传热 Chapter2

dΦ − 2πL(ti − t∞ ) 1 / λro − 1 / hro2 =0= 2 dro ln(ro / ri ) 1 + λ ro h
(
)
→ ro = λ / h = rcrit Critical radius
2-7 Heat-source system 1. Problem is shown in right fig. 2. Governing eq
& ∂ 2t ∂ 2t ∂ 2 t Φ ∂t = a 2 + 2 + 2 + ∂x ∂τ ∂y ∂z ρc
t1
t2
Φ x
or
d t =0 2 dx
2
∆x
Boundary condition：
x = 0, t = t1 ; x = ∆x, t = t 2
That is
d 2t =0 2 dx
t 2 − t1 t3 − t 2 t 4 − t3 Φ = −λ A A = −λ B A = −λC A ∆x A ∆xB ∆xC t1 − t 4 ∆x A / λ A A + ∆xB / λB A + ∆xC / λC A
Solving the three equations simultaneously
Φ=
Generally
Φ=
t1 − t n +1
∑ ∆x / λ A
i =1 i i
n
Analogy between heat and electricity For the plane wall with constant thermal conductivity ∆t ∆t Φ = λA = ∆x ∆x /(λA) ∆x Thermal resistance R= λA thermal potential difference Heat flow = thermal resistance
& Φ − k − L = hθ L k
= c1 = 0
x =0
& ΦL θL = h
&L & L2 Φ Φ Q =− + c2 h 2λ
& L2 Φ L & Φ ∴ c2 = + 2λ h
Then
& & q qL 2 2 θ = (L − x ) + 2k h
When h→∞, tw =t∞, the temperature distribution is the same as before
d dt dr r dr = 0 r = ri , t = t i r=r t=t o, o
∂ 2 t 1 ∂t + =0 2 ∂r r ∂r
t = c1 ln r + c2
r ti − to The temperature profile t = ti + ln ln(ro / ri ) ri
& d 2t Φ + =0 2 dx λ
Φ T∞ λ
Set
θ = t −tf
& d 2θ Φ + =0 2 dx λ dθ x = 0, =0 dx dθ x = L, - λ = hθ dx
T∞
then
& dθ Φ = − x + c来自百度文库 dx λ
& dθ Φ 2 x + c1 x + c2 θ =− dx 2λ The 2nd boundary condition
dt λ (t ) = c dx
λ (t )dt = cdx
∫
T
T1
π (t )dt = cx
t − t1 t − t1
∫
t
t1
λ (t ) dt = cx
cx
(t − t1 )λ = cx
t = t 1+
λ
c=
t 2 = t1 +
c∆x
λ
λ (t 2 − t1 )
∆x
x t = t1 + (t 2 − t1 ) ∆x
dt / dx = c1
t1 = c2
Then
t 2 = c1∆x + c2
h1 (t A − t1 ) = −λc1
h2 (t 2 − tf2 ) = −λc1
set
∆t = t A − t B
c1 =
− ∆t ∆x +
solving
λ
h1
+
λ
h2
1 ∆t c2 = t A − h1 ∆x + 1 + 1 λ h1 h2
2-8 Cylinder with heat source The differential eq. is
& ∂ 2 t 1 ∂t Φ + + =0 2 r ∂r λ ∂r
& ∂ 2 t ∂t ∂ ∂t Φr r 2+ = r = − ∂r ∂r ∂r ∂r λ
The boundary conditions
t − to x = t w − to L
2
t − to t w − to t − t w x − = = −1 t w − to t w − to t w − to L
2
t − tw x = 1− to − t w L
2
The 3rd kind of boundary condition
ti = c1 ln ri + c2 to = c1 ln ro + c2
dt According to Fourier’s law: Φ = −λA dr We know: to − ti 1 dt A = 2πrL = dr ln(ro / ri ) r
Then
2πλL(ti − to ) ti − t o 1 to − ti Φ = −λ 2πrL = = ln(ro / ri ) r ln(ro / ri ) ln(ro / ri ) 2πλL ln( ro / ri ) set: R th = 2πλ L
dt dx = c1 =
x =0
− ∆t ∆x +
λ
h1
+
λ
h2
∆t dt Φ = −λA = 1 ∆x 1 dx + + h1 A λA h2 A
Set:
Φ = kA∆toverall
1 k= = 1 1 R value + + h1 λ h2 1 ∆x
In accordance with eq above, k should be
4πλ (ti − to ) Φ= 1 / ri − 1 / ro Convection boundary condition
