Dr. Abdulsallam Al- Shboul

Environmental Design

May 8, 2013

Energy saving Architecture: redesigning existed building façade and existed

inner space.

Abstract

Solar radiation plays a main role in affecting any inner space as a heat. The design itself and the

ratio of openings related with the affection of heat gain as in this research will be considered.

In this research we will analyze a case study and see how is the solar radiation effect that case

and we will work on redesigning ideas to reach better solutions.

1. Introduction

Buildings in Jordan are rarely built to

use energy efficiently, despite its high costs

owners don’t know the differences based on

working with an energy saving design, the

reasons for this kind of lack of interest is

affected by the massive market failure that

doesn’t give an importance for saving

energy.

What has to be in consideration to have a

better energy saving designs is to work on a

design education and an elimination of

perverse incentives for designers and

engineers.

Buildings use a large amount of energy

and electricity to be cooled, ventilated, lit

and equipped in inefficient ways not because

anyone was venal or stupid, but because

they are faithfully did their job responding

to the incentives they saw.

Changes can be done with working on

ways to let the social understands the effect

that a studied design can give, which is a

better energy saving design.

On this study, we will work on proving

how a normal design can be redesigned with

the same outlines and be an energy saving

design.

We took an existed case study at

Philadelphia University which is located on

the road between Amman and Jarash cities.

See Fig (1) and Fig (2)

Fig (1) The map of Jordan with the cities Amman and

Jarash shown on it.

With zooming in to the top view of

Philadelphia University we can easily

recognize the place of the building of the

research case study.

See Fig (3)

Fig (3) Philadelphia University’s top view with the

case study Architectural Department building shown

with the red.

Our case study will be architectural

drawing classrooms at the Department of

Architecture’s building at Philadelphia

University (Amman-Jarash) these

classrooms are designed on a simple shaped

south oriented facade. See Fig (4)

Fig (2) The main road between Amman and Jarash

shows the exact place of Philadelphia University with

the red sign.

Fig (4) 3D drawing shows the mass of the building

and two classroom spaces with the red color shown

on the south facade facing the slop.

2. Methodology of work

We will first study how the existed cases

are affected by the solar radiation on

December (21 Dec.) and we will count the

hours of Insolation by using the sun path

diagram then by knowing the insolation

hours we will use the monthly hourly mean,

st. dev. , max, and min of global radiation on

horizontal surface table of Amman (as we

found out that there is no big differences

between Amman’s table and Jarash’s table)

to find out the energy radiated from the sun

on December and count how the classrooms

gain and lose heat reaching the thermal

comfort in the classrooms.

Then we will work on redesigning

alternatives that will be suggested to find out

the ideal classroom design.

2.1. How the existed cases are affected

by the solar radiation.

The two cases are facing a slope of

mountain in the south façade. See Fig (5)

and Fig (6)

Fig (5) the two cases shown facing the slope of the

mountain – south façade and the angles of calculating

the shadow of the slope

Fig (6) the elevation of the south façade of the

building with the two cases shown

2.1.1. First case: a classroom with

a 14 meters width dimension on the

façade and 10 meters depth. See

Fig (7)

Fig (7) plan of the first case classroom with a 14

meter width dimension on the façade and 10 meters

depth

From the top view of the building we

found the effect of the mountain’s slope on

the center point of the façade of the

classroom (middle window) first with

finding the horizontal angles

which are (16.35o, 46.59

o). See Fig (8)

Fig (8) the top view of the first case with the

horizontal angles taken from the mountain slope

And from the section we got the vertical

angles . See Fig (9)

Fig (9) section shows the vertical angels affecting the

center point from the mountain’s slope

Results of the azimuth and latitiude

angles between the point are taken from the

façade and the casting edges of the slope

which may make shaded on the elevation are

shown in Table 1. See table (1)

Case 1 16.35 16.59

46.59 11.62 Table (1) : azimuth and latitiude angles of the first

case

And from both horizontal and vertical

angles we drew the effect of the mountain’s

slope at the façade on the sun path diagram.

See Fig (10)

Fig (10) the sun path diagram shows the shadow of

the mountain’s slope

After knowing that the solar radiation effect on the first case with no shading hours we will count

the energy reached from the solar radiation by using the monthly hourly mean, st. dev. , max, and

min of global radiation on horizontal surface table.

Energy reached from the solar radiation = 2072 watt (from the table)

Energy absorbed to the classroom space = (2072 / 1.33) * Cos (32) = 1321.17 watt

ــــــــــــــــــــــــــــــــــــ

Calculating heat gain and heat loss on the first case

Heat gain of the classroom space

Q wall = A*U*(Tin – Tout) = 58*0.49*(21-8.8) = 346.724 watt

As we assume that Tin is 21

Q glass = A*U*(Tin – Tout) = 12*6.7*(21-8.8) = 980.88 watt

Sensible heat for 23 persons = 220*23 = 5060 watt

Total Heat gain - Q total = 6387.6*1.3 = 8303 watt

As we multiplied by 1.3 as it is what we gain from other effectors (lights, etc. ...)

