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Scholars' Mine Scholars' Mine
Masters Theses Student Theses and Dissertations
1949
The application of the relaxation method to the solution of The application of the relaxation method to the solution of
problems involving the flow of fluids through porous media problems involving the flow of fluids through porous media
Alexander Antoine Zwierzchowski
Follow this and additional works at: https://scholarsmine.mst.edu/masters_theses
Part of the Mechanical Engineering Commons
Department: Department:
Recommended Citation Recommended Citation Zwierzchowski, Alexander Antoine, "The application of the relaxation method to the solution of problems involving the flow of fluids through porous media" (1949). Masters Theses. 4784. https://scholarsmine.mst.edu/masters_theses/4784
This thesis is brought to you by Scholars' Mine, a service of the Missouri S&T Library and Learning Resources. This work is protected by U. S. Copyright Law. Unauthorized use including reproduction for redistribution requires the permission of the copyright holder. For more information, please contact [email protected].
THE APPLICATION OF THE RELAXATION METHOD TO THE SOLUTION
OF PROBLEMS INVOLVING THE FLOW OF FLUIDS
THROUGH POROUS MEDIA
BY
ALEXANDER A. ZWIEMC HCN/SKI
A
THESIS
submitted to the faculty of the
SCHOOL OF MINES AND METALLURGY OF THE UNIVERSITY OF JaSSOUR!
in partial fulfillment of the work required for the
Degree ot
MASTER OF SCIE~E IN MEX::HANICAL ENGINEERING
Rolla, 140.
1949
-, I··Approved by <'~ C;~
Pi'Ofeasor of Meohanical En in eting/
i
AC KNONLEDGEMENTS
The author is indebted to Professor A. J. Miles for his
timely suggestions and assistance in the solution of the
probl m.
11
iii
PREFACE
The purpose of this investigation is to determine the feasibility
of applying the relaxation method of mathematical analysis to the
solution of problems in rheology. The relaxation method was first
introduced by R. V. Southwell(l) who applied it to the problem of
electrical conducitivity. It Was later used by U. • Emmons(2) to solv
two-dimensional heat conduction problems.
The steady-state flow of fluids through porous media Can be
difficult or impossible by the usual analytical methods even with rather
simple boundary conditions. The relaxation method promises to provide
an easier method of solution and to provide a means of solving som of
the problems not yet solved. It is the object of this thesis to
establish the feasibility of the relaxation method rather than to solve
any new problem. It finds a maximum usefulness in reservoir mechanics
of petroleum engineering.
(1) Southwell, R. V., Proc. Boyal Society, Series A, Vol. 168,PP. 317-350 (1938)
(2) Emmons, H. ., Transactions A.S.M.I., Vol. 63, No.6,pp. 607-61; (Aug. 1943)
CONTENTS
Acknowledgements ... •' .
Preface , .
List of illustrations •••••••.••••..••••••.••.•....•.••.••••..
List of tables ..•..•.••.•••••.••.•...••••••••••••••••••••••••
List of plates ••.•.•••••••.•.••••••••••••••..•.••••••.••••••.
Introduction - A brief discussion of the correspondence
between Heat Conduction and the Flow of
Fluids through Porous Media •••••••••••••••••••
Body - Problem 1 - Solution of simple Radial flow probl m
using the relaxation method of mathematioal analysis ••
Problem 2 - The relaxation method of mathematical
analysis applied to 8 more lengthy Radial flo
problem .
Problem 3 - Solution of Square drainage area prabl
iv
Page
ii
iii
v
vi
vii
1
3
12
by the r laxation method of thematical nalyai us!
a quare network ot flaw ••••••••••••.••••••••••••••••• 21
Problem 4 - Solution of Circular drainage area problem
by the relaxation method of mathematical analysis using
a rectangular network of flow•••••••••••••••••••••..••
Conclusions ••••••.•••.••••••••••••••.•...••••••••.•..••••••••
S \1II1D'lSry- ••••••••••••••••••••••••••••••••••••••••••••••••••••••
Bibliography .. "! .
29
31
39
40
LIST OF ILLUSTRATIONS
y
Fig.
1. Sketch of circular drainage area of problem 1••.••••••••
2. Sketch of circular drainage area of problem 2•••••••••••
3· Sketch of square drainage area of problem 3••••.••••••••
Page
3
12
21
4. Sketch of circular drainage area with rectangular
network of flow of problem 4....................... 295. Enlarged cut view of fig. 4............................. 30
LIST OF TABLES
Table No.
1. Tabulated results and intervening steps in order of
calculation of problem 1 •••••••••••••••••••••••••
2. Tabulated results and intervening steps in order of
calculation of problem 2, method 1••••••••••••••••
3. Tabulated results and intervening steps in order ot
calculation of problem 2, method 2••••••••••••••••
4. Comparison of correct and obtained results in tabular
form problem 2 .•..•••.••••••...•.•..•......•••.•••
,. Tabulat d results and intervening steps in order ot
calculation of problem 3••••••••••••••••••••••••••
6. Tabulated results and intervening steps in order of
calculation ot problem 4••••••••••••••••••••••••••
vi
Page
9
16
18
20
2;
33
vii
LIST OF PLATES
Plate No.
