第三章 带对流边界的一维热传导
有对流边界的一维热传导
3.1 引言（概述） 对于涉及热交换的系统，传热分析的一项基本任务是求出系统中的温度分布，因为它在系统设计和热通量计算中起着基础作用。然而，这项任务能否达到目的，在很大程度上取决于温度分布的空间和时间特征以及系统的规定边界条件
在热力系统中，温度一般可表示为空间坐标
(
x
,
y
,
z
)
(
x
,
y
,
z
)
(x,y,z) (x, y, z)
(
t
)
(
t
)
(t) (t)
T
=
T
(
x
,
y
,
z
,
t
)
T
=
T
(
x
,
y
,
z
,
t
)
T=T(x,y,z,t) T=T(x, y, z, t) 如果特定问题中的温度只在
(
x
,
y
,
z
)
(
x
,
y
,
z
)
(x,y,z) (x, y, z) 在本章中，我们将讨论具有对流边界的固体内部的一维热传导，包括稳态和瞬态问题。虽然所选的例子看似非常简单，但肯定有助于我们理解传热原理和掌握分析方法。此外，我们还会发现，许多具有重要技术意义的传热问题都可以利用本章所提供的信息进行建模。通过一些典型的例子，我们将展示这些模型在工程系统分析和设计中的应用。
3.2 通过固体壁的稳态热传导 在本节中，我们首先讨论通过单层固体墙壁（包括平面墙壁、圆柱形墙壁和球形墙壁）的热传导，然后扩展到多层固体的情况。接下来，我们考虑了壁面对流，提出了两种流体通过隔墙的传热速率方程，即串联热阻模型，并讨论了该模型的应用，甚至在一些非稳态系统中的应用。最后，我们以内部热源为例，演示了如何处理传导问题。
3.2.1 单层固壁 考虑如图 3.2-1 所示的平面或圆柱形墙壁。 假设墙壁表面分别保持恒温
T
1
T
1
T_(1) T_{1}
T
2
(
T
1
≠
T
2
)
T
2
T
1
≠
T
2
T_(2)(T_(1)!=T_(2)) T_{2}\left(T_{1} \neq T_{2}\right) (a) 平面墙 (b) 圆柱形壁 图 3.1-1 穿过实心墙的一维热传导 对于平面壁，在
y
y
y y
z
z
z z
θ
θ
theta \theta
z
z
z z
T
=
T
(
x
)
or
T
=
T
(
r
)
T
=
T
(
x
)
or
T
=
T
(
r
)
T=T(x)quad" or "quad T=T(r) T=T(x) \quad \text { or } \quad T=T(r) 因此，可以在厚度为
d
x
d
x
dx \mathrm{d} x
d
r
d
r
dr \mathrm{d} r
Q
i
n
−
Q
o
u
t
=
0
Q
i
n
−
Q
o
u
t
=
0
Q_(in)-Q_(out)=0 Q_{i n}-Q_{o u t}=0 为了便于比较，我们列出了从建立热平衡方程到最终给出平面壁和圆柱壁的传导率方程的并行步骤如下。
流程 平面墙 圆柱形墙壁
热平衡方程 Heat Balance
equation | Heat Balance |
| :--- |
| equation |
q
x
A
−
(
q
x
A
+
∂
q
x
A
∂
x
d
x
)
=
0
q
x
A
−
q
x
A
+
∂
q
x
A
∂
x
d
x
=
0
q_(x)A-(q_(x)A+(delq_(x)A)/(del x)(d)x)=0 q_{x} A-\left(q_{x} A+\frac{\partial q_{x} A}{\partial x} \mathrm{~d} x\right)=0
q
r
A
r
−
(
q
r
A
r
+
∂
q
r
A
r
∂
r
d
r
)
=
0
,
A
r
=
2
π
r
q
r
A
r
−
q
r
A
r
+
∂
q
r
A
r
∂
r
d
r
=
0
,
A
r
=
2
π
r
q_(r)A_(r)-(q_(r)A_(r)+(delq_(r)A_(r))/(del r)(d)r)=0,A_(r)=2pi r q_{r} A_{r}-\left(q_{r} A_{r}+\frac{\partial q_{r} A_{r}}{\partial r} \mathrm{~d} r\right)=0, A_{r}=2 \pi r
T
T
T T Differential
equation of T | Differential |
| :--- |
| equation of $T$ |
d
q
x
d
x
=
0
→
d
d
x
(
k
d
T
d
x
)
=
0
d
q
x
d
x
=
0
→
d
d
x
k
d
T
d
x
=
0
((d)q_(x))/((d)x)=0rarr((d))/((d)x)(k((d)T)/((d)x))=0 \frac{\mathrm{~d} q_{x}}{\mathrm{~d} x}=0 \rightarrow \frac{\mathrm{~d}}{\mathrm{~d} x}\left(k \frac{\mathrm{~d} T}{\mathrm{~d} x}\right)=0
d
r
q
r
d
r
=
0
→
d
d
r
(
k
r
d
T
d
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)
=
0
d
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q
r
d
r
=
0
→
d
d
r
k
r
d
T
d
r
=
0
((d)rq_(r))/((d)r)=0rarr((d))/((d)r)(kr((d)T)/((d)r))=0 \frac{\mathrm{~d} r q_{r}}{\mathrm{~d} r}=0 \rightarrow \frac{\mathrm{~d}}{\mathrm{~d} r}\left(k r \frac{\mathrm{~d} T}{\mathrm{~d} r}\right)=0
对于
k
=
k
=
k= k=
d
2
T
d
x
2
=
0
d
2
T
d
x
2
=
0
(d^(2)T)/((d)x^(2))=0 \frac{\mathrm{d}^{2} T}{\mathrm{~d} x^{2}}=0
d
d
r
(
r
d
T
d
r
)
=
0
d
d
r
r
d
T
d
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=
0
((d))/((d)r)(r((d)T)/((d)r))=0 \frac{\mathrm{~d}}{\mathrm{~d} r}\left(r \frac{\mathrm{~d} T}{\mathrm{~d} r}\right)=0
边界条件 Boundary
conditions | Boundary |
| :--- |
| conditions |
T
|
x
=
0
=
T
1
,
T
|
x
=
δ
=
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2
T
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=
0
=
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,
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x
=
δ
=
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2
T|_(x=0)=T_(1),T|_(x=delta)=T_(2) \left.T\right|_{x=0}=T_{1},\left.T\right|_{x=\delta}=T_{2}
T
|
r
=
R
1
=
T
1
,
T
|
r
=
R
2
=
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T
r
=
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1
=
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r
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=
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T|_(r=R_(1))=T_(1),T|_(r=R_(2))=T_(2) \left.T\right|_{r=R_{1}}=T_{1},\left.T\right|_{r=R_{2}}=T_{2}
温度分布 Temperature
distribution | Temperature |
| :--- |
| distribution |
T
=
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1
−
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1
−
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2
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δ
x
T
=
T
1
−
T
1
−
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δ
x
T=T_(1)-((T_(1)-T_(2)))/(delta)x T=T_{1}-\frac{\left(T_{1}-T_{2}\right)}{\delta} x
T
=
T
1
−
(
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−
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)
δ
R
m
ln
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1
