معادله پخش هدایت گرما:
ρ
c
∂
T
∂
t
=
k
(
∂
2
T
∂
x
2
+
∂
2
T
∂
y
2
)
{\displaystyle \rho c{\frac {\partial T}{\partial t}}=k({\frac {{{\partial }^{2}}T}{\partial {{x}^{2}}}}+{\frac {{{\partial }^{2}}T}{\partial {{y}^{2}}}})}
شرایط دائمی:
∂
T
∂
t
=
0
→
∂
2
T
∂
x
2
+
∂
2
T
∂
y
2
=
0
{\displaystyle {\frac {\partial T}{\partial t}}=0\to {\frac {{{\partial }^{2}}T}{\partial {{x}^{2}}}}+{\frac {{{\partial }^{2}}T}{\partial {{y}^{2}}}}=0}
T
(
x
,
y
)
=
?
{\displaystyle T(x,y)=?}
∂
2
T
∂
x
2
+
∂
2
T
∂
y
2
=
0
T
(
x
,
0
)
=
T
0
T
(
x
,
H
)
=
T
0
+
A
sin
(
π
x
/
L
)
T
(
0
,
y
)
=
T
0
T
(
L
,
y
)
=
T
0
{\displaystyle {\begin{aligned}&{\frac {{{\partial }^{2}}T}{\partial {{x}^{2}}}}+{\frac {{{\partial }^{2}}T}{\partial {{y}^{2}}}}=0\\&T(x,0)=T0\\&T(x,H)=T0+A\sin(\pi x/L)\\&T(0,y)=T0\\&T(L,y)=T0\\\end{aligned}}}
تغییر متغییر:
θ
=
T
−
T
0
{\displaystyle \theta =T-T0}
∂
2
θ
∂
x
2
+
∂
2
θ
∂
y
2
=
0
θ
(
x
,
0
)
=
0
θ
(
x
,
H
)
=
A
sin
(
π
x
/
L
)
θ
(
0
,
y
)
=
0
θ
(
L
,
y
)
=
0
{\displaystyle {\begin{aligned}&{\frac {{{\partial }^{2}}\theta }{\partial {{x}^{2}}}}+{\frac {{{\partial }^{2}}\theta }{\partial {{y}^{2}}}}=0\\&\theta (x,0)=0\\&\theta (x,H)=A\sin(\pi x/L)\\&\theta (0,y)=0\\&\theta (L,y)=0\\\end{aligned}}}
نسبت به شرایط مرزی x همگن میباشد
جداسازی متغییرها:
θ
=
X
(
x
)
.
Y
(
y
)
{\displaystyle \theta =X(x).Y(y)}
X
″
X
=
−
Y
″
Y
=
−
λ
2
{\displaystyle {\frac {{X}^{''}}{X}}=-{\frac {{Y}^{''}}{Y}}=-{{\lambda }^{2}}}
X
″
+
λ
2
x
=
0
X
(
0
)
=
X
(
L
)
=
0
{\displaystyle {\begin{aligned}&{{X}^{''}}+{{\lambda }^{2}}x=0\\&X(0)=X(L)=0\\\end{aligned}}}
X
(
x
)
=
a
sin
(
λ
x
)
+
b
cos
(
λ
x
)
X
(
0
)
=
0
→
b
=
0
X
(
L
)
=
0
→
X
(
L
)
=
a
n
sin
(
λ
n
L
)
=
0
→
λ
n
=
n
π
L
{\displaystyle {\begin{aligned}&X(x)=a\sin(\lambda x)+b\cos(\lambda x)\\&X(0)=0\to b=0\\&X(L)=0\to X(L)={{a}_{n}}\sin({{\lambda }_{n}}L)=0\to {{\lambda }_{n}}={\frac {n\pi }{L}}\\\end{aligned}}}
X
n
(
x
)
=
a
sin
(
λ
n
x
)
{\displaystyle {{X}_{n}}(x)=a\sin({{\lambda }_{n}}x)}
Y
n
″
−
λ
n
2
Y
n
=
0
Y
n
(
y
)
=
c
n
sinh
(
λ
n
y
)
+
d
n
cosh
(
λ
n
y
)
y
n
(
0
)
=
0
→
d
n
=
0
→
Y
n
(
y
)
=
c
n
sinh
(
λ
n
y
)
{\displaystyle {\begin{aligned}&{{Y}_{n}}^{''}-{{\lambda }_{n}}^{2}{{Y}_{n}}=0\\&{{Y}_{n}}(y)={{c}_{n}}\sinh({{\lambda }_{n}}y)+{{d}_{n}}\cosh({{\lambda }_{n}}y)\\&{{y}_{n}}(0)=0\to {{d}_{n}}=0\to {{Y}_{n}}(y)={{c}_{n}}\sinh({{\lambda }_{n}}y)\\\end{aligned}}}
θ
n
(
x
,
y
)
=
X
n
(
x
)
.
