Laplace transform解传热偏微分方程

Cooling of a Sphere in Contact with a Well-Stirred Fluid

A homogeneous solid sphere of radius R, initially a t a uniform temperature T1 T 1 , is suddenly immersed at time t=0 t = 0 in a volume Vf V f of well-stirred fluid of temperature T0 T 0 in an insulated tank. It is desired to find the thermal diffusivity αx=ksρsC^ps α x = k s ρ s C ^ p s of the solid by observing the change of the fluid temperature Tf T f with time. We use the following dimensionless variables
Dimensionless solid temperature:

Θs(ξ,τ)=T1TsT1T0(1) (1) Θ s ( ξ , τ ) = T 1 − T s T 1 − T 0

Dimensionless fluid temperature:
Θs(ξ,τ)=T1TsT1T0(2) (2) Θ s ( ξ , τ ) = T 1 − T s T 1 − T 0

Dimensionless radial coordinate:
ξ=rR(3) (3) ξ = r R

Dimensionless time:
τ=αstR2(4) (4) τ = α s t R 2

Solution:

For the solid sphere,
Heat transfer equation

Θsτ=1ξ2ξ(ξ2Θsξ)(5) (5) ∂ Θ s ∂ τ = 1 ξ 2 ∂ ∂ ξ ( ξ 2 ∂ Θ s ∂ ξ )

Initial condition
At τ=0 τ = 0 , Θs=0 Θ s = 0 (6) (6)
Boundary condition
At ξ=1 ξ = 1 , Θs=Θf Θ s = Θ f (7) (7)
At ξ=0 ξ = 0 , Θs=Finite Θ s = F i n i t e (8) (8)

For the fluid,

dΘfdτ=3BΘsξξ=1(9) (9) d Θ f d τ = − 3 B ∂ Θ s ∂ ξ | ξ = 1

in which
B=ρfC^pfVfρsC^psVs B = ρ f C ^ p f V f ρ s C ^ p s V s

Initial condition
At τ=0 τ = 0 , Θf=1 Θ f = 1 (10) (10)

Laplace transformation

Laplace transform function

F(s)=0f(t)estdt(11) (11) F ( s ) = ∫ 0 ∞ f ( t ) e − s t d t

Apply Laplace transform on the left part of Eq.5

L{Θsτ}=0Θsτesτdτ=Θsesτ0+s0Θsesτdτ=s0Θsesτdτ(12) L { ∂ Θ s ∂ τ } = ∫ 0 ∞ ∂ Θ s ∂ τ e − s τ d τ = Θ s e − s τ | 0 ∞ + s ∫ 0 ∞ Θ s e − s τ d τ (12) = s ∫ 0 ∞ Θ s e − s τ d τ

Define
Θ¯¯¯¯(s)=0Θsesτdτ(13) (13) Θ ¯ ( s ) = ∫ 0 ∞ Θ s e − s τ d τ

we can get
L{Θsτ}=sΘ¯¯¯¯(s)(14) (14) L { ∂ Θ s ∂ τ } = s Θ ¯ ( s )

Apply Laplace transform on the right part of Eq.5
L(1ξ2ξ(ξ2Θsξ))=01ξ2ξ(ξ2Θsξ)esτdτ=1ξ20ξ(ξ2Θsξ)esτdτ=1ξ2ddξ(ξ20dΘsdξesτdτ)=1ξ2ddξ(ξ2ddξ0Θsesτdτ)=1ξ2ddξ(ξ2ddξΘ¯¯¯¯(s))(15) L ( 1 ξ 2 ∂ ∂ ξ ( ξ 2 ∂ Θ s ∂ ξ ) ) = ∫ 0 ∞ 1 ξ 2 ∂ ∂ ξ ( ξ 2 ∂ Θ s ∂ ξ ) e − s τ d τ = 1 ξ 2 ∫ 0 ∞ ∂ ∂ ξ ( ξ 2 ∂ Θ s ∂ ξ ) e − s τ d τ = 1 ξ 2 d d ξ ( ξ 2 ∫ 0 ∞ d Θ s d ξ e − s τ d τ ) = 1 ξ 2 d d ξ ( ξ 2 d d ξ ∫ 0 ∞ Θ s e − s τ d τ ) (15) = 1 ξ 2 d d ξ ( ξ 2 d d ξ Θ ¯ ( s ) )

Hence,the equation 5 becomes the format
sΘ¯¯¯¯(s)=1ξ2ddξ(ξ2ddξΘ¯¯¯¯(s))(16) (16) s Θ ¯ ( s ) = 1 ξ 2 d d ξ ( ξ 2 d d ξ Θ ¯ ( s ) )