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T
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−
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2
δ
R
m
ln
r
R
1
T=T_(1)-((T_(1)-T_(2)))/(delta)R_(m)ln((r)/(R_(1))) T=T_{1}-\frac{\left(T_{1}-T_{2}\right)}{\delta} R_{m} \ln \frac{r}{R_{1}}
Heat flux | Heat flux |
| :--- |
q
x
=
−
k
d
T
d
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=
k
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δ
q
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−
k
d
T
d
x
=
k
T
1
−
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2
δ
q_(x)=-k((d)T)/((d)x)=k((T_(1)-T_(2)))/(delta) q_{x}=-k \frac{\mathrm{~d} T}{\mathrm{~d} x}=k \frac{\left(T_{1}-T_{2}\right)}{\delta}
q
r
=
−
k
d
T
d
r
=
k
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T
1
−
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)
δ
R
m
r
q
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=
−
k
d
T
d
r
=
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T
1
−
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2
δ
R
m
r
q_(r)=-k((d)T)/((d)r)=k((T_(1)-T_(2)))/(delta)(R_(m))/(r) q_{r}=-k \frac{\mathrm{~d} T}{\mathrm{~d} r}=k \frac{\left(T_{1}-T_{2}\right)}{\delta} \frac{R_{m}}{r}
热传导率 Rate of heat
conduction | Rate of heat |
| :--- |
| conduction |
Q
x
=
q
x
A
=
A
k
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T
1
−
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2
)
δ
Q
x
=
q
x
A
=
A
k
T
1
−
T
2
δ
Q_(x)=q_(x)A=Ak((T_(1)-T_(2)))/(delta) Q_{x}=q_{x} A=A k \frac{\left(T_{1}-T_{2}\right)}{\delta}
Q
r
=
q
r
A
r
=
A
m
k
(
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1
−
T
2
)
δ
Q
r
=
q
r
A
r
=
A
m
k
T
1
−
T
2
δ
Q_(r)=q_(r)A_(r)=A_(m)k((T_(1)-T_(2)))/(delta) Q_{r}=q_{r} A_{r}=A_{m} k \frac{\left(T_{1}-T_{2}\right)}{\delta}
Processes Plane wall Cylindrical wall
"Heat Balance
equation" q_(x)A-(q_(x)A+(delq_(x)A)/(del x)(d)x)=0 q_(r)A_(r)-(q_(r)A_(r)+(delq_(r)A_(r))/(del r)(d)r)=0,A_(r)=2pi r
"Differential
equation of T" ((d)q_(x))/((d)x)=0rarr((d))/((d)x)(k((d)T)/((d)x))=0 ((d)rq_(r))/((d)r)=0rarr((d))/((d)r)(kr((d)T)/((d)r))=0
For k= const (d^(2)T)/((d)x^(2))=0 ((d))/((d)r)(r((d)T)/((d)r))=0
"Boundary
conditions" T|_(x=0)=T_(1),T|_(x=delta)=T_(2) T|_(r=R_(1))=T_(1),T|_(r=R_(2))=T_(2)
"Temperature
distribution" T=T_(1)-((T_(1)-T_(2)))/(delta)x T=T_(1)-((T_(1)-T_(2)))/(delta)R_(m)ln((r)/(R_(1)))
"Heat flux" q_(x)=-k((d)T)/((d)x)=k((T_(1)-T_(2)))/(delta) q_(r)=-k((d)T)/((d)r)=k((T_(1)-T_(2)))/(delta)(R_(m))/(r)
"Rate of heat
conduction" Q_(x)=q_(x)A=Ak((T_(1)-T_(2)))/(delta) Q_(r)=q_(r)A_(r)=A_(m)k((T_(1)-T_(2)))/(delta) | Processes | Plane wall | Cylindrical wall |
| :--- | :--- | :--- |
| Heat Balance <br> equation | $q_{x} A-\left(q_{x} A+\frac{\partial q_{x} A}{\partial x} \mathrm{~d} x\right)=0$ | $q_{r} A_{r}-\left(q_{r} A_{r}+\frac{\partial q_{r} A_{r}}{\partial r} \mathrm{~d} r\right)=0, A_{r}=2 \pi r$ |
| Differential <br> equation of $T$ | $\frac{\mathrm{~d} q_{x}}{\mathrm{~d} x}=0 \rightarrow \frac{\mathrm{~d}}{\mathrm{~d} x}\left(k \frac{\mathrm{~d} T}{\mathrm{~d} x}\right)=0$ | $\frac{\mathrm{~d} r q_{r}}{\mathrm{~d} r}=0 \rightarrow \frac{\mathrm{~d}}{\mathrm{~d} r}\left(k r \frac{\mathrm{~d} T}{\mathrm{~d} r}\right)=0$ |
| For $k=$ const | $\frac{\mathrm{d}^{2} T}{\mathrm{~d} x^{2}}=0$ | $\frac{\mathrm{~d}}{\mathrm{~d} r}\left(r \frac{\mathrm{~d} T}{\mathrm{~d} r}\right)=0$ |
| Boundary <br> conditions | $\left.T\right\|_{x=0}=T_{1},\left.T\right\|_{x=\delta}=T_{2}$ | $\left.T\right\|_{r=R_{1}}=T_{1},\left.T\right\|_{r=R_{2}}=T_{2}$ |
| Temperature <br> distribution | $T=T_{1}-\frac{\left(T_{1}-T_{2}\right)}{\delta} x$ | $T=T_{1}-\frac{\left(T_{1}-T_{2}\right)}{\delta} R_{m} \ln \frac{r}{R_{1}}$ |
| Heat flux | $q_{x}=-k \frac{\mathrm{~d} T}{\mathrm{~d} x}=k \frac{\left(T_{1}-T_{2}\right)}{\delta}$ | $q_{r}=-k \frac{\mathrm{~d} T}{\mathrm{~d} r}=k \frac{\left(T_{1}-T_{2}\right)}{\delta} \frac{R_{m}}{r}$ |
| Rate of heat <br> conduction | $Q_{x}=q_{x} A=A k \frac{\left(T_{1}-T_{2}\right)}{\delta}$ | $Q_{r}=q_{r} A_{r}=A_{m} k \frac{\left(T_{1}-T_{2}\right)}{\delta}$ |
其中，对于圆柱形壁，
δ
=
R
2
−
R
1
δ
=
R
2
−
R
1
delta=R_(2)-R_(1) \delta=R_{2}-R_{1}
R
m
R
m
R_(m) R_{m} 根据上述结果，可以得出 (1) 对于跨壁厚的一维稳态传导，平面壁内的温度分布是
x
x
x x
r
r
r r (2) 对于平面壁，热通量
q
x
q
x
q_(x) q_{x}
Q
x
Q
x
Q_(x) Q_{x}
A
r
A
r
A_(r) A_{r}
(
=
2
π
r
L
)
(
=
2
π
r
L
)
(=2pi rL) (=2 \pi r L)
q
r
q
r
q_(r) q_{r}
r
r
r r
Q
x
Q
x
