Chapter+2+-+Converter+calculation


 * __Chapter 2.__**

Fe - Iron is 94.9kg C - carbon is 4.55kg Si - Silicon is 0.25kg Mn - Magnesium is 0.3 kg
 * 1. Calculate the mass (kg) of each element.**

In the converter process we can make steel with the composition : Fe with 0.05%C and 0.1% Mn. 2% of Iron 4.5% of Carbon 0.2% of Magnesium All of Silicon
 * 2. How many kg of each element are oxidised (take Fe-efficiency into account) of the 100 kg hot metal?**


 * 3. How many kg of steel are produced?**

98% of Fe becomes steel (2%) oxidised. So, 0.98 X 94.9Kg = __= 93Kg__ 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) à Total steel produced: 93 + 0.23 + 0.03 = __93.26Kg.__

The amount of energy that is produced from oxygen is: From Fe: 4.3 * 1.898 = 8.1614 Mj From C: 11.7 * 4.5 = 52.65 Mj From Si: 33.9 * 0.25 = 8.475 Mj From Mn: 7.4 * 0.2 = 14.8 Mj
 * 4. Calculate the amount of energy that is generated during the process.**

Yearly, 64000MJ of natural gas is used in each household because 2000x32MJ=64000MJ (m3).
 * 5. How much energy is used in natural gas a year per household ?**


 * 6. (Missing for now)**

1kg C = 11.7Mj 4.55kg of C, 4.5kg C is oxidised so 0.05kg becomes steel 4.5 x 11.7 = 52.65Mj 45% is waste so 55% useful, as a decimal is 0.55 0.55 x 52.65Mj = 28.96Mj left by C 28.96 + 18.11(oxidised Fe,Si&Mn) = 47.07Mj
 * 7.If 45% of the heat generated by the C combustion will leave the process through the waste gasses what will be the temperature of the liquid steel?**

Starting temperature = 1350 An extra 4707 = 47070000Kj of energy 47070000/690 = 68217.4 68217.4/93(how much steel to be produced) = 733.5 Degree increase So 1350 + 733.5 = 2083.5 Degrees Celsius

Start temp = 20 So 1650 - 20 = 1630 Degrees Celsius To heat 1kg by 1 degree = 690J So 1630 x 690 = 1124700J is needed to heat the scrap To melt it, it needs to have 271Kj = 271000j/Kg So 1124700 + 271000 = 1395700 J/Kg is needed to heat and melt the scrap Which = 1395.7 Kj
 * 8. If we want to melt 1kg of scrap and heat it to 1650°C. How much energy is required?**

The total energy created = 70.76 Mj 70.76 - 23.69 = 47.07 Mj 47.07 - 19.28 = 27.79 So 27.79/1.39 = 19.9 kg of scrap can be melted.
 * 9.How many kgs of scrap can be melted with the 100kg of hot metal making steel at 1650°C?**

Well out of the 100 Kg 93.2 is made into steel. 19.9 + 93.2 = 113.1 Kg of steel.
 * 10. How many kg of steel are produced?**

From 100 Kg we can make 19.9 kg of scrap and 93.2 kg of steel 320/113.1 = 2.829 x 93.2 = 263.7 tons of hot metal needed 320/113.1 = 2.829 x 19.9 = 56.3 tons of scrap needed
 * 11. If we want to make 320 tons of this steel how many tons of hot metal and scrap are required?**

320 tons density = 7000kg/m cubed 1000 kg in 1 ton So there is 7 ton / metre cubed of steel made Volume = 320/7 = 45.7m cubed.
 * 12. How many cubic meters of steel are that?**

The surface area = Pi r squared So 3.1415 x 4 = 12.56cm squared The volume = 45.71m cubed 45.71 / 12.56 = 3.69m So the height must be 3.69m plus 0.5m for spillage = 4.19m.
 * 13. 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?**

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