Advanced Thermal Analysis: Modelling of Distribution of Heat on a One-Dimensional Body (Research Paper Sample)

Advanced Thermal Analysis
ME 530
Student’s Name
Professor
Date of Submission
Nomenclature
cp-specific heat capacity at constant pressureJK
I-current Amps
L-length of slab [m]
R-resistance ohms
S-source term
t-time [sec]
n-natural numbers [unitless]
T-temperature [K]
Ttr-transient temperature
Tst-steady state temperature
TC-ambient temperature [K]
α-thermal diffusivity m2sec
ρ-density kgm3
κn-Eigenvalue nπL
Abstract
The aim of this project is to analytically solve the problem of thermal diffusion in one-dimensional conduction using Cartesian domain subjected to a steady source of heat due to a direct current (DC) Ohm’s heating. The current is connected to a slab, which lead to heat being transferred by conduction. The slab is of length, L, and has an ambient temperature Tc at x = 0 and x = L. There is no heat transfer due to convection. The slab has a constant conductivity along its length. The problem was solved and analyze to give a Fourier series solution
Introduction
One-dimensional thermal diffusion in the Cartesian domain subjected to uniform heat generation due to direct current (DC) Ohm’s heating for a slab of length L and subjected to an ambient temperature TC at x=0 and x=L with constant properties, energy generation, no convection and constant conductivity. The problem was solved and analyzed to give a Fourier series solution.
Problem Formulation
The problem can be modelled as shown below.
1294765139700 z = Hz = 0///////////////////LTcTCx00 z = Hz = 0///////////////////LTcTCx
2169795806450025184108763000
244157550800S00S167894042545H00H
Figure 1: One-dimensional Thermal Diffusion in the Cartesian Domain subjected to uniform heat.
The governing equation is,
 T t+V’ T=α 2T+SρCp

(1)

The governing equation is obtained by applying Fourier equation as follows
In figure 1, the temperature at the boundaries are set to an ambient temperature of TC.
Also, it’s shown that the top and bottom of the slab are insulated resulting no heat transfer occurs in the z-direction. Equation 2 is the heat source over the density and specific heat.
G=SρCp

(2)

We assume constant thermal conductivity. Then, we apply the assumption to the governing equation as the following,
 T z=0

(3)

 2T z2=0

(4)

 T y=0

(5)

 2T y2=0

(6)

Therefore, the equation becomes
 T t=α 2T x2+G

(7)

The below equation shows the relationship between resistance and voltage,
W=V×I=I×R×I=I2R

(8)

The source is uniformly distributed, as in the conversion from electrical to thermal energy (Ohmic heating):
S=I2RV= I2RL3 Wm3

(9)

Where V = Volume of the slab
V=L×L×L
Method: Steady State solution
The equation must be solved subject to specific boundary and initial conditions in order to determine T(x, t). Equation 3 can now be used to produce finite difference equations approximating the heat conduction problem. The equation may be integrated twice to obtain the general solution.
Tx,t=Tstx+Ttrt,x
The temperature at any given point has two components, i.e., the steady state and the transient temperature.

(10)

Subject to the boundary condition
x=0 : T=Tc
x=L :T=Tc
* The temperatures at the margins are the ambient temperatures
 2T t2= 2T x2+ 2T y2+ 2T z2

(11)

Tst0+Ttrt,0=Tc→Tst0=Tc

(12)

Tx,0=Tstx+Ttrt,0

(13)

The initial condition
t=0 :T=Tc
Ttr0,x=Tc-Tstx
We have uniform heat generation (S per unit volume) within the domain. The general heat conduction equation reduces to,
0=kρCp 2T x2+SρCp

(14)

→  2T x2+Sk=0

(15)

Integrating the above equation, we obtain the general solution as
 Tstx t=-Skx+C1

(16)

Tst=-S2kx2+C1x+C2

(17)

Both surfaces are maintained at a common temperature: (Refer Fig. 1) and the prescribed boundary conditions,
x=0 : T=Tc
x=L :T=Tc
Using the boundary conditions as explained, the constants of integration take the values,
C2=Tc ; C1=S2kL
Where C1 and C2 are constant of integration and can be determined according to boundary condition. The solution gives temperature distribution and heat transfer in a plane wall.
Tst=-S2kx2+S2kLx+Tc

