BUILDING SCIENCE AND SERVICES E-PORTFOLIO

PROJECT 1

LIGHTNING AND VAC DESIGN FOR COMFORT

BUILDING SCIENCE AND SERVICES ASSIGNMENT 1

NAME : MADELINE OTTOMAN STUDENT ID : 0345443 TUTOR : MS SHARON TEH

BACHELOR OF ARTS (HONS) INTERIOR ARCHITECTURE

1. SITE ANALYSIS

Figure 1.1 Figure 1.2

The site that I have chosen for this project is my own room (Figure 1.1) and my apartment kitchen, (Figure 1.3 and 1.4) which is located in U Residence, Taylor’s Hostel. The room that i have stayed in is a Standard Single type, where in 1 apartment each person will get their own rooms, however we are required to share bathroom and kitchenette. In my room, it has a 1 ceiling fan, 1 air conditioner, 1 window for ventilation. For the lightning, it contains only 1 fluorescent light for the study desk and 1 wall lamp (Figure 1.2). In this room, I almost do my homework in the study desk and take a rest on the bed.

Figure 1.3

Figure 1.4

On the other hand, this is the kitchen which is located outside of my room (Figure 1.3 and 1.4) .In this kitchen, I always do my cooking for lunch and dinner as well as cutting my groceries for cooking.

Figure 1.5

Figure 1.6

Figure 1.7

Figure 1.8

Figure 1.9

In my room , they have 3 electrical switch sockets where 2 of them are located below the shelf ( Figure 1.5) , whereas 1 of it is located below the study desk ( Figure 1.6) . Overall, it also has 4 switches for the lightning. The first and second switch which is next to the door is switches for the wall lamp and the ceiling fan ( Figure 1.7 ), then the one near the window is the switches for the study desk ( Figure 1.8 ).The last of the switches that is located next to the air conditioner controller ( Figure 1.9 ) is also for light up the wall lamp.

Figure 1.10

Figure 1.11

In the kitchen, there are 2 switch sockets above the microwave ( Figure 1.10), 3 consecutive switches above the dining table which are used for the ceiling fan, kitchen lighting and corridor lighting ( Figure 1.11)

2. PROBLEMS AND SOLUTIONS OF THE SITE

2.1. PROBLEM :

Firstly, when i get into this room I realize that this room needs additional lighting for light up every inch of the room, because in this room they only provided one fluorescent light that contains cool white illumination for the study desk. In my opinion, it is not good for the room because it only lights up half of the room, whereas the rest of it feels so dark. Apart from that, when I am searching for something inside my wardrobe, I cannot see anything in my wardrobe because of the door itself blocking the light to the wardrobe and in that area they don't install lights at all. ( Figure 2.1)

Figure 2.1

SOLUTION :

I wanted to add additional lightning such as fluorescent lighting to light every single inch of my room. The reason why I chose the fluorescent lighting is because my room is very compact, and there is a ceiling fan spinning above the ceiling. If I use another lightning such as a downlight, i am afraid that when I turn the light on, the lightning will be blocked by the fan itself. In this case the color of the illumination will be cool white, and I will place the lightning at the edge part of my room.

2.2. PROBLEM :

Secondly, while I lived here for a while, I feel that the air conditioner is too cold for me ( Figure 2.2). Typically I have a skin allergy when the temperature is too extreme, a spot that is the same as a mosquito bite will appear throughout my body. In this case, I would like to change my air conditioner to make sure that the temperature of the room stays the same throughout the time.

Figure 2.2

SOLUTION :

Due to the extreme temperature caused by the air conditioner, I have decided to change the air conditioner into the inverter mode because it can make the room temperature stable. Apart from that, it also has a feature of dehumidification for the room. The reason why I chose this air conditioner is because since my room is very compact, there is also a high possibility for my room to be humid. If this is the case that i choose to ignore, my room will grow mold inside the room.

