Turunkan persamaan difusi panas, Persamaan 2.29, untuk koordinat bola dengan memulai dari volume kendali diferensial yang ditunjukkan pada Gambar 2.13.
$$\frac{1}{r^2} \frac{\partial}{\partial r} \left(kr^2 \frac{\partial T}{\partial r} \right) + \frac{1}{r^2 \sin^2 \theta} \frac{\partial}{\partial \phi} \left(k \frac{\partial T}{\partial \phi} \right) + \frac{1}{r^2 \sin^2 \theta} \frac{\partial}{\partial \theta} \left(k \sin \theta \frac{\partial T}{\partial \theta} \right) + \dot{q} = \rho c_p \frac{\partial T}{\partial t}$$
Definisi Geometri Volume Kendali Diferensial
Tinjau volume kendali diferensial dalam koordinat silindris dengan dimensi:
- Arah radial ($r$): panjang $dr$
- Arah polar/zenit ($\theta$): panjang busur $r \ d \theta$
- Arah azimutal ($\phi$): panjang busur $r \sin \theta \ d \phi$
Maka volume kendali ($dV$) adalah:
$$dV = (dr)(r \ d \theta)(r \sin \theta \ d \phi)$$
$$dV = r^2 dr \ \sin \theta d \theta \ d \phi$$
Luas penampang diferensial untuk masing-masing permukaan adalah:
- $A_r = r^2 \ \sin \theta d \theta \ d \phi$ (luas permukaan tegak lurus sumbu-$r$
- $A_\theta = r dr \ \sin \theta \ d \phi$ (luas permukaan tegak lurus sumbu-$\theta$)
- $A_\phi = r dr \ d \theta$ (luas permukaan tegak lurus sumbu-$\phi$)
Prinsip Konservasi Energi
Berdasarkan Hukum Pertama Termodinamika untuk volume kendali:
$$\dot{E}_{in} – \dot{E}_{out} + \dot{E}_{gen} = \dot{E}_{st}$$
- Laju Energi Masuk ($\dot{E}_{in}$):
$$\dot{E}_{in} = q_r + q_\theta + q_\phi$$
- Laju Energi Keluar ($\dot{E}_{out}$):
$$\dot{E}_{out} = q_{r + dr} + q_{\theta + d \theta} + q_{\phi + d \phi}$$
- Laju Pembangkitan Energi ($\dot{E}_{gen}$):
$$\dot{E}_{gen} = \dot{q} \cdot dV$$
$$\dot{E}_{gen} = \dot{q} (r^2 dr \ \sin \theta d \theta \ d \phi)$$
- Laju Penyimpanan Energi ($\dot{E}_{st}$):
$$\dot{E}_{st} = \rho \cdot dV \cdot c_p \frac{\partial T}{\partial t}$$
$$\dot{E}_{st} = \rho c_p (r^2 dr \ \sin \theta d \theta \ d \phi) \frac{\partial T}{\partial t}$$
Ekspansi Deret Taylor untuk Laju Perpindahan Panas
Menggunakan ekspansi deret Taylor (dengan mengabaikan orde yang lebih tinggi), laju perpindahan panas yang keluar dari permukaan dapat dinyatakan sebagai:
- Arah Radial ($r$):
$$q_{r + dr} = q_r + \frac{\partial q_r}{\partial r} dr$$
$$q_r – q_{r + dr} = -\frac{\partial q_r}{\partial r} dr$$
- Arah Polar ($\theta$):
$$q_{z + d \theta} = q_\theta + \frac{\partial q_\theta}{\partial \theta} d \theta$$
$$q_\theta – q_{\theta + d \theta} = -\frac{\partial q_\theta}{\partial \theta} d\theta$$
- Arah Azimutal ($\phi$):
$$q_{\phi + d\phi} = q_\phi + \frac{\partial q_\phi}{\partial \phi} d \phi$$
$$q_\phi – q_{\phi + d \phi} = -\frac{\partial q_\phi}{\partial \phi} d \phi$$
Penerapan Hukum Fourier
Berdasarkan Hukum Fourier, laju konduksi panas pada masing-masing arah adalah:
- Pada arah $r$:
$$q_r = -k A_r \frac{\partial T}{\partial r} = -k (r^2 dr \ \sin \theta d \theta \ d \phi) \frac{\partial T}{\partial r}$$
- Sehingga:
$$q_r – q_{r + dr} = -\frac{\partial}{\partial r} \left[-k (r^2 \ \sin \theta d \theta \ d \phi) \frac{\partial T}{\partial r} \right] dr = \sin \theta \frac{\partial}{\partial r} \left(kr^2 \frac{\partial T}{\partial r} \right) dr \ d \theta \ d \phi$$
- Pada arah $\theta$:
$$q_\theta = -k A_\theta \frac{\partial T}{r \partial \theta} = -k (r dr \ \sin \theta \ d \phi) \frac{1}{r} \frac{\partial T}{\partial \theta}$$
- Sehingga:
$$q_\theta – q_{\theta + d \theta} = -\frac{\partial}{\partial \theta} \left[-k (r dr \ \sin \theta \ d \phi) \frac{1}{r} \frac{\partial T}{\partial \theta} \right] d \theta = \frac{\partial}{\partial \theta} \left( k \sin \theta \frac{\partial T}{\partial \theta} \right) dr \ d \theta \ d \phi$$
- Pada arah $\phi$:
$$q_\phi = -k A_\phi \frac{\partial T}{r \sin \theta \partial \phi} = -k (r dr \ d \theta) \frac{1}{r \sin \theta} \frac{\partial T}{\partial \phi}$$
- Sehingga:
$$q_\phi – q_{\phi + d\phi} = -\frac{\partial}{\partial \phi} \left[-k (r dr \ d \theta) \frac{1}{r \sin \theta} \frac{\partial T}{\partial \phi} \right] d \phi = \frac{1}{\sin \theta} \frac{\partial}{\partial \phi} \left( k \frac{\partial T}{\partial \phi} \right) dr \ d \theta \ d \phi$$
Substitusi ke Neraca Energi Total
Substitusikan seluruh komponen ke dalam persamaan neraca energi $(\dot{E}_{in} – \dot{E}_{out}) + \dot{E}_{gen} = \dot{E}_{st}$:
$$\sin \theta \frac{\partial}{\partial r} \left(kr^2 \frac{\partial T}{\partial r} \right) dr \ d \theta \ d \phi + \frac{\partial}{\partial \theta} \left( k \sin \theta \frac{\partial T}{\partial \phi} \right) dr \ d \theta \ d \phi + \frac{1}{\sin \theta} \frac{\partial}{\partial \phi} \left( k \frac{\partial T}{\partial \phi} \right) dr \ d \theta \ d \phi + \dot{q} (r^2 dr \ \sin \theta d \theta \ d \phi) = \rho c_p (r^2 dr \ \sin \theta d \theta \ d \phi) \frac{\partial T}{\partial t}$$
Bagi seluruh ruas dengan volume diferensial $dV = r^2 dr \ \sin \theta d \theta \ d \phi$:
$$\frac{1}{r^2} \frac{\partial}{\partial r} \left(kr^2 \frac{\partial T}{\partial r} \right) + \frac{1}{r^2 \sin \theta} \frac{\partial}{\partial \theta} \left( k \sin \theta \frac{\partial T}{\partial \phi} \right) + \frac{1}{r^2 \sin^2 \theta} \frac{\partial}{\partial \phi} \left( k \frac{\partial T}{\partial \phi} \right) + \dot{q} = \rho c_p \frac{\partial T}{\partial t}$$
