EPORTFOLIO BUILDING SCIENCE

Bachelor of Science (Honours) in Architecture / (Architecture) BUILDING SCIENCE

Name:CARISSA VASHTI WIRAWAN…………………………………………………….

ID: 0346416…………………………………

LECTURER: ……MS.SHARON ………………………………………………………….

ASSIG NMENT 1

The place I chose was one of the rooms in the Dlatour Apartment in building B. The occupants already have a fixed position for bed and wardrobe but they feel uncomfortable with the placement of the lights, air conditioner,and storage area. In the condition of the occupant's room, a room with less comfortable lighting is a problem in arranging the house. In addition to causing visibility to be limited if there is insufficient lighting, a room with lighting that is too bright or not well lit also affects a person's mood when resting or working.

In the resident's apartment unit, the strongest light will flow from around 12 noon to 3 pm which makes the room hotter and brighter. On the other hand, because the occupant's room does not have a window, the light entering the room is not too significant. Therefore, to get good lighting in the occupants' room, the use of proper lighting is needed in this room.

Based on the luxmeter, when the room is turned off at the strongest sun hour, the luxmeter shows the number 8 lux.

In the occupant's room using slim downlights in the bedroom and bathroom area.

125mm diameter slim downlight was installed In the bedroom and bathroom area

Using 125mm diameter slim Downlight with warm white color

When using the air conditioner, the user wants to change the air conditioner to a Daikin FTKG-T SERIES (R32) wall mounted air-conditioner model which provides 1 horsepower. The previous air-conditioner was a Hitachi RAS-F10CF with 1 horsepower. The occupants wanted to replace the air-conditioner because in the previous air conditioner, when the air conditioner is turned on it often emits a damp smell.

Hitachi RAS-F10CF

DAIKIN FTKG-T SERIES(R32)

The occupants want to change the placement on the layout of the sockets next to the table and mattress, to reduce the danger of short circuits and also make it easier when they want to use items that require electricity.

Dissatisfactory Issues Comfort in lighting is an important thing in a room. The quality of the lighting produced by the lighting system will determine the comfort of the room user. If the quality of lighting is not adequate, it will clearly hinder the activities of space users. The eye health of space users is influenced by the quality of the lighting produced by the lighting system.

The occupant of the selected room is a student who requires proper lighting and electric socket settings. However, there is no lamp near the table to make it easier for residents when working on assignments or while reading.Based on Lux meter the brightness of the light fixture in the sleeping area at night is approximately 106 Lux, which is not suitable for eye comfort level.

The two pictures show a comparison of the luxmeter in the wardrobe area when the lights are turned off and when they are turned on.

The Lux meter app measured the total lux in wardobe area when switching on the light

The Lux meter app measured the total lux in wardobe area when switching off the light

The correct installation of the air conditioner is important, because installing the air conditioner in the wrong place can cause discomfort and also the air conditioner to not perform optimally. In the current air conditioner placement, the air conditioner is placed on top of the wardrobe, this placement causes dust in the cupboard to fly because it is exposed to wind gusts from the air conditioner and also placing the air conditioner near the mattress causes discomfort while sleeping.

When using the room, the occupants feel uncomfortable with the use of lights. The use of lights that will make the room bright if it is turned on at night. On the other hand, because the occupant's room does not have a window, the condition of the room becomes too dark if the lights are turned off.

The Lux meter app measured the total lux in the bedroom when switching on the light

The Lux meter app measured the total lux in the bedroom when switching off the light

Occupants are willing to move sockets for convenience and to avoid the danger of a short circuit.

The relocation of the socket next to the mattress is used to avoid a hazard in the event of an electrical short while in bed or to prevent electric shock if accidentally touching the socket while it is at a nearby outlet. The socket is moved near the table because of the electrical socket located on the lower wall. Placing the socket at the bottom of the wall makes the table unable to be brought closer to the wall.Changing the table placement close to the wall makes the area to enter the room wider.

When using the lights at night, residents feel uncomfortable if the lights are turned on at the same time while sleeping. Conversely, if both lights are turned off, the room becomes too dark, no light enters because there are no windows in the occupants' rooms. The use of a switch that can turn on both lights directly is not appropriate, so the occupants of the room want to replace the switch so that the lights are separated.

Comfort In the new room arrangement, there are several changes to the arrangement of lights and air conditioning. In this arrangement, the occupant does not want changes to the arrangement of furniture such as wardrobes, beds and tables.

