Module 3 - Benjamin-Mills

SS2.3
GETTING THE ‘HEAT BALANCE’ RIGHT
Part 2: How much scrap steel is needed?
This involves a lengthy calculation: we have done some of it for you, and you will
do only the last part.
Look back at the ‘heat balance’ equation on the previous page. Some of the
energy given out by the oxidation reactions is lost from the converter to the
surroundings through normal cooling. It can be estimated that approximately
25 ¥ 10 6 kJ were lost by radiation and convection in this way.
The rest of the energy released by the oxidation reactions is used to raise the
temperature of the converter contents. We can calculate how much energy is
required to raise each of the contents of the converter from their initial
temperatures to the final temperature if we know how much of each substance
was added and its specific heating capacity. The results of such calculations are
shown in Table 2.
Material Quantity added Energy absorbed
to converter
/10 6 kJ
Lime 11.9 tonnes 18.4
Calcined dolomite 9.49 tonnes 16.5
Oxygen gas 16.8 ¥ 10 3 m 3 20.8
Molten iron 278 tonnes 86.3
Scrap steel ? ?
Table 2 Energy absorbed in heating up the
contents of the converter
c
Using the information above, calculate how much scrap steel was
needed in order to achieve the tapping temperature of 1738°C.
Assume that the steel added was at 20°C and that the specific
heating capacity of steel is 0.83 ¥ 10 3 kJ t –1 K –1 . (This value has been
averaged out over the temperature range and allows for the enthalpy
change of fusion on melting. It gives the amount of energy which
must be supplied to raise the temperature of 1 tonne of steel by 1K.)
The energy absorbed by the steel is related to the rise in
temperature in the following way:
energy absorbed = mass ¥ specific heating capacity ¥ temperature rise
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„ Salters Advanced Chemistry 2000 – see Copyright restrictions