Y
n
(
y
)
→
θ
(
x
,
y
)
=
∑
n
=
1
∞
e
n
sin
(
n
π
x
L
)
.
sinh
(
n
π
y
L
)
θ
(
x
,
H
)
=
A
sin
(
π
x
L
)
→
θ
(
x
,
H
)
=
∑
n
=
1
∞
e
n
sin
(
n
π
x
L
)
.
sinh
(
n
π
H
L
)
=
A
sin
(
π
x
L
)
T
(
x
,
y
)
=
T
0
+
A
sin
(
π
x
L
)
.
sinh
(
π
y
L
)
sinh
(
π
H
L
)
{\displaystyle {\begin{aligned}&{{\theta }_{n}}(x,y)={{X}_{n}}(x).{{Y}_{n}}(y)\to \theta (x,y)=\sum \limits _{n=1}^{\infty }{{e}_{n}}\sin({\frac {n\pi x}{L}}).\sinh({\frac {n\pi y}{L}})\\&\theta (x,H)=A\sin({\frac {\pi x}{L}})\to \theta (x,H)=\sum \limits _{n=1}^{\infty }{{e}_{n}}\sin({\frac {n\pi x}{L}}).\sinh({\frac {n\pi H}{L}})=A\sin({\frac {\pi x}{L}})\\&T(x,y)=T0+A\sin({\frac {\pi x}{L}}).{\frac {\sinh({\frac {\pi y}{L}})}{\sinh({\frac {\pi H}{L}})}}\\\end{aligned}}}
T
(
x
,
y
)
=
?
{\displaystyle T(x,y)=?}
∂
2
T
∂
x
2
+
∂
2
T
∂
y
2
=
0
T
(
x
,
0
)
=
T
0
T
(
x
,
H
)
=
T
0
+
A
sin
(
π
x
/
L
)
T
(
0
,
y
)
=
T
0
T
(
L
,
y
)
=
T
0
+
y
(
H
−
y
)
{\displaystyle {\begin{aligned}&{\frac {{{\partial }^{2}}T}{\partial {{x}^{2}}}}+{\frac {{{\partial }^{2}}T}{\partial {{y}^{2}}}}=0\\&T(x,0)=T0\\&T(x,H)=T0+A\sin(\pi x/L)\\&T(0,y)=T0\\&T(L,y)=T0+y(H-y)\\\end{aligned}}}
تغییر متغییر :
θ
=
T
−
T
0
{\displaystyle \theta =T-T0}
∂
2
θ
∂
x
2
+
∂
2
θ
∂
y
2
=
0
θ
(
x
,
0
)
=
0
θ
(
x
,
H
)
=
A
sin
(
π
x
/
L
)
θ
(
0
,
y
)
=
0
θ
(
L
,
y
)
=
y
(
H
−
y
)
{\displaystyle {\begin{aligned}&{\frac {{{\partial }^{2}}\theta }{\partial {{x}^{2}}}}+{\frac {{{\partial }^{2}}\theta }{\partial {{y}^{2}}}}=0\\&\theta (x,0)=0\\&\theta (x,H)=A\sin(\pi x/L)\\&\theta (0,y)=0\\&\theta (L,y)=y(H-y)\\\end{aligned}}}
θ
(
x
,
y
)
=
θ
1
(
x
,
y
)
+
θ
2
(
x
,
y
)
{\displaystyle \theta (x,y)={{\theta }_{1}}(x,y)+{{\theta }_{2}}(x,y)}
∂
2
θ
1
∂
x
2
+
∂
2
θ
1
∂
y
2
=
0
θ
1
(
x
,
0
)
=
θ
1
(
0
,
y
)
=
θ
1
(
L
,
y
)
=
0
θ
1
(
x
,
H
)
=
A
sin
(
π
x
L
)
{\displaystyle {\begin{aligned}&{\frac {{{\partial }^{2}}{{\theta }_{1}}}{\partial {{x}^{2}}}}+{\frac {{{\partial }^{2}}{{\theta }_{1}}}{\partial {{y}^{2}}}}=0\\&{{\theta }_{1}}(x,0)={{\theta }_{1}}(0,y)={{\theta }_{1}}(L,y)=0\\&{{\theta }_{1}}(x,H)=A\sin({\frac {\pi x}{L}})\\\end{aligned}}}
∂
2
θ
2
∂
x
2
+
∂
2
θ
2
∂
y
2
=
0
θ
2
(
x
,
0
)
=
θ
2
(
x
,
H
)
=
θ
2
(
0
,
y
)
=
0
θ
2
(
L
,
y
)
=
y
(
H
−
y
)
{\displaystyle {\begin{aligned}&{\frac {{{\partial }^{2}}{{\theta }_{2}}}{\partial {{x}^{2}}}}+{\frac {{{\partial }^{2}}{{\theta }_{2}}}{\partial {{y}^{2}}}}=0\\&{{\theta }_{2}}(x,0)={{\theta }_{2}}(x,H)={{\theta }_{2}}(0,y)=0\\&{{\theta }_{2}}(L,y)=y(H-y)\\\end{aligned}}}
θ
=
θ
1
+
θ
2
{\displaystyle \theta ={{\theta }_{1}}+{{\theta }_{2}}}
برای تعیین انتقال حرارت در شکل‌های هندسی مختلف ضریب شکل تعریف می شود. در جدول زیر ضریب شکل (s) در حالت‌های مختلف آمده است.
q
=
s
k
(
T
1
−
T
2
)
{\displaystyle q=sk\left({{T}_{1}}-{{T}_{2}}\right)}
با توجه به شکل زیر توزیع دما را بدست بیاورید .
ρ
c
∂
T
∂
t
=
k
(
∂
2
T
∂
x
2
+
∂
2
T
∂
y
2
)
{\displaystyle \rho c{\frac {\partial T}{\partial t}}=k({\frac {{{\partial }^{2}}T}{\partial {{x}^{2}}}}+{\frac {{{\partial }^{2}}T}{\partial {{y}^{2}}}})}
شرایط دائمی:
∂
T
∂
t
=
0
→
∂
2
T
∂
x
2
+
∂
2
T
∂
y
2
=
0
{\displaystyle {\frac {\partial T}{\partial t}}=0\to {\frac {{{\partial }^{2}}T}{\partial {{x}^{2}}}}+{\frac {{{\partial }^{2}}T}{\partial {{y}^{2}}}}=0}
∂
2
T
∂
x
2
+
∂
2
T
∂
y
2
=
0
T
(
x
,
0
)
=
T
0
T
(
x
,
H
)
=
T
0
+
T
1
sin
(
π
x
/
L
)
T
(
0
,
y
)
=
T
0
T
(
L
,
y
)
=
T
0
+
T
1
sin
(
π
x
/
L
)
{\displaystyle {\begin{aligned}&{\frac {{{\partial }^{2}}T}{\partial {{x}^{2}}}}+{\frac {{{\partial }^{2}}T}{\partial {{y}^{2}}}}=0\\&T(x,0)=T0\\&T(x,H)=T0+T1\sin(\pi x/L)\\&T(0,y)=T0\\&T(L,y)=T0+T1\sin(\pi x/L)\\\end{aligned}}}
مـــــــــــعـــــــا د لــــــــــــه 1 :
∂
2
T
1
∂
x
2
+
∂
2
T
1
∂
y
2
=
0
T
1
(
x
,
0
)
=
0
T
1
(
x
,
