C3T+group2

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Members of the group: Trung Thijs Lucas Ingrid Floris Fahad Daisy

__Question 1 (page 7): Calculate the mass of each element__ Fe 94.9 % = 94.9 kg of Iron C 4.55 % = 4.55 kg of Carbon Si 0.25 % = 0.25 kg of Silicon Mn 0.33 % = 0.3 kg of Manganese

__Question 2 (page 7): How many kg of each element are oxidised (take Fe-effieciency into account) of the 100 kg of hot metal?__ Fe= 2% / 100 = 0.02 0.02 * 94.9kg = Fe 1.898kg C4.55kg - 0.05kg = C 4.5kg Si 0.25kg - 0kg = Si 0.25kg Mn 0.3kg - 0.1kg = Mn 0.2kg

__Question 3 (page 7): How many kg of steel are produced?__ Fe = 1.898kg C = 4.5kg Si = 0.25kg Mn = __0.2kg +__ 6.848kg 100kg - 6.848kg = 93.152kg of steel is produced.

__Question 4 (page 7): Calculate the amount of energy that is generated during the process per 100kg hot metal.__ Fe = 1.898kg * 4.3MJ = 8.1614MJ C = 4.5kg * 11.7MJ = 52.65MJ Si = 0.25kg * 33.9MJ = 8.475MJ Mn = 0.2kg * 7.4MJ = __1.48MJ+__ 70.7664MJ

__Question 5 (page 8): How much energy is used in natural gas a year per household__ 2000m³ x 32MJ = 64000MJ/m³

__Question 6 (page 8): the energy from how many kg of hot metal is equivelant to the energy used in natural gas of one household?__ 64000MJ/m³ / (70.7664MJ / 100) = 90438.40014kg of steel.

__Question 7 (page 8): if 45% of the heat generated bij the C combustions will leave the process through the waste gasses, what will be the temperature of the liquid steel?__ The amount of C which is oxidised is 4.5kg (4.55%of 100kg)-0.05kg= 4.5kg). 11.7 amount of jheat generated (MJ) by oxidation per Kg 11.7MJx 4.5kg= 52.65MJ 52.65 is 100%, but you need 55% 52.65:100x55= 28.9575MJ=amount energy generated and warms up the liquid steel 28.96Mjx1000000=289560000J

70.7664MJ-52.65MJ=18.1164MJ= is the amount of energy generated exept the Carbon (because we alreay calculated that) 18.1164*1000000= 18116400 J + 289560000 J = 47072400 J = 47.07 MJ 47.07 * 1000000 : 690 = 68217.391 : 93.152 = 732.32342 °C + 1350 = 2082.3 °C

__Question 8 (page 8):If we want to melt one kilogram of scrap and heat it to 1650°C how much energy is required?__ Heat capacity of steel = 690J (kg °C) Latent heat of steel 271 kJ/kg 1650°C-20°C = 1630°C 1630°C * 690J + 271000J/kg = 1395700J

__Question 9 How many of scrap can be melted with the 100 kg of hot metal making steel of 1650 °C?__ ( 47.0739 - 19.282464 ) :1.3957 = 19.91218 kg of scrap

__Question 10 How many kg of steel are produced?__ The total amount of steel produced is 93.251 + 19.91218 = 113.0642 kg of steel.

__Question 11 If we want to make 320 tons of this steel, how many tons of hot metal and scrap are required?__ For 320 ton we need: 320 : 113.0642 * 93.251 = 263.64 ton of hot metal 320 : 113.0642 * 19.91218 = 56.36 ton of scrap

__Question 12 How many cubic metres of steel are that?__ 320 ton steel with density 7000 kg / m³ = 7 ton/m³

__Question 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?__ Gives 320 : 7 = 45.71 m³ Surface of ladle bottom = pi r² = 3.1415 * (4:2)² = 12.57 m² Ladle height 0.5 m + 45.71 : 12.57 = 4.14 meter