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)=T1−TsT1−T0(1) (1) Θ s ( ξ , τ ) = T 1 − T s T 1 − T 0
Dimensionless fluid temperature:
Θs(ξ,τ)=T1−TsT1−T0(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
∂Θs∂τ=1ξ2∂∂ξ(ξ2∂Θs∂ξ)(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=−3B∂Θs∂ξ∣∣∣ξ=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 transform function
F(s)=∫∞0f(t)e−stdt(11) (11) F ( s ) = ∫ 0 ∞ f ( t ) e − s t d t
Apply Laplace transform on the left part of Eq.5
L{∂Θs∂τ}=∫∞0∂Θs∂τe−sτdτ=Θse−sτ∣∣∣∞0+s∫∞0Θse−sτdτ=s∫∞0Θse−sτdτ(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)=∫∞0Θse−sτdτ(13) (13) Θ ¯ ( s ) = ∫ 0 ∞ Θ s e − s τ d τ
we can get
L{∂Θs∂τ}=sΘ¯¯¯¯(s)(14) (14) L { ∂ Θ s ∂ τ } = s Θ ¯ ( s )
Apply Laplace transform on the right part of Eq.5
L(1ξ2∂∂ξ(ξ2∂Θs∂ξ))=∫∞01ξ2∂∂ξ(ξ2∂Θs∂ξ)e−sτdτ=1ξ2∫∞0∂∂ξ(ξ2∂Θs∂ξ)e−sτdτ=1ξ2ddξ(ξ2∫∞0dΘsdξe−sτdτ)=1ξ2ddξ(ξ2ddξ∫∞0Θse−sτdτ)=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¯¯¯¯f−1=−3BdΘ¯¯¯¯sdξ∣∣∣ξ=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=C1ξsinhs√ξ+C2ξcoshs√ξ(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=C1ξsinhs√ξ(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(sinhs√−s√coshs√)(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/ssinhs√−3Bs(sinhs√−s√coshs√)(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/ssinhs√−3Bs(sinhs√−s√coshs√)Bs(sinhs√−s√coshs√)=1s+31/s(sinhs√−s√coshs√)Bssinhs√−3sinhs√+3s√coshs√=1s+31/s(tanhs√−s√)(Bs−3)tanhs√+3s√=1s+3(1/s√)tanhs√−1(Bs−3)s√tanhs√+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√)tanhs√−1(Bs−3)s√tanhs√+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)
L−1{Θ¯¯¯¯f}=L−1{1s}+3L−1{(1/s√)tanhs√−1(Bs−3)s√tanhs√+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+L−1{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+6B∑k=1∞exp(−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(x2dydx−a2y)=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
