to reduce the load heat load on the building
during summer and we want to maximize the heat transfer to the building during winter
it is always beneficial to keep windows doors etcetera on south side okay. So that the radiation
part will be small in summer and it will be large in winter. So your cooling capacity
requirement will be small and the heating capacity requirement also will be small okay.
So this is actually this is a principle generally used in passive cooling and heating techniques
okay. Use the south wall properly and of course these results hold good for northern hemisphere
okay. Because the altitude is twenty three degree north the results will be.
If you are talking about southern hemisphere okay, and so for this ASHRAE model in fact
if you remember I have mentioned in the last class that this model assumes that the sky
is cloud less okay. But if the sky is cloudy sometimes what is done is a clearness index
value is used to calculate the incident direct solar radiation okay. The clearness index
value is one if the sky is clear and it will be less than one if the sky is cloudy and
the clearness index value for different reasons during different seasons are available okay.
So from that data you can calculate what is the radiation during a cloudy day okay.
Now next let us look at solar radiation through fenestration first. So all what is fenestration
refers to any glazed apertures in a building such as glass door windows skylights etcetera.
That means it refers to all those transparent surfaces such as doors windows skylights etcetera
okay. All these are called as fenestration we need fenestration all buildings have fenestration
why do we need fenestration it is required. Because it provides day light heat and outside
air and fenestration also provides visual communication to the outside world it also
improves aesthetics and finally it provides a escape route in case of fires in low rise
buildings because of all these factors almost all buildings will have some amount of fenestration.
That means you will have some amount of glass windows or glass doors etcetera okay. Because
of the four reasons mentioned here.
Now let us see what is the affect of this on the cooling and heating loads heat transfer
due to solar radiation through transparent surfaces is distinctly different from heat
transfer opaque surfaces. This is one important thing to note okay, what is the difference
radiation incident on an opaque surface for example a wall or a roof is partly absorbed
and the remaining is reflected back because it is opaque surface. So its transitivity
is zero. So whatever radiation is incident either it is absorbed or it is reflected.
That means or it is reflected that means it is partly observed and partly reflected and
out of the observed part only a fraction of it is transferred into the building this aspect
we will discuss in the next lecture. So ultimately whenever radiation is incident
on an opaque building some of it is observed and some of it is reflected back and out of
the observed portion only a part if finally transferred to the building. That means only
a part of the radiation incident on a opaque surface finally becomes a load on the building
okay. This is as far as the opaque surface is concerned. Now let us look at a transparent
surface for a transparent surface a major part of the radiation incident on a transparent
surface is directly transmitted into the building while the rest is observed and reflected back
we this from our basic physics that all transparent surfaces have high transmissivity okay.
So whenever radiation is incident on the surface most of it is transmitted through the surface
right transmitted through the glass and only a small part of it is observed and a small
part of it is reflected back. So this is the major difference between the transparent surface
and opaque surface and this has a huge affect on the heating and cooling loads okay. First
let me, let us look at a reference surface and see what happens to the radiation okay.
This is a distribution of solar radiation on a clear plate glass okay. On a clear plate
glass clear plate glass has these optical properties it has a transmittivity of point
eight it has an observitity of point one two it has reflectivity of point zero eight okay.
So these are the values for the solar radiation solar radiation these are the values now since
it has a tranmittivity of point eight if hundred percent of solar radiation is incident on
this window okay. Which is nothing but your window made of clear flat glass plate glass
eighty percent of it is directly transmitted to the indoors okay.
Because it has as transmittivity of point eight right and twelve percent of it is observed
by the glass while eight percent of it is reflected back. Okay. This is based on the
optical properties of the clear glass now of the twelve percent absorb what happens
to this twelve percent the radiation observed by the glass this will increase the temperature
of the glass. That means glass temperature increases due to absorption okay. Absorption
of radiation right once the glass temperature increases it reject some heat by convection
to the outside and some part of it is rejected to the indoors okay. For example typical values
are this particular glass will reject about eight percent to the outside and about four
percent to the indoors. So final you see that out of the hundred percent
solar radiation incident on the surface eighty percent plus four percent that is eighty four
percent is directly transmitted to the building or to the indoors while sixteen percent is
transmitted to the outside okay. This is for a clear plate glass and these values will
be different for different type of glasses okay.
Now what is the importance of fenestration and glaze surfaces, why do we have to consider
this seriously? Fenestration or glaze surfaces contribute a major part of cooling load of
a building. So the energy transfer due to fenestration depends on the characteristics
of the surface and its orientation, weather and solar radiation conditions. So careful
design of fenestration can reduce the building energy consumption considerably okay. So if
you are, if you design it properly then fenestration can help you introducing the initial and running
cost. But if you do not design it properly then you will have to pay both in terms of
initial and running cost. For example if you design the fenestration properly that means
if put the glass windows doors etcetera, properly on proper direction and with proper orientation
you will find that the radiation transmitted to the building during summer can be reduced
very much whereas it can very much increased during winter.
As a result the cooling load on the building can be reduced during summer and the heating
load on the building can be reduced in winter okay. So the required cooling and heat capacities
will be less so your initial and running cost will be less okay. So if you are that is if
you are using the fenestration properly okay. Instead of that if you are putting glass left
and right just to increase the ethictics or anything you will find that both heating and
as well as cooling loads will increase okay. Ultimately you have to pay for this in terms
of initial and running cost okay. So they, that, why it is very important to understand
the importance of fenestration and the issues involved in the design of fenestration for
buildings.
Now let us look at simple model for calculating solar radiation passing through a transparent
surface okay. So this model is based on these assumptions the assumptions are like this.
So this is valid under the assumption that the transmittivity, that is a tau value and
the absorptivity that is alpha value of the fenestration is same for both direct as well
as diffuse radiation. This is an important point to be remembered in actual case you
find that the transmittivity and absorptivity of the glass will be different for direct
radiation and diffuse radiation okay the values will be different and the transmittivity and
absorptivity values of direct radiation okay. It depends strongly on the angle of incidence
right. So this is again a very important point to be remembered the transmittivity and absorptivity
of the glass for direct radiation is not a constant okay.
So it is a function of the angle of incidence and as we have seen the angle of incidence
itself various it varies with the altitude, it varies with our angle, it varies with declination
and with orientation etcetera right. So the, based on this the transmittivity and absorptivity
of the glass also varies whereas the transmittivity and absorptivity of the diffuse radiation
remains more or less constant okay. However the model that I am going to present which
is taken actually from ASHRAE hand books, it is based on the assumption that the transmittivity
and absorptivity values are same or constant. And they are same for both direct radiation
as well as diffuse radiation. However this assumption is justified because normally we
do the calculations for the peak load conditions okay. And at peak load conditions you find
that the contribution of direct radiation is much larger compared to the direct diffuse
radiation, I am sorry. So direct radiation is much larger than the
diffuse radiation. So the, if you are assuming that the tau values and alpha values are same
for both, you will not be making a huge error okay. And besides it simplifies the procedure
however it is not necessary that you have to assume them to be same okay. With a little
bit of extension you can take different values of tau and alpha for diffuse radiation and
direct radiation and do the calculation separately okay.
So under these assumption you can write the solar radiation transmitted to the building
Q subscript sg okay, is equal to A within multiplied by within brackets tau into I subscript
t plus n multiplied by alpha multiplied by I subscript t here, as I said Qsg okay, is
the radiation transmitted to the building and the units are watts is SI units okay.
And A is the area of the surface exposed to radiation in meter square It is the total
radiation incident on the surface. So this includes both direct plus diffuse okay. Of
course you can also include reflected radiation normally reflected radiation will be small
part. So this includes the total radiation incident
on the surface and tau is the, as you know is the transmittivity of the glass alpha is
the absorptivity of the glass. And what is n? n is a new parameter and this is the fraction
of absorbed radiation transferred to the indoors by conduction and convection. In fact I was
mentioning that when the glass absorbs radiation its temperature increases. So because the
temperature between the glass and the surroundings they will be heat transfer due to convection
okay. So this factor n takes into account this heat transfer that means the heat transfer
due to convection because the temperature difference between the glass and the surrounding
air okay.
And it can be very easily shown that at steady state the value of N is given by U divided
by h naught where U is the overall heat transfer coefficient okay and h naught is the external
heat transfer coefficient. So U is the overall heat transfer coefficient for a flat glass
one by U you know that the expression one by U is one by hi plus one by h naught plus
delta x by k wall okay. This is the resistance of the wall this, the convictive resistance
on the outside this is the convective resistance of the inside.