1. Pressure Distribution in Radial flow - problem 1 •••••••••
2. Pressure Distribution in Radiel flow - problem 2,
method 1•........ " .
3. Pressure Distribution in Radial flow - problem 2,
method 2 " " "••• of •••••••••
4. Pressure Distribution in Rectangular flow along west or
east direction from sink - problem 3••••••••••••••••
Page
11
17
19
26
,. Pressure Distribution in Rectangular flow along southeast
direction from sink - problem 3..................... 27
6. Pressure Distribution in Rectangular flow along east-
south-east direction from sink - problem 3.......... 28
7. Pressure Distribution in Rectangular flow due north
from sink - probl m. 4 .•••••••••••....•••.••..•••••••
8. Pressure Distribution in Rectangular flow along northeast
direction from sink - problem 4•••••••••••••••••••••
9. Pressure Distribution in Rectangular tlow of all points
on fig. , - problem 4 •••••••••••••..•••••.••••••••.•
34
35
36
INTRODlX:TION
1
Before illustrating the use of the relaxation method, the
correspondence between heat conduction and the flow of an incompressible
liquid through a porous media will be shown; since the problem of
applying the relaxation method to flow of fluids was attempted when it
was seen that M. W. Emmons(3) has applied it euccessfully to heat
conduction problems.
Heat Conduction
Temperature T
Thermal Conductivity k
Rate of Heat Transfer Q • -ItA fIsothermal Surtace T =Constant
steady state Flow
Pressure P
Permeability !Viscosity u
Velocity Vector V • - ~ ~
Equipressure Surfaoe P - Cons ·t
From above comparison it is seen that Fourier's La is similar to
D'Arcy's Law, the only difference being that temperature is substituted
by pressure.
Also from Laplace's equation, in heat conduction:
= "
and in fluid flow, introducing the velocity potential pt
::.0
(3) Emmons, Ope cit-po iii.
2
Therefore, since the similarity between heat conduction and fluid
flow is clearly shown, it is concluded that the relaxation method will
also apply to problems involving fluid flow.
The main idea in solving steady fluid flow problems by the
relaxation method is that fluid mass in any closed system Can be neither
created nor destroyed. In other words at steady flow conditions the
total quantity, Q, of fluid at any given point at any given instant of
time must add up to zero e.g. 8S much fluid is flowing towards the point
as away from it.
The above idea is the basis for the solution of fluid problems
involving steady state conditions. Essentially the procedure followed
is to assume values of pressure at all points whose preeaure 1s to be
determined, and to calculate Q at any particular point using D'Arcy's
Law. For steady flow Q - 0, so reassign values of pressure until Q
approaches very closely or equals zero.
Four problema involving radial and rectangular flow actually show
the feasibility of using the relaxation method to solve problems of.
similar nature. The radial flow problems can be actually checked by
formula as will be shown subsequently.
PROBLE 1
SOLUTION OF SIMPLE RADIAL FLCYN PROBLEM
USING THE RELAXATION METHOD OF
MATHEMATICAL ANALYSIS
3
As an illustration of the use of the Relaxation Method, consider the
following two-dimensional problem•
.,.,,'
Fig. 1
Given: well bore 100 feetpressure 0 psi
drainage radius 1200 feetpressure 1100 psi
To find pressure at radius of 300', 600', and 900'.
1. Assume straight line variation of pressure with drainag
radius.
2. Flo is radial only.
3. Since drainage area is symmetric 1 in all directions, take
a sample equivalent pipe on any radius 88 shown above.
4
Q·-!A~u ~R
where k = permeability
u • viscosity
A • cross-sectional area of flow
AP = difference in pressuree
AR • difference in radiuses along flow lines
let ! =1 for particular problemu
A N R R· aritbmetic average value of radius
A • 2~R (for unit thickness)27fR p
Q=-AbR
For particular problem
Q::: ,0 I
21T'R P == .2 7T (200) 6 P ... ',:;'8 AP= -.0() , /01> .3-- 6R ;;'00
1 0.2 :t 300' Q.."o I = :2 ". (""5"0) A P'" tf 44 L> P
300
.2 ~ '" 'ooQ<joo' = ;;J.7T' (7So) ~ P == 15.7 ~r>
300
3 0 .. ='1~{)':l.o I ;::. ;1. 7T" (; OS-D) L:."P= :l~.o 6,) P3 (J Q
as shown on Fig. 1
From plate No. 1 assuming straight line variation of pressur with
drainage radius,
P @ 100' =a psi
p @ 3°0' • 200 psi
p @ 600' • 5'00 psi
P @ 900' • 800 psi
For steady flow conditions, total Q • o at any point.
