1100 1050 1060 1235 Pure Aluminum Coil for PP Cap

1100 1050 1060 1235 Pure Aluminum Coil for PP Cap

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Loading Port:: Shanghai

Payment Terms:: TT OR LC

Min Order Qty:: 5 m.t.

Supply Capability:: 5000 m.t./month

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1100 1050 1060 1235 Pure Aluminum Coil for PP Cap

Packaging & Delivery

Product Specification
Grade	1000 Series: 1050 1060 1100 3000 Series: 3003 3004 3105 5000 Series: 5052 5605,5083 6000 Series: 6061 6063 8000 Series: 8011 8021 8079
Thickness	0.20-8.00mm
Width	2400mm max.
Grade	1000,3000,5000,6000,8000series
Coil ID	75mm, 150mm, 200mm, 300mm, 400mm, 508mm or negotiable
Coil weight	1000-5000kgs
Coil OD	1700mm max.
Temper	O, H12, H14, H16, H111, H22 ,H24, H26, H28,T4, T6,etc.

Temper
F	processing state
H	strain hardening state
O	annealing
T	heat treatment state
H112	pure state of strain hardening, adjustment has been made to degree of strain hardening and annealing
T4	solid solution treatment and natural efficiency to achieve sufficient stability condition
T5	artificial aging condition of the reentry after high temperature thermal cooling
T6	artificial aging state after solid solution treatment

Chemical Composition
Grade	Si	Fe	Cu	Mn	Mg	Cr	Ni	Zn	Al
1050	0.25	0.4	0.05	0.05	0.05	-	-	0.05	99.5
1060	0.25	0.35	0.05	0.03	0.03	-	-	0.05	99.6
1070	0.2	0.25	0.04	0.03	0.03	-	-	0.04	99.7
1100	Si+Fe:0.95		0.05-0.2	0.05	-	-	0.1	-	99
1200	Si+Fe:1.00		0.05	0.05	-	-	0.1	0.05	99
1235	Si+Fe:0.65		0.05	0.05	0.05	-	0.1	0.06	99.35
3003	0.6	0.7	0.05-0.2	1.0-1.5	-	-	-	0.1	remains
3004	0.3	0.7	0.25	1.0-1.5	0.8-1.3	-	-	0.25	remains
3005	0.6	0.7	0.25	1.0-1.5	0.20-0.6	0.1	-	0.25	remains
3105	0.6	0.7	0.3	0.30-0.8	0.20-0.8	0.2	-	0.4	remains
3A21	0.6	0.7	0.2	1.0-1.6	0.05	-	-	0.1	remains
5005	0.3	0.7	0.2	0.2	0.50-1.1	0.1	-	0.25	remains
5052	0.25	0.4	0.1	0.1	2.2-2.8	0.15-0.35	-	0.1	remains
5083	0.4	0.4	0.1	0.40-1.0	4.0-4.9	0.05-0.25	-	0.25	remains
5154	0.25	0.4	0.1	0.1	3.1-3.9	0.15-0.35	-	0.2	remains
5182	0.2	0.35	0.15	0.20-0.50	4.0-5.0	0.1	-	0.25	remains
5251	0.4	0.5	0.15	0.1-0.5	1.7-2.4	0.15	-	0.15	remains
5754	0.4	0.4	0.1	0.5	2.6-3.6	0.3	-	0.2	remains
6061	0.40-0.8	0.7	0.15-0.40	0.15	0.8-1.2	0.04-0.35	-	0.25	remains
6063	0.20-0.6	0.35	0.1	0.1	0.45-0.9	0.1	-	0.1	remains
6082	0.7-1.3	0.5	0.1	0.40-1.0	0.6-1.2	0.25	-	0.2	remains
6A02	0.50-1.2	0.5	0.20-0.6	Or Cr0.15-0.35	0.45-0.9	-	-	0.2	remains
8011	0.50-0.9	0.6-1.0	0.1	0.2	0.05	0.05	-	0.1	remains

1100 1050 1060 1235 Pure Aluminum Coil for PP Cap

Q:Can aluminum coils be used in automotive manufacturing?: Yes, aluminum coils can be used in automotive manufacturing. Aluminum is a lightweight and versatile material that offers various advantages in the automotive industry. It is commonly used to make parts like body panels, engine blocks, wheels, and heat exchangers. The use of aluminum coils in automotive manufacturing allows for improved fuel efficiency, as the reduced weight of the vehicle contributes to lower energy consumption. Additionally, aluminum has excellent corrosion resistance, making it suitable for use in various weather conditions. Overall, the use of aluminum coils in automotive manufacturing provides numerous benefits, making it a popular choice for many car manufacturers.

Q:show step by step: First write out the formula reaction: 4Al + 3O2 --- 2Al2O3 Next, find the number of moles of oxygen gas needed to oxidize 3.42 g of aluminium completely. From the cchemical equation above, 3 mol of oxygen is used to oxidize completely 4 mol of aluminium. Calculate the no. of moles of aluminium used: no. of moles of aluminium used = 3.42 g / molar mass of Al = 3.42 g / 26.98 g mol^-1 = 0.127 mol Use the mole ratio concept to find the no. of moles of oxygen required: no. of moles of oxygen required : no. of moles of aluminium used = 3 : 4 no. of moles of oxygen required / 0.127 = 3 / 4 no. of moles of oxygen required = (3 / 4)(0.127) = 0.095 mol Next, use the gas equation PV = nRT [P: Pressure of gas in Pa; V: Volume of gas in m^3; n: no. of moles; R, gas constant, 8.314 J mol^-1 K^-1; T, Temperature of gas in K] -Convert 0.975 atm to Pa: 0.975 atm = (0.975 x 101325) Pa = 98791.875 Pa -Convert 25 oC to K: 25 oC = (25 + 273) K = 298 K -Substitute into the equation PV = nRT and solve it: (98791.875)(V) = (0.095)(8.31)(298) (98791.875)(V) = 235.2561 V = 235.2561 / 98791.875 = 0.002381 m^3 = (0.002381 x 10^3) L = 2.381 L