So you can calculate overall heat transfer coefficient and h naught is the external heat
transfer coefficient okay. So if you know the overall heat transfer coefficient external
heat transfer coefficient properties of the glass and if you can estimate the total radiation
incident on the surface. As you have shown in the example last example you can calculate
what is the solar energy transmitted to the building okay, through a fenestration.
Now what we can do is, from the last expression we can write Qsg like this okay, A and take
out A and you can also take out It. So you can write Qsg is A into within brackets It
multiplied by within bracket tau plus alpha U by h naught as i said tau is the transmittivity
alpha is absorptivity U by h naught is your fraction n okay. Now what is done is, a single
sheet clear window glass okay single sheet clear window glass is taken as a reference
and a factor called solar heat gain factor or SHGF is defined as follows okay. So taking
the single sheet clear window as reference we define a solar heat gain factor as this
is nothing but total radiation incident on the surface multiplied by within brackets
tau plus alpha U by h naught okay. And the subscript stands for subscript ss
stands for singles single sheet clear window glass. That means it is for the reference
glass okay. You will see immediately what is the advantage of defining this, the advantage
is that the maximum solar heat gain factor values for different altitudes months and
orientations have been found and tabulated. For example ASHRAE hand books gives the maximum
solar heat gain factors for different altitudes months and orientation etcetera. So once you
know the solar heat gain factor maximum solar heat gain factor of or from the table for
given altitude for a given orientation for a given a day then you can calculate what
is the energy transmitted into the building if the glass is made of reference glass. Because
the energy transmitted is nothing but area multiplied by SHGF right. So let me show a
typical table adopted from ASHRAE hand books.
Okay this is the, this table gives maximum solar heat gain factor for sunlit glass okay.
Located at thirty two degrees north latitude and the units here are watt per meter square
okay. For example for month December okay, orientation of the surface if the window is
facing north or window is shaded then you find that the maximum solar heat gain factor
is sixty-nine watt per meter square in December, it is seventy five watt per meter square in
during January and November it is eighty five during February and October and hundred during
march and September like that. Similarly for north east and north west. That means the
window is facing north east or north west direction then during December this is the
value during January and November this is the value February this is the value like
that okay. Like that it gives the solar heat gain factor maximum solar heat gain factor
for all orientation for different orientations okay.
The eight orientations and also for horizontal orientation for all these we are assuming
that the glass is vertical okay. That is the tilt angle is ninety degrees okay. That means
on a vertical wall right and this data is valid for thirty-two degrees north latitude
ASHRAE hand books gives this similar data for other latitude also. For example twenty-four
degrees forty degrees like that right. So from this table you can calculate what is
the maximum solar heat gain factor. For example I would like to find out during the month
of July, I want to find out what is the solar heat gain factor through a south facing window
okay. That value is given by two thirty right during east or west facing window it is six
eighty five like that we can find out the SHGF maximum.
One interesting thing you can absorb here is the benefit of putting the window on the
south side. As I was mentioning in the last example you can see that for south side windows
okay. The maximum solar heat gain factor that means radiation transmitted into the building
because of solar radiation through the glass is very small during summer whereas it is
large during winter okay. What we want is, we would like to heat the buildings in winter
using solar radiation okay. And we do not want any solar radiation into the building
during summer right. Because that way you can keep the building cooler during summer
and you can keep the building warmer during the winter.
So using the solar radiation right, so this is very beneficial and this is very beneficial
only when you put the windows on the south side. Because you see for the south side there
is practically not much radiation during summer whereas there is lot of radiation during winter
okay. Whereas for other directions like east west you find that is almost same during throughout
the year. So you should not put look ta the table for east west the solar radiation is
almost constant throughout the year okay. May some twenty thirty percent variation is
there. That means you should never keep windows on east or west direction because on east
or west directions during summer. You will receive lot of radiation. Of course during
winter also you receive radiation but not as much as you receive if you put the window
on the south side okay.
So this kind of hand book, a data are available in the hand books.
Now this data is for a standard reference glass. How about other glasses for fenestrations
other then the reference SF glass a shading coefficient is defined okay. This shading
coefficient is defined such that the heat transfer due to solar radiation is given by
Qsg multiply that is equal to area A okay multiplied by SHGF maximum of the standard
glass multiplied by the shading coefficient. So finally you find that shading coefficient
is nothing but the heat transmitted through the actual glass divided by the heat transmitted
through a standard glass okay. That is how the shading coefficient is defined
and the shading coefficient depends upon the type of the glass and also on the type of
the internal shading devices okay. You can use a wide variety of internal shading devices.