When the occupants of the room want to turn on the lights, there are additional switches, so residents need to press both switches if they want a brighter room. Conversely, if at night, residents can only turn on one of the lights. The use of a bed site lamp can actually be an alternative for residents' problems, but due to limited space, the use of separate switches is the most suitable alternative.

When the occupants of the room want to open the closet and take clothes, the occupants feel that the closet area lacks lighting so it is less comfortable to choose clothes. it is necessary to install a light in the wardrobe area, by adding a cabinet LED light strip with a sensor, this light does not require a power outlet to turn it on, this LED cabinet uses type AA batteries, and is also equipped with sensors that are ready to help you if you want to take their clothes off.

A) “Stottaled”cabinet lighting strip w sensor, battery-operated white

Change of Air-Cond model to DAIKIN FTKG-T SERIES(R32) B) Daikin R32 inverter Wall-Mounted FTKG- T series Air Conditioner

C) Changing the lamp from a Slim Down Light to a Panel Light because the light emitted from the Panel light is more evenly distributed in all directions, while for the Slim Down Light it uses an inside light bulb so that the distribution of lights is more concentrated in one point.

D) Philips 72016 4.5W Cool white LED Desk Light

E) Simon Switch i7 701089-46 13A Universal Switch Socket Outlet Golden Champagne

F) Simon Switch 701022-46 i7 6A 2 Gang 2 Way Golden Champagne

Energy Calculation Area of the room

= 2,8 x 3,4 = 9,52m2

Heat load (BTU) multiply with 700

=9,52 x 700 = 6.664

Number of occupants,multiply with 500 = 500 Total Wattage from electrical equipment =LED Downlight +LED Desk light =(9Wx2)+5W =23W Estimated heat from electrical equipment, multiply with 4,25 =23Wx 4,25 =97,75 Total heat Calculation = 6.664+500+97,75 =7.261,75 Air conditioner sizing =7.261,75 :9800 =0,7409949 =1HP

A) Light Fixtures Item

Old Design Product Name Slim Downlight LPDL12RE02W 12W Illumination: warm white

Calculation Power = 12w=0.012kw Per day= 0,012 x 7 hours =0.084 Per Month=0.084x30 =2,52kwh 5 lights =2,52x5 =12,6 Bill Costing : Prorated Block(kwh)x Rate(RM) =12,6x0.037 =0,4662

B)Light Fixtures Item

Old Design Product Name PHILIP Meson LED Downlight 59449 Illumination: Warm White

Calculation Power = 9w=0.009kw Per day= 0,009 x 7 hours =0,063 Per Month=0.063x30 =1,89 kwh 5 lights =1,89x5 =9,45 Bill Costing : Prorated Block(kwh)x Rate(RM) =9,45x0.037 =0,34965

Net Savings per Month =0,4662- 0,34965 =0,11655

Item A-1

References

Sources 29, 31 33, 35, Jalan Bpu 6, Bandar Puchong Utama, 47100 Puchong, Selangor

A-2

A-G-3A, Kuchai Exchange, 43, Jalan Kuchai Maju 13, 58200 Kuala Lumpur, Malaysia.

A-3

2A, Jalan Cochrane, Maluri, 55100 Kuala Lumpur, Wilayah Persekutuan Kuala Lumpur

A-4

A-G-3A, Kuchai Exchange, 43, Jalan Kuchai Maju 13, 58200 Kuala Lumpur, Malaysia.

A-5

A-G-3A, Kuchai Exchange, 43, Jalan Kuchai Maju 13, 58200 Kuala Lumpur, Malaysia.

B-1

Unit G-G-5 & Unit G-G-6 Plaza Arkadia No. 3 Jalan Intisari Perdana, Desa Parkcity, 52200 Kuala Lumpur

CAD DRAWING (OLD DESIGN)

