APPLICATION OF BUILDING SCIENCE
AR343 Building Science 01 Semester July 2020
Prepared by: Lai Zhi Zhen 1001747427 Prepared for: Mr Chan Chiew Chuen
C O N T E N T
Task 1.0 1.1 Sectional Cut of walls & U value 1.2 Propose new existing wall 1.3 Comparison
Task 2.0
X
2.1 Calculation of MRT 2.2 Thermal comfort Y
2.3 Propose a solution
Task 3.0 3.1 Sun Path Diagram 3.2 Draw and Comparison
Task 4.0 4.1 Design of external shading device
4.2 Passive Ventilation 4.3 Conclusion & references
01 02 03
04
GA DWG
LOCATION: NEXT TO UCSI UNIVERSITY, TAMAN CONNAUGHT, CHERAS, KUALA LUMPUR.
W E S T
SOUTH
E L E V A T I O N
ELEVATION
E A S T
NORTH
E L E V A T I O N
ELEVATION
SECTION
X-X
TASK 0 1
1.1
LOCATION OF SELLECTED ENCLOSED SPACE AND WALLS
I. WALL BETWEEN INDOOR AND OUTDOOR SPACE (MAKE LAB – GARDEN)
II. WALL BETWEEN INDOOR AND INDOOR SPACE
(MAKE LAB – WASHROOM)
Ground floor plan
SECTIONAL CUT OF WALL DETAILS WITH SPECIFICATION & DIMENSIONS
SECTIONAL
I. WALL BETWEEN INDOOR AND OUTDOOR SPACE (MAKE LAB – GARDEN)
MAKE LAB
II. WALL BETWEEN INDOOR AND INDOOR SPACE (MAKE LAB – WASHROOM)
GARDEN
U-VALUE
MAKE LAB
WASHROOM
U – VALUE = ___________________ 1
U – VALUE = _______________________ 1
Rso + R1 + R2 + R3 + Rsi
Rso + R1 + R2 + R3 + R4 + Rsi
= ________________________ 1 Rso + ( b )1 + ( b ) 2+ ( b ) 3 + Rsi k k k
= _____________________________ 1 Rso + ( b )1 + ( b ) 2+ ( b ) 3+ ( b ) 4+ Rsi k k k k
=
=
1 ______________________________ 0.02 0.055 + ( ) + ( 0.11 ) + ( 0.02 ) + 0.123 0.5 0.84 0.5 = ____________________________ 1
0.055 + ( 0.02) + ( 0.11) + ( 0.02) + ( 0.01 ) + 0.123
0.5
0.5
1.3
0.055 + 0.04 + 0.131 + 0.04 + 0.00769 + 0.123
=
= ______ 1 0.389
THERMAL CONDUCTIVITIES
0.84
= _____________________________ 1
0.055 + 0.04 + 0.131 + 0.04 + 0.123
U – VALUE = 2.571 W/m2 °C
1 ______________________________
1 ______ 0.3966
U – VALUE = 2.521 W/m2 °C
SURFACE RESISTANCE
BRICKWORK = 0.84 W/m˚C
OUTSIDE = 0.055 m2 ˚C/W
PLASTER = 0.50 W/m˚C
INSIDE = 0.123 m2 ˚C/W
CEMAMIC TILES = 1.3 W/m˚C
R =
b (MATERIAL THICKNESS)_______ k (MATERIAL THERMAL CONDUCTIVITY)
1.2
PROPOSED ADDITION LAYERS OF WALL
MALAYSIA IS A TROPICAL COUNTRY IN WHICH USUALLY HOT AND HUMID THROUGHOUT THE YEAR. IN MY KNOWLEDGE, THE WAY TO KEEP THE SPACE COOL DURING THE HEAT OF SUMMER IN MALAYSIA CLIMATE IS INSULATING OR AIRING. CAVITY WALL IS ONE OF THE GOOD OPTION TO IMPROVE THE EXISTING WALL AS CLEAR CAVITY CONSISTS OF AIR SPACE AND FILLED CAVITY WALL CONSISTS OF INSULATION.
