<!--:en-->Cool and Economical with Lightweight Concrete<!--:--><!--:id-->Adem dan Hemat dengan Beton Ringan<!--:-->

Cool and Economical with Lightweight ConcreteAdem dan Hemat dengan Beton Ringan

By: Ir. Elisa Haryonugroho1

Convenience inside a house/building is very important. Convenience enables occupants to do their activities well and more productively. Determining factors of convenience are temperature, humidity, air circulation. Convenience can be attained by at least 2 things: good architectural planning and choosing the right materials. Convenience issue becomes crucial when the outer wall is facing westward. For several cases, air conditioning can be achieved by controlling passive air, for example by making window openings. However, for buildings in urban areas, it’s often difficult to do this. One of the obstacles is the polluted air and a dense surrounding.

Other possible solution to address air conditioning problem is using lightweight concrete wall. To see how much efficiency we can benefit from using lightweight concrete wall, below is an explanation of electricity cost saving if we use active air conditioning.

For a comparative illustration, we are going to use a room of 3m x 4m with 3m height to the ceiling. The first room use lightweight concrete roof plate and lightweight concrete block walls (Autoclaved Aerated Concrete) with PM-200 plaster/render (Picture 1). The second room of the same size uses conventional concrete plate and terracotta brick wall with cemment-sand palster. The calculation will measure the heat that passes through the wall material and roof plate, where the heat energy from the outside will be reduced by the material. This makes the room inside to become cooler (Picture 2). This reduction of heat depends on the material’s capacity to resist heat. The process is similar to that of electricity energy, that will decrease after passing a lamp or other electrical equipments (Picture 3).

ADEM dan HEMAT DENGAN DINDING HEBEL

ADEM dan HEMAT DENGAN DINDING HEBEL

ADEM dan HEMAT DENGAN DINDING HEBEL

Lightweight concrete walls

Terracotta brick walls

LIGHTWEIGHT CONCRETE’s density  (ρ) = 575 kg/m3
Block’s heat conductivity (λ) = 0,1575 W/(m.K)
λ of render = 0,35 W/(m.K)
Thermal Resistance (R) = d/λ
R of lightweight concrete = 0,125/0,1575=0,79

Terracotta Brick’s Density (ρ) = 1.500 kg/m3
Brick’s heat conductivity (λ) = 0,5 W/(m.K)
λ of plaster = 1,4 W/(m.K)
Thermal Resistance ( R) = d/λ
R of terracotta brick =0,09/0,5=0,18

The higher its ‘Thermal Resistance’, the better its heat insulation capacity.

LIGHTWEIGHT CONCRETE
Block

Terracotta Brick

R of inner render = d/λ
R of inner render  = 0,01/0,35=0,0286
R of outer render = 0,01/0,35=0,0286
Total R = 0,79+0,0286+0,0286 = 0,847
R of plaster = d/λ
R of inner plaster = 0,025/1,4=0,018
R of outer plaster =  0,025/1,4=0,018
Total R = 0,18+0,018+0,018 = 0,216

Inner air layer Thermal Resistance (Rsi) dan outer air layer Thermal Resistance (Rse)

The lower the air film’s ‘thermal transmittance’ on the surface of walls, the better its heat insulation capacity.

Thermal transmittance (U)= 1/(Rsi+R+Rse)
U beton ringan = 1/(0,13+0,847+0,04) = 0,983
Thermal transmittance (U)= 1/(Rsi+R+Rse)
U of lightweight concrete = 1/(0,13+0,847+0,04) = 0,983
The lower its ‘thermal transmittance’, the better its capcaity of heat insulation
LIGHTWEIGHT CONCRETE Roof Panel Concrete Plate

LIGHTWEIGHT CONCRETE Roof Panel

Concrete Plate

R screed PM-600 = d/λ
R = 0,02/0,35 = 0,057
R plester = d/λ
R = 0,02/1,4 = 0,014
R of LIGHTWEIGHT CONCRETE Panel = d/λ
R = 0,125/0,8 = 0,69
Total R = 0,057+0,69 = 0,747
R of concrete plate  = d/λ
R = 0,125/2,1 = 0,06
Total R = 0,014+0,06 = 0,074
Thermal transmittance (U)= 1/(Rsi+R+Rse)
U of lightweight concrete panel = 1/(0,13+0,747+
Thermal transmittance (U)= 1/(Rsi+R+Rse)
U of concrete roof = 1/(0,13+0,074+0,08) = 3,52

Energy Passing Through the LIGHTWEIGHT CONCRETE Wall and LIGHTWEIGHT CONCRETE Floor Panel

Energy Passing through the Terracotta Brick Wall and Concrete Floor (Q)

Wall size (A of wall) = 39,3 m2
Roof size (A of roof) = 12 m2
Outside average temperature 33° C , inside temperature 28° C, the temperature difference  (at) = 5° C
Using assumption of room occupancy in 10 working hours

