Nicole,+Safron,+Karis

·  Nicole (nicoleee-marieee) ·   Lindsey (linzi22) **__Chapter 1 __** ·     What are the main three raw materials used to make iron? The main three raw materials used to make iron ore are, coke and limestone. ·   Where do the raw materials come from and how are they transported to the plant? The raw materials come from Major producers in Australia, Brazil, China, Russia and India. They are transported to the plant by rail. __· __    How and why are the raw materials processed before they are used in the blast furnace? Raw materials are processed before they are used in the blast furnace to get rid of the oxygen, huge amounts of air blast in at the bottom of the furnace ·   What role does each of these three raw materials play in the iron-making process? Iron ore is a metal, coke reacts with the iron ore because iron ore is mainley iron oxide and mixes with carbon. to create iron. __· __    What can you say about the ecological effects of using these raw materials (for the country where they come from and for the means of transportation)? The ecological effects of using raw materials are damages the environment, costs a lot of money and causes pollution. __<span style="color: black; font-family: Calibri; msofareastfontfamily: Symbol; msolist: Ignore; msobidifontsize: 12.0pt; msobidifontfamily: Symbol; msothemecolor: text1; msofareastlanguage: EN-GB;">· __    <span style="msofareastfontfamily: 'Times New Roman'; msobidifontfamily: Arial; msothemecolor: text1; msofareastlanguage: EN-GB;">How can the bad side effects be reduced? Use local materials and use the cheapest transport which gives off the least pollution.
 * <span style="font-size: 12pt; color: black; line-height: 115%; font-family: 'Calibri','sans-serif'; mso-bidi-font-family: Tahoma; mso-themecolor: text1; mso-ascii-theme-font: minor-latin; mso-hansi-theme-font: minor-latin;">Our Group: **
 * <span style="font-size: 12pt; color: black; line-height: 115%; font-family: 'Calibri','sans-serif'; mso-bidi-font-family: Tahoma; mso-themecolor: text1; mso-ascii-theme-font: minor-latin; mso-hansi-theme-font: minor-latin;">Much is said nowadays about the consequences of human activity for the environment. **

__ <span style="font-family: Calibri; msofareastfontfamily: Symbol; msolist: Ignore; msobidifontsize: 12.0pt; msobidifontfamily: Symbol; msothemecolor: text1; msofareastlanguage: EN-GB;">· __    <span style="msofareastfontfamily: 'Times New Roman'; msobidifontfamily: Arial; msothemecolor: text1; msofareastlanguage: EN-GB;">How many kilograms are oxidised ? <span style="msofareastfontfamily: Arial; msolist: Ignore; msothemecolor: text1; msofareastlanguage: EN-GB;">•   •<span style="color: black; mso-fareast-font-family: +mn-ea; mso-bidi-font-family: +mn-cs; mso-themecolor: text1;">Fe efficiency = 98% so 2% gets oxidised. • <span style="color: black; font-family: Calibri; msofareastfontfamily: +mn-ea; msobidifontfamily: +mn-cs; msothemecolor: text1; msofareastlanguage: EN-GB;">2% of 94.9 = 0.02 X 94.9 = 1.898Kg. • <span style="color: black; font-family: Calibri; msofareastfontfamily: +mn-ea; msobidifontfamily: +mn-cs; msothemecolor: text1; msofareastlanguage: EN-GB;">4.5% of C is oxidised • <span style="color: black; font-family: Calibri; msofareastfontfamily: +mn-ea; msobidifontfamily: +mn-cs; msothemecolor: text1; msofareastlanguage: EN-GB;">0.2% Mn is oxidised • <span style="color: black; mso-fareast-font-family: +mn-ea; mso-themecolor: text1;"> All Si is oxidised. •So how many Kg of steel produced from the 100Kg mixture? •98% of Fe becomes steel (2%) oxidised. So, 0.98 X 94.9Kg = **__<span style="font-size: 12pt; color: black; line-height: 115%; font-family: 