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.