2.3. PROBLEM :

In the kitchen, the lightning appeared in a warm white illumination. In this case it is not good because I feel like it can trick my eyes on looking at something. For example, when i was cooking in the kitchen, for example i pan fried a chicken meat, i cannot see clearly whether it is already cooked or it is undercooked, since the appearance of the meat under this lightning makes the meat looks a bit of pink color ( Figure 2.3 ). In this situation, I always light up the torch of my phone to see clearly whether the meat is already cooked well.

Figure 2.3

SOLUTION :

The illumination of the lightning will be changed into a cool white color to ensure the safety of the kitchen as well as cooking procedure inside the apartment.

2.4. PROBLEM :

I would like to replace the wall lamp switch because I feel that it is not suitable for me to turn on the wall lamp (warm yellow white color) when I first enter my room. For references, the switches of the wall lamp are located near the door ( Figure 2.4 ). So when I first came in I could only access to this light to turn on the light of the room. I feel that the switches of the wall lamp are not suitable to be placed next to the door because firstly the room is very compact and small, and a person might not look very clear in every inch of the room when they enter the room itself.

SOLUTION :

For references, the wall lamp switches is a two way switch where we only need to turn either 2 switches to turn on the wall lamp. The switches are located each above the bed and next to the door. In this case, I would like to change the position of the wall lamp switch into a one way switch only at the back of the bed Then, I rearrange the switches path for another lightning applied into the room

2.5. PROBLEM :

Lastly, I would like to change the placement of the socket below my study desk ( Figure 2.5) For this socket I know that it is purposely for charging a laptop, however for me it is hard for me to reach this socket as I need to bend over to charge my laptop charger. Apart from that, I also feel that this socket is useless to locate below the study desk.

Figure 2.5

SOLUTION

:

I wanted to change the location of the socket into the upper part of the wall to ensure that it can be seen by myself and also can be used when the rest of the sockets are being used.

3. VAC CALCULATION

Since I was planning to change my Air conditioner, I will calculate how much hp of the air conditioner that is suitable for my room.

Brand Name : York Bio Shields Model : G series

Specification :

Model Indoor Unit : YWM09G HP : 1.0 hp

Cooling Capacity : 9000 Btu/hr Rated (min-max) : 2.64 kW

Running Current : 4.3 A Dimension indoor ( H x W x D) = 260 x 799 x 198 Dimension outdoor ( H x W x D) = 495 x 600 x 245

Ceiling Fan = 20 W, Wall Light = 15 W, Study desk Light ( Fluorescent Lightning ) = 36 W, Laptop = 180 W, Iphone X = 7.5 W, Apple watch = 20 W, Hair dryer 1000 W Total = 20 W + 15 W + 36 W + 180 W + 7.5 W + 20 W + 1000 W = 1278.5 W

1278.5 W x 3.5 = 4474.75 Btu/hr 4)

4816 + 500 + 4474.75 = 9790.75 Btu/hr

Brand Name : Fluorescent Lightning LED T8 Tube Daylight 18 W (120 CM)

Specification :

- Dimension of the lightning : 0.6M = 600 mm - Wattage : 9W - 6500K = 1820 lumen - One way switch -Illumination = Daylight -Bulb type = T8 2.

5.

(RENDERED IN SKETCHUP)

KITCHEN AREA ( 4 downlights has already added above the kitchen )

3BLADECEILINGFAN 2 LOCATEDINMYROOMANDKITCHEN

FLUORESCENTLIGHT FIXTURE 3 2WARMWHITEILLUMINATION KITCHEN),1COOLWHITE ILLUMINATIONINSTUDYDESK

SWITCHSOCKET 6 1BELOWTHESTUDYDESK

ONEWAYSWITCH 3 LOCATEDINMYROOMANDKITCHEN

TWOWAYSWITCH 6 LOCATEDINMYROOMANDKITCHEN

CARDINSERTERFOR ACTIVATEAC 1 LOCATEDINMYROOM AIRCONDITIONER 1 LOCATEDINMYROOM

FLUORESCENTLIGHT

SWITCHSOCKET 6ONEWAYSWITCH 3 LOCATEDINMYROOMANDKITCHEN TWOWAYSWITCH 5 LOCATEDINMYROOMANDKITCHEN

CARDINSERTERFOR

PROJECT 2

INDOOR COMFORT DESIGN AND INVENTION

INDOOR COMFORT DESIGNAND INVENTION

NAME : MADELINE OTTOMAN STUDENT ID : 0345443 TUTOR : MS SHARON TEH

Temperature 1 = 7 AM

25℃ = 77 ℉

At this time, the temperature is 25 because the air and the weather is very cold and the sun has not shone throughout the area.