5
Calculate Q with assumed pressures at particular pointe, and it Q1 0
assume other values of pressure until Q=°with correct values of pressure.
a.) At point where r • 300'
Flowing in (+) Q • 9.44 (500-200) • T 2832 unite300 600
Flowing out (-) Q • 6.28 (200-0) • -1256 units100 300
Totel Q • 1" 2832-1256 • + 1576 units
b. ) At point where r l: 600·
Flowing in (+) Q • 15.7 (BOO-Soo) • +4110 units600 900
Flowing out (-) Q • 9.44 (500-200) • -2832 units300 600
Total Q • +4710..2832 .... 1878 units
c.) At point where r • 900'
Flowing in (+) Q = 22.0 (1100-800)='" 6600 units900 1200
Flowing out (-) Q =15.7 (800-500) • -4110 units600 900
Total Q • +6600-4710 • ... 1890 unit
From above it is s en that 8sumed Ta1uee of pressure re inoorrect
for Q , zero for all three points, as Q is not zero for any point.
Therefore using the method of relaxation, that is continually
assuming values ot pressures for a point until equivalent Q values re
qual to or approach zero.
This is shown on table No. 1 where all results are tabulat d in
their order of calculation.
To check results obtained, compare with values of pressure calculated
trom the formula for redial tlo to a sink 88 shown on page 10.
Calculations
Given Po • 0 psi
PI • 200 psi
P2 • 500 psi
P3 = 800 psi
P4 =1200 psi
step 0
Ql • 9.44 (500-200) -6.28 (200-0) =+ 1570
Q2 I: 15.7 (800-500) -9.4-4 (500-200) • +1880
Q3 = 22.0 (1100-800) -15.7 (800-S00) • + 1890
Step I
Change pressure ~3 Bnd make it 8001" 100 • 900 psi
Q3 =22.0 (1100-900) -15.7 (900-500) • -1880
Q2 =15.7 (900-500) -9.44 (500-200)· 34;0
Step II
Change pressure ~ and make it 500+ 200 • 700 psi
Q2 • 15.7 (900-700) -9.44 (700-200) =-1588
'h • 9.44 (700-200) -6.28 (200-0) • 1" 3468
Q3
• 22.0 (1100-900) -15.7 (900-700) • + 1260
Step III
Change pr 8sure OJ. nd make it 200+ 200 =400 psi
Ql • 9.44 (700-400) -6.28 (400-0) aT 318
Q2 • 15.7 (900-700) -9.44 (700-400) -;- 302
6
Step IV
Change pressure C3 and make it 900+ 50 = 950 psi
Q3 • 22.0 (1100-950) -15.7 (950-700) = -635
Q2 : 15.7 (950-700) -9.44 (700-400) -+1087
step V
Change pressure ~2 nd make it 700+80 = 780 psi
Q2 a 15.7 (950-780) -9.44 (780-400) =-824
Ql • 9.44 (780-400) -6.28 (400-0) : +1081
Q3· 22.0 (1100-950' -15.7 (950-780) a+621
Step VI
Change pressure C3 and make it 950-r 20 - 970 psi
Q3 =22.0 (1100-970) -15.7 (970-780)- -133
Q2 • 15.7 (970-780) -9.44 (780-400) • -510
step VII
Change pressure Cl and make it 400 + 70 a 470 pai
Ql z 9.44 (780-470) -6.28 (470-0) • -27
Q2 • 15.7 (970-780) -9.44 (780-470) a + 150
Step VIII
Change pr saure 2 and make it 780-r 8 • 788 psi
Q2 =15.7 ('70-788) -9.44 (788-470) : -51
Q3 =22.0 (1100-910) -15.7 (970-788) : -7
Ql = 9.44 (785-470) -6.28 (470-0) -+ 48
7
Step IX
Change pressure ~1 and make it 470 T 4 • 474 psi
Ql =9.44 (788-474) -6.28 (474-0) • -t 15
Q2 • 15.7 (970-788) -9.44 (188-474) .of 13
Step X
Change pressure @2 and make it 788 -t-l I: 189 psi
Q2 • 15.7 (970-789) -9.44 (789-474) • -12
Ql • 9.44 (189-474) -6.28 (474-0) • ..,. 24
Q3 : 22.0 (1100-970) -15.7 (910-789) =+ 8
Step XI
Change pressure ~1 and make it 4741" 2 • 476 psi
Q1 • 9.44 (789-476) -6.28 (476-0) : -7
QZ • 1,.7 (970-789) -9.44 (789-476) • ..,. 7
8
Tabulated Results
Step QO Po Q1 PI Q2 P2 Q3 P3 Q4 P4-
0 0 +1570 200 +1880 500 +1890 800 lloo1 +3 50 -1880 +1002 +3468 -1588 +200 +12603 +318 +200 +3024- +1087 -63, +50, +1073 -824- +80 +6216 -510 -133 +207 -27 +70 +1,08 -+48 -51 +8 -79 +15 +4- +13
10 -+24 -12 +1 +811 -7 -t2 +7-
Fin 1 Op i 476psi 789psi 970psi noop 1Pressure
Table 1
9
10
It is seen from plate No. 1 that the relaxation method applied to
problem No. 1 resulted in getting values of pressures for point 1, 2,
and 3 very close or identical to thoBe calculated from formula (1) for
radial flow to a well.