'Calibri','sans-serif'; mso-fareast-font-family: +mn-ea; mso-bidi-font-family: 'Times New Roman'; mso-themecolor: text1; mso-ascii-theme-font: minor-latin; mso-hansi-theme-font: minor-latin; mso-bidi-theme-font: minor-bidi;">= 93Kg __** <span style="color: black; mso-fareast-font-family: +mn-ea; mso-themecolor: text1;"> •0.05% of the carbon becomes steel = 0.05 X 4.55 = 0.23Kg (Rest Ox) •So, mass of carbon oxidised = 4.55 -0.23 = 4.32Kg •0.1% of Mn becomes steel = 0.1 X 0.3Kg = 0.03Kg (0.27Kg Oxidised) <span style="color: black; mso-fareast-font-family: +mn-ea; mso-bidi-font-family: +mn-cs; mso-themecolor: text1;"> <span style="color: black; font-family: Wingdings; mso-ascii-font-family: Calibri; mso-char-type: symbol; mso-symbol-font-family: Wingdings; mso-ascii-theme-font: minor-latin; mso-hansi-font-family: Calibri; mso-hansi-theme-font: minor-latin;"> <span style="font-family: Calibri; msofareastfontfamily: +mn-ea; msochartype: symbol; msosymbolfontfamily: Wingdings; msobidifontfamily: 'Times New Roman'; msothemecolor: text1; msofareastlanguage: EN-GB; msoasciifontfamily: Calibri; msochartype: symbol; msosymbolfontfamily: Wingdings; msoasciithemefont: minor-latin; msohansifontfamily: Calibri; msohansithemefont: minor-latin;">à <span style="color: black; mso-fareast-font-family: +mn-ea; mso-bidi-font-family: +mn-cs; mso-themecolor: text1;"> Total steel produced: 93 + 0.23 + 0.03 = __93.26Kg.__ •__Amount of energy generated from oxidation?__ •From Fe: 4.3 X 1.898 = •From C: 11.7 X 4.5 = •From Si: 33.9 X 0.25 = •From Mn: 7.4 X 0.2 = Add all 4 figures to give total = **__<span style="font-size: 12pt; color: black; line-height: 115%; font-family: 'Calibri','sans-serif'; mso-fareast-font-family: +mn-ea; mso-bidi-font-family: +mn-cs; mso-themecolor: text1; mso-ascii-theme-font: minor-latin; mso-hansi-theme-font: minor-latin;">70.6Mj __**<span style="color: black; mso-fareast-font-family: +mn-ea; mso-bidi-font-family: +mn-cs; mso-themecolor: text1;">
 * __ Chapter 2__**

•__How much energy is used from natural gas in a house in a year?__ **<span style="font-size: 12pt; color: black; line-height: 115%; font-family: 'Calibri','sans-serif'; mso-bidi-font-family: Arial; mso-ascii-theme-font: minor-latin; mso-hansi-theme-font: minor-latin;">2000 X 32MJ ** <span style="msofareastfontfamily: Arial; msolist: Ignore; msothemecolor: text1; msofareastlanguage: EN-GB;">• 64000MJ  <span style="color: black; font-family: Calibri; msofareastfontfamily: +mn-ea; msobidifontfamily: +mn-cs; msothemecolor: text1;"> <span style="color: black; mso-fareast-font-family: +mn-ea; mso-bidi-font-family: Arial; mso-themecolor: text1;"> we want to compare it to how much energy is actual given out from the oxidation of the hot steel mix! •<span style="color: black; mso-fareast-font-family: +mn-ea; mso-bidi-font-family: +mn-cs; mso-themecolor: text1;">100Kg of mixture gave 70.6MJ of energy. <span style="color: black; font-family: Calibri; msofareastfontfamily: +mn-ea; msothemecolor: text1;">• <span style="color: black; font-family: Calibri; msofareastfontfamily: +mn-ea; msobidifontfamily: +mn-cs; msothemecolor: text1; msofareastlanguage: EN-GB;">So the number of Kg that would give 64000MJ is……. <span style="color: black; mso-fareast-font-family: +mn-ea; mso-bidi-font-family: Arial; mso-themecolor: text1;"> <span style="color: black; mso-fareast-font-family: +mn-ea; mso-themecolor: text1;"> 64000 / 70.6 = __906Kg

