Com base em seus conhecimentos e na equação seguinte, responda:
\frac{\partial \psi}{\partial t}+(v \cdot \nabla) \psi=\dot{\psi}_{G}+\delta(\nabla \cdot \nabla \psi)-\psi^{\prime}(\nabla \cdot v)
Reescreva a equação (1) em coordenadas esféricas para transferencia de calor.

Passo 1

O enunciado pede para reescrever, em coordenadas esféricas (r,\theta,\varphi), a equação de transporte-convecção-difusão com geração e termo dilatacional
\frac{\partial \psi}{\partial t} + (v\cdot\nabla)\psi = \dot{\psi}_{G} + \delta\,(\nabla\cdot\nabla\psi) - \psi'(\nabla\cdot v).
Para isso precisamos das fórmulas em esféricas de
  1. derivada convectiva (v\cdot\nabla)\psi,
  2. laplaciano \nabla^2\psi,
  3. divergência de v, isto é, \nabla\cdot v.

Passo 2

Em coordenadas esféricas, com velocidade v=(v_{r},v_{\theta},v_{\varphi}), a derivada convectiva é
(v\cdot\nabla)\psi = v_{r}\,\frac{\partial \psi}{\partial r} + \frac{v_{\theta}}{r}\,\frac{\partial \psi}{\partial \theta} + \frac{v_{\varphi}}{r\sin\theta}\,\frac{\partial \psi}{\partial \varphi}.

Passo 3

O laplaciano de \psi em esféricas vale
\nabla^2\psi =\frac{1}{r^{2}}\frac{\partial}{\partial r}\Bigl(r^{2}\frac{\partial \psi}{\partial r}\Bigr) +\frac{1}{r^{2}\sin\theta}\frac{\partial}{\partial \theta}\Bigl(\sin\theta\,\frac{\partial \psi}{\partial \theta}\Bigr) +\frac{1}{r^{2}\sin^{2}\theta}\,\frac{\partial^{2}\psi}{\partial \varphi^{2}}.
E a divergência de v fica
\nabla\cdot v =\frac{1}{r^{2}}\frac{\partial}{\partial r}\bigl(r^{2}v_{r}\bigr) +\frac{1}{r\sin\theta}\frac{\partial}{\partial \theta}\bigl(v_{\theta}\sin\theta\bigr) +\frac{1}{r\sin\theta}\frac{\partial v_{\varphi}}{\partial \varphi}.

Passo 4

Basta inserir esses três resultados na equação original, obtendo em esféricas
\frac{\partial \psi}{\partial t} +v_{r}\frac{\partial \psi}{\partial r} +\frac{v_{\theta}}{r}\frac{\partial \psi}{\partial \theta} +\frac{v_{\varphi}}{r\sin\theta}\frac{\partial \psi}{\partial \varphi} =\dot{\psi}_{G} +\delta\, \Biggl[ \frac{1}{r^{2}}\frac{\partial}{\partial r}\Bigl(r^{2}\frac{\partial \psi}{\partial r}\Bigr) +\frac{1}{r^{2}\sin\theta}\frac{\partial}{\partial \theta}\Bigl(\sin\theta\,\frac{\partial \psi}{\partial \theta}\Bigr) +\frac{1}{r^{2}\sin^{2}\theta}\frac{\partial^{2}\psi}{\partial \varphi^{2}} \Biggr] -\psi'\, \Biggl[ \frac{1}{r^{2}}\frac{\partial}{\partial r}(r^{2}v_{r}) +\frac{1}{r\sin\theta}\frac{\partial}{\partial \theta}(v_{\theta}\sin\theta) +\frac{1}{r\sin\theta}\frac{\partial v_{\varphi}}{\partial \varphi} \Biggr].

Resposta

\frac{\partial \psi}{\partial t} +v_{r}\frac{\partial \psi}{\partial r} +\frac{v_{\theta}}{r}\frac{\partial \psi}{\partial \theta} +\frac{v_{\varphi}}{r\sin\theta}\frac{\partial \psi}{\partial \varphi} =\dot{\psi}_{G} +\delta\, \Biggl[ \frac{1}{r^{2}}\frac{\partial}{\partial r}\Bigl(r^{2}\frac{\partial \psi}{\partial r}\Bigr) +\frac{1}{r^{2}\sin\theta}\frac{\partial}{\partial \theta}\Bigl(\sin\theta\,\frac{\partial \psi}{\partial \theta}\Bigr) +\frac{1}{r^{2}\sin^{2}\theta}\frac{\partial^{2}\psi}{\partial \varphi^{2}} \Biggr] -\psi'\, \Biggl[ \frac{1}{r^{2}}\frac{\partial}{\partial r}(r^{2}v_{r}) +\frac{1}{r\sin\theta}\frac{\partial}{\partial \theta}(v_{\theta}\sin\theta) +\frac{1}{r\sin\theta}\frac{\partial v_{\varphi}}{\partial \varphi} \Biggr].

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