where P =pressure @ radius R
Pa z known pressure @ radius Re• 1100 psi
Pw =pressure at face of well bore• 0 psi
Re =drainage radius where pressure is knownI: 1200 f t
Rw = radius of ell• 100 feet
'P =1100- 0 .....
t.... r<)..u. 1:1.00;'00 /00
+0 == 4 o4~ ilIA .1L100
By Relaxation Method
@lR = 300'
@R =600'
@It • 900'
PJ. • 480 psi
P2 .: 784 psi
P3 • 970 psi
PI • 476 p i
P2 = 789 psi
P3 • 970 psi
Comparing actual values ith valu s obtained by usi g the relaxation
method, it is concluded that the relaxation method applied to flow of
fluids will give fairly accurate results tor the problem considered.
It must be noted that had other values of pressure been as umed,
the results would have been slightly altered, as will be shown in
problem No.2.
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PROBLEM 2
THE RELAXATION METHOD OF MATHEMATICAL ANALYSIS
APPLIED TO
A MORE LENGTHY RADIAL FLC1N PROBLEM
s a further illustration of the use of the relaxation method,
consider the followi g two-dimensional radial flow problem.
3.2.0 f'si.
Fig. 2
12
Given w 11 bore 8 inchespressure 100 psi
drainage radius 660 feetpressure 320 pei
Cheek pressure @ points 5',10',25',50',100',200', and 400'from bor hoI •
From D'Arcy'e Lew
where !. 1 (Assumed)u
=2"R (for unit thickness)
For particular problem
13
;l7f(2.3a) a P '" 1.01T",AP
4.~1!>
.. 3.0 '7J.A P
= :lff(J7.!!J. D P z: 3.0 -,r DP;2 ~-
= ,0 I.:100 400
= ,:2""'(150) ~ P "3.0?T ~ P100
:: ;l7f (3 0D) ~ P 5< 3.0 71' c. P.2 DO
= :2 7T' (SJo) L:;) '? : 4. 0" .". .A P:4 '0
As previously Q is ~ for souree (flow into point)
and Q i8 - for sink (flow out of point)
14
From plate No. 2 assuming straight line eriation of pressur ith
drainage radius,
P @1/3' 100 psi
p @5' 101 psi
P @10' 104 psi
P @25' 108 psi
p @50' 116 psi
P @100' 132 psi
P @200' 165 p i
P @400' 232 psi
P @660' 320 psi
Results are shown in ccompenying table No. 2 and plate No.2.
It will be shown that if other than straight line variation of
pressur with drainage radius i assumed, using the relaxation thod
to determine pr saure, will result in very slight (negligibl ) or no
change in final results, 8S is shown on t ble No.4.
From plate No. 3 assuming curv d line variation of pressure with
drainage radius,
P @1/3' 100 psi
p @5' 160 psi
P @10' 200 psi
P C1l25' 220 psi
P 60' 240 psi
P @lOO' 260 psi
P 200' 280 psi
P 00' 300 psi
P @660' 320 psi
Results are shown on table No. 4 and plate o. 3.
Again, the relaxation method applied to th pr c ding r dial flo
problem gives results of pressure very close or identical to those
caleulated from formula (1) for radial flow to a well, as shown below.
15
R"en Rw
where P =320 psi
p.. • 100 psi
Re • 660 feet
Rw • 4 inches • 1/3 teet
p = 3.2 0 - 100 ,£n .!i ~ / 0 0 =.::J If "en 3 R + 10 0/} "'~O 'I~
~n-
'iB
Tabulated Results
Step Qo Po QI PI Q2 P2 Q3 P3 Q4 P4 Q5 P, Q6 P6 Q7 P7 Qa Pa0 100 +8 101 -2 104 +17 108 +24 116 +51 132 +102 165 +156 232 3201 +312 -338 +702 +381 -348 i"1l0 -83 +384 -339 +120 1"124 -t377 -336 +120 +21
5 +251 -206 +110 -66 +293 -252 +95 +127 -7 +75 -278 .,.24 -39 +10 +429 -9 -18 +10 +22
10 t-jO -12 +6 +911 -9 +l +312 -11 -7 +7 +913 -tl5 -3 +2 -314 +2 -3 +3 +315 -16 +5 -3 -516 -4 -3 -417 i'3 -3 +1 +318 -2 -3 +1 +
Final lOOps! I73psi 196psi 225psi 246psi 265psi 28,psi 305psi 320psiPressure
Table 2
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Step QO Po Q1 PI Q2 P2 Q3 P3 Q4 P4 QS Ps Q6 P6 Q7 P7 Qa Pa
0 100 +60 160 -74 200 +14 220 0 240 0 260 0 280 +21 300 3201 -20 +20 -12 -5 -7 +4 +123 +2 -6 +3 +94 +6 -15 +4 +125 -6 -6 +3 +306 -t'9 -5 +7 0 -3 +2 +18 +5 0 +1 +39 +2 0 +1 +3
10 0 -5 -1311 -9 +3 -3 -712 -1 -2 -313 -5 -1 -1 0
Final 100psi 173psi 197psi 224pBi 244psi 264psi 285psi 305psi 320psiPressure
Table 3
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Tabulated Pressure Values
Correct Metbod Relaxation Method
Firat ssumption First assumptionstraight line curved line
Radius From formula variation variation
5' 171 psi 173 psi 173 psi
10' 198 psi 196 psi 197 psi
25' 225 psi 225 psi 224 psi
50' 245 psi 246 psi 244 psi
100' 265 psi 265 psi 264 psi
200' 285 psi 285 psi 285 psi
400' 305 p i 305 psi 305 psi
Tabl 4
20
PR BLEM 3
SOLUTION OF SQUARE DRAINAGE AREA PROBLEM BY
THE RELAXATIUN METHOD OF MATHEMATICAL ANALYSIS
USING A SQUARE NE'lWORK OF FLOR
Statement of Problem
a.) Use rectangular network instee ot radial as indicated
below.
