Basic heat transfer using EAI TR-10 electronic analog computer
by Michael Epifanio Hilario
A thesis submitted to the Graduate Faculty in partial fulfillment of the requirements for the degree of
MASTER OF SCIENCE in Mechanical Engineering
Montana State University
© Copyright by Michael Epifanio Hilario (1965)
Abstract:
The electronic analog computer is a very useful problem solving tool to the engineering student. The
Montana State University Engineering Department has two analog computers for the use of the faculty
and students. These are the EAI (Electronic Associates, Incorporated) TR - 48 and the smaller EAI TR
- 10. The Mechanical Engineering Department has 4 TR -10 patchboards and the necessary wires and
resistors to set up computer problems. Heat transfer is one field of engineering, among others, in which
this department will use the analog computer for problem analysis at the undergraduate and graduate
level.
An introductory course in analog computer programming should enable the student to use the computer
for various problem solutions even without a strong background in electronics or servomechanisms.
This thesis is presented as a guide to the student wishing to utilize the analog computer for heat transfer
work. It demonstrates general problem analysis and some necessary procedures for obtaining solutions
for some of the basic heat transfer equations.
The literature search in preparation of this thesis left the impression that there has not been much work
in heat transfer using the analog computer at the undergraduate level. Rather, it appears that
analog-simulation systems; such as the resistance analog, conductive-solid analog, and
conductive-liquid analog, have been used more often for analysis.
EiASIC HEAT TRANSFER USING EAI TR-IO
ELECTRONIC ANALOG COMPUTER
MICHAEL EPIFANIO HILARIO
A t h e s i s s u b m itte d to th e G ra d u a te F a c u lty in p a r t i a l
f u l f i l l m e n t o f th e re q u ire m e n ts f o r th e d e g re e
or
MASTER OF SCIENCE
ln
M ech an ical E n g in e e rin g
A pproved:
H.
H ead, M ajor D ep k ^tm en t>
C hairm an, E xam ining Com m ittee
D ean, G rad u ate D iv is io n
MONTANA STATE UNIVERSITY
Bozeman, M ontana
A u g u st, 1965
- iii-
Acknowledgment
A p p r e c ia tio n and th a n k s a r e e x te n d e d to D r. H. F. M u l i i k i n 9 Head
o f th e M ech an ical E n g in e e rin g D e p a rtm e n t, f o r h i s h e lp and a d v ic e .
-iv TABLE OF CONTENTS
g ag e
T i t l e Page
V ita
o
o
©
o
o
o
o
o
e
e
e
o
o
e
e
o
g
o
e
o
o
o
o
o
o
e
e
e
o
o
e
e
t
p
o
e
e
Acknowledgment
0
0
0
0
0
0
0
0
0
9
0
0
0
0
0
L i s t o f F ig u r e s
0
0
0
0
0
0
0
0
0
0
0
0
9
0
0
A b s tr a c t
o
o
o
o
o
o
e
e
e
o
C h a p te r I - I n t r o d u c t i o n
o
0
e
9.
0
0
0
0
9
0
0
o
o
e
e
e
o
i
e
ii
0
0
9
0
e
0
0
0
0
0
0
0
0
0
0
9
0
0
O O
0
0
O O
0
O
0
0
O
O
O
9
0
0
0
0
0
EAI TR-IO E l e c t r o n i c A nalog Computer
C h a p te r 3
S te a d y - S t a te H eat C o n d u c tio n :
C h a p te r 4
H eat T r a n s f e r by th e T ech n iq u e o f Lumped P a ra m e te rs
C h a p te r 5
Two T ec h n iq u es f o r th e S o lu tio n o f P a r t i a l
D i f f e r e n t i a l E q u a tio n s .
0
0
0
0
-
0
0
0
0
0
0
0
0
0
O
O
O
O
O
O
0
9
0
0
0
0
9
0
0
6
0
0
0
0
0
I
5
7
17
0
C h a p te r 6 - A nalog S o lu ti o n o f an U n s te a d y - S ta te H eat C o n d u ctio n
P robIem
0
0
v ii
0
U niform C r o s s - S e c tio n F in
O O
iii
v
0
C h a p te r 2
0
'9
0
0
0
31
0
C h a p te r 7
O n e-D im en sio n al; U n s te a d y - S ta te H eat Flow P ro b lem
39
C h a p te r §
A nalog Computer S im u la tio n o f a P lu g -F lo w System
45
A ppendix
L i t e r a t u r e C o n su lte d
0
0
0
0
0
0
9
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
6
0
0
6
0
0
0
o
0
o
0
o
0
o
o
0
o
0
0
52
o
o
69
-v LIST OF FIGURES
F ig u re No.
U niform C r o s s - S e c tio n Rod F in
e
e
o
Q
o
o
o
e
e
o
o
o
o
o
A nalog Computer D iagram f o r Rod F in P roblem
Computer S o lu tio n o f Rod F in P ro b lem
S o lu tio n o f Rod F in P ro b lem .
e
e
o
o
e
°
e
Page No
I
8
, 2
14
o
o
o
o
o
e
e
•
3
15
e
*
4 .
16
O
5
20
•
6
21
e
e
e
e
e
e
e
e
e
O ne-D im ensio n al H eat T r a n s f e r A cro ss a
P la n e -C o o lin g o f a S p h ere e e o e
e
o
o
e
e
e
e
o
A nalog Com puter D iagram f o r C o o lin g S p h ere P ro b lem
Com puter S o lu tio n o f C o o lin g S p h e re P ro b lem .
o,
e
o
q
o
e
•
7
22
F i n i t e D if f e r e n c e s - H e a t T r a n s f e r i n a S lab
0
9
0
0
9
0
6
8
29
9
30
9
10
35
A nalog Computer D iagram f o r U n s te a d y - S ta te H eat
C o n d u ctio n i n a S lab
O
11
36
T em p eratu re - Time:
H eat C o n d u ctio n i n S lab
°
12a
37
T em p eratu re - Time:
H eat C o n d u ctio n i n S lab
°
12b
38
0
13
42
14
43
*
15
44
0
16
49
°
S e p a r a tio n o f V a r ia b le s , .
U n s te a d y ^ S ta te H eat C o n d u ctio n
O
0
0
0
0
0
0
O ne-D im ensio n al H eat T r a n s f e r i n an Aluminum Rod
»
Com puter D iagram f o r O n e-D im en sio n al H eat Flow i n an
Aluminum Rod e » o o • o o
S o lu tio n o f O ne-D im en sio n al H eat Flow i n an
Aluminum Rod
O
O
O
O
O
•
Gas F low ing i n ap I n s u l a t e d P ip e
o
e
e
o
o
e
o
o
o
e
o
e
o
o
o
o
17
50
0
0
0
0O
CO
- t*“ 4~
Computer D iagram f o r P lu g -F lo w S im u la tio n
e
51
19
65
P lu g -F lo w S im u la tio n - R esponse o f Tg0 to Changes
I " Tg i
0
0
0
0
0
B a s ic Computer Symbols
O
6
0
0
O
6
O
O
6
O
0
O
0
0
0
6
-v i—
LIST OF FIGURES ( c o n tin u e d )
F ig u re No.
B a s ic Com puter Symbols
. . . . . . . . . .
-Page No.
. . .
20
66
S o lu tio n o f Example P ro b lem . . . . . . . .
. . .
21
67
Example P roblem s - Diagram m ing F undam entals
. . .
22
68
-v ii-
A b stra C t
The e l e c t r o n i c a n a lo g com puter i s a v e ry u s e f u l p ro b lem s o lv in g to o l
t o th e e n g in e e r in g s t u d e n t . The M ontana S t a t e U n iv e r s ity E n g in e e rin g
D ep artm en t h a s two a n a lo g co m p u ters f o r th e u s e o f th e f a c u l t y and s tu d e n ts .
These a r e th e EAI ( E l e c t r o n i c A s s o c i a t e s , I n c o r p o r a te d ) TE - 48 and th e
s m a lle r EAI TE - 10. The M ech an ical E n g in e e rin g D ep artm en t h a s 4 TE -1 0
p a tc h b o a rd s and th e n e c e s s a r y w ir e s and r e s i s t o r s to s e t up com puter
p ro b le m s. H eat t r a n s f e r i s one f i e l d o f e n g i n e e r in g , among o t h e r s , in
w hich t h i s d e p a rtm e n t w i l l u se th e a n a lo g com puter f o r p ro b lem a n a l y s i s
a t th e u n d e r g r a d u a te and g r a d u a te l e v e l .
An i n t r o d u c t o r y c o u rs e i n a n a lo g co m p u ter program m ing sh o u ld e n a b le
th e s tu d e n t t o u s e th e com puter f o r v a r io u s p ro b lem s o l u t i o n s even w ith ­
o u t a s tr o n g b ack g ro u n d i n e l e c t r o n i c s o r serv o m ec h an ism s. T h is t h e s i s
i s p r e s e n te d as a g u id e to th e s tu d e n t w is h in g to u t i l i z e th e a n a lo g
co m p u ter f o r h e a t t r a n s f e r w ork. I t d e m o n s tra te s g e n e r a l p ro b lem a n a l y s i s
and some n e c e s s a r y p ro c e d u re s f o r o b ta in in g s o l u t i o n s f o r some o f th e
b a s i c h e a t t r a n s f e r e q u a t io n s .
The l i t e r a t u r e s e a r c h in p r e p a r a t i o n o f t h i s t h e s i s l e f t th e im p re s s io n
t h a t t h e r e h a s n o t b een much work in h e a t t r a n s f e r u s in g th e a n a lo g com­
p u t e r a t th e u n d e rg ra d u a te l e v e l . E a t h e r , i t a p p e a rs t h a t a n a lo g - s im u la ­
t i o n s y s te m s ; su ch as th e r e s i s t a n c e a n a lo g , c o n d u c t iv e - s o li d a n a lo g , and
c o n d u c t i v e - l i q u i d a n a lo g , h av e b een u se d more o f te n f o r a n a l y s i s =
CHAPTER I
INTRODUCTION
T h ere a r e two ty p e s o f e l e c t r o n i c co m p u ters i n g e n e r a l u s e in i n d u s t r y
f o r p u rp o s e s o f r e s e a r c h , d a t a p r o c e s s in g , c o n t r o l s y s te m s , to name a few .
T hese a r e th e a n a lo g com p u ter and th e d i g i t a l co m p u ter.
p u t e r i s a b le t o s t o r e p ro b lem d a t a and i n s t r u c t i o n s .
The d i g i t a l com­
Through v a r io u s r e ­
l a t e d u n i t s , th e d i g i t a l f i n a l l y s u p p lie s th e r e s u l t s o f c o m p u ta tio n s .
G e n e r a lly , th e d i g i t a l co m p u ter p e rfo rm s o n ly a d d i t i o n and t h e r e i s no
r e l a t i o n s h i p b etw een th e com puting tim e and th e p ro b lem v a r i a b l e s .
On th e
o th e r h a n d , a n a lo g co m p u ters do n o t c o u n t i n term s o f n u m b ers, b u t u se
c o n tin u o u s ly v a r i a b l e d a t a .
P ro b lem s o l u t i o n by th e a n a lo g com puter i s
a c c o m p lish e d by a n a lo g y ; th e co m p u ter i s programmed so t h a t i t s c i r c u i t
e q u a tio n s hav e th e same m a th e m a tic a l form as th e e q u a t i o n s . o f th e p ro b lem .
Thus v o lt a g e s r e p r e s e n t v a r io u s p h y s ic a l q u a n t i t i e s su ch as d i s t a n c e , tem­
p e r a t u r e , f o r c e , d is p la c e m e n t, and so o n .
The v o lta g e m a g n itu d e s a r e r e ­
l a t e d to th e p h y s i c a l q u a n t i t y m a g n itu d es by a s c a l e f a c t o r .
The a n a lo g
co m p u ter p e rfo rm s i t s o p e r a t io n s i n a c o n tin u o u s p r o c e s s w ith r e s p e c t to
tim e ; t h e r e f o r e , t h e r e i s a r e l a t i o n s h i p b etw een th e a c t u a l com puter ru n
tim e and th e p ro b lem tim e o r some o t h e r in d e p e n d e n t v a r i a b l e .
A nalog co m p u ters make i t p o s s i b l e to s im u la te e n t i r e s y ste m p ro b le m s ,
co m p o n en ts, o r s u b sy s te m s .
I t may b e t h a t p a ra m e te rs o f a sy ste m w ould be
d i f f i c u l t , e x p e n s iv e , o r tim e consum ing t o a c t u a l l y change i n r e a l e q u ip ­
m en t.
H owever, i f s im u la te d on an a n a lo g c o m p u te r, p a ra m e n te rs co u ld be
v a r i e d by s im p ly u s in g a S iM in. a w e ll d e s ig n e d a n a lo g c i r c u i t .
P ro c e s s
e f f e c t s t h a t ta k e h o u rs o r ev en days can b e s im u la te d on an a n a lo g com­
p u t e r so t h a t d a t a can be. o b ta in e d i n a m a tte r o f s e c o n d s .
