1 _____________________________________ Two Dimensional Convection-Diffusion ____________________________________ Extension to 2-D Convection-Diffusion Extension of 1-dimensional convection-diffusion formulation to 2-dimensional convection-diffusion is straightforward. We will discuss 2-dimensional convection-diffusion formulation in terms of x-y coordinates. Extension to axisymmetric and polar coordinates will be similar to conduction cases. S y J x J t y x (1) where the convection-diffusion in x-y coordinates are y - v = J x - u = J y x (2) In Eq.(2), u is the velocity in x-direction and v is the velocity in y-direction. Integrating Eq.(1) over a control volume as shown in Fig.1, we obtain b + a + a + a + a = a S S N N W W E E P P (3) where 0 , + |) P A(| D = a e e E e F (4a) 0 , + |) P A(| D = a w w W w F (4b) 0 , + |) P A(| D = a n n N n F (4c) 0 , + |) P A(| D = a s s S s F (4d)
Transcript
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_____________________________________
Two Dimensional Convection-Diffusion
____________________________________
Extension to 2-D Convection-Diffusion
Extension of 1-dimensional convection-diffusion formulation to 2-dimensional
convection-diffusion is straightforward. We will discuss 2-dimensional convection-diffusion
formulation in terms of x-y coordinates. Extension to axisymmetric and polar coordinates will be
similar to conduction cases.
Sy
J
x
J
t
yx
(1)
where the convection-diffusion in x-y coordinates are
y - v = J
x - u = J
y
x
(2)
In Eq.(2), u is the velocity in x-direction and v is the velocity in y-direction.
Integrating Eq.(1) over a control volume as shown in Fig.1, we obtain
b + a + a + a + a = a SSNNWWEEPP (3)
where
0, + |)PA(| D = a eeE eF (4a)
0,+ |)PA(| D = a wwW wF (4b)
0, + |)PA(| D = a nnN nF (4c)
0,+ |)PA(| D = a ssS sF (4d)
2
t
yx = a
0
P0P
(4e)
0
P
0PC a +y x S = b (4f)
yxS - a + a + a + a + a = a P0PSNWEP (4g)
y )u ( = F ; y )u ( = F wwee (4h)
x) v( = F ; x ) v( = F ssnn (4i)
)x(
y = D ;
)x(
y = D
w
ww
e
ee
(4j)
)y(
x = D ;
)y(
x = D
s
ss
n
nn
(4k)
and
)|P|0.1 - (1 , 0 = |)PA(|5
(4l)
Local Peclet number is defined by P=F/D.
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Figure 1 Control volume in x-y coordinates
Axisymmetric and Polar Coordinates
2-D axisymmetric convection-diffusion in cylindrical coordinate is governed by
S = x
J +
y
J
y
1+
t
xy
(5)
where
x - u = J
)y
- v(y = J
x
y
(6a;6b)
are the convection-diffusion flux along the radial (y) and axial (x) direction, respectively.
In the polar coordinate system, convection-diffusion equation is
S = J
x
1 +
x
J
x
1 +
t
x
(7)
where
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y
x
1 - v = J
)x
- u x(= J
y
x
(8a; 8b)
In these equations x is the radial direction and y is the angular direction measured positive in the
counter clockwise direction and u is the radial velocity and v is the angular velocity.
Integration of Eqs.(5) and (7) will yield finite volume representations similar to Cartesian
case with appropriate changes associated with the geometry changes as was demonstrated in the
conduction formulation. Details are left as an exercise. Diffusion and convection fluxes in
cylindrical axisymmetric flow are
)y(
x y k = D ;
)x(
y y k = D
n
nn
n
e
Pe
e
(9a; 9b)
x y ) v( = F ; y y )u ( = F nnnPee (10a; 10b)
In the polar coordinates, they are
)y(x
x k = D ;
)x(
y x k = D
nP
nn
e
eee
(11a; 11b)
x ) v( = F ; y x )u ( = F nneee (12a;.12b)
Figure 2 Velocity components in the cylindrical and polar coordinates
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2-D Convection-Diffusion Program
By modifying 1-D convection-diffusion program, a general 2-D convection-diffusion
program can be easily obtained. This program is named "conv2d.m" which solves 2-D Cartesian
(geo=1), cylindrical axisymmetric (geo=2) and polar coordinates (geo=3) problems with a
known 2-dimensional velocity field.
Thermally Developing Pipe Flow
In a previous chapter, thermally-fully-developed flow was analyzed by using 2-
dimensioanl steady conduction formulations. To include the entrance region in the calculation,
convection-conduction equation must be used. For a circular pipe, the governing differential
equation is Eq.(5) with
0 = S and C
k = , T =
P
(13)
Consider a circular pipe with hydrodynamically fully developed laminar flow. A radially
and axially constant wall heat flux or constant wall temperature is imposed beginning at x=0. At
x=0, temperature of the fluid is known. At x=L, it can be assumed that conduction flux in the
axial direction is much smaller than the convection flux in the axial direction and conduction
flux in the radial direction, i.e.,
y
Tk ; Tu C
x
Tk P
(14)
Thus
0 = x
T
L
(15)
For a laminar flow, velocity profile is parabolic. Thus
])r
r( - [1u = u(r)
2
0
m ax (16)
where umax is the centerline velocity and r0 is the radius of the pipe. The radial velocity is zero
because of hydrodynamically fully developed assumption.