Chapter 2 Part 2 :__ __•__  __<span style="color: black; font-family: Calibri; msofareastfontfamily: +mn-ea; msobidifontfamily: +mn-cs; msothemecolor: text1; msofareastlanguage: EN-GB;">How many Kg of each element are oxidised? __<span style="color: black; font-family: Calibri; msofareastfontfamily: +mn-ea; msobidifontfamily: Arial; msothemecolor: text1;"> FE- efficiency is 98%. So, 98% of 94.9Kg of iron is converted = 93Kg •  <span style="color: black; mso-fareast-font-family: +mn-ea; mso-bidi-font-family: +mn-cs;">Only 0.05% of the carbon used goes in to the steel. • <span style="color: black; mso-fareast-font-family: +mn-ea; mso-bidi-font-family: +mn-cs;">So how much is oxidised? • <span style="color: black; font-family: Wingdings; mso-char-type: symbol; mso-symbol-font-family: Wingdings; msofareastfontfamily: +mn-ea; msoasciifontfamily: Calibri; msobidifontfamily: +mn-cs; msochartype: symbol; msosymbolfontfamily: Wingdings; msoasciithemefont: minor-latin; msohansifontfamily: Calibri; msohansithemefont: minor-latin;">à <span style="color: black; font-family: Calibri; msofareastfontfamily: +mn-ea; msobidifontfamily: +mn-cs;"> 0.05% X 4.55Kg = 0.2275Kg goes into the steel, 4.32Kg Ox. Repeating this for Mn, 0.1% of 0.3Kg = 0.03Kgused, 2.7Kg Ox. __<span style="color: black; font-family: Calibri; msofareastfontfamily: +mn-ea; msobidifontfamily: +mn-cs;">How much energy is used per year? __ <span style="color: black; font-family: Calibri; msofareastfontfamily: +mn-ea; msobidifontfamily: +mn-cs;">2000 X 32 = 64,000MJ (Mega Joules) __<span style="font-size: 12pt; color: black; font-family: Calibri; msofareastfontfamily: +mn-ea; msobidifontfamily: +mn-cs;">How many Kg of hot metal need to be produced to release this much energy? __ <span style="font-size: 12pt; color: black; font-family: Calibri; msofareastfontfamily: +mn-ea; msobidifontfamily: +mn-cs;">If 45% of the heat generated by the reaction is lost when the waste gases carbon dioxide & monoxide leave, __<span style="font-size: 12pt; color: black; font-family: Calibri; msofareastfontfamily: +mn-ea; msobidifontfamily: +mn-cs;">what will the remaining temperature of the hot metal mixture? __ <span style="font-size: 12pt; color: black; font-family: Calibri; msofareastfontfamily: +mn-ea; msobidifontfamily: +mn-cs;">When 1 Kg of carbon is oxidised it generates 11.7Mj Out of the 4.55Kg of carbon in the mixture, 0.05Kg becomes part of the steel- this means 4.5Kg is oxidised. So, 4.5Kg X 11.7Mj = **__52.65Mj__** But as 45% leaves as waste energy, this leaves 55% of that energy in the reaction: 55% = 0.055 as a decimal so 0.55 X 52.65Mj = **__28.96Mj left by C__.** When all the oxidations took place (including Fe, Si & Mn) there was 70.76Mj energy released. Energy from oxidation from these 3= 8.16 + 8.47 + 1.48= **__18.11Mj__** Add this to heat from the carbon that is not lost= 18.11 + 28.96= **__47.07Mj__** So, we calculated heat generated from the reaction once we lost heat by loosing hot gases like carbon dioxide. This = 47.07Mj. Now we need to work out what temperature this EXTRA heat would causes our metal mixture to increase to. Starting temperature of mixture 1350 degrees Celsius. 47.07**__M__**j of extra heat released= 47070000j of energy. 47070000j / 690 (the energy needed to heat 1kg of steel by 1 degree) = **__68217.4__** But there are 93kg of steel produced so: 68217.4 / 93 = 733.5 degrees __increase__, So final temperature= 1350 (start temp) + 733.5= **__2083.5 degrees Celsius.__** __<span style="font-size: 12pt; color: black; font-family: Calibri; msofareastfontfamily: +mn-ea; msobidifontfamily: +mn-cs;">How much energy is needed to melt 1Kg of scrap and then heat it to 1650 degrees? __ <span style="font-size: 12pt; color: black; font-family: Calibri; msofareastfontfamily: +mn-ea; msobidifontfamily: +mn-cs;">Start temp of scrap is 20 degrees. This means we need to heat it by 1650 – 20= 1630 degrees. Energy required to heat 1Kg of steel by 1 degrees is 690J. So, 1630 X 690 = **__1124700joules of energy is required to heat the scrap.