Q:Are aluminum coils compatible with other materials?: Indeed, other materials can be used in tandem with aluminum coils. Aluminum, being a versatile and extensively utilized substance, readily blends with various materials for different purposes. Take, for instance, the utilization of aluminum coils alongside copper tubing in air conditioning and refrigeration systems to amplify heat transfer efficiency. Furthermore, aluminum coils can be coated or layered with polymers or paints to furnish supplementary safeguard against corrosion, enhance visual appeal, or elevate specific performance attributes. Consequently, aluminum coils exhibit compatibility with a multitude of materials, rendering them apt for an array of applications within industries like construction, automotive, aerospace, and beyond.

Q:How are aluminum coils different from steel coils?: Aluminum coils differ from steel coils in terms of their material composition, weight, corrosion resistance, and conductivity. Aluminum coils are made of aluminum, while steel coils are made of steel. Aluminum is a lighter metal compared to steel, making aluminum coils lighter in weight. Moreover, aluminum coils have higher corrosion resistance due to the formation of a protective oxide layer on the surface. In terms of conductivity, aluminum coils have higher electrical conductivity compared to steel coils.

Q:How are aluminum coils tested for surface hardness?: Aluminum coils are typically tested for surface hardness through a variety of methods, including the use of a hardness tester or durometer. These instruments measure the resistance of the coil's surface to indentation or penetration by applying a specific amount of force. The results are then compared to a standardized scale to determine the surface hardness of the aluminum coil.

Q:How are aluminum coils protected against transportation damage?: The prevention of transportation damage to aluminum coils is ensured by manufacturers and transportation companies through various measures. Primary among these measures is the packaging and storage of the coils in robust crates or pallets, which act as a barrier against external impact and guarantee stability during transit. These crates, which are typically constructed from wood or metal, are specifically designed to withstand the challenges of transportation. To provide further protection against damage, the coils are often enveloped in protective materials like plastic or stretch film. This supplementary layer of packaging serves to prevent scratches, abrasions, and moisture damage that may occur during handling and transportation. In addition, the coils are frequently secured within the crates or pallets using straps or bands, thereby averting any potential movement or collision with other objects. In order to create a cushioning effect within the packaging, manufacturers frequently employ padding materials such as foam or cardboard inserts. These inserts absorb shock and vibration during transit, thereby reducing the likelihood of damage caused by impacts or sudden movements. Transportation companies also play a critical role in ensuring the secure transportation of aluminum coils. They utilize specialized handling equipment, such as forklifts or cranes, to minimize the risk of mishandling and dropping the coils. Furthermore, they meticulously load and secure the coils within shipping containers or trucks, ensuring proper bracing and immobilization to prevent shifting or rolling during transportation. In conclusion, the protection of aluminum coils from transportation damage is achieved through the utilization of robust packaging, protective wraps, padding materials, and appropriate handling techniques. These measures aim to minimize the risk of scratches, dents, and other forms of damage caused by impacts, vibrations, or mishandling during transit.

Q:Can aluminum coils be used in marine environments?: Marine environments can indeed utilize aluminum coils. Aluminum is a favored option for marine applications because of its exceptional resistance to corrosion. When exposed to air, it develops a protective oxide layer on its surface, which aids in halting further corrosion. In marine settings, such as those with saltwater or high humidity, this oxide layer proves particularly effective. Moreover, aluminum's lightweight nature facilitates easier handling and installation in marine applications. Additionally, it possesses remarkable durability and can withstand challenging conditions, including exposure to UV rays and extreme temperatures. Consequently, aluminum coils find common usage in diverse marine applications, including boat hulls, shipbuilding, offshore structures, and other marine equipment.

Q:What is the typical coefficient of thermal expansion for aluminum coils?: The typical coefficient of thermal expansion for aluminum coils is around 23.6 x 10^-6 per degree Celsius.

Q:Are aluminum coils suitable for thermal insulation?: No, aluminum coils are not suitable for thermal insulation. Aluminum is a highly conductive material, meaning it readily transfers heat. This makes it ineffective in preventing heat transfer between different areas or maintaining a desired temperature within a space. For thermal insulation purposes, materials with low thermal conductivity, such as fiberglass, mineral wool, or foam insulation, are preferred as they minimize heat transfer and provide better insulation.

Q:... of aluminum atoms?(mole = 6.022x10^23): Wonderful first answer, but I am going to try anyways. For 1 mole, you need 27 grams of aluminum. The density of aluminum is 2700 kg/m^3, which means that 2.7 grams is contained in 1 cm^3 We know, Volume = Length * Breadth * Thickness 10 = L * (30.48) * Thickness So, just convert the thickness that you have into cm, and then use it in the above equation. --edited from here-- The atomic radius is 125 pM So, you have each atom having 250 pM Total thickness = 250 * 10^-12 * 92000 m = 0.0023 cm Use this in the equation : 10 = L * 30.48 * 0.0023 L = 142 cm

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