Q of lightweight concrete = U x A x at x T
Q of lightweight concrete = 0,983×39,3x5x10 = 1.931
Q of lightweight concrete = 1,9 kWh
Q of terracotta brick = U x A x at x T
Q of terracotta brick =  2.59×39,3x5x10 = 5.089,35 
Q of terracotta brick = 5,1 kWh
Q of lightweight concrete panel = U x A x at x T
Q of lightweight concrete panel = 1,04x12x15x10 = 1,04x12x15x10 = 1.872 Wh
Q of lightweight concrete panel = 1,87 kWh
Q of terracotta brick = U x A x at x T
Q of concrete roof = 3,52 x12x15x10 = 6.336 Wh
Q of concrete roof =  6,3 kWh
Total Q = 1,9 + 1,87 = 3,77 kWh
Usage in 30 days
Q = 3,77 kWh x 30 = 113 kWh
Total Q = 5,1 + 6,3 = 11,4 kWh
Usage in 30 days
Q = 11,4 kWh x 30 = 342 kWh

Using the assumption of household usage with the category  R-1 2200 VA

Block I     0 –20 kWh = 390,- x 20  =    7.800,-
Block II   21-60 kWh = 445,- x 60   =   26.700,-
Block III  61 kWh ≤   = 495,- x 33   =   16.335,-
Total cost for 1 month = Rp 50.835,-
Block I     0 –20 kWh = 390,- x 20  =    7.800,-
Block II   21-60 kWh = 445,- x 60   =   26.700,-
Block III  61 kWh ≤   = 495,- x 262 = 129.690,-
Total cost for 1 month = Rp 164.190,-

Cost saving attained by using LIGHTWEIGHT CONCRETE material = Rp 164.190,– Rp 50.835,- = Rp 113.335,-

By using LIGHTWEIGHT CONCRETE material, a room of 3m x 4m can save electricty cost up to Rp 113.335,- per month. That’s worth saving, isn’t it?

1The writer is an architect, member of IAI and HDIIOleh: Ir. Elisa Haryonugroho1

rumah-adem

Kenyamanan di dalam ruang sangat penting. Karena dengan kenyamanan, penghuni dapat melakukan berbagai kegiatannya dengan baik dan lebih produktif. Faktor-faktor yang mempengaruhi kenyamanan adalah suhu, kelembaban, pergerakan udara. Sedangkan kenyamanan dapat diciptakan melalui setidaknya 2 hal yaitu perencanaan arsitektur yang baik dan pemilihan bahan yang tepat. Persoalan kenyamanan menjadi sangat krusial manakala dinding luar menghadap ke arah Barat. Untuk beberapa kasus pengkondisian udara dengan melakukan kontrol udara pasif, misalnya dengan mengadakan pembukaan beberapa jendela sehingga terjadi cross ventilation, masih dimungkinkan. Tetapi untuk bangunan-bangunan di daerah urban seringkali sulit. Kesulitan membuat pembukaan dinding untuk jendela ini disebabkan antara lain polusi udara yang tinggi dan kondisi lingkungan bangunan yang padat.

Solusi lain yang dapat menjawab persoalan pengkondisian udara adalah penggunaan dinding beton ringan. Nah, untuk mengetahui seberapa besar efisiensi yang diperoleh berikut ini dipaparkan penghematan listrik yang didapat seandainya menggunakan pengkondisian udara aktif.

Sebagai perbandingan ruang berukuran 3m x 4m dengan tinggi plafon 3m ruang pertama menggunakan plat atap beton ringan BETON RINGAN dan dinding blok beton ringan (Autoclaved Aerated Concrete) BETON RINGAN dengan plesteran/render PM- 200. (Gambar 1). Sedang ruang kedua yang sama ukurannya menggunakan plat beton konvensional dan dinding batu bata dengan plesteran semen-pasir. Pengukuran dilakukan terhadap panas yang melalui material dinding dan plat atap. Dimana energi panas dari luar akan ditahan oleh material, sehingga ruang dalam menjadi berkurang panasnya. (Gambar 2). Berkurangnya panas ini, tergantung dari kemampuan material menagahan panas. Proses ini sama halnya dengan energi listrik yang melalui tahan (R ) yang dapat berupa lampu atau peralatan listrik lainnya, maka setelah melalui tahanan tersebut daya listrik akan berkurang. (Gambar 3)

ADEM dan HEMAT DENGAN DINDING HEBEL

ADEM dan HEMAT DENGAN DINDING HEBEL

ADEM dan HEMAT DENGAN DINDING HEBEL

Dinding Beton Ringan
Dinding Bata
Berat jenis blok BETON RINGAN (ρ) = 575 kg/m³
Konduktifitas panas blok (λ) = 0,1575 W/(m.K)
λ render = 0,35 W/(m.K)
Thermal Resistance (R) = d/λ
R beton ringan= 0,125/0,1575=0,79
Berat jenis bata (ρ) = 1.500 kg/m³
Konduktifitas panas bata (λ) = 0,5 W/(m.K)
λ plester = 1,4 W/(m.K)
Thermal Resistance ( R) = d/λ
R bata=0,09/0,5=0,18
Semakin tinggi nilai ‘Thermal Resistance’, semakin baik kemampuan insulasi panas.

