Solid cylinder with internal heating.

A long cylinder is initially at zero temperature and the boundary at $r=b$ is maintained at zero temperature. Find the temperature in the cylinder resulting from spatially uniform internal energy generation, $g_0$ (W/m$^3$).

The temperature satisfies the following equations:

$$\frac{1}{r} \frac{\partial}{\partial r} \left[r \frac{\partial T}{\partial r} \right] + \frac{g_0}{k} = \frac{1}{\alpha} \frac{\partial T}{\partial t}; \; \; 0 < r < b$$ (18)

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

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

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

This case is number R01B0T0G1. The GF solution equation for the temperature contains only the internal heating term:

$$T(r,t) = \frac{\alpha}{k} \int_{\tau=0}^t \int_{r^{\prime}=0... ...d r^{\prime},\tau) \; 2\pi r^{\prime} \; d r^{\prime} \;d \tau$$ (19)

The large-time form of the GF for this case is given by

$$G_{R01}(r,t \mid r^{\prime },\tau ) = \frac{1}{\pi b^{2}} \sum_{m=1}^{\infty }\exp \left[ -\beta _{m}^{2}\alpha (t-\tau )/b^{2}\right]$$ (20)

$$\times \frac{J_{0}(\beta _{m}r/b) J_{0}(\beta _{m}r^{\prime }/b) } {\left[ J_{1}(\beta _{m}) \right] ^{2}}$$ (21)

where $J_0$ and $J_1$ are Bessel functions and the eigenvalues given by $J_{0}(\beta _{m})=0$. The two integrals in the above solution will be considered one at a time. The spatial integral over $r^{\prime}$ acts on one $J_0$ term, and it may be simplified by the substitution $x=\beta _{m}r^{\prime }/b$ and then evaluated as shown below:

$$\int_{r^{\prime}=0}^b J_{0}( \beta _{m}r^{\prime }/b ) \; 2\pi r^{\prime} dr^{\prime} = 2 \pi \frac{b^2}{\beta_m^2} \int_{x=0}^{\beta_m} J_{0}( x) \; x \; dx$$

$$= 2 \pi \frac{b^2}{\beta_m^2} \left[ x J_1(x) \right]_{x=0}^{\beta_m}$$

$$= 2 \pi \frac{b^2}{\beta_m^2} \left[ \beta_m J_1(\beta_m) -0 \right]$$

The time integral acts only on the exponential term of the series for the GF, given by

$$\int_{\tau=0}^t e^{-\beta _{m}^{2} \alpha (t-\tau )/b^2} \; ... ...= \frac{b^2}{\beta_m^2 \alpha}[1- e^{-\beta_m^2 \alpha t/b^2}]$$ (22)

Then the above two integrals can be combined with the entire GF solution to give:

$$T(r,t)=2\frac{g_0b^2}{k} \sum_{m=1}^{\infty} [1- e^{-\beta_m... ...t/b^2}] \frac{J_{0}( \beta_{m}r/b) } {\beta_m^3 J_1(\beta_m) }$$ (23)

For numerical evaluation, the steady term should be computed separately and substituted for the slowly converging series term (the term without the exponential). When the steady-state portion of the solution is substituted (see below), the temperature is given by:

$$T(r,t)= \frac{g_0b^2}{4k} \left[ 1-(r/b)^2 \right] - 2\frac{... ... t/L^2} \frac{J_0 ( \beta _{m}r/b) } {\beta_m^3 J_1(\beta_m) }$$ (24)

This form of the solution converges rapidly for $\alpha t/b^2 >0.01$.

Steady Solution. Next the steady solution for the solid sphere with internal heating will be derived. The steady temperature satisfies the following differential equation:

$$\frac{1}{r} \frac{\partial}{\partial r} \left[r \frac{\partial T}{\partial r} \right] = -\frac{g_0}{k} \; \; 0 < r < b$$ (25)

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

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

The general solution may be found by integrating the differential equation twice:

$$T(r) = \frac{g_0 r^2}{4k} + C_1 \log r + C_2$$

The constants of integration $C_1$ and $C_2$ may be determined by applying the boundary condtions. For $T$ to be bounded at $r=0$ requires $C_1=0$. At $r=b$ the boundary condition $T=0$ then determines $C_2$. Then the steady temperature is given by

$$T(r) = \frac{g_0 b^2}{4k} \left[ 1 - \frac{r^2}{b^2} \right]$$