b.) Given 40 acre spacing or 1320 feet on each side.
21
c.) Pressure @ boundaryPressure @ sink
600 peio pei
/ 3;20 ~e-t
i-JII s/n .t:~ 0nj-,
0t'( (/
m......{.
~
d
c
b
()..
o . I~ (> ~f':"
Fig. 3
3
22
Assumptions
1. .) Since figure is symmetrical, to determine values ot
pressure at all points indicat d (intersection of lines shown
on figure 6), it will be only necessary to determine pressure
at all points on lower right corner bounded by lines a-g and 0-6.
b.) Again, since upper right part of cut figure i symmetrical
to lower left part, pressures are calculated only at half or
given points, .g. point £,2 and e,l are symmetrical to eaoh
other as are £,3 and d,l etc.
2. Consider all rlow in any spacing, concentrated in a pipe shown
as vertical and horizontal lines. For example: segment d,O
d,l denotes pipe hieh inelude flow from 10 er half of
pacinI d,O - e,O - a, - d,l and upper half of segment e,O
d,O - d,2 - 0,2.
3. ssume values of pressure for giTen points and consider
horizontal and vertical flow, a shown by flow network.
4. Determine Q using assumed pressures, and if Qf °at particular
point, relax by increasing or decreasi g pressures until the
lue of Q equals or pproachea very nearly to zero.
From D' rcy's Lew
where ! • 1 (Assumed)u
=constant (same for all segments)
.:)x. constant (equal lengths of segments)
L) P:II difference in pres ure between t..,o points
where
f =funct ion of
23
24
5
Determination of Q @ point ,3.
step 0
From plate No. 3 sswne pressures as follows:
@ 8,3 P • 390 psi
o e,4 p • 460 psi
d,3 p • 410 psi
o 8,2 p : 330 psi
@ f,3 p = 350 p i
• (460-390) + (410-390) + (330-390) + (350-390)
m -10 units
All result in the order of calculation are shown on table No.5.
Plates No.4, No.5 and No. 6 sho plotting of preeeur vs. distance
in teet from sink.
TABLE ;
Tabulated Results
Qg,l Pg,l Qg,2 Pg,2 Qg,3 Pg,3 Qg,4 Pg,4 Qg,5 Pg,5 Qr,1 Pr,1 Qr,2 Pr ,2 Qr,3 Pr ,3 Qr,4 Pr ,4 Qr,5 Pr,5
417psi519psi466psi393psi 563psi
-20 520+30'
+100 t
-20" +30+;'"
+15 aC.-t2;.Jr'
-15'" -+10-5"-+5"
+10'"_2S7'
0 11 +3
of 1' '''+3
'S"
-+4 7f
~1'"+2'.
47bpsi 524psi
-10 3;0 + 10 430+130" +100"+2607 +200"-140"+100 -SO' +70-70' -;0"+10''a' 0"-+2;"''' +1;'&'- 35·" +15 +30~~-2;i' -10 +10_lOi? -+ 5"+5,s, +15 J '
+1;1' -25"+10+23"11 -15'"
_94"'1 +8 _8"r~.J, Ofl
: 3""'" -20S'" +5+6.$"1 _11 t7
sr ~
- 6t.' +3 -14"~3" -11 Co'-2 -9
07' _1' '1
4-27S' _15'S' +2
+97' _13 7Co
-1 rJ, +1 -12''''0'" 0'" -3
'-3" ...,1"+1" -1 -2"
+10 270+180.2.+300'-220~+130- 120f'
-10"+5.l.z.