As you know very well, as I want to shade the window from radiation you can use curtains
you use venetian blinds roller rollers like that okay. So the shading coefficient depends
upon what kind of internal shading device you are using and it also depends upon what
kind of glass your using whether it is a double or a single glass etcetera okay. And typical
values of shading coefficient for different types of glass with different types of internal
shading devices have been measured and are tabulated for example in ASHRAE hand books
okay let me show a typical table.
Again this is taken from ASHRAE hand books this table gives shading coefficient for different
types of glass and internal shading for example for type of glass for a single glass okay.
Single sheet regular glass this is the standard glass if it has a thickness of three mm and
if does not have a no internal shading the shading coefficient is one the shading coefficient
one because this glass is taken as the reference okay. So for this glass the Q is simply A
into SHGF maximum that you get from the table right. However if you are using internal shading
internal shading device. For example if you are using venetian blinds and the venetian
blinds are of medium type you find that the shading coefficient reduces to point six four
and if you are using light venetion blinds then the shading coefficient becomes point
five five. If you are using roller shades and dark roller shades shading coefficient
is point five nine if it is light roller shade it is point two five this is for a single
glass similarly for different types of glass. For example if you are using a double glass
a regular double glass of three mm thickness you find that without internal shading the
shading coefficient is point nine whereas for the single glass which one and with internal
shading these are the values right. Other types of glasses are, you can also have heat
absorbing glass of six mm thickness you find that are heat absorbing glass for six mm thickness
the shading coefficient without internal shading is point seven. That means thirty percent
less compared to a regular glass okay. So that how if you have this kind of information
then you can select the proper shading coefficient value from these tables and once you know
the shading coefficient value you can calculate what is the heat transmitted into the building
through this glass okay.
Now heat transferred through the glass due to solar radiation can be reduced considerably
using suitable internal shadings okay. You might have noticed that when you are using
internal shadings. The shading coefficient is less than one okay. That means amount of
heat transmitted is less than one okay. So with this we know everybody knows that if
you want to reduce the radiation you can simply put a curtain or you can simply use a venetian
blinds okay. There by you can cut down the amount of radiation enter entering into the
building okay. So that is good as far as the cooling load
is concerned. Of course there is a negative aspect to this when you are using a dark curtain
or venetian blinds etcetera; the light that is entering into building also gets reduced
okay. So light is not sufficient then you may have to use some artificial lighting so
you have to pay for the artificial lightings. So again you have to see the balance you have
to balance between the requirement for the light and requirement for reduction in the
cooling load okay and from the type of the sunlit glass its location and orientation
and the type of internal shading one can calculate the maximum heat transfer rate due to solar
radiation. Let me give a small example this is very simple example.
We have to calculate the maximum heat transfer rate through a one point five meter square
area unshaded regular double glass facing south during June and December without and
with internal shading oaky. The internal shading is light venetian blinds and the location
is thirty two degrees north right. So the data given is you have to calculate during
June and December okay. So first let us do the calculation for June for the month of
June the SHGF max from table is one ninety watt per meter square okay. So this is solar
heat gain factor for the standard glass okay. Then using the values of shading coefficients
from the table that heat transfer rate is given by this si. The expression Q is A into
SHGF max into shading coefficient if you are not using any shading coefficient any internal
shading sorry. Then you find that area is one point five meter square and SHGF max is
one ninety watt per meter square and shading coefficient for the double glass is point
nine from the table. So you find that Qsg is two fifty six point
five watt on June okay, and with internal shading. That means if you are using venetian
blinds then for the same date right. Find that the maximum energy transmitted is again
given by the same formulae area is same SHGF of max is also same but the shading coefficient
is different now it is point five one okay. So find that the total or at the rate at which
energy is being transmitted the maximum transmission rate is given by one forty-five point three
five watt. So from these you can see that with venetian blinds you can reduce the energy
transmitted considerably it is almost fifty percent compared to without internal shading
and if you do the same thing for December okay.