Perspective view

ASSIGN MENT IIA

W1 W6

W1

= 8.93 x 10.5 =93.765ft2

W2

=10.7 x10.5 =112.35ft2

W3

=14.8x10.5 =155.4ft2

W4

=10.4 x10.5 =109.2 ft2

W5 W2 W4

W4 WITH WINDOW =109.2- (9.35X7.35 =109.2-68,73 = 40.47 W5

=5.35 x10.5 =56.175 ft2

W3

W5 WITH DOOR

=56.175 –(8.2X4.6) =56.175-37.72 =18,45

W6

= 5.45 x 10.5

W6 WITH WINDOW

=57.225ft2 =57.225- (4.86 X4.93) =57.225- 23,96 = 33,265

W1

R value concrete

0.08

DAY 1 07.00 INTERIOR EXTERIOR DIFF TEMP

77.18 73.40 3.78 DAY 2

07.00 INTERIOR EXTERIOR DIFF TEMP

75.2 76.64 1.44 DAY 3

07.00 INTERIOR EXTERIOR DIFF TEMP

71.06 73.76 2.7 DAY 4

07.00 INTERIOR EXTERIOR DIFF TEMP

69.44 73.76 4,32

DAY 1 12.00 INTERIOR EXTERIOR DIFF TEMP DAY 2 12.00 INTERIOR EXTERIOR DIFF TEMP DAY 3 12.00 INTERIOR EXTERIOR DIFF TEMP DAY 4 12.00 INTERIOR EXTERIOR DIFF TEMP

DAY 1 83.84 84.74 0.9

17.00 INTERIOR EXTERIOR DIFF TEMP

78.8 80.24 1.44 DAY 2

82.76 82.40 0,36

17.00 INTERIOR EXTERIOR DIFF TEMP

84.56 84.38 0.18

17.00 INTERIOR EXTERIOR DIFF TEMP

78.08 80.60 2.52 DAY 3 81.68 82.22 0.54 DAY 4

85.28 86.36 1.08

17.00 INTERIOR EXTERIOR DIFF TEMP

72.14 82.76 10.62

07.00 12.00 17.00

07.00 12.00 17.00

07.00 12.00 17.00

07.00 12.00 17.00

HEAT TRANSFER CALCULATION DAY 1-W1 Q = 93.7 X 3.78/0.08 =4.427,3 Btu/hr Q = 93.77 X 0.9/0.08 = 1054,9 Btu/hr Q = 93.77 X 1.44/0.08 = 1.687,86 Btu/hr HEAT TRANSFER CALCULATION DAY 2-W1 Q = 93.77 X1.44/0.08 = 1.687,86 Btu/hr Q = 93.77 X0.36/0.08 = 421,96 Btu/hr Q = 93.77 X2.52/0.08 = 2.953,7 Btu/hr HEAT TRANSFER CALCULATION DAY 3-W1 Q = 93.77 X 2.7/0.08 = 3.164,7 Btu/hr Q = 93.77 X 0.18/0.08 = 210,9 Btu/hr Q = 93.77 X 0.54/0.08 = 632,94 Btu/hr HEAT TRANSFER CALCULATION DAY 4 -W1 Q = 93.77 X 4.32/0.08 = 5.063,5 Btu/hr Q = 93.77 X 1.08/0.08 =1.265,8 Btu/hr Q = 93.77 X 10.62/0.08 = 12.447,9 Btu/hr

W2

R value concrete

0.08

DAY 1 07.00 INTERIOR EXTERIOR DIFF TEMP

DAY 1 78.26 78,62 0.36

DAY 2 07.00 INTERIOR EXTERIOR DIFF TEMP

74.12 72.32 1,8 DAY 3

07.00 INTERIOR EXTERIOR DIFF TEMP

70.7 72.5 1.8 DAY 4

07.00 INTERIOR EXTERIOR DIFF TEMP

68.54 66.20 2.34

12.00 INTERIOR EXTERIOR DIFF TEMP DAY 2 12.00 INTERIOR EXTERIOR DIFF TEMP DAY 3 12.00 INTERIOR EXTERIOR DIFF TEMP DAY 4 12.00 INTERIOR EXTERIOR DIFF TEMP