OPTION 01
CLEAR CAVITY WALL
BRICKWALL
THE AIR TRAPPED BETWEEN THE BRICKWALL AND BLOCK BEING POOR CONDUCTOR OF HEAT SUCH AS A BARRIER TO HEAT TRANSFER. IT IS LOWER IN THERMAL CONDUCTIVITY, HENCE THIS HAS A POSITIVE INFLUENCE ON THE SPACE COMFORT.
AIR SPACE BLOCKWORK THERMAL BOARD PLASTER
OPTION 02
FILLED CAVITY WALL
BRICKWALL INSULATION BLOCKWORK PLASTER
THERMAL RESISTANCE OF AIR SPACE: 10-20MM = 0.14 W/m˚C 20-50MM = 0.17 W/m˚C
IN THIS OPTION, THE ADDITION OF INSULATION LAYER IS MINIMIZING THE AMOUNT OF HEAT TRANSFER INTO THE SPACE IN OUR TROPICAL CLIMATE. IT ALSO RESTRICT THE MOISTURE FROM PASSING THROUGH THE WALL WHICH IS VERY SUITABLE TO USE IN OUR HUMID CLIMATE.COMPARE TO OPTION 1, OPTION 2 HAS LOWER THERMAL RESISTANCE WHICH SHOWN BELOW, SO THE FILLED CAVITY WALL WILL BE CHOSEN AS THE FINAL PROPOSED WALL IN THIS PROJECT.
THERMAL RESISTANCE OF FIBREGLASS INSULATION: FIBREGLASS INSULATION = 0.037 W/m˚C
COMPARISON BETWEEN THE EXISTING WALL AND THE PROPOSED WALL PROPOSED WALL
EXISTING WALL
390MM THK. CAVITY WALL WITH 150MM THK. FIBREGLASS INSULATION AND 20MM THK. PLASTERING AT ONE SIDE
SECTIONAL
110MM THK. BRICKWALL WITH 20MM THK PLASTERING ON BOTH SIDE
MAKE LAB
GARDEN MAKE LAB
U-VALUE
GARDEN
U – VALUE = ___________________ 1
U – VALUE = _______________________ 1
Rso + R1 + R2 + R3 + Rsi
Rso + R1 + R2 + R3 + R4 + Rsi
= ________________________ 1 Rso + ( b )1 + ( b ) 2+ ( b ) 3 + Rsi k k k
= _____________________________ 1 Rso + ( b )1 + ( b ) 2+ ( b ) 3+ ( b ) 4+ Rsi k k k k
=
=
1 ______________________________ 0.02 0.055 + ( ) + ( 0.11 ) + ( 0.02 ) + 0.123 0.5 0.84 0.5 = ____________________________ 1
0.055 + ( 0.11) + ( 0.15 ) + ( 0.11) + ( 0.02 ) + 0.123
0.84
= =
= ______ 1 0.389
THERMAL CONDUCTIVITIES
0.037
0.65
0.5
1 ____________________________ 0.055 + 0.131 + 4.054 + 0.169 + 0.04 + 0.123
0.055 + 0.04 + 0.131 + 0.04 + 0.123
U – VALUE = 2.571 W/m2 °C
1 _______________________________
1 ______ 4.572
U – VALUE = 0.219 W/m2 °C
SURFACE RESISTANCE
BRICKWORK = 0.84 W/m˚C
OUTSIDE = 0.055 m2 ˚C/W
BLOCK WORK = 0.65 W/m˚C
INSIDE = 0.123 m2 ˚C/W
PLASTER = 0.50 W/m˚C FIBREGLASS INSULATION = 0.037 W/m˚C
R =
b (MATERIAL THICKNESS)_______ k (MATERIAL THERMAL CONDUCTIVITY)
COMPARISON AND EXPLAIN THE DIFFERENCE BETWEEN THE EXISTING WALL AND PROPOSED WALL
1.3
PROPOSED WALL
EXISTING WALL
390MM THK. CAVITY WALL WITH 150MM THK. FIBREGLASS INSULATION AND 20MM THK. PLASTERING AT ONE SIDE
SECTIONAL
110MM THK. BRICKWALL WITH 20MM THK PLASTERING ON BOTH SIDE
MAKE LAB
GARDEN GARDEN
PLASTER
U-VALUE
CHOICE OF MATERIAL
PLASTER
MAKE LAB
BRICKS
AERATED BLOCK
THIN LAYER OF PLASTER LAYER AND HIGH THERMAL CONDUCTIVITIES ALLOW THE HEAT TRANSMITTED EASILY. SPECIAL ADDITION MATERIAL : FIBREGLASS BRICKWALL
HIGHER THERMAL CONDUCTIVITIES MAKES THE HEAT TRANSFER EASILY INTO THE SPACE.