Also, apply Laplace on the boundary conditions. They are

At ξ=1 ξ = 1 , Θ¯¯¯¯s=Θ¯¯¯¯f Θ ¯ s = Θ ¯ f (17) (17)
At ξ=0 ξ = 0 , Θ¯¯¯¯s=Finite Θ ¯ s = F i n i t e (18) (18)
The same scenario for Eq.9, its Laplacian form is

sΘ¯¯¯¯f1=3BdΘ¯¯¯¯sdξξ=1(19) (19) s Θ ¯ f − 1 = − 3 B d Θ ¯ s d ξ | ξ = 1

Here s s is the transform variable.
The general solution of Eq.16 is
Θ¯¯¯¯s=C1ξsinhsξ+C2ξcoshsξ(20) (20) Θ ¯ s = C 1 ξ sinh ⁡ s ξ + C 2 ξ cosh ⁡ s ξ

Substitution of boundary conditions into Eq. 20 gives C2=0 C 2 = 0 .
Θ¯¯¯¯s=C1ξsinhsξ(21) (21) Θ ¯ s = C 1 ξ sinh ⁡ s ξ

Substitution of Eq.(21) into Eq.(19) to solve Θ¯¯¯¯f Θ ¯ f ,
Θ¯¯¯¯f=1s+3C1Bs(sinhsscoshs)(22) (22) Θ ¯ f = 1 s + 3 C 1 B s ( sinh ⁡ s − s c o s h s )

Using the boundary condition Eq.(17) to solve C1 C 1 , it gives
C1=1/ssinhs3Bs(sinhsscoshs)(23) (23) C 1 = 1 / s sinh ⁡ s − 3 B s ( sinh ⁡ s − s cosh ⁡ s )

Plug Eq.(23) into Eq.(22),
Θ¯¯¯¯f=1s+31/ssinhs3Bs(sinhsscoshs)Bs(sinhsscoshs)=1s+31/s(sinhsscoshs)Bssinhs3sinhs+3scoshs=1s+31/s(tanhss)(Bs3)tanhs+3s=1s+3(1/s)tanhs1(Bs3)stanhs+3s(24) Θ ¯ f = 1 s + 3 1 / s sinh ⁡ s − 3 B s ( sinh ⁡ s − s cosh ⁡ s ) B s ( sinh ⁡ s − s c o s h s ) = 1 s + 3 1 / s ( sinh ⁡ s − s c o s h s ) B s sinh ⁡ s − 3 sinh ⁡ s + 3 s cosh ⁡ s = 1 s + 3 1 / s ( tanh ⁡ s − s ) ( B s − 3 ) tanh ⁡ s + 3 s (24) = 1 s + 3 ( 1 / s ) tanh ⁡ s − 1 ( B s − 3 ) s tanh ⁡ s + 3 s

finally
Θ¯¯¯¯f=1s+3(1/s)tanhs1(Bs3)stanhs+3s(25) (25) Θ ¯ f = 1 s + 3 ( 1 / s ) tanh ⁡ s − 1 ( B s − 3 ) s tanh ⁡ s + 3 s

Apply inverse Laplace transform on Eq(28)
L1{Θ¯¯¯¯f}=L1{1s}+3L1{(1/s)tanhs1(Bs3)stanhs+3s}(26) (26) L − 1 { Θ ¯ f } = L − 1 { 1 s } + 3 L − 1 { ( 1 / s ) tanh ⁡ s − 1 ( B s − 3 ) s tanh ⁡ s + 3 s }

to get
Θf=1+L1{N(s)D(s)}(27) (27) Θ f = 1 + L − 1 { N ( s ) D ( s ) }

where
N(0)D(0)=1/31+B(28) (28) N ( 0 ) D ′ ( 0 ) = 1 / 3 1 + B

N(sk)D(sk)=2B9(1+B)+B2b2k(29) (29) N ( s k ) D ′ ( s k ) = 2 B 9 ( 1 + B ) + B 2 b k 2

where
bk=3bk3+Bb2k(30) (30) b k = 3 b k 3 + B b k 2

Hence, the solution will be obtained as
Θf=B1+B+6Bk=1exp(b2kτ)9(1+B)+B2b2k=1(31) (31) Θ f = B 1 + B + 6 B ∑ k = 1 ∞ exp ⁡ ( − b k 2 τ ) 9 ( 1 + B ) + B 2 b k 2 = 1


1.
1x2ddx(x2dydxa2y)=0(1) (1) 1 x 2 d d x ( x 2 d y d x − a 2 y ) = 0
Solution:
y=C1xsinhax+C2xcoshax(2) (2) y = C 1 x sinh ⁡ a x + C 2 x cosh ⁡ a x

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