DAY 1 84.02 83.84 0.18

17.00 INTERIOR EXTERIOR DIFF TEMP

82.40 82.94 0.54

17.00 INTERIOR EXTERIOR DIFF TEMP

78.44 80.24 1.8 DAY 2 78.08 80.24 2.16 DAY 3

84.56 84.38 0.18

17.00 INTERIOR EXTERIOR DIFF TEMP

81.86 80.96 0.9 DAY 4

85.10 85.10 0

17.00 INTERIOR EXTERIOR DIFF TEMP

72.32 83.12 10.8

07.00 12.00 17.00

07.00 12.00 17.00

07.00 12.00 17.00

07.00 12.00 17.00

HEAT TRANSFER CALCULATION DAY 1-W2 Q = 112.35 X 0.36/0.08 =505,58 Btu/hr Q = 112.35 X0.18/0.08 = 252,8 Btu/hr Q = 112.35X1.8/0.08 = 2.527,8 Btu/hr HEAT TRANSFER CALCULATION DAY 2-W2 Q = 112.35 X1.8/0.08 = 2.527,8 Btu/hr Q = 112.35 X0.54/0.08 = 758,36 Btu/hr Q = 112.35 X2.16/0.08 = 3.033,45 Btu/hr HEAT TRANSFER CALCULATION DAY 3-W2 Q = 112.35 X 1.8/0.08 = 2.527,8 Btu/hr Q = 112.35 X 0.18/0.08 = 252,78 Btu/hr Q = 112.35 X 0.9/0.08 = 1.263,9 Btu/hr HEAT TRANSFER CALCULATION DAY 4 -W2 Q = 112.35 X 2.34/0.08 = 3.286,2 Btu/hr Q = 112.35 X 0/0.08 =0 Q = 112.35 X 10.8/0.08 =15.167,25 Btr/hr

W3

R value concrete

0.08

DAY 1 07.00 INTERIOR EXTERIOR DIFF TEMP

DAY 1 77.90 73.94 3.96

DAY 2 07.00 INTERIOR EXTERIOR DIFF TEMP

74.48 76.28 1,8 DAY 3

07.00 INTERIOR EXTERIOR DIFF TEMP

70.34 73.58 3.24 DAY 4

07.00 INTERIOR EXTERIOR DIFF TEMP

68.72 73.58 4.86

12.00 INTERIOR EXTERIOR DIFF TEMP DAY 2 12.00 INTERIOR EXTERIOR DIFF TEMP DAY 3 12.00 INTERIOR EXTERIOR DIFF TEMP DAY 4 12.00 INTERIOR EXTERIOR DIFF TEMP

DAY 1 83.66 84.38 0.72

17.00 INTERIOR EXTERIOR DIFF TEMP

82.40 82.58 0.18

17.00 INTERIOR EXTERIOR DIFF TEMP

78.26 80.6 2.34 DAY 2 78.08 80.24 2.16 DAY 3

84.92 84.02 0.90

17.00 INTERIOR EXTERIOR DIFF TEMP

82.04 82.94 0.9 DAY 4

85.10 85.82 0.72

17.00 INTERIOR EXTERIOR DIFF TEMP

72.5 82.4 9.9

07.00 12.00 17.00

07.00 12.00 17.00

07.00 12.00 17.00

07.00 12.00 17.00

HEAT TRANSFER CALCULATION DAY 1-W3 Q = 155.4 X 3.96/0.08 = 7.692,3 Btu/hr Q = 155.4 X0.72 /0.08 =1398,6 Btu/hr Q = 155.4 X 2.34/0.08 = 4.545,45 Btu/hr HEAT TRANSFER CALCULATION DAY 2-W3 Q = 155.4 X1.8/0.08 = 3.496,5 Btu/hr Q = 155.4 X0.18 /0.08 = 349,65 Btu/hr Q = 155.4 X2.16/0.08 = 4.195,8 Btu/hr HEAT TRANSFER CALCULATION DAY 3-W3 Q = 155.4 X 3.24/0.08 = 6.293,7 Btu/hr Q = 155.4 X 0.9/0.08 = 1.748,25 Btu/hr Q = 155.4 X 0.9/0.08 = 1.748,25 Btu/hr HEAT TRANSFER CALCULATION DAY 4 -W3 Q = 155.4 X 4.86/0.08 =9.440,5 Btu/hr Q = 155.4 X 0.72/0.08 =1.398,6Btu/hr Q = 155.4 X 9.9/0.08 =19.230,75 Btu/hr

W4 R value concrete R value single pane

0.08 0.91

DAY 1 07.00 INTERIOR EXTERIOR DIFF TEMP

78.62 78.80 0.18 DAY 2

07.00 INTERIOR EXTERIOR DIFF TEMP

75.02 75.02 0 DAY 3

07.00 INTERIOR EXTERIOR DIFF TEMP

71.42 78.8 7.38 DAY 4

07.00 INTERIOR EXTERIOR DIFF TEMP

07.00 12.00 17.00

HEAT TRANSFER CALCULATION DAY 1-W4 Q = 40.47 X 0.18/(0.08+0.91) = 7,28/0.99 = 7,35 Btu/hr Q = 40.47 X0.18/(0.08+0.91) = 7,29/0.99 = 7,36 Btu/hr Q = 40.47 X 1.62/(0.08+0.91) = 65,5614 /0,99 = 66,22 Btu/hr

69.08 78.08 9

DAY 1 12.00 INTERIOR EXTERIOR DIFF TEMP DAY 2 12.00 INTERIOR EXTERIOR DIFF TEMP DAY 3 12.00 INTERIOR EXTERIOR DIFF TEMP DAY 4 12.00 INTERIOR EXTERIOR DIFF TEMP

D 83.84 84.02 0.18

17.00 INTERIOR EXTERIOR DIFF TEMP

D 82.94 84.20 1,26

17.00 INTERIOR EXTERIOR DIFF TEMP

84.38 87.44 3.06

17.00 INTERIOR EXTERIOR DIFF TEMP

D

D 85.28 93.20 7.92

17.00 INTERIOR EXTERIOR DIFF TEMP

07.00 12.00 17.00

07.00 12.00 17.00

07.00 12.00 17.00

HEAT TRANSFER CALCULATION DAY 2-W4 Q = 40.47X0/(0.08+0.91) =0 Q = 40.47X1.26/(0.08+0.91) = 50,99/0.99 =51,5 Btu/hr Q = 40.47 X15.48/(0.08+0.91) = 626,47 /0.99 =632,8 Btu/hr HEAT TRANSFER CALCULATION DAY 3-W4 Q = 40.47 X7.38/(0.08+0.91) = 298,6/0.99 =301,68 Btu/hr Q = 40.47x3.06/(0.08+0.91) = 123,8/0.99 =125,09 Btu/hr Q = 40.47 X1.8/(0.08+0.91) =72,84 /0.99 =73,58 Btu/hr HEAT TRANSFER CALCULATION DAY 4 -W4 Q = 40.47 X9/(0.08+0.91) = 364,2/0.99 =367,9 Btu/hr Q = 40.47 X7.92/(0.08+0.91) =320,52/0.99 =323,76 Btu/hr Q = 40.47 X12.96/(0.08+0.91) = 524,49/0.99 =529,78 Btu/hr

W5

R values Concrete R value metal door

0.08 2.2

DAY 1 07.00 INTERIOR EXTERIOR DIFF TEMP

77.90 78.80 0.9 DAY 2

07.00 INTERIOR EXTERIOR DIFF TEMP

75.02 75.02 0 DAY 3

07.00 INTERIOR EXTERIOR DIFF TEMP

71.6 78.8 7.2 DAY 4

07.00 INTERIOR EXTERIOR DIFF TEMP

07.00 12.00 17.00

69.62 78.08 8.46

DAY 1 12.00 INTERIOR EXTERIOR DIFF TEMP DAY 2 12.00 INTERIOR EXTERIOR DIFF TEMP DAY 3 12.00 INTERIOR EXTERIOR DIFF TEMP DAY 4 12.00 INTERIOR EXTERIOR DIFF TEMP

HEAT TRANSFER CALCULATION DAY 1-W5 Q= 18.45 X0.9/(0.08+ 2.2) =16,6/2,28 =7,28 Btu/hr Q=18.45 x 0.36/(0.08 +2.2) = 6,64/2.28 = 2.91 Btu/hr Q =18.45 X1.8/(0.08+2.2) = 33,21/2,28 =14,57 Btu/hr