Q_(x) Q_{x} (3) 对于平面墙壁和圆柱形墙壁，甚至球形墙壁（见问题 3.5），它们的传导热流率都可以用均匀方程表示为
Q
=
k
A
m
(
T
1
−
T
2
)
δ
=
(
T
1
−
T
2
)
δ
/
k
A
m
Q
=
k
A
m
T
1
−
T
2
δ
=
T
1
−
T
2
δ
/
k
A
m
Q=kA_(m)((T_(1)-T_(2)))/(delta)=((T_(1)-T_(2)))/(delta//kA_(m)) Q=k A_{m} \frac{\left(T_{1}-T_{2}\right)}{\delta}=\frac{\left(T_{1}-T_{2}\right)}{\delta / k A_{m}} 其中，
A
m
A
m
A_(m) A_{m}
x
x
x x
r
r
r r
R
m
R
m
R_(m) R_{m}
for plane wall:
A
m
=
A
for cylindrical wall:
δ
=
R
2
−
R
1
,
R
m
=
(
R
2
−
R
1
)
ln
(
R
2
/
R
1
)
,
A
m
=
2
π
L
R
m
for spherical wall:
δ
=
R
2
−
R
1
,
R
m
=
R
1
R
2
,
A
m
=
4
π
R
m
2
for plane wall:
A
m
=
A
for cylindrical wall:
δ
=
R
2
−
R
1
,
R
m
=
R
2
−
R
1
ln
R
2
/
R
1
,
A
m
=
2
π
L
R
m
for spherical wall:
δ
=
R
2
−
R
1
,
R
m
=
R
1
R
2
,
A
m
=
4
π
R
m
2
{:[" for plane wall: ",A_(m)=A],[" for cylindrical wall: ",delta=R_(2)-R_(1)","R_(m)=((R_(2)-R_(1)))/(ln(R_(2)//R_(1)))","quadA_(m)=2pi LR_(m)],[" for spherical wall: ",delta=R_(2)-R_(1)","R_(m)=sqrt(R_(1)R_(2))","A_(m)=4piR_(m)^(2)]:} \begin{array}{ll}
\text { for plane wall: } & A_{m}=A \\
\text { for cylindrical wall: } & \delta=R_{2}-R_{1}, R_{m}=\frac{\left(R_{2}-R_{1}\right)}{\ln \left(R_{2} / R_{1}\right)}, \quad A_{m}=2 \pi L R_{m} \\
\text { for spherical wall: } & \delta=R_{2}-R_{1}, R_{m}=\sqrt{R_{1} R_{2}}, A_{m}=4 \pi R_{m}^{2}
\end{array} 需要注意的是，方程 (3.2-9) 适用于导热系数
k
k
k k
k
k
k k
k
k
k k 例 3.2-1 导热系数可变的热传导 在许多情况下，热导率
k
k
k k
k
=
k
0
[
1
+
β
(
T
−
T
0
)
]
k
=
k
0
1
+
β
T
−
T
0
k=k_(0)[1+beta(T-T_(0))] k=k_{0}\left[1+\beta\left(T-T_{0}\right)\right] 其中
k
0
k
0
k_(0) k_{0}
T
0
;
β
T
0
;
β
T_(0);beta T_{0} ; \beta
T
T
T T
Q
Q
Q Q 解：继续方程 (3.2-4)
d
d
x
(
k
d
T
d
x
)
=
0
d
d
x
k
d
T
d
x
=
0
(d)/((d)x)(k((d)T)/((d)x))=0 \frac{\mathrm{d}}{\mathrm{~d} x}\left(k \frac{\mathrm{~d} T}{\mathrm{~d} x}\right)=0 将
k
(
T
)
k
(
T
)
k(T) k(T)
k
0
(
(
T
−
T
0
)
+
β
2
(
T
−
T
0
)
2
)
=
C
1
x
+
C
2
k
0
T
−
T
0
+
β
2
T
−
T
0
2
=
C
1
x
+
C
2
k_(0)((T-T_(0))+(beta)/(2)(T-T_(0))^(2))=C_(1)x+C_(2) k_{0}\left(\left(T-T_{0}\right)+\frac{\beta}{2}\left(T-T_{0}\right)^{2}\right)=C_{1} x+C_{2} 应用边界条件
T
|
x
=
0
=
T
1
,
T
|
x
=
δ
=
T
2
T
x
=
0
=
T
1
,
T
x
=
δ
=
T
2
T|_(x=0)=T_(1),T|_(x=delta)=T_(2) \left.T\right|_{x=0}=T_{1},\left.T\right|_{x=\delta}=T_{2} 得出的温度分布形式为
