Question 6 of 9: BOF Cold-Scrap Melting Heat Balance

Question 6: BOF Cold-Scrap Melting Heat Balance (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Basis: 1 tonne (1000 kg) of charge, scrap fraction $x=0.28$. Scrap heats from $T_{room}=20\,{}^{\circ}\text{C}$ through melting at $T_m=1520\,{}^{\circ}\text{C}$ to bath temperature $T_{bath}=1600\,{}^{\circ}\text{C}$: $C_{p,s}=450$, $C_{p,l}=717$ J/kg K, $L=271{,}000$ J/kg. Heat comes from the hot-metal (liquid steel) sensible heat, $C_{p,l}=717$ J/kg K.

Find. Heat absorbed per tonne of scrap heated/melted to 1600°C; the hot-metal temperature rise (and resulting turn-down temperature) had that scrap fraction not been added.

Approach. Three-stage sensible+latent heat balance for the scrap (solid heating, melting, liquid superheat) gives the MJ/tonne figure directly; energy conservation then says that same heat, if not absorbed by scrap, stays in the hot-metal bath and raises its sensible temperature instead.

1. Stage 1 — heat solid scrap from room temperature to its melting point. Per tonne (1000 kg) of scrap: $$Q_1=mC_{p,s}\Delta T=(1000)(450)(1520-20)=675\times10^6\ \text{J}=675\ \text{MJ}$$
2. Stage 2 — latent heat of melting. $$Q_2=mL=(1000)(271{,}000)=271\times10^6\ \text{J}=271\ \text{MJ}$$
3. Stage 3 — superheat the melted scrap from $T_m$ to bath temperature. $$Q_3=mC_{p,l}\Delta T=(1000)(717)(1600-1520)=57.4\times10^6\ \text{J}=57.4\ \text{MJ}$$
4. Total heat absorbed per tonne of scrap. $$Q_{total}=Q_1+Q_2+Q_3=675+271+57.4$$ $$\boxed{Q_{total}=1003\ \text{MJ/tonne scrap}}$$
5. If the 28% scrap fraction had not been added. Per tonne of total charge, scrap absorbs $Q_{scrap}=0.28\times1003=280.9$ MJ, drawn from the sensible heat of the remaining $0.72\times1000=720$ kg of liquid hot metal. Removing that heat sink lets the same heat raise the hot-metal temperature instead: $$\Delta T=\dfrac{Q_{scrap}}{m_{hotmetal}C_{p,l}}=\dfrac{280.9\times10^6}{(720)(717)}$$ $$\boxed{\Delta T=544\ ^{\circ}\text{C}\ \Rightarrow\ T_{final}\approx1600+544=2144\,{}^{\circ}\text{C}}$$

This 544°C excursion is a theoretical energy-balance result testing the concept — scrap as a heat sink limiting turn-down temperature — rather than a physically-observed operating point: in practice a real BOF would never be allowed to reach 2144°C (well above practical refractory and metallurgical limits), which is exactly why the 28% cold-scrap addition, or an equivalent coolant, is a mandatory part of the charge balance.