DAY 1 84.38 84.02 0.36

17.00 INTERIOR EXTERIOR DIFF TEMP

78.8 77 1,8 DAY 2

83.48 84.20 0,72

17.00 INTERIOR EXTERIOR DIFF TEMP

84.92 87.44 2.52

17.00 INTERIOR EXTERIOR DIFF TEMP

77.54 93.20 15.66 DAY 3 82,94 84.02 1,08 DAY 4

85.64 93.20 7.56

17.00 INTERIOR EXTERIOR DIFF TEMP

72.68 86 13.32

07.00 12.00 17.00

07.00 12.00 17.00

07.00 12.00 17.00

HEAT TRANSFER CALCULATION DAY 2-W5 Q = 18.45X0/(0.08+2.2) =0 Q = 18.45 X0.72/(0.08+2.2) = 13,28 /2.28 =5,83 Btu/hr Q = 18.45 X 15.66/(0.08+2.2) =288,9 /2.28 =126,72 Btu/hr HEAT TRANSFER CALCULATION DAY 3-W5 Q=18.45 X7.2/(0.08+2.2) = 132,84/2,28 = 58,2 Btu/hr Q =18.45 X2.52/(0.08 +2.20) = 46,49/2,28 =20,39 Btu/hr Q = 18.45 X1.08/(0.08+2.2) = 19,92/2,28 =8,74 Btu/hr HEAT TRANSFER CALCULATION DAY 4 -W5 Q =18.45 X8.46/(0.08+2.2) = 156,08/2,28 =68,459 Btu/hr Q =18.45 X7.56/(0.08 +2.2) =139,48/2,28 = 61,17Btu/hr Q = 18.45 X13.32 /(0.08+2.2) =107,78 Btu/hr

W6

R value concrete R value single pane

0.08 0.91

DAY 1 07.00 INTERIOR EXTERIOR DIFF TEMP

77.90 78.80 0.9 DAY 2

07.00 INTERIOR EXTERIOR DIFF TEMP

75.02 75.02 0 DAY 3

07.00 INTERIOR EXTERIOR DIFF TEMP

71.6 78.8 7.2 DAY 4

07.00 INTERIOR EXTERIOR DIFF TEMP

07.00 12.00 17.00

69.62 78.08 8.46

DAY 1 12.00 INTERIOR EXTERIOR DIFF TEMP DAY 2 12.00 INTERIOR EXTERIOR DIFF TEMP DAY 3 12.00 INTERIOR EXTERIOR DIFF TEMP DAY 4 12.00 INTERIOR EXTERIOR DIFF TEMP

HEAT TRANSFER CALCULATION DAY 1-W6 Q=33.27X 0.9/(0.08+0.91) = 29,94/0,99 = 30,24 Btu/hr Q=33.27x 0.36/(0.08+0.91) = 11,97 /0,99= 12,09 Btu/hr Q =33.27 X 1.8/(0.08+0.91) = 59,89 /0.99 =60,49 Btu/hr

DAY 1 84.38 84.02 0.36

17.00 INTERIOR EXTERIOR DIFF TEMP

78.8 77 1,8 DAY 2

83.48 84.20 0,72

17.00 INTERIOR EXTERIOR DIFF TEMP

84.92 87.44 2.52

17.00 INTERIOR EXTERIOR DIFF TEMP

77.54 93.20 15.66 DAY 3 82,94 84.02 1,08 DAY 4

85.64 93.20 7.56

17.00 INTERIOR EXTERIOR DIFF TEMP

72.68 86 13.32

07.00 12.00 17.00

07.00 12.00 17.00

07.00 12.00 17.00

HEAT TRANSFER CALCULATION DAY 2-W6 Q =33.27 X0/(0.08+0.91) =0 Q = 33.27 X0.72/(0.08 +0.91) = 23,95/0.99 =24,19 Q = 33.27 X15.66/(0.08 +0.91) = 521/0.99 = 526,27 Btu/hr HEAT TRANSFER CALCULATION DAY 3-W6 Q = 33.27 X7.2/(0.08+0.91) = 239,5/0,99 = 241,96 Btu/hr Q = 33.27X 2.52/(0.08 +0.91) = 83,84 /0.99 =84,68 Btu/hr Q = 33.27 X1.08/(0.08 +0.91) = 35,93 /0,99 =36,29 Btu/hr HEAT TRANSFER CALCULATION DAY 4 -W6 Q =33.27 X8.46 /(0.08+0.91) =281,46/0.99 =284,3 Btu/hr Q =33.27 X7.56/(0.08+0.91) =251,52/0.99 =254,06Btu/hr Q = 33.27 X13.32 /(0.08 +0.91) =447,63 Btu/hr

ASSIGN MENT IIB

Heat transfer causes items to become the same temperature. Heat always moves from a warmer place to a cooler place, so cold items heat up to room temperature while warm items cool down to room temperature. The transfer of heat occurs consistently to make everything the same temperature. The primary reason system designers must limit or minimize heat loss is to maintain process control.In some homes, the shiny material is combined with insulation to create a reflective insulation system that doubles up on reducing heat transfer.

