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Numerical modeling example. Hot steel slab. Rolling mill. A simple s teel reheat furnace model – pg. 102-107. Reheat furnace. Final product. The problem:.
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Numerical modeling example Hot steel slab Rolling mill A simple steel reheat furnace model – pg. 102-107 Reheat furnace Final product
The problem: • Steel companies use models to simulate the batch heating of steel slabs (~4 hours, reaching 1450-1550K) prior to the rolling mill. This software is built into their process control systems. • Two-dimensional models often suffice for this purpose (a balance between computing speed and model detail). • The reheat furnace may contain many steel slabs and their heating progress must be monitored and paced through the furnace consistent with downstream operations in the rolling mill. • Different zones of the furnace provide different heating strategies over the heating cycle.
A full model would involve: • Implicit finite volume formulation. • Variable properties – thermal conductivity and heat capacity are a function of temperature and steel composition. • The phase change in steel structure from BCC to FCC at 910C. • Surface oxidation of the steel to the various oxides – occurs more rapidly above 570C. FeO0.95 (dominant oxide) has a heat of formation (a generation term) and low thermal conductivity (good insulator).
Our treatment will involve: • Heat a steel slab from ambient temperature to a relatively uniform ~1400 K over a 4 hour period. • A one-dimensional model will be used. • Constant properties, no phase change and no scale growth. • Explicit formulation – spreadsheet demonstration. • Implicit formulation – can be coded in matlab or VBA (Excel add-in feature).
The slab … • Dimensions: • 10 m long • 1 m wide • 0.3 m thick • Steel Properties: • k = 35 W/m-K • CP = 473.3 J/kg-K • r = 7820 kg/m3 • Model heat transfer through the 0.3 m dimension • “Convective” heat transfer based on a furnace gas temperature of 1400 K • Heat transfer symmetry on both sides of the slab; h = 200 W/m2-K • Centreline of the slab has a zero gradient boundary condition (q = 0)
Problem formulation and set up: Three different types of cells (explicit form): • Boundary next to the hot combustion gases – radiation modeled as a convective process. One cell only. • An array of interior cells. Many cells. • The centre plane boundary condition – plane of symmetry with q = 0. One cell only.
Cell adjacent to furnace gases: q = hA(T - TP) Accumulation within cell P = Input to cell P = Output from cell P = Generation within cell P = 0 Accumulation = Input – Output + Generation
Alternative surface cell formulation: Evaluate the cell P temperature as “true” surface value Place the P node at the surface Adjust the cell volume (optional) Accumulation term is now half the original value used previously
General interior cells: Accumulation within cell P = Input to cell P = Output from cell P = Generation in cell P = 0
Centreline Cell: Accumulation within cell P = Input to cell P = Output from cell P = 0 Generation in cell P = 0
Some computational considerations: • We want to cover the slab region 0 < x < 0.15 m • With x = 0.01 m we will require 15 cells Explicit scheme stability criteria: Time steps of t = 2.5 – 5 s will meet this criteria To cover the 3-4 h time required for the batch heating process we will need ~3000 – 4000 time steps!
The set of equations involve 15 cells with cell 1 adjacent to the furnace gases through to cell 15 on the centre symmetry plane of the slab Equation for cell 1: Equations for cell 2 – 14: Equation for cell 15: