Dimulai dengan volume kendali diferensial berbentuk kulit bola, turunkan persamaan difusi panas untuk sistem koordinat radial bola satu dimensi dengan pembangkitan panas internal. Bandingkan hasil Anda dengan Persamaan 2.29.
$$\frac{1}{r^2} \frac{\partial}{\partial r} \left(kr^2 \frac{\partial T}{\partial r} \right) + \frac{1}{r^2 \sin^2 \theta} \frac{\partial}{\partial \phi} \left(k \frac{\partial T}{\partial \phi} \right) + \frac{1}{r^2 \sin^2 \theta} \frac{\partial}{\partial \theta} \left(k \sin \theta \frac{\partial T}{\partial \theta} \right) + \dot{q} = \rho c_p \frac{\partial T}{\partial t}$$
Analisis Geometri Bangun Ruang
Tinjau sebuah volume kendali diferensial berbentuk kulit bola konsentris (spherical shell) dengan parameter sebagai berikut:
- Jari-jari dalam: $r$
- Ketebalan radial: $dr$
- Jari-jari luar: $r + dr$
Kita hitung luas penampang dalam
$$A_{r} = 4 \pi (r^2)$$
$$A_{r} = 4 \pi r^2$$
Kita hitung luas penampang luar
$$A_{r + dr} = 4 \pi (r + dr)^2$$
$$A_{r + dr} = 4 \pi (r^2 + 2rdr + dr^2)$$
$$A_{r + dr} = 4 \pi r^2 + 8 \pi rdr + 4 \pi dr^2$$
Kita hitung volume kendali diferensial
$$dV = A_{r} \cdot dr$$
$$dV = 4 \pi r^2 \cdot dr$$
$$dV = 4 \pi r^2 dr$$
Perhitungan Semua Energi
Kita hitung energi yang masuk
$$\dot{E}_{in} = q_r$$
$$\dot{E}_{in} = -k A_{r} \frac{\partial T}{\partial r}$$
$$\dot{E}_{in} = -k (4 \pi r^2) \frac{\partial T}{\partial r}$$
Kita hitung energi yang keluar
$$\dot{E}_{out} = q_{r + dr}$$
$$\dot{E}_{out} = -k A_{r} \frac{\partial T}{\partial r} + \frac{\partial}{\partial r} \left(-k A_{r} \frac{\partial T}{\partial r} \right) dr$$
$$\dot{E}_{out} = -k (4 \pi r^2) \frac{\partial T}{\partial r} + \frac{\partial}{\partial r} \left(-k (4 \pi r^2) \frac{\partial T}{\partial r} \right) dr$$
Kita hitung energi yang dibangkitkan
$$\dot{E}_{gen} = \dot{q} \cdot dV$$
$$\dot{E}_{gen} = \dot{q} (4 \pi r^2 dr)$$
Kita hitung energi yang tersimpan
$$\dot{E}_{st} = \rho \cdot c_p \cdot dV \cdot \frac{\partial T}{\partial t}$$
$$\dot{E}_{st} = \rho \cdot c_p \cdot (4 \pi r^2 dr) \cdot \frac{\partial T}{\partial t}$$
Penerapan Hukum Fourier
Berdasarkan Hukum Pertama Termodinamika untuk kondisi transien:
$$\dot{E}_{in} – \dot{E}_{out} + \dot{E}_{gen} = \dot{E}_{st}$$
$$-k (4 \pi r^2) \frac{\partial T}{\partial r} – \left(-k (4 \pi r^2) \frac{\partial T}{\partial r} + \frac{\partial}{\partial r} \left(-k (4 \pi r^2) \frac{\partial T}{\partial r} \right) dr \right) + \dot{q} (4 \pi r^2 dr) = \rho c_p (4 \pi r^2 dr) \frac{\partial T}{\partial t}$$
$$\frac{\partial}{\partial r} \left(k (4 \pi r^2) \frac{\partial T}{\partial r} \right) dr + \dot{q} (4 \pi r^2 dr) = \rho c_p (4 \pi r^2 dr) \frac{\partial T}{\partial t}$$
$$\frac{\partial}{\partial r} \left(kr^2 \frac{\partial T}{\partial r} \right) (4 \pi dr) + \dot{q} (4 \pi r^2 dr) = \rho c_p \frac{\partial T}{\partial t} (4 \pi r^2 dr)$$
$$\color{blue} \frac{1}{r^2} \frac{\partial}{\partial r} \left(kr^2 \frac{\partial T}{\partial r} \right) + \dot{q} = \rho c_p \frac{\partial T}{\partial t}$$
Persamaan umum 1D dalam koordinat bola (Persamaan 2.29):
$$\color{blue} \frac{1}{r^2} \frac{\partial}{\partial r} \left(kr^2 \frac{\partial T}{\partial r} \right) \color{black} + \frac{1}{r^2 \sin^2 \theta} \frac{\partial}{\partial \phi} \left(k \frac{\partial T}{\partial \phi} \right) + \frac{1}{r^2 \sin^2 \theta} \frac{\partial}{\partial \theta} \left(k \sin \theta \frac{\partial T}{\partial \theta} \right) \color{blue} + \dot{q} = \rho c_p \frac{\partial T}{\partial t}$$
Karena gradien suhu pada arah azimut ($\frac{\partial T}{\partial \phi} \approx 0$) dan polar ($\frac{\partial T}{\partial \theta} \approx 0$) diabaikan, maka suku kedua dan ketiga bernilai nol. Suku sisanya secara tepat persis sama dengan hasil yang diturunkan di atas.
