Emma,+Kate,+Eva,+Jayne,+Kirsty

 ** __Members of our Group__ ** Kate Emma Eva Jayne Kirsty ** __Chapter 1:__ ** · The 3 main raw materials used to make iron are; Iron Ore, Coke, Limestone · The raw materials for the plant come from Iron Ore – Asia, China, Brazil + India these are transported through trains and boats Coke – China, India + England these are transported through trains Limestone – India this is transported by boats · The raw materials are processed before they are used in the blast furnace. Iron Ore needs to be smelted before it is put in the furnace Coke needs to be fused together with water, cold tar and cold gas at high temperatures before it can be used. Limestone process consists of · Role of each of these raw materials in the iron-making process is Iron Ore – is needed because it has iron in Coke – this makes the furnace burn at higher heat Limestone – this makes the impurities in the iron ore melt · The ecological effects of using these raw materials is probably bad for the environment because of the amount of transportation needed for each country, this is causing a lot of carbon dioxide to be released in the atmosphere therefore contributing to global warming These negative effects of transportation and production could be reduced by using less transportation. ** __Chapter 2:__

** **The mass (kg) of each element in 100kg of hot metal;**
 * [[image:http://c1.wikicdn.com/i/edit.png width="128" height="37" caption="Edit This Page" link="http://longfield.wikispaces.com/page/edit/Chapter2"]]Chapter 2

· Fe – 94.9 · C – 4.55 · Si – 0.25 · Mn – 0.3 **In the converter process we make steel with the composition;** Fe with 0.05% C and 0.1% Mn.

**The amount of kg of each element oxidised taking Fe – efficiency into account; which is 98%** **(so 2% of the iron is oxidised during the process.)**

· Fe = 0.98% x 94.9kg = 93kg (94.9kg – 93kg = 1.9kg oxidised ) · C = 0.05% x 4.55kg 0.23kg (4.55kg – 0.23kg 4.32kg oxidised ) · Mn = 0.1% x 0.3kg = 0.03kg (0.1kg - 0.03kg = 0.27kg oxidised )  **The amount of steel produced in kg 93kg + 0.23kg + 0.03kg = 93.26kg **  • From Fe: 4.3 X 1.9 = 8.17 • From C: 11.7 X 4.32 = 50.54 • From Si: 33.9 X 0.25 = 8.47 • From Mn: 7.4 X 0.27 = 1.99
 * The amount of energy that is generated during the process is:**

**The amount of energy used in natural gas a year per household is:** 2000m ³  x 32 Mg = 64 000 Mg <span style="font-size: 12pt; line-height: 115%; font-family: 'Arial','sans-serif'; mso-fareast-font-family: Calibri; mso-ansi-language: EN-GB; mso-fareast-theme-font: minor-latin; mso-fareast-language: EN-US; mso-bidi-language: AR-SA;">³ <span style="font-size: 12pt; line-height: 115%; font-family: 'Arial','sans-serif'; mso-fareast-font-family: 'Times New Roman'; mso-ansi-language: EN-US; mso-fareast-language: EN-US; mso-bidi-language: AR-SA;"> **If 45% of the heat generated by the C combustions will leave the process through the waste gasses, the temperature of the liquid steel will be: ** **An average Dutch household uses 2000 m3 natural gas a year. Combustion of natural gas produces 32 MJ/m3.**

How much energy is used in natural gas a year per household? = 32 X 2000= 64000 MJ

=64000/70.7664= 904.4 kg
 * The energy produced from how many kg of hot metal is equivalent to the energy used in natural gas of one household?**


 * The combustions of Carbon generated carbon monoxide and carbon dioxide which leave the process as waste gasses.**


 * If 45% of the heat generated by the C combustions will leave the process through the waste gasses what will be the temperature of the liquid steel?** = 2083.5 degrees celsius

=1395700 / 1000= 1395.7Kj
 * If we want to melt 1 kg of scrap and heat it to 1650°C how much energy is required?**


 * How many kg of scrap can be melted with the 100 kg of hot metal making steel of 1650°C?** = 19.9kg


 * How many kg of steel are produced?** = 113.1kg

56.3 tons of scrap
 * If we want to make 320 tons of this steel how many tons of hot metal and scrap are required? =** 263 .7 tons of hot metal and


 * How many cubic meters of steel are that?** 320 divided by 7 = 45.7m cubed


 * Suppose a steel ladle has an internal diameter of 4 meters. What is the required height of the ladle if the steel surface must be 50 cm below the top of the ladle?**

3.1415 X 4 = 12.56 cm<span style="font-size: 12pt; line-height: 115%; font-family: 'Arial','sans-serif'; mso-fareast-font-family: Calibri; mso-ansi-language: EN-GB; mso-fareast-theme-font: minor-latin; mso-fareast-language: EN-US; mso-bidi-language: AR-SA;">². (bottom surface of ladle)
 * The surface of the bottom of the ladle is**=pi r². Thus: 3.1415 X 2²=

3.69m<span style="font-size: 12pt; line-height: 115%; font-family: 'Arial','sans-serif'; mso-fareast-font-family: Calibri; mso-ansi-language: EN-GB; mso-fareast-theme-font: minor-latin; mso-fareast-language: EN-US; mso-bidi-language: AR-SA;">³
 * The volume form our last question was 45.71m cubed. Height of cylinder**=45.71 / 12.56=
 * Then the height of the ladle must be 3.69m + 0.5m for spillage** = 4.14 ||

Chapter 3:

Chapter 4: