Dimulai dengan volume kendali diferensial berbentuk selongsong silindris, turunkan persamaan difusi panas untuk sistem koordinat radial silindris satu dimensi dengan pembangkitan panas internal. Bandingkan hasil Anda dengan Persamaan 2.26.
$$\frac{1}{r} \frac{\partial}{\partial r} \left(kr \frac{\partial T}{\partial r} \right) + \frac{1}{r^2} \frac{\partial}{\partial \phi} \left(k \frac{\partial T}{\partial \phi} \right) + \frac{\partial}{\partial z} \left(k \frac{\partial T}{\partial z} \right) + \dot{q} = \rho c_p \frac{\partial T}{\partial t}$$
Analisis Geometri Bangun Ruang
Tinjau volume kendali diferensial berbentuk selongsong silindris sesumbu (coaxial cylindrical shell) dengan parameter berikut:
- Jari-jari dalam: $r$
- Ketebalan radial: $dr$
- Jari-jari luar: $r + dr$
- Panjang aksial: $L$
Kita hitung luas penampang dalam
$$A_{r} = 2 \pi r \cdot L$$
$$A_{r} = 2 \pi r L$$
Kita hitung luas penampang luar
$$A_{r + dr} = 2 \pi (r + dr) \cdot L$$
$$A_{r + dr} = 2 \pi r L + 2 \pi dr L$$
Kita hitung volume kendali diferensial
$$dV = A_{r} \cdot dr$$
$$dV = 2 \pi r L \cdot dr$$
$$dV = 2 \pi r L dr$$
Perhitungan Semua Energi
Kita hitung energi yang masuk
$$\dot{E}_{in} = q_r$$
$$\dot{E}_{in} = -k A_{r} \frac{\partial T}{\partial r}$$
$$\dot{E}_{in} = -k (2 \pi r L) \frac{\partial T}{\partial r}$$
Kita hitung energi yang keluar
$$\dot{E}_{out} = q_{r + dr}$$
$$\dot{E}_{out} = -k A_{r} \frac{\partial T}{\partial r} + \frac{\partial}{\partial r} \left(-k A_{r} \frac{\partial T}{\partial r} \right) dr$$
$$\dot{E}_{out} = -k (2 \pi r L) \frac{\partial T}{\partial r} + \frac{\partial}{\partial r} \left(-k (2 \pi r L) \frac{\partial T}{\partial r} \right) dr$$
Kita hitung energi yang dibangkitkan
$$\dot{E}_{gen} = \dot{q} \cdot dV$$
$$\dot{E}_{gen} = \dot{q} (2 \pi r L dr)$$
Kita hitung energi yang tersimpan
$$\dot{E}_{st} = \rho \cdot c_p \cdot dV \cdot \frac{\partial T}{\partial t}$$
$$\dot{E}_{st} = \rho \cdot c_p \cdot (2 \pi r L dr) \cdot \frac{\partial T}{\partial t}$$
Penerapan Hukum Fourier
Berdasarkan Hukum Pertama Termodinamika untuk kondisi transien:
$$\dot{E}_{in} – \dot{E}_{out} + \dot{E}_{gen} = \dot{E}_{st}$$
$$-k (2 \pi r L) \frac{\partial T}{\partial r} – \left(-k (2 \pi r L) \frac{\partial T}{\partial r} + \frac{\partial}{\partial r} \left(-k (2 \pi r L) \frac{\partial T}{\partial r} \right) dr \right) + \dot{q} (2 \pi r L dr) = \rho c_p (2 \pi r L dr) \frac{\partial T}{\partial t}$$
$$\frac{\partial}{\partial r} \left(k (2 \pi r L) \frac{\partial T}{\partial r} \right) dr + \dot{q} (2 \pi r L dr) = \rho c_p \frac{\partial T}{\partial t}$$
$$\frac{\partial}{\partial r} \left(kr \frac{\partial T}{\partial r} \right) (2 \pi L dr) + \dot{q} (2 \pi r L dr) = \rho c_p \frac{\partial T}{\partial t} (2 \pi r L dr)$$
$$\color{blue} \frac{1}{r} \frac{\partial}{\partial r} \left(kr \frac{\partial T}{\partial r} \right) + \dot{q} = \rho c_p \frac{\partial T}{\partial t}$$
Persamaan umum 3D dalam koordinat silinder (Persamaan 2.26):
$$\mathbf{\color{blue}\frac{1}{r} \frac{\partial}{\partial r} \left(kr \frac{\partial T}{\partial r} \right) \color{black} + \frac{1}{r^2} \frac{\partial}{\partial \phi} \left(k \frac{\partial T}{\partial \phi} \right) + \frac{\partial}{\partial z} \left(k \frac{\partial T}{\partial z} \right) \color{blue} + \dot{q} = \rho c_p \frac{\partial T}{\partial t}}$$
Karena gradien suhu pada arah azimut ($\frac{\partial T}{\partial \phi} \approx 0$) dan aksial ($\frac{\partial T}{\partial z} \approx 0$) diabaikan, maka suku kedua dan ketiga bernilai nol. Suku sisanya secara tepat persis sama dengan hasil yang diturunkan di atas.
