Solid sphere with convection.

Find the temperature in a solid sphere of radius $b$, initially at elevated temperature $T_0$, that is suddenly immersed in a fluid at constant temperature $T_{\infty}<T_0$. The heat transfer coefficient for the process is a constant value, $h$. The temperature satisfies the following equations:

$$\frac{1}{r^2} \frac{\partial}{\partial r} \left[r^2 \frac{\partial T} {\partial r} \right] = \frac{1}{\alpha} \frac{\partial T}{\partial t}; \; \; 0 < r < b$$ (26)

$$\left[ -k \frac{\partial T}{\partial r} \right]_{r=b} = h[T\mid_{r=b} - T_{\infty}]$$

$$T(r=0,t) \mbox{is bounded.}$$

$$T(r,t=0) = T_0$$

As stated, this is case RS03B1T1. Note that the convection boundary condition provides that the heat flux at $r=b$ will be positive for $T(r=b) - T_{\infty}>0$. There are two driving terms, however one of them can be made homogeneous by suitable choice of a normalized temperature. (Generally it is better to zero out boundary conditions in favor of initial conditions but the purpose of this example is to demonstrate the convection boundary term.) Let $\theta(r,t) = T(r,t)-T_0$. Then the differential equation is unchanged, the initial condition is $\theta(r,t=0) = 0$, and the boundary condition may be written in standard form as

$$\left. k \frac{\partial \theta}{\partial r} \right\vert _{r=b}+ h\theta \mid_{r=b} = h \theta_{\infty} \nonumber$$

Where $\theta_{\infty} = T_{\infty}-T_0$. The temperature is given by the boundary-heating term of the GF solution equation in the form

$$T(r,t) - T_0 = \frac{\alpha}{k} \int_{\tau=0}^t h (T_{\inft... ...; G_{RS03}(r,t \mid r^{\prime}=b,\tau) \; 4 \pi b^2 \; d \tau$$ (27)

The large-time GF for this case is given by:

$$G_{RS03}(r,t \mid r^{\prime },\tau ) = \frac{1}{2\pi b r r^{\prime}} \sum_{m=1}^{\infty}\exp \left[ -\beta _{m}^{2}\alpha (t-\tau )/b^{2}\right]$$ (28)

$$\times \frac{ \beta_m^2+B^2 }{\beta_m^2+B^2 +B} \sin(\beta_m r/b) \sin(\beta_m r^{\prime}/b)$$

where the eigenvalues are found from $\beta_m \cot \beta_m = -B$ and where $B=(hb/k-1)$. The time integral is easily evaluated to give

$$T(r,t) - T_0 = \frac{2h(T_{\infty}-T_0)b^2 }{kr} \sum_{m=1}^{\infty} [1-e^{-\beta _{m}^{2}\alpha t/b^{2}}]$$ (29)

$$\times \frac{ \beta_m^2+B^2 }{\beta_m^2(\beta_m^2+B^2 +B)} \sin(\beta_m r/b) \sin(\beta_m)$$

For numerical evaluation the steady-state term of the series should be replaced by its constant value. As $t \rightarrow \infty$, the sphere takes on the fluid temperature. That is, $\theta(r \rightarrow \infty,t) \rightarrow (T_\infty - T_0)$. Substitute this constant value in place of the steady-state portion of the above series to find the following result:

$$T(r,t) = T_{\infty} + \frac{2h(T_0 - T_{\infty})b^2 }{kr} \sum_{m=1}^{\infty} e^{-\beta _{m}^{2}\alpha t/b^{2}}$$ (30)

$$\times \frac{ \beta_m^2+B^2 }{\beta_m^2(\beta_m^2+B^2 +B)} \sin(\beta_m r/b) \sin(\beta_m)$$

Note the sign change.