So you find that the SHGF of max for December is seven ninety five watt meter square. So
without internal shading Qsg is ten seventy three point two five watts with internal shading
this is six hundred eight point one seven five watt okay. Again you can see here that
because this is a south facing glass the energy transmitted during summer is much less where
whereas it is much higher during winter. Of course, obviously during winter you should
not use any internal shading because you want this whereas during summer you should use
internal shading because you don't want this okay.
Now let us look at effect of external shading. So for we can been assuming that the window
is not shaded externally. That means the full area of the window is exposed to radiation
okay. But most of the times we find that most of the windows will have some kind of an external
shading okay. For example in over hang right this over hang provides external shading that
means not all the area of the window will be exposed to solar radiation part of it is
exposed and part of it will be in shade okay. So this will have a bearing on the heat transferred
to the building okay. So let us see what is the effect of this.
The solar radiation incident on a glazed window can be reduced considerably by using external
shadings. For example over hangs of course the external shading can also be provided
by let us say a tree or an adjacent building right. But here am I assuming that it is based
on the over hangs okay. So by proper design of the over hangs it is possible to block
the solar radiation during summer and allow it into the building during winter okay. There
by you can reduce the cooling load and also the heating load. So what is the affect of
over hangs as we know that over hangs reduce the area of windows exposed to solar radiation
and thereby reduce the heat transmission into the building due to direct radiation okay.
Let me show this.
So this is the window oaky. So window as the dimensions of height H and width W so area
of the window is given by H into W right and area exposed to solar radiation is equal to
this without over hang okay. Because the entire area is exposed to solar radiation. But if
I am using a overhang with inset that is I am using a overhang with inset this the overhang
with inset the red hatched portion. So you have the certain length of it certain it has
certain width and it has certain depth okay. Depth of the inset is this okay, and width
is this right. So if you are using an overhang you find that, this hatched area at this particular
incident when the sun is at this particular position this area is not exposed to direct
radiation that means this is in shade okay. Only this much area is exposed to solar radiation
oaky. That means without over hang the entire area
is exposed to direct radiation whereas width over hang only the small portion is exposed
to direct radiation and this small portion is given here as multiplication of x into
y okay and at any point you can calculate x and y because this is related to your solar
geometry. For example this is related to your altitude angle beta and it is also related
to your surface azimuth angle surface solar azimuth angle alpha okay. What is the relation?
You can find that at any point x is given by this relation x is equal to w minus d into
tan alpha whereas y is equal to H minus d into tan beta by cos alpha. As I said beta
is your altitude angle alpha is your wall solar azimuth angle. Whereas is the width
of the window which is equal to width of the overhang and d is the depth of the inset that
means this dimension okay. So if you know the dimensions of the overhang, what is the
width of it? What is the depth of the inset and if I want to find at any particular incident
how much area of the window is shaded, how much area of the window is exposed? Then you
can calculate it very easily if you can calculate the altitude and wall solar azimuth angles
okay. Let me give a small example here.
We have to calculate the energy transmitted into the building at three pm okay. On July
twenty first due to solar radiation through a south west facing window made of regular
single glass. That means it is made of the reference glass the dimensions of the window
are, height is two meters and width is one point five meters and depth of the inset d
is point three meters okay. Now from the given data that means at three pm that means hour
angle is forty five degrees July twenty first you can calculate what is the declination
on July twenty first and it is facing south west. So your zeta is forty five degrees okay.
So from the hour angle zeta and declination you can calculate the altitude angle and if
you do the calculation you will find that the altitude angle works out to be forty-eight
point two three degrees oaky. Similarly you can also calculate the surface solar azimuth
angle which works out to be thirty nine point eight degrees. So this is the first step you
have to calculate these angles once you calculate this angles you can calculate the dimensions
x and y okay x and y is the length and height of the unexposed area I am sorry of the exposed
area right. So x is given as w minus d into tan alpha where w is the width of the window
which is given as one point five meter okay and d is the depth of the inset that is given
as point three oaky. So you find that x is equal to one point two four nine meters and
y is equal to h minus d into tan beta by cos alpha where h is two meters d is point three
tan beta is forty eight point two three and alpha is thirty-nine point four eight seven.
So if you substitute this value you find the y is equal to one point five six two okay.
