Smooth/Corrugated/Stucco Embossed Aluminium Coil

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Loading Port:
Shanghai
Payment Terms:
TT OR LC
Min Order Qty:
2.5
Supply Capability:
5000 m.t./month

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Item specifice:

Grade: 3000 Series Surface Treatment: Embossed,Mill Finish Shape: Roll
Temper: O-H112 Application: Decorations

Product Description:

Pipeelines Covered Roofing Aluminium Coil


Product Information

1.  Alloy: 1100, 1145, 1050, 1060, 1070, 3003, 3013, 3005, 3A21,4343,4045, 5052, 5082, 5083, 5086,6061,8011 etc.

2.  Temper: H12.H14.H16.H18.H22.H24.H26.H32.H34.H36.H38.O etc.

3.  Thickness: 0.2mm -- 7mm

4.  Width:   under2550mm

Remarks: Specific requirement of alloy, temper or specification can be discussed at your request.

Application: 

Building, curtain wall, ceiling, panels, transformers, food packaging, air conditioning, condenser, air filter, refrigerators, washing machines, solar energy, automobile manufacturing, ship manufacturing, machinery manufacturing, electric equipment such as cosmetic packaging, machinery manufacturing industry, can also be used in power plants, chemical anti-corrosion insulation in petrochemical industry, etc.

Characteristics:

Resistant to weather corrosion ,durability, formability, self Cleansing


Packaging & Delivery

Packaging detail: Sea Worthy Wooden pallet

Delivery detail: About 25 days

Smooth/Corrugated/Stucco Embossed Aluminium Coil


Company Profile

CNBM International Corporation, China National Building Materials (Group) Corporation, is one of the largest companies in China building material & equipment industry, with 42,800 employees and sales in 2005 of US Dollar 4.395 billion. In 2006, China National Building Material Company Limited was listed on Hong Kong Stock Market with the stock code as 3323. 
Aluminium Products have been our featured products. We have specialized in aluminium products for about a decade and have sold our good quality aluminium products to the worldwide.  

Smooth/Corrugated/Stucco Embossed Aluminium Coil


CNBM World Wide

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Smooth/Corrugated/Stucco Embossed Aluminium Coil

Smooth/Corrugated/Stucco Embossed Aluminium Coil

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FAQ

Q: Do you provide free samples?

A: Yes, free samples will be sent to you on freight at destination.

Q: Can I get your latest products catalogue?

A: Yes, it will be sent to you in no time.

Q: What is the MOQ?

A: 2.5 tons

Q: What are your payment terms?

A: We accept L/C, T/T.

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Q:can you weld metal to aluminum?
i don't see why not becasue volvo uses aluminum exhaust systems.
Q:Removing electrons from Aluminum?
One approach to this can use the density of aluminum to find the mass of the1 cm³ block. The density of aluminum is 2.70-g/cm³ so your cube has a mas of 2.70-g. The number of aluminum atoms in this block is: 2.7-g Al x (1 mol Al / 27.0-g Al) x (6.023 X 10²³ atoms Al / 1 mol Al) = 6.023 x 10²² atoms Al. Each Al, as you pointed out, contains 13 electrons so we have 6.023 x 10²² atoms Al x 13 electron/atom = 7.83 x 10²³ electrons. 1 x 10¹² pC = 1 C and 1 C = 6.24 x 10¹⁸ electrons=== 1.5 pC x (1C / 1 x 10¹² pC) x ( 6.24 x 10¹⁸ electrons / 1 C) = 9.36 x 10⁶ electrons a.) % removed = 9.36 x 10⁶ / 7.83 x 10²³ x 100% = 1.2 x 10⁻¹⁵ % b.) Each electron has a mass of 9.11 x 10⁻²⁸-g , so the total mass removed =9.11 x 10⁻²⁸-g/elec x 9.36 x 10⁶ electrons = 8.53 x 10⁻¹⁴-g. Thus the % decrease is (8.53 x 10⁻¹⁴-g/ 2.7-g) x 100% = 3.16 x 10⁻¹² %
Q:What can you put on aluminum to make it permanently black?
Dad is on the right track....maybe. There are blackeners for aluminum you can get from gunshops. Used to touch up scratched or worn black anodized finish on aluminum gun parts. Never used it in large areas. Best regards
Q:will aluminum paste work the same as aluminum powder in the mixture for thermite?
The aluminum paste may or may not work. It depends on what the solvent is that the aluminum is dissolved in. Why not get a block of aluminum and a grinder, and make your own aluminum dust?
Q:Can I bake aluminum foil inside of clay?
Assuming you mean polymer clay - aluminum foil is fine and is commonly used. In fact you almost have to use it to have uniform thickness because solid will crack from shrinking during baking. Fired ceramic clay, nope - aluminum foil melts much too low a temp.
Q:Heat, vaporization, and melting points of aluminum.?
Calculate the quantity of energy required to heat 1.58 mol of aluminum from 33°C to its normal melting point in KJ- 1.58 mol x 26.98 g/mol x 0.902 J/g°C x (658 - 33)°C x 1 kJ/1000 J = 24.03 kJ Calculate the quantity of energy required to melt 1.02 mol of aluminum at 658°C In KJ- 1.02 mol x 26.98 g/mol x 3.95 kJ/g = 108.7 kJ Calculate the amount of energy required to vaporize 1.02 mol of aluminum at 2467°C In KJ- 1.02 mol x 26.98 g/mol x 10.52 kJ/g = 289.5 kJ
Q:How does aluminum ingot be processed to aluminum coil (aluminum sheet)?
First it should be made into blank and then put on the rolling mill.
Q:Can you get aluminum poisoning?
Aluminium can accumulate in the brain possibly causing alzheimers. If you inhaled it it would poison you and damage your lungs, if you got it in your body it would damage your skin, possibly cause an infection and the cut might not heal. Aluminium is a solid so how could you accidently get it in your body unless you have been melting it etc? If it wasn't solid it would burn you if you got it on your skin. You can only inhale vapours not a metal. If you ate it then it would probably damage or clog up your insides otherwise just pass out in a stool, depends how much you ate. People eat from aluminium and touch it every day so it is not dangerous to be in contact with.
Q:Food containing aluminium?
complex aspect. research using the search engines. it can assist!
Q:aluminum bar resistance in physics?
Resistance equals resistivity times length divided by cross-sectional area. R = ρ∙l/A The resistivities at 20°C are aluminum ρ = 2.82×10⁻⁸Ωm copper ρ = 1.72×10⁻⁸Ωm So the aluminum bar with rectangular cross section has a resistance of R = ρ∙l/(a∙b) = 2.82×10⁻⁸Ωm ∙ 3.8m / (0.01m ∙ 0.05m) = 2.1432×10⁻⁴Ω The resistance of copper wire with circular cross section is given by R = ρ∙l/(π∙d²/4) = 4∙ρ∙l/(π∙d²) Hence a wire of same resistance as the aluminum bar has a length of l = R∙π∙d² / (4∙ρ) = 2.1432×10⁻⁴Ω ∙ π ∙ (0.0015m)² / (4∙1.72×10⁻⁸Ωm) = 0.0220m = 2.2cm

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