Problem description: Water is flowing steadily through a 5 cm-dia circular pipe. A constant wall heat flux is
applied between x=0 and x=15 m. We would like to obtain a steady temperature distribution in
the water and the local Nusselt number variation along the axis of pipe. Assume constant
thermodynamic properties of water and neglect the thickness of the pipe. Inlet air temperature is
300 K. The results presented in non- dimensional form do not depend on inlet and boundary heat
flux and fluid properties.
User written functions: Geo is 2 since cylindrical axisymmetric case. Twenty five unequal control volumes in the
axial direction, which is clustered near the entrance and 20 unequal control volumes in
the radial direction are used. Time step is taken to be very large to obtain a steady state
solution. Depending on the index (PWRN), grid stretching in any direction can be easily
accomplished. The thermodynamic properties of water at about 300 K are prescribed.
These properties can be made temperature dependent without difficulties if desired. Axial
velocity profile is calculated by Eq.(12) with centerline velocity of 0.0034 m/s and radial
velocity is set to zero. Initial temperature of the fluid is 300 K. Final solution is of course
independent of initial temperature.
Boundary condition at x=0 is given at 300 K. Thus bx0(j)=1 and te(j,1)=300 for all j's. At
x=L, conduction flux is assumed to be negligible. This amounts to zero conduction flux
at x=L. Thus bx1(j)=2 for all j's. At y=0, symmetric boundary condition is applicable.
Thus, by0(i)=2 for all i’s. At y=ymax, a constant wall heat flux (qw) is given. Thus,
by1(i)=2 and qx1c(i)= -qw and qx1p(i)=0 for all i’s.
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There are no source terms in the present example.
Local Nusselt number along the axis is calculated to compare with the known data.
k
)(2 hx = uN
,2
0i
i
iM
r
(17)
where hxi is the local convective heat transfer coefficient given by
T - T
=hxim,,2
''
i
iM
wq
(18)
and Tm,i is the mean fluid temperature at each ith location,
yyTu
u2
1
2 = T jjj ,,
2M+
1j=2
maxmax
im, iij
y (19)
And wall heat flux is calculated by
(20)
Non-dimensional temperature is defined by
(21)
Results: The non-dimensional temperature distribution and local Nusselt number are shown in
Fig. 5. Non-dimensional temperature distribution becomes independent of axis away
from the entrance region as expected. Local Nusselt number is large in the developing
region and approaches to the value of fully developed value of 4.36.
Comments:
In this example a constant wall heat flux boundary is used to compare the numerical
results with the known exact solutions to validate the accuracy of numerical formulations.
In numerical calculation, we can apply variety of complex boundary conditions for which
exact solutions cannot be found easily. These boundary conditions can be axially non-
uniform and can also be time dependent as well.
clear all close all clc %conv2d_input_sample.m %trnasient, 2-dimensional convection-diffusion
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%nonuniform diffusivity and sources. Finite volume formulation %using matlab program. (By Dr. S. Han, October 15, 2007) %modified March, 2013 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % function nux_conv2d % geometry selection:geo=1(cartesian, x,y), geo=2(cylindrical axisymmetric,
x=z,y=r), % geo=3 (polar ,x=r,y=theta (in radian)) %select the geometry geo=2; %specify the number of control volumes n=25;%number of control volumes in x-direction m=20;%number of control volumes in y-direction np1=n+1;np2=n+2;np3=n+3;mp1=m+1;mp2=m+2;mp3=m+3; maxiter=300;%maximum iteration number %assign time step and maximum time tstop=40; %time to stop calculation dt=1.0e10; %time increment, dt=1.0e10 for steady state calculation mwrite=0; %time iteration to print the results. If dt>tstop set mwrite=0 re=1.0;%relaxation coefficient %define calculation domain tl=15;%length of the pipe %clustered grid small=1.0e-10; pwrn=0.5;% cluster x=0 (pwrn<1.0), cluster near x=xmax(pwrn>1) for i=2:np2 temp=1-(i-2)/(np2-2); x(1,i)=tl*(1-temp^pwrn)+small; end x(1,1)=x(1,2)-small; x(1,np3)=x(1,np2)+small; for i=1:np2 dx(1,i)=x(1,i+1)-x(1,i); end % if geo=3 , y is angular direction and must be in radian height=0.025;%radius %clustered grid in y-direction small=1.0e-10; pwrn=1.5;% cluster y=0 (pwrn<1.0), cluster near y=ymax(pwrn>1) pwrn=3; for j=2:mp2 temp=1-(j-2)/(mp2-2); y(j,1)=height*(1-temp^pwrn)+small; end y(1,1)=y(2,1)-small; y(mp3,1)=y(mp2,1)+small; for j=1:mp2 dy(j,1)=y(j+1,1)-y(j,1); end %assign y-coordinate y(1,1)=0; % for j=1:mp2 y(j+1,1)=y(j,1)+dy(j,1); end %prescribe intitial temperatures for all control volumes for j=1:mp2
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for i=1:np2 te0(j,i)=295; te(j,i)=te0(j,i); tep(j,i)=te(j,i); %predicted value at the next iteration end end %specify velocity u=zeros(mp2,np3); v=zeros(mp3,np2); ymax=y(mp3,1); umax=0.0034; for i=1:np3 for j=1:mp2 yp=0.5*(y(j,1)+y(j+1)); u(j,i)=umax*(1-(yp/ymax)^2);%parabolic profile end end for i=1:np2 for j=1:mp3 v(j,i)=0.0; % y-direction velocity end end %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 properties for i=1:np2 for j=1:mp2 tk(j,i)=0.613;%conductivity of water ro(j,i)=999; cp(j,i)=4179; end end %prescribe boundary conditions %initialize fluxes qx0c=zeros(mp2,1);qx0p=zeros(mp2,1);qx0=zeros(mp2,1); qx1c=zeros(mp2,1);qx1p=zeros(mp2,1);qx1=zeros(mp2,1); qy0c=zeros(1,np2);qy0p=zeros(1,np2);qy0=zeros(1,np2); qy1c=zeros(1,np2);qy1p=zeros(1,np2);qy1=zeros(1,np2); %bx0(j)=1,2,3(known temperature, known flux, periodic) at x=0 %bx1(j)=1,2,3 (same) at x=xmax %by0(i)=1,2,3(known temperature, known flux, periodic) at y=0 %by1(i)=1,2,3 (same) at y=ymax %at x=0 & x=xmax for j=1:mp2 bx0(j)=1; te(j,1)=300; end for j=1:mp2
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bx1(j)=2; end %at y=0 & y=ymax for i=1:np2 by0(i)=2; end for i=1:np2 by1(i)=1; te(mp2,i)=700;%constant wall temp % by1(i)=2; %constant wall heat flux % qy1c(i)=-100; % qy1p(i)=0; % qy1(i)=qy1c(i)+qy1p(i)*te(mp2,i); end %evaluate sourec terms for i=1:np2 for j=1:mp2 sp(j,i)=0; sc(j,i)=0; end end %%%%%%%%%%%%%%%% conv2d_invariant; %%%%%%%%%%%%%%%%%%%%%%% 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? if iwrite>mwrite %print the results fprintf('iteration number is %i \n',iter) fprintf('time is %9.3f\n',t) iwrite=0; end iwrite=iwrite+1; end %this end goes with while t<tstop%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%OUTER LOOP % % calculate the local and average Nusselt number along the axis nux_conv2d_new;%calculate local Nussel number
%plot the result subplot(1,2,1) figure2d_conv(geo,x,y,m,n,te,theta); for i=1:np2 xc(i)=0.5*(x(1,i)+x(1,i+1)); end subplot(1,2,2) plot(xc,nux,'-') grid on
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title('local Nusselt number vs x') xlabel(' axis, m'), ylabel(' Nux') % save nux_laminar_pipe nux
Figure 5 Non-dimensional temperature distribution (left) and
local Nusselt number in thermally developing laminar circular
pipe flow with constant wall temperature (right)
Thermally-developing Turbulent Pipe Flow
A thermally-developing turbulent pipe flow can be analyzed by adding increased
thermal conductivity due to turbulent velocity field to the physical conductivity. The simplest
model for turbulence is the mixing length hypothesis. In this model, turbulent viscosity is
determined by
y
uLt
2
Where the mixing length L is represented by
non-dimensional temperature contour
x(m)
y(m
)
5 10 15
0.005
0.01
0.015
0.02
0.025
0 5 10 15 203.5
4
4.5
5
5.5
6
6.5
7local Nusselt number vs x
axis, m
Nux
12
271
5.0
42
5.0Re026.0
)(
06.008.014.0),(
26/exp1
,
meanw
w
u
yHy
H
y
H
yHyH
yyD
yHyDL
Turbulent thermal conductivity is then calculated by
ttpt Ck Pr/
where Prt is the turbulent Prandtl number (0.9-1.0).
Problem description:
The geometry of the pipe and the fluid properties are those used in previous example
except the velocity. The maximum velocity is increased to 0.43 m/s to have Reynolds number
of about 16000. A hydrodynamically fully-developed turbulent velocity profile is represented
by
n
H
yuyu
/1
max 1)(
where n=6.6 is used. H is the radius and y is measured from the axis.
User written functions:
User supplied functions are similar to the previous example of laminar flow except the
part for calculating the turbulent thermal conductivity using mixing length model. In order to
increase the accuracy a grid stretching in the radial direction is also used to give smaller grid
near the solid wall.
