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it take? The next example artificial) example of a spherical chicken! Example 10.3 The spherical chicken A spherical chicken of radius $a$ at initial temperature $T_0$ is placed into an oven at temperature $T_1$ at time $t = 0$ (see Figure 10.3). The boundary conditions are that the oven is at temperature $T_1$ so that $T(a, t) = T_1$, (10.30) and the chicken is originally at temperature $T_0$, so that $T(r, 0) = T_0$. (10.31) We want to obtain the temperature as a function of time at the centre of the chicken, i.e. $T(0, t)$. Solution: We will show how we can transform this to a one-dimensional diffusion equation. This is accomplished using a substitution $T(r, t) = T_1 + \frac{B(r, t)}{r}$, (10.32) where $B(r, t)$ is now a function of $r$ and $t$. This substitution is motivated by the solution to the steady-state problem in eqn 10.29 and of course means that that we can write $B$ as $B = r(T - T_1)$. We now need to work out some partial differentials: $\frac{\partial T}{\partial t}$

          it take? The next example
artificial) example of a spherical chicken!
Example 10.3
The spherical chicken
A spherical chicken of radius $a$ at initial temperature $T_0$ is placed into an oven at temperature $T_1$ at time $t = 0$ (see Figure 10.3). The boundary conditions are that the oven is at temperature $T_1$ so that
$T(a, t) = T_1$,
(10.30)
and the chicken is originally at temperature $T_0$, so that
$T(r, 0) = T_0$.
(10.31)
We want to obtain the temperature as a function of time at the centre of the chicken, i.e. $T(0, t)$.
Solution: We will show how we can transform this to a one-dimensional diffusion equation. This is accomplished using a substitution
$T(r, t) = T_1 + \frac{B(r, t)}{r}$,
(10.32)
where $B(r, t)$ is now a function of $r$ and $t$. This substitution is motivated by the solution to the steady-state problem in eqn 10.29 and of course means that that we can write $B$ as $B = r(T - T_1)$.
We now need to work out some partial differentials:
$\frac{\partial T}{\partial t}$

it take? The next example
artificial) example of a spherical chicken!
Example 10.3
The spherical chicken
A spherical chicken of radius a at initial temperature T0 is placed into an oven at temperature T1 at time t = 0 (see Figure 10.3). The boundary conditions are that the oven is at temperature T1 so that
T(a, t) = T1,
(10.30)
and the chicken is originally at temperature T0, so that
T(r, 0) = T0.
(10.31)
We want to obtain the temperature as a function of time at the centre of the chicken, i.e. T(0, t).
Solution: We will show how we can transform this to a one-dimensional diffusion equation. This is accomplished using a substitution
T(r, t) = T1 + (B(r, t))/(r),
(10.32)
where B(r, t) is now a function of r and t. This substitution is motivated by the solution to the steady-state problem in eqn 10.29 and of course means that that we can write B as B = r(T - T1).
We now need to work out some partial differentials:
(∂ T)/(∂ t)

Added by Andrea G.

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