Temperature 2 = 12 PM

29℃ = 84.2 ℉

At this time, the sun has started shining throughout the area but it is not as intense and bright that has a glare.

Temperature 3 = 5 PM

30℃ = 86 ℉

At this time, the sun has started to shine brightly throughout the area because it is already evening when the sun has started to sunset.

1 3 ExternalTemperature of the Other Wall

Wall 2

Firstly, the next room beside wall 2 is a room where my roommate stays Most of the time she turns on the air conditioner at night, and sometimes she only uses the ceiling fan to cool out her room

26℃ = 78.8 ℉

Wall 3

This is the wall where it faces the entrance of the bathroom as well as the corridor of the apartment In this corridor I can access directly to the kitchen Since the kitchen has a ceiling fan, the fan is turned on at any time I feel that it also depends on the weather to feel the cold and warmth of the space

7AM 24℃ = 75.2 ℉

12 PM

27℃ = 80.6 ℉ 5 PM

The room that is attached next to the wall 4, there is no one living inside the room Apart from that, based on my understanding, if an empty room does not open any air conditioner as well as ceiling fan, it is possible that the temperature is very hot and the air is humid. However, since the room is surrounded by reinforced concrete, the temperature might be fairly cold, which means it is not too hot and too cold

2. ELEVATION of each of the walls ( CAD and pictures ) along with the CALCULATIONS of heat transfer rate

Formula : Q = (A/R) * ΔT

Where :

Q = Heat Flow ( Btu/hr )

A=Area ( ft²)

R = Resistance ( ft²-hr-℉ / Btu)

ΔT=Temperature Differential ( ℉ )

Additional Information regarding to the calculation :

1 foot = 30 48 cm = 304 8 mm

1 inch = 2 54 cm = 25 4 mm

Data : -Materials of the wall : -Reinforcement Concrete ( R = 0 08 )

- Single Glass Window ( R = 0 91 )

Th

R total = R1 + R2 = ( 0 08 * 6 49 ) + ( 0 91 * 1 18 ) = 0 5192 + 1 073= 1 6992 = 1 60 ft²-hr-℉ / Btu

- DImension of the wall :

L1 = 2540 mm = 8 33 feet W1 = 3026 mm = 9.92 feet

Area (A1) = L1 * W1= 8 33 ft * 9 92 ft = 82 63 ft²

- DImension of the window :

L2 = 1180 mm = 3 87 feet W2 = 895 mm = 2.93 feet

Area (A2) = L2 * W2= 3 87 ft * 2 93 ft = 11 33 ft²

TotalArea of the Wall =A1 -A2 = 82 63 ft² - 11 33 ft² = 71.3 ft²

7AM

- T1 (WeatherTemperature ) = 77℉ ;T2 ( InternalTemperature ) = 82 4 ℉

ΔT= 82 4 - 77 = 5.4 ℉

Calculation :

Q = (A/R) * ΔT = ( 71.3 / 1.60 ) * 5.4 = 241 Btu/hr

12 PM

- T1 = 84 2℉ ;T2 = 82 4 ℉

ΔT= 84 2 - 82 4 = 1.8 ℉

Calculation :

Q = (A/R) * ΔT = ( 71.3 / 1.60 ) * 1.8 = 80.2 Btu/hr

5 PM

- T1 = 86℉ ;T2 = 82.4 ℉

ΔT= 86 - 82 4 = 3.6 ℉

Calculation :