-2 -
A lth o u g h a know ledge o f e l e c t r o n i c s i s v e ry v a lu a b le to a p e rs o n
u t i l i z i n g th e a n a lo g c o m p u te r, i t i s n o t n e c e s s a r y .
c o u rs e i n c i r c u i t a n a l y s i s w i l l s u f f i c e .
An i n t r o d u c to r y
The e n g in e e r in g s tu d e n t w i l l
f in d t h a t u s in g an A nalog computer i s fairly e a s y and very applicable to
e n g in e e r in g p ro b le m s .
F o r ex am p le, m e c h a n ic a l v i b r a t i o n d i f f e r e n t i a l
e q u a tio n s h av e tim e as th e in d e p e n d e n t v a r i a b l e and th e a n a lo g com puter
1
:
'
/ "
p ro d u c e s s o l u t i o n s t o d i f f e r e n t i a l e q u a tio n s w ith th e in d e p e n d e n t v a r i ^
a b le b e in g tim e .
T h is th e n in v o lv e s th e d i r e c t s o l u t i o n o f th e e q u a f ld n s .
Many h e a t t r a n s f e r p ro b lem s a l s o hav e tim e as th e in d e p e n d e n t v a r i a b l e
and th e a n a l y s i s and s o l u t i o n o f th e s e p ro b lem s i s f a i r l y s t r a i g h t f o r w a r d .
Any d is c u s s io n o f h e a t t r a n s f e r s o l u t i o n s w ould n o t be co m p lete w ith ­
o u t i n c lu d in g th e s o l u t i o n o f p a r t i a l d i f f e r e n t i a l e q u a t io n s .
The m athe­
m a tic a l d e s c r i p t i o n o f any c o n tin u o u s m edia u s u a ll y in v o lv e s p a r t i a l
d i f f e r e n t i a l e q u a t i o n s , i . e . , d i f f e r e n t i a l e q u a tio n s h a v in g d e r i v a t i v e s
w ith r e s p e c t to more th a n one in d e p e n d e n t v a r i a b l e .
S in c e th e e l e c t r o n i c
a n a lo g com p u ter can i n t e g r a t e w ith r e s p e c t t o o n ly one v a r i a b l e , nam ely
tim e , i t can o n ly b e u se d to s o lv e o r d in a r y d i f f e r e n t i a l e q u a tio n s .
Thus i t i s n e c e s s a r y t o c o n v e r t a p a r t i a l d i f f e r e n t i a l e q u a tio n to one
o r more o r d in a r y d i f f e r e n t i a l e q u a tio n s i n o r d e r to s o lv e i t on th e a n a lo g
co m p u ter.
T h ere a r e two p o p u la r te c h n iq u e s f o r d o in g t h i s :
( I ) th e m ethod
o f s e p a r a t i o n o f v a r i a b l e s , le a d in g to an e ig e n v a lu e p ro b le m ; ( 2) th e
f i n i t e d i f f e r e n c e m ethod w h e re in d e r i v a t i v e s w ith r e s p e c t . t o a l l v a r i a b l e s
b u t one a r e r e p la c e d by f i n i t e - d i f f e r e n c e a p p r o x im a tio n s .
The f i r s t m ethod
w i l l o n ly b e d is c u s s e d and th e seco n d m ethod w i l l be d e m o n stra te d by a n a ly ­
s i s and a n a lo g com puter s o l u t i o n .
-3 -
The b a s i s o f a n a lo g com puter o p e r a t io n i s p r e s e n te d i n th e A ppendix.
No a tte m p t i s made to p ro v e o r e x p l a in any o f th e co m p u ter o p e r a tio n s on
‘
th e b a s i s o f e l e c t r o n i c th e o r y . The r e a d e r i s in s te a d r e f e r r e d to one o f
th e t e x t s l i s t e d i n th e l i t e r a t u r e c o n s u lt e d .
The o p e r a t io n o f th e e l e c ­
t r o n i c a n a lo g co m p u ter i t s e l f i s p r e s e n te d i n C h a p te r 2 .
a n a lo g com p u ter was u sed e x c l u s i v e l y .
The EAI TR-IO
T h is i s a r e l a t i v e l y s m a ll com puter
and th u s i s somewhat li m i t e d i n th e s i z e o f p ro b lem s w hich can be s o lv e d .
The l i m i t i n g f a c t o r i s th e number o f o p e r a t i o n a l a m p l i f i e r s a v a i l a b l e f o r
th e p ro b lem s o l u t i o n .
The answ er^ from th e com puter a r e o b ta in e d i n th e form o f o b s e r v a tio n s
o r r e c o r d in g s o f c o n tin u o u s tim e v a r i a b l e s .
I f th e p ro b lem i s s t a t i c , t h a t
i s , i f th e an sw ers h av e one and o n ly one v a lu e when th e s o l u t i o n h a s b een
r e a c h e d , a l l th e v a r i a b l e s w i l l have c e a se d t o change and t h e i r v a lu e s can
be re a d w ith a s im p le v o lt m e t e r .
In m ost c a s e s , h o w ev er, th e p ro b lem s o l ­
u t i o n in v o lv e s v a r i a b l e s t h a t change c o n tin u o u s ly w ith tim e and a r e r e c o r d ■ ■
’
ed c o n tin u o u s ly i n tim e . The p ro b lem s o l u t i o n s p r e s e n te d h e r e i n w ere r e ­
co rd ed on a tw o -d im e n s io n a l X -Y p l o t on p a p e r m e a su rin g 10 in c h e s x 15
in c h e s th e n r e - p l o t t e d on s ta n d a r d 8 1 / 2 . in c h x 11 in c h p a p e r .
R eg ard in g
th e s o l u t i o n o f an e q u a tio n o r a s e t o f e q u a t io n s , i t s h o u ld be u n d e rs to o d
'I
t h a t v e ry seld o m i s j u s t one n u m e ric a l answ er s o u g h t.
,
.
R a th e r a b e t t e r
u n d e r s ta n d in g o f th e s i t u a t i o n o r s y ste m b e in g s tu d ie d i s more im p o r ta n t.
A l o g i c a l q u e s tio n a r i s e s as to when t o u se a d i g i t a l com puter and
when t o u s e an a n a lo g co m p u ter.
B a s i c a l l y , th e f o llo w in g s ta te m e n ts con­
c e r n in g when to u se a d i g i t a l com puter c o u ld be s t a t e d .
-4 +
1.
p ro b lem s in v o lv in g l a r g e am ounts o f ta b u l a t e d d a t a in
d is c r e te - n u m b e r form .
2.
Problem s w ith p r e c i s e in p u t d a t a r e q u i r i n g p r e c i s e s o l u t i o n s
S i m i l a r l y , th e f o llo w in g s ta te m e n ts a p p ly c o n c e rn in g when t p u s e an e l e c ­
t r o n i c a n a lo g co m p u ter.
1.
S o lu tio n o f s im u lta n e o u s d i f f e r e n t i a l e q u a tio n s .
2.
When an a n a lo g y o f a m e c h a n ic a l, o r o th e r ty p e o f sy ste m i s
r e q u ir e d .
T h a t i s , a m a th e m a tic a l model w i l l r e p r e s e n t th e
p h y s ic a l p ro b le m , th u s s o l u t i o n o f th e e q u a tio n s w i l l be
a n a lo g o u s to th e p h y s ic a l sy ste m .
CHAPTER 2
EAI TR - 10 ELECTRONIC ANALOG COMPUTER
The TR - 10 p a tc h b o a rd c o n ta in s a l l o f th e com puter com ponent con­
n e c t io n s w hich a r e e x p la in e d ii} th e A ppendix.
The p a tc h b o a rd i s remov­
a b le from th e com puter so t h a t f u l l u se can be made o f th e co m p u ter.
P roblem s can be " p a tc h e d 1* o r w ire d b e f o r e p u t t i n g th e p a tc h b o a rd on th e
co m p u ter; th e n p o te n tio m e te r s e t t i n g s can be made when th e o p e r a t o r i s
re a d y f o r th e p ro b lem s o l u t i o n .
The c o n t r o l p a n e l o f th e TR-10 c o n ta in s th e fo llo w in g c o n t r o l sw itch es.
Mode C o n tro l S w itc h ; p o s i t i o n s a r e RESET, HOLD, and OPERATE.
R eset
The programmed p ro b lem i s a t th e c o n d ito n s c o rre s p o n d ­
in g to tim e z e r o .
H old
T h is p o s i t i o n w i l l s to p and h o ld th e p ro b lem s o l u t i o n
i n s t a n t l y f o r o b s e r v a ti o n .
O p e ra te
T h is w i l l s t a r t th e p ro b lem s o l u t i o n .
O p e ra tio n can
- s t a r t from th e r e s e t o r h o ld p o s i t i o n .
M eter Mode S e l e c t o r S w itc h ; p o s i t i o n s a r e POT BUS, NULL, V .M ., AMP, BAL.
P o t Bus
T h is p o s i t i o n i s u sed to s e t v a lu e s on p o t s .
T h is i s a
p r e c i s i o n p o te n tio m e te r w hich i s b r id g e d so t h a t when th e
p o t w hich i s b e in g s e t re a c h e s th e same v a lu e as th e
- . ' 1 VI
.V
\ V
- ! .a ’1
■
I
[I ‘
’
p r e c i s i o n p o te n tio m e te r , th e r e a d in g on th e m e te r w i l l
show a n u l l i n d i c a t i o n .
N u ll
A p o s i t i o n u sed f o r s e t t i n g p o te n tio m e te r s o th e r th a n
th e ones w hich a r e found on th e M ontana S t a t e U n iv e r s ity
a n a lo g co m p u ter.
“,6 “*
VoMo
T h is c o n n e c ts th e m e te r f o r u s e as a v o |t i m e t e r = The
VoMo s w itc h was n o t u se d f o r any p ro b le m s «
Amp
T h is c o n n e c ts th e m e te r as a v o l t m e t e r .t o th e a m p l i f i e r
s e l e c t o r s w itc h so t h a t th e o u tp u t v o lt a g e o f each
a m p l i f i e r pen b e re a d on th e m e te r »
T h is i s u sed as a
ch eck to d e te rm in e w h at a m p l i f i e r i s o v e rlo a d e d
10
v o l t s ) when th e r e i s an o v e rlo a d in d ic a tio n o
B al
Used to b a l a n c e , t o z e r o , th e s t a b i l i z e r o u tp u ts o f a l l
a m p lifie rs o
The im p o r ta n t s w itc h p o s i t i o n s f o r p ro b le m s o l u t i o n a r e th e p o t bus
and amp s e t t i n g s »
s e ttin g s o
The p o t bus i s n e c e s s a r y i n o r d e r to o b ta in a c c u r a te p o t
The amp s e t t i n g e n a b le s th e o p e r a t o r to r e a d th e v o lta g e o u tp u t
o f eac h a m p l i f i e r .
The com p u ter w i l l hpm d i s t i n c t l y a u d ib ly when t h e r e i s
an a m p l i f i e r w h ich i s o v e r lo a d e d .
T h is i s a b r i e f p r e s e n t a t i o n o f th e more im p o r ta n t c o n t r o l com ponents
o f th e TR-IO a n a lo g c o m p u te r,
A more e x t e n s i v e e x p la n a tio n o f c o n t r o l s can
b e found in th e o p e r a t o r ’ s handbook w h ich i s alw ays lo c a te d i n th e com p u ter
room.
The o p e r a t o r ’ s handbook w i l l a l s o i l l u s t r a t e how t o make th e p a t c h ­
b o a rd c o n n e c tio n s r e q u ir e d from th e co m p u ter d ia g ra m .
The c o n n e c tio n s a r e
v e ry e a s y to make; th e handbook b e in g e a s y to fo llo w and a l s o , th e o p e r a t o r
soon le a r n s th e b a s i c c o n n e c tio n s .
The m ost im p o rta n t a s p e c t o f u s in g th e
a n a lo g com p u ter i s to o b t a i n th e c o r r e c t com puter d ia g ra m .
CHAPTER 3
STEADY-STATE HEAT CONDUCTION;
UNIFORM CROSS-SECTION FIN
C o n s id e r a f i n o f u n ifo rm c i r c u l a r c r o s s - s e c t i o n . F ig u re I , whose b a s e
i s a tta c h e d t o a p la n e s e c t i o n w hich i s m a in ta in e d a t th e c o n s ta n t te m p er­
a t u r e Tjj .
The f i n ex ch an g es h e a t a lo n g i t s s u r f a c e w ith a f l u i d a t a b u lk
te m p e ra tu re T3 .
A h e a t b a la n c e f o r a d i f f e r e n t i a l e le m e n t o f th e f i n m ust
c o n s id e r h e a t flo w i n t o th e e le m e n t by c o n d u c tio n , h e a t g e n e r a te d by th e
e le m e n t, h e a t flo w o u t o f th e ele m e n t by c o n d u c tio n , and th e h e a t c a p a c it y
o f th e e le m e n t.