+20"S"-40,21+15_30.21'
-15'"_TrJ
_2$"J
+1£"+4'"-f'6""....2" 1
't-l+3'"+5"+6'"+21'J +1+3"!+2"+1'.t"
343psi
-100 210+200'-280(,+120_20'+10,a.,+40~'
Oa"+lO+12""O~ +3
+2sr
+4 '"'+6"+2,f +1+4'.1-2 rt+2'.r -1
561psi
+40 500-r130"-70$' +;0-10'·+5'&'
+2;'"II'
-I; +10
-~-2+4n+6"+2'1 +1+1"
+60 400+200'-160' +90-110·
+30'_30" +15_1;~1
+5.1'+2;17+3;'Y_;J' +10+2.1'.1
+12$'\ot"? +3
...3""+77.r_11 ' +2
0'77
-H"-;"_1'7 -1
+100 300+270"-290' +140-200f
0'+15'U.+30.21-30t~ T15
ou"+10"+26.s"1_2.r.a +7-+3·'+61'1
+12'"0'2. +3
+2'"+4
7'
0" +1+2'.a
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+140 200+290'-390~+170-250'
-tiO'+-25.&1-35~~ +15- 20.u'
+10'"-+16 ".+23.f'.&.
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+2GO 100-380' +150- 210.a.
+30·_302) +15-1501a
+5.&r-19"" +6-144""_
8r 'l
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+1" -3_1'"
Final 270 psiPressure
Qe,2 Pe ,2 Qe,3 Pe ,3 Qe,4 Pe,4 Qe,5 Pe,5 Qd,3 Pd,3 Qd t4 Pd,4 Qd t5 Pd,5 Qc,4 PCt4 Qc,5 PCt5 Qb,5 Pb,5
Final 457psiPressure
569psi 524psi ;76psi 584psi 592psi
+60 560+100 11
0"61'25+20"'"_4'" +6+2$1l
-+4 to
Of' ·n
o 550i"20 11
+70,r-10" +20+1~·+2,''1-1;011'-+10-3-"+3
011
+7'1"-tIl-I'
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;68psi
+80 5001'140'"-60'''' +50,f-20+20s'ao+40·""-8~'+12
+121/</-4'" +4+21"Di'6"<1_2 '0 +2-t2"-2'"
0"_2,J
+10 540+35·'" .+65"-15" 1'20+~If
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+15 tJ'If/~
-1 +4+2~o
+4$'1+6~"i
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+1"-1',(0"
+1'"0"
551psi
-;0 4900"
+90'''-,(,.
-30 +30_10"+40'''+;5"-25'·+20_10"
0'-1+12'"-28¥1+10-211/~
-16'"-12"-S'I'-6'"_14,<i -+2_U'.s"_9 1
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-13 101
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-3'"+1" -10'"
+120 410+280 1
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-BoIS'+90-20"+10~'+;0.1-_10" +1;
+611j1+26"1/+6~' +;
+12"T16'oJ+4(J +3+8 '1
-+12"0'" +3_4"0" -1
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+ 2'Y -1
o 530-t30"+80'1-20'" ....25
0"-n,'"+25'1/-15~.r+10-;:"+2"S'+6'1"+94""-nS1 +2+3"+;,a.+1
7' +1
+31"1_If" +1
o 460+70'
+150''''co J)-..,10 i'50
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536psi498psi
-10 390+90'
+200"-120 /"- +80
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+3;""-25~1 +15-10~o
+5~'
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o 330+260 .,-18011+110-20''''i'10.z,1+40'&'_20"t +15- 4'11
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NOTE: small numbers to the immediate right of values of ~ denote step number (order or calCUlation).
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PROBLE 4
SOLUTION OF CIRCULAR DRAINAGE AREA PROBLEM BY
THE REJ~TION METHOD OF MATHEMATICAL ANALYSIS
USING A ~TANGULAR NETNORK OF FLOi
29
Finally consider the follo ing problem in olving vertical and
horizontal flow, or a rectangular network appli d to flow fro a
circular boundary to a sink."(I
~r--- 1'00' r--.r---....
// 16"••' "'~ "0• ..../ JilO' f\.
/ 3u' \/ ;J," ~
\110.'
01-<",. I,••'"'-;., ,
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J\\ V
\ /'\ V
~"" VV
""'- [::/"
Fig. 4
Given: sink pressure 0 psi
draina e radius 660 f atpressure 600 psi
To find pressures at all points i dicated (intersections of
vertical and horizontal flow lines).
30
Assumptions
1. Assume str ight line v riation of pressure with dr in ge radius
at all points (shown on plate No.7).
2. a.) Given circular figure is symmetrical, ao consider only one
quarter of circle as shown be10 •
b.) Upper left corner is symmetrical to lower left, so consider
only ower right corner.
3
~o'
,... r-........r---...'~
s-,,' i'..,g0",
"40.' I\.
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1/0.' \"...."". ,
I~'·' J,,' "II.' ~... ' '01/10.
c
(J.o
h
d
f
Fig. 5
3. Flow is Be shown above in arrows e.g. for point c,l to sinkr
direction and path of flo is along se ants c,l - e,O, e,O - b,O,
b,O - a,O and c,l - b,l, b,l - a,l, a,l - a,O.
From D'Arcy's Law
where k • 1 (Assumed)U
=constant (same for all egments)
L!.X = variable
: 100 feet for all segments excepting tho asegments indicated on figure 7.
AP= difference in preae re between two pointswhere Q is considered.
31
32
Sample Calculation
Determination of Q @ point g,2
Step 0
From plates No.7, No.8 snd No.9 8Sum d pressures follow:
g,2 p. 515 psi
C g,l p. 553 psi
f,2 p a 490 psi
boundary p =600 p i
From Q • ~ P , it is noted that Q is inversely proportional to&>x
distance of flow AX • For all points excepting those adjacent to the
boundary, the lengt of path of flow 1s 100 feet. So we may 8sume it
is a constant except for points in question.
hen determining Q at point g,2, we note that vertical distance to
boundary is 28 feet while horizontal distance to boundary is 75 feet.
Since we are not interested in units of Q, but in just its value,
we may neglect entirely ~x for all segments aqua Ing 100 feet in lengt
and use a proportionality constant in case length 1 100 feet.
e.g. at point g,2
• +16 units
s before pressur s ar increased or dec rea ed until all values ot
Q appro ch very nearly or ar qual to roo All results In the order
of calculation are shown on table No.6.
Plates No.7, No.8 and No.9 show plotting of pressure .s. distance
in t at from sink.
TABLE 6
Tabulated Results
Q Pg,O g,O Q Pg,l g,l Q P Q P Q Pg,2 gt2 f,O ftO f,l r,l Qr,2 Pr ,2 Qr,3 Pr,3 Qr,4 Pf',4
+17 545 +16 553 +16 575 +20 454 -+16 464 +16 490 +19 530 +4 583+l07/S' +96/~ -+-86J7 +150,0 +141" +116'''''' +109 / +59'01-33~' +30 +-125
J4 +116.:/-210',{' t90 +231,r +196'" +178 17 +99"
+27,J'~,
+12.z.% +15 -50" -89" "'"80 _8411 +70 +13'P' +40 +5" +10-22J.l +30+3
J' +5 -7.31 -2" 2 _20~o -19 /7 -44" i"22 " +1141
+6"" ,;. + -12.1; +11.11 _29.2.3. -21/1 +6 +1~ +1-2 0_3"D +2 +4 ~. _7Jf -4-18.rS'" -11 Jl -1", +2tl'2. +61.6 +5-¥/l -6 iI' +6 _,~o 0"
+8'1 +90$" _3S"7 +1 'II_2',2 +2 -3..t"1 +3 Or, +3'"
_1,"0 ....2'.1Finel ;82p!i ;a5ps! 592pd ;47psi ;;Opei 560psi 576psi 594pBiPressure
Qe,O Pe,O Qe,l Pe,l Qe,2 Pe,2 Qe,3 Pe,3 Qa,4 Pe,4 Qd,O Pd,O Qd,l Pd,l Qd,2 Pd,2
+2; 363t17, I.
-345/0
+130-95"_;'s'",
+10 .l.a.
+201.2-12J 'I ....8+2~S'
t4S
'i-10.('7-6.s1" -+4_3.J"'-1 "
Final 505psiPressure
+21 375i-171'+301,Q-199" +12;_99',z-19"-4~'
+14"3....22"1(_6J~ +7
O·P+6-t1·+9s'S'
_3$"7 +3+1 t'Y+2'·
S10psi
+20 407+16rf'+285"-115'" +100
-25,1+45.1
+57"-15" +18-8~-4 n-l uo+2'"+3
7J
525psi
+17 455+137<;+23"'~-123'~ +90-68''1-28"_10 3 '_2J
'+4 'II
,oJ-4 ,I' +2-3'1--1
547pai
+20 514+200,3-20'" -+55
O,f+24/1"+6'"+2~t' "'1
570pei
+27 273+187"-413' +150-1137
+17'"+37'""-23"'t' ."'15
+7'"+15",0-2,'.... +10-17'-{_,~7
-1 '13+1 sa+2rl1"8....r
'(. +2+4.)"Y-4" +2_2"
0'10
452psi
+32 287+212'"+362(,-2387 1'150-98'+27"+42,2f-18--' +15-8.t.'"'431
+14'+21"S"
;)7 +6- 3<1'1+1~
+9..rJ_3:$ +3_1:1'+21'7+3 .. ,+5"+1,6 -+1+2 71
+3,a462psi
+26 328+186,s-1"336
7
-224 r' -+140-104'
-4·....i"11~+26.%1-22" +12_4JJ
J'+10"+15.f~-1 tI'1' +4+2'1'+4.r6
+7"+3'" H+470
+67 .....