__** **__<span style="font-size: 12pt; color: black; line-height: 115%; mso-fareast-font-family: +mn-ea; mso-bidi-font-family: +mn-cs;">But __**<span style="font-size: 12pt; color: black; line-height: 115%; mso-fareast-font-family: +mn-ea; mso-bidi-font-family: +mn-cs;"> in doing this, we also need to take it through its melting point and turn it into a liquid. This needs **__more__** energy again! <span style="font-size: 12pt; color: black; font-family: Calibri; msofareastfontfamily: +mn-ea; msobidifontfamily: +mn-cs;">Latent heat (energy needed to melt 1 Kg) = 271Kj = 271000J/Kg. Now we add the energy needed to heat it and the energy needed to melt it: 1124700 + 271000= **__1395700J is needed to heat and melt 1Kg of scrap!__** <span style="font-size: 12pt; color: black; font-family: Wingdings; mso-char-type: symbol; mso-symbol-font-family: Wingdings; msofareastfontfamily: +mn-ea; msoasciifontfamily: Calibri; msobidifontfamily: +mn-cs; msochartype: symbol; msosymbolfontfamily: Wingdings; msoasciithemefont: minor-latin; msohansifontfamily: Calibri; msohansithemefont: minor-latin;">à **__<span style="font-size: 12pt; color: black; line-height: 115%; mso-fareast-font-family: +mn-ea; mso-bidi-font-family: +mn-cs;"> 1395700 / 1000 = 1395.7Kj __**<span style="font-size: 12pt; color: black; line-height: 115%; mso-fareast-font-family: +mn-ea; mso-bidi-font-family: +mn-cs;"> __<span style="font-size: 12pt; color: black; font-family: Calibri; msofareastfontfamily: +mn-ea; msobidifontfamily: +mn-cs;">How many Kg of scrap can be melted with 100Kg of hot metal if we need the temperature to be 1650 degrees? __ <span style="font-size: 12pt; color: black; font-family: Calibri; msofareastfontfamily: +mn-ea; msobidifontfamily: +mn-cs;">The total energy generated was 70.76Mj But remember we lost around 23Mj of energy when the hot gases of carbon dioxide escaped so the remaining energy is: 70.76 – 23.69 = **__47.07Mj of energy left in mixture.__** **__<span style="font-size: 12pt; color: black; line-height: 115%; mso-fareast-font-family: +mn-ea; mso-bidi-font-family: +mn-cs;">47.07 – 19.28 (energy used) = 27.79Mj __**<span style="font-size: 12pt; color: black; line-height: 115%; mso-fareast-font-family: +mn-ea; mso-bidi-font-family: +mn-cs;"> <span style="font-size: 12pt; color: black; font-family: Calibri; msofareastfontfamily: +mn-ea; msobidifontfamily: +mn-cs;">Now we need to divide this by the energy needed to melt 1 Kg of scrap = 1.39Mj (if our last calculation for melting scrap was correct) <span style="font-size: 12pt; color: black; line-height: 115%; font-family: 'Times New Roman','serif'; mso-fareast-font-family: 'Times New Roman'; mso-themecolor: text1;"> •  **__ How many Kg of steel can we then produce if we did this? __** <span style="font-size: 12pt; color: black; line-height: 115%; font-family: 'Times New Roman','serif'; mso-fareast-font-family: 'Times New Roman'; mso-themecolor: text1;"> •    Well, from our 100Kg hot metal mixture we can make 93.2Kg of steel (remember the rest is oxidised). <span style="font-size: 12pt; color: black; line-height: 115%; font-family: 'Times New Roman','serif'; mso-fareast-font-family: 'Times New Roman'; mso-themecolor: text1;"> •   And we can add 19.9Kg of scrap to this, so, 93.2 + 19.9 = **__113.1Kg of steel!