FIBREGLASS ACTS AS AN IMPORTANT ROLE. IT MINIMIZES THE HEAT TRANSFER TO THE ROOM AND HELPS TO KEEP THE ROOM COOL.
U – VALUE = 2.571 W/m2 °C
U – VALUE = 0.219 W/m2 °C
DUE TO THE THINKNESS AND HIGHER THERMAL CONDUTIVITIES OF MATERIAL USED, THE U-VALUE IS HIGHER. THEREFORE THE HEAT CAN BE TRANSMITTED IN AND OUT EASILY TO THE SPACE
DUE TO THE ADDITION OF THE INSULATION LAYERING AND THE LOWER THERMAL CONDUCTIVITES OF MATERIAL USED, THE UVALUE IS GETTING LOWER. THEREFORE, IT CAN HELPS TO KEEP THE TEMPERATURE OF THE SPACE AND MAINTAIN THE THERMAL COMFORT DURING THE HOT TIME. IN THIS PROJECT, WE CAN SEE THAT INSULATION IS PLAYING AN IMPORTANT ROLE TO CONTROL THE THERMAL PERFORMANCE OF THE SPACE.
TASK 0 2
2.1
MEAN RADIANT TEMPERATURE (MRT) AVERAGE TEMPERATURE FOR THE WALL FACING TO THE NORTH IS 36˚C, SOUTH IS 40˚C,
EAST IS 34˚C, AND WEST IS 45˚C.
POINT A MRT A = (36 X 146) + (34 X 110) + (40 X 53) + (45 X 51) _____________________________________ 360
A
= _______________________ 5256 + 3740 + 2120 + 2295 360 = ______ 13411 360 MRT A = 37.25°C
POINT B MRT B = (36 X 102) + (34 X 62) + (40 X 97) + (45 X 100) _____________________________________ 360 = 3672 + 2108 + 3880 + 4500 _______________________ 360
B
= 14160 ______ 360 MRT B = 39.33°C
MRT OF POINT A = 37.25°C MRT OF POINT B = 39.33°C
2.2
COMPARISON OF MRT VALUE DIFFERENCES BETWEEN POINT A & B
POINT B
POINT A
MRT VALUE
A B
MRT = 37.25°C (LOWER)
MRT = 39.33°C (HIGHER)
• THE MRT VALUE OF POINT A IS LOWER COMPARE TO POINT B.
COMPARE TO POINT B.
• THE WALLS FACING NORTH AND EAST
• THE WALLS FACING SOUTH AND WEST
HAVE LOWER AVERAGE TEMPERATURE
HAVE HIGHER AVERAGE TEMPERATURE
WHICH
WHICH
AFFECTS
THE
RATE
OF
AFFECTS
THE
RATE
OF
CONVECTIVE AND EVAPORATIVE BODY
CONVECTIVE AND EVAPORATIVE BODY
HEAT LOSS.
HEAT LOSS. THEREFORE THERE IS MORE
THEREFORE THERE IS
LESS HEAT TRANSFER TO POINT A.
DIFFERENCE
• THE MRT VALUE OF POINT A IS HIGHER
HEAT TRANSFER TO POINT B.
