1
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2-Dimensional Transient Conduction ____________________________________________________________________________
We have discussed basic finite volume methodology applied to 1-dimensional steady and
transient conduction. It was noted that steady state formulation is a special case of transient
formulation and that transient numerical model does not require any significant changes over the
steady state model.
When the temperature variation in a system under consideration is appreciable in more
than one space dimensions, conduction becomes multidimensional. Majority of our discussion in
extending 1-D formulation to multidimensional formulation will be limited to 2-D conduction in
Cartesian coordinates (x, y) for the simplicity of presentation. Two dimensional conduction in
other orthogonal coordinate systems, such as cylindrical and polar coordinates, are straight
forward.
Finite Volume Equation
The general form of two dimensional transient conduction equation in the Cartesian
coordinate system is
Following the procedures used to integrate one dimensional transient conduction equation, we
integrate Eq.(1) over a control volume as shown in Figure 1.
Integrating the second term, we have
CT
t=
x(k
T
x) +
y(k
T
y) + S
(1)
0
1
s
n
w
e
t
t
0
1
s
n
w
e
t
t
P0
1
s
n
w
eP1
P0
P P0
0
1
0
1
( CT
t) dtdxdydz = C
T
tdt dxdydz
= CT
tdz dy dx = C(
T - T) t(1)
= C (T - T )
tt
t y x
x y
(2)
2
Figure 1 Control volume in the x-y coordinate
Integrating the third term, we get
Finally, integrating the linearized source term, we obtain
0
1
0
1
0
1
t
t
0
1
s
n
w
e
0
1
s
n
w
e
t
t
s
n
e w e wt
t
eE P
e
wP W
w
x(k
T
x) dxdydzdt = dz d(k
T
x) dydt
(1) (kT
x) - (k
T
x) dydt = (1) (k
T
x) - (k
T
x) dt
= ( )( ) kT - T
( x )- k
T - T
( x )
= y
y t
(3)
0
1
t
t
0
1
w
e
s
n
nN P
n
sP S
sy(k
T
y) dydxdzdt = k
T - T
( y )- k
T - T
( y )
x t
(4)
0
1
t
t
0
1
s
n
w
e
C P C P P(S + S T) dxdydzdt = (S + S T ) x y t (5)
3
Collecting terms, we get a finite volume representation of Eq.(1):
where
and
In deriving these expressions, the fully implicit scheme is used to evaluate the average values
during a given time step t as discussed.
Inspection of Eq.(6) shows that temperature TP is influenced by temperatures in the
neighboring four control volumes, TE, TW, TN and TS plus temperature at the immediate previous
time level, T0. Extension of 1-D to 2-D formulation is simply to add influences coming from two
additional control volumes in the y-direction. All coefficients appearing in Eq.(6) are positive.
In evaluating interface conductivities at the north and south sides of control volume, we
employ the same method as discussed for the east and west interfaces.
xk+xk
ykk2=
)x(
yk=a
1i+j,i1i+,
j1i+,,
e
eE
ij
jij
(9a)
P P E E W W N N S Sa T = a T + a T + a T + a T + b (6)
Ee
e
Ww
w
a =k
( x ); a =
k
( x )
y y
(7a; 7b)
Nn
n
Ss
s
a =k
( y ); a =
k
( y )
x x
(7c; 7d)
P0
C P0
P0a =
C; b = S + a T
x y
tx y (7e; 7f)
P E W N S P0
Pa = a + a + a + a + a - S x y (7g)
4
Figure 2 Index notations in x-y coordinate
5
Tj+1,i • N
(n)
x
Tj,i-1 • W x Tj,i • P x Tj,i+1• E
(w) (e)
x (s)
j (y) Tj-1,i • S
I (x)
Figure 2 continued
and
Expressions for aW and aS are similar to Eq.(9a) and Eq.(9b), respectively.
Solution Method
If we write Eq.(6) for all i’s (2 i N+1) and j’s (2 j M+1) , there result in (N x
M) simultaneous equations for (N x M) unknowns. A direct solution of this large sparse matrix
could take considerable computing time. Since many conduction problems are inherently
nonlinear due to number of factors as discussed in previous chapter, many iteration are usually
needed before reaching a converged solution. Therefore direct solution of nonlinear problem
may require too excessive computing time and is not recommended.
yk+yk
xkk2=
)y(
xk=a
1j+i,ji,1
i,1,
n
nN
jj
ijij
(9b)
t
yxC=a=a
ji,,0,
0P
ijij
ij
(9c)
TaxS=b 0,
0,+iC ji, ijijjy (9d)
P E E N S P0
Pa = a + a + a + a + a - S i, j x yi j (9e)
6
An alternative method is to use an alternate-direction-implicit (ADI) method [1]. The
most common practice is to use TDMA to solve dependent variable along one direction of spatial
coordinate implicitly while treating the dependent variable in the remaining spatial coordinate
explicitly. Fig. 3 shows such an approach called "line-by line" method. In the first step (Fig.3a),
TDMA is used to solve temperatures along the x-direction implicitly while treating temperatures
in the adjacent y-direction as a part of source terms. Thus
Using the index notation, Eq.(10) can be expressed as
Eq.(11) is solved for all j’s (2 j M+1) sweeping in the y-direction (Fig. 3a). After
completing y-direction sweep, we may now treat y-direction implicitly (Fig.3b) to speed up the
convergence. That is
or, in index notation
for all i’s (2 i N+1) (x-direction sweep). Sweeping along the x- and y-direction continues
until the solution converges within a given time step. To bring a faster convergence, it is
recommended that sweeping should start from the boundaries where perturbations originate. In
this way, computation will bring the changes into the calculation domain faster resulting in a
faster convergence.
Boundary Conditions
Treatment of boundary conditions for 2-dimensional conduction is similar to those of 1-
dimensional conduction. In addition to boundary conditions at x=0 and x=L, boundary
conditions are needed at y=0 and y=H, where L and H are the length and height of 2-
dimensional domain. At the boundaries, either one of three boundary conditions, known
temperatures, known heat flux and periodic or combination of these conditions are to be
prescribed. Heat flux is in general nonlinear function of boundary temperature. Detailed
P P E E W W N*
N*
S S*a T = a T + a T + a T + a T + b (10)
d+Tc+Tb=Ta ,1,,1,,j ,, ijijijijijiij (11)
P P N N S S E*
E*
W*
W*a T = a T + a T + a T + a T + b (12)
d+Tc+Tb=Ta ,1,j-,1,j+j,,, ijiijiiijij (13)
7
discussions on the treatment of boundary conditions along the x-direction are presented
previously.
Figure 3 Line-by-line TDMA Solution Scheme
8
. Following the same procedures, we can treat boundary conditions along the y-direction.
If the boundary temperatures are known, boundary conditions are given by
T1,i= T0
TM+1,i = TH (14)
where T0 and TH are given boundary temperatures at y=0 and y=H, respectively at “i” in x-
direction.
If a boundary heat flux is given at y=0, as shown in Fig. 4, energy balance shows that
]qk
y[
2
1+T=T
"
,2
2i,21, y
i
i
(15)
where is the heat flux along y-direction at y=0.
y
T1,i T2,I T3,i
x
• • •
Figure 4 Energy balance at y=0
Similarly, at y=H, we have
where heat flux is positive if directed to the positive y-direction.
]qk
y[
2
1-T=T y
1,M+
1M+1,M+2,M+
i
ii (16)
9
Treatment of all nominal boundary conditions on the surface of 2-dimensional geometry
can be handled for all types of boundary conditions; known temperature, known heat flux and
periodic in both x- and y-directions.
Sometimes in the 2-dimensional conduction, one has to deal with internal boundary
conditions to handle the effects of islands as shown in Fig. 5. The temperatures of the interior
region or regions are known. This could happen if interior region carries fluid and has a very
large convective heat transfer coefficient such that temperature at the interface between the fluid
and the solid is equal to the fluid temperature. We do not break up the calculation domain into
many regions and treat the effects of island as nominal boundary for the divided calculation
domains. Using the concept of interfacial conductivity and source term linearization, we can
handle islands very effectively [1]. We let the conductivity of the island be very large so as to
ensure temperature in the island prevails all the way to the interface between the island and the
control volumes adjacent to it. Next let the linearlized source term be represented by
where Tisland is the temperature of the island region.
If the temperature at the interface between the fluid and solid is different from the fluid
temperature, Eq.(17) can not be used. This is the case when the convective coefficient is
moderate and the temperature at the interface is different from the fluid temperature. A more
general treatment of such cases is presented in a later chapter [4].
Figure 5 Island region with known temperatures
Other 2-D Orthogonal Coordinate Systems
C20
island P20S = 10 T and S = -10 (17)
10
Three-dimensional transient conduction equation in the cylindrical coordinate is given by
where r is the radial, z, axial and , angular coordinate, respectively as shown in Fig. 6.
Axisymmetric Conduction
First consider a 2-D conduction for the axisymmetric case. For this case, temperature
variation is independent of angular angle () and Eq.(18) reduces to
Figure 6 Cylindrical coordinate system
Finite volume representation of Eq.(19) can be obtained by integrating the equation about
a control volume as defined in Fig. 7. This is left as an exercise. The resulting finite volume
equation is
where
CT
t=
1
r r(rk
T
r) +
1
r(k
1
r
T) +
z(k
T
z) + S
(18)
CT
t=
1
r r(rk
T
r) +
z(k
T
z) + S
(19)
P P E E W W N N S Sa T = a T + a T + a T + a T + b (20)
11
)z(
rk=a;
)z(
rk=a
w
Pw
W
e
Pe
E
rr (21a,b)
)r(
rk=a;
)r(
rk=a
s
ssS
n
nnN
zz (21c,d)
t
zrrp
C=a
0P
(21e)
zrrp S-a+a+a+a+a=a P0PSNWEP (21f)
and
Ta+S=b 0P
0PC zrrp (21g)
Comparing the finite volume equations of axisymmetric case and those of Cartesian
coordinate case, we observe the difference is due to geometry of control volume as shown in
Fig. 8. With some modifications in evaluating diffusion conductance and source term, therefore,
we can treat both situations similar way.
Figure 7 Control volume in axisymmetric case
12
Figure 8 Comparison of control volumes in x-y and z-r coordinates
Two dimensional conduction program for the axisymmetric case can be obtained from a
2-dimensional Cartesian program by noting the difference in the control volume shapes.
Polar Coordinate
In the polar coordinate, Eq.(18) reduces to
Integrating Eq.(22) over a control volume as shown in Fig. 9 we obtain
where
S+)T
r
1(k
r
1+)
r
T(rk
rr
1=
t
TC
(22)
P P E E W W N N S Sa T = a T + a T + a T + a T + b (23)
)r(
rk=a;
)r(
rk=a
w
wwW
e
eeE
(24a,b)
)(
k=a;
)(
k=a
s
s
S
n
nN
pp r
r
r
r (24c,d)
13
t
rrp
C=a
0P (24e)
rrS-a+a+a+a+a=a PP0PSNWEP (24f)
and
In these expressions, angles are in radians.
Again we observe that minor changes in the coefficients and source terms due to the
geometry change are all needed to modify the program written for the Cartesian coordinate
system. A general-purpose 2-dimensional transient as well as steady conduction program that can
be used for all three coordinate systems can be created.
One additional consideration is needed for the boundary condition in the polar coordinate.
The conduction flux along the -direction in the polar coordinate is
Taking energy balance at the fictitious control volume (Fig.10) at =0, we have
Solving for T1,i, we have
Likewise at =max , we have
Ta+S=b 0P
0PC rrp (24g)
T
r-=Jk
(25)
)+(2
1
T-Tk=J=q=q
21
1,,2
,2ondc
p
ii
i
r
(26)
q
k2
1+T=T
2,
2
2,1,
i
p
ii
r (27)
q
k2
1-T=T
1,M+
1M+
1,M+2,M+
i
p
ii
r (28)
14
Again heat flux along the positive angular direction is taken positive heat flux.
Figure 9 Control volume in polar coordinate
Figure 10 Energy balance at the boundary (=0)
15
General Purpose 2-D Conduction Program
A general purpose 2-dimensional conduction program (cond2d.m) that can handle three
orthogonal geometries is created by extending cond1d.m, that is the general purpose program
created in previous section. Extension of cond1d.m to cond2d.m is straightforward. Cond2d.m
has the same structure and follows the same logic in handling boundary conditions and solution
schemes. No new additional ideas are required. Listing of cond2d.m is shown below and
functions associated with the following example is in webpage.
Familiarity with cond2d.m can be best obtained by applying the program to few example
problems.
Listing of cond2d.m %cond2d.m %transient, 2-dimensional conduction (thickness is assumed to be 1) %,nonuniform condutivity and sources. Finite volume formulation %using matlab program. (By Dr. S. Han, Feb 15, 2007). modified on May 2012. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %functions called in main program % function grid2d_modified % function inital2d % function propty2d % function boundy2d_modified % function source2d % function solve2d_modified (function tdma or ctdma is called in this
function) % function convcheck2d % % geometry selection:geo=1(cartesian, x,y), geo=2(cylindrical axisymmetric,
x=z,y=r), % geo=3 (polar ,x=r,y=theta (in radian)) % clear all close all clc %specify the geometry geo=3; %specified cyclic boundary or not %x-direction. 0=non periodic, 1=periodic iperiodic=0; %y-direction. 0=non-periodic, 1=periodic jperiodic=0; %specify the number of control volumes n=10;%number of control volumes in x-direction m=10;%number of control volumes in y-direction np1=n+1; np2=n+2; np3=n+3;
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mp1=m+1; mp2=m+2; mp3=m+3; maxiter=100;%maximum iteration number %assign time step and maximum time tstop=2500; %time to stop calculation dt=1.0e10; %time increment, dt=1.0e10 for steady state calculation only mwrite=0; %time iteration to print the results. If dt>tstop set mwrite=0 re=1;%relaxation coefficient %define calculation domain (problem dependent) [x,y,dx,dy]=grid2d(geo,m,n,iperiodic,jperiodic); %prescribe intitial temperatures for all control volumes (problem dependent) [te,tep,te0]=inital2d(m,n); %%%%%%%%%%%%%%% %time loop begins here t=0;%starting time plotte=[te]; %save data for plot iwrite=1;%printout counter, iwrtie<mwrite means skip print out while t<tstop % calculation continues until t>tstop%%%%%%%%%%%%%%%OUTER LOOP %iteration for convergence in each time step iter=0; iflag=1; %iflag=1 means convergence is not met %iteration loop for the convergence while iflag==1 % end is at the end of program ***************INNER LOOP %prescribe thermal conductivity, density and specific heat (problem %dependent) [tk,ro,cp]=propty2d(m,n); %prescribe boundary temperature (problem dependent) [te,bx0,qx0c,qx0p,qx0,bx1,qx1c,qx1p,qx1,... by0,qy0c,qy0p,qy0,by1,qy1c,qy1p,qy1]=boundy2d(geo,te,tk,dx,dy,m,n,t,dt); %evaluate sourec terms (problem dependent) [sp,sc]=source2d(geo,te,m,n,x,dx,y,dy,t); %evaluate the coefficients and source and solve equations by function tdma %or ctdma (THIS FUNCTION NEED NOT BE CHANGED !!!!) [te,qx0,qx1,qy0,qy1]=solve2d(geo,x,dx,y,dy,tk,te,tep,te0,ro,cp,dt,m,n,sp,sc,r
e,...
bx0,qx0c,qx0p,qx0,bx1,qx1c,qx1p,qx1,by0,qy0c,qy0p,qy0,by1,qy1c,qy1p,qy1); %check the convergence(DO NOT CHANGE!!) [iflag,iter,tep]=convcheck2d(te,tep,m,n,iter); if iter>maxiter break %if iteration goes beyond maximum set iteration number stop the
computation end end % this end goes with the while iflag==1 at the top************INNER LOOP %solution converged within time step or iter > maxiter %advance to the next time level t=t+dt; %increase time %reinitialize dependent variable for i=1:np2 for j=1:mp2 te0(j,i)=te(j,i);%new temperature becomes old temperature tep(j,i)=te0(j,i); end end %write the results at this time?
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if iwrite>mwrite %print the results fprintf('iteration number is %i \n',iter) fprintf('time is %9.3f\n',t) % disp('temperatures are') % fprintf('%9.3f\n',te) iwrite=0; end iwrite=iwrite+1; end %this end goes with while t<tstop%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%OUTER LOOP %plot the result figure2d(geo,x,y,m,n,te);%2-D contour plot %additional calculations %energy balance qx_0=0; qx_1=0; qy_0=0; qy_1=0; for j=1:mp2 if geo==1 dae=dy(j); daw=dae; end if geo==2 dae=0.5*(y(j)+y(j+1))*dy(j); daw=dae; end if geo==3 dae=x(np3)*dy(j); daw=x(1)*dy(j); end qx_0=qx_0+qx0(j)*daw; qx_1=qx_1+qx1(j)*dae; end for i=1:np2 if geo==1 dan=dx(i); das=dan; end if geo==2 dan=y(mp3)*dx(i); das=y(1)*dx(i); end if geo==3 dan=dx(i); das=dan; end qy_0=qy_0+qy0(i)*das; qy_1=qy_1+qy1(i)*dan; end qin=-qy_1 qout=-qy_0-qx_0+qx_1
18
__________________________________________________________________________
Example 1
__________________________
Problem description:
Two dimensional, transient conduction in a rectangular geometry as shown in Fig.11
subject to the following initial and boundary conditions is considered [2].
T(x,y,0) = T1
T(0,y,t) = T1 T(W,y,t) = T1
T(x,0,t) = T1 T(x,H,t) = T2
T1=100 0C T2=200 0C
Figure 11 Two-dimensional conduction in x-y coordinates
For example applications for axisymmetric cylindrical problems, inner radius is assumed to be
0.5 m and polar angles from 0 to 2 radians.
Results:
Only the steady state solutions are considered. Temperature contour plots for three cases
are shown below figures. Energy balance shows good agreement for all cases.
T2
T1
W
H
x
y
T1
T1
19
temperature contour
x(m)
y(m
)
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0.2
0.4
0.6
0.8
1
1.2
1.4
temperature contour
x(m)
y(m
)
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0.6
0.8
1
1.2
1.4
1.6
1.8
2
20
Figure 12 Temperature Contour Plots for Example 1:
(Top) the Cartesian coordinate case, (middle) axisymmetric case with inner radius 0.5 m and
outer radius 2m, and (bottom) polar coordinate case with inner radius 0.5 m, outer radius 2m
and polar angles from 0.5 to 2 radians. Boundary conditions are the same for all three cases.
References
1. Patankar, S. V., Numerical Heat Transfer and Fluid Flow, Hemisphere Pub., 1980.
2. Janna, William, S., Engineering Heat Transfer, PWS publishers, 1986.
3. Incropera, F. P., and Dewitt, D. P., Fundamentals of Heat Transfer, 3rd Ed., John Wiley
& Sons, 1990.
4. Patankar, S. V., Computation of Conduction and Duct Flow Heat Transfer, Innovative
Research, Inc., 1991.
5. Arpaci, V. S., Conduction Heat Transfer, Addison-Wesley, 1966.
temperature contour
x(m)
y(m
)
-0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 1.2
0.4
0.6
0.8
1
1.2
1.4
21
Problems
1. Derive finite volume equation for axisymmetric case, Eq. (20).
2. Derive finite volume equation for polar coordinate, Eq.(23).
3. The diagram shows an element of electrical rotor used in an electrical drill. Energy is
generated in the epoxy-impregnated region by current dissipation and in the metal region by
induced current. Internally generated energy is dissipated by convection to keep the
temperature in the rotor below certain limit. The material properties and energy generation
rates are:
Epoxy-impregnated
winding
Insulating retainer Metal
Density (lbm/ft3) 350 58 490
Conductivity
(Btu/hr.ft.F)
0.8 0.05 26
Specific heat
(Btu/lbm.F)
0.11 0.485 0.11
Heat generation
(W/cm3)
6.3 0.0 7.54
(a) Calculate the transient temperature in the rotor when h=100 Btu/hr.ft2.R and Tinitial=70
F, which is the room temperature. The melting temperature of the epoxy is 400 F. How
long will it take to reach this temperature?
(b) Pot steady state temperature distribution and take energy balance for case with h=1000.
22
4. A simplified representation for cooling in very large-scale integration (VLSI) of microelectronics
is shown in the sketch. A silicon chip is mounted in a dielectric substrate, and one surface of system
is cooled convectively, while the remaining three surfaces are insulated. Before the machine is turned
on system is in thermal equilibrium with the surrounding air at 20 0C. When the machine is turned
on, the electric power dissipation in the chip is uniform at 107 W/m3 and the forced convective
coefficient is 500 W/m2.K. [3]
(a) Calculate time dependent temperature distributions until steady state is reached. How long will it
take to reach the steady state?
(b) Is the maximum temperature in the system below 85 0C, the maximum allowable temperature,
during the operation?
(c) Take an energy balance at the steady state.
(d) Plot temperature distribution of the steady state.
23
5. Consider a cylindrical stainless steel (AISI 302) fin with a 5 cm-diameter and 20 cm long. The
base temperature is 500 K and the remaining surface of the fin is subject to convection and radiation
except the end of the fin, which is insulated. The emissivity of the fin is 0.22, density, 8055 kg/m3,
and specific heat, 480 J/kg.K. The fin is initially at the surrounding air temperature of 300 K.
Assume a two dimensional conduction in axial and radial directions.
(a) The heat transfer coefficient is made of two parts, hc (natural convection) and hr (radiation).
Discuss methods to find these values.
(b) Calculate the time dependent temperature distribution in the fin and determine the time at which
steady state is reached.
(c) Plot the temperature distributions for the steady state.
(d) Plot hc and hr as a function of fin axis for the steady state.
(e) Take an energy balance for the steady state.
L=27mm
H=12mm
Air at 20 0C and
Hf=500 W/m2K
L/3
H/4
Chip, kc=50 W/m.K
Substrate, ks=5 W/m.K
Insulation Symmetry line
24
6. Reconsider the fin problem described in Problem 7. Instead of a natural convection, a forced
convection is applied. The air velocity is 10 m/s. Repeat the calculations and answer all questions as
posed in Problem 7.
7. Both ends of a solid rod of radius R and length 2L are insulated. Half way between the the ends of
this rod a metal sleeve of negligible thickness and length 2l rotates with constant angular velocity, .
The pressure between the sleeve and the rod is p, and the coefficient of dry friction is . Energy
dissipation by friction is q''=pV , where V=R. Heat transfer from the surface to the
surrounding fluid is by forced convection on the surface of the metal sleeve and natural convection
plus radiation from the rod. Assume emissivity of the surface to be 0.3 for both surfaces. Calculate
the steady state temperature distribution in the rod. [5]
25
26
8. Two solid cylinders made of different materials are rotating reverse with a constant angular
velocity and are pressed suddenly by a force P as shown in the diagram. Energy generated by the
solid friction ( PV ) is conducted through the solids and to the surrounding fluid by convection.
Calculate the steady-state temperature in the solids. [5]
L=L’=1.3 cm, R= 15 cm, 21 = 8000 kg/m3, Cp1=Cp2=420 J/kg.K, k=35 W/m.K,
k’=1.5k1, h1=2200 W/m2K, h2=1.5h1, Tfluid =25 0C, ω=1500 rpm, Tinitial=25 0C, 34 /10 mJP and V=rω is the linear velocity at radius r, m/s.
(a) Write the appropriate heat conduction equation and discuss boundary conditions?
(b) Calculate transient conduction for the duration of 5 min. Use 10 by 10 control volumes and
1.0 sec. Plot steady state temperature distribution. Where and what is the maximum
temperature?
(c) Plot transient temperature at the corner of the right cylinder as a function of time and estimate
the time needed to reach a steady-state.
(d) At the steady state, take the energy balance.
27
9. A long bar of rectangular cross section is made of two materials as shown. All three sides are
exposed to convective environment and the bottom surface is at constant temperature. Determine the
heat transfer rate per unit depth from the bottom that is removed by the cooling water channel. Plot
temperature contours and heat flux vectors [3].
Conductivity of material A is 15.0 W/m.K and B is 1.0 W/m.K. W=300 mm, LA=50 mm, LB=100
mm. T0=0 0C and hf=30 W/m2 K and Tf=100 0C.
W
LA
LB
T0
Cooling water channel
A
B
Hf, Tf