Blok
BETON RINGAN

Batu bata

R render dalam = d/λ
R render dalam = 0,01/0,35=0,0286
R render luar = 0,01/0,35=0,0286
R total = 0,79+0,0286+0,0286 = 0,847
R plesteran = d/λ
R plesteran dalam = 0,025/1,4=0,018
R plesteran luar = 0,025/1,4=0,018
R total = 0,18+0,018+0,018 = 0,216

Thermal Resistance lapisan udara dalam (Rsi) dan lapisan udara luar (Rse)

Semakin rendah nilai ‘thermal transmittance’ film udara di permukaan dinding,
semakin baik kemampuan insulasi panas.
Thermal transmittance (U)= 1/(Rsi+R+Rse)
U beton ringan = 1/(0,13+0,847+0,04) = 0,983
Thermal transmittance (U)= 1/(Rsi+R+Rse)
U bata = 1/(0,13+0,216+0,04) = 2,59
Semakin rendah nilai ‘thermal transmittance’, semakin baik kemampuan insulasi panas.
Panel Atap BETON RINGAN Plat Beton

Panel
BETON RINGAN

Plat
Beton

R screed PM-600 = d/λ
R = 0,02/0,35 = 0,057
R plester = d/λ
R = 0,02/1,4 = 0,014
R panel BETON RINGAN = d/λ
R = 0,125/0,8 = 0,69
R total = 0,057+0,69 = 0,747
R plat beton = d/λ
R = 0,125/2,1 = 0,06
R total = 0,014+0,06 = 0,074
Thermal transmittance (U)= 1/(Rsi+R+Rse)
U panel beton ringan = 1/(0,13+0,747+0,08) = 1,04
Thermal transmittance (U)= 1/(Rsi+R+Rse)
U atap beton = 1/(0,13+0,074+0,08) = 3,52
Energi yang Mengalir Melalui Dinding BETON RINGAN dan Panel Lantai BETON RINGAN (Q) Energi yang Mengalir Melalui Dinding Bata dan Plat Lantai Beton (Q)
Luas dinding (A dinding) = 39,3 m²
Luas atap (A atap) = 12 m²
Rata-rata suhu luar 33° C , suhu di dalam ruang ditentukan 28° C, jadi selisih suhu (ªt) = 5° C
Asumsi pemakaian ruang pada jam kerja 10 jam
Q beton ringan = U x A x ªt x T
Q beton ringan = 0,983×39,3x5x10 = 1.931,6 Wh
Q beton ringan = 1,9 kWh
Q bata = U x A x ªt x T
Q bata = 2.59×39,3x5x10 = 5.089,35 Wh
Q bata = 5,1 kWh
Q panel beton ringan = U x A x ªt x T
Q panel beton ringan = 1,04x12x15x10 = 1.872 Wh
Q panel beton ringan = 1,87 kWh
Q atap beton = U x A x ªt x T
Q atap beton = 3,52 x12x15x10 = 6.336 Wh
Q atap beton = 6,3 kWh
Q total = 1,9 + 1,87 = 3,77 kWh
Pemakaian selama 30 hari
Q = 3,77 kWh x 30 = 113 kWh
Q total = 5,1 + 6,3 = 11,4 kWh
Pemakaian selama 30 hari
Q = 11,4 kWh x 30 = 342 kWh
Asumsi pemakaian pada rumah tangga dengan kategori R-1 2200 VA
Blok I 0 –20 kWh = 390,- x 20 = 7.800,-
Blok II 21-60 kWh = 445,- x 60 = 26.700,-
Blok III 61 kWh ≤ = 495,- x 33 = 16.335,-
Biaya total untuk 1 bulan = Rp 50.835,-
Blok I 0 –20 kWh = 390,- x 20 = 7.800,-
Blok II 21-60 kWh = 445,- x 60 = 26.700,-
Blok III 61 kWh ≤ = 495,- x 262 = 129.690,-
Biaya total untuk 1 bulan = Rp 164.190,-
Penghematan listrik dengan bahan beton ringan BETON RINGAN = Rp 164.190,– Rp 50.835,- =
Rp 113.335,-

Dengan menggunakan bahan beton ringan BETON RINGAN, maka ruang kerja ukuran 3mx4m dapat menghemat pemakaian listrik sebesar Rp 113.335,- per bulan. Lumayan bukan ?!

1Penulis adalah arsitek anggota IAI dan HDII

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