a) At the interface, one boundary condition is that the temperatures of both materials are equal,
T_1 (0, t) = T_2 (0, t)
The other boundary condition is that the heat current densities j_{Q1} = −κ_1 ∂_x T_1 \hat{x} and j_{Q2} = −κ_2 ∂_x T_2 \hat{x} are equal as well,
κ_1 \frac{∂T_1 (0, t)}{∂x} = κ_2 \frac{∂T_2 (0, t)}{∂x}
where \hat{x} is the unit vector along the x-axis.
b) The heat diffusion equation in block 1 reads,
\frac{∂T_1 (x, t)}{∂t} = λ_1 \frac{∂^2T_1 (x, t)}{∂x^2}
The partial derivatives of the function T_1 (x, t) have to be recast in terms of the partial derivatives of the function f_1 (η_1) where η_1 (x, t) = x_2/λ_1t . The partial derivatives of the function η_1 (x, t) = x_2/λ_1t are,
\frac{∂η_1}{∂t} = – \frac{x^2}{λ_1t^2} = – \frac{η_1}{t} and \frac{∂η_1}{∂x} = \frac{2x}{λ_1t} = \frac{2η_1}{x}
Since f_1 (η_1) = T_1 (x, t) , the first-order partial derivatives of the function T_1 (x, t) are recast in terms of the first-order derivative of the function f_1 (η_1) as,
\frac{\partial T_1}{\partial t} = \frac{df_1}{d\eta _1}\frac{\partial \eta _1}{\partial t} = -\frac{x^2}{\lambda _1t^2} \frac{df_1}{d\eta _1} = – \frac{\eta _1}{t} \frac{df_1}{d\eta _1}
\frac{\partial T_1}{\partial x} = \frac{df_1}{d\eta _1}\frac{\partial \eta _1}{\partial x} = \frac{2x}{\lambda _1t} \frac{df_1}{d\eta _1} = \frac{2\eta _1}{x} \frac{df_1}{d\eta _1}
The second-order partial derivative of the function T_1 (x, t) is recast in terms of the second-order derivative of the function f_1 (η_1) as,
\frac{\partial^2 T_1}{\partial x^2} = \frac{\partial}{\partial x}\biggl(\frac{2x}{\lambda _1t}\frac{df_1}{d\eta _1} \biggr) = \frac{2}{\lambda _1t} \frac{df_1}{d\eta _1} + \frac{2x}{\lambda _1t} \frac{d^2f_1}{d\eta ^2_1} \frac{\partial \eta _1}{\partial x} = \frac{2\eta _1}{x^2} \frac{df_1}{d\eta _1} +\frac{4\eta ^2_1}{x^2} \frac{d^2f_1}{d\eta ^2_1}
Thus, the heat diffusion equation becomes,
-\frac{\eta _1}{t} \frac{df_1}{d\eta _1} = \frac{2\lambda _1\eta _1}{x^2} \frac{df_1}{d\eta _1} + \frac{4\lambda _1\eta ^2_1}{x^2} \frac{d^f_1}{d\eta ^2_1}
Using the definition of the dimensionless function η_1, this differential equation is recast as,
4\eta _1\frac{d^2f_1}{d\eta ^2_1} + (\eta _1+2)\frac{df_1}{d\eta _1} =0
Here, we introduce a function g_1 (η_1) as the derivative of f_1 (η_1) with respect to η_1,
g_1 (η_1) = \frac{df_1}{dη_1}
Thus, the differential equation becomes,
4 η_1 \frac{dg_1 (η_1)}{dη_1}+ (η_1 + 2) g_1 (η_1) = 0
which is recast as,
\frac{dg_1(\eta _1)}{g_1(\eta _1)} = – \Bigl(\frac{1}{4}+\frac{1}{2\eta _1} \Bigr)d\eta _1
When integrating from η_0 to η_1 we find,
\int_{g_0}^{g_1(\eta _1)}{\frac{d\acute{g}_1(\acute{\eta }_1 ) }{\acute{g}_1(\acute{\eta }_1 ) } } =- \int_{\eta _0}^{\eta _1}{\Bigl(\frac{1}{4}+ \frac{1}{2\acute{\eta }_1 } \Bigr) }d\acute{\eta } _1
where g_0 = g_1 (η_0) . The solution is,
\ln \Bigl(\frac{g_1(\eta _1)}{g_0}\Bigr) =-\frac{1}{2} \ln \bigl(\frac{\eta _1}{\eta _2}\bigr) -\frac{1}{4} (\eta _1-\eta _0)
which is recast as,
\ln \Biggl(\frac{g_1(\eta _1)\eta ^{1/2}_1}{g_0 \eta ^{1/2}_0}\Biggr) = -\frac{1}{4}(\eta _1-\eta _0)
and implies that,
g_1 (η_1) = \frac{A_1}{η^{1/2}}\exp \bigl(-\frac{\eta _1}{4}\bigr)
where the constant A_1 = g_0 η^{1/2}_0 \exp (η_0/4) . When integrating from η_0 to η_1 we find,
f_1(\eta _1)= \int_{\eta _0}^{\eta _1}{\acute{g}_1(\acute{\eta }_1 )d\acute{\eta }_1 } = A_1\int_{\eta _0}^{\eta _1}{\frac{1}{\acute{\eta }^{1/2}_1 } } \exp\Bigl(-\frac{\acute{\eta }_1 }{4}\Bigr)d\acute{\eta }_1
Using the change of variable,
\nu _1 = \frac{\eta ^{1/2}_1}{2} = \frac{x}{2\sqrt{\lambda _1t} } then d\nu _1=\frac{d\eta _1}{4\eta ^{1/2}_1}
and taking into account the definition h_1 (ν_1) = f_1 (η_1) , we can recast the solution as,
h_1 (ν_1) = B_1 \int_{\nu _0}^{\nu _1}{\exp (-\acute{\nu }^2_1 )} d\acute{\nu _1}
where the constant B_1 = 4 A_1 , which can be rewritten as,
h_1 (ν_1) =C_1 +B_1 \int_{\nu _0}^{\nu _1}{\exp (-\acute{\nu }^2_1 )} d\acute{\nu _1}
where the constant C_1 is given by,
C_1 = -\int_{\nu _0}^{\nu _1}{\exp (-\acute{\nu }^2_1 )} d\acute{\nu _1}
The error function erf (ν), defined as,
erf (ν) = \frac{2}{\sqrt{\pi } } \int_{0}^{\nu }{\exp (-s^2)} ds
is an odd function, i.e. erf (−ν) = −erf (ν) such that erf (0) = 0 and erf (∞) = 1. The derivative of the error function erf (ν) is given by,
\frac{d erf (ν)}{dν} =\frac{2}{\sqrt{\pi } } \exp(−ν^2)
Thus, using the error function we find,
h_1 (ν_1) = C_1 + D_1 erf (ν_1)
where D_1 = (\sqrt{\pi }/2) B_1 . Since h_1 (ν_1) = T_1 (x, t) and ν_1 = x/2\sqrt{λ_1t } , we have,
T_1 (x, t) = C_1 + D_1 erf\Bigl(\frac{x}{2\sqrt{\lambda _1t} }\Bigr) where x ≤ 0
Likewise, we obtain the temperature profile in block 2,
T_2 (x, t) = C_2 + D_2 erf\Bigl(\frac{x}{2\sqrt{\lambda _2t} }\Bigr) where x ≥ 0
c) At the interface, i.e. x = 0, the first boundary condition, T_1 (0, t) = T_2 (0, t) = T_0 , is satisfied provided that,
C_1 = C_2 = T_0
The blocks are long enough so that the temperatures at the end of each block is, at all time, equal to its initial temperatures. This is written as,
T_1 = T_1 (−∞, t) = T_0 + D_1 erf (−∞) = T_0 − D_1
T_2 = T_2 (∞, t) = T_0 + D_2 erf (∞) = T_0 + D_2
Hence, the temperature profiles can be recast as,
T_1 (x, t) = T_0 + (T_0 − T_1) erf \Bigl(\frac{x}{2\sqrt{\lambda _1t} }\Bigr) and x ≤ 0
T_2 (x, t) = T_0 + (T_2 − T_0) erf \Bigl(\frac{x}{2\sqrt{\lambda _2t} }\Bigr) and x ≥ 0
The spatial derivative of the temperature profiles are given by,
\frac{∂T_1 (0, t)}{∂x} = (T_0-T_1)\frac{d}{d\nu } \bigl(erf(\nu )\bigr)\mid _{\nu =0}\frac{\partial}{\partial x} \Bigl(\frac{x}{2\sqrt{\lambda _1t} }\Bigr) \mid _{x=0} = \frac{T_0-T_1}{\sqrt{\pi \lambda _1t} }
\frac{∂T_2 (0, t)}{∂x} = (T_2-T_0)\frac{d}{d\nu } \bigl(erf(\nu )\bigr)\mid _{\nu =0}\frac{\partial}{\partial x} \Bigl(\frac{x}{2\sqrt{\lambda _2t} }\Bigr) \mid _{x=0} = \frac{T_2-T_0}{\sqrt{\pi \lambda _2t} }
The second boundary condition, i.e. κ_1 ∂_x T_1 (0, t) = κ_2 ∂_x T_2 (0, t) , is recast as,
\kappa _1\frac{T_0-T_1}{\sqrt{\pi \lambda _1t} } = \kappa _2\frac{t_2-T_0}{\sqrt{\pi \lambda _2t} }
It has to hold for all times t after the interface has reached temperature T_0 . According to relation (11.36),
λ ≡ \frac{κ}{c_e} (11.36)
λ_1= \frac{κ_1}{c_1} and λ_2= \frac{κ_2}{c_2}
Thus,
\frac{T_1-T_0}{T_0-T_2} =\frac{\kappa _2}{\kappa _1} \sqrt{\frac{\lambda _1}{\lambda _2} } = \frac{\kappa _2}{\kappa _1}\sqrt{\frac{\kappa _1c_2}{\kappa _2c_1} } = \sqrt{\frac{\kappa _2c_2}{\kappa _1c_1} } = \frac{E_2}{E_1}
which implies that,
T_0 = \frac{E_1 T_1 + E_2 T_2}{E_1 + E_2}