Q = (A/R) * ΔT = ( 71 3 / 1 60 ) * 3 6 = 160 42 Btu/hr

In conclusion, the highest heat transfer rate that is happening in wall 1 is at 7AM, which is 241 Btu/hr

- Dimension of the wall :

L1 = 2710 mm = 8 89 feet W1 = 3026 mm = 9.92 feet

Area = L1 * W1= 8 33 ft * 9 92 ft = 82.6 ft²

7AM

T1 = 78.8℉ ;T2 = 82.4 ℉ ΔT= 82 4 - 78 8 = 3.6 ℉

Calculation : Q = (A/R) * ΔT = ( 82 6 / 0 5192 ) * 3 6 = 573 Btu/hr

12 PM

T1 = 78.8℉ ;T2 = 82.4 ℉

ΔT= 82 4 - 78 8 = 3.6 ℉

Calculation : Q = (A/R) * ΔT = ( 82 6 / 0 5192 ) * 3 6 = 573 Btu/hr

T1 = 78 8℉ ;T2 = 82 4 ℉

ΔT= 82.4- 78.8 = 3.6 ℉

Calculation :

Q = (A/R) * ΔT = ( 82 6 / 0 5192 ) * 3 6 = 573 Btu/hr

In conclusion, the heat transfer rate throughout the day is stay the same, which is 573 Btu/hr

Thickness : Wall 165 mm 6 49 inch

- Solid Wood Door = 30 mm = 1 18 inch

R total = R1 + R2 = ( 0 08 * 6 49 ) + ( 3 03 * 1 18 ) = 0 5192 + 3 5754 = 4.09 ft²-hr-℉ / Btu

- Dimension of the wall :

L1 = 2540 mm = 8 33 feet W1 = 3026 mm = 9.92 feet

Area = L1 * W1= 8 33 ft * 9 92 ft = 82 6 ft²

- Dimension of the door :

L2 = 2045 mm = 6 70 feet W2 = 949 mm = 3.11 feet

Area = L2 * W2= 6 70 ft * 3 11 ft = 20 84 ft²

TotalArea of the Wall =A1 -A2 = 82 6 ft² - 20 84 ft² = 61.76 ft²

7AM

- T1 = 75.2 ℉ ;T2 = 82.4 ℉ ΔT= 82 4 - 75 2 = 7.2 ℉

Calculation :

Q = (A/R) * ΔT = ( 61 76 / 4 09 ) * 7 2 = 109 Btu/hr

- T1 = 80 6 ℉ ;T2 = 82 4 ℉

ΔT= 82.4 - 80.6 = 1.8 ℉

Calculation :

Q = (A/R) * ΔT = ( 61 76 / 4 09 ) * 1 8 = 27.18 Btu/hr

5 PM

- T1 = 77 ℉ ;T2 = 82 4 ℉

ΔT= 82 4 - 77 = 5.4 ℉

Calculation :

Q = (A/R) * ΔT = ( 61 76 / 4 09 ) * 5 4 = 81 54 Btu/hr

In conclusion, the highest heat transfer rate that is happening in wall 3 is at 7AM, which is 109 Btu/hr

Note in regards to the wall :

In this matter the wall 4 structure is not the same as the where there is a built-in wardrobe at the wall Apart from that, the wall B, C and f the wardrobe structure For the wall E part of the wardrobe. In this drawing, I single wall that is simplified from the

Note : For Wall C, the wall is connected to support the structure of the wardrobe, it just the contractor take the middle point of the wall to create a hinge mechanism for the wardrobe door panel

Pictures of the wall :

WallA

Wall B ( Side of the wardrobe, Exterior part )

Wall C ( Outside of the wardrobe ) Wall C ( Inside of the wardrobe )

