東京大学 工学系研究科 機械工学専攻 2019年8月実施 機械工学第1部 問題1 II Author 祭音Myyura (Based on SN's answer refined with GPT 6 Astra)
Description 図 1-3 に示すように，半径 r r r l l l λ \lambda λ I I I T 0 T_0 T 0 r r r l l l T ( x ) T(x) T ( x ) x x x x x x
この金属細線のある位置における電気抵抗率を ρ \rho ρ Δ x \Delta x Δ x I 2 ρ Δ x / ( π r 2 ) I^2\rho\Delta x/(\pi r^2) I 2 ρ Δ x / ( π r 2 ) λ \lambda λ T 0 T_0 T 0
はじめに，ρ \rho ρ ρ 0 \rho_0 ρ 0
(1) この金属細線が真空中に置かれている場合を考える。T ( x ) T(x) T ( x ) T ( x ) T(x) T ( x )
(2) この金属細線が，バルク温度が T 0 T_0 T 0 T ( x ) T(x) T ( x ) x x x h { T ( x ) − T 0 } h\{T(x)-T_0\} h { T ( x ) − T 0 } h h h
次に，この金属細線の電気抵抗率 ρ \rho ρ T ( x ) T(x) T ( x )
ρ ( T ( x ) ) = ρ 0 [ 1 + β { T ( x ) − T 0 } ] (1) \rho(T(x))=\rho_0[1+\beta\{T(x)-T_0\}]
\tag{1} ρ ( T ( x )) = ρ 0 [ 1 + β { T ( x ) − T 0 }] ( 1 ) ここで，β \beta β
(3) この金属細線が，バルク温度が T 0 T_0 T 0 T ( x ) T(x) T ( x ) x x x h { T ( x ) − T 0 } h\{T(x)-T_0\} h { T ( x ) − T 0 } h h h
(4) 設問 (3) で求めた熱伝導方程式を解き，T ( x ) T(x) T ( x )
(5) この金属細線が真空中に置かれている場合を考える。金属細線の電極間の電気抵抗 R R R
R = ∫ − l / 2 l / 2 ρ π r 2 d x (2) R=\int_{-l/2}^{l/2}\frac{\rho}{\pi r^2}\,\mathrm dx
\tag{2} R = ∫ − l /2 l /2 π r 2 ρ d x ( 2 ) ( R − R 0 ) / R 0 (R-R_0)/R_0 ( R − R 0 ) / R 0 λ \lambda λ R 0 = ρ 0 l / ( π r 2 ) R_0=\rho_0l/(\pi r^2) R 0 = ρ 0 l / ( π r 2 )
题目描述
图 1-3 中，均匀材料的金属细线半径为 r r r l l l λ \lambda λ I I I T 0 T_0 T 0 r ≪ l r\ll l r ≪ l T ( x ) T(x) T ( x ) x x x
某位置的电阻率为 ρ \rho ρ Δ x \Delta x Δ x I 2 ρ Δ x / ( π r 2 ) I^2\rho\Delta x/(\pi r^2) I 2 ρ Δ x / ( π r 2 ) λ \lambda λ T 0 T_0 T 0 ρ 0 \rho_0 ρ 0
(1) 细线在真空中时，写出 T ( x ) T(x) T ( x )
(2) 细线在主体温度为 T 0 T_0 T 0 x x x h { T ( x ) − T 0 } h\{T(x)-T_0\} h { T ( x ) − T 0 } h h h
再考虑电阻率依赖温度，满足式 (1)，其中 β \beta β
(3) 细线置于主体温度 T 0 T_0 T 0 h { T ( x ) − T 0 } h\{T(x)-T_0\} h { T ( x ) − T 0 } h h h
(4) 解 (3) 的导热方程，求 T ( x ) T(x) T ( x )
(5) 细线置于真空中，电极间电阻 R R R ( R − R 0 ) / R 0 (R-R_0)/R_0 ( R − R 0 ) / R 0 λ \lambda λ R 0 = ρ 0 l / ( π r 2 ) R_0=\rho_0l/(\pi r^2) R 0 = ρ 0 l / ( π r 2 )
Kai A = π r 2 A=\pi r^2 A = π r 2 P = 2 π r P=2\pi r P = 2 π r θ = T − T 0 \theta=T-T_0 θ = T − T 0