(18)

Tst=S2kL-xx+Tc

(19)

For steady state equation,
Tst=S2kL-xx+Tc

(20)

Transient solution
Solving for transient,
 Tstx t+ Ttrt,x t= ” 2Tstx x2+” 2Ttrt,x x2

(21)

→  Ttr t=α 2Ttr x2

(22)

Transient Boundary Conditions
x =L : Ttr=0
x=0 ‶ Ttr=0
Transient Initial Conditions
t=0 : Tstx+Ttr0,x=Tc
t=0 ‶ Ttr(0,x)=Tc-Tst(x)
Transient Solution- Separation of Variables
Ttrt,x=θtXx

(23)

θtXx=αθ(t)X''x ÷α Xxθt

(24)

Since the left hand side is a function of X only and the right-hand side of the below equation is a function of θ only, both sides must be equal to a separation constant, κ, i.e.,
1αθtθt= X''xXx=-κ2

(25)

1αθθ=-κ2

(26)

1θdθdt=-α κ2

(27)

dθθ= -κ2α dt

(28)

Since κ cannot be zero, C2 must be zero and the equation becomes
lnθ=-κ2α t+C2

(29)

θ=eC3-κ2”t =eC3e-κ2αt=Ae-κ2”t

(30)

θ=Ae-κ2α t

(31)

1x d2xdx2=-κ2

(32)

d2Xdx2=-κ2X

(33)

d2xdx2+κ2X=0

(34)

λ2+κ2=0

(35)

Then, we obtain the equation,
λ2=-κ2

(36)

=±-k2 =±-1

(37)

λ κ= ±i

(38)

X=B1eλ1x+B2eλ2x = B1eiκx+B2e-iκx

(39)

Therefore, Euler identities,
eiφ=cosφ+isinφ

(40)

e-iφ=cosφ-isinφ

(41)

For the real values,
B2=B1*

(42)

B1=B1r+iB1i

(43)

Complex Conjugate
B1*=B1r-iB1i

(44)

Then,
B1+B1*=2B1r

(45)

B1B1*=B1r2-i2B1i2=B1r2+B1i2

(46)

Solving for X as the following,
X=B1eλ1x+B2eλ2x=B1eiκx+B1*e-iκx

(47)

X=B1cosκx+iB1sinκx+B1*cosκx+iB1*sinκx

(48)

X=B1+B1*cosκx+iB1+B1*sinκx

(49)

X=2B1rcosκx+i22B1isinκx

(50)

X=2B1rcosκx-2B1isinκx

(51)

X=b1sinκx+b2cosκx

(52)

Boundary & Initial Conditions
x=0 : Ttr = 0= X0θt , X0=0
L=0 : Ttr = 0= XLθt , XL=0
Ttr=0=XLθt→XL=0

(53)

t=0 : Ttr=Tc-Tstx
Therefore, the equation becomes
x=0 : 0=b1’0+b2’1=b2
x=L : 0=b1sinκx+0
From both equations we can see that,
b2=0
b1sinκx=0
sinκL=0

(54)

κL=n π

(55)

κ=n πL for n=1,2,3, …

(56)

Therefore,
Xn=b1nsin(κnX)

(57)

The solution of transient,
Ttr=n=1„θtXnx=n=1„Cne-κn2αtsinnπLx

(58)

Integrating both sides,
0LsinmπLxTc-Tstxdx=n=1„Cn0LsinnπLx sinmπLxdx

(59)

Orthogonally Condition
0LsinnπLxsinmπLxdx=0 for m≠nL2 for m=n

(60)

Then, the solution becomes,
0LsinmπLx-S2kL-xxdx=n=1„Cn0LsinnπLx sinmπLxdx

(61)

0LsinmπLx-S2kL-xxdx=n=1„CnL2 for m=n

(62)

L2Cn=0LsinnπLx-SxL2k+Sx22kdx

(63)

L2Cn= -SL2k0LxsinnπLxdx+S2k0Lx2sinnπLxdx