R = 0 08 * 6 49 = 0 5192 ft²-hr-℉ / Btu

- WallAData : LA= 1735 mm = 5 69 ft WA= 3026 mm = 9 93 ft

Area of WallA = LA*WA= 5.69 ft * 9.93 ft = 56.5 ft²

- Wall B Data : LB = 485 mm = 1 59 ft WB = 3026 mm = 9.93 ft

Area of Wall B = LB*WB = 1 59 ft * 9 93 ft = 15 8 ft²

- Wall C Data : LC = 140 mm = 0.46 ft WC = 3026 mm = 9 93 ft

Area of Wall C = LC*WC = 0.46 ft * 9.93 ft = 4.56 ft²

- Wall D Data : LD = 485 mm = 1 60 ft WD = 3026 mm = 9 93 ft

Wall D ( Side of the wardrobe, Interior part ) Wall E ( Interior of the wardrobe )

Area of Wall D = LD*WD = 1.60 ft *9.93 ft = 15.89 ft²

- Wall E Data : LE = 835 mm = 2 74 ft WE = 3026 mm = 9 93 ft

Area of Wall E = LE*WE = 2 74ft * 9 93ft = 27 20 ft²

TotalArea for Wall 4 =A+ B + C + D + E = 56.5 ft² + 15.8 ft² + 4.56 ft² + 15.89 ft² + 27.20 ft² = 119.95 ft²

7AM

- T1 = 86℉ ;T2 = 82 4 ℉

ΔT= 86 - 82 4 = 3 6 ℉

Calculation :

Q = (A/R) * ΔT = ( 119 95 / 0 5192 ) * 3 6 = 832 Btu/hr

12 PM

- T1 = 86℉ ;T2 = 82 4 ℉

ΔT= 86 - 82.4 = 3.6 ℉

Calculation :

Q = (A/R) * ΔT = ( 119 95 / 0 5192 ) * 3 6 = 832 Btu/hr

5 PM

- T1 = 84 2℉ ;T2 = 82 4 ℉

ΔT= 84 2 - 82 4 = 1 8 ℉

Calculation :

Q = (A/R) * ΔT = ( 119 95 / 0 5192 ) * 1 8 = 415.8 Btu/hr

In conclusion, the highest heat transfer rate that is happening in wall 4 is at 7AM and 12 PM, which is 832 Btu/hr

Overall of the Calculation, the Highest HeatTransfer Rate that is happening between the 4 walls is = WALL4 at 7AM and 12 PM ( 832 Btu/hr)

TIME WALL1 WALL2 WALL3 WALL4

27.18 Btu/hr 832 Btu/hr 5:00 PM 160.42 Btu/hr 573 Btu/hr 15

81.54Btu/hr 415.8 Btu/hr

1.INTRODUCTION

1.1 SITE PROBLEM

BasedonmycalculationthatIhavecountedonassignment2a,mostofthehighesttransfer heatthatIfoundisstartingfrom7am.Thisisbecausethesunstartedtoriseinthemorning anditshinesintomyroom.Theroutineofopeningthecurtaininthemorningisamustinmy familytomakesurethereissunlightgetintomyroomaswellasopeningthewindowfor circulation.OverallIenjoytheglarethatcomesintomyroomhoweverformeitistoomuch heatandglareandIcannotenjoytheoutsideviewthatmuchbecauseofthesunlightcoming directlyintomyroom.Sincemyviewwasthelake,sometimesthesunlightcanreflectthe light from the lake itself, and I mostly got too much heat into the room.

DESIGN IDEAS

Formyproduct,IwouldliketomakeafoldingcurtainwhereIcanblockouttheglareintomy room,butatthesametimeitgivesmeabitofglaretoentermyroom.InthisdesignIgot inspirationfromasunshadecarwhereitcanblockthesunlightfromgettingintothecar,as wellasasuncatcher,whichisatransparentglassthatcanreflectcolorfromthesunlight.For myproduct,itfunctionsasadevicetoblockouttheheatandglaregetintomyroom. Besidesblockingtheheatandglare,italsopenetratessunlightfromtheholeoftheheat insulator.

Figure 1.3

2.Design Sketches

Figure 1.4 Figure 1.5

SKETCHESAND DESIGN DEVELOPMENTSTEPBYSTEPIN MAKINGTHE PRODUCT

1.Firstly, i tried out the panels that i wanted to make in sketchup.This is ensure that my product works well on my window as well as i can visualized on how the panel will turned out at the end of the day.

