Toplotna provodljivost: Matematički model i rešenje

𝑅1 = 1.25𝑥10−2
𝑅2 = 𝑅1 + 12.5𝑥10−3
𝑘 = 204
ℎ = 60
𝑒0 = 5𝑥10−3
𝑇𝑤 = 200
𝑇𝑎 = 20
𝑒(𝑟) = 𝑒0 (1 −
𝑒𝑝(𝑟) =
𝑟 − 𝑅1
)
𝑅2 − 𝑅1
𝑒0
𝑟 − 𝑅1
(1 −
)
2
𝑅2 − 𝑅1
2
𝑑
𝑑
− (𝑞𝑟 ∗ 𝑟 ∗ 𝑒(𝑟)) − 2√1 + ( (𝑒(𝑟))) ∗ ℎ(𝑇 − 𝑇𝑎) == 0
𝑑𝑟
𝑑𝑟
𝑞𝑟 = −𝑘
𝑑𝑇
𝑑𝑟
𝑇[𝑅1] = 0
𝑇 ′ [0.99999𝑅2] = 0
𝑇 = 𝐹𝑖𝑟𝑠𝑡[𝑇./𝑁𝐷𝑆𝑜𝑙𝑣𝑒[{ }, 𝑇, {𝑟, 𝑅1,0.99999 ∗ 𝑅2}]