The following approximate expression for *Φ*
_{
S
}, the heat flux entering the upper and lower surfaces of the strip, was derived in [5].

$$\begin{array}{*{20}l} \Phi_{S}&= \frac{\epsilon_{S}\sigma\left(T_{W}^{4}-T_{1}^{4}\right)}{1+\frac{\epsilon_{S}(1-\epsilon_{W})}{\epsilon_{W}}\frac{w}{p}}, \end{array} $$

(12)

where *ε*
_{
S
}≈0.2 and *ε*
_{
W
}≈0.9 are the emissivities of the strip and furnace materials respectively, *σ*=5.670×10^{−8}Wm^{−2}K^{−4} is the Stefan–Boltzmann constant and *p* is the sum of height and width of a cross-section of the space inside the furnace. The temperature of the furnace walls and heating elements is assumed to be the same and is given by *T*
_{
W
}. We note that this flux does not vary across the width of the strip. Smaller emissivities are associated with more reflective surfaces, which lead to a greater amount of reflection of radiant heat energy arriving at a surface.

*Φ*
_{
E
}, the heat flux absorbed at the edges, is expected to be greater than *Φ*
_{
S
} because the steel strips are formed by cold rolling of steel which results in a rougher, less reflective surface at the edges.

We limit our analysis to the steady state operation of the furnace. This simplifies the analysis because it allows us to approximate the heat flux *Φ*
_{
S
} using the power supplied to the heating elements. For the non-steady state operation, one needs to take into account the heat dynamics that occur near the inner surface of the furnace walls, which are coupled to the dynamics of radiant heat transfer and heat transfer within the steel strip. For steady state operation, one can simply use the fact that the furnace walls are very good insulators and neglect the heat lost through them.

We thus seek steady state solutions of Eqs. (8), (10) and (11). In order to get a closed-form expression for the solution, we assume in Section 2.1 that *ρ*
_{
S
}, *C*
_{
S
}, *k*, *Φ*
_{
S
} and *Φ*
_{
E
} are all constant. We analyse the more general case for which these quantities are not constant in Section 2.2.

### 2.1 The case of constant *ρ*
_{
S
}
*C*
_{
S
}, *k*, *Φ*
_{
S
} and *Φ*
_{
E
}

In terms of the dimensionless variables \(\tilde {x}=x/L\), \(\tilde {y}=y/w\), \(\tilde {T}=T \frac {hv \rho _{S} C_{S}}{\Phi _{S} L}\), Eqs. (8), (10) and (11) become

$$\begin{array}{*{20}l} &\frac{\partial \tilde{T}}{\partial \tilde{x}}=\delta \frac{\partial^{2} \tilde{T}}{\partial \tilde{y}^{2}}+2, \end{array} $$

(13)

$$\begin{array}{*{20}l} &\tilde{T}(0,\tilde{y})=\tilde{T}_{0}, \end{array} $$

(14)

$$\begin{array}{*{20}l} &\frac{\partial \tilde{T}}{\partial \tilde{y}}(\tilde{x}, \pm 1/2)=\pm \frac{h}{w} \frac{\Phi_{E}}{ \Phi_{S}}\frac{1}{\delta}. \end{array} $$

(15)

Here, \(\tilde {T}_{0}=T_{0} \frac {hv \rho _{S} C_{S}}{\Phi _{S} L}\) and *δ* is given by (5). In these equations, \(0<\tilde {x}<1\) and \(-1/2<\tilde {y}<1/2\).

We note that *h*/*w* and *δ* happen to be of the same order of magnitude for this industrial application, so the non-dimensional flux term in (15) is of order 1. This indicates that boundary edge heating is significant. However, *δ* is small, so we expect that the temperature of parts of the strip not close to the edges satisfies

$$\begin{array}{*{20}l} &\frac{\partial \tilde{T}_{1}}{\partial \tilde{x}}=2, \end{array} $$

(16)

$$\begin{array}{*{20}l} &\tilde{T}_{1}(0,\tilde{y})=\tilde{T}_{0}, \end{array} $$

(17)

which immediately gives

$$\tilde{T}_{1}(\tilde{x},\tilde{y})=\tilde{T}_{0}+2\tilde{x}. $$

We seek the *steady state solution* of the whole system (13)–(15), so we set \(\tilde {T}=\tilde {T}_{0}+2\tilde {x}+\tilde {T}_{2}\) and see that \(\tilde {T}_{2}\) must satisfy

$$\begin{array}{*{20}l} &\frac{\partial \tilde{T}_{2}}{\partial \tilde{x}}=\delta \frac{\partial^{2} \tilde{T}_{2}}{\partial \tilde{y}^{2}}, \end{array} $$

(18)

$$\begin{array}{*{20}l} &\tilde{T}_{2}(0,\tilde{y})=0, \end{array} $$

(19)

$$\begin{array}{*{20}l} &\frac{\partial \tilde{T}_{2}}{\partial \tilde{y}}(\tilde{x}, \pm 1/2)=\pm \frac{h}{w} \frac{\Phi_{E}}{ \Phi_{S}}\frac{1}{\delta}. \end{array} $$

(20)

We expect \(\tilde {T}_{2}\) to remain close to zero, except in boundary layers near \(\tilde {y}=\pm 1/2\), so we write \(\tilde {y}=\delta ^{1/2}\zeta -1/2\). The scale factor *δ*
^{1/2} for this *inner variable*
*ζ* is chosen so that heating near the edge \(\tilde {y}=-1/2\) is given by the equations

$$\begin{array}{*{20}l} &\frac{\partial \tilde{T}_{2}}{\partial \tilde{x}}=\frac{\partial^{2} \tilde{T}_{2}}{\partial \zeta^{2}}, \end{array} $$

(21)

$$\begin{array}{*{20}l} &\tilde{T}_{2}=0, \quad \text{for} \tilde{x}=0, \end{array} $$

(22)

$$\begin{array}{*{20}l} &\left.\frac{\partial \tilde{T}_{2}}{\partial \zeta}\right|_{\zeta=0}=-\frac{h}{w} \frac{\Phi_{E}}{ \Phi_{S}}\frac{1}{\delta^{1/2}}. \end{array} $$

(23)

Outside this boundary layer, the solution must match the *outer solution* and for this we use the simple matching condition \(\tilde {T}_{2}\to 0\) as *ζ*→*∞*. The solution is easily obtained by taking the Laplace Transform with respect to the \(\tilde {x}\) variable,

$$F(s,\zeta)=\int_{0}^{\infty} \tilde{T}_{2}(\tilde{x},\tilde{y})e^{-s\tilde{x}}\,d\tilde{x}, \quad \text{with} \tilde{y}=\delta^{1/2}\zeta-1/2. $$

Equation 21 then gives *sF*=*F*
_{
ζ
ζ
}, from which we find that

$$F(s,\zeta)=\frac{h}{w} \frac{\Phi_{E}}{ \Phi_{S}}\frac{1}{\sqrt{s^{3} \delta}}e^{-\sqrt{s} \zeta}. $$

This gives

$$ \tilde{T_{2}}= \frac{1}{\sqrt{\delta}}\frac{h}{w} \frac{\Phi_{E}}{ \Phi_{S}}\psi(\zeta,\tilde{x})=\frac{1}{\sqrt{\delta}}\frac{h}{w} \frac{\Phi_{E}}{ \Phi_{S}}\psi\left(\frac{\tilde{y}+1/2}{\sqrt{\delta}},\tilde{x}\right), $$

(24)

where

$$ \psi(\zeta,\tilde{x})= \zeta\left(\text{erf}\left(\frac{\zeta}{2\sqrt{\tilde{x}}}\right)-1\right) + 2\sqrt{\frac{\tilde{x}}{\pi}}\exp\left(\frac{-\zeta^{2}}{4\tilde{x}}\right), $$

(25)

and erf represents the error function,

$$\text{erf}(z)=\frac{2}{\sqrt{\pi}}\int_{0}^{z} e^{-s^{2}}\,ds. $$

A similar expression approximates the boundary layer near \(\tilde {y}=1/2\). One can combine the inner and outer solutions to obtain a composite approximation for the steady state solution of (13)–(15), valid for −1/2≤*y*≤1/2:

$$\tilde{T}\,=\,\tilde{T}_{0}+2\tilde{x}+\frac{1}{\sqrt{\delta}}\frac{h}{w} \frac{\Phi_{E}}{ \Phi_{S}}\left(\!\psi\!\left(\frac{\tilde{y}+1/2}{\sqrt{\delta}},\tilde{x}\right) \,+\,\psi\!\left(\!\frac{1/2-\tilde{y}}{\sqrt{\delta}},\tilde{x}\right)\right) $$

In terms of the original variables, one finds that the boundary layer penetrates to a distance

$$2\sqrt{\frac{k_{S} x}{v \rho_{S} C_{S}}} $$

and the increased temperature at the edge is

$$\Phi_{E}\sqrt{\frac{x}{\pi k_{S} v \rho_{S} C_{S}}}. $$

We use these equations to plot an example of increased strip temperature near an edge in Fig. 1. The calculations for the figure use *Φ*
_{
E
}=*Φ*
_{
S
} but in fact we expect *Φ*
_{
E
}>*Φ*
_{
S
} because the upper and lower surfaces are very smooth and are thus expected to have a lower emissivity than the edge surfaces. Thus we expect the edge temperatures to be greater than those shown in the graphs. Also used in the calculations are the values *k*
_{
S
}=50 Wm ^{−1}
*K*
^{−1}, *C*
_{
S
}=500 JK ^{−1}
*Kg*
^{−1}, *T*
_{0}=573 K, *ρ*
_{
S
}=7854 Kgm ^{−3}, *h*=0.5 mm, *w*=0.5 m, *v*=2 ms ^{−1}, *L*=100 m, *Φ*
_{
E
}=*Φ*
_{
S
}=1500 Wm ^{−2}.

### 2.2 The case of variable *ρ*
_{
S
}, *C*
_{
S
}, *k*, *Φ*
_{
S
} and *Φ*
_{
E
}

In this section, we wish to follow the analysis of Section 2.1, but now in the more realistic setting of variable *ρ*
_{
S
}, *C*
_{
S
}, *k*, *Φ*
_{
S
} and *Φ*
_{
E
}. In practice, the large temperature variation within annealing furnaces requires that we take into account the temperature variation, especially of *C*
_{
S
} and *k*
_{
S
}, and to a lesser extent, *ρ*
_{
S
}. The variation of *C*
_{
S
} and *k*
_{
S
} with temperature, shown in Figs. 2 and 3, is taken from data in [4]. Figure 3 shows that heat conductivity is approximated well using linear regression:

$$ k_{S}=73.9823-0.0437 T, $$

(26)

and Fig. 2 shows the *C*
_{
p
} data approximated by an interpolating quartic:

$$ \begin{aligned} C_{S} &=345 - 0.504333T + 0.004895T^{2} \\ &- 9.06667{\times10}^{-6}T^{3} + 5.5{\times10}^{-9}T^{4}. \end{aligned} $$

(27)

To allow for such variations, we assume that *ρ*
_{
S
}, *C*
_{
S
} and *k*
_{
S
} are known functions of the strip’s temperature. Further, because our system is at equilibrium, *Φ*
_{
S
} and *Φ*
_{
E
} are assumed to be known functions of *x* which can be calculated by measuring the power supplied to the heating elements in the vicinity of a distance *x* along the furnace.

The form of Eq. (1) is only valid for constant diffusivity *k*
_{
S
}. With variable *k*
_{
S
}, we must instead write

$$ \rho_{S} C_{S}\left(\frac{\partial u}{\partial t}+v \frac{\partial u}{\partial x}\right)=\nabla \cdot (k_{S} \nabla u). \quad \text{in} \mathcal{S}\times(0,\infty). $$

(28)

Consequently, instead of (8), we have

$$ \rho_{S} C_{S}\left(\frac{\partial T}{\partial t}+v \frac{\partial T}{\partial x}\right) = \frac{\partial}{\partial y}\left(k_{S}\frac{\partial T}{\partial y}\right)+\frac{2\Phi_{S}}{h}. $$

(29)

Hence the steady state temperature must satisfy

$$\begin{array}{*{20}l} &\rho_{S} C_{S} v \frac{\partial T}{\partial x} = k_{S}(T)\frac{\partial^{2} T}{\partial y^{2}}+k_{S}'(T)\left(\frac{\partial T}{\partial y}\right)^{2}+\frac{2\Phi_{S}}{h}, \end{array} $$