__** <span style="font-size: 12pt; color: black; line-height: 115%; font-family: 'Times New Roman','serif'; mso-fareast-font-family: 'Times New Roman'; mso-themecolor: text1;"> •  **__ If someone wanted us to make 320 tons of steel, how many tons of hot metal and scrap are required? __** <span style="font-size: 12pt; color: black; line-height: 115%; font-family: 'Times New Roman','serif'; mso-fareast-font-family: 'Times New Roman'; mso-themecolor: text1;"> •    We can make 113.1Kg of steel from 100Kg mixture and 19.9Kg of scrap so: <span style="font-size: 12pt; color: black; line-height: 115%; font-family: 'Times New Roman','serif'; mso-fareast-font-family: 'Times New Roman'; mso-themecolor: text1;"> •   320 / 113.1 = 2.829 X 93.2 = __263.7 tons__ of hot metal needed <span style="font-size: 12pt; color: black; line-height: 115%; font-family: 'Times New Roman','serif'; mso-fareast-font-family: 'Times New Roman'; mso-themecolor: text1;"> •   320 / 113.1 = 2.829 X 19.9 = __56.3 tons__ of scrap needed. <span style="font-size: 12pt; color: black; line-height: 115%; font-family: 'Times New Roman','serif'; mso-fareast-font-family: 'Times New Roman'; mso-themecolor: text1;"> •   CHECK WORKING: 263.7 + 56.3= **__320 tons__** <span style="font-size: 12pt; color: black; line-height: 115%; font-family: 'Times New Roman','serif'; mso-fareast-font-family: 'Times New Roman'; mso-themecolor: text1;"> •   How many cubic metres will 320 tons take up? <span style="font-size: 12pt; color: black; line-height: 115%; font-family: 'Times New Roman','serif'; mso-fareast-font-family: 'Times New Roman'; mso-themecolor: text1;"> •   320 tons has density of 7000kg / m cubed. <span style="font-size: 12pt; color: black; line-height: 115%; font-family: 'Times New Roman','serif'; mso-fareast-font-family: 'Times New Roman'; mso-themecolor: text1;"> •   There is 1000kg in 1 ton. <span style="font-size: 12pt; color: black; line-height: 115%; font-family: 'Times New Roman','serif'; mso-fareast-font-family: 'Times New Roman'; mso-themecolor: text1;"> •   So the mass of the steel is 7 ton / metre cubed. <span style="font-size: 12pt; color: black; line-height: 115%; font-family: 'Times New Roman','serif'; mso-fareast-font-family: 'Times New Roman'; mso-themecolor: text1;"> •  **__ Volume = 320 / 7 = 45.7m cubed. __**  <span style="font-size: 12pt; color: black; line-height: 115%; font-family: 'Times New Roman','serif'; mso-fareast-font-family: 'Times New Roman'; mso-themecolor: text1;"> •    A steel ladle has an internal diameter of 4m. <span style="font-size: 12pt; color: black; line-height: 115%; font-family: 'Times New Roman','serif'; mso-fareast-font-family: 'Times New Roman'; mso-themecolor: text1;"> •   What must we make the height of the ladle if we want the surface of the steel mixture to be 50cm (0.5m) below the height of it? <span style="font-size: 12pt; color: black; line-height: 115%; font-family: 'Times New Roman','serif'; mso-fareast-font-family: 'Times New Roman'; mso-themecolor: text1;"> •   For this we are assuming the ladle is a cylinder shape – not a dish shape. <span style="font-size: 12pt; color: black; line-height: 115%; font-family: 'Times New Roman','serif'; mso-fareast-font-family: 'Times New Roman'; mso-themecolor: text1;"> •   The surface of the bottom of the ladle is = pi r squared. <span style="font-size: 12pt; color: black; line-height: 115%; font-family: 'Times New Roman','serif'; mso-fareast-font-family: 'Times New Roman'; mso-themecolor: text1;"> •   So: 3.1415 X (4 / 2) squared = 3.1415 X 2 squared <span style="font-size: 12pt; color: black; line-height: 115%; font-family: 'Times New Roman','serif'; mso-fareast-font-family: 'Times New Roman'; mso-themecolor: text1;"> •   =3.1415 X 4 = __12.56 cm squared.