2.Then, I cut the aluminium foil panel for the dimension of 15.8 x 16.7 according to my window.And i also proceed to cut it on aluminium foil foam

3.Due to the time limitation as well as the size requirement for the product which is 100 square inch to 150 square inch, i decided to limit my each panel into 5 inches x 5 inches.Apart from that, i also limit my panels from 16 into 6 panels only ( 3 panel horizontally, 2 panel vertically ). In this case i recut my aluminium foil and foam panels into the according size.

4.Next, i made an opening in the shape of star. I took the remaining cardboard and experimenting the size of the star to make sure it suits well to the panel. When i feel satisfied with the size, i proceed to cut out the shape of the star.

I ended up trace the star into the centre of the material and i cut it with cutter to create a hole on the center of the panel.

5.I cut out the transparent plastic paper in the size of 9.5 cm x 9.5 cm to fit into the star opening.

7.LAYERING PROCESS

6.Next i cut out the bubble wrap according to the size of the panel.

Firstly, paste the transparent plastic sheet, cover with bubble wrap, aluminium foil, bubble wrap and lastly the aluminium foil foam.

Repeat the same process throughout the remaining panels. In this case i arrange it according to the amount that i want, as well as arrange according to the colors. In this prototype i use primary color of plastic sheet.

8.I also arrange the color scheme of the wooden beads to make it more fun and interesting.

9.JOINERIES ( WIRINGAND ROPE )

For the wiring of the panels, I cut an aluminium wire approximately 8 cm, then I cut it out with pliers.After that, i bend it over and shape it into an eye on the edge of the wire itself. I make around 17 eye hooks attached to the side of the panels.

Next,Itieduptheropeintotheplastichook,andImeasuredaround10cmforcuttingthe rope. I inserted the beads that I had arranged just now and I tied it back to the panel.

I repeat this process to join the panels below as well as the next other panels.

10.FRAME PANEL( EVAFOAM )

Icutouttheorigamipaperabout0.7cmx13.5cmtocoverupthesurfaceoftheeva foamtoensurethevarietyofthesurfaceoftheevafoam.Rightafterthat,icutout somefoamonthesamesizeofthepaper.Ipastetheorigamipapertothefoam,and thenstickitintotheedgeofthepanels.Ileavesomeedgestomakesuretheother eva foam frames can stick to each other.

FINALPRODUCTOFTHE PANELANDTHE FRAME

FINALPRODUCTOFTHE JOINERIES WITHTHE PANEL

Total Costing for each of the Materials and the usage

EVAFoam ( Cactus Craft ) =

Area of the EVAfoam = 59 4 cm x 42 cm = 2494 8 cm^2

Area used = 38 5 x 16 8 = 646 8 cm^2

Price of Eva Foam = 5 40

Total Price = ( 646 8 x 5 40) / 2494 8 = 1.4 RM

Aluminium Foil = 75 sq feet = 69677 sq cm for 8 10 983 04 cm^2 = ?

(983 04 x 8 10 ) / 69677 = 0.11 RM

Aluminium Foil Foam ( DIY) = RM 1.20

Used around 38 4 x 25 6 = 983 04 cm^2

Price = 0.10 RM

- Origami Design Paper ( Daiso ) 60 sheets / 5 60 = RM 10 71

Used 1 = RM 10.71

- Transparent Plastic Paper ( Cactus Craft ) 1.20 each Used 3 = 1.20 x 3 = RM 3.60

- Bubble Wrap :Agift from a friend

- Rope ( DIY) = RM 2

- Aluminium Wire = Recycle ( Get it from previous IAD project )

- Cardboard ( Daiso ) = 5 90 for 10 sheets (A4 paper )