(30)

$$\begin{array}{*{20}l} &T(0,y)=T_{0}, \quad -w/2<y<w/2, \end{array} $$

(31)

$$\begin{array}{*{20}l} &\pm k_{S} \frac{\partial T}{\partial y}(x,\pm w/2)=\Phi_{E}, \quad 0<x<L. \end{array} $$

(32)

As before, the small diffusivity indicates that for *y* not close to ±*w*/2, *T*(*x,y*)≈*T*
_{1}(*x*), where *T*
_{1} satisfies the ordinary differential equation

$$ \rho_{S}(T_{1})C_{S}(T_{1})v T_{1}'(x)=\frac{2\Phi_{S}(x)}{h},\quad T_{1}(0)=T_{0}. $$

(33)

As before, it is useful to consider a new dimensionless variable \(\tilde {x}\), this time chosen to make the coefficients of Eq. (32) more similar to those of the constant coefficient case. We do this by choosing \(\tilde {x}\) to be the solution to

$$L\frac{d\tilde{x}}{dx}= \frac{k_{S}(T_{1})\rho_{S}(T_{0})C_{S}(T_{0})}{k_{S}(T_{0})\rho_{S}(T_{1})C_{S}(T_{1})}, \quad \tilde{x}(0)=0, $$

where *T*
_{1}=*T*
_{1}(*x*). We also let \(\tilde {y}=y/w\).

In terms of these variables, Eq. (30) takes the form

$$ \begin{aligned} \frac{\!k_{S}(\!T_{1}\!)\!\rho_{S}\!(\!T)C_{S}(T)}{k_{S}(\!T)\!\rho_{S}(\!T_{1})C_{S}(\!T_{1})}\frac{\partial T}{\!\partial \tilde{x}}\,=\,\delta\frac{\partial^{2} T}{\partial \tilde{y}^{2}}\,+\,\delta \frac{k_{S}'(T)}{k_{S}(T)}\!\left(\!\frac{\partial T}{\partial \tilde{y}}\!\right)^{2}\!\,+\,\frac{2\Phi_{S} k_{S}(\!T_{0}\!)L}{h v \rho_{S}(\!T_{0})C_{S}(\!T_{0})k_{S}(\!T)}, \end{aligned} $$

(34)

where *δ* is again given by Eq. (5), but with *k*
_{
S
}, *ρ*
_{
S
} and *C*
_{
S
} evaluated at temperature *T*
_{0}. Finally, we choose a dimensionless temperature

$$\tilde{T}=T\frac{h v \rho_{S}(T_{0})C_{S}(T_{0})}{\overline{\Phi}_{S} L}, $$

where \(\overline {\Phi }_{S}\) is the average of *Φ*
_{
S
},

$$\overline{\Phi}_{S}=\frac{1}{L}\int_{0}^{L} \Phi_{S}(x)\,dx. $$

Equation 34 becomes

$$ \begin{aligned} \frac{k_{S}(T_{1})\rho_{S}(T)C_{S}(T)}{k_{S}(T_{0})\rho_{S}(T_{1})C_{S}(T_{1})}\frac{\partial \tilde{T}}{\partial \tilde{x}}&= \\ \frac{k_{S}(T)}{k_{S}(T_{0})}\delta\frac{\partial^{2} \tilde{T}}{\partial \tilde{y}^{2}}&+\delta \frac{\overline{\Phi}_{S} L}{h v \rho_{S}(T_{0})C_{S}(T_{0}\!)} \frac{k_{S}'(T)}{k_{S}(T_{0})}\!\!\left(\!\frac{\partial \tilde{T}}{\partial \tilde{y}}\!\right)^{2}\!\,+\,\frac{2\Phi_{S}}{\overline{\Phi}_{S} }\!. \end{aligned} $$

(35)

The dimensionless solution of Eq. (35) corresponding to the solution *T*
_{1} of Eq. (33) is given by

$$\tilde{T}_{1}(\tilde{x})=\frac{T_{1}(x)}{\overline{T}}, \quad \text{where} \quad \overline{T}=\frac{\overline{\Phi}_{S} L}{h v \rho_{S}(T_{0})C_{S}(T_{0})}, $$

and this corresponds to a solution outside boundary layers.