• THE LOCATION OF POINT A HAS LOWER
• THE LOCATION OF POINT A HAS HIGHER
HUMIDITY COMPARE TO POINT B
HUMIDITY
COMPARE
TO
POINT
B.
BECAUSE THERE IS LESS WARM AIR
BECAUSE OF THE TOILET NEXT TO THE
AROUND IT.
SPACE. THEREFORE THE WARMER AIR
• THE WINDOW RIGHT NEXT TO POINT A
HOLD MORE AMOUNT OF WATER VAPOR.
ALLOW GOOD AIR VENTILATION AND
• THERE IS NO OPENINGS NEAR TO POINT
SUFFICIENT AIR MOTION TO FLUSH
B, SO THAT THERE IS INSUFFICIENT AIR
OUT THE BODY HEAT SINCE HEAT CAN
MOTION THAT CAUSED THE STUFFINESS
TRANSFER
AND AIR STRATIFICATION WHICH DO
BY
CONVECTION
EVAPORATION IN AIR MOTION.
AND
NOT ALLOWS TO FLUSH OUT BODY HEAT.
2.3
WAYS OF HEAT TRANSFER AND PROPOSED SOLUTION
WAYS OF HEAT TRANSFER OF THE SELECTED ENCLOSED SPACE
01
CONDUCTION
02
CONVECTION
THE TRANSFER OF HEAT THROUGH SECTION OF WALLS AND CEILINGS (SOLID MATERIAL) TO THE COOLER SPACE.
THE TRANSFER OF HEAT BY MOVING AIR. WARM AIR
RISES
AND
TRANSFER
HEAT
UPWARD
TO
THE
CEILING AND ROOF.
THE TRANSFER OF HEAT
03 RADIATION
IN
THE
FORM
OF
ELECTROMAGNETIC WAVES.
HEAT
TRANSFERRED THE
ROOF
IS FROM
TO
THE
CELING AND FINALLY TO THE SPACE.
PROPOSED SOLUTION OF REDUCING DIRECT HEATING INTO THE SPACE
TASK 0 3
3.1
STEREOGRAPHIC SUN PATH DIAGRAM
SUMMER SOLSTICE AZIMUTH - 8° N ALTITUDE - 71°
SUMMER SOLSTICE
VERNAL EQUINOX
VERNAL EQUINOX AZIMUTH – 138° SE ALTITUDE - 85°
WINTER SOLSTICE
WINTER SOLSTICE AZIMUTH -176° S ALTITUDE - 62°
SHADOW ANGLE PROTRACTOR
SUMMER SOLSTICE HSA - 0° VSA - 0°
VERNAL EQUINOX HSA - 30° VSA -83°
WINTER SOLSTICE HSA - 68° VSA - 75°
Time 1400
IN THIS REPORT, THE SUN PATH DIAGRAM CHOSEN IS AT KUALA LUMPUR, MALAUSIA. THE SELECTION OF TIME IS AT 2PM NOON ON SUMMER SOLSTICE, VERNAL EQUINOX AND WINTER SOLSTICE. THE ORIENTATION OF THE CHOSEN OPENING IS FACING DIRECTLY TO 108° ESE, IT IS MORE TOWARDS EAST. THEREFORE, THE SUNLIGHT ABLE TO PENETRATE IN THE SPACE DURING MORNING TIME.