H
)
=
0
T
1
(
0
,
y
)
=
0
T
1
(
L
,
y
)
=
T
´
s
i
n
(
π
y
H
{\displaystyle {\begin{aligned}&{\frac {{{\partial }^{2}}T_{1}}{\partial {{x}^{2}}}}+{\frac {{{\partial }^{2}}T_{1}}{\partial {{y}^{2}}}}=0\\&T_{1}(x,0)=0\\&T_{1}(x,H)=0\\&T_{1}(0,y)=0\\&T_{1}(L,y)={\acute {T}}sin({\frac {\pi y}{H}}\\\end{aligned}}}
مـــــــــــعـــــــا د لــــــــــــه 2 :
∂
2
T
2
∂
x
2
+
∂
2
T
2
∂
y
2
=
0
T
(
x
,
0
)
=
T
0
T
(
x
,
H
)
=
T
0
+
T
1
sin
(
π
x
/
L
)
T
(
0
,
y
)
=
T
0
T
(
L
,
y
)
=
T
0
{\displaystyle {\begin{aligned}&{\frac {{{\partial }^{2}}T_{2}}{\partial {{x}^{2}}}}+{\frac {{{\partial }^{2}}T_{2}}{\partial {{y}^{2}}}}=0\\&T(x,0)=T0\\&T(x,H)=T0+T1\sin(\pi x/L)\\&T(0,y)=T0\\&T(L,y)=T0\\\end{aligned}}}
جواب کلی بصورت زیر است :
T
(
x
,
y
)
=
T
1
(
x
,
y
)
+
T
2
(
x
,
y
)
{\displaystyle T(x,y)=T_{1}(x,y)+T_{2}(x,y)}
کل نرخ انتقال حرارت را در سطح پایینی (y=0) را بدست آورید.
q
″
=
−
k
(
∂
T
∂
y
)
{\displaystyle q''=-k({\frac {{\partial }T}{\partial {y}}})}
q
y
″
=
−
k
T
1
sin
(
π
x
L
)
.
cosh
(
π
y
L
)
sinh
(
π
H
L
)
{\displaystyle {\begin{aligned}q''_{y}=-kT_{1}\sin({\frac {\pi x}{L}}).{\frac {\cosh({\frac {\pi y}{L}})}{\sinh({\frac {\pi H}{L}})}}\\\end{aligned}}}
در معادله بالا y=0 میگذاریم:
q
y
″
=
−
k
T
1
sin
(
π
x
L
)
.
π
L
sinh
(
π
H
L
)
{\displaystyle {\begin{aligned}q''_{y}=-kT_{1}\sin({\frac {\pi x}{L}}).{\frac {\frac {\pi }{L}}{\sinh({\frac {\pi H}{L}})}}\\\end{aligned}}}
q
=
∫
0
L
q
y
″
d
x
{\displaystyle q=\int _{0}^{L}q''_{y}\,dx}
q
=
−
k
T
1
.
(
π
L
)
sinh
(
π
H
L
)
∫
0
L
sin
(
π
x
L
)
d
x
{\displaystyle q=-kT_{1}.{\frac {({\frac {\pi }{L}})}{\sinh({\frac {\pi H}{L}})}}\int _{0}^{L}{\sin({\frac {\pi x}{L}})}\,dx}
⇒
q
=
−
k
T
1
S
i
n
h
(
π
H
L
)
{\displaystyle {\begin{aligned}&\\&\Rightarrow q={\frac {-k{{T}_{1}}}{Sinh\left({\frac {\pi H}{L}}\right)}}\\\end{aligned}}}
نرخ انتقال حرارت را در شکل زیر بیابید.
حل :
q
=
K
S
Δ
T
S
=
2
π
D
1
−
D
4
Z
=
2
π
∗
0.5
(
m
)
1
−
0.5
(
m
)
4
∗
1
(
m
)
=
3.6
m
⇒
q
=
0.2
(
w
m
.
k
)
∗
3.6
(
m
)
∗
80
(
k
)
=
57.6
W
{\displaystyle {\begin{aligned}&q=KS\Delta T\\&S={\frac {2\pi D}{1-{\frac {D}{4Z}}}}\\&={\frac {2\pi *0.5\left(m\right)}{1-{\frac {0.5\left(m\right)}{4*1\left(m\right)}}}}=3.6m\\&\Rightarrow q=0.2\left({\frac {w}{m.k}}\right)*3.6\left(m\right)*80\left(k\right)=57.6W\\\end{aligned}}}
در مرکز قطعه‌ای به طول L=2m با مقطع عرضی چهارگوش به ضلع w=1m سوراخی به قطر d=0.25m در امتداد طول قطعه مته شده است. رسانندگی گرمایی قطعه k=150w/m.k است. عبور سیال از سوراخ باعث می‌شود سطح داخلی در T=75 و سطح خارجی آن در T=25 بمانند. آهنگ انتقال گرما در قطعه چقدر است؟
حل : با توجه به شکل های بالا داریم :
s
=
(
2
π
L
L
n
(
1.08
W
D
)
)
=
2
π
∗
2
ln
(
1.08
1
0.25
)
=
8.59
m
⇒
q
=
s
k
(
T
1
−
T
2
)
=
8.59
m
∗
150
w
m
.
k
(
75
−
25
)
∘
c
=
64.4
k
w
{\displaystyle {\begin{aligned}&s=\left({\frac {2\pi L}{Ln\left(1.08{\frac {W}{D}}\right)}}\right)\\&={\frac {2\pi *2}{\ln \left(1.08{\frac {1}{0.25}}\right)}}=8.59m\\&\Rightarrow q=sk\left({{T}_{1}}-{{T}_{2}}\right)=8.59m*150{\frac {w}{m.k}}{{\left(75-25\right)}^{\circ }}c=64.4kw\\\end{aligned}}}
آهنگ انتقال گرما را برای استوانه تک دمای افقی در داخل یک محیط نیم‌نامتناهی اگر T2 وT1 به ترتیب برابر با صفر درجه ساتیگراد و ۱۰۰ درجه سانتیگراد باشد، بدست آورید.
q
=
k
s
(
T
1
−
T
2
)
i
f
L
>>
D
T
h
e
n
S
=
2
π
L
cosh
−
1
(
2
Z
D
)
i
f
Z
>
3
2
D
T
h
e
n
S
=
2
π
L
L
n
(
4
Z
D
)
Z
=
1
m
>
3
2
D
=
0.015
m
⇒
S
=
2
π
L
L
n
(
4
Z
D
)
⇒
q
=
0.02
×
170
×
100
=
340
w
{\displaystyle {\begin{aligned}&\\&q=ks({{T}_{1}}-{{T}_{2}})\\&if{\begin{matrix}{}&L>>D&Then&S={\frac {2\pi L}{{{\cosh }^{-1}}({\frac {2Z}{D}})}}\\\end{matrix}}\\&if{\begin{matrix}{}&Z>{\frac {3}{2}}D&Then&S={\frac {2\pi L}{Ln({\frac {4Z}{D}})}}\\\end{matrix}}\\&Z=1m{\begin{matrix}>{\frac {3}{2}}D=0.015m&\Rightarrow &S={\frac {2\pi L}{Ln({\frac {4Z}{D}})}}&\Rightarrow {\begin{matrix}q=0.02\times 170\times 100=340w&{}\\\end{matrix}}\\\end{matrix}}\\\end{aligned}}}