5.90 / 10 = RM 0.59

Used 2 sheets = 0 59 x 2 = RM 1.18

Plastic Hook = RM 2 2/10 = 0 20 RM each

Used 3 = 0.20 x 3 = 0.60 RM

Total Price = 1.4+ 0.11 + 0.10 + 10.71 + 3.60 + 2 + 0.59 + 1.18 = RM 19.69

DESIGN VALUE OF MYPRODUCT

Design Value = (Total price / audience ) -Totalprice of Product

= (788/26) - 19.69 = 30.30 - 19.69 = RM 10.61

In conclusion, the design value that i got form the audience is RM 10.61

Afterexperiencingmakingaproductandpresentit livetotheaudience,ifeelproudofmyselfthatican createasolutionformyroom.Atthefirsttimeiwas verynervousandscarewhethermyworkdoesnot meetmyexpectation.ButonceIgetthroughthe processandexperimentmore,IfeellikeIamsatisfied onhowIhavereachthisfar.Duringpresentation,I havetriedmybesttospeakinfrontofalotofpeople anddemonstratemywork.Regardingtothis assignment,Ihavelearnedalotofpotentialand possibilitiestoblockandreducetheheatandglare. Besidesofitsfunctionality,Ifeelthaticanexplorea lotofcreativitytoinsertintomyproductandialso haveknowntheknowledgeandbenefitsofthe materialsthatIhaveusedthroughoutmakingthe product.IalsofeelhappythatIcanlistentomy friend’sproductandiamsoamazedandinspiredon how my friends come up with something for this project.

FINAL PROJECT

DESIGN OF BUILDING SERVICES

IAD 3 INTERGRATED)

BUILDING SCIENCEAND SERVICES

INTERGRATED FINALPROJECT

NAME : MADELINE OTTOMAN

STUDENT ID : 0345443 TUTOR : MS SHARON TEH

BACHELOR OFARTS IN INTERIORARCHITECTURE

1 SITE SELECTIONAND QUALITATIVE DESCRIPTION OF INCORPORATED BUILDING SCIENCE

Figure 1.1

This site is located inTamarind Square, Cyberjaya, where our site that is used for our project is the treehouse that is located at the north court of the tamarind square itself My site selection for this assignment is the 4th floor, where they have a lounge ( Figure 1.1 ) for the visitor to sit and relax in there, and an outdoor music performance ( Figure 1 2 ) where visitors can watch someone play the piano and enjoy the music in that area

SPECIFICACTIVITIESAND BUILDING SERVICES OF SELECTEDAREA

Lounge area is for the visitor to relax and get some cup of coffee and cake, while outdoor music performances is for the pianist to play the piano to express their feelings and for the visitor to enjoy the music itself

1) Determine SpaceArea

85122995 6369 sq mm = 85 sq m

85 sq foot x 700 = 59500 Btu/Hr

1) Occupants

25 person x 500 = 12500 Btu/hr

2) Heat from Electrical Equipment

Coffee machine = 950 W, Oven =800 W, Refrigerator = 780 W, Freezer = 100 W, =

Total = 950 W + 800 W + 780 W + 100 W = 2630 W 2630 W x 3 5 = 9205 Btu/hr

3) Total of Btu/hr from previous calculation

59500 + 12500 + 9205 = 81205 Btu/hr

NewAir Conditioner Sizing = 81205 / 9800 = 8.2 hp

8 5 hp = 2 5 x 3 = 3 air conditioner to be used on that space

LIGHT ILLUMINANCE

E ( lux ) = Luminous Flux ( lm ) /A

A= 85 sq m

Total Lights = 7 downlights ( 548 lumen each )

7 (548) / 85 sq m = 3836/85 = 45 lm/ sq m

Duration used for the lightning = 7 hours

Monthly Energy used for the lightning = 0 009 x 7 x 30 = 1.89 kWhr

7 lightning = 7 x 1.89 = 13.23 kWhr

Monthly energy used for the VAC

8 hp = 5965.6 watt = 5.965

5 965 W x 7 hour x 30 days = 1252.65 kWhr

Rear set drain = Standard 3 5 inches for opening, Size = 75 x 41 x 22 ( Lx W x D ), Material = 304 Stainless Steel

Rate of electricity for business = 0.388 kWH

Water supply drainage plan and Intergrated drawings (A3 size paper )