Heat loss of the classroom space

Q wall = A*U*(Tin – Tout) = 58*0.49*(21-8.8) = 346.724 watt

As we assume that Tin is 21

Q glass = A*U*(Tin – Tout) = 12*6.7*(21-8.8) = 980.88 watt

Q slap borders = A*F*(Tin – Tout) = 140*0.55*(21-8.8) = 939.4 watt

F=edge loss factor: 0.81 for no edge insulation, 0.55 for edge insulation

Q airflow around windows = 0.018*q*(Tin – Tout) = 0.018*(0.1*2*2)*(21-8.8) = 0.087

0.087*3windows = 0.26 watt

q=total infiltration airflow volume in cubic meter per hour

Q slab = F.A = 2*140 = 280 watt

F=conduction loss factor: 2 for floor slabs, 4 for basement walls

Total heat loss - Q total = 2547 watt

2.1.2. Second case: a classroom

with a 10 meters width dimension

on the façade and 14 meters depth.

See Fig (11)

Fig (11) plan of the second case classroom with a 10

meter width dimension on the façade and 14 meters

depth

From the top view of the building we

found the effect of the mountain’s slope on

the center point of the façade of the

classroom (middle window) first with

finding the horizontal angles which are

(37.91o, 30.31

o). See Fig (12)

Fig (12) the top view of the second case with the

horizontal angles taken from the mountain slope

And from the section we got the vertical

angles. See Fig (13)

Fig (13) section shows the vertical angels affecting

the center point from the mountain’s slope

Results of the azimuth and latitiude

angles between the point are taken from the

façade and the casting edges of the slope

which may make shaded on the elevation are

shown in Table 2. See table (2)

Case 2 37.91 13.69

30.31 14.57

Table (2): azimuth and latitiude angles of the second

case

And from both horizontal and vertical

angles we drew the effect of the mountain’s

slope at the façade on the sun path diagram.

See Fig (14)

Fig (14) the sun path diagram shows the shadow of

the mountain’s slope

After knowing that the solar radiation effect on the second case with no shading hours we

will count the energy reached from the solar radiation by using the monthly hourly mean, st. dev.

, max, and min of global radiation on horizontal surface table.

Energy reached from the solar radiation = 2072 watt

Energy absorbed to the classroom space = (2072 / 1.33) * Cos (32) = 1321.17 watt

ــــــــــــــــــــــــــــــــــــ

Calculating heat gain and heat loss on the first case

Heat gain of the classroom space

• Q wall = A*U*(Tin – Tout) = 38*0.49*(21-8.8) = 227.164 watt

As we assume that Tin is 21

• Q glass = A*U*(Tin – Tout) = 12*6.7*(21-8.8) = 980.88 watt

• Sensible heat for 23 persons = 220*23 = 5060 watt

Total Heat gain - Q total = 6268*1.3 = 8148 watt

As we multiplied by 1.3 as it is what we gain from other effectors (lights, etc. ...)

Heat loss of the classroom space

• Q wall = A*U*(Tin – Tout) = 38*0.49*(21-8.8) = 227.164 watt

As we assume that Tin is 21

• Q glass = A*U*(Tin – Tout) = 12*6.7*(21-8.8) = 980.88 watt

• Q slap borders = A*F*(Tin – Tout) = 140*0.55*(21-8.8) = 939.4 watt

F=edge loss factor: 0.81 for no edge insulation, 0.55 for edge insulation

• Q airflow around windows = 0.018*q*(Tin – Tout) = 0.018*(0.1*2*2)*(21-8.8) = 0.0806

0.0806*3windows = 0.25 watt

q=total infiltration airflow volume in cubic meter per hour

• Q slab = F.A = 2*140 = 280 watt

F=conduction loss factor: 2 for floor slabs, 4 for basement walls

Total heat loss - Q total = 2427 watt

A table as result of studying the two cases below showing the differences between them

with the values shown. See Table (3)

Table (3): Values of the angles of the two cases

2.1.3. First conclusion

After being through the excited cases

and studying the effect of the mountain’s

slope (surroundings) and the effect of the

solar radiation on the facades we can

came out with knowing the better design

of the two cases which is the one with

the 10 meters dimension on the elevation

and with the depth 14 meters. See Fig

(15)

Fig (15) the better case of the two cases we studied

We took this case in to consideration to

redesign it reaching the best energy saving

design.

I global (watt)

Solar Energy (watt)

Insulation efficiency

( hours/day )

Heat gain (watt)

Heat loss (watt)

Case 1 2072 1321.17 10 hours 8303 2547

Case 2 2072 1321.17 10 hours 8148 2427

3. Redesigned facades

In this part of the research we took the

selected case to find an improvement to

have a better design.

The redesign idea is making a recessed

façade which may decrease the solar

efficiency hours on the elevation.

We considered three dimensions of

depth on the façade (resisted) to study their

effect on our case (2m, 4 m, 6 m) we will

study the effect when using each dimension

to find which alternative will be the best to

use.

The effect will be with having less solar

efficiency hours.

3.1. Studying the three alternatives of

the redesigned facades

The three alternatives are done with

three different depths as shown in Fig (16)

Fig (16) the three alternatives are done with three

different depths (2, 4and 6 meters)

The classroom with 10 meters width on

the façade and 14 meters depth is taken first

with 2 meters recess. See Fig (17)

Fig (17) plan of case (A) of the alternatives design

with a 2 meters recess

From the top view of the building we

found the effect of the walls on the center

point of the façade of the classroom (middle

window) first with finding the horizontal

angles and vertical angles taken from

section. See Table (4)

Table (4) contained azimuth and latitude angles

between the points we took on the façade and the

casting edges of the walls which may make shaded

on this elevation.

Calculation for the other two redesigned

alternatives are done as we did for the first

alternative and been organized in a table and

drawn on the sun path diagrams.

See Fig (18), Fig (19) and table 4

Case A 66.97 61.74

88.57 61.74

Fig (18) plans of the three alternatives as each alternative with the angles on it.

Table (4) values of the angles calculated from both top view and section of each case.

Fig (19) sun path diagrams for three cases

Case A 66.97

88.57

36.36

38.65

Case B 50.48

88.57

31.63

38.65

Case C 39.23

88.57

26.83

38.65

Table (5) the results of calculations for three cases.

3.2. Second conclusion

As a result of studying the three cases

the main factors that we considered is

reaching thermal comfort spaces by

decreasing the heat gain which effected by

the insulation efficiency. See Table (5)

The best case of the three cases (A, B,

and C) was case (C) with the 6 meters

recessed façade which get the minimum

solar energy. See Fig (20)

Fig (20) plan of the best case that has a lower solar

energy effect on its façade

4. The ratio of openings

In this step we will study the ratio of

openings between the opened area and total

area of the elevation of our space and we

will study how this ratio will effect on the

solar energy and will effect also on

difference between heat gain and heat loss

reaching to the best design which has the

best thermal comfort .See Fig (21)

The ratio of openings existed in our

case which we selected is calculate as:

Total area of elevation = 50 m2

Area of opening = 12 m2

The ratio = 12 \ 50 = 24 %

We suggested another two ratios to show how

it effect on the ratio on the thermal comfort as

shown on the table (6). See Table (6) and Fig (22)

Table (6) the suggested ratios

I global (watt)

Solar Energy (watt)

Insulation efficiency

( hours/day )

Heat gain (watt)

Heat loss (watt)

Case A 2072 1321.17 10 hours 8148 2427

Case B 1699 1083 6 hours 8148 2427

Case C 1238 789.4 4 hours 8148 2427

Case a Case b Case c

Opening m2 12 4 16 Elevation m2 50 50 50

0.24 0.08 0.32

Fig (21) plan and path diagram of the best case which has the lowest solar energy affecting the façade.

Fig (22) suggested openings drawn on elevations as a three cases (case a – case b – case c).

Table (2) showing the calculation of the three cases to prove what case was the best to reach of thermal comfort

I global (watt)

Solar Energy (watt)

Insulation efficiency

( hours/day )

Heat gain

(watt)

Heat loss (watt)

Case a 1238 789.4 4 hours 8303 2547

Case b 1238 789.4 4 hours 6371 1822

Case c 1238 789.4 4 hours 8542 2118

4.1. Third conclusion

From the results in tables and from the

chart we found out that Case (b) the best

case to reach thermal comfort in the

classroom because it has a smallest

difference between heat loss and heat gain,

so we can make a relation between the ratio

of openings and the thermal comfort defined

as:

Thermal comfort = heat gain – heat loss

Thermal comfort 1 / ratio of opening

(this result is just with the case of SOUTH

oriented facades)

5. Final Conclusion

After being through studying first the cases of Philadelphia university and being out with the better

case to redesign and after finding out the best alternative of three alternatives we reached the best case

with the best openings on it which is as shown in the Figers below. See Fig (23) and Fig (24)

Fig (23) the best elevation design to reach a thermal comfort

Fig (24) the best plan design to reach a thermal comfort

We found out that the best orientations for the classrooms is to have the small side on the façade

and we found out that if the heat gain was high we can redesign the whole façade to recessed facades

to make shade that decreases the solar radiation heat gain with the consideration of daylight and

structure.

It’s better to minimize the openings on the southern elevation to have a lower difference between

heat gain and heat loss to reach a better thermal comfort.

6. Recommendations

We do advise further studies about another solutions to decrees the differences between the

heat gain and heat loss reaching a better thermal comfort spaces.

7. References

http://encyclopedia2.thefreedictionary.com/Thermal+Comfort

http://www.wausauwindow.com/education/daylighting/daylighting.pdf

http://en.wikipedia.org/wiki/Solar_gain

http://personal.cityu.edu.hk/~bsapplec/solar3.htm

http://ar.wikipedia.org/wiki/%D9%83%D9%88%D8%AF%D8%A7%D8%AA_%D8%A7%D9%84%D8%AA%D

8%B5%D9%85%D9%8A%D9%85_%D8%A7%D9%84%D8%A8%D9%8A%D8%A6%D9%8A