(1)
単位長さの熱収支は λ A T ′ ′ + I 2 ρ 0 / A = 0 \lambda A T''+I^2\rho_0/A=0 λ A T ′′ + I 2 ρ 0 / A = 0
T ( x ) = T 0 + I 2 ρ 0 2 λ A 2 ( l 2 4 − x 2 ) . \boxed{T(x)=T_0+\frac{I^2\rho_0}{2\lambda A^2}
\left(\frac{l^2}{4}-x^2\right)}. T ( x ) = T 0 + 2 λ A 2 I 2 ρ 0 ( 4 l 2 − x 2 ) . (2)
λ A T ′ ′ + I 2 ρ 0 A − h P ( T − T 0 ) = 0 . \boxed{\lambda A T''+\frac{I^2\rho_0}{A}-hP(T-T_0)=0}. λ A T ′′ + A I 2 ρ 0 − h P ( T − T 0 ) = 0 . (3)
θ ′ ′ − b θ = − a , a = I 2 ρ 0 λ A 2 , b = h P λ A − a β , θ ( ± l / 2 ) = 0. \begin{gathered}
\boxed{\theta''-b\theta=-a},\\
a=\frac{I^2\rho_0}{\lambda A^2},\\
b=\frac{hP}{\lambda A}-a\beta,\\
\theta(\pm l/2)=0.
\end{gathered} θ ′′ − b θ = − a , a = λ A 2 I 2 ρ 0 , b = λ A h P − a β , θ ( ± l /2 ) = 0. (4)
対称性から奇関数の項はなく、
θ ( x ) = { a b [ 1 − cosh ( b x ) cosh ( b l / 2 ) ] , b > 0 , a 2 ( l 2 4 − x 2 ) , b = 0 , a k 2 [ cos ( k x ) cos ( k l / 2 ) − 1 ] , b = − k 2 < 0. \boxed{\theta(x)=
\begin{cases}
\dfrac ab\left[1-\dfrac{\cosh(\sqrt b\,x)}{\cosh(\sqrt b\,l/2)}\right],&b>0,\\[6pt]
\dfrac a2\left(\dfrac{l^2}4-x^2\right),&b=0,\\[6pt]
\dfrac a{k^2}\left[\dfrac{\cos(kx)}{\cos(kl/2)}-1\right],&b=-k^2<0.
\end{cases}} θ ( x ) = ⎩ ⎨ ⎧ b a [ 1 − cosh ( b l /2 ) cosh ( b x ) ] , 2 a ( 4 l 2 − x 2 ) , k 2 a [ cos ( k l /2 ) cos ( k x ) − 1 ] , b > 0 , b = 0 , b = − k 2 < 0. b < 0 b<0 b < 0 k l < π kl<\pi k l < π
(5)
真空では h = 0 h=0 h = 0 k = I ρ 0 β / λ / A k=I\sqrt{\rho_0\beta/\lambda}/A k = I ρ 0 β / λ / A
R R 0 = 1 l ∫ − l / 2 l / 2 [ 1 + β θ ( x ) ] d x = tan ( k l / 2 ) k l / 2 . \frac R{R_0}=\frac1l\int_{-l/2}^{l/2}[1+\beta\theta(x)]\,dx
=\frac{\tan(kl/2)}{kl/2}. R 0 R = l 1 ∫ − l /2 l /2 [ 1 + βθ ( x )] d x = k l /2 tan ( k l /2 ) . R − R 0 R 0 = tan ( I l 2 π r 2 ρ 0 β λ ) I l 2 π r 2 ρ 0 β λ − 1 . \boxed{\frac{R-R_0}{R_0}
=\frac{\tan\left(\dfrac{Il}{2\pi r^2}\sqrt{\dfrac{\rho_0\beta}{\lambda}}\right)}
{\dfrac{Il}{2\pi r^2}\sqrt{\dfrac{\rho_0\beta}{\lambda}}}-1}. R 0 R − R 0 = 2 π r 2 I l λ ρ 0 β tan ( 2 π r 2 I l λ ρ 0 β ) − 1 . Reference 