In s t e a d y - s t a t e h e a t t r a n s f e r th e h e a t c a p a c it y o f th e
e le m e n t i s u n im p o rta n t and i f th e f i n i s h ig h l y c o n d u c tin g th e te m p e ra tu re
a lo n g th e f i n i s a f u n c tio n o f th e a x i a l d i s t a n c e a lo n e .
Now a h e a t b a l ­
ance f o r a d i f f e r e n t i a l ele m e n t in su ch a f i n w i l l b e c o n s id e r e d in o r d e r to
d e r iv e an e q u a tio n f o r th e te m p e ra tu re d i s t r i b u t i o n a lo n g i t s le n g th .
H eat
flow s by c o n d u c tio n i n t o one f a c e o f th e e le m e n t w h ile h e a t flow s o u t o f
th e o p p o s ite fa c e by c o n d u c tio n and from th e s u r f a c e by c o n v e c tio n .
In
th e a b se n c e o f e n e rg y g e n e r a tio n w ith i n th e ele m e n t and u n d e r s t e a d y - s t a t e
c o n d i tio n s :
R ate o f h e a t flow by
c o n d u c tio n i n t o e l e m ent a t x
dT.
-kA dx
=
-kA d T +
R ate o f h e a t flo w by
c o n d u c tio n o u t o f e l e - +
ment a t (x + dx)
d _ f-kA &
dx + ..»
"dx \
dx ,
R a te o f h e a t flo w
by c o n v e c tio n from
su rfa c e
+ hAg
2
-kA £Ex = - kA £Ex -kA U E x dx + h (Pdx) (Tx - Ta )
dx
dx
dx2
(Tx - Ta )
Ta = 100 F
P erim eter —\
A = (tTd) dx
\
dx H —
/
Tx=Q= 200 F
0 .7 0 6 f t
Ta = 100 F
/
F igu re I .
Uniform C ro ss-S ectio n Rod F in
-9 -
(1 )
Where
q = R ate o f h e a t flo w
k = Therm al c o n d u c t iv it y
h = H eat t r a n s f e r c o e f f i c i e n t
As = S u r f a c e a r e a
A = A rea o f f i n c r o s s - s e c t i o n
P = P e r im e te r
L et
T = CTx - Ta )
Ta = C o n s ta n t
T h e re fo re
and
^ 2Tx
d 2r
dx2
dx2
e q . I can be w r i t t e n
In d e p e n d e n t V a r ia b le O th e r Than Time
The in d e p e n d e n t v a r i a b l e o f e q . 2 i s d i s t a n c e r a t h e r th a n tim e .
T h is
does n o t mean t h a t i s i s im p o s s ib le to s o lv e i t on th e a n a lo g co m p u ter, i t
does mean a tr a n s f o r m a tio n betw een time- and d is ta n c e m ust b e d e f in e d .
The
d e g re e o f th e c o rre s p o n d e n c e betw een th e d i s t a n c e x and th e tim e t w i l l
depend on th e u n i t s i n w hich th e e q u a tio n i s
w ritte n .
I f d i s t a n c e x i s in
- 10™
i n c h e s , a p o s s i b l e c o rre s p o n d e n c e may be I in c h = I seco n d o r i f x i s in
f e e t , th e c o rre s p o n d e n c e may b e I f o o t = I s e c o n d , and so f o r t h .
e rin g eq,
of x.
C o nsid­
2 , th e u n i t s o f p , h , k , and A m ust be c o n s i s t e n t w ith th e u n i t s
The r e l a t i o n o f tim e and
d i s t a n c e f o r th e s o l u t i o n o f th e ro d f i n
h e a t t r a n s f e r e q u a t io n , e q , 2 , can be w r i t t e n :
x ( f t ) = a , t (sec)
it=a
a ls o
x = f(t)
T .= g (x )
By u s in g th e c h a in r u l e tw ic e and s u b s t i t u t i n g
dT . dx
dt
dt
dt
d t2
dT
dx
dx
= a r_ d ZdT
[_dx W x
dx
dt
S - - S - it
d2r
dt 2
and
d2T
a 2 d%
2
(3 )
A nalog Com puter S o lu tio n o f a S te a d y - S ta te H eat C o n d u c tio n P roblem
A n a ly s is o f th e u n ifo rm c r o s s - s e c t i o n ro d f i n w i l l c o n tin u e by con­
s i d e r i n g an exam ple p ro b le m .
P ro b lem :
An e n g in e i s a i r - c o o l e d by c i r c u l a r - r o d f i n s .
ro u n d in g am b ien t a i r te m p e r a tu r e i s 100 F.
The s u r ­
The te m p e ra tu re
-
11-
o f th e end o f th e f i n t h a t i s e n t i r e l y in th e a i r i s 200F.
D eterm in e th e te m p e ra tu re d i s t r i b u t i o n a lo n g th e le n g th o f
th e f i n .
U
_
r-
The p ro b lem i s i l l u s t r a t e d i n F ig u re I .
BtU
h r ft2 F
■
L et
I in c h d ia m e te r
L = 0 .7 0 6 f t
E q u a tio n 2 h as b een found ti
th e ro d f i n .
I-,et
<5^ h v - Frj-- F
I
9.6, ^ 2 ' 1
Then
(4 )
A s u b s t i t u t i o n m ust be made
v a r i a b l e o th e r th a n tim e .
w i l l b e s u b s t i t u t e d i n t o Eq
(4)
d h = mT
dx2
a2 d t2
or
(5)
d2I . a 2,
d t2
I n o r d e r t o o b ta in a n u m e ric a l v a lu e f o r a , t i s a r b i t r a r i l y s e l e c t e d
as I seco n d o f p ro b lem tim e .
a = * = Oi T O O t
t
I seco n d
a 2 = ( 0 .7 0 6 ) 2 = 0 .5 0
.
-1 2 -
U sin g d o t n o t a t i o n f o r th e d e r i v a t i v e s so t h a t ^2% and 4 ^ w i l l be T and
d t2
d£
and T r e s p e c t i v e l y r e s u l t s i n :
( 6)
T = a 2mT = 0 .5 0 ( 9 .6 ) T = 4 ,8 0 T
T h is th e n i s th e e q u a tio n w hich m ust be s o lv e d on th e a n a lo g co m p u ter.
The o n ly b o u n d ary c o n d itio n i s :
Tkg^
= (200 - 1 0 0 )F = IOOF
T hree s t e p s w i l l be fo llo w e d b e f o r e w ir in g th e com puter p a tc h b o a r d .
a re :
These
( I ) m a g n itu d e s c a l i n g , (2 ) tim e s c a l i n g , (3 ) draw a pompI e t e com­
p u t e r d ia g ra m .
From know ledge o f th e p ro b le m , i t i s p o s s i b l e to e s tim a te
maximum te m p e ra tu re s and p ro b lem d u r a t i o n .
I t s h o u ld b e k e p t i n mind t h a t
s h o u ld an e s ti m a t e be w ro n g , i t i s an e a sy t h i n g to change th e s c a l e f a c ­
t o r s by ch a n g in g th e p o t s e t t i n g s .
t h a t Tmax = 1000F.
F or t h i s p ro b lem , i t w i l l be assumed
T h e r e f o r e , on a " p e r v o l t " b a s i s ^ th e s c a l e f a c t o r
w ill be:
I O v - 1000F
and th e s c a le d v a r i a b l e i s :
[ 0 .0 1
I v o l t = IOOF
The com puter tim e c o n s ta n t i s :
B — —*«•
t
w here
t c = com puter tim e
t = a c t u a l p ro b lem tim e.
F o r th e ro d f i n p ro b le m , a s a t i s f a c t o r y v a lu e o f com puter ru n tim e i s 10
s e c o n d s , so t h a t B = 10.
-1 3 -
In t h i s p ro b le m , T i s a f u n c tio n o f x so t h a t t h e r e i s no p roblem
tim e in v o lv e d i n t h i s s t e a d y - s t a t e c o n d i t i o n , b u t a v a lu e o f B m ust be
s e l e c t e d t o g iv e a s a t i s f a c t o r y com puter tim e .
The com p u ter d ia g ra m i s shown in F ig u r e 2 .
The i n i t i a l c o n d i tio n , o r
known b o u n d ary v a l u e , i s a t th e ex p o sed end o f th e ro d f i n w here x = 0 .
The com puter d ia g ra m i s a s o l u t i o n t o E q u a tio n 6 .
on th e com p u ter d ia g ra m and r e p r e s e n t s T.
A v o lta g e m ust be g e n e ra te d to
r e p r e s e n t d i s t a n c e x o r tim e t on th e X a x i s .
The v o lt a g e i s g e n e ra te d
by i n t e g r a t i n g a v o lt a g e as i s shown i n F ig u re 2 .
w i l l b e g e n e r a te d in 10 s e c o n d s .
The o u tp u t Y i s shown
A v o lta g e o f 10 v o l t s
An arm s c a l e s e t t i n g on th e X - Y
p lo t­
t e r o f I v o l t / i n c h w i l l g e n e r a te 10 in c h e s on th e p l o t t i n g p a p e r in 10
seconds.
I n o r d e r t o show th e com puter s o l u t i o n ( v o l t s and tim e ) and th e ,c o n ­
v e r s i o n to p ro b le m s o l u t i o n (te m p e r a tu re and d i s t a n c e ) , b o th g rap h s a r e
shown a s F ig u re 3 and F ig u r e 4 r e s p e c t i v e l y .
I v o l t = 100 F
The c o n v e rs io n f a c t o r s u s e d :
(p e r v o lt b a s is )
I seco n d o f p ro b lem tim e = 0 .7 0 6 f e e t
10 seco n d s o f com puter tim e = I seco n d o f p ro b lem tim e
The com puter tim q was 10 s e c o n d s ; t h e r e f o r e , th e p ro b lem tim e was I s e c o n d .
T h is I seco n d r e p r e s e n t e d 0 .7 0 6 f e e t o f ro d f i n le n g th .
-Ik -1 0 v
I n i t i a l C o n d itio n :
I v = 100 F
I(O )= 100 F
I v o l t in p u t
-1 0 v
F ig u r e 2«,
A nalog Computer D iagram f o r Rod F in Problem
-5 1 Computer Time (seco n d s)
F ig u re 3»
Computer S o lu tio n o f Rod F in Problem
■x ( d is ta n c e i n f e e t a lo n g ro d f i n )
x = a»t j
F ig u re U.
S o lu tio n o f Rod F in Problem
t = t c/B
x = (0 „0 7 0 6 )* tc
CHAPTER 4
HEAT TRANSFER BY THE TECHNIQUE OF LUMPED PARAMETERS
T h is i s p ro b a b ly th e s im p le s t te c h n iq u e a v a i l a b l e f o r th e tr e a tm e n t o f
th e t r a n s i e n t exchange o f th e rm a l e n e rg y betw een a s o l i d and th e f l u i d in
w hich i t i s im m ersed.
In t h i s ap p ro ac h th e te m p e ra tu re g r a d ie n ts on th e
i n t e r i o r o f th e s o l i d a r e n e g le c te d and th e s o l i d i s assum ed to be a t a
u n ifo rm te m p e r a tu r e .
Such a c o n d itio n i s ap p ro ac h ed when a s o l i d o f h ig h
th e rm a l d i f f u s i v i t y t r a n s f e r s h e a t to a f l u i d th ro u g h a f il m t h a t p r e s e n ts
a h ig h r e s i s t a n c e t o h e a t flo w .
T h is s i t u a t i o n i s som etim es c h a r a c te r i z e d
as one i n w hich th e f i l m r e s i s t a n c e i s c o n t r o l l i n g .
U nder th e s e c o n d itio n s ^
th e r a t e a t w hich th e i n t e r n a l en e rg y o f th e body d e c r e a s e s may be e q u a te d
to th e r a t e a t w hich e n e rg y i s t r a n s f e r r e d a c ro s s th e b o u n d ary o f th e s o l i d .
I n F ig u re 5 a d e t a i l e d d e r i v a t i o n o f t h i s ty p e i s p r e s e n te d as an exam p le.
G e n e ra lly th e e q u a tio n w hich r e s u l t s from t h i s te c h n iq u e shows th e r a t e o f
c o o lin g o f th e s o l i d i s p r o p o r t i o n a l t o th e d i f f e r e n c e b etw een th e mean
te m p e ra tu re o f th e s o l i d and th e b u lk te m p e ra tu re o f th e s u rro u n d in g f l u i d ;
(7)
w here
dTi
dt
- K d 1 - T2)
T1 = mean te m p e r a tu r e o f th e c o n d u c tin g s o l i d
T2 = b u lk te m p e ra tu re o f th e am b ien t f l u i d
K = a c o n s ta n t o f p r o p o r t i o n a l i t y .
I n tr o d u c in g th e e x c e s s te m p e r a tu r e , T - T 1 - T 2 , th e s o l u t i o n to t h i s
e q u a tio n i s :
( 8)
T ( t ) = T(O) e~Kt
The t r a n s i e n t b e h a v io r o f th e mean te m p e ra tu re o f th e s o l i d w i l l depend on
th e c o n s ta n t K w hich i n t u r n i s r e l a t e d to such g e o m e tric f a c t o r s as th e
—1 8 -
s u r f a c e a r e a and volum e and to su ch h e a t t r a n s f e r p a ra m e te rs as th e f il m
c o e f f i c i e n t and th e s p e c i f i c h e a t o f th e s o l i d .
A p ro b lem i n w hich E q u a tio n *7 w i l l a p p ly may be s t a t e d i n o r d e r to
i l l u s t r a t e th e com puter s o l u t i o n .
P ro b lem :
A m e t a l - c a s t i n g o f volum e V and s u r f a c e a r e a As , i n i t i a l l y
a t 900 F i s s u d d e n ly p la c e d i n an atm o sp h ere o f OF.
U sing
K v a lu e s o f 0 .3 and 0 . 7 , l e t us d e te rm in e th e te m p e ra tu re
o f th e c a s t i n g a s a f u n c ti o n o f tim e .
S in c e th e d i f f e r e n t i a l e q u a tio n i s a lr e a d y a f u n c tio n o f tim e , E q u a tio n 7
can b e s c a le d and a co m p u ter d ia g ra m draw n.
The maximum v a lu e s w hich can
be assum ed f o r m ag n itu d e s c a l i n g a r e
Tmax = 900 F
-
'
,
Tmav = 300 F /se c o n d
•
T h e r e f o re , th e s c a le d v a r i a b l e s on th e b a s i s t h a t 10 v o l t s = 900 F w i l l be
The te m p e ra tu re d ro p i s v e ry r a p id w ith th e K v a lu e s w hich a r e u s e d , t h e r e ­
f o r e , 2 seco n d s o f p ro b lem tim e w i l l be c o n s id e r e d .
The com puter tim e can
b e e x te n d e d to 10 seco n d s by u s in g a tim e s c a l e f a c t o r .
t c = Bt
B = & . = 10 seco n d s = 5
t
2 seco n d s
The in p u t t o th e i n t e g r a t o r m u st, h o w ev er, be d iv id e d by 5 .
c o n d i tio n i s T(O) - 900F.
in te g r a to r is u sed .
f o llo w s :
The i n i t i a l
S in c e 10 v o l t s e q u a l 900F, 10 v o l t s in p u t to th e
The e q u a tio n f o f th e com puter d ia g ra m can be found as
—1 9 —
T = -K T
(9)
( 10)
The c o m p lete com puter d ia g ra m i s shown i n F ig u re 6 , a lo n g w ith th e com puter
d ia g ra m w hich w i l l g e n e r a te th e tim e f u n c ti o n .
shown i n F ig u r e 7.
The co m p u ter s o l u t i o n i s
- 20-
As (S u rfa c e A rea) T]_ = T em perature o f S phere
O F = T em perature o f
S u rro u n d in g F lu i d
(T1 - Tf )
C o n sta n t
Where:
H eat T r a n s fe r
C o e f f ic ie n t
S p e c if ic h e a t
F ig u r e 5 .
S u rfa c e a re a
V = Volume
O ne-D im ensional H eat T r a n s f e r A cross a S u rfa c e — C ooling
o f a S phere
- 21-
-IO v
I n i t i a l C o n d itio n
-SK 900
Pen
Arm
-IO v
0 .2
L
F ig u r e 6 .
A nalog Computer Diagram f o r C o o lin g S phere Problem
E xcess T em perature (Tf l u i d - Tsp h e re
Time (seco n d s)
F ig u re 7«
Computer S o lu tio n o f C ooling Sphere Problem
CHAPTER 5
TWO TECHNIQUES FOR THE SOLUTION OF PARTIAL DIFFERENTIAL EQUATIONS
Many h e a t t r a n s f e r p ro b lem s r e q u i r e th e s o l u t i o n o f p a r t i a l d i f f e r e n ­
t i a l e q u a t io n s .
The s o l u t i o n m ust s a t i s f y th e p a r t i a l d i f f e r e n t i a l eq u a­
t i o n s and th e p e r t i n e n t b o u n d ary c o n d itio n s d e f in i n g th e p ro b lem .
The
e l e c t r o n i c a n a lo g com puter can s o lv e p a r t i a l d i f f e r e n t i a l e q u a tio n s o n ly
by c o n v e r tin g them t o o r d in a r y d i f f e r e n t i a l e q u a t io n s .
a re two m ethods o f a c c o m p lish in g t h i s c o n v e r s io n .
G e n e r a lly , th e r e
Thqy a r e ; ( I )
t i o n o f v a r i a b l e s le a d in g t o an e ig e n v a lu e p ro b lem , and ( 2)
se p a ra ­
r e p la c in g
th e p a r t i a l d e r i v a t i v e s by f i n i t e d i f f e r e n c e s .
Of th e two a p p r o a c h e s , th e f i n i t e d i f f e r e n c e m ethod i s m ost o f te n
u sed .
The o t h e r m ethod i s n o t alw ays a p p l ic a b l e s in c e th e v a r i a b l e s
som etim es c a n n o t be s e p a r a t e d ;
e q u a t io n s .
t h i s i s e s p e c i a l l y t r u e w ith n o n - l i n e a r
A ls o , th e sum m ation o f a l l o f th e e ig e n f u n c tio n p r o d u c ts i s
in v o lv e d and t h i s r e q u i r e s many a n a lo g com p u ter co m p o n en ts.
F or th e p u rp o se o f o b ta in in g th e o n e -d im e n s io n a l h e a t c o n d u c tio n
e q u a t io n , c o n s id e r th e s la b o f u n ifo rm c r o s s s e c t i o n w h ich i s shown in
F ig u re 8a ,
The s la b i s i n s u l a t e d a t a l l s u r f a c e s e x c e p t th e ends so t h a t
•
•
'
h e a t flow i s in th e x d i r e c t i o n a lo n e .
'
I f a h e a t b a la n c e i s w r i t t e n f o r
th e d i f f e r e n t i a l e le m e n t o f volume A d x , th e o n e -d im e n s io n a l h e a t conduc­
t i o n e q u a tio n may be d e r iv e d .
The t o t a l h e a t flo w in g i n t o th e
e le m e n ta l
volum e e i t h e r in c r e a s e s th e a v e ra g e te m p e r a tu r e o f th e volum e o r flow s
th ro u g h th e e le m e n t:
—2 4 —
- kA£ L + kA £ U
-K
+[K
A
il 4
(A d *> 7T
*
(kA
£)£] - C
/=(A
d
x
)jl
- k A i r + k A J ? ■» K A - ^ I d x
= C z , (A d s) J X
kA^ d
x =c/>A
d
x
kA
dxA
=c/3 A
or:
(ID
^ Xa
= f
AL
H
o n e -d im e n s io n a l
h e a t c o n d u c tio n
e q u a tio n
F i n i t e - D i f f e r e n c e A p p ro x im atio n
I t can be se e n t h a t E q u a tio n 11 c o n ta in s two in d e p e n d e n t v a r i a b l e s —
l o c a t i o n and tim e .
T h e r e f o r e , an e x a c t s o l u t i o n on th e e l e c t r o n i c a n a lo g
com puter i s n o t p o s s i b l e , h o w ev er, i n s t e a d o f com puting te m p e ra tu re a t a l l
x v a l u e s , th e te m p e ra tu re T can be com puted a t c e r t a i n s t a t i o n s alo n g x .
F ig u re 8b .
T h is i s p o s s i b l e by u s in g th e f i n i t e d i f f e r e n c e te c h n iq u e and
w i l l re d u c e E q u a tio n 11 to a sy ste m o f f i r s t - o r d e r o r d in a r y d i f f e r e n t i a l
e q u a tio n s .
When s o lv e d s im u lta n e o u s ly , th e y w i l l p ro v id e an ap p ro x im ate
s o lu tio n .
I t m ust be p o in te d o u t t h a t i t i s p o s s i b l e to d iv id e th e r i g h t s i d e ,
» in t o d is c r e t e i n t e r v a l s , A t , how ever, s in c e th e a n alog computer
-2 5 -
i n t e g r a t e s w ith r e s p e c t to tim e , i t i s a d v an tag eo u s to d iv id e d is ta n c e x
i n t o d i s c r e t e i n t e r v a l s and i n t e g r a t e w ith r e s p e c t to tim e .
C e n tr a l f i n i t e - d i f f e r e n c e s w i l l be u sed in tr a n s f o r m in g
i n t o f i n i t e - d i f f e r e n c e form .
C o n sid e r th e te m p e ra tu re - d is ta n c e cu rv e o f
F ig u re 8b , w ith th e A x in c re m e n ts show n.
F i r s t th e r e s p e c t i v e s lo p e s o f
th e g e n e r a l cu rv e a t th e s t a t i o n s (n + 1/ 2) and (n - 1/ 2) .
c*"1"
T T .n + k
<)T
T rt+, - T n
Ax
T n ~ Tn-i
n - Vz
Xl-jS i n c e i s
AX
th e r a t e o f change o f s l o p e , i t fo llo w s t h a t :
i!l
=
- 4 1
In+'/%
A ^
s u b s titu tin g
Tn-vi ~Tn _Tn ~ TnAX
ax
^T
AX
( 12)
Trt-H - 2 In ^ In-I
£ T
b<3'
"
( 6 *) =
S u b s t i t u t i n g E q u a tio n 12 i n t o E q u a tio n 11 y i e l d s a g e n e r a l e q u a tio n f o r
th e Ot *1 s t a t i o n .
dT
('3 )
d-fc h
_
_k
Tn+i
2 Tn 3- T m —1
(A X )^
“26-
Now, th e tim e d e r i v a t i v e o f Tn i s a t o t a l d e r i v a t i v e s in c e x i s f ix e d a t
(n°A%) f o r Tn .
E q u a tio n 13 c o n ta in s o n ly one in d e p e n d e n t v a r i a b l e (tim e )
and i s in a form w hich can b e s o lv e d on th e a n a lo g co m p u ter.
The e q u a tio n
r e p r e s e n t s an a r r a y o f s im u lta n e o u s e q u a t io n s ; th e s i z e o f th e a r r a y i s
e q u a l to th e number o f in c re m e n ts w hich may b e a r b i t r a r i l y s e l e c t e d a lo n g
th e x a x i s .
S e p a r a tio n o f V a r ia b le s
As an exam ple o f th e s o l u t i o n o f E q u a tio n 11, l e t u s c o n s id e r th e
p ro b lem o f th e s la b o f th ic k n e s s
when b o th s u r f a c e s a r e m a in ta in e d a t
z e ro te m p e ra tu re and th e i n i t i a l tim p e r a tu r e d i s t r i b u t i o n i s f ( x ) .
T h at
i s , th e p ro b lem w here
and
k
(H)
dt
r
. T (o ,t) = 0
T (8 , t ) = 0
T (x ,o ) = f ( x )
To s e p a r a t e th e v a r i a b l e s l e t us assum e t h a t
T ( x , t ) = F (x) G (t)
and by s u b s t i t u t i o n i n th e p a r t i a l d i f f e r e n t i a l e q u a t io n , i t may be
shown t h a t th e f u n c tio n s m ust s a t i s f y th e f o llo w in g o r d in a r y d i f f e r e n t i a l
e q u a tio n s :
(14)
F " (x ) + p 2 F (x) = 0
(15)
G '( t ) +
pc,
G (t) = 0
.
The s o l u t i o n s o f th e s e e q u a t io n s , a f t e r a p p ly in g th e b o u n d ary c o n d itio n s
a re ,
Fn (x ) = An s i n px
,
/I Tf _
w here p = - j - a t n
= n ( t) - Bn e - V'’a *
w here Vn =
-2 7 -
Hence f o r any v a lu e o f n , th e fo llo w in g e x p r e s s io n s a t i s f i e s b o th th e
t r a n s i e n t h e a t c o n d u c tio n e q u a tio n and th e b o u n d ary c o n d itio n s o f th e
exam ple p ro b lem .
L e t t i n g Cn = AnBn
Not o n ly i s t h i s e x p r e s s io n a s o l u t i o n o f e q u a tio n 11 and o f th e b o u n d ary
c o n d i t i o n s , b u t th e sum o v e r a l l v a lu e s o f n o f e x p r e s s io n s o f t h i s t^jke
is a lso a s o lu tio n .
The s o l u t i o n m ust s a t i s f y th e i n i t i a l c o n d itio n
T (x , 0) = f ( x ) , t h e r e ­
f o r e , i t fo llo w s t h a t
w here Fn i s th e e ig e n f u n c tio n c o rre s p o n d in g to th e e ig e n v a lu e n .
At t h i s
p o i n t , th e e x p r e s s io n may b e . re c o g n iz e d to be th e F o u r ie r s in e s e r i e s e x ­
p a n s io n and
I t may b e p o s s i b l e to s o lv e th e t r a n s i e n t h e a t c o n d u c tio n e q u a tio n
by u s in g th e s e p a r a t i o n o f v a r i a b l e s te c h n iq u e on th e a n a lo g co m p u ter.
W ith o u t s e e k in g a n u m e ric a l s o l u t i o n , b u t c o n s id e r in g th e p ro b lem o n ly in
term s o f th e s t e p s n e c e s s a r y to a c h ie v e a com puter s o l u t i o n , we may p ro c e e d
as f o l l o w s .
S e t up com puter d ia g ra m s . F ig u r e s 9a and 9 b , w hich w i l l g iv e
s o l u t i o n s to E q u a tio n s 14 and 15.
The b o u n d ary v a lu e F (O ), F ig u re 9 a , i s
Il
-2 8 -
p la c e d on th e seco n d i n t e g r a t o r .
The p2 p o te n tio m e te r i s a d ju s te d u n t i l
th e i n t e g r a t o r o u tp u t s a t i s f i e s th e b o u n d ary c o n d i t i o n s .
A number o f
e ig e n v a lu e s d e te rm in e d by th e v a lu e s o f th e p p o te n tio m e te r s e t t i n g s may
be found and eac h one t h a t w ill,, s a t i s f y th e; bo u n d ary c o n d itio n s y i e ld s
an e i g e n f u n c tio n Fn .
To d e te rm in e th e Bn v a l u e s , a number o f c i r c u i t s
su ch as F ig u re 9c a r e p a tc h e d a t th e same tim e and th e p r e v io u s ly
d e te rm in e d e ig e n f u n c tio n s Fn m u l t i p l i e d by p o te n tio m e te r s r e p r e s e n t in g
th e v a lu e s o f Bn .
These p r o d u c ts a r e th e n summed and th e Bn p o te n tio m e te r s
a d j u s t e d by t r i a l and e r r o r u n t i l th e o u tp u t o f th e summer a p p ro x im ate s
th e i n i t i a l te m p e r a tu r e d i s t r i b u t i o n , f ( x ) ; F ig u re 9 c .
I t i s o b v io u s t h a t th e s e p a r a t io n .o f v a r i a b l e s te c h n iq u e u s in g th e
a n a lo g com p u ter in v o lv e s many s te p s and becomes q u i t e i m p r a c ti c a l even i f
th e s e r i e s i s s e v e r e l y tr u n c a te d .
T h is ap p ro ac h h a s o th e r s e v e r e l i m i t a ­
t i o n s , p a r t i c u l a r l y w here th e o r d e r o f th e e q u a tio n i s h ig h and in v o lv e s
more th a n two v a r i a b l e s .
Much eq u ip m en t i s n eed ed f o r t h i s m ethod o f
s o l u t i o n and c o n s id e r a b le o p e r a to r judgm ent i s r e q u ir e d .
The EAI TR-IO
w ould n o t b e an a d e q u a te a n a lo g com puter f o r th e s e p a r a t i o n o f v a r i a b l e s
m ethod.
-2 9 -
H eat flo w
(a )
H eat flo w i n a s la b
AT = T,
(b)
F ig u r e 8 .
T em perature a t i n t e r v a l s o f a x
F i n i t e - D i f f e r e n c e s —H eat Flow i n a S la b
-3 0 -
(a )
S o lu tio n o f F
(b)
S o lu tio n o f G1 +
~ 0
-T (x .O )
(c )
F ig u r e 9 .
I n i t i a l -tem p eratu re d i s t r i b u t i o n
S e p a r a tio n o f V a ria b le s
CHAPTER 6
ANALOG SOLUTION OF AN■UNSTEADY-STATE HEAT CONDUCTION PROBLEM
The exam ple p ro b lem w hich fo llo w s w i l l d e m o n s tra te th e u s e o f f i n i t e d i f f e r e n c e s and th e s o l u t i o n o f th e u n s t e a d y - s t a t e one d im e n s io n a l h e a t
e q u a t io n , E q u a tio n 11.
P roblem :
A s t e e l s la b i n i t a l l y a t a te m p e ra tu re o f 100 F i s su d d e n ly
ex p o sed to s u rro u n d in g s a t a te m p e ra tu re su ch t h a t th e f a c e s
o f th e s la b a r e a t a c o n s ta n t te m p e ra tu re o f OF.
The s la b
i s c o n s id e r e d to be i n f i n i t e betw een th e two p la n e s and th e
te m p e r a tu r e v a r i e s w ith d i s t a n c e x as shown i n F ig u re 10.
D eterm ine th e te m p e r a tu r e d i s t r i b u t i o n a lo n g th e x d i r e c t i o n .
The u n s te a d y s t a t e h e a t t r a n s f e r i n th e s la b can be e x p r e s s e d by E q u a tio n
11.
JT _ _k JfT
d-t _ 9> an
A f i n i t e - d i f f e r e n c e . .. s u b s t i t u t i o n . E q u a tio n 14, i s made so t h a t
dTn
dt
=
k
Cp ( A x ) Z
( Tn+1 ~ 2Tn + Tn-1>
The f o llo w in g v a lu e s w i l l be u sed f o r th e p ro b lem s o l u t i o n .
k = 26 B tu
hr ft F
P = 489 l b / f t 3
C = 0 .1 2 B tu /lb F
x = L -8
in c h e s
x = l in c h = 1 /1 2 f t .
Le£
m = C f(A x )Z
:
—3 2 —
Then
B tu
26 h r f t F_________
I b 1F )
.________________
C4 8 9
£ t2 >
(3 * 0 0 S g )
m = 0 .0 1 7 7 s e c - l
The s la b w i l l b e d iv id e d i n t o 8 in c re m e n ts ( A x) o f I in c h .
S in c e b o th
fa c e s o f th e s la b a r e ex p o sed to th e same te m p e r a tu r e , by symmetry as
shown i n F ig u re IOs
T0 = T8 = 0 F
T l = T7
.
T2
=
T6
T3 - T 5
T^ = c e n t e r o f s la b
T h e r e f o r e , th e p ro b lem s o l u t i o n n eed o n ly p r o v id e th e te m p e ra tu re s T l; T2 ,
T3 » ant^ ^ 4 ;
lo c a t i o n s
, Xg, X5 , apd
re s p e c tiv e ly .
These te m p era­
t u r e s w i l l p r o v id e a te m p e ra tu re p r o f i l e as a f u n c tio n o f tim e th ro u g h o u t
th e s l a b .
The b o u n d ary c o n d itio n s a r e , f o r T = T ( x ,t )
T (x ,0 ) = Tn
0 <x < L
T (0 ,t) = 0
t > 0
T ( L ,t) = 0
x = L
The maximum v a lu e o f T i s 100 F; t h e r e f o r e , th e s y s te m e q u a tio n s w i l l
b e s c a le d so t h a t 10 v o l t s o u tp u t w i l l e q u a l 100F.
The s c a le d sy stem
e q u a tio n s w i l l be
0 .0 1 7 7
Ty = OF
-3 3 -
The i n i t i a l c o n d itio n (tim e z e ro ) a t each s t a t i o n i s Tn = IQO F.
S in c e 10 v o l t s = 100 F , each i n t e g r a t o r m ust have 10 v o l t s in p u t (100 F ) ,
as shown in F ig u re 11,
In some c a s e s , i t may be n e c e s s a r y to u se a t r i a l and e r r o r method
to tim e s c a l e th e p ro b lem .
For t h i s p ro b le m , i t was found t h a t 2000
seco n d s w ere r e q u ir e d b e f o r e th e te m p e ra tu re a t any l o c a t i o n in th e s la b
n e a r l y re a c h e s 0 F.
T h e r e f o r e , th e tim e s c a le f o r th e p ro b lem was
5 seco n d s co m p u ter tim e = 500 seco n d s p ro b lem tim e
B
tc
t
5
500
0.01
T h is means t h a t th e com puter s o l u t i o n ta k e s p la c e 100 tim e s f a s t e r th a n th e
a c tu a l s o lu tio n .
-3 4 -
P o ts 3 , 5 , 9 , 13, (F ig u r e 11) m ust be d iv id e d by B ( 0 .0 1 ) .
T his
a p p l ie s w henever th e p ro b lem tim e i s s c a le d to a com puter tim e (R ef.
A p p e n d ix )o
The in p u t m u l t i p l i e r i s
0 .0 1 7 7
0 .0 1
I . 77
The p o ts can o n ly b e s e t to a v a lu e e q u a l to o r l e s s th a n u n i t y , so th e
p o ts m ust be u se d w ith a 10 m u l t i p l i e r .
The p o t i s s e t a t 0 .1 7 7 and t h i s
i s m u l t i p l i e d by 10 to o b ta in 1 .7 7 .
The tim e f u n c tio n g e n e r a to r , F ig u re 11, f o r g e n e r a tin g th e tim e on th e
X-Y p l o t t e r i s s e t to p ro d u c e an o u tp u t o f O . l t .
0 .1 t = 2 v o l t s .
T h u s, i f t = 20 s e c o n d s ,
In o th e r w o rd s, 2 v o l t s a r e g e n e r a te d i n 20 s e c o n d s .
By
s e t t i n g th e "arm s c a l e " s e t t i n g o f th e X-Y p l o t t e r a t 0 .5 v o l t s / i n c h , th e
2 v o l t s w i l l be p l o t t e d o v e r a d i s t a n c e o f 4 in c h e s and s in c e th e tim e
s c a l e f a c t o r i s 1 0 0, th e 4 in c h e s w h ich a r e p l o t t e d i n 20 seco n d s w i l l
I
r e p r e s e n t 2000 seco n d s o f a c t u a l p ro b lem tim e .
In d ro p p in g from a te m p e ra tu re o f 100 F to 0 F , th e m ag n itu d e s c a l i n g
i s such t h a t a ra n g e o f 10 v o l t s i s c o v e re d .
The "p en s c a l e " s e t t i n g o f
th e X-Y p l o t t e r i s s e t a t I v o l t / i n c h so t h a t 10 in c h e s a r e p l o t t e d f o r th e
10 v o l t s (100 F) o f th e p ro b lem .
T h u s, f o r th e Y -a x is
I v o l t = I in c h = 10 F
The com puter s o l u t i o n i s shown i n F ig u re 12a and F ig u r e 12 b .
-ii-
Face
L o c a tio n :
F ig u r e 1 0 .
U n s te a d y -S ta te H eat C onduction
-3 6 -
-IO v
-IO v
-IO v
F ig u r e 1 1 .
A nalog Computer Diagram f o r U n s te a d y -S ta te
H eat C onduction i n a S lab
-3 7 100
B4
6o
•H
1000
1500
Time (seco n d s)
F ig u r e 1 2 a .
T em p eratu re-T im e.
H eat C on d u ctio n i n S la b
-3 8 -
Time (seco n d s)
F ig u r e 1 2 b ,
T em p eratu re-T im e.
H eat C o n d u ctio n i n S la b
CHAPTER 7
ONE DIMENSIONAL. UNSTEADY-STATE HEAT FLOW PROBLEM
A one d im e n s io n a l, u n s t e a d y - s t a t e h e a t c o n d u c tio n p ro b lem can s e r v e as
an exam ple in o r d e r to a g a in d e m o n s tra te th e m ethod o f f i n i t e - d i f f e r e n c e s
in s o lv in g p a r t i a l d i f f e r e n t i a l e q u a tio n s .
One d im e n s io n a l h e a t flow can
be d e f in e d by th e e q u a tio n
2 1
(H)
-
_k
tit
£ T
c/>
The alum inum r o d . F ig u r e 13, w i l l be d iv id e d i n t o 4 in c re m e n ts so t h a t
A x w i l l e q u a l 1 /4 f e e t .
Thig w i l l p r o v id e 4 s t a t i o n s in th e ro d a t w hich
te m p e ra tu r e s w i l l b e fo u n d .
The ro d i s i n i t i a l l y i n th e rm a l e q u ilib r iu m a t
100 F and i s assum ed to be p e r f e c t l y i n s u l a t e d e x c e p t a t th e l e f t e n d .
The
u n s te a d y h e a t - f lo w s t a t e in th e ro d w i l l be in tr o d u c e d by re d u c in g th e tem ­
p e r a t u r e o f th e ex p o sed end to 0 F.
in F ig u re 13»
The p a ra m e te rs o f th e ro d a r e g iv e n
N ote t h a t k i s in seco n d s r a t h e r th a n in h o u r s .
T his w i l l
f a c i l i t a t e th e tim e s c a l i n g .
L et
kx _
H =
k.
„
c/Xaxr
0, OisH sec.
M
The s u b s t i t u t i o n o f E q u a tio n 12 i n t o E q u a tio n 11 g iv e s th e e q u a tio n
T T '
'
2 T" + T- 1>
The maximum v a lu e o f Tn i s 100 F , t h e r e f o r e , th e s c a le d v a r i a b l e s a r e
f Tn 1
_100
9
_ io q J
For th e 4 s t a t i o n s b e in g c o n s id e r e d , th e s c a le d s y ste m e q u a tio n s w hich
a re to be s o lv e d s im u lta n e o u s ly a r e
-4 0 -
N o tic e t h a t a l t e r n a t e e q u a tio n s have a l t e r n a t e s i g n s .
T h is i s so t h a t
no s ig n -c h a n g in g a m p l i f i e r s a r e r e q u i r e d , s in c e th e p h ase o f each i n t e g r a t ­
in g a m p l i f i e r o u tp u t i s c o r r e c t f o r th e r e s p e c t i v e in p u ts to a p re c e d in g o r
fo llo w in g a m p l i f i e r .
I t was found t h a t to tim e s c a le th e s o l u t i o n .
10 s e c com puting tim e = (B) 1000 s e c p ro b lem tim e
B = 0 .0 1
The i n i t i a l c o n d itio n s a re
T (x , 0) = 100 F
t h e r e f o r e , s in c e 10 v o l t s = 100 F , 10 v o l t s in p u t m ust be th e i n i t i a l
c o n d itio n a t a l l 4 i n t e g r a t o r s .
The com puter d ia g ra m , a lo n g w ith th e tim e g e n e r a tin g d ia g ra m , i s shown
in F ig u re 14.
N o tic e t h a t a m u l t i p l i e r o f 2 i s r e q u ir e d a t many i n p u t s .
T h is can be o b ta in e d w ith o u t a p o t p r e c e d in g a 10 m u l t i p l i e r by t h i s t e c h n iq u e :
In o th e r w o rd s, 2 in p u ts from th e same s o u rc e r e s u l t in a m u l t i p l i c a t i o n o f
tw o.
—4 1 —
The com puter s o l u t i o n i s shown in F ig u re 15.
b a s i c p ro b lem may be acc o m p lish e d v e ry e a s i l y .
Mapy v a r i a t i o n s o f th e
The i n i t i a l te m p e ra tu re
d i s t r i b u t i o n a lo n g th e ro d can be changed by a d j u s t i n g th e a p p r o p r ia te
i n i t i a l c o n d itio n s ( v o l t a g e s ) .
O th e r c o m b in a tio n s o f k /c ^ j
i n v e s t i g a t e d by m e re ly ch a n g in g p o t s e t t i n g s .
v a lu e s may be
-Ll2 -
toof-
I n s u la te d
0.25 f e e t
Aluminum M a te r ia l
P r o p e r tie s:
0.0^7 sec ft F
I n i t i a l C o n d itio n s
T ( 0 ,t) = O F
T (x ,0 ) = 100 F
F ig u re 1 3 .
O ne-D im ensional H eat T r a n s f e r i n an Aluminum Rod
F ig u re lU«
+10 v
-10 v
Computer D iagram f o r O ne-D im ensional H eat Flow
i n an Aluminum Rod
AVJ
I
+10 v
-1 0 v
10 v o l t s g e n e ra te d i n 10 seconds
Arm s e t t i n g o f I v o l t / i n c h
10 v o l t s i n 10 in ch es
-1 0 v
To Arm
1000
Time ( s e c o n d s )
F ig u r e 1 5 .
S o lu tio n o f O ne-D im ensional H eat Flow i n a n Aluminum Rod
CHAPTER 8
ANALOG COMPUTER SIMULATION OF A PLUG FLOW SYSTEM
I n d e v e lo p in g m a th e m a tic a l m odels f o r com bined f l u i d f lo w - h e a t t r a n s ­
f e r d e v i c e s , i t h a s b een fo u n d , a c c o rd in g to S. T. B a l l , I n s tr u m e n ts and
C o n tro l S yste rn s, t h a t th e p lu g flo w c o n c e p t y i e l d s b e t t e r r e s u l t s th a n
o th e r a p p ro x im a tio n s f o r t u r b u l e n t flo w i n p ip e s o r tu b e s .
In th e p lu g
flo w s y ste m , i t i s assum ed t h a t th e r e i s no m ix in g o r h e a t C o n d u ctio n in
th e ; f l u i d i n th e d i r e c t i o n o f flo w , and p e r f e c t m ix in g p e r p e n d ic u la r to
• : .
th e flow p a t h .
I f p e r f e c t m ix in g i s assum ed, c o m p u ta tio n s may be made w ith
a v e ra g e f l u i d te m p e r a tu r e s r a t h e r th a n l o c a l v a lu e s and th e a n a l y s i s i s
c o n s id e r a b ly s i m p l i f i e d s in c e o r d in a r y d i f f e r e n t i a l e q u a tio n s d e s c r ib e th e
'i'.
sy ste m .
S. T. B a ll s t a t e s t h a t i n l i q u i d flo w , th e t r a n s p o r t tim e s o r d i s t a n c e v e l o c i t y la g s o f th e l i q u i d s a r e u s u a ll y s i g n i f i c a n t i n d e te rm in in g th e
t r a n s i e n t re s p o n s e c h a r a c t e r i s t i c s .
U s u a lly s o l u t i o n s a r e f o r o n ly one
p a r t i c u l a r c a s e b e c a u se g e n e r a liz e d s o l u t i o n s a r e im p r a c t i c a l due to th e
l a r g e num ber o f v a r i a b l e s in v o lv e d .
I n gas s y s te m s , th e t r a n s p o r t tim es
a r e n e g l i g i b l e com pared to th e th e rm a l l a g s , so g e n e r a liz e d s o l u t i o n s a r e
o b ta in e d more r e a d i l y .
To i l l u s t r a t e th e u s e o f th e a n a lo g com puter in p ro b lem s o f t h i s t y p e ,
a flo w regim e i n a s e c t i o n o f i n s u l a t e d p ip e w i l l be i n v e s t i g a t e d .
The
re s p o n s e o f th e f l u i d o u t l e t te m p e ra tu re t o s te p changes i n th e i n l e t tem­
p e r a t u r e to an i n s u l a t e d p ip e w i l l be s im u la te d .
C o n s id e r th e c a s e o f gas flo w in g th ro u g h a s e c t i o n o f i n s u l a t e d p ip e
w hich h a s s i g n i f i c a n t h e a t c a p a c i t y . F ig u r e 16.
Assume t h a t a l l o f th e
mass o f t h e p i p e i s c o n c e n tra te d , a t a p o in t w hich i s a t some mean
-4 6 -
te m p e ra tu re and t h a t th e t r a n s p o r t tim e o f th e gas i s n e g l i g i b l e .
i s s o , th e s e c t i o n - l e n g t h h a s to be s m a ll.
I f th is
A ls o , in d e v e lo p in g th e sy ste m
e q u a t i o n s , th e gas mean te m p e ra tu re i s assum ed to be e q u a l t o 1 /2 o f th e
sum o f th e i n l e t gas te m p e ra tu re and o u t l e t gas te m p e r a tu r e .
T hus, th e
le n g th h as to be s m a ll i n o r d e r t h a t th e v a lu e o f th e gas mean te m p e ra tu re
b e an a p p ro x im a te ly a c c u r a te a s s u m p tio n .
The sy ste m e q u a tio n s a re
Gas h e a t b a la n c e
^
P ip e h e a t b a la n c e
A dding
. S u b tr a c t in g
InA
L.T?
Tci )
-4 7 -
And
T1 =T1/ + h^V 9C3 Crp-T3)
The p re c e d in g e q u a tio n s can be r e l a t e d and c o n n e c te d a c c o rd in g to th e com­
p u t e r d ia g ra m o f F ig u re 17.
T here w i l l n o t be any tim e o r m ag n itu d e s c a l ­
in g , r a t h e r th e p o te n tio m e te r s w i l l be s e t m e re ly to show th e re s p o n s e o f
th e sy ste m to ch an g es o f i n l e t gas te m p e r a tu r e .
N ote t h a t th e s e t t i n g o f p o t I r e p r e s e n t s th e r e c i p r o c a l tim e c o n s ta n t
f o r p ip e te m p e r a tu r e ch an g es and i s in d e p e n d e n t o f p ip e le n g th b e c a u se th e
h e a t t r a n s f e r a r e a A and p ip e mass M a r e b o th p r o p o r t i o n a l to le n g th .
The
o u tp u ts o f p o ts 2 and 3 ea c h r e p r e s e n t o n e - h a lf o f th e t o t a l change i n gas
te m p e r a tu r e , w hich i s p r o p o r t i o n a l to th e d i f f e r e n c e i n th e mean gas and
p ip e te m p e r a tu r e s .
p ip e le n g th .
The c o e f f i c i e n t s o f p o ts 2 and 3 a r e p r o p o r ti o n a l to
The t r a n s i e n t c h a r a c t e r i s t i c s o f th e s y ste m can be c o m p le te ly
d e s c r ib e d when j u s t two q u a n t i t i e s a r e s p e c i f i e d
- fliE
T - T a
tim e - c o n s ta n t tp in seco n d s
- - I T m )
2 lW Cg /
d im e n s io n le s s
I t was fou n d t h a t th e v a lu e o f th e s e c t i o n le n g th a f f e c t e d th e r e s u l t s
c o n s id e r a b ly .
The v a lu e n = Q .50 was s e l e c t e d f o r a l l o f th e com puter r u n s «
The p ip e tim e c o n s ta n t was s e t a t a v a lu e o f tp = 1.0=
t h i s v a lu e r e s u l t e d i n a s lo w e r p ro b lem r e s p o n s e .
A d e c re a s e o f
T h is i s b e c a u se th e d e­
c r e a s e co u ld o n ly be a f f e c t e d by th e h e a t t r a n s f e r c o e f f i c i e n t .
T h is means
t h a t no change i n flo w v e l o c i t y was c o n s id e r e d s in c e a change in v e l o c i t y
w ould a f f e c t th e h e a t t r a n s f e r c o e f f i c i e n t h .
The o u t l e t gas te m p e ra tu re . I g 0 was a f f e c t e d by a change in th e i n l e t
—4 8 -
gas te m p e ra tu re T g i-
T h is can be s e e n i n F ig u re 18.
The 100%, Y a x i s , o f
F ig u re 18 r e p r e s e n t s th e maximum re s p o n s e o f Tg0 a t J.00% in p u t ; on th e
TR-10 , t h i s w ould be 10 v o l t s ^ n p u t a t T g i.
2, 4 , 6 , 8 v o lts in p u t.
The o th e r c u rv e s r e p r e s e n t
W ith n u m e ric a l v a lu e s and p r o p e r s c a l i n g , a
n u m e ric a l s o l u t i o n c o u ld b e fo u n d .
The com puter d ia g ra m w ould rem ain th e
s ame „
T h is p ro b lem d e m o n s tra te s how sy ste m s o f r e l a t e d e q u a tio n s can be
m odeled on th e a n a lo g co m p u ter.
■insulation
VI T=L. g a s xaass f l o w r a t e
It/see
—ga s s p e c i f i c h e a t B t u / l b F
Tg0 =Tgas o u t l e t t e m p e r a t u r e F
Tg^= gas i n l e t temperature F
h — h e a t t r a n s f e r c o e f f i c i e n t B t u / s e c f t 2I
A
— h e a t t r a n s f e r a r ea f t
2
Tp ~ p i p e mean t e m p e r a t u r e
F
T g - ga s mean t e m p e r a t u r e F
Wp = mass o f p i p e
lb
C0 = p i p e s p e c i f i c h e a t
t
F ig u re 1 6 .
=- tim e
Gas F low ing i n an I n s u l a t e d P ip e
B tu /lb F
-5)0-
-IO v
g°"^gi
F ig u r e 1 7 .
Computer D iagram f o r P lu g -F lo w S im u la tio n
-5 1 -
•H
Eh
UO
<1
-S
UO
Bh IOO
O
<U
W
g
S'
CD
P3
80
O
60
Eh
cS
20
0
F ig u r e I S 0
P lug-F low S im u la tio n - R esponse o f T9o to Changes i n Tg:
APPENDIX
v< !.><
• ■'
APPENDIX
B a s i c a l l y , th e e l e c t r o n i c a n a lo g com puter s o lv e s e q u a tio n s s in c e i t
p e rfo rm s m a th e m a tic a l o p e r a tio n s w hich r e s u l t i n s o l u t i o n s o f th e e q u a tio n s
u sed t o r e p r e s e n t a c t u a l system s^
To s o lv e th e s e e q u a t io n s , i t i s
n e c e s s a r y f o r th e e l e c t r o n i c a n a lo g co m p u ter to be a b le to p e rfo rm s p e c i f i c
m a th e m a tic a l o p e r a tio n s on d i r e c t v o lt a g e s and to i n t e r c o n n e c t th e s e
v a r io u s o p e r a t io n s .
I t i s th e p u rp o se o f t h i s s e c t i o n t o p r e s e n t th e e le m e n ts o f th e a n a ­
lo g com puter and t h e i r o p e r a t io n to th e e x t e n t t h a t th e o p e r a t o r w i l l be
a b le to s e t up com puter d ia g ra m s .
The o b v io u s o p e r a tio n s w hich a re n e c e s ­
s a r y f o r th e s o l u t i o n o f a d i f f e r e n t i a l e q u a tio n a r e :
1.
to m u l tip l y
2.
to i n t e g r a t e
3.
to o b ta in th e sum o f 2 o r more v a r i a b l e s .
The b a s i c e l e c t r o n i c com ponents w hich p e rfo rm th e s e o p e r a t io n s a r e th e m ul­
t i p l i e r o r a t t e n u a t o r , th e i n t e g r a t o r n e tw o rk , and th e o p e r a t i o n a l a m p li­
fie r.
The o p e r a t i o n a l a m p l i f i e r i s th e m ost v e r s a t i l e com puting u n i t
w ith i n th e co m p u ter.
T h ere a r e o th e r com puter com ponents w hich p e rfo rm more
s p e c i a l i z e d o p e r a tio n s =
F or exam ple th e
d io d e f u n c tio n g e n e r a t o r , th e
lo g d io d e f u n c tio n g e n e r a to r , th e v a r i a b l e d io d e f u n c tio n g e n e r a to r , and
o th e rs .
These a r e d e s c r ib e d as n eed ed i n th e s o l u t i o n o f p ro b le m s .
OPERATIONAL AMPLIFIER
The b a s i c com puter com ponent i s th e a m p l i f i e r .
By a s u i t a b l e c h o ic e
o f r e s i s t o r s and c o n n e c tio n s to o th e r co m ponents, many r e l a t i o n s can be
o b ta in e d .
-5 4 -
Io
S ig n ch an g in g
The o u tp u t v o lta g e w i l l alw ays be o p p o s ite in s ig n t o th e i n p u t .
W henever an a m p l i f i e r i s u s e d , th e s ig n w i l l alw ays c h an g e .
The com puter
sym bol i s shown i n F ig u re 19 ( a ) .
2.
M u l t i p l i c a t i o n by 10
I f th e fe e d b a c k r e s i s t o r h a s te n tim e s th e v a lu e o f th e in p u t r e s i s ­
t o r , th e n th e o u tp u t w i l l b e o p p o s ite i n s ig n and te n tim e s th e in p u t
v o lta g e .
3o
The com p u ter sym bol i s shown i n F ig u re 19 ( b ) .
Summation
The a m p l i f i e r can hav e two o r more in p u t v o lta g e s and th e o u tp u t w i l l
be th e sum o f th e in p u t v o l t a g e s .
The com puter symbol i s Shown in
F ig u re 19 ( c ) .
4.
S u b tr a c t io n
By u s in g two a m p l i f i e r s , one as a s ig n ch an g e r and th e second as a
summer, two v o lt a g e s can be s u b t r a c t e d .
The method and com puter symbol
a r e shown i n F ig u re 19 ( d ) .
ATTENUATOR OR POTENTIOMETER
The a t t e n u a t o r p r o v id e s th e c a p a b i l i t y to m u ltip ly a v a r i a b l e v o lta g e
by a c o n s ta n t whose v a lu e l i e s betw een z e ro and u n i t y .
f o r th e a t t e n u a t o r o r p o t i s shown i n F ig u r e 20 ( a ) .
The com puter sym bol
The number i n s i d e o f
th e p o t sym bol r e p r e s e n t s th e number o f th e com puter p o te n tio m e te r w hich
was u s e d .
I0
M u l t i p l i c a t i o n by a c o n s ta n t C b etw een - I and -1 0
T h is n e c e s s i t a t e s th e com bined u se o f a p o t and an a m p l i f i e r .
The
com puter sy m b o l, F ig u re 20 ( b ) , shows t h a t th e p o t i s s e t to C/10 o f th e
-5 5 -
d e s ir e d v a lu e .
The d e n o m in a to r o r 10 i s due to th e m u l t i p l i c a t i o n f a c t o r
o f 10 w hich i s s e t on th e a m p l i f i e r .
2.
D iv is io n by a c o n s ta n t B
T h is com puter o p e r a t io n can be t r e a t e d as m u l t i p l i c a t i o n by 1/B .
INTEGRATING NETWORK
The i n t e g r a t o r n etw o rk i s c o n n e c te d a c r o s s a h ig h g a in d -c a m p l i f i e r
and th e n e c e s s a r y in p u t r e s i s t o r s a r e a p p lie d t o t h a t n e tw o rk .
I f a seco n d
o r d e r d i f f e r e n t i a l e q u a tio n i s to be s o lv e d , two i n t e g r a t o r s a r e r e q u ir e d
and i f o n ly a f i r s t o r d e r d i f f e r e n t i a l e q u a tio n i s to b e s o lv e d , o n ly one
in te g r a to r is needed.
shown i n F ig u re 20 ( c ) .
The com puter sym bol f o r an i n t e g r a t i n g n etw o rk i s
The p o t p r e c e d in g th e i n t e g r a t o r may o r may n o t be
n eed e d .
REFERENCE VOLTAGE
At v a r io u s p o in t s i n a p ro b le m , c o n s ta n t v o lta g e s o f d i f f e r e n t v a lu e s
a re re q u ire d .
The EAI TR-10 a n a lo g com puter u se s p lu s and minus 10 v o l t s
as s o u rc e s f o r a l l co m p u ter s i g n a l v o l t a g e s .
p a tc h p a n e l .
These can be found on th e
To p ro d u ce c o n s ta n t v o lta g e s o th e r th a n p lu s o r minus 10
v o l t s , th e p o ts a r e u se d a lo n g w ith a r e f e r e n c e v o lta g e s o u r c e .
F or e x ­
am ple, i f a r e f e r e n c e v o lta g e o f 5 v o l t s i s r e q u i r e d , u s e p lu s 10 v o l t s
r e f e r e n c e and th e n c o n n e c t t h i s to a p o t w hich i s s e t to 0 .5 0 .
T h is i s th e
same as m u l tip l y in g 10 v o l t s x 0 .5 0 = 5 v o l t s .
The r e f e r e n c e v o lt a g e i s alw ays u se d as an i n i t i a l c o n d i tio n o r as a
f o r c in g f u n c ti o n .
The b a s i c e le m e n ts o f th e co m p u ter have b een l i s t e d a lo n g w ith th e com­
p u t e r d ia g ra m sy m b o ls,
A more co m p lete l i s t o f sym bols u se d to r e p r e s e n t
—5 6 -
com puter d e v ic e s can be found i n th e com puter o p e r a t o r ’ s handbook=
The p o te n tio m e te r o r a t t e n u a t o r , th e summer, and th e summing i n t e g r a ­
t o r n etw o rk a r e th e th r e e b a s i c l i n e a r e le m e n ts r e q u ir e d f o r th e s o l u t i o n
o f sy stem s o f l i n e a r d i f f e r e n t i a l e q u a t i o n s .
S o lv in g n o n - l i n e a r e q u a tio n s
r e q u i r e s th e a d d i t i o n a l com ponents o f m u l t i p l i c a t i o n , f u n c ti o n g e n e r a tio n ,
l i m i t i n g and s w itc h in g .
H ow ever, th e p r e s e n t d is c u s s io n w i l l c o n s id e r
o n ly l i n e a r d i f f e r e n t i a l e q u a tio n s .
Computer D iagram Example
A s im p le l i n e a r d i f f e r e n t i a l e q u a tio n exam ple can b e s t i l l u s t r a t e th e
m ethod o f o b ta in in g a com puter d ia g ra m u s in g com puter sy m b o ls.
tu d e s c a l i n g . o r tim e s c a l i n g w i l l be u se d in t h i s ex am p le.
No m agni­
I t m ust b e r e ­
membered t h a t no f a m i l i a r i t y w ith th e com puter i s r e q u ir e d f o r th e r e p r e ­
s e n t a t i o n o f th e com puter s e tu p by means o f a com puter d ia g ra m .
C o n s id e r th e d i f f e r e n t i a l e q u a tio n
4 % + dx + % = 0
d t?
dt
U sing d o t n o t a t i o n
x + x -f- x = 0
S te p I .
S o lv e f o r th e h i g h e s t - o r d e r d e r i v a t i v e
6
X
=
-
X
-
X
Sometimes a " b lo c k d ia g ra m " . F ig u re 21 ( a ) , w i l l be h e l p f u l f o r p r e lim in a r y
a n a ly s is .
T h is d ia g ra m ig n o r e s a l l s ig n s and v a lu e s and i s u sed to show
th e " flo w " o f th e p ro b lem .
S te p 2 .
I n t e g r a t e th e h i g h e s t - o r d e r d e r i v a t i v e and a l l s u b se q u e n t d e r i v a d e riv a tiv e s .
F ig u re 21 ( b ) .
N o tic e th e s ig n c h a n g e s .
-5 7 -
S te p 3.
M u ltip ly each te rm by th e c o n s ta n ts g iv e n i n th e e q u a tio n . F ig u re
21 ( c ) .
S te p 4 ,
A p o t i s p la c e d b e f o r e eac h i n t e g r a t o r .
Add th e term s as g iv e n in th e e q u a t io n .
m ust be ad d ed .
S te p 5 .
A summer i s u s e d ..
At t h i s p o i n t , - ( x + x )
F ig u re 21 ( d ) .
S a t i s f y th e e q u a tio n by c l o s i n g th e lo o p as g iv e n by th e e q u a tio n .
I n F ig u re 21 ( d ) , x h a s b een found t o be th e o u tp u t o f summer B .
T h e r e f o r e , i t can be c o n n e c te d i n t o i n t e g r a t o r C.
T h is c lo s e s
th e lo o p .
' M agnitude S c a lin g
I t i s im p o r ta n t t o remember t h a t th e com puter w i l l e s t a b l i s h m athe­
m a tic a l r e l a t i o n s n o t b etw een th e o r i g i n a l v a r i a b l e s o f a p ro b lem , b u t
b etw een v o lt a g e s s im u la tin g th e s e v a r i a b l e s .
S in ce th e l i m i t i n g v a lu e o f
th e TR-IO a n a lo g co m p u ter i s t 10 v o l t s , th e v o lta g e s m ust b e a rra n g e d so
t h a t th e y do n o t ex ceed th e s e v a l u e s .
M agnitude s c a l i n g w i l l s e t up a r e l a t i o n s h i p b etw een th e v o lta g e ob­
s e rv e d as an o u tp u t and th e p ro b lem v a r i a b l e .
T here a r e v a r io u s m ethods o f
m ag n itu d e s c a l i n g ; h o w ev er, o n ly th e m ethod u sed in th e p ro b lem s i s p r e ­
s e n te d h e r e .
I f i n a p ro b lem in v o lv in g some v a r i a b l e T, th e v a lu e o f T v a r i e s from
0 F to 1000 F , th e n I v o l t on th e co m p u ter co u ld r e p r e s e n t 100 F.
I f th is
i s s o , th e p ro b lem v a r i a b l e F w i l l v a ry b etw een 0 F and 1000 F and th e com­
p u t e r v o lt a g e w i l l v a ry b etw een 0 v o lts , and +10 v o l t s .
I t i s d e s i r a b l e to
have th e com puter v o lt a g e re a c h i t s maximum v o lta g e when th e v a r i a b l e F
re a c h e s i t s maximum.
I f th e com puter can c o v e r th e maximum v o lta g e ran g e
( t 10 v o l t s ) , i t w i l l m in im ize th e e f f e c t s o f n o i s e , r e s o l u t i o n e r r o r s ,
—5 8 —
and com pu ter d r i f t .
The maximum v a lu e s o f th e p h y s ic a l v a r i a b l e can be s c a le d on a " p e r
v o l t " b a s i s so t h a t
____________ 10 v o l t s ______________
max. v a lu e o f p h y s ic a l v a r i a b l e
T h is i s c a l l e d th e m a g n itu d e s c a l e f a c t o r .
In th e exam ple c i t e d , th e mag­
n it u d e s c a l e f a c t o r w ould be
—1-Q-..3I— =
1000 F
0 .0 1 V
F
I v o l t = 100 F
A n o th er way o f o b ta in in g a m ag n itu d e s c a l e f a c t o r w ould be t o w r ite th e
s c a l e f a c t o r a s su ch
I j- I
IlOOOj
10 v = 1000 F
T h is sim p ly means t h a t 10 v o l t s = 1000 F.
I t i s & m a tte r o f k e e p in g t r a c k
o f th e v o lt a g e - v a r i a b l e c o rre s p o n d e n c e .
The f o llo w in g exam ples show how .to d e te rm in e th e m a g n itu d e s c a le
fa c to rs:
P h y s ic a l V a r ia b le
v e lo c ity
X m
a x
= 200 f t / s e c
R ate o f te m p e r a tu r e change
S c a le d V a r ia b l e - " p e f v o l t "
10 x
200
f
10 T
100
S c a le d V a r ia b le
lira]
m
nC
[4]
K
Tmax = 100 F /s e c
T em p eratu re
Tmax = 800 F
10 T = FjrJI
1000
[ iooJ
A lth o u g h th e maximum te m p e ra tu re i s e x p e c te d to be 800 F , a maximum o f
1000 F was u sed to o b ta in a s g a le f a c t o r .
T h is i s p e r m i s s i b le and m e re ly
g iv e s a num ber ( 1000) w h ich i s e a s i e r to w ork w ith .
The s c a le d v a r i a b l e s w hich w i l l a p p e a r a s v o lta g e s i n th e com puter
-5 9 -
c i r c u i t a r e i d e n t i f i e d by u s in g s q u a re b r a c k e ts ab o u t them .
GAIN FACTOR - EXAMPLE PROBLEM I
1I n o r d e r to c o n s t r u c t th e s c a le d com puter d ia g ra m , th e " g a in f a c t o r "
m ust be in c lu d e d a t th e in p u t to each o f th e i n t e g r a t o r s .
T h is g a in f a c ­
t o r i s n e c e s s a r y and r e s u l t s from th e d i f f e r e n c e in s c a l e f a c t o r s w hich
a re u sed .
The fo llo w in g h y p o t h e t i c a l co m p u ter d ia g ra m a n a l y s i s can
i l l u s t r a t e th e g a in f a c t o r as w e ll as th e m ethod o f o b ta in in g th e d ia g ra m .
Suppose t h a t th e maximum v a lu e s o f th e v a r i a b l e s o f a p ro b lem a r e :
xTiiax = 32 f t / s e c 2
0
Xmax = 8 f t / s e c
Xmax = 2 f t
The e q u a tio n i s
DO
6
X+ 3 X + 1 6 X = 0
The m ag n itu d e s c a l e f a c t o r s a r e
X_
_32_
9
Pr
X
, T i'
_8_
L2J
and th e s c a le d e q u a tio n i s
32
+ ( 3 ) ( 8 ) X + (1 6 )(2 )
8
0
I n o r d e r to m a in ta in th e e q u a tio n , th e te rm s o f th e e q u a tio n a r e m u l t i p l i e d ,
o r d iv i d e d , by th e same v a lu e as i s u se d in o b ta in in g th e s c a l e f a c t o r s .
S o lu tio n o f th e e q u a tio n f o r th e h ig h e s tr - o r d e r d i f f e r e n t i a l :
The com pu ter d ia g ra m , F ig u r e 22 ( a ) , shows th e m u l t i p l i c a t i o n by a f a c t o r
o f 4 a t th e in p u t t o each i n t e g r a t o r .
C o n sid e r th e e q u a tio n to be s o lv e d
—6 0 —
by th e f i r s t i n t e g r a t o r :
— X = — y XT d ' t - -h Y. C*)
o
U sing s c a l e f a c t o r s
- M - -/«[£]-!«
8
A g e n e r a l m ethod to f in d th e v a lu e o f th e " g a in " n e e d e d i s :
S c a le d c o e f f i c i e n t o f th e (n - I )
o rd e r d i f f e r e n t i a l
-JL
9
S c a le d c o e f f i c i e n t o f th e
n fc^ o r d e r d i f f e r e n t i a l
Exam ple:
2 » 100
The g a in f a c t o r w hich m ust be u sed as an in p u t to th e i n t e g r a t o r i s 50.
T h is g a in f a c t o r v a lu e does n o t c o n s id e r th e tim e s c a l e f a c t o r w hich w i l l be
d is c u s s e d i n th e n e x t s e c t i o n .
I t w i l l be s e e n t h a t a la r g e g a in f a c t o r can
b e re d u c e d w ith t h e a p p r o p r ia te tim e s c a l e f a c t o r .
Time S c a lin g
I t w i l l o f te n happen t h a t th e r a t e a t w hich a p r o c e s s s im u la tio n ru n s
i s u n s u i t a b l e ; i t may b e to o f a s t f o r th e r e c o r d in g d e v ic e to fo llo w o r to o
slow f o r th e co n v e n ie n c e o f th e p e rs o n i n t e r e s t e d in th e an sw er.
In e i t h e r
c a s e , i t i s n e c e s s a r y to a l t e r th e r a t e a t w hich th e s im u la tio n r u n s ; t h i s
i s done by a te c h n iq u e c a b le d tim e s c a l i n g .
The i d e a l d u r a ti o n o f a s i n g l e ru n on th e com puter i s somewhere from
10 to 60 s e c o n d s .
Too lo n g a s o l u t i o n tim e may in tr o d u c e some e r r o r due to
in te g ra to r d r i f t .
A lso th e com puter o p e r a t o r d e s i r e s r e s u l t s in a r e a s o n ­
a b le tim e .
The c a p a c it y o f th e r e c o r d in g eq u ip m en t to fo llo w th e s o l u t i o n
-6 1 -
p la c e s a low er l i m i t on th e s o l u t i o n tim e .
To rem ain w ith i n th e s e
l i m i t a t i o n s , a change i n tim e can be in tr o d u c e d by r e l a t i n g p h y s ic a l o r r e a l
tim e to a n a lo g com puter tim e by a tim e s c a l e f a c t o r B, so t h a t
t c - Bt
t c = com puter tim e i n seco n d s
t = r e a l tim e i n seco n d s
I f B = I , th e co m p u ter i s o p e r a t in g i n r e a l tim e .
I f B - C l , th e com puter s o l u t i o n i s sp eed ed u p ; th in g s hap p en f a s t e r on th e
com puter th a n in th e p h y s ic a l sy ste m .
I f B /» I , th e com puter s o l u t i o n i s slow ed down; th in g s h ap p en s lo w e r on th e
com puter th a n in th e p h y s ic a l sy ste m .
Through m a th e m a tic a l d e d u c tio n , i t can be p ro v en t h a t tim e s c a l i n g
can be acc o m p lish e d by m u l tip l y in g e v e ry in p u t to an i n t e g r a t o r by 1/B .
X -Y
P lo tte r
The o u tp u t o f th e p ro b lem may be o b s e rv e d on an o s c il lo s c o p e o r r e ­
co rd e d on a s t r i p - c h a r t r e c o r d e r d r on an X-Y p l o t t e r .
X-Y p l o t t e r was u sed e x c l u s i v e l y .
s ta tio n a r y p a p e r.
In t h i s r e p o r t , th e
The X-Y p l o t t e r u se s a m oving pen and
T h is d e v ic e h a s two v o lt a g e i n p u t s , e n a b lin g one to p l o t
one v a r y in g v o lta g e a g a i n s t a n o th e r .
The "Arm" o u tp u t moves alo n g a
h o r i z o n t a l l i n e and th e "P en" moves v e r t i c a l l y d e p en d in g on v o lta g e f o r i t s
d e fle c tio n .
The in p u t f o r th e Pen i s th e o u tp u t o f th e d ep en d e n t v a r i a b l e
o f th e p ro b lem .
The o u tp u t o f th e d e p e n d e n t v a r ia b ly can come from any
l o c a t i o n i n th e co m p u ter c i r c u i t .
a m p lifie r.
T h at i s , i t can be th e o u tp u t o f any
The in p u t t o th e Arm w i l l d e te rm in e th e r a t e o f h o r i z o n t a l t r a v ­
e l and w i l l r e p r e s e n t th e in d e p e n d e n t v a r i a b l e tim e .
The v o lta g e r e p r e s e n t ­
in g tim e i s g e n e r a te d by an i n t e g r a t o r and d r iv e s th e Arm on th e X-Y p l o t t e r
—
a d i s ta n c e p r o p o r t i o n a l to tim e .
62
—
F or exam ple
■Q— I >—^ *"/f
-IOv
I f B = 10, th e n
-^c
J
I
-t >
'
I^
I f t c = 10 s e c o n d s , th e n 10 v o l t s w i l l be g e n e r a te d in 10 seco n d s o f a c t u a l
com puter o p e r a t in g tim e .
The "Arm S c a le " s e t t i n g on th e p l o t t e r w i l l co n ­
t r o l th e le n g th o f t r a v e l o f th e p en .
I f th e s e t t i n g i s s e t to I v o l t /
in c h , th e p l o t t e r w i l l g e n e r a te th e above 10 v o l t s in 10 in c h e s and 10
seco n d s.
Summary-Example
P ro b lem 2
To sum m arize th e te c h n iq u e s o f o b ta in in g a co m p lete a n a lo g com puter
d ia g ra m , a s e c o n d - o r d e r d i f f e r e n t i a l e q u a tio n w i l l be c o n s id e r e d as an e x ­
am ple.
(1 8 )
X
1
' + 8x ' + IOOx = f ( t )
x ( 0) = 20 in c h e s
x '( 0 ) = -5 in c h /s e c o q d
Assume maximum v a lu e s o f
xm = 2 in c h e s
X
1 m
=
20 in c h /s e c o n d
x ' ' m = 200 in c h /s e c o n d ^
The m a g n itu d e s c a l e f a c t o r s a re
10 v o l t s = 2 in c h
10 v o l t s = 20 i n / s e c
-
63
-
10 v o l t s = 200 in / s e c ^
S o lv in g E q u a tio n 18 f o r th e h ig h e s t o r d e r d i f f e r e n t i a l , s u b s t i t u t i n g th e
m agnitude s c a le d v a r i a b l e s , and m aking p r o p e r a d ju s tm e n ts t o keep th e
e q u a tio n t r u e :
( 18)
- 8 x ' - 100 x + f ( t )
2 0 0 [ f ^ ] ” ( ' 8) ( 20) [ f Q ™
( 100H 2) L ? ] + f ( t )
[S ]--0-80[firJ-[f]
+ m i
200
T h is e q u a tio n can now be u sed to draw th e com puter d ia g ra m . F ig u re 22 ( b ) .
The tim e s c a l e f a c t o r B can be n e g le c te d d u r in g t h i s p r o c e s s b u t m ust be
added in th e com puter c i r c u i t .
To o b ta in th e n e c e s s a r y g a in f a c t o r b etw een
10
T h is g a in c o u ld be o b ta in e d by an i n t e g r a t o r in p u t m u l t i p l i c a t i o n o f 10.
A tim e s c a l e f a c t o r o f B = 10 would slow down th e s o l u t i o n and a ls o d e ­
c r e a s e th e g a in . F ig u re 22 ( c ) .
I t s h o u ld be remembered t h a t th e im p o rta n t p h ase o f p ro b lem s o lv in g
w ith th e a n a lo g com puter i s to o b ta in a c o r r e c t d ia g ra m so t h a t th e p a tc h
b o a rd can be w ire d c o r r e c t l y .
Of l e s s e r im p o rtan ce i s th e e s tim a tio n o f
maximum m a g n itu d es and ru n d u r a tio n s in c e th e s e v a lu e s can u s u a ll y be
—6 4 - '
changed by m e re ly ch an g in g th e p o te n tio m e te r s e t t i n g s .
-6 5 -
(a)
S ign Changing
(b)
M u ltip lic a tio n by 10
(c)
Summation
(d)
S u b tr a c tio n
F ig u r e 1 9 .
B asic Computer Symbols
-6 6 -
(a )
P o te n tio m e te r (P o t)
(b )
M u lt i p l i c a t i o n Between
(c )
I n t e g r a t i n g Network
- I and -10
± IOn
(d)
F ig u r e 20.
R e fe re n c e V oltag e
B asic Computer Symbols
-6 7 -
(a)
Block D iagram
Th . x
*0
X
(b)
lO
S te p 2
—
In te g ra tio n
-"
(c)
S te p 3 — M u l t i p l i c a t i o n
r> -
o —c H
--------------O
(d)
F ig u r e 21.
S te p U - 5 — Computer Diagram
S o lu tio n o f Example Problem
-6 8 -
(a )
Gain F a c to r — Example Problem I
(b)
Computer D iagram f o r Example Problem 2
10/B = I
10/B = I
(c )
F ig u r e
R educing Gain F a c to r b y Time S c a lin g
Example Problem s
Diagramming F undam entals
LITERATURE. CONSULTED
Howe, R. M ., S o lu tio n o f P a r t i a l D i f f e r e n t i a l E q u a tio n s , U n iv e r s ity
o f M ich ig an .
J a c k s o n , A. S . , A nalog C o m p u tatio n , M cG raw -H ill Book Company, I n c . ,
New Y ork, 1960.
J e n n e s s , R. R ., A nalog C o m putation and S im u la tio n :
A lly n and B acon, I n c . , B o s to n , 1965.
L a b o ra to ry A pproach,
K r e i t h , F ra n k , P r i n c i p l e s o f H eat T r a n s f e r , Second E d i t i o n , I n t e r n a t i o n a l
T extbook Company, S c r a n to n , P e n n s y lv a n ia , 1959.
Mull i k i n , H. F . , H eat T r a n s f e r , M ontana S t a t e U n iv e r s it y , 1964.
PACE TR - 10 A nalog Computer O p e r a t i o n s H andbook, E l e c t r o n i c s A s s o c ia te s
I n c , , Long B ran ch , New J e r s e y .
M agazines
B a l l , S. T . , "P lu g -F lo w S y ste m s" , I n s tr u m e n t s . a n d .C o n tr o l S y ste m s, Volume
36, No. 2 , F e b ru a ry , 1963.
«rTkTC UNIVERSITY LIBRARIES
3 1762 10014318 7
N378
co p . ?
H i l a r i o , M. E.
B a sic h e a t t r a n s f e r u s in g
EAI TR-IO . . .
NAMC ANO
t
A D D w gSS