"13 .H+3"'1
486psi
331psi
Qb,1 Pb,l
+74 129+374'-266.% +1601"94"
_6.&.J -t"2"-10.z.I"+30.t.f
_231 +8+8,r
'It'+18",-20 +51"6-zSl .... 2
$"J-t-4,r-4 0 +2
0"+2 1
'
260pe1
Qb,O ~ ,0
+-76 911
-524 +150_204.a.-44'
+6a.t.'1
+26.a.Y-6 +8+2"t'
+18:~-24jJ +5+2
+12""1/1-4 +4_2.f"f
+2rl"-3s-t{+5"+9""+ll1 +2
444pei
Qc,2 Pc,2
+32 258+392~-248 t-160+32
Y_28027 +1-~~
+16J1y....
-4 "" +5+4.p-+14",
+2r .a +3-#-8
OS'3 "'2U
+2+4 71
...6'J
+2 ' '1 +1
403psi
Qe,1 Pc,1
...40 204....200."
+:360 '-360: +180-200_507
_2,.l 3
-5~"+lO~"+2;~7-I;:: +10-7+IJ1
37+7iJ."
+12+-16 '13
'It'-4oJ" +5+1,,1+4+6~+8.('1-4 S'.t +3_2oj
rI-1,(",{'
+2+4,1+6,r+2 70 +1+3 11
+41
"379psi
Qc,O Pe,O
+44 182+194
'-446.3 "'160-86"+64"_16,2~ +20
_1.1+7.zT
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-;IJ. +8+;~r
-tIO-13-6 +4+4+'S'+8'"
s'40;,Z. 1"2
+6S"o+2sY +l.rt,+'4 "+6_21.1 -t"2
010
'+2 7
'
Final 5lopsire89ure
NOTE: small numbers to the immediate right of values of Q denote step number (order of calculation).,
Qd,3 Pd ,3
+22 386+3021'-178' +120
-t2/J
+26'1_63 ' +8+2 oi'l+4 c.J
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CONCLUSIONS
37
From the results obt ined in the solution of probl me involving
steady radial flow to a well, whose solutions are known, it may be seen
that the mathematioal analysis of relaxation can be successfully employed
to the solution of similar problema.
Comparison of results obtained with those obtained by formula show
negligible differences. It must be noted that the length of tim to
solve any given problem will depend primarily on two factors, namely:
1. values of pressure first assumed, and 2. degree and manner in Which
pressures are increased or decreased at particular points. This is
clearly shown in the solution of problem No.2.
From giv n results, it may be seen that the value of Q not always
equals zero, ev n though the particular problem is co.nsidered completed.
An attempt to bring all values of Q down to zero can be undertaken if
more accurate values of pressure are required. otherwi a, depending on
the given data, values of Q slightly great r than or leBa than zero will
result in correct pressur s with one or t most two pounds.
The solutions to problems No. 3 and No. 4 Cannot be checked by
formula, but they may be considered correct ae 1. the curves drawn ar
quite satisfactory and 2. proof that the method will work, from problems
No.1 and No.2.
Problem No. 4 tends to show that no matter what shape of drainage
area one has, one can suocessfully spply the relaxation method of
mathsmatical analysis a surning any desired form of network. The drainage
area in problem No. 4 has a circular boundary, but a rectangular or
rather a square net ork of flo wa assumed and results wers very
sati factory.
38
In conel sion it can be definitely stated that the mathematical
analysi of relaxation will apply to the solution of problems involvin
steady flo of fluids through porOUB media.
StnolARY
39
A very brief review of the similarity between the 8ol~tion of heat
conduction problema and that of flow of fluids through poro~s media is
made.
The steps folIo ed in the solution of a problem by use of the
relaxation method are shown at the same time solving two radial rl~
problems which are checked for correctness using the formula for radial
flow to a ell.
Two mar problems are solved using the relaxation method, one
involving horizontal and vertical flow in a rectangular boundary, and the
other a circular boundary.
40
BIBLIOGRAPHY
1. Books:
a. Musket, M., The flow of homogeneous fluids through porous edia.
1st ed. N. Y., MCGraw-Hill, 1937. pp. 55-74, 121-140, 149-156.
b. Ing reoll, L. R., Zobel, O. J., and Ingersoll, A. C., Heat
conduction. 1st ed. N. Y., McGraw-Hill, 1948. pp. 1 , 213-216.
2. Periodica s:
a. Christopherson, B. A., and Southwell, R. V., Relax tioD methods
applied to engineering probl ms Ill. Problems involving two
independent veriables. Proceeding of the Royal Society,
series A, Vol. 168, pp. 317-350 (1938).
b. Emmons, H. W., The numerical solution of heat-conduction
problems. Transactions of the A.8.M.E. Vol. 63, No.6,
pp. 607-615 (August 1943).
e. Miles, A. J., and Stephenson, E. A., Pressure distribution in
oil and gas reservoirs by membrane analogy. Amer. rnst. of
Min. and Met. Engrs. Tech. Pub., No. 919 (May, 1938).
VIT
The author was born January 1, 1924, at Istanbul, Turkey. His
high-school education was completed in 1940 at Robert Academy in
Istanbul, Turkey. In 1945 he graduated from Robert College Engineering
School with a B. S. degree in Mechanical Engineering. He entered the
Missouri School of Mines in September, 1947.