As in Section 2.1, we set \(\tilde {y}=\delta ^{1/2}\zeta -1/2\); *ζ* is our boundary layer variable near the edge \(\tilde {y}=-1/2\). We also write \(\tilde {T}=\tilde {T}_{1}(\tilde {x})+\tilde {T}_{2}(\tilde {x},\zeta)\). Rewriting the boundary condition (32) at this edge in terms of the new variables gives

$$ \left.\frac{\partial \tilde{T}_{2}}{\partial \zeta}\right|_{\zeta=0}=-\frac{h}{w} \frac{\Phi_{E} k_{S}(T_{0})}{ \overline{\Phi}_{S} k_{S}(T)}\frac{1}{\delta^{1/2}}. $$

(36)

It is very desirable for the industrial application that edge heating is very small. This is consistent with our observation that the physical parameters happen to be such that *h*/*w*=*O*(*δ*), and thus the right-hand-side of (36) is *O*(*δ*
^{1/2}). In any case, we assume that

$$\frac{h}{w}\delta^{-1/2}=\epsilon $$

is small and we write \(\tilde {T}_{2}(\tilde {x},\zeta)=\epsilon \theta (\tilde {x},\zeta)+O(\epsilon ^{2})\). This allows us to use a first order Taylor approximation to *ρ*
_{
S
}(*T*), expanded about the point *T*=*T*
_{1},

$$\rho_{S}(T)=\rho_{S}(T_{1})+\epsilon \rho_{S}'(T_{1})\overline{T}\theta(\tilde{x},\zeta)+O(\epsilon^{2}). $$

We use similar approximations for *C*
_{
S
}(*T*) and *k*
_{
S
}(*T*).

Recalling that \(\partial /\partial \tilde {y}=\delta ^{-1/2} \partial /\partial \zeta \), we expand Eq. (35) up to *O*(*ε*) to find that *θ* must satisfy

$$ \left(\frac{\rho_{S}'(T_{1})}{\rho_{S}(T_{1})}+\frac{C_{S}'(T_{1})}{C_{S}(T_{1})}\right)\frac{\partial T_{1}}{\partial \tilde{x}}\theta+\frac{\partial \theta}{\partial \tilde{x}}=\frac{\partial^{2} \theta}{\partial \zeta^{2}}. $$

(37)

Equation 37 may be simplified by setting

$$ \chi=\frac{\rho_{S}(T_{1})C_{S}(T_{1})}{\rho_{S}(T_{0})C_{S}(T_{0})}\theta, $$

(38)

and we find that *χ* satisfies

$$ \frac{\partial \chi}{\partial \tilde{x}}=\frac{\partial^{2} \chi}{\partial \zeta^{2}}. $$

(39)

The flux boundary condition (36) translates to

$$ \left.\frac{\partial \chi}{\partial \zeta}\right|_{\zeta=0}=-\frac{\Phi_{E}}{\overline{\Phi}_{S}}\frac{k_{S}(T_{0})\rho_{S}(T_{1})C_{S}(T_{1})}{k_{S}(T_{1})\rho_{S}(T_{0})C_{S}(T_{0})}=f(\tilde{x}). $$

(40)

*χ* must also satisfy an “initial” condition, *χ*(0,*ζ*)=0, and a matching condition, \(\chi (\tilde {x},\zeta)\to 0\) as *ζ*→*∞*.

The solution of this system, readily found by use of the Laplace transform, is

$$ \chi(\tilde{x},\zeta)=\int_{0}^{\tilde{x}} g(\zeta,\sigma)f(\tilde{x}-\sigma)\,d\sigma, $$

(41)

where \(g(\zeta,\tilde {x})=\frac {\partial \psi }{\partial \tilde {x}}=e^{-\zeta ^{2}/4\tilde {x}}/\sqrt {\pi \tilde {x}}\) and *ψ* is given by Eq. (25).

In summary, we have found that there is a boundary layer near the edges of the strip. Outside the boundary layer, the temperature *T*
_{1} of the strip, at a distance *x* along the furnace, may be found by solving the ordinary differential Eq. (33). With *T*
_{1}(*x*), we may then calculate \(f(\tilde {x})\) from (40) and then *χ* from (41). This gives us *θ* from (38). The actual perturbation to the temperature near the edge *y*=−*w*/2 is given by

$$T_{2}=\overline{T}\tilde{T_{2}}=\overline{T}\epsilon \theta(\tilde{x},\zeta). $$