CALIFORNIA STATE UNIVERSITY, NORTHRIDGE
FLUID FLOW SIMULATION
BY FINITE DIFFERENCE METHODS
ON A DIGITAL COMPUTER
A graduate project submitted in partial satisfaction of
the requirements for the degree of Master of Science in
Engineering
by
Alan Vincent von Arx
May 1982
The Graduate Project of Alan Vincent von Arx is approved:
(Leonard Spunt)
(Laurence S. Caretta, Chairman)
California State University, Northridge
ii
TABLE OF CONTENTS
Page
INTRODUCTION . . .
1
CURRENT ALGORITHMS
6
The Mixing Problem
6
Upwind Difference Modeling Problem
9
SOLUTIONS.
15
Solution to the Mixing Problem
15
Solution to the Upwind Modeling Problem
23
DISCUSSION
33
USER GUIDANCE
37
CONCLUSION
39
REFERENCES
40
APPENDICIES
41
Appendix A:
Appendix B:
Sample Problem - Computer
Output . . . . .
41
Exact Solution to a Pipe
Flow Problem with a Step
Change as Input
58
;;;
LIST OF FIGURES
FIGURE
PAGE
1. Modeling a Nonlinear Problem Using Discrete Nodes 8
2.
Eulerian Solution to a 40-Node Flow Network
10
3.
Eulerian Solution to a 4-Node Flow Network
11
4.
Upwind and Central Difference Algorithms
12
5,
Comparison of Upwind and Central Difference
Solutions in a Simple Heat Exchanger Problem
14
6.
Models for 40 and 4 Node Networks
17
7.
Lagrangian Solution to a 40-Node Flow Network
18
8.
Lagrangian Solution to a 4-Node Flow Network
19
9.
Lagrangian Solution Technique
21
10.
Eulerian Solution to a 40-Node Problem with
a Ramp Inlet Condition
24
11.
Eulerian Solution to a 4-Node Problem with
a Ramp Inlet Condition
25
12.
Lagrangian Solution to a 40-Node Problem with
a Ramp Inlet Condition
26
13.
Lagrangian Solution to a 4-Node Problem with
a Ramp Inlet Condition
27
14.
Three node Flow Problem for Comparing Upwind
and Central Difference Techniques
30
15.
Solutions to a 1-D Convection - diffusion Problem 31
16.
Comparison of a Typical Critical Parameter Using
Various Solution Techniques
iv
35
LIST OF TABLES
TABLE
PAGE
A Comparison of Eulerian And Lagrangian
Methods
1
22
NOt~ENCLATURE
A = area (ft 2 )
C, Cp, Cv
=
specific heat (BTU/lbm/°F)
E = energy (BTU)
h
=
heat transfer coefficient (BTU/sec/ft 2;°F)
k
=
thermal conductivity (BTU/sec/ft/°F)
k0
=
forward difference multiplier
M = mass (lbm)
m = mass flowrate (lbm/sec)
P
=
Peclet Number
Q = internal heat generation (BTU/sec)
q = internal heat generation per unit volume (BTU/sec/ft 3 )
T
=
temperature (°F)
Told' f, Tnew = temperature at the beginning, middle and end of
a time step (°F)
t = time (sec)
U = overall heat transfer coefficient (BTU/sec/ft 2;°F)
u = velocity (ft/sec)
V = volume (ft 3 )
v
x, y, z
=
cartesian coordinates (ft)
Y = thermal admittances (BTU/sec/°F)
P = density (lbm/ft 3 )
T
= time constant (sec)
Other
* used in equations to denote multiplication
** used to denote exponentiation
V gradient operator
~
difference operator
vi
ABSTRACT
FLUID FLOW SIMULATION
BY FINITE DIFFERENCE METHODS
ON A DIGITAL COMPUTER
by
Alan Vincent von Arx
Master of Science in Engineering
This report analyzes problems that involve flui,d-fiow boundary
conditions in a general heat conduction program.
The difficulty in
the problem stems from adjusting a finite difference method that is
used in solving conduction problems to be used directly with fluid
flow.
The capabilities and limitations associated with two types of
solution aigorithms, Eulerian and Lagrangian, are presented.
The
Eulerian method which uses a stationary controi voiume approach, is
simplest and preferred.
The Lagrangian method, in which the controi
volume moves vdth the fluid, is more complex but can generate a more
accurate solution in certain circumstances.
The finite difference
techniques used, cailed upwind and central differencing, can affect
the accuracy and the stability of the solution.
The report also contains a user guidance section which provides
suggestions to those who wish to solve a fluid flow problem by using
a conduction-based computer code.
vii
INTRODUCTION
Fluid flow simulation may be an important part of concern in
heat transfer.
When computer codes such as the Thermal Analyzer
Program (TAP) (Ref. 1) are used, the capability of modeling fluid
transport is necessary in many applications.
cally used to solve conduction problems.
The TAP code is basi-
It has been modified to
accomodate fluid flow boundary conditions.
Because I have found
that the accuracy and limitations crf fluid transport models are not
fully understood by most users, I chose this problem as the topic of
this project.
There are three main topics which are covered:
1.
Current Algorithms And Their Associated Problems
2.
Revised Algorithms And Their New Problems
3.
Conclusions and Recommendations
The first topic discusses the methods that are currently in use for
fluid transport problem solving.
These methods will be limited to
the techniques available in the Thermal Analyzer Program (TAP) which
is in use at the Energy Systems Group of Rockwell International
(Ref. 1).
This TAP code is a proprietary computer program which
evolved over a few generations to its current state.
Although the
listing of it is not available to the public, its basic heat balance
equations and iteration scheme are well publicized, and earlier versions of this code are generally available.
1
2
Basically, the TAP code solves the equation:
where T is the temperature (deg F),
x, y, z are the cartesian coordinates (feet)
although other coordinate systems may be used,
k is the thermal conductivity (BTU/sec/ft/deg F),
q is the internally generated heat (BTU/sec/ft**3),
Pis the density (lbm/ft**3),
C is the specific heat (BTU/lbm/deg F), and
t
1s
time (sec).
The differential equation can be thought of as
E
into a
node
where
1
E
out of
a node
E1 refers to energy.
represents a unit of volume.
be
11
lumpedil into this node.
+
E
internally
generated
=
E
absorbed
(stored)
A node is used to describe a point that
The mass of this volume is assumed to
Thus a continuous system can be modeled
as a series of inter-connected nodes.
This allows the use of a finite
number of difference equations to represent the differential equation.
3
The first term of the differential equation is then rewritten in
the difference form
(\7 (k\7T))D.V
=
D.y_D.z
k
[cr x+l ,y, z - Tx,y,z ) -
D.x
+k
D.xD.z
D.y
+k
D.xb.y
( \7 (k \7 T) ) Do X D,y Do Z ( Tx,y,z - Tx-l,y,z)]
[(T x,y+l,z - Tx,y ,z ) - (T x,y,z - Tx,y-l,z)]
[(T x,y,z+l - Tx,y ,z )- (T x,y ,z - Tx,y,z-1)]
D.z
The thermal admittance in the x-direction can be defined as
Y
x
=
kAx
D.x
=
kb.y_D. z
D.x
The admittances in the y and z-directions would have similar expressions.
Rewriting the rest of the differential equation, which represents
a continuous system, into the familiar discrete representation, we
have
r
L
j=l
Ti
Y.. (T.-T.) + qV.1 = PCV.1D.
-1J
J
1
D.t
where i refers to the node being considered and
j
refers to the neighboring node(s).
Y.. refers to the admittance that connects node i and
1J
node j. (Note that this subscript was rewritten since
it no longer refers to any particular direction.)
V.1 refers to the volume of material that the node represents.
A steady-state problem can be solved by dropping the time dependent
term.
4
The net heat into a node is computed using the average temPerature during a time step.
scheme (Ref. 2)
whe~e
This is the Crank-Nicholson iteration
the heat balance on a node is performed with
the average temperatures between the old and new times.
It is neces-
sarily iterative but unconditionally stable in that there is no maximum allowable time step.
(There are however limitations in that the
accuracy will be affected when the time step is too large.)
The
equation is now written as:
r
""Y .. (T.-T.) + qV.
~
lJ
J
1
1
=
PCV.
T.1
1
(
j=l
- T.1
new~ t ol d
)
where.T = Tnew +Told
2
Solving for T.
1
new,
= 2T.1
r
where
T;
'(Y .. T.) + qV. + T.
=
j~
· '!J J·
r
1
1
old
(2PCV./~t)
1
'(Y .. ) + 2 PCV./~t
~
lJ
1
j;:;j
So far, the equations have dealt with heat transfer by conduction
only.
When a problem involves fluid flow (convection), there are
additional terms of the form P*u*C*dT/dx
(~tJhere
u is the fluid velo-
city in the x-direction). The convection differential terms are then
discretized to be used in a finite difference scheme.
This will allow the user to model fluid flow boundary conditions
5
\>!hich occur in convection problems.
It is not intended to solve for
the thermal profile in the boundary layer.
The current algorithm for solving fluid flow has some inaccuracies associated with it.
Depending on the nature of the problem
and modeling variables, these inaccuracies may or may not be of
significance.
The next section of this report will propose revised algorithms
for solving fluid transport problems.
Although these are more accu-
rate representations of the actua·l differential equation, they too
have limitations.
The final section will discuss programming considerations which
should he.lp the user identify potential modeling problems and some
guidance in avoiding them.
CURRENT ALGORITHMS
There are two problems that are identified with the current
fluid transport model.
The first occurs in transient problems only,
and the second occurs in steady-state as well as time dependent problems.
THE MIXING PROBLEM
The first problem occurs when one has a string of flow nodes in
which a step change in the boundary condition occurs at the inlet of
the stream.
(For the sake of simplicity, we will consider a 1-dimen-
sional problem only.)
Neglecting conduction in the direction of flow
(which is a good assumption anyway when the velocities are high), it
would take a finite amount of time for the WaVe
11
11
to reach the exit.
However, because each node acts like a mixing tank
11
11
the wave will
jump ahead of itself, and a response at the outlet will be noted
ahead of time.
This mixing tank analogy can be demonstrated by considering the
basic heat balance equation:
E
into a
node
-E
=E
+E
out of
a node
internally
generated
absorbed
Neglecting internal heat generation and considering only transport
heat as heat that enters or leaves (i.e. an enthalpy flux), we have:
6
7
where T.1n is the average temperature of the fluid entering
the node during the time step, dt;
Toutis the average temperature of the fluid leaving
the node during the time step, dt;
dT is the change in the fluid 1 s temperature over the time
step, dt.
m is the fluid flow rate (lbm/sec);
Cp and Cv are the fluid's specific h~ts (BTU/lbm°F)
In the difference form, this is written as:
yf (Ti-l- Ti)
PCvVi
=
Tine~ Tiold
t:.t
where Yf refers to the flow admittance which is the product of
m and CP and
T.1- 1 is the fluid node upstream of node T.1
This works like a mixing tank because the exit temperature of a node
happens to be the bulk temperature of the node (See Figure 1).
.
In
other words, a given amount of heat (me p (T.1n -Tou t)dt) will enter the
node over the time step and mix:: with the average nodal heat content
11
resulting in some temperature change.
This can also be thought of
as the wave (the step change in inlet temperature) just enters the'
node, it instantaneously mixes with the fluid contained in the node,
and a quantity of fluid (equal to the volume that entered) is discharged.
What has happened is that a fractional amount of heat
entering was instantly transported to the exit.
Depending on the
size of the node, the flowrate, and the timestep, the wave will
8
T
T2-t-----~-.
TRUE SOLUTION
~
I
T1 ---------·-I
(
direction
of flow
1---r-----,
I
i
I
(
X
T
T2+--+----i
node
i-1
AS MODELLED
,_....;__ _ _...:..__. This is the integrated
1
'
average temperature
over the nodal volume.
Despite its nonlinearity,
it is assumed constant
over the length of the
node. (Note how the exit
X
temperature of node i has
node
node
deviated
from the true
i
i+l
va 1ue.)
Figure 1 -Modelling Nonlinear Problems
Using Discrete Nodes
9
11
travel through 11 the node faster than it can in reality.
This phenomenon is demonstrated in Figure 2 which shows what
happens to a step change in the inlet of a 100 foot pipe.
The velo-
city is 5 ft/sec so it should take 20 seconds to reach the exit.
As
Figure 2 clearly shows, the nodal temperatures are changing before
the wave front reaches the node.
(Axial conduction is neglected.)
Figure 3 shows what happens when one simplifies the model by going
from 40 nodes to 4 nodes axially.
The error is quite noticable.
This will be referred to the mixing problem since the fluid does
not really move as a slug.
11
Instead, it mixes like in a string of
mixing tanks. 11
UPWIND DIFFERENCE MODELING PROBLEM
The second problem to be investigated in this report comes from
an upwind differencing method.
the temperature of the
fl~id
As stated in the previous section,
which flows from node i to node i+1
(see Figure 1) is assumed to be the average temperature at node i.
The temperature of the entering fluid is therefore the temperature
of the upstream node.
Figure 4 shows this and compares it to a
central differencing method which intuitively will yield a more
accurate heat balance.
The central difference method actually shows
the average nodal temperature to be located at the centroid of the
node.
The exiting temperature will be at somewhere between the
average nodal temperature and the average downstream nodal temperature.
With the upwind difference method, it is not clear if the
nodal temperature is actually representing the average temperature
10
_I
CASE t-O.
600
F'lRTY f'aE F1..Cl4 t£nm< L611'13 ELlER IAN I'£THD - !00' Plf'Etlll>'5 FPS
SlEP a-iAIG:-I ~'LET I ItS ICINE Til'£ = 0 .!:EEC I Til'£ SlEP = 0. !:EEC
I
L
TAf'+ - 5
-*"
-'L
!'B)
~~
$0
v
510
L
I
0
F
I
r
400
~
0
lJ
v
v
I
I
-
v
5
/
/v
v
:/
v
I
....
~
~
10
v
v
_g'_
vs
1/
..%
~
~
l?
1/
'!'10
I
I
.I
I
~
15
II"
I/
ll
....... ,_a--
~
,.¥'
~
gr"
1
j
'1211
.!!'"'
,_.$'
I
I
'!8l
v
.........
/
/
I
'lQ)
/
~
f~ v-
....,...
/
I
5llll
0
E
G
~
k>"'
v
,/
1/
T
H
-
v
I
520
E
p
.X
v
""
'""*"" ~
e-X"'
v
F
u
f-*" ~
~
(-II-
/
/
!,-sa'
~
.!"
P<-
1ST
r-m
II
• =
~ N:n
!I
p= p.m ~
p= illU V"L f.E
20
Tlf'E-!£tl:Nl3
Figure 2 - Eulerian Solution to A 40-node Flow Network
25
11
_j
FCl.R NIE FLQ.l
600
CAS£ I'll.
:I
I
I
'
I
I
L
5:0
D
T
E
-
I
500
D
E
G
F
I
v
;<
v
./
1/
v
I
r
I
I
r I
40 0
v
~
I
I I
i
I
I
~ rk
I
rtl
'
I
I
I
v
v
I
~v
v
......--
v
/
/
v
?
~ ~; ~ ·""' ~
""'
v
v
I !
I
I
..;- ~
L-
"'"'
v f-Y'
I
~
~
I
?'
v
v
;.¥
.<Y
v
ir""
I
.-""
I
$
V'
,-""
,?"
.-""
v
.--""
/
I
,X
I
II
~
I
I
i
lO
15
=liST !Nnj I/
. :1~1~
l/
'"='~!
I
iO
5
'"'
~
,-""
_.,.-'
,-"'
~
.,.... ~
:?'
..JiY'
L
iI
I I
~
...... ~
%'
.,.:?'
I
I.
I
....- ,.¥
v
/
I
!
rii
'
I
I
v
I
1
I
I
,/
I
44 0
/
!
I
z r--
v
I
L
TAP+- S
O.~C
Tll"£ S1EP =
/
!
480
5.~C I
I
II
I
F
M
p
LSIN3 ELllR!AN t-£nm - 100' P!PEDV=5 FPS
I
I
I
I
y
~
S1EP G-IANI-It-l.ET I f£SICEN::£ Ttl"£ =
I
20
Ttl"£-5EaNlS
Figure 3 - Eulerian Solution to A 4-node Flow Network
=1~ V4LE
30
12
T.
J
T.J- 1
I
f
>f.·
,_
T.1
T. 1
Qin
T.+.
1 1
Qout
.
.
UPWIND
DIFFERENCE
ivtETHOD
(flow from i-1 to i)
Qabsorbed
=
.C:J·
1
hA(T.-T.) + mC T. 1 - mC T. =pVC - J 1
p 1' p 1
v 6t
Jj+l
T.J- 1
f J
T.,_ 1
T.
1
. 1-'2
CENTRAL
DIFFERENCE
METHOD
T.1
=
Qabsorbed
where T.+
1 = T. +
1 '2
1
T1. -'2
1
= T.+
1 '2
1
~hA(T.-T.)
J
1
for the upstream node
Figure 4 - Upwind and Central Difference Algorithms
13
or the exiting temperature.
Figure 5 shows a simple steady-state heat exchanger problem
which compares the two methods.
yield misleading results.
As shown, the upwind method can
This problem will become even more com-
plex in models of varying node sizes and heat transfer properties.
14
400,..
1
I
T
hotout
Thot
= constant
T
hotin
r
I
3solI
J
T
e
m
I
p
e 36o-l-I
r
a
t
u 340
r
e
+
t
0
6.
L::.
Exact Solution
Central Difference
Upwind Difference
6.
OF
320
/::,.
col din
300
0
2
!
T
col din
r
4
8
6
Distance from Inlet - feet
I
£
I
I
I
\oldin
300 OF
u
Thot.
ln
400 OF
A=
mCp =
r
~
10
!
r
-
10 BTU/hr/ft 2!'1=
1 ft 2/node
50 BTU/hr/'F
Figure 5 - Comparison of Upwind and Central Difference
Solutions in a Simple Heat Exchanger Problem
Thoti n
SOLUTIONS
Aside from the fact that the obvious solution to any fluid transport is to have finer nodal mesh, one still needs to recognize when
his/her model is an accurate simulation.
In this section, various
solutions to these modeling problems will be discussed along with
their limitations.
SOLUTION TO THE MIXING PROBLEM
There are two basic ways to model a problem which involves a
moving fluid.
The first is the Eulerian approach in which one examines
the fluid from a fixed point.
This can also be thought of as the
"control volume" method where mass, energy, and momentum are conserved
over a stationary region in space.
The second method is the Lagrangian approach in which the point
of reference moves with the fluid.
This technique will be used to
solve the errors that were obtained as shown in Figures 2 and 3.
The algorithm is simple in its concept but it can become complex
from a "bookkeeping" standpoint.
E
into a
node
-E
Going back to our original equation,
+E
out of
a node
=
internally
generated
E
absorbed
the first two terms involve heat which crosses the nodal boundaries.
This can be heat that is transfered by conduction, convection, radiation, or transport.
(Note that convection here refers to heat trans-
fered across a boundary layer by molecular and eddy conduction.
15
16
Earlier, it was used to denote heat transfer by fluid transport.)
However, when the Lagrangian approach is used, there is no longer any
heat transfer due to the fluid flowing in and out of the control
volume (or node).
This is caused by the fact that the nodal boundaries
are moving with the fluid.
Therefore, the P*Cp*u*dT/dx drops out.
Instead, the node will have to be tracked as it moves down the flow
stream.
One algorithm which has been developed will b_e discussed at this
point.
Consider the models shown in Figure 6.
These nodal networks
were used to calculate the temperature response to
a step
change in the
flow stream inlet using the Eulerian method shown in Figures 2 and 3
(a 1isting of the TAP input data is shown in Appendix A). However, when
the Lagrangian method is used as shown in Figures 7 and 8, one can see
significant improvements.
model.
This is particularly true for the four node
Each of these plots (Figures 2,- 3, 7 &8) show some data points
which represent the exact solution for the temperature response of the
wave front which were calculated by hand as shown in Appendix B.
These
data points can be used as reference points to note the accuracy of
the various curves.
The Lagrangian method is performed as follows:
1.
Calculate the time step necessary for the fluid to
move from one node to the next.
u
=
dx/dt
.
=
.
m/( P *A); Y = m * Cp;
C = P*A*dx*Cp
where u is the fluid velocity (ft/sec),
dx is the length of the node (ft),
dt is the time step (sec),
17
51
52
53
54
55
/ 88
89
90
i--.....-------.
Outer wa 11 nodes
<
1
2
4
3
5
f.;.-
10 0
101 102
Inner wall nodes
\
103 /'104 105
140
I
Inlet
Temperature
I
51
1
I
2
53
..
54
3
4
·~
101
""=:-
~-1ode 1
52
.
l 00
I
-."2.5'
40 Node
Fluid nodes
T03
102
Outer wall nodes
~
..
Inner wall nodes
To 4
Fluid nodes
25'
-
4 Node
~~ode i
-· . 2
Afl ow =l , n
u=5 ft/sec
h=.001 BTU/(s in2 F)
t1 r wa 11 =. 2 i n
Figure 6 - Models For 40 and 4 Node Networks
18
L
FCHTY NXE FLO-I N':ll-m< LS!N:l I..ACFAI\G!AN 1-ETHD - 100' PlPEtiV=6 FPS TAft+ - 5
CA!:£ N:l.
600
SlEP CWHE-!N..ET I
1
~li:EJICE
Tlt-E = O.!'E':C I Tlt-£ STEP= O.!'E':C
~
5D
-lb
.,.. ~ ~
:EO
,X'
~
~~
~!'-"
~ :..---
~ "/
xv
Sill
r
L
u
l
,
/
/
v/
500
I
..... ~
..... ~
I
J;(_
d'/
I
I
/
/
1-'
I
/
=1ST
• =
/
/
.._.
·"'
C=
5
10
15
co
C5
TIHE-£ElXN:S
Figure 7 - Lagrangian Solution to a 40-node Flow Network
(Broken lines show Eulerian solution from Fig. 2)
rm
~ rm
1-.rn
0=
0
('Y
p-; V'"
!/
I
~
~
~'/
lI I/
I
I
/
/
l/"1
lfv
I
4'10
v/
%~
I
~
v/
~
I
I
400
.,.. ~ ~
/
I /
181
'1211
0
/
vI
I
460
t:::'
... ~
~~
--
~f-'"
/
/ /
I
D
E
G
r
y
I
T
E
-
?"' /
/
I
520
D
H
p
~~
/
I
p-. ~·
II'
II
....... ;.,.,.
lam ~
ITIU
VALfE
I)
r
19
_J
Fo..R r-uE FLQ;
CASO t-0.
600
l
I
I
I
I
I
I
~
v
?
j
u
I
0
j
I
520
E
/
I
.I
If I/
0
E
G
F
I;
500
'leJ
I
I
~!/
I
~~~
~.:::.
I
5
I
v
10
q= ....-
p
...- .,%'
v
I
v
/
I/
!;a"'
~
/
!,)"
v
,(.,
,)"
.,. .....
/
1.....i.-"'
/
/
/,
/
/
~
I
v'J
.I
1/
I
/
I
I
II
IX= 1ST
i
I
•
I
'
/
--
l%
/
/
/
15
,.., v
6-J'f
I/
J._,-b-a-·
'""
1:.¥ b%
!;.%
/
/
~
l.--
_.
_._. 1---
b..%
,%'
v
l k
..>"
I
,____
,_7
l
r=+ +-!-+-
I__.1....i
l-.o--
~
1
~
I
/
v IJ
-~ l-!-· - -+-- ~-~ ~-
1'"
-:::::: _,..
I/
'
V/
1
II
1--
--t-
_yl
_$'
I
·v
VI
/
[/
I
i
L·
Tftf'+ - 5
0.5SEC
p... v
I v
Vp
.,;
I
/
v
-t
"
I
I
I
I i
v""
v
!'I
k
I
~
I II I
K..lt
vv
I
I
I
/
v
/
I
II
I
I
/
5.ca:C I Til"£ Slt:P
/
~
II
j
IJ'
I/
v
/
~
I
kJ
LX
v
i·
II
0
I
I/
I
40 011
iT
v
/
.X
~
I
I
1/
T
M
p
v
/
F
L
I 1
I
I
L611-0 l...AffiANJli'N i'E1HlJ - 100" P!F£DV=5 FPS
I f£51CEN:E Til"£
I
1
I I
F-.<.
~
~-!N..ET
Slt:P
1
. I
11
=e~IKUJ
':•.......1 ~
\n = \o_rn!ET
I
20
I~ 1/
10 = ~
25
Til"£-:ECCNl5
Figure 8 - Lagrangian Solution to a 4-node Flow Network
(Broken lines show Eulerian solutionfrom Fig. 3)
Note: The slight oscillations in the Lagrangian solution
comes from the time step being less than the residence
time. The algorithm has a little difficulty when the
fluid flows only a fraction through the node during a
time step. A larger one will correct this. •
V4l£
30
20
m is the mass flowrate (lbm/sec),
is the density (lbm/ft**3),
A is the cross sectional area (ft**2),
y is the flow admittance (BTU/sec/deg F), and
c
is the nodal heat capacity (BTU/deg F).
Therefore, dt = C/Y.
2.
Move the fluid node downstream by half a node with
respect to the stationary nodes.
3.
Perform a heat balance on the fluid node using the
following equation, and the notation of Figure 9.
1/2 L Y,.
.
J
1J
('f.-f.)+ 1/2 L Y.+ 1 .+ 1(T.+ 1-T.)+Q. = PV.C (T.
-T.
)
J 1
j +1 1 , J
J
1
1
1 v 1 new
1 01 d
~t
where Ti is the fluid node being iterated upon,
Ti+l is the fluid node downstream from Ti'
T. represents any node connected toT.,
J
1
Tj+l represents any node connected to T1+1 '
Y.. is the admittance between T, and T,,
1J
1
J
Yi+ 1,j+l is the admittance between Ti+l and Tj+ 1 '
G; is the internal heat generation for node i, and
PVCv is the nodal capacitance for node i.
4.
Compare average temperature of the fluid node just
calculated with the prior value from the last iteration.
(If this is the first iteration, compare with the initial
5.
guess, T = T01 d.)
Once all fluid and non-fluid nodal temperature converge
(the maximum change from one iteration to the next is less
21
T.
-1
T.
J
\+
~
direction
of flow -
·~
,..b
.I
T.1-J., Ti
1+
-·
.
~
Ti+1 T.+,
1 '-
--
Beginning of time
step; t=t
0
Iterate on
temperatures;
t=t 0+.1t/2
T.+1 T.+2
End of time step;
T.
T.
1- 1
T.
1
Ti+1
•----;>~ t=t +~t
0
[Ti+ 1J fri+ 21 - - Redefine fluid
L ~
temperatures for
the next time step
Figure 9
Lagrangian Solution Technique
22
than a predetermined amount), the temperatures at the
end of the time step are calculated.
For non-fluid nodes,
new = 2*f - Told
For the fluid nodes, they are advanced another half of
T
a timestep to thier new position.
,
6.
The fluid nodes are assigned new temperatures to
reflect the fact that the fluid has moved to the next
nodal position.
11
7.
This is what was refered to as the
bookkeeping 11 part mentioned earlier.
Update the time and repeat for the new time step.
As expected, the Lagrangian method will take more computer time
due to its greater complexity.
Table 1 shows some comparison for
four different cases whose results are shown in Figures 2, 3, 7 and 8.
An interesting note is that just because a model
Table 1
Comparison of Eulerian and Lagrangian Methods
Method
4 node flow string
40 node flow string
Eulerian
0.40 sec CPU
1. 04 sec CPU
Lagrangian
0.50 sec CPU
2.48 sec CPU
has 10 times the number of nodes, it doesn•t take 10 times the CPU
time.
Also note that these 4 cases each had the same time step for
comparison purposes.
Actually, the models with fewer nodes could use
a larger time step without any impact on the thermal results.
By
looking at this, one can conclude that the greater accuracy of the
Lagrangian method is partially offset by using more nodes with the
23
Eulerian method.
Another limitation with the Lagrangian approach not shown in
these simple models is the time step restriction in that it has to be
limited to the nodal residence time.
This makes the problem more
complex when a model has different node sizes, branching flow,
changing flow rate, etc.
Additonal program logic would then be
required which in turn will take additional CPU time.
A final consideration to be taken into account is the nature
of the transients.
The Lagrangian approach has its advantage in that
it can simulate the discontinuity in a temperature profile due to a
step change in the inlet condition.
However, many real world problems
do not have instantaneous changes.
Rather, the boundary condition can
be simulated as a ramp of some sort.
Figures 10 thru 13 show that
for the 4 and 40 node flow strings, the effect on the outlet temperature due to the Eulerian and Lagrangian techniques is reduced with a
reduction in the rate of change of the inlet boundary temperature.
SOLUTION TO THE UPWIND MODELING PROBLEM
The next logical question might be as to the reason for the
upwind differencing methods.
The answer lies in the fact that the
central differencing method can under certain conditions lead to
stability or accuracy problems.
The following discusses two ways in
which central differencing can lead to problems.
Potential instabilities may result when using central difference
methods.
This has been shown by analysis which is shown in Ref. 3.
What this indicates is the following stability criterion:
24
_j
~'ETH'.O
FCRlY N:li FLCI' I>ET\-I:A< LGit-13 flLER!tN
600
CASE NJ.
RAt1'UIO=Ist:C
I
F£SIC£!-U: Tlt-E
I
I
= 0.55:C
- IOO' P!F£CV=6 FFS
Tlt-E STEP
I
-,I
I
~
.,-
I
I
l.-/
,......~
I
1/
F
L
vD
I
5:0
T
E
H
p
-
I
500
E
G
F
480
I
440
v
1 - f-
v.
I
-"'-
v
X
/
/
-R--
I
~
II
I
...-"'
J:>.
ll
v
/
~~
v
v
v
v
I
I
I
v
I
/
II
I
v
,/"'
/
I
I
I
v
_L
I
I
v
;,..v ~II.
,/
v
I
I
460
I
y¥
v
v
I
I
40 0
I
l
v
D
?!'
~
~~
~~
~~
,_-Y
I/
L
TAf'+ - S
O.!';g:C
v
v
v
v
v
1/
ll
I
,(
/
/
•=
/
I~
~IlL
c = 31
~3/1
I
0 =OJ ~
IO
30
Figure 10- Eulerian Solution to a 40-node Problem
with a Ramp Inlet Condition
40
50
25
_j
FCLR i'a:£ FLO. f£1'.-m< l.Glt-0 EU..£RIAN I'£THD - 100' PIPEOV=5 FPS
CPff t-0.
600
RAI'RJIOCEGI!EC
I
F£Sle8'U: Til'£
I
= 5.UOC:C
Til'£ SiEP
I
L
TAf>+ - S
= 0.5!£C
I
~
/
1/
u
I
0
I
520
E
50 0
0
E
G
F
J
460
I
J
0
v
v
v v
ll
I/
l!
v
v v
l/
~...""
/
]/"
sz(
lvff
v
v
v
7
f("
v
v
v
0
10
/
,/'
/
1/
•=
-
l< ~ !Hr 6::!' l0r ~ Y"'
40 0
,./
!/'
/
7
v
le/
/
,('
v
7
Vi
./
/'
./
v
.,...-
7
I
h--~
/
v
l/
I
'!80
v
v
~
,.... ~
v
/
I
I
M
,_..$-
lY
/
1/
T
p
./
...&-'
. . . l-¥
v
F
L
v
v VI
v
L
I
~ ~I
I'
~II'
~
~.._j,<;
0=3'
p N:1f 31'
I
0=~~
30
40
Figure 11- Eulerian Solution to a 4-node Problem
with a Ramp Inlet Condition
\
50
26
_j
600
R.o>WDIO CEG/SEC
t
RESII::EN:E Tli-E
I
= O.Cff:C
I
T\1-£ S1EP
I
v f.>-
v
I
u
-
F
I
_L
480
/
l!
I
460
1/
L
/
I
I
I
0
.I
/
__(
tO
I
I
/
v
'/
%
v
r.l
/
,;
I
I
I
o!
I
/
v
v
,/
v
v
v
20
/
I
/
I
I
440
I
~
I
I
I
?
%
,;
I
I
0
K"
v
~
yrV'
./"'
I
I
so0
E
G
v
sr
I
I
/
T
. ..s-' ~
v
5:0
E
M
p
I
,;
1/
L
~~
/
I
F
"""
/
,/
= O.Cff:C
.....--
I
I
0
I
'-
FffiTY r-aE FLCW N::ll·lR< USII\G LAGW!JI.'N I"ETKil - 100' P\F£DV=6 Fi'S TW+ - S
CI>E£ t-0.
•=
I
l ..r-uk-J.I
f'U£ \/
$;.--c.
al= ~~
O=IJ
30
Til-E~
Figure 12- Lagrangian Solution to a 40-node Problem
with a Ramp Inlet Condition
3/
so
27
_j
FUR NIE FLCW 1-ETW:A< L61N:l LAGflPNJ!AN
Cl>ff. N:l.
600
I
RAI1'010i:£GISCC
I
v
v
u
1
T
E
-
r
500
v
E
G
F
I
J
440
v
/
v
v
400
0
!/"
~
10
~·
/
v'""
V'
I/
_/
v
v
;I
!./"
./'
of
_/
v
/
ll
I/"
/
~
L.-Y
I
7
«'
;;I
I
I
I
I
ll
./
v
I
I
~
;-!Or
_;I
/
l)
l-Y
;.A- ~
I
I
/
I
~
v
I
/
460
,_.y
I
v
I
I
I
J
480
I
v
D
_y
.,/
II
I
::eo
.
/
v
1/
D
M
p
_/
v
L
TAF'+ - S
= O.SSCC
v"'
I
F
L
- 100' PlF£D"v=5 FFS
Tli"E S1EP
I
v~
v
f....
5130
~EllW
= 5.=c
F£Slt:BU: Tli"E
I
51'
Vjy~
• = I~ ·
/
20
NIE
11
I
0=31' )~31
0 =--,:rr
30
Til-E~
Figure 13 - Lagrangian Solution to a 4-node Problem
with a Ramp Inlet Condition
'+0
50
r
28
m
k
M
0
> 0
T
where m is the mass flow rate (lbm/sec)
M is the mass of the fluid per node (lbm)
k0 is the forward difference multiplier
(k 0
=~implies
central difference method)
(k 0 = 0 implies upwind difference method)
T
is the heat transfer time constant
=
UA/Cp
UA is the heat transfer rate; bulk-to-bulk
fluid (BTU/sec/deg F)
Cp is the specific heat of the fluid (BTU/lbm/°F)
This indicates that instabilities may result under low flow conditions
(m is small) when using central differencing methods.
Of course,
there are a few obvious solutions:
(~y
1.
Decrease the node size
decreasing M and A) and
2.
Decrease k0 by going to the upwind difference method.
3.
Add conduction between the fluid nodes.
The last is, of course, a very easy method to get rid of the instabilities.
However, there is still another source of error with the
central difference method which is discussed in the following.
This other source of error results when conduction is added in
the direction of the flow stream.
Consider our basic heat transfer
equation in one dimension only:
P *C *dT + P*C *u*dT
P dt
P
dx
=
d (k*dT) + q
dx
dx
29
where Pis the density (lbm/ft**3)
Cp is the specifice heat of the fluid (BTU/lbm/deg F)
T is the temperature (deg F)
t is time (sec)
u is the velocity of the fluid in the x-direction
(ft/sec)
k is the thermal conductivity (BTU/sec/ft/deg F)
q is the internal heat generation (BTU/sec)
If we consider steady-state without a source term, it is simplified to:
P*C *u*dT
P dx
=
d (k*dT)
dx dx
The left hand side can be thought as the convection (or flow) term
whereas the right hand side can be thought as the conduction (or
diffusion) term.
It has been shown (Ref. 4) that the central dif-
ference scheme will again lead to errors depending on the ratio of the
convection to conduction terms.
Figure 14 shows a simple three node
model of a one-dimensional flow problem.
By finite differencing, the
differential equation becomes:
{P*Cp*u)
(~)
(T3-T1)
=
k*T3-2*T2-Tl))/dx
If the cell Peclet Number is defined as:
P = (P*Cp*u*dx)/k,
we can describe the behavior of the solution.
exact solutions for various values of P.
Figure 15 shows the
Note that when P is 0
(conduction only), the solution is a a linear profile.
large, the upstream temperature will predominate.
When P is
As for the value
of T2 calculated by the central difference method, Figure 15 shows what
30
T1
-c
Figure 14 - Three Node Flow Problem for Comparing
Upwind and Central Difference Techniques
31
T31
-------··--~.
p -1
i
I
I
T
).
P=1
Exact solution for
T vs
X
1
...
p 1
T2
0
1.2
.8
""'\ Centra 1
r-----Upwind/
Exact
T2
L
X
,
Difference
Solutions by
various schemes
for T2 vs P
(T 1=0.0)
(T =1.0)
3
.......
'
.4
-
0
\
-10
-5
0
5
10
p
Figure 15 - Solutions to 1-D
*
Convection-diffusion Problem
* from Ref. 4
32
happens as the Peclet Number (P) increases.
exact solution is tremendous.
The deviation from the
However, the upwind differencing scheme
will once again yield stable results.
The reason for this can be seen in Figure 15.
The difference
equation will result in a linear temperature profile between any
two adjacent grid points.
Of course, the large Peclet Number cases
(P>>1) is highly non-linear which
vo~ill
lead to errors.
The solutions to this include:
1.
Reduce the grid size to better approximate the
non-linear profile,
2.
Use a higher ordered equation to describe the
non-linearity, or
3.
Use the upwind scheme which yields bounded results.
These two cases demonstrate why the upwind differencing technique is used in spite of the errors that may result.
Instabilities
and/or errors can result under very low flow conditions (i.e. a high
time constant).
The upwind difference method yields reasonably
accurate temperatures while being calculationally stable.
DISCUSSION
It has been demonstrated that there are indeed more sophisticated techniques to derive the solution to a fluid transport
problem than those which are currently used.
Errors occured when a
highly non-linear temperature profile was approximated by a linear
one.
A solution in both cases was to use more nodes to refine the
grid size to better simulate the non-linear profile with a piecewise linear profile.
Also, higher
to simulate the non-linearity.
order equations could be used
However, there is a trade-off to be
considered which is the computational time required to:
1.
Calculate the temperatures for the additional nodes or
2.
Calculate the temperatures using equations with more
terms in them.
The first choice is desirable from the simplicity of the logic
standpoint.
In general, sophisticated algoithms are not as prevalent
today due to higher storage capabilities and reduced computer costs.
Another consideration which needs to be investigated when
evaluating the requirement for a highly accurate solution algorithm
is the accuracy of the heat transfer correlations.
These, when based
on test data, are only accurate to a certain extent--typically 20 to
25 percent.
With this in mind, it may do no good to use a more
complex (and likewise, expensive) iteration scheme.
The impact of
this depends on the value of the heat transfer coefficient as well
as the thermal properties of the material forming the fluid boundaries.
33
34
Of course, this is no excuse to over-simplify the problem.
A
thermal analyst must use good judgement in developing any heat
transfer model.
These sophisticated methods have another limitation and that is
the type of problem they•re good for.
For instance, the Lagrangian
method for modeling slug flow works well for a step change in the
inlet fluid boundary condition.
However, most real world problems
involve a temperature ramp of some sort.
Errors due to slug flow
are mitigated along with the ramp rate.
In the final analysis, the effect of any errors due to modeling
limitations depends on what is desired from the analysis.
In the
pipe flow problem considered here, it is true that there may be some
discrepancies in the fluid temperature histories.
However, it your
only interested in the temperature drop across the pipe wall which
can be related to a thermal stress, then the error caused by the
modeling assumptions depend on the true wall
~ T.
~T
verses the calculated
Figure 16 shows the values of the pipe wall .LlT as a function
of time, and it compares the Lagrangian and Eulerian techniques as
well as the effect of the node size.
ramp verses a step.
Also shown is the effect of a
Although the ramp is steep (10°F/sec) one can
already see some improvement.
One might conclude that the apparently
severe errors in the fluid temperatures have only a minor impact on
the final desired answer.
Of course, it would be a difficult task
to quantify the degree of error in the fluid temperature as well as
other parameters if we were interested in them since they are dependent on the thermal properties and dimension (i.e. Biot Numbers).
35
For Step Input
Lagrangian
4 nodes
For Ramp Input
(10 F/sec)
Lagrangian
40 nodes
Lagrangian
40 nodes
/Eulerian
nodes
/4o
lQj_
Eulerian
4 nodes
i
-l.1
8+
1
'4tt
2
0
10
20
Time -
30
40
seconds
Figure 16 - Comparison of a Typical Critical Parameter
Using Various Solution Techniques
50
36
Errors from modelling fluid flow are a function of many variables
such as the nature of the transient, node size, heat transfer properties, type or accuracy of answer desired.
The best way to avoid
these errors is to understand the capabilities and limitations of
the modeling program and the basic heat transfer principles involved.
The following section contains some basic suggestions to guide the
user in modelling fluid flow.
..
USER GUIDANCE
The first step is to look for the potential problems.
There
are two primary ones to be concerned with:
1.
Very low flow
2.
Severe transients
One can recognize a low flow condition when the product of the flow
.
rate and the specific heat (mCp)· is on the same order of magnitude
or less than the other admittances into a node.
When using the TAP
code (Ref. 1), this can be done by perusing the node summary at the
beginning of the run.
(A sample run is shown in Appendix A.)
Rec-
ognizing a severe transient is a bit more difficult in that there
aren•t any exact rules that one can use.
As an initial suggestion,
users should be concerned if a thermal boundary condition is changing
significantly during a single time step (i.e. 10% of the overall
temperature change).
Steps should be taken to insure accuracy such
as smaller time step or a reasonably fine node grid.
The second step is to select the computational algorithm.
The
recommended method is to take the straight forward Eulerian technique
and use a generous number of nodes.
If the alternate method is
chosen because of problem restrictions, then an effort should be
made to keep the fluid nodes as uniform in size as possible in order
~o
give each node the same residence time.
The final step is to try to determine how accurate the analysis
37
38
by selecting the critial parameter(s) and estimating their sensitivity to a more detailed analysis.
analysis•.
One way is to do an •asymptotic
This is done by studying the effect in the critical para-
meter when a change is made in the number of nodes or the iteration
scheme selected.
In other words, one has a reasonably accurate model
when further refinements produce little or no improvement in the
solution.
If this would be expensive because of the complexity of
the model, then a.simplified approximation exposed to the boundary
condition will often yield useful results.
CONCLUSION
The objectives of this project were 1) to analyze problems in
modelling fluid flow in a conduction-based code and
2) to summarize
the results for general users of one such code-TAP.
The fluid modelling was analyzed in terms of the existing literature and two specific problems were identified - mixing and upwind.
Directions were given to guide users to:
1.
Identify potential problems
2.
Overcome them.
This should help the user to appreciate the applications and limitations of the simple fluid flow models used for boundary conditions
in TAP and similar conduction codes.
39
REFERENCES
1.
A. V. VonArx, Thermal Analyzer Program (TAP)
Description and Users r~anuai
N036CPM620000
Rev. A, May 14, 1981, Energy Systems Group,
Canoga Park, California
11
'II
2.
Carnahan, Luther & Wilkes, Applied Numerical
t•1ethods, John ~Jil ey & Sons, 1969
3.
Willcox, Subsea Nuclear Power Transient
Mode 1 Computer Code, N652TI 140018, fv!ay 16, 1977,
Energy Systems Group, Canoga Park, California
\~.
t~.
11
11
4.
S. V. Patankar, Numericai Heat Transfer and
Fluid Flow, 1980, McGraw-Hill
40
APPENDIX A
Sample ProblemComputer Output
41
All NI:W INPUT IJATA t-OR THIS RUN
CASE
fAP4 - S
1
00000100
*
***
1
2
•
STAI<
001 - HlLES t; STOHC CCf!HENT.S
FORTY NCDI: FLCW NET~ORK USI~G EULERIA~ ~ETt-Ct - 100 1 PlPE~V=5 FPS
STEP CHANbE· INLET I RESIDENCE TIME = O.~SEC I TIME STEP= 0.5SEC
2.5 FEET PER NOCE, 100 fEEl ICTAL
FLOk AREA = 1 1~**2
PIPE WAll= ;2°~THICK, ~=10, RHC=490, CP=.ll
H = 518 I.0016TLISEC/OEGf/lN••2l
SURFACE AREA= 106.35 1~**2 PER 2.5 1 LE~Gih
10
••• STAR 010- STATIC
y
y
•
1 y
51 y
11
40
90
0
0
.o
.o
.a
.a
ttt
y 101 y 140
20
•
1
.o
T 10•)
0
1 T 40
T
T 51 T 90
T 101 T 14•J
y
1 v 40
y
51 y 90
y 10 1 y 140
c 1 c 40
0
0
600.()0
400.00
400.00
0
400.00
I)
.10635
.14380
2.0833
.39050
-1
0
0
0
NET~CRK OESCRIFTIG~
1
51
STAR 011 - FLCh NETkCRK
.o
ttt
T 101 J
T
1 1
STAR 020 -
.a
.o
.o
.o
.o
.o
.o
.o
T 100 T 101
lNITIALIZATIC~
1
1
1
1
1
l
l
OOC01300
00001400
1
& CO~STA~T CATA
0
0
0
1
0
ll
0
0
0
0
0
0
0
0
0
1
1
0
0
0
0
c
0
c
0
0
00001000
CCCOllCC
COC012CO
OESCRlPllC~
0
0
0
CQQ(;Q2CO
001100300
CCCu0400
CCCQC5CO
00000600
00000700
CGOCC80C
COC009CO
0
0
c
0
0
0
1
l CO~~fCTIC~-~All
1 CC~CLCIIC~-fiPE
1 ~OCT•CP fl(~ ~·s
1 1/2 kUL ~CCE
OOC01500
OOCC160C
OOC0170C
00001800
G0001Sil0
00002000
OGG02100
00002200
..p,.
N
c
c
51
101
30
•
c
c
90
140
J
0
.o
.o
• 39050
1.0417
c
0
0
0
0
0
1 D <tJ
1 E 40
4
1
5
8
l
4
5
10
•
51
71
Ci2
qz
'i2
0
60
.o
.c
.o
.o
.a
1.0000
l.OOCO
l.COCO
1.0000
1.00CO
••*
/
l'lOT
PLOT
PlOT
PLOT
PLOT
PlOT
PLCT
PLCT
PLOT
PLCT
PLOT
PlOT
PlOT
PlCT
PLCT
(;0002300
00002400
STAR a60 -
STATf~fNTS
51
1
1
1
1
10
10
1
c
0
0
~ARIABlE
0
31
2
3
4
5
CURVE NO. 10
.o
s.ooc
toc.o
520 .c
10.00
15.0C
20.00
LABEl IS
-1.0000
1 ~\All DEllA J
.l MAll llH JA T
1
0
0
0
1
l
Cf
CF
CF
Cf
Cf
CF
400.00
c. a
400.00
c.o
o.c
o.o
600.00
30.00
600.00
30.00
zc.co
30.01)
f
472. 1
443.2
426.C
Ol Cl.TLE T
STA~
070- SPECIAL
CCNVERGENCE lEST
(~
CCI\I~Cl
~AlE
00002600
OOC021GC
OOOC2800
00002900
OCCOJCGO
000031CO
PlGI CRl DATA
Y AXIS lABEl IS flUID TtMP-DEG f 1\11~ liMITS
X AXIS lABEl IS TIHE-SECCNCS
illltt LIMITS
kltt-1 Llli ITS
Y <\XIS UBEl IS lu\ll HMF-CfG f
X AXIS lABEl IS 11 fiE- SECCNCS
k1H1 lli!ITS
Y AXIS LABEl IS kALL OT CEG f
k IT tt L Hl!ll S
X AXIS lABEL IS ll H-SEC£NCS
WIH llliiTS
CURIIE NO. 1 lASH IS lSl t-CLJE 1/4
CURVE NC. 2 LABEL IS 2NC NODE 1/2
CuRVE NO. 3 LABEL IS 3Rt f\CCE 3/4
CURVE NO. 4 lABEL 1 S CUllET
CURVE NC. 5 LASH IS 1S1 ~ODE 1/4
ClJR VE NC. 6 li\BH IS 21\C f\CCE 1/2
z, CURVE NC. 7 LABEL IS 3RC t-CDE 3/4
2t CURVE NO. 8 LABEL IS 4H 1\00E OL I
1t CLRIIE NC. 9 LABEL IS IRLI: ~ ALUE
HUD 1EHP- OEG
PT. NO.
II ME· SECCNOS
•••
4
1 1
1 c
E
1 110
T 10
0 40
J
1,
1,
2,
2t
3,
3,
lt
lt
lt
lt
2t
2,
1
PLOT
II
1
I flL!C 1\0DE
~0002500
••• STAR OJO- FLNCTICI\
1 D
2 E
0
0
Of
Oil003200
OOOC3.300
OOQC34ClC
COC03500
OCC03600
QQOQ31CC
OOCQJSCC
00003900
QQCG 40 0 C
COCC41CC
CCC04200
CCCQ430C
OOQQHQO
00004500
CCCC46CQ
00Cil46QQ
0Cillil46QO
CCCC47CC
ilOCC480C
OC00490C
CGC05CCC
OOQ05Hl0
CO~STAI\TS
TEMPE~AIU~f
ChAI\Gf
00005300
/
..j:::,
w
•
6
1.0000
17
80
• SJOOO
+1;
OL~P Y,T.c,,, ' DUH~IES AT END Lf
If lNFUl, O~ERRIDES CALCULAlEO OThEJA
***
T 10 T 40
0
T 100
T 60 T 90
0
0
0
T 110 T 140
0
D
D
E
•
10 D
J
40
0
10
40.
;)
90
e
0
0
0
n
1
0
1
0
1
J
080-
SJ~R
.o
.o
.a
.o
.o
.o
.o
.o
.o
.o
.a
.a
.a
.o
.u
.c
.c
.o
a
0
0
0
0
0
0
a
Q
••• SlAR
•
0
0
99
0
0
0
0
PRl~JOLT
.o
09a-
.soooo
***
ELAPSt~
0
CPL JlHE fCR
*** ELAPSEC CPL
lJ~E
fCA
00005~00
OOC055CO
OQCGSilOC
SPECifJCAJlC~S
10
1
10
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
10
0
a
c
0
0
0
0
a
0
PRl~TCLT
0
.a
30.000
C~SE
£
STOP
0
0
0
0
0
a
l
10
1
OOQ05100
00005800
00005900
"ClQilCOC
OO,Oil100
CCCOil2CO
QQCCil3CO
OCC06~00
10
OCCOfSCC
OCCQUCC
1
1
OOC067ill0
Tl~ES
0
0
l~PLT
=
O. SEC
•u
A~Jlh
=
0.0 Sf:C
***
OJCQ6800
OOCCil'ICO
~
..p.
,
SUMMA~V
NOOE
NO. TVP E
GKUUP C
1
2
3
4
5
29
30
31
32
TEMPERATURE
en
Of NUCE CCNNECTIOS
CAPAC I IAt\CE
1
1
1
1
1
1
1
1 -
l
4CO.'l0
400.00
400.00
400.00
400.00
400.00
400.00
400.00
400.00
.39050
.39050
.39050
.39050
• 39050
.39050
.39050
.39050
.39050
RATE
TE~PS.
ARE
AT TIME=
~ALUES
HEAT GEI\ER.
iCJ
fiNITE-CAPACITY NUDES WHCSE
A~t
'QJ
1\CGE
CO~I\EC. f GR
NO~
lYPE
Jelf\EC
OETER~INEt
av
c.o
~EAT
1
1
1
.25015
.10635
.14360
1.51111
2
52
53
1
1
.25015
.10635
.143f:IC
1.5611
3
53
4
54
1
1
.25015
.106.35
.14380
1. 5611
104
54
1
55
5
55
l
.ts015
.10t35
.14360
!.5611
105
12'1
2'1
1
'· 5611
H
7S
1
.25015
.1Jtl5
.14380
30
l
l
.25015
.11lt35
.14 3e0
1.56lll
130
60
131
81
1
1
.25015
.10635
.14380
1.56U
31
.25015
1.5611
1
102
52
Hl3
.o
.o
.o
lC/SUM'fl
1. 5611
1
51
.o
.o
l.IME CUSI.
.25015
.10t35
.14360
101
-s1
.a
.o
l'fl
BALANCES
.o
.o
AGHUIANUS
eo
81
..j:::>
()1
33
1
400.0J
34
1
400. JJ
3)
l
36
1
31
1
38
39
40
51
52
53
1
400 • .JJ
400.1)0
400.01)
400.0\l
1
4JO.OO
1
400.00
1
1
1
4JJ.JO
400.00
400.0J
.3905J
.39J50
• 390 50
.390 5u
.39J51l
.390 50
.39050
.39C 50
.3905J
.39050
• 390 50
.o
.a
.o
.o
.o
.a
.o
.o
.a
132
32
641
82
1
1
.10f35
.14380
33
83
1
1
.25015
.10t35
.14380
1.5oU
13 3
a3
34
84
1
.25015
.10f35
.14380
1.5611
134
84
35
a5
1
.25015
.104:35
.14380
1. 5611
13 5
85
3t
86
1
1
.25015
.10f35
.H3aO
1.5611
llf
86
31
87
1
1
.25015
.10f35
• 14 380
1.5611
131
81
3E
1
a8
1
.25015
.104:35
• 14 380
1.5611
138
8e
H
as
1
1
.25015
• 10 f3 5
.14 380
1.5611
13S
as
c;o
40
90
1
1
.25015
.10f35
• 14 380
1.5611
140
51
1
• 14 3SO
• lit 380
2. 7156
1
52
1
• 14 3 80
.14380
2.115t
2
.llt380
2.115t
3
53
1
.o
.o
1
l
.llt380
..j:::>
Q)
,
400.00
54
55
87
aa
69
90
101
102
103
1l4
105
1
1
1
400 .oo
400.00
400.00
.39050
• 39050
.39050
.39050
4->J .Jo
.39050
~
400.00
.390 50
l
400.00
1
1
l
1
1
400.00
41Ji).JQ
400.00
400.00
1.0417
1.0417
1.0417
1.0417
1o0'tl7
.a
.l't3BO
.llt380
2.115tl
.H360
• llt3 80
2. 7156
i
!
.l't 380
.14380
2.7156
81
as
1
.14380
.14380
2. 115t
31:
ac;
1
.14 380
.14380
2.1156
39
sc
1
.14.380
.14:380
2.1156
40
1
!C 1
102
1
0
0
Z.UIS6
.l0t35
.2.0E33
2 .Of33
.H5H
l
100
-102
2
102
103
1
0
0
2.18'iit
.l0f35
2.06!3
2.0833
.lt757't
2
101
- 103
3
103
104
1
0
Q
2 .l€'ii6
.10f35
2.0833
2.1JE33
.47514
3
102
-104
4
lO't
10!
1
0
2 .1B'i6
.10t35
2.01:33
2 .OEB
.U514
4
10 3
-10!:
5
10 !l
1
2.18'i6
.Hlt35
l.OE33
.475H
5
104
It
5lt
'
5
5!:
31
.o
.a
.o
.o
.a
.o
.o
.a
.o
c
.o
c
+»
""-.1
,
106
101
137
136
139
1
1
1
1
1
4\)Q.()Q
400.1)0
4GO.OO
400.00
400.il0
-lOt
lOt
a
t
105
-101
6
lOt
101
1
.10635
0
2.0633
2.0633
1
1
1
106
-108
101
106
0
0
31
31
!
136
-138
131
136
0
0
36
131
3S
1
138
0
- l]Cj
13f»
0
3'i
3CJ
138
.a
1.0411
1
4JO.OO
.415H
2.11iCJ6
.10t35
2.0633
z.0633
.H511t
2.1896
.10635
2 .oe33
2.0833
.4l5H
13S
140
a
2ol8CJ6
.10635
2.0833
2.0633
.ltl514
1
0
1
13S
Q
2.1S'>t
.1063 5
2.0f33
.ltHH
40
140
- 101
101
0
2.0SH
.a
1.0411
.o
l.O'tl7
1.0417
•.IJ
.a
1.0411
ItO
SOURCE
..>I{OUP S
lOJ
-1
NCO~S
6JO .00
WHOSE TEHPS
0
A~f
GIVEN
.a
.c
***
.415 H
2.1896
.10635
2.0f.33
2.oe33
-14C
140
2.1856
.o
1.01tl7
2.u6H
El~SPEC
CPU IIHE fOR
SEI~P
=
.c
O. SEC •••
-J::.
00
TIME
T 10
T 60
T 110
0
10
E 10
TIME
o.o
:
=
=
"'
"'
"'
:
400. OJ
400.00
4 oc. 00
.o
•c
=
TIME
-=
10
60
"'
400.ao
-=
400~00
-
=
40J.J8
4.8828E-OJ
4. 8828E- 03
T
T
r uo
0
E
10
10
TIME
T
10
60
T 110
11'1
0
E 10
T
TIME
=
"'
=
=
=
r
20
T 70
1 120
D 2J
E 20
u.5ooo
T 10
T 60
T 110
0
10
E 10
400.00
4J.:J .oo
400.01
4.88281:-04
4.ll828E-u4
=
=
:
1 20
T 70
T 120
0 20
E 20
1.3().:10
;
=
;
400.04
400.00
400.48
3.27l5E-02
3.2 715E-02
2.0000
=
-
400.00
400.00
40C. CO
= .o
= .o
CEL TA T=
=
=
=
=
"'
40Q.oa
4Ca.OO
400.00
.a
.a
r 20
T 70
T 12J
D
20
E
20
==
=
-=
;
= .o
.soooo
CEllA T=
T
T
20
10
f
120
c;
20
E
20
=
=
=
-=
=
CELT A J=
30
I ac
1 130
D 3a
E 30
T
400.00
400.00
400.00
.o
.o
.50000
=
=
=
=
T 30
T SJ
T 13G
D 30
E 30
"'
Cf
..
=
=
=
Cf
T
30
l
80
r 130
c 30
E 30
=
=
=
:
Cf
~~.
T 30
T 80
T 130
D 30
E 30
~0.
=
4
itOO.OO
T
40
4CO.CO
4CC.OO
T 90
ITEAATIC~S
llERAI!L~S
4CO.• CO
400.00
4GC.CO
IlE~AJICNS
400.00
400.00
400.00
=
=
"'
.;c
=
=
.o
Cf
I
40
90
I
T l'tO
0 4a
40
.a
.a
;
0
E
.a
.c
=
~C.
400.00
400.00
400.00
.c
.c
~0.
.sooou
400.00
40a.co
40C.aa
.o
Cf ITERATICNS =
~C.
.5oaoo
DELTA T=
1. 500 0
;
.50000
OEl TA J:
IIE~AJICNS
r 140
0 40
E 40
=
"'
=
=
=
:
"':
400.00
400.00
400.00
.a
.o
T 100
400.aa
4ao.ca
400.00
.o
T laO
'tOO.CO
400.00
4aO.CO
I 100
=
"'
=
.:
:
600.00
.o
.o
.o
6CO.OO
.o
= .a
= .a
=
=
=
= .a
4
r 40
T 90
T 140
0 40
E 40
=
4
r
'tO
T 90
T litO
0 '40
E
40
=
"
"'
=
=
=
.o
;
"'
= .o
:
=
=
400.00
400.CO
400.00
= .a
= .o
=
"'
1 100
=
=
tco.oo
.c
.o
.c
600.00
.c
= .o
= .o
.j:>.
~0
11)
60
T 110
D 10
E 10
=
=
=
TIME
=
10
T 6J
T 110
=
T
T
-
=
400.17
400.02
4\) 1. 94
.14941
.14941
l
2()
T 70
T 120
D 20
E 20
40C.CO
400.00
400.00
.;
=
=
=
=
.a
.o
CELT A T=
2.50JQ
30
T 80
T 130
c 30
E 30
1
.5COOO
400.00
400.00
4CQ.OO
.::
=
"'
"'
.a
.a
=
r teO
T 90
T 140
0 40
40
f
CF JTERATICNS"'
~0.
=
=
=
=
"'
400.00
400.00
400.00
.o
l 100
=
:::
"'
"'
.o
600.00
.c
.a
.a
4
ill>
T
=
=
10
10
=
TIME
=
T
10
60
T 110
D 10
E 10
;:
T
.::
TIME
-
D
E
T
Ll
60
T 110
D 10
E 10
T
TIME
1 11.1
T 60
=
.;;
;:
.::
=
=
=
=
.
=
::
40(). 58
400.u8
405. 73
.50098
.500'18
l
=
'
.::
20
70
T 120
0 20
E 20
3.00()0
'tOl. 56
400.25
413.25
1. 3130
1.3130
=
T
20 =
70 . ::
T 120 =
0 20 ;:
E 20 =
DEL U
20
70
I 120
c 20
E 20
T
T
T=
=
=
=
=
=
IJELT A T=
1
20
T
70
=
=
1
c
E
30
=
=
30
8C
T 130
.a
= .a
Cf
::
1
=
c
30
E
30
=
=
30
80
1 130
r; 30
E 3a
'
3C
80
4GO.OO
4CO.CO
4CC.OO
.a
T 40
T 90
T 140
0 40
E 40
=
=
:0
"'
=
::
4CO.OO
400.00
400.00
.G
.o
Of ITERATICNS
..
=
4CO.OO
400.00
"'
;:
=
400 .·co
4CC.OO
400.CO
I 100
.o
= .o
=
=
=
=
=
=
400.00
400.00
400.GO
I 100
'tO
=
90
T 140
0 4C
E 40
=
=
400.00
400.00
400.00
' 100
T
r
= 6uD.OO
= .a
= .a
::
.a
=
t:co.ao
"'
"'
.o.
4
T 40
T 90
r 140
c 40
E 40
Cf ITERA11CNS =
~0.
l
T
JTE~ATIC~S
= .o
~0.
T
400.00
400.00
4CO.OO
:::
T
.50000
400.01
4uO.OJ
J(j
=
~C.
.50000
400.00
400.00
400.C2
1.9531E-CJ
1.9531E-C3
3Q
1 80
T 130
.5ooao
400.00
400.00
400.00
4. f828E-C4
4.E828E-C4
T
4 .OOOJ
406.56
401.31
=
.:
Of:LTA T=
3.5000
403.46
400.63
425.25
2. 82 96
2.8296
400.00
40\l.OO
400.00
4. E828E-C4
4.8828E-C4
.o
.o
.Q
..
.a
"'
600.00
4
=
.G
=
.o
400.00
4CO.CC
=
4
T
40
=
T
tOO
"'
.Q
= .o
=
::
T 100
..
=
.o
600.00
.o
Ul
0
T 110
10
E 10
.:
TIME
=
10
T 60
T 110
0 10
E 10
=
=
D
T
·=
=
=
=
TIME
=
T 10
T 60
T 110
0 10
E 10
=
TIME
.::
10
60
=
1 110
0 10
=
T
T
E
=
=
=
=
=
=
10
=
TIME
=
441.22
5.1953
5.1953
T 120
D 2il
E 20
4.51)()0
410.98
402.l:l
459.51
8.3628
8.3t.28
19 .5\)i)J
400.09
1.3242E-CJ
l. 3242E· C3
T 130
1: 30
E 30
=
4J0.02
T
30
'
::
~oo;oo
T
BC
=
=
it00.26
2.2949E-C2
2.29it9E-C2
' 130
D 30
E 30
20
70
1 120
D 20
E 20
=
Cfl TA T=
20
r 70
r 1.20
D 20
E 20
I
=
=
=
=
=
DELTA 1=
T
20
70
T 120
I)
20
E 20
1
=
=
=
=
=
DELTA T=
400.Cl7
itOC.01
400.t5
E.201ZE· 02
6.201ZE-C2
I
30
T 80
1 130
0 30
E 30
.50000
488.68
46f.tl
54 0. 21
20.055
21.6<;5
.50001)
Cf ITERAIICNS
'
30
so
T 130
0 30
E 30
~0.
4CO.OO
400.00
400.00
=
=
=
= .c
.o
=
CF ITERAIICNS
=
=
400.00
itCO.OO
400.00
=
= .o
,. .c
r Ho
0
f
40
40
=
4
T 40
T 90
T 140
0 40
E 40
=
1
'
40
':lu
431.66
lt22.lt7
4E2.S8
ItO
'T 140
90
=
15.1S3
15.1S3
0
E
40
40
=
3
=
=
CF
ITE~A11CNS
400.ou
=
teoo.co
=
=
=
=
=
.o
.o
:z
1 100
400.00
400.00
=
=
600.00
=
.c
.o
.o
.::
600.00
"'
.a
.o
.o
.c
:
=
:::
.::
=
400.00
40C.CO
itCO.CO
T 100
=
.o
.o
.o
.::
600.00
.
=
.o
.a
3
=
=
=
=
=
4
1 litO
0 40
E 40
Cf 11ERA11CNS =
~U.
J
.o
.c
=
~0.
.50000
ltCO.OO
=
=
~0.
.50000
T
19 .001)0
541.J9
525.01
580.18
16.J81
29.661
-
=
CEllA T=
5.0000
416.58
404.50
418.04
12.086
12.086
=
T
=
=
=
ltC6.at
lt02.86
it23.01
lt.OOH
:::
~.003'i
=
I 100
.o
= .o
= .a
""
U1
......
"
T
=
10
60
T 110
D 10
E 10
=
=
TIME
=
T 10
T 60
T llO
J
10
E 10
=
=
TIME
=
T 10
T 60
T 110
=
=
T
=
=
=
=
=
=
I)
10
E
10
=
=
TIME
=
T 10
T 60
T 110
D 10
E 10
=
=
=
T1 ME
=
T lJ
T 60
T llO
D 10
=
=
=
=
=
=
543.41
521.92
58l.J5
15 ~ 4<J2
29.661
T 20
70
T 120
D 20
E 20
l
20
70
T 120
0 20
E 20
1
T
20.5000
54 7. 78
533.41
582.68
14.372
29.661
20
T 70
T 120
D 20
E 20
'
=
=
=
=
=
=
=
=
=
=
=
DELTA T=
T
20
l
10
T l2J
D 20
E 20
=
=
=
=
=
CELT A T=
21.50il0
551.84
5 38. 51
584. 17
13.326
=
DELTA T=
21.0000
549.S!i
536.01
583.44
13.840
29.661
-=
CEL TA T=
20.0000
545.b4
530. 71
581.88
14.923
29.661
-=
=
r
20
70
120
0 20
l
J
=
=
=
=
491.S6
H2.28
542.38
19.681
21.6'i5
T 3C
T 8CJ
1 130
0 30
E 30
.50000
495.16
H5.S7
544.41
19.2S4
21.6'i5
479.38
546.48
18.900
21.6'i5
=
=
'
"'
0
30
=
E
3C
..
Of
T
::
1
"'
=
=
=
30
J 80
1 130
0 30
E 30
=
=
..=
=
30
80
T 130
c 30
r
'
lTE~ATlONS
447.10
431.25
4Cjj4.21
16.451
16.451
=
450.c;8
l.t34.30
4S1.~5
16.687
16.687
=
=
=
=
454.23
437.38
500.73
1t.849
=
=
=
=
=
40S."a2
403.67
426.sc;
"· 7500
4.7500
1 100
H0.14
404.t1
HO.Ci2
5.53CJ
5.5303
1 !CO
412.03
4a5.1Q
435.C4
6o32S6
6.32Et
1 100
414.07
406.<;4
43S.lS
7.127<;
7.1279
J 100
H6.24
408.33
443.33
7.Cjjl36
1 lOG
= cco.oo
= .(j
= .o
"'
.a
.=
600.00
3
I
40
T 90
T 140
0 40
E 40
=
=
=
=
=
=
.a
.a
"'
=
600.00
.a
=
.a
3
40
T 9C
T 140
r.: 40
E 40
I
=
=
=
=
=
=
.G
=
.o
=
6CO.Cl0
3
40
T 90
T 140
T
0
ItO
E
40
Cf ITERATIONS'"
~C.
T
40
90
1 140
0 40
E 40
T
Cf ITERATICNS =
~0.
T
444.38
428.25
4Cj(l.68
16.130
16.130
=
~0.
30
80
T 130
0
30
E 30
441.03
42!5.32
486.<;5
15.114
15.714
Cf ITERATICNS =
30
8a
T 13a
.50000
5.:14.30
486.19
550.30
18.102
=
=
T
.50000
501.33
482.83
548.43
18.502
21.6'i5
"'
~0.
.50000
49S.28
=
"'
=
=
,.
=
""
.c
.c
.c
..
t.IJO.OO
:
:
~
T ItO
T 90
T 140
0 l.tO
=
=
=
::
""
==
=
.c
.o
.a
(j]
N
LO
=
TIME
=
E
T
10
T 60
T 110
D . 10
E 10
TIME
T
T
10
60
=
.:
=
=
=
=
=
=
10
10
=
.:
=
TIME
=
T 110
0
E
T 10
T 60
T 110
D 10
E 10
=
=
=
=
=
29.661
E
22.0000
553. 15
540.92
584.86
12.830
29.661
)\). J <l·Jil
576.02
569.13
592.66
6. 8936
29.661
=
DELTA T=
T 20
T
10
T 120
D 20
E 20
T
T
21)
70
1 120
D 20
E 20
507.19
489.~'1
=
=
=
55t.ll
11.100
21.6t;5
510.Cl
=
~53.26
=
=
~92.11
17.300
21.695
30
80
1 13C
()
30
E 30
T
1
T
3C
80
1 130
D 30
E 30
'
·-
54"4. 36
532.59
'573.18
.:
ll.7l:8
=
2l.b95
3()
80
1 130
IJ
30
E 10
ItO
Cf 11ERA11CNS =
3
..=
.:
=
=
lt57.1t4
41t0.49
503.78
16.947
16.947
OF ITERATICNS
=
=
=
=
=
46G.t1
H3.t2
506.11
16.992
16.9"2
lt[j
'T 140
90
T
0
E
=
=
=
=
=
=
503.02
48E. 24
51tl.31
14.113
16.992
40
ItO
40
90
T l'tO
0 40
E 40
'
7.9136
=
H8.52
it09.85
447.40
e.t7C9
8.67G9
1 100
T Hill
=
lt20.'i0
411.51
lt51.39
9.3882
"'
9.38B2
=
lt60.31
Ht.25
498.U
14.0ts
-=
"'
=
=
"'
te:o.oo
"'
=
.c
.o
.c
=
600.00
=
=
=
"'
.o
"' .o
= .a
=
3
40
9C
T 140
D 40
E 40
I
T
=
3
J
Gf ITERATICNS =
~0.
T
T
E
16.8~9
:
~0.
.50000
=
"'
30
~C.
.50()iJJ
=
=
E
.50000
=
DEll A I=
T 20
I
70
T 120
D 20
E 20
21.6«;5
=
DElTA T=
22.50JJ
555.59
543.21t
585.53
12.351
29.661
20
=
=
=
=
1~.ou
T 1u0
= 600.00
= .c
.o
=
= .a
(.)'1
(.,.)
vARIABLE
CU~P
AT
TI~E
=
3.COOOOE•Ol
ADM I lT ANCES
0
5
10
15
20
25
30
35
40
50
55
6a
65
70
75
80
85
90
100
1 05
LlO
115
120
12 5
130
135
140
.a
.10635
• 1()63 5
.10635
.10635
.10 63 5
• 10 63 5
.10635
.10635
.a
.14380
.1438a
.14380
.14380
.14380
.14380
.14380
.14380
.o
2.0833
2.0833
2.0833
2.0833
2.0833
2.0833
2 .il833
• 1063 5
.10635
.10635
.10635
• 1063!:
.10635
.10635
.1063!:
.o
.10635
.10635
.10t35
.10t35
.10f35
.10t35
.10t35
.10t35
.o
LOUT IU-lS
.14380
.1H8C
.1438G
.1438C
.1438C
.14380
.143BC
.14380
45
LCCA TICNS
2.083:
2.0833
2.0833
2.0833
2.0833
2.0833
2.0833
2.0833
'i5 TliRCloGii
.o
T"~OUGI1
.UlBO
• 11t380
.lltJBO
.14380
.143SO
.14380
.14380
.11t38C
.o
~.0833
2.0633
2.0833
2.0833
2.0833
2.0833
2.01!33
2.0833
.10635
.10635
.10635
.10635
.10635
.10635
.10t35
.10635
.!0635
.10635
.10635
.10635
.10635
.10635
.!Ot35
.!OflS
4c; EQLAL ZE:RO
.14380
.14380
.14380
.14380
.14380
.14380
.14380
.14380
.11t380
.14380
.14380
.14380
.14380
.14380
.1'1380
.14380
99 fQIJAL .ZEIIO
2.0833
2.0833
2.0833
2.0833
2.0833
2.0833
2.0833
2.01!33
2.0833
2.0833
2;.0833
2.0833
2.08H
2.ClE33
2.0833
2.CEJ3
.c
.o
.o
.c
2.083~
TEHPERATLRES
0
5
10
15
20
25
30
.o
586.8S
576.02
561.76
5 44. 3f
524.46
503.02
593.18
5S4.9fi
513.44
558.51
540.5~
520.U
498.6 ~
591.80
582.95
570.12
555.1!5
536.t5
516.00
5'ii0.2'1
5S0.7.8
567.86
551.66
532.66
49~.27
lt89.9Q
511.11
sce.t. 5
518.41
564.81
548.06
528.60
507.37
48.5. 51t
(YJ
.f:.•
35
40
481. 21
460.31
.c
LOCATlCNS
50
55
60
65
70
75
80
85
90
100
105
110
115
120
12 5
130
13 5
140
• ·J
582.40
56'1.13
552.34
532.59
510.81
488.24
4t:6.26
446.25
600.00
597.4S
592.66
584.90
513.78
559.17
541.11
520.81
498.61
472.t:d
476.92
.o
45 ThRCl.(Jii
588.EC
571.53
562.81
544.75
524.06
5Cl.Ba
H9.3C
457.95
5Ci0.38
580.04
566.C4
548.60
528.3t
506.3.(
41E3. H
41f2.0t
.IC
LOCAl ICNS
599.6 5
5Cj6.7:
5H.3E
5 82.9 5
5H.l;
555.8!:
537.40
516.47
.a
'iS
H~Ol.Gii
59S.2l
595.87
ses.c;4
580.87
568.35
552.3«;
533.38
512.01
4t8.4S
.a
4S EQl.Al ZERO
586.68
sH.8e
55S.45
540.H
51S.6S
44i7.28
414.9C
453.94
.o
9S EQLAl ZERO
sse .73
51:Jit.91
s u.3c;
578.65
565.43
548.81
529.21
507.63
ltlt'l.36
.o
sa~.u
512.07
H5.4i6
536.14
515.21
lt92 .75
'tJ0.55
45C.G3
.,Q
ssa.u
SCJ3.84
SU.ll
sH.za
562.37
545.12
525.08
503 .lit
NODAL \IC LUME S
LCCA TICN S
0 ltlRCl.Gii
140 EQuAL ZERC
CAPi\C IT IES
a
5
10
15
20
25
30
35
4()
50
55
60
65
70
.o
.39050
.31:J050
• 39050
.39050
.39050
• 39050
.39050
.39()50
.o
.3905.:1
.3'1050
• 3'i050
• 39050
.39050
.;9osc
.;9csc
.39050
• 39050
.39050
.39050
• 3905C
.a
LCCU llNS
$3905C
.3CJ05C
.39050
.39C5C
.3905C
.39050
.39050
.39C50
.39050
.3SC5C
.3'i050
.39050
.39050
.a
45 HRCLH
.31i050
.3905il
.39050
.3'i050
.3SC50
.39050
.39050
.3905a
.39050
.39050
.3S050
.39050
.39050
.c
.3Sa50
.39050
.3SG50
.3Sa50
.39050
.3'i050
.3'i050
.3'i050
.c
4'i EQuAL lERG
.39050
.3c;050
.39050
.39050
.39050
.39050
.3'i050
.3Sa50
.3'i05C
.3'i050
Ul
c;,
75
80
85
90
100
105
110
ll5
120
125
130
135
140
.39J50
.39050
.39050
.39050
.3905C
.39050
.3905C
.o
LOCAll(NS
l.OH 1
•a
1.0411
1.0 417
1.0411
1.0411
1.0417
1.0417
1.0411
1.0411
1.0417
1.0417
1.0417
1.0411
1.0411
1.0411
1.0417
.39C50
• l'iO 50
.3'i050
.o
.39050
.3S05C
.3'i050
.o
'i5 HRCLGh
1.0417
1. 0411
1.0411
1.0411
l.C417
1.0411
1.0411
1.0411
'iS EQl.Al ZERO
0 ThROLGh
140 El.lLAl lERG
1.0417
1.0411
1o0411
1.0411
1.0411
l.04l7
1.0411
1.0417
.J9u5o
• 390 50
.39050
.o
1.0417
1.0417
1.C411
1.0417
l.O't 11
1.C'tll
1.C417
1.0411
HEAT GENEUTICtl
LOCATICNS
D
0
5
10
15
20
25
30
35
40
.o
35.939
29.661
25.098
21.695
19.058
30
35
16.992
40
ARR~Y
.a
5
10
15
20
25
IIALLES
13.SH
3.1S53
5.4185
7.90.38
10.3'15
12.5S2
14 .19'i
14. ec;o
14.S6C
14.128
2.H9e
4.9463
7.39H
s .. c;cs1
12.191
4.4863
6.8936
9.4155
11.768
13.645
14.773
14.<:l51
14.068
E
0
ARR~Y
15.375
14.068
14.i;61
3.6094
5.SQl'i
8. 410t
1C.86fl
12.97C
14.421
15.002
1'1.551
't.u~tou
t.3'i40
fl.'H55
11.326
13.322
14.US
1'1.997
H.Hl
VALLES
43.051t
H. 50S
28.617
24.355
21. 10 3
1a. 597
16.641
'tl.083
33.1H
27.t65
23.63t
20.560
lS.C'.lt
l4.EZCi
18.113
16.298
39.207
.H.882
26.175
22.939
20.031
11.155
31.43S
30. 14S
25.Sll
22 •. 308
19.530
11.37(1
lS.'HS
ts.tn
14.572
l't.331
SPECIAL CCNSTANlS
tYI
0)
57
Q'
0
I
loY
00
om
o-
.·•
or-
• •ao
~--
I
"'
0
I
w
00
0
00
0
CIO
00
0
Cl
_. .........
•
• 0
•
it
•
•
....•
Cl
Cl
.c:
Q
IX.
g
OOCIU'I
• • • •
~
a:
...
0
w
-...
X
::J
..,
"'
~
u
•
w
au
OCI
ou
oo
-•
...
.,
0
~
.
oQg
•
<C
d
w
·•...
•
.
<3'
-4
.. ..
0'1
0011\00
,..,..
._...
58
APPENDIX B
Exact Solution To A Pipe
Flow Problem With A Step Change
As Input
59
RESPONSE TO A STEP CHANGE
To calculate the response of the leading edge of a wave having gone
through a step change, assume that a slice of fluid always sees the
wall at the initial temperature.
(By the time the wall responds,
the leading edge of fluid will have moved on.)
sees
Do a heat balance on the fluid element just behind the leading edge;
q = hA(T-Tw) = CpV~~
where T = temp of fluid
Tw = 400°F
T = To = 600°F at t = 0
~~~~w~ e -~~~) t ~
t
0 sec
5
10
15
20
T
600°F
520
472
443
426
e -.1021t
h
.001 _BT_U__
sec°F i n2
A = l7Td
v = 7T/4d 21
d = 1.1284 11
P = 60 lbm/ft 3
C = 1 BTU/lbm°F
=