clear all close all clc %conv2d_input_turbulent.m %transient, 2-dimensional convection-diffusion %nonuniform diffusivity and sources. Finite volume formulation %using matlab program. (By Dr. S. Han, October 15, 2007) %modified on March 2013 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % function nux_conv2d % geometry selection:geo=1(cartesian, x,y), geo=2(cylindrical axisymmetric,
x=z,y=r), % geo=3 (polar ,x=r,y=theta (in radian))
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%select the geometry geo=2; %specify the number of control volumes n=25; m=20;%number of control volumes in y-direction np1=n+1;np2=n+2;np3=n+3;mp1=m+1;mp2=m+2;mp3=m+3; maxiter=300;%maximum iteration number %assign time step and maximum time tstop=40; %time to stop calculation dt=1.0e10; %time increment, dt=1.0e10 for steady state calculation mwrite=0; %time iteration to print the results. If dt>tstop set mwrite=0 re=1.0;%relaxation coefficient %define calculation domain tl=15;%length of the pipe %clustered grid small=1.0e-10; pwrn=0.5;% cluster x=0 (pwrn<1.0), cluster near x=xmax(pwrn>1) for i=2:np2 temp=1-(i-2)/(np2-2); x(1,i)=tl*(1-temp^pwrn)+small; end x(1,1)=x(1,2)-small; x(1,np3)=x(1,np2)+small; for i=1:np2 dx(1,i)=x(1,i+1)-x(1,i); end % if geo=3 , y is angular direction and must be in radian height=0.025;%radius %clustered grid in y-direction small=1.0e-10; pwrn=1.5;% cluster y=0 (pwrn<1.0), cluster near y=ymax(pwrn>1) pwrn=3; for j=2:mp2 temp=1-(j-2)/(mp2-2); y(j,1)=height*(1-temp^pwrn)+small; end y(1,1)=y(2,1)-small; y(mp3,1)=y(mp2,1)+small; for j=1:mp2 dy(j,1)=y(j+1,1)-y(j,1); end %assign y-coordinate y(1,1)=0; % for j=1:mp2 y(j+1,1)=y(j,1)+dy(j,1); end %prescribe intitial temperatures for all control volumes for j=1:mp2 for i=1:np2 te0(j,i)=295; te(j,i)=te0(j,i); tep(j,i)=te(j,i); %predicted value at the next iteration end end %specify velocity u=zeros(mp2,np3);v=zeros(mp3,np2);
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ymax=y(mp3,1); %umax=10; umax=0.342; tindex=6.6; for i=1:np3 for j=1:mp2 yp=0.5*(y(j,1)+y(j+1)); u(j,i)=umax*(1-(yp/ymax))^(1/tindex); end end for i=1:np2 for j=1:mp3 v(j,i)=0.0; % y-direction velocity end end %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 for i=1:np2 for j=1:mp2 tk(j,i)=0.613;%conductivity of water ro(j,i)=999; cp(j,i)=4179; end end visl=855e-6;% for water tkl=tk;%laminar conductivity %add turbulent conductivity using mixing length model [vist,tkt,rex]=mixinglength_conv(n,m,x,y,dx,dy,ro,cp,u,v,visl);%as in note % prt=0.9; tkt=vist.*cp/prt; tk=tkl+tkt;%total conductivity %prescribe boundary temperature %initialize fluxes qx0c=zeros(mp2,1);qx0p=zeros(mp2,1);qx0=zeros(mp2,1); qx1c=zeros(mp2,1);qx1p=zeros(mp2,1);qx1=zeros(mp2,1); qy0c=zeros(1,np2);qy0p=zeros(1,np2);qy0=zeros(1,np2); qy1c=zeros(1,np2);qy1p=zeros(1,np2);qy1=zeros(1,np2); %bx0(j)=1,2,3(known temperature, known flux, periodic) at x=0 %bx1(j)=1,2,3 (same) at x=xmax %by0(i)=1,2,3(known temperature, known flux, periodic) at y=0 %by1(i)=1,2,3 (same) at y=ymax %at x=0 & x=xmax for j=1:mp2 bx0(j)=1; te(j,1)=300; end
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for j=1:mp2 bx1(j)=2; end %at y=0 & y=ymax for i=1:np2 by0(i)=2; end for i=1:np2 by1(i)=1; te(mp2,i)=700; % by1(i)=2; % qy1c(i)=-100; % qy1p(i)=0; % qy1(i)=qy1c(i)+qy1p(i)*te(mp2,i); end %evaluate sourec terms for i=1:np2 for j=1:mp2 sp(j,i)=0; sc(j,i)=0; end end
%evaluate the coefficients and source and solve equations by function tdma conv2d_invariant %%%%%%%%%%%%%%%%% 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? if iwrite>mwrite %print the results fprintf('iteration number is %i \n',iter) fprintf('time is %9.3f\n',t) iwrite=0; end iwrite=iwrite+1; end %this end goes with while t<tstop%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%OUTER LOOP %additional calculation % calculate the local and average Nusselt number along the axis % tk=tkl; [nux,nuavg,tmean,uavg,theta]=nux_conv2d(geo,m,n,te,x,dx,y,dy,tk,u); nux(1)=nux(2);%to eliminate artificially large value at x=0 %plot the result subplot(1,2,1) figure2d_conv(geo,x,y,m,n,te,theta);
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for i=1:np2 xc(i)=0.5*(x(1,i)+x(1,i+1)); end subplot(1,2,2) plot(xc,nux,'-') grid on title('local Nusselt number vs x') xlabel(' axis, m'), ylabel(' Nux') %
Results:
Local Nusselt number shows much higher value than the laminar case in the developed
region, about 100 compared with 4.36 due to the increased turbulent mixing. Experimental
data shows Nusselt number for fully developed turbulent pipe flow can be represented by
In the present case it gives 101.8 and is in an excellent agreement with the numerical result.
The thermal entry length of turbulent pipe flow is very short and approximately 10 times of
pipe diameter, 0.5 m in the present example.
temperature contour
x(m)
y(m
)
5 10 15
0.005
0.01
0.015
0.02
0.025
0 5 10 15 2095
100
105
110
115
120
125
130
135
140local Nusselt number vs x
axis, m
Nux
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Figure 6 Non-dimensional temperature distribution (left) and
local Nusselt number in thermally developing turbulent circular
pipe flow with constant wall heat flux (right)
Circular Couette Flow
Heat transfer rates between two surfaces are significantly affected by the flow motion
between the surfaces. Flow between two concentric circular pipes depends on the rotational
speed of the cylinder. If the rotational speed is not large, flow speed can be considered linear
having the maximum velocity at the surface of the rotating cylinder. Thus the angular velocity of
the fluid between two concentric cylinders, assuming inner cylinder is rotating, is
R - R
R -r - 1 V = V
12
1max, (22)
If there is no flow motion, heat transfer rate is determined by the conduction alone,
neglecting radiation effects. It is interesting to observe dominant effects of convection in such a
simple flow situation. Exact solution of convective heat transfer in Couette flow when the inner
and outer surfaces are maintained at different uniform temperatures can be obtained. A slight
variation in the boundary conditions, however, makes it difficult to obtain an exact solution. We
will examine one such problem.