Φ conv = hA(t w − t∞ ) = t w − t∞ 1 / hA
For an electric-resistance analogy, 1/hA is the the convective resistance
For hollow cylinder
t A − tB Φ= ln(ro / ri ) 1 1 + + hi Ai 2πλL ho Ao
The heat transfer area is not the same for both fluids. The overall heat transfer coefficient based on the inside area of the tube is: 1 ki = Ai ln( ro / ri ) Ai 1 1 + + hi Ai 2πλ L Ao ho The overall heat transfer coefficient based on the outside area of the tube is: 1 ko = Ao Ao ln(ro / ri ) 1 + + hi Ai 2πλL ho For either the plane wall or cylindrical coordinate system:
If k varies with T according to some linear relation
λ = λ0 (1 + bt )
b 2 2 q=− (t 2 − t1 ) + 2 (t 2 − t1 ) ∆x
λ0 A
[2 − 2]
Theoretical method
d dt λ (t ) dx = 0 dx
2-5 The overall heat transfer coefficient
d 2t =0 2 dx
x = 0, t = t1 ; x = ∆x, t = t 2
dt h1 (t A − t1 ) = −λ dx dt h2 (t 2 − t B ) = −λ dx
x =0
x = ∆x
t = c1 x + c2
Multi-layer cylindrical walls
2πL(t1 − t 4 ) Φ= ln(d 2 / d1 ) / λ1 + ln(d 3 / d 2 ) / λ2 + ln(d 4 / d 3 ) / λ3
For n-layer cylindrical wall t1 − t n +1 2πL(t1 − t n +1 ) Φ= n = n ln(d i +1 / d i ) ln(d i +1 / d i ) ∑ 2πλ L ∑ λ i =1 i =1 i i Spheres
t ( 0) = C 2 = t o
& Φ 2 C2 = t w + L = to 2λ
& Φ 2 t ( L) = − L + C2 = t w 2λ
The temperature distribution is
& Φ 2 t − to = − x 2λ & Φ 2 t w − to = − L 2λ
Chapter 2 第二章
Steady-State Conduction-One dimension 一维稳态导热
§ 2-1 Introduction § 2-2 The plane wall (slab) & 1．Condition: λ=const. Φ = 0 ２．Governing equation
∆toverall Φ= ∑ Rth
The electrical analogy may be used to solve complex problems involving series and parallel thermal resistances
2-3 Insulation and R value R value Classification of insulation materials
dt λ A(t 2 − t1 ) Φ = −λ A = − dx ∆x
If
λ = λ0 (1 + bt )
Then
λ ⇒ t = 0.5(t1 + t 2 )
Multi-layer wall
Φ
T1
∆x A
∆xB ∆xC
T2
kA kB
T3 T 4
kC
T1
T2
T3
T4
The heat flux is
x = 0, t = t1
x = ∆x, t = t 2
t (δ ) = t 2 = c1∆x + c2
t ( x) = c1 x + c2
t (0) = t1 = c2
c 2 = t1 t 2 − t1 c1 = ∆x
t 2 − t1 t= x + t1 ∆x
dt t 2 − t1 = dx ∆x dt λA Φ = −λA = (t1 − t 2 ) dx ∆x [2 − 1]
∆t R= Φ/ A
2-4 Radial system Cylinder Problem: 1-D steady-state, constant k without heat generation
& ∂ 2 t 1 ∂t 1 ∂ 2 t ∂ 2 t Φ ∂t = a 2 + + 2 + 2 + 2 ∂r ∂τ r ∂ r r ∂φ ∂ z ρc
kA = 1 / ∑ Rth = 1 / Rth,overall
2-6 Critical thickness of insulation
The heat transfer rate is
2πL(ti − t∞ ) Φ= ln(ro / ri ) 1 + λ ro h
The maximization condition is
& d 2t Φ + =0 2 dx λ
3. Boundary condition
t = tw
at
x = ±L
dt = 0 at dx
x=0
4. integrating
& dt Φ = − x + C1 dx λ
& Φ 2 t=− x + C1 x + C2 2λ
dt dx
= C1 = 0
x =0