Azimuth
Altitude
HSA
VSA
Summer Solstice
8° N
71°
0°
0°
Vernal Equinox
138° SE
85°
30°
83°
Winter Solstice
176° S
62°
68°
75°
3.2
COMPARISON OF THE SHADOW CAST
21ST DECEMBER
WINTER
21ST JUNE
SUMMER
EXISTING WINDOW DESIGN
HSA = 0° VSA = 0°
PLAN
ELEVATION
SECTION
HSA = 68° VSA = 75°
PLAN
ELEVATION
SECTION
PROS AND CONS
OF THE EXISTING WINDOW DESIGN FOR SUMMER SOLSTICE AND WINTER SOLSTICE CONS
• ALLOW MORE VISION
• THE
• ALLOW AIR VENTILATION OF THE SPACE
SOLSTICE
WINTER
POSITION
OF
THE
WINDOW AT THIS TIME IS NOT ABLE TO CAPTURE ANY NATURAL LIGHT
SOLSTICE
SUMMER
PROS
• ARTIFICIAL
LIGHTS
ARE
REQUIRED TO THE SPACE
• PROVIDE NATURAL LIGHT PENETRADE
INTO
THE
SPACE • ALLOW MORE VISION • ALLOW AIR VENTILATION OF THE SPACE
• THE DIRECT LIGHT CANNOT BE CONTROLLED AS THERE IS NO ANY SHADING DEVICE • CAUSES OF DESTROY ON THE FURNITURE DUE TO EXPOSION TO THE UV LIGHT • THE ROOM TEMPERATURE INCREASE EASILY
PROPOSED SOLUTION TO IMPROVE THE COMFORT LEVEL OF THE SPACE
TASK 0 4
4.1
DESIGN OF EXTERNAL SHADING DEVICE
DESIGN DEVELOPMENT
DESIGN INTENTION
FINAL DESIGN OF SHADING DEVICE
ALLOW LIGHT PENETRADE INTO THE SPACE TO MINIMIZE THE USAGE OF ARTIFICIAL LIGHT OF THE SPACE.
AVOID DIRECT LIGHT PENETRADE ON HUMAN EYE LEVEL.
PROMOTE GOOD AIR VENTILATION & CROSS VENTILATION TO IMPROVE THE THERMAL COMFORT OF THE SPACE. ALLOW CAPTURE OF THROUGH THE OPENING.
VIEW
COMPARISON OF THE SHADOW CAST
21ST DECEMBER
WINTER
21ST SEPTEMBER
EQUINOX
WITH THE EXTERNAL SHADING DEVICE
HSA = 30° VSA = 83°
PLAN
ELEVATION
SECTION
HSA = 68° VSA = 75°
PLAN
ELEVATION
SECTION
4.2
IMPLICATIONS OF FENESTRATION
DESIGN: - 2 OPERABLE WINDOW: LARGE INLET AND SMALL OUTLET TO ACHIEVE CROSS VENTILATION - USE STACK VENTILATION TO CREATE A CHIMNEY EFFECT
CROSS VENTILATION: - THE EXISTING OPENING WITH THE PROPOSED EXTERNAL SHADING DEVICES (LARGE INLET) ALLOW THE AIR FLOW IN TO THE SPACE AND THE ADDITIONAL AWNING OPENING IS PROPOSED AS AN OPPOSITE SIDE OF OPENINGTO ENCOURAGE CROSS VENTILATION. THE AIR VELOCITY AT THE LARGE INLET WILL BE LOWER BUT THE VOLUME OF AIR PASSING IN UNIT TIME WILL BE HIGHER. - THE VERTICAL LOUVER INSTALL AT THE INLET IS ADJUSTABLE TO CHANNEL THE WIND IN AND CONTROL THE DIRECTION OF THE WIND. -
SINCE THE HOT AIR RISES AND COLD AIR SINKS, THE PLACEMENT OF THE OPENING IS TO CARRY THE HOT AIR OUT OF THE SPACE AS LOWER INLET ALLOWS CHILL AIR TO ENTER AND HAVE OPTIMUM FLOW.