Example 3 ______________________________________________
Problem description: A long concentric cylinder with radii, 0.1 and 0.2 m, contains mercury. Inner cylinder is
rotating at a linear velocity of 1 cm/s. Temperature at the one-half of the outer surface is at 600 K
and the other half at 300 K. Inner surface is insulated. Such a device can be used to remove
energy from one compartment to another in variety of engineering devices. Assume constant
thermodynamic properties at the mean temperature and neglect the gravity effects.
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Figure 7 Heat transfer in rotating concentric cylinders
User written functions: We set geo=3 since a polar coordinate problem. Only steady state solution is sought.
Thus a large time step is used and TSTOP is smaller than time step. Of course, we can
easily obtain transient solution as well by selecting an appropriate time step. A grid
system with 20 uniform grid in the radial direction and 20 uniform grid in the angular
direction is used.
Tthe grid system is designed to handle a periodic boundary condition along the
angular direction. There is no need for fictitious boundary control volumes in the
periodic boundary condition. Thus
22M+1M+1 = ; =
Angular direction grid arrangement is shown in the figure below.
Figure 8 Periodic grid in the angular direction
Radial direction grid distribution is typical, uniform grids with fictitious boundary cells.
19
Properties of mercury at 450 K are prescribed. Also, angular velocity is calculated by
Eq.(22). Initial temperature is 450 K.
Boundary is insulated at the inner cylinder. Thus, bx0(j)=2, for all j’s. At the outer
cylinder, bx1(j)=1, te(j,np2)=600 for j=2,..,11 and te(j,np2)=300 for j=12,...,21. Because
of the periodic boundary, by0(i)=by1(i)=3 for all i’s. There are no source terms.
Total heat transfer rate from the hot surface per unit length of the pipe is estimated by
j2
11j=
2j=
total R)(1 = Q jqx
Results: Temperature contour plot shows that due to the rotation of the inner cylinder, cold fluid is
brought up closer to the hot cylinder surface in the right upper region and hot fluid is
moved closer to the cold surface in the left lower region resulting in higher heat transfer
rate compared with a pure conduction case. The total heat transfer rate with the rotation is
32170 W/m and 5828 W/m without rotation, about5.5 times higher.
The effects of convection on the heat transfer rate depend on the magnitude of convection
strength and conduction strength. For the given geometry and boundary conditions, this
parameter may be defined as
R - R
Tk
T C =
conduction
convection = Pec
12
avgPm ax,
V
For the present example, Pec=253. If Pec is increased by increasing the rotational speed,
or any other changes in the properties and geometry, heat transfer rate will be increased.
Same effects will be observed regardless of the type of fluids used between the cylinders.
clear all close all clc %conv2d_input_coutte.m %transient, 2-dimensional convection-diffusion %nonuniform diffusivity and sources. Finite volume formulation %using matlab program. (By Dr. S. Han, October 15, 2007) %modified March 2013 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % geometry selection:geo=1(cartesian, x,y), geo=2(cylindrical axisymmetric,
x=z,y=r), % geo=3 (polar ,x=r,y=theta (in radian)) %select the geometry geo=3;
20
%specify the number of control volumes n=20;%number of control volumes in x-direction m=20;%number of control volumes in y-direction np1=n+1;np2=n+2;np3=n+3;mp1=m+1;mp2=m+2;mp3=m+3; maxiter=300;%maximum iteration number %assign time step and maximum time tstop=40; %time to stop calculation dt=1.0e10; %time increment, dt=1.0e10 for steady state calculation mwrite=0; %time iteration to print the results. If dt>tstop set mwrite=0 re=1.0;%relaxation coefficient %define calculation domain r1=0.1; r2=0.2; w=r2-r1;% delx=w/n; dx=ones(1,np2); dx=delx*dx; dx(1)=1.0e-10;% dx(mp2)=1.0e-10; % %assign x-coordinate x(1)=r1; % for i=1:np2 x(i+1)=x(i)+dx(i); end % if geo=3 , y is angular direction and must be in radian height=2*pi;%360 degrees dely=height/m; dy=ones(mp2,1); dy=dely*dy; dy(1)=dy(mp1);%periodic in y-direction dy(mp2)=dy(2); % %assign y-coordinate y(1,1)=0; % for j=1:mp2 y(j+1,1)=y(j,1)+dy(j,1); end %prescribe intitial temperatures for all control volumes for j=1:mp2 for i=1:np2 te0(j,i)=450; te(j,i)=te0(j,i); tep(j,i)=te(j,i); %predicted value at the next iteration end end %specify velocity u=zeros(mp2,np3); v=zeros(mp3,np2); ymax=y(mp3,1); vmax=0.01;%inner cylinder rotation speed r1=0.1; r2=0.2; w=r2-r1; for j=1:mp3 for i=1:np2
21
t1=(i-1)/np1; v(j,i)=vmax*(1-t1);%linear profile end end for i=1:np3 for j=1:mp2 u(j,i)=0.0; % x-direction velocity end end %%%%%%%%%%%%%%% %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 for i=1:np2 for j=1:mp2 tk(j,i)=10.4;%conductivity of mercury ro(j,i)=13000;%density of mercury cp(j,i)=135;%specific heat of mercury end end %prescribe boundary temperature %initialize fluxes qx0c=zeros(mp2,1);qx0p=zeros(mp2,1);qx0=zeros(mp2,1); qx1c=zeros(mp2,1);qx1p=zeros(mp2,1);qx1=zeros(mp2,1); qy0c=zeros(1,np2);qy0p=zeros(1,np2);qy0=zeros(1,np2); qy1c=zeros(1,np2);qy1p=zeros(1,np2);qy1=zeros(1,np2); %bx0(j)=1,2,3(known temperature, known flux, periodic) at x=0 %bx1(j)=1,2,3 (same) at x=xmax %by0(i)=1,2,3(known temperature, known flux, periodic) at y=0 %by1(i)=1,2,3 (same) at y=ymax %at x=0 & x=xmax for j=1:mp2 bx0(j)=2; end for j=1:mp2 bx1(j)=1; if j<=11 te(j,np2)=600;%hot wall else te(j,np2)=300;%cold wall end end %at y=0 & y=ymax for i=1:np2 by0(i)=3;%periodic by1(i)=3;%periodic end %evaluate sourec terms
22
for i=1:np2 for j=1:mp2 sp(j,i)=0; sc(j,i)=0; end end %solve equations conv2d_invariant %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 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? 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_coutte(geo,x,y,m,n,te); %calculate heat transfer rate sum=0; r2=0.2; for j=2:11 dsum=qx1(j)*r2*dy(j); sum=sum+dsum; end disp (['heat transfer rate (W/m)=' num2str(sum)])
23
Fig. 9 Temperature distribution with inner cylinder rotation
Conjugate Heat Transfer
When a heat transfer calculation domain includes both solid and liquid in motion, one has
to solve purely conduction equation for the solid and convection-diffusion equation for the
liquid. Heat transfer problem involving both conduction and convection is often called a
conjugate heat transfer problem. It has been an accepted procedure to solve solid and liquid
region separately and match the solution at the interface, which acts like a nonlinear boundary
for both regions. Such standard analysis is difficult for transient case. Further if thermodynamic
properties, such as conductivity, specific heat and density are different for liquid and solid,
solution becomes even more difficult to obtain even in steady state case.
In the framework of present numerical method, density and conductivity at each control
temperature contour
x(m)
y(m
)
-0.15 -0.1 -0.05 0 0.05 0.1 0.15
-0.15
-0.1
-0.05
0
0.05
0.1
0.15
24
volume can be arbitrary and thus no additional considerations are required. Also conduction and
convection problems can be solved by convection-conduction formulation simply setting zero
velocity in the solid regions. However, spatial variation of specific heat must be properly
handled to solve conjugate heat transfer problems.
In a previous section, it was briefly mentioned that if specific heat is not constant, energy
equation written in terms of enthalpy is more convenient to use. Two dimensional energy
equation in Cartesian coordinates written in terms of enthalpy is
hS= )y
h
C
k -h v(
y + )
x
h
C
k -h u (
x +
t
h
PP
(23)
where the specific heat, CP, is not constant. Integration of Eq.(23) as shown in previous section
will give a finite volume formulation in 2-dimensions. conv2d.m contains this formulation in the
Cartesian, cylindrical axisymmetric and polar coordinates.
Since the dependent variable is enthalpy, not temperature, boundary conditions must be
prescribed in terms of enthaply. If we set enthalpy h=0 at T=0 and assuming CP is not a function
of temperature (or an average value), then
T C =h P (24)
If heat flux is given at a boundary, say at x=0 , energy balance requires that
)x
T( k- = T q + q = q
ww1PC
This is the expression that has been used to linearize the nonlinear boundary conditions when
temperature is the dependent variable. In convection-diffusion formulation, q'' represents a
virtual flux (not just conduction flux).
Figure 10 Energy balance at x=0
Since the dependent variable is enthalpy not temperature, linearized boundary condition should
be modified. By multiplying CPw to the above equation, we have
25
)x
T(C k- =T C q + C q = Cq"
wPw1PPPCP wwww
And
)x
h( k - = h )q( + )q( = q
ww1hPhC
''
h (25)
where
q = )q( ; C q = )q( ; Cq = PhPPChCP
''
11
hq
Eq.(25) is the linearized flux equation to be used to handle nonlinear boundary conditions when
enthalpy is the dependent variable. It should be noted that actual heat flux is q'', not ''
User written functions: geo=2 since cylindrical coordinates. A steady state is sought, hence a large time step. A
25 control volumes in the axial direction and 12 control volumes (10 for the fluid and 2
for the solid) in the radial direction. Including boundary control volumes in the radial
direction, j=1,..,11 for the fluid and j=12,..,14 for the solid. Grids are clustered in the
entrance region.
Material properties in the fluid are those of Example 1. In the solid, steel properties are
used. Note that te(j,i) denotes enthalpy (h) not temperature in this problem. A fully-
developed velocity profile is prescribed in the fluid region. Initial temperature of the
fluid and solid is 300 K. Thus the
initial condition is
T*C = h ,,, ijijPij
where Tj,i=300 K.
At x=0, insulated boundary for the solid. Thus bx0(j)=2 for j=12,13 and 14. For the fluid,
a known temperature of incoming fluid at 300 K is given. Thus, bx0(j)=1, and
te(j,1)=300*cp(j,1) for j=1,...,11. At the axis of the pipe, boundary is symmetric. Thus,
by0(i)=2 for all i's. At the surface of the pipe, constant wall heat flux (qw) is assumed.
Thus by1(i)=2,
0 = )(1 ; ),2(C*q- = )(1 Pwipqyimpicqy
for all i's. Note that the actual heat flux entering through the pipe surface is
),2(C
)(1 = q
P
wimp
iqy
At x=L, solid pipe is insulated and conduction flux is neglected in the fluid since
convection is dominant. Thus, bx1(j)=2 for all j's. There are no sources.
The mean temperature of the fluid is calculated and the Nusselt number along
the axis is calculated by
k
D h = uN
T - T
Q = h
f
xx
m0
0x
27
where T0 is the interface (between the solid and fluid) temperature and Q0 is the interface
heat flux. D is the inner diameter of the pipe.
Results:
The presence of a solid wall with a large conductivity introduces significant conduction
along the axial direction in the solid and influences interface temperature and interface
heat transfer rate. Fig. 11 shows the presence of pipe increased local Nusselt number
slightly in most of pipe except very near the entrance where the presence of wall
decreased local Nusselt number. This is due to the increased interface temperature with
large conductivity of steel. The conductivity ratio of solid to fluid is about 100 for this
example and the influence of solid is not significant.
clear all close all clc %conv2d_input_conjugate.m %trnasient, 2-dimensional convection-diffusion %nonuniform diffusivity and sources. Finite volume formulation %using matlab program. (By Dr. S. Han, October 15, 2007) %select the geometry geo=2; %te=h in this problem %specify the number of control volumes n=25;%number of control volumes in x-direction m=12;%number of control volumes in y-direction np1=n+1;np2=n+2;np3=n+3;mp1=m+1;mp2=m+2;mp3=m+3; maxiter=300;%maximum iteration number %assign time step and maximum time tstop=40; %time to stop calculation dt=1.0e10; %time increment, dt=1.0e10 for steady state calculation mwrite=0; %time iteration to print the results. If dt>tstop set mwrite=0 re=1.0;%relaxation coefficient %define calculation domain tl=15;%total length of the pipe %tl=15; %clustered grid small=1.0e-10; pwrn=0.5;% cluster x=0 (pwrn<1.0), cluster near x=xmax(pwrn>1) for i=2:np2 temp=1-(i-2)/(np2-2); x(1,i)=tl*(1-temp^pwrn)+small; end x(1,1)=x(1,2)-small; x(1,np3)=x(1,np2)+small; for i=1:np2 dx(1,i)=x(1,i+1)-x(1,i); end % if geo=3 , y is angular direction and must be in radian h=0.025;%radius hs=0.002; %steel thickness for j=2:mp1 if j<= (m-1)
28
dy(j,1)=h/(m-2); else dy(j,1)=hs/2; end end dy(1,1)=1.0e-10; dy(mp2,1)=1.0e-10; %assign y-coordinate y(1,1)=0; % for j=1:mp2 y(j+1,1)=y(j,1)+dy(j,1); end %prescribe thermal conductivity, density and specific heat for i=1:np2 for j=1:mp2 if j<=(m-1) tk(j,i)=0.613;%conductivity of water ro(j,i)=999; cp(j,i)=4179; else tk(j,i)=65;%conductivity of steel ro(j,i)=7854; cp(j,i)=434; end end end %prescribe intitial temperatures for all control volumes for j=1:mp2 for i=1:np2 te0(j,i)=300*cp(j,i);%enthlpy is the variable te(j,i)=te0(j,i); tep(j,i)=te(j,i); %predicted value at the next iteration end end %specify velocity ymax=y(mp3,1); umax=0.0034; for i=1:np3 for j=1:mp2 yp=0.5*(y(j,1)+y(j+1)); u(j,i)=umax*(1-(yp/ymax)^2);%parabolic profile end end for i=1:np2 for j=1:mp3 v(j,i)=0.0; % y-direction velocity end end %%%%%%%%%%%%%%% %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;
29
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 for i=1:np2 for j=1:mp2 if j<=(m-1) tk(j,i)=0.613;%conductivity of water ro(j,i)=999; cp(j,i)=4179; else tk(j,i)=65;%conductivity of steel ro(j,i)=7854; cp(j,i)=434; end end end %prescribe boundary temperature %initialize fluxes qx0c=zeros(mp2,1);qx0p=zeros(mp2,1);qx0=zeros(mp2,1); qx1c=zeros(mp2,1);qx1p=zeros(mp2,1);qx1=zeros(mp2,1); qy0c=zeros(1,np2);qy0p=zeros(1,np2);qy0=zeros(1,np2); qy1c=zeros(1,np2);qy1p=zeros(1,np2);qy1=zeros(1,np2); %bx0(j)=1,2,3(known temperature, known flux, periodic) at x=0 %bx1(j)=1,2,3 (same) at x=xmax %by0(i)=1,2,3(known temperature, known flux, periodic) at y=0 %by1(i)=1,2,3 (same) at y=ymax %at x=0 & x=xmax for j=1:mp2 if j<=(m-1) bx0(j)=1; te(j,1)=300*cp(j,1);%enthalpy of entering fluid else bx0(j)=2;%treat the steel part insulated end end for j=1:mp2 bx1(j)=2;%treat insulated and convection dominated end %at y=0 & y=ymax for i=1:np2 by0(i)=2;%symmetric at the axis end for i=1:np2 by1(i)=2;%constant wall flux qy1c(i)=-10*cp(mp2,i);%in terms of enthalpy qy1p(i)=0; qy1(i)=qy1c(i)+qy1p(i)*te(mp2,i); end % %evaluate sourec terms for i=1:np2 for j=1:mp2 sp(j,i)=0; sc(j,i)=0; end
30
end % conv2d_invariant % 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? 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 % te=te./cp;%convert to temperature % calculate the local and average Nusselt number along the axis nux_conv2d_new_conjugate;%using interface temperature % %plot the result subplot(1,2,1) figure2d_conv(geo,x,y,m,n,te,theta); for i=1:np2 xc(i)=0.5*(x(1,i)+x(1,i+1)); end subplot(1,2,2) %plot(xc,nux,'x'); plot(xc,nux,'') grid title('local Nusselt number vs x') xlabel(' axis, m'), ylabel(' Nux-conjugate') % nux_conjugate=nux; % load nux_laminar_pipe;%this is the case without steel pipe nux_laminar=nux; figure(2) plot(xc,nux_laminar,'-ok',xc,nux_conjugate,'-xk') xlabel('axis, m'),ylabel('Nux') legend ('without steel pipe','with steel pipe')
31
Figure 11 Effects of solid wall on Nux
Problems
1. An incompressible fluid flows steadily between two infinite large parallel plates as shown in
the figure. The flow is laminar. You may use any fluid (air or water) with constant properties.
The results will not depend on the fluid or its properties as long as you present the results in
non-dimensional forms.
For the hydrodynamically fully developed flow:
(a) Numerically obtain a steady state velocity profile and compare with the known exact
solution,
22 48
1)( yh
dx
dpyu
(b) Calculate fRe using the numerical results obtained in part (a) and compare with the
known value of 96.0.
For the thermally developing flow:
0 2 4 6 8 10 12 14 164
4.5
5
5.5
6
6.5
7
7.5
8
axis, m
Nux
without steel pipe
with steel pipe
32
Using the velocity obtained in part (a), now consider thermally developing flow. The wall
boundary condition on both surfaces is a constant temperature.
(c) Calculate the thermally developing temperature field. Plot the non-dimensional
temperature distribution.
(d) Calculate the local Nusselt number and plot Nux along the x-axis. Does the local Nusselt
number approach the known value of 7.54 in the thermally fully developed region?
(e) Repeat the thermally developing flow calculation with a constant wall heat flux boundary
condition.
2. A laminar flow between two concentric cylinders as shown in the figure is considered in this
problem.
(a) Determine the hydrodynamically-fully developed flow and compare with known exact
solution.
r
a
ab
bara
dx
dpru ln
ln4
1)(
2222
(b) Using the velocity field obtained in part (a), calculate the thermally-developing flow in the
annular space. Consider both cases of constant wall heat flux and constant wall
temperature.
h
y
x
constTorconstq ww
r=a r
constTorconstq oo "
33
3. Rework Example 10.1 using Tw=constant instead of constant wall heat flux.
4. Rework Example 10.2 with an uniform energy generation in the core region.
5. Rework Example 10.3 with constant wall temperature instead of constant wall heat
flux.
6. Reconsider Example 10.4a using cylindrical geometry rather than Cartesian geometry.
7. Rework Problem 2 with a uniform heat generation in the inner cylinder.
8. Reconsider Example 10.1 for turbulent flow case. Determine the hydrodynamically-
fully-developed flow using appropriate mixing length model for turbulent viscosity.
Assume turbulent Prandtl number is 1.0 for the thermally-developing flow calculation.
r=
b
x
constTorconstq ii "
34
The figure shows a simplified model of electronic circuit board cooled by a hydro-dynamically
fully developed laminar flow in a parallel-wall channel of fixed length, L. The walls of the
channel are insulated. The board substrate has a sufficiently high conductivity such that the
board temperature is uniform at TW. The air flow is driven by a fixed pressure ΔP, and the air
temperature at the inlet is T0.
H
X=L
Air
T0
∆P
x
insulation
insulation
Circuit board at TW
substrate
y
35
The channel spacing H must be selected in such a way that the thermal conductance defined by
0/ TTq W
is the maximum. q is the energy removed by the air flow per unit height. That is
Hqq /
where q is the total energy rate removed from the board by the air stream per unit depth.
L=0.1 m, ΔP=3.0e-3 Pa, T0=300 K, Tw=500 K. Assume air properties remain constant at 300 K.
The fully developed velocity profile for laminar flow between parallel plates is given by
))((2
1)( Hyy
dx
dPyu
Use numerical simulation to determine the optimal spacing H which gives the maximum value of
the thermal conductance for the given operating conditions.
Present your investigation in a report form. Use any report format you choose. But it should
include as minimum following sections: introduction, method of analysis (equations used,
boundary conditions, etc), results and conclusions. There are no page limit but make it concise
and yet clearly presented so that your peer can understand what you did. Present the results in
tabular or graphical form.
(Project is due on 11:00 am, 12th
Dec. You must work on this project independently.)
Repeat the problem with a fixed wall heat flux instead of fixed circuit board temperature. Use
Tw=Th, where Th is the board temperature at x=L to define the thermal conductance. Why is the
maximum thermal conductance higher with constant wall temperature case? Is the increase in
heat transfer significant enough to justify the use of a high conductivity substrate?
Consider both cases with temperature dependent properties. Are the results different? (Bejan,