So now you can calculate what the transmitted energy because of solar radiation is. So that
is now given by SHGF max multiplied by the shading coefficient multiplied by the exposed
area okay, is exposed area is x into y. So that is given by one point two four nine into
one point five six two okay. This multiplied by this and SHGF max from your ASHRAE table
works out to be two thirty watt per meter square okay. Because this thirty two degree
north latitude okay and the shading coefficient is one shading coefficient is one because
this is a standard glass. So you find that the radiation transmitted
is four forty-eight point seven watts okay. If you do not use the overhang you will find
that the energy transmitted is w into h into SHGF of max that is six ninety watts okay.
You can see that with over hang it is about four forty-nine with without overhang it is
six ninety. So there is a considerable reduction in the energy transmitted to the, into building
because of the absence of overhang. So over hang is beneficial right.
Now complete shading of the window in summer and complete unshading in winter is possible
use separation between the top of the window and overhang okay. So ideally we would like
to completely shade it in summer because we do not want any heat in summer whereas we
want lot of heat in winter so you want complete unshading okay. So complete shading and unshading
is possible using what is known as the separation is nothing but the distance between the top
of the window and the overhang and an infinite combinations of overhang width and separation
dimensions can provide complete shading in summer and complete unshading in winter okay.
so let me show this.
Okay, one thing I would like to mention here is that the position of the sun will be varying
continuously okay. So when I am saying that this over hang can completely shade this window
it can only complete shade at a particular point okay. It cannot complete shade all the
time right. Because the position of the sun varies continuously. So normally what you
have to do is you have to find out the point at t\which the load is likely to be peak okay.
And the design overhang in such a way that during summer the window is completely shaded
at this particular incident oaky and during winter it is completely unshaded.
So this is one thing you must remember again all these dimension and all will be varying
between latitude to latitude. Because the solar geometry is varying from latitude to
latitude and it also varies from the orientation to orientation okay. These things again you
have to keep in mind. Now you can see from this figure that suppose you have this dimension
w naught and this is what is known as separation okay. So this is your window if you provide
this separation and this mush width of the overhang. Because this is your over hang right
you can completely shade this window because the sunlight is falling at this point. So
the entire portion this entire thing is in shade okay whereas the, whatever is below
this is unshaded. So no solar radiation is entering into the
indoors which are on this side right. If you take this w naught and this S you can also
take for example this w naught okay and this S. Then also you can completely shade the
window or you can take this w naught and this S oaky. Again you can completely shade the
window. That means a large number of combinations of w naught and S are possible which can result
in the complete shading or unshading of the window okay. Of course there are certain limitations
of fixed overhang.
So far we have been discussing about fixed over hang it they have certain limitation
what are the limitations any overhang provides protection against direct solar radiation
only okay. It cannot provide any protection against diffuse or reflected radiation and
you find that because of the variation of the transmittivity with incident angle during
the peak summer time you find that for a vertical surface only about forty percent of the total
solar radiation consists of the contribution of direct radiation okay. That means about
forty percent of the total during summer is because of the direct solar radiation and
provision of overhang can only reduce this right. That means it can only handle the forty
percent rest sixty percent cannot be handled by the overhang. Because rest sixty percent
consists of diffuse as well as reflected radiation and overhang cannot reduce diffuse and reflected
radiations okay. External over shading right and you find that sometimes over hang can
work in a negative manner. For example the overhang may actually reflect the ground radiation
on to the windows right. So it is a bit it may look like a paradoxic,
that you provide an overhang. You find that there is more heat in the room oaky. This
happens because if you are ground is highly reflective right then first the radiation
will be reflected from the ground on to the overhang and from the overhang on to the wall
on to the window right. So ultimately because of these reflections you will find that there
is lot of radiation entering into the building not through direction radiation but through
reflected radiation okay. Then over hang is working in a negative manner okay. So this
one of the limitations of the fixed over hang and during mornings and evenings when the
sun, so low in the sky that over hangs can provide only minimum protect okay. So they
are not really very useful during mornings and evenings and over hangs are truly affective
for windows facing thirty to forty five degrees of south okay, if over hangs are not very
useful on east and west direction. For example why because during on east and
west direction the maximum radiation occurs during morning for east and during evening
for west and during morning for east and during evening for west the position of the sun is
very low right when the position is of the sun is very low. That means the latitude angle
is very low the overhang cannot block the direct radiation okay. So it is not of no
use the only way of blocking solar radiation east and west faces during morning and evening
is to use something else like a wall or a tree or an adjacent building right. But not
overhang this is the, another limitation of overhang right in spite of all these limitations
overhangs are widely used because they also provide protection against rain. So overhangs
are highly recommended okay. So this is as far as the effect of solar radiation
and how to calculate the cooling and heating loads this requires little bit of extension.