4.3
CONCLUSION
IN CONCLUSION, I HAVE LEARNT ABOUT THE IMPORTANTANCE OF THE CHOICE OF MATERIALS WHICH HIGHLY AFFECTS THE THERMAL PERFORMACE OF THE BUILDING THROUGH COMPARISON OF THE CALCULATION OF U-VALUE AND MRT VALUE BETWEEN DIFFERENT MATERIAL USED. IN OUR MALAYSIA TROPICAL COUNTRY, THE USAGE OF BAD CONDUCTOR MATERIAL IS MORE RECOMMENDED TO IMPROVE THE THERMAL COMFORT OF THE BUILDING. BY GOING THROUGH SUN PATH PROCESS, I HAVE LEARNT THAT EVERY MONTH HAVE THEIR DIFFERENT ANGLE OF SHADOW CAST IN ORDER TO HELP ON DESIGN OF SHADING DEVICE. SHADING DEVICE PLAY AN IMPORTANT ROLE OF THE BUILDING WHICH DOES NOT ONLT PROVIDE THE SHADE, BUT ALSO THE WAY OF HOW USER INTERACT WITH NATURAL LIGHT, CAPTURE VIEW AND ALLOWING THE WIND ENTER THE SPACE AND EVEN IT IS IMPORTANT TO THE BUILDING FAÇADE TOO. BY GOING THROUGH THE DESIGN EVOLUTION OF SHADING DEVICE, I HAVE LEARNT HOW TO ACHIEVE THE BALANCING BETWEEN AESTHTIC FORM AND FUNCTIONAL SHADING DEVICE IN ORDER TO CONTROL A GOOD THERMAL COMFORT OF THE SPACE. BASED ON MY UNDERSTANDING OF PASSIVE VENTILATION, THE IMPLICATION OF FENESTRATION IS VERY IMPORTANT TO MAINTAIN THE THERMAL COMFORT OF THE SPACE. AS THE WIND VENTILATION CAN MAINTAIN MINIMUM AIR QUALITY, REMOVING HEAT AND OTHER POLLUTANTS AND FACILITATING AIR MOVEMENT TO ENHANCE THERMAL COMFORT. THE ORIENTATION, SIZE AND POSITION OF OPENING ALSO HAS A BEARING ON THE EFFECTIVENESS OF THIS FORM OF
VENTILATION. WIND VENTILATION IS THE MOST COMMON AND OFTEN LEAST EXPENSIVE FORM OF PASSIVE VENTILATION TO BE USED IN OUR COUNTRY. IN OVERALL, I HAVE PRACTICE AND EVALUATE THE KNOWLEDGE APPLICATION OF PHYSICAL AND BEHAVIORAL SCIENCE IN BUILDINGS IN MY BUILDING DESIGN FROM THE SUBJECT ARCHITECTURE DESIGN STUDIO 3 (AR315).
REFERENCE EARTHEASY PUBLISHED ON JAN 4, 2020 RETRIEVED FROM https://learn.eartheasy.com/guides/natural-home-cooling/ ENERGUIDE.BE PUBLISHED ON APRIL 16, 2020 RETRIEVED FROM https://www.energuide.be/en/questions-answers/how-can-i-keep-my-home-cool-in-summer/236/ CONNECTUS PUBLISHED ON AUG 28, 2016 RETRIEVED FROM https://connectusfund.org/9-advantages-and-disadvantages-of-cavity-wall-insulation TANISHA AGARWAL PUBLISHED ON MAY 1, 2017 RETRIEVED FROM https://www.slideshare.net/TanishaAgarwal1/cavity-walls-75561973 CLEAR PUBLISHED ON DEC 6, 2015 RETRIEVED FROM https://www.new-learn.info/packages/clear/index.html GUN FAISAL PUBLISHED ON APRIL 23, 2016 RETRIEVED FROM https://www.researchgate.net/figure/External-Shading-Devices_tbl1_303503284 CLEAR PUBLISHED ON MAY 17, 2014 RETRIEVED FROM https://www.newlearn.info/packages/clear/thermal/buildings/passive_system/passive_cooling/natural_ventilation/design.ht ml AMAR PUBLISHED ON AUGUST 4, 2015 RETRIEVED FROM https://indiansustainability.wordpress.com/2015/08/04/importance-of-cross-ventilation/ ROOPA CHIKKAIGI PUBLISHED ON MAR 23, 2017 RETRIEVED FROM https://www.slideshare.net/roopachikkalgi/natural-ventilation-73526859
Program:
Bachelor of Science (Hons) in Architecture
Course Code and Name:
AR 343 Building Science 1
Semester
2020-07
Student's Name and ID:
Lai Zhi Zhen 1001747427
Lecturer:
Mr. Chan Chiew Chuen
Date of Submission:
November 13, 2020.
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