(
T
1
−
T
)
(
1
+
β
2
(
T
1
+
T
−
2
T
0
)
)
=
(
T
1
−
T
2
)
(
1
+
β
2
(
T
1
+
T
2
−
2
T
0
)
)
x
δ
T
1
−
T
1
+
β
2
T
1
+
T
−
2
T
0
=
T
1
−
T
2
1
+
β
2
T
1
+
T
2
−
2
T
0
x
δ
(T_(1)-T)(1+(beta)/(2)(T_(1)+T-2T_(0)))=(T_(1)-T_(2))(1+(beta)/(2)(T_(1)+T_(2)-2T_(0)))(x)/( delta) \left(T_{1}-T\right)\left(1+\frac{\beta}{2}\left(T_{1}+T-2 T_{0}\right)\right)=\left(T_{1}-T_{2}\right)\left(1+\frac{\beta}{2}\left(T_{1}+T_{2}-2 T_{0}\right)\right) \frac{x}{\delta} 当
β
=
0
β
=
0
beta=0 \beta=0
T
=
T
1
−
(
T
1
−
T
2
)
δ
x
T
=
T
1
−
T
1
−
T
2
δ
x
T=T_(1)-((T_(1)-T_(2)))/(delta)x T=T_{1}-\frac{\left(T_{1}-T_{2}\right)}{\delta} x 当
β
≠
0
β
≠
0
beta!=0 \beta \neq 0
[
1
+
β
(
T
−
T
0
)
]
2
=
[
1
+
β
(
T
1
−
T
0
)
]
2
(
1
−
x
δ
)
+
[
1
+
β
(
T
2
−
T
0
)
]
2
x
δ
1
+
β
T
−
T
0
2
=
1
+
β
T
1
−
T
0
2
1
−
x
δ
+
1
+
β
T
2
−
T
0
2
x
δ
[1+beta(T-T_(0))]^(2)=[1+beta(T_(1)-T_(0))]^(2)(1-(x)/( delta))+[1+beta(T_(2)-T_(0))]^(2)(x)/( delta) \left[1+\beta\left(T-T_{0}\right)\right]^{2}=\left[1+\beta\left(T_{1}-T_{0}\right)\right]^{2}\left(1-\frac{x}{\delta}\right)+\left[1+\beta\left(T_{2}-T_{0}\right)\right]^{2} \frac{x}{\delta} 或者，表示
T
m
=
(
T
1
+
T
7
)
/
2
T
m
=
T
1
+
T
7
/
2
T_(m)=(T_(1)+T_(7))//2 T_{m}=\left(T_{1}+T_{7}\right) / 2
[
1
+
β
(
T
−
T
0
)
]
2
=
[
1
+
β
(
T
1
−
T
0
)
]
2
−
[
1
+
β
(
T
m
−
T
0
)
]
2
β
(
T
1
−
T
2
)
x
δ
1
+
β
T
−
T
0
2
=
1
+
β
T
1
−
T
0
2
−
1
+
β
T
m
−
T
0
2
β
T
1
−
T
2
x
δ
[1+beta(T-T_(0))]^(2)=[1+beta(T_(1)-T_(0))]^(2)-[1+beta(T_(m)-T_(0))](2beta(T_(1)-T_(2))x)/(delta) \left[1+\beta\left(T-T_{0}\right)\right]^{2}=\left[1+\beta\left(T_{1}-T_{0}\right)\right]^{2}-\left[1+\beta\left(T_{m}-T_{0}\right)\right] \frac{2 \beta\left(T_{1}-T_{2}\right) x}{\delta} 根据这个表达式，可以很容易地求得温度梯度
d
T
d
x
=
−
(
T
1
−
T
2
)
δ
[
1
+
β
(
T
m
−
T
0
)
]
1
+
β
(
T
−
T
0
)
=
−
(
T
1
−
T
2
)
δ
k
m
k
d
T
d
x
=
−
T
1
−
T
2
δ
1
+
β
T
m
−
T
0
1
+
β
T
−
T
0
=
−
T
1
−
T
2
δ
k
m
k
(dT)/((d)x)=-((T_(1)-T_(2)))/(delta)([1+beta(T_(m)-T_(0))])/(1+beta(T-T_(0)))=-((T_(1)-T_(2)))/(delta)(k_(m))/(k) \frac{\mathrm{d} T}{\mathrm{~d} x}=-\frac{\left(T_{1}-T_{2}\right)}{\delta} \frac{\left[1+\beta\left(T_{m}-T_{0}\right)\right]}{1+\beta\left(T-T_{0}\right)}=-\frac{\left(T_{1}-T_{2}\right)}{\delta} \frac{k_{m}}{k} 