But how to calculate the energy transmitted into the building due to solar radiation first
through opaque surfaces and next through transparent surfaces again the for the transparent surfaces
with and without external shading so from the discussion we can calculate all these
things.
Now let us look at another aspect of cooling and heating load calculation that is ventilation
and infiltration air inside a space show okay, should be pure to ensure healthy and comfortable
living conditions however conditioned air normally consist of several pollutants okay.
So what are these pollutants? these pollutants are orders various gasses such as carbon dioxide
and volatile organic compounds or VOCs and particulate matter okay, see, you find that
any conditioned space will not be hundred percent pure.
But it will consist of these impurities and if these impurities if the concentration of
these impurities goes beyond a certain level you find that the indoor environment is not,
neither it is healthy nor it is comfortable for the occupants okay. So these pollutants
are due to internal and as well external sources internal sources are human beings appliances
etcetera whereas the external sources are the outside air itself. And indoor air quality
abbreviation is IAQ can be controlled by the removal of the contaminants in the air or
by diluting the air. That means you can maintain the purity of the air inside a conditioned
space either by removing the contaminants or by diluting the air okay.
There are you can use the both of these techniques now with reference to this indoor air quality.
What is ventilation, what is the purpose of ventilation? The purpose of ventilation is
to dilute the air inside the conditioned space okay. There by maintain required indoor air
quality and ventilation is defined as the supply of fresh air to the conditioned space
either by natural or by mechanical means for the purpose of maintain acceptable indoor
air quality okay. So the whole purpose of ventilation is to maintain required indoor
air quality which is very much essential for comfortable and healthy living conditions
and this consist of supplying fresh air and either by natural means. That means either
by natural ventilation or by mechanical mean that means either by using an exhaust fans
or blowers etcetera okay. This is the definition of ventilation and
ventilation air generally comprises of fresh outdoor air and some amount of re-circulated
air that is treated to maintain acceptable indoor air quality okay. So the ventilation
air need not to be all outdoor air it can be a part of outdoor air and some part of
it can be re-circulated when you are using re-circulated air for ventilation purpose
you have to treat it first then use it for ventilation okay. And if the out outdoor air
is not pure that means outdoor air itself is dirtied consists of lot of dust then it
also has got to be treated before supplying to the space.
Though the amount of fresh air required for breathing is quite small that means its amount
of fresh air required is about point two liter per second per person from breathing point
of view the actual requirement of fresh air is large as ventilation air in addition to
providing oxygen for breathing also has to serve the following purposes. So only one
of the purposes of ventilation is to provide oxygen. In addition to providing oxygen ventilation
air also as to dilute the orders inside the occupied space this is very important to a
socially acceptable level okay. So odor dilution is a important function of ventilation second
important function is to maintain the carbon dioxide concentration at a satisfactory level
and the third practical requirement is that it should be able to pressurize the escape
routes in the event of fire okay. So these are the other important functions of ventilation
air.
Estimate, let us look at the estimation of minimum outdoor air required for ventilation
how do we find how much air is required for ventilation from energy conservation of point
of view it is important to choose the ventilation requirement suitably as ventilation is one
of the major components of system load s oaky. So it is very important to choose it properly
you cannot have lot of ventilated air then you have to pay in terms of running cost the
amount of air required for ventilation purposes depends on several factors such as application
activity level extent of cigarette smoking etcetera.
So several factors affect the amount of required air okay. And based on several studies extending
over several years guidelines for minimum ventilation requirements have been established
for example one of the guidelines is ASHRAE standard sixty two bar nineteen eighty nine.
Let me give a small example of this ASHRAE standard what is it say this is a typical
outdoor air requirements for ventilation adopted. From ASHRAE standards you can see that for
offices if the occupancy level is seven people for hundred meter square floor area and if
it is smoking zone you require ten liter per second per person and if it is a non smoking
zone you require two point five liter per second per person okay.