This prototype is inspired by the system to reduce heat transfer in a thermos flask.

Material needed :

Food container

Aluminum Foil

Bubble Wrap

Foam

Modeling Board

Aluminum food foam

Double Tape

Styrofoam

Function: Aluminum Foil

: effective insulating material because it doesn’t radiate heat out into the environment Aluminum foil Foam : thermal insulation,sound insulation Bubble wrap : air bubles trapped inside plastic sheets, it serve as a good insulator Foam : high thermal resistance Styrofoam : trap the air in small pocket,blocking the flow of heat energy

Layer Aluminum Foil :0.38 M X 0.48 M Aluminum food foam :0.19 M X 0.16 M

Bubble Wrap : 0,22M X 0,2 M Foam : 0,23M X 0,21 M Bubble Wrap : 0,24M X 0,22 M

Aluminum food foam : 0,25M X 0,23 M Styrofoam : 0,52M X 0,38 M

Modeling board : 0,52M X 0,38 M

Work in progress : Wrap the food container with aluminum foil

For the second layer, wrap the container with aluminum foam

For the third layer, wrap the container with bubble wrap

For the fourth layer, wrap the container with foam

For the fifth layer, wrap the container with bubble wrap

For the sixth layer, wrap the container with foam

wrap the container with styrofoam

wrap the container with modeling board Experiments on the prototype were carried out using hot water for

3 minutes -In the use of warm water, which is placed on the prototype and left for 3 minutes, can prevent heat transfer, the water remains warm

Cost Food Container : Cost : RM 1.96

Aluminum food foam Cost : RM 2.95 (0.4M X 1.0M=0.424 M2 ) Price /area = 0.424/2.95 = 0,143 RM Cost : 0,44 M x0,39 M = 0.1716m2 0,1716 x 0.143 = 0.0245 RM

Bubble Wrap Cost : RM 1.9 (1M X 1M =1M2) Price /area = 1/1.90 = 0.53 RM Cost : 0,47 M x 0,4 M = 0.18 0,18 X 0.53 = 0.0996 RM

Foam

Cost : RM 5.50 (0.5M X 0.7M =0.35M2) Price /area = 0.35/5.5 = 0.064 RM Cost : 0,23M X 0,21 M = 0.0483 0,0483 X 0.064 = 0.003 RM Aluminum Foil Cost : RM 5.5 (0.25M X 18M =4.5M2)

Price /area = 4,5/5.50 = 0.81 RM Cost : = 0.38 M X 0.48 M =0.1824 0.1824 x0.81=0,0328 RM

Modeling Board Cost : RM 6,8 (0.5M X 0.75M =0.375M2) Price /area = 0.375/6.8 = 0,1 RM Cost : = 0,52M X 0,38 M =0.1976 0.1976 X 0.1=0,01976 RM

Styrofoam Cost : RM 6,8 (0.6M X 1.1M =0.66M2) Price /area = 0.66/6.8 = 0.1 RM Cost : = 0,52M X 0,38 M =0.1976 0.1976 X 0.1=0,01976 RM

Double tape Cost : RM 3 (0,024 M x 8 = 0.192 ) Price /area = 0.192/3 = 0.064 RM Cost : = 0,024 x 2 = 0.048 0.048 X 0.064=0,003

Total Cost : RM 2,63876 = RM 2,7 Design Value = 5 - 2,7 = RM 2.3 Market Value average = 45+25+30+30+20+25+40+35+30+25+15+40+20+40+35+30+30 17 =515 17 = 30,29 Design Value = Market value average - Total cost = 30,29 - 2,7

= 27,59

Reflection During the learning period with the theme of products that can reduce heat transfer, presenting a prototype task can help us better understand and learn about what material are and are not good at reducing heat transfer.In this prototype assignment, I used several materials that can reduce heat transfer, such as styrofoam, aluminum foil, foil foam, modeling board as a substitute for gypsum material, and bubble wrap. All the materials I use have their function and use in reducing heat transfer. The use of aluminum foil which functions as an effective insulating material because it does not emit heat to the environment, use foil foam because foil foam is thermal insulation,sound insulation. It was interesting to see the various models brought by all the class participants. From there, I can see various kinds of products resulting from my classmate's thoughts on the theme of products that can reduce heat transfer.

Making prototypes in addition to making students understand more can also train the creativity of each student to think of an idea from the themes and lecture