__ (bottom surface of ladle) <span style="font-size: 12pt; color: black; line-height: 115%; font-family: 'Times New Roman','serif'; mso-fareast-font-family: 'Times New Roman'; mso-themecolor: text1;"> •   The volume form our last question was 45.71m cubed. <span style="font-size: 12pt; color: black; line-height: 115%; font-family: 'Times New Roman','serif'; mso-fareast-font-family: 'Times New Roman'; mso-themecolor: text1;"> •   Height of cylinder = 45.71 / 12.56 = 3.69m cubed. <span style="font-size: 12pt; color: black; line-height: 115%; font-family: 'Times New Roman','serif'; mso-fareast-font-family: 'Times New Roman'; mso-themecolor: text1;"> •   So the height of the ladle must be 3.69m + 0.5m for spillage= **__4.14m.__** <span style="font-size: 12pt; color: black; line-height: 115%; font-family: 'Times New Roman','serif'; mso-fareast-font-family: 'Times New Roman'; mso-themecolor: text1;"> •   How long does it take to flow along it? <span style="font-size: 12pt; color: black; line-height: 115%; font-family: 'Times New Roman','serif'; mso-fareast-font-family: 'Times New Roman'; mso-themecolor: text1;"> •   Time = distance / speed = 6m / 0.1m/s = **__60 seconds!__** <span style="font-size: 12pt; color: black; line-height: 115%; font-family: 'Times New Roman','serif'; mso-fareast-font-family: 'Times New Roman'; mso-themecolor: text1;"> •   In this section we try to remove as many inclusions as possible. These would set in the steel and make it weak. <span style="font-size: 12pt; color: black; line-height: 115%; font-family: 'Times New Roman','serif'; mso-fareast-font-family: 'Times New Roman'; mso-themecolor: text1;"> •   So, inclusions have only 60 seconds to rise up out of the mixture. <span style="font-size: 12pt; color: black; line-height: 115%; font-family: 'Times New Roman','serif'; mso-fareast-font-family: 'Times New Roman'; mso-themecolor: text1;"> •   What is minimum rising velocity to get out in 60 seconds? <span style="font-size: 12pt; color: black; line-height: 115%; font-family: 'Times New Roman','serif'; mso-fareast-font-family: 'Times New Roman'; mso-themecolor: text1;"> •   Height of section is 0.78m. <span style="font-size: 12pt; color: black; line-height: 115%; font-family: 'Times New Roman','serif'; mso-fareast-font-family: 'Times New Roman'; mso-themecolor: text1;"> •   Speed= distance / time = 78cm / 60 seconds = **__1.3cm/ s__** <span style="font-size: 12pt; color: black; line-height: 115%; font-family: 'Times New Roman','serif'; mso-fareast-font-family: 'Times New Roman'; mso-themecolor: text1;"> •   So, for an inclusion to rise up out of the mixture it must travel at a speed (rising velocity) of at least 1.3cm/second <span style="font-size: 12pt; color: black; line-height: 115%; font-family: 'Times New Roman','serif'; mso-fareast-font-family: 'Times New Roman'; mso-themecolor: text1;"> •   (if it were at the bottom of the tank) <span style="font-size: 12pt; color: black; line-height: 115%; font-family: 'Times New Roman','serif'; mso-fareast-font-family: 'Times New Roman'; mso-themecolor: text1;"> •  **__  Looking at the table which inclusion sizes will rise out of the mixture? __** <span style="font-size: 12pt; color: black; line-height: 115%; font-family: 'Times New Roman','serif'; mso-fareast-font-family: 'Times New Roman'; mso-themecolor: text1;"> •   Inclusions of size 0.1 and less only travel at 0.39 cm/s far less that 1.3cm/s **__so these wont make it__** <span style="font-size: 12pt; color: black; line-height: 115%; font-family: 'Times New Roman','serif'; mso-fareast-font-family: 'Times New Roman'; mso-themecolor: text1;"> •  **__ BUT all those of size 0.2mm and above do get out! __**
 * __<span style="font-size: 12pt; color: black; font-family: Calibri; msofareastfontfamily: +mn-ea; msobidifontfamily: +mn-cs;">So 27.79 / 1.39 = 19.9Kg of scrap can be added and melted in the reaction. __**<span style="font-size: 12pt; color: black; line-height: 115%; mso-fareast-font-family: +mn-ea; mso-bidi-font-family: +mn-cs;">
 * __Chapter 3__**