And for operation theatres you require high levels of ventilation and the occupancy will
be twenty persons per hundred meter square and normally smoking is not allowed but still
you require high ventilation air. Because you have to maintain high level of purity
and for lobbies where the occupancy is slightly higher thirty persons per hundred meters square
floor area and if smoking is allowed you require about seven point five liter per second per
person. And if the smoking is not there then two point five liters per second.
And for class room for example where the occupancy is still higher hundred person per fifty persons
per hundred meters square floor area and normally smoking is not allowed. And the required ventilation
is eight liter per second per person okay. And for meeting places these are the values
so one thing you can notice here is that as the occupancy level increases the required
amount of outdoor air are increases. Similarly if you are allowing smoking then outdoor air
also requirement also increases okay.
So based on, if you have this kind of data you can decide what is the outdoor air requirement.
Let us see how to estimate infiltration. Now, so for I have been discussing about ventilation
and let us see what is infiltration and how to estimate infiltration loads first of all
infiltration it is defined as the uncontrolled entry of untreated outdoor air directly into
the conditioned space okay.
So you have to carefully note the different between infiltration and ventilation both
involve supply of outdoor air. Whereas ventilated air is in complied in a controlled manner
infiltration is uncontrolled entry okay. You have no control on that and infiltration of
outdoor air takes place. Because of two effects one is what is known as wind effect and the
other one is what is known as stack effect. As the name implies the wind effect means
infiltration due to the wind blowing over the building.
So whenever wind blows over the building pressure difference are created because of these pressure
differences outside air enters into the building and air from the building leaves the building
okay. This is what is known as wind effect and the second effect is what is known as
stack effect this is nothing but the entry of outdoor air because of the effect. So the
affect is created because of the temperature difference between the inside and outside
whether it is summer or winter the inside temperature will be different from the outside
temperature you have a some temperature difference. So because of this temperature difference
there will some density differences between the outside air and inside air and if your
building have some openings. Because of the density differences there will be entry of
outdoor air into the building and some amount of or same amount of air leaves the building
okay. So this is what is known as stack effect. In commercial buildings even though infiltrations
supplies outdoor air in commercial buildings effect are made to minimize infiltration.
Because it is uncontrolled and unreliable you cannot make any design depending upon
infiltration because it depends upon unreliable factor such as wind and stack okay.
So generally it is reduced and so what are measures to reduce this, these measures? As
you know very well are the use of vestibules or revolving doors use of air curtains building
pressurization if you pressurize the building. And if the inside pressure is higher, then
the outside pressure no outdoor air can entry. So there by you can reduce infiltration and
also you can use proper ceiling of doors windows etcetera thereby you can reduce the infiltration
loads.
Now the estimation of the exact amount of infiltration is very difficult as it depend
on several factors such as the type and age of the building indoor and outdoor air conditions
outdoor outside wind velocity and direction etcetera okay. So analytical estimation is
very difficult there is several methods are used one method is what is known as a air
changes per hour in this method depending upon the building type whether it is lose
building medium building or tight building infiltration rates are specified or rated
empirically to win velocity and temperature difference. So if you know the wind velocity
and temperature difference using the empirical equations you can calculate the infiltration
rates and data is also available for infiltration rate to different types of windows doors etcetera
okay.
Now let us look at heating and cooling loads due to ventilation and infiltration due to
ventilation and infiltration buildings gain sensible and latent energy in summer and lose
sensible and latent energy in winter. So you have to note here that the energy transfer
is both in the form of sensible as well as latent modes okay. And the energy is gained
in summer and it is lost in winter and sensible and latent heat transfer rates are simply
given by Q sensible is m dot Cp into delta t which is, you can write in terms of volumetric
flow rate that is v naught into density into Cp and latent heat transfer rate that is this
okay. Here hfg is the latent heat of ha vaporization m dot o is the air flow rate due to infiltration
or ventilation and w o and w i are outside moisture content and inside moisture content
and t naught and t i are the outside dry bulb temperature and inside dry bulb temperature.
And as I have written here m dot and v naught are mass and volumetric flow rates dye to
infiltration and ventilation. So if you know the infiltration ventilation
rates and outdoor and indoor condition you can calculate what is the essensible heat
transfer because of ventilation because of ventilation infiltration what is the latent
heat transfer rate because of infiltration and ventilation okay. So that is how we can
calculate but one of the major difficult is to find out the infiltration rates okay. At
this point I stop this lecture and we will continue this discuss in the next lecture.
Thank you.
