The effect of high viscosity on the flow around a cylinder and around a sphere

MISSING IMAGE

Material Information

Title:
The effect of high viscosity on the flow around a cylinder and around a sphere
Series Title:
NACA TM
Physical Description:
29 p. : ill. ; 27 cm.
Language:
English
Creator:
Homann, F
United States -- National Advisory Committee for Aeronautics
Publisher:
NACA
Place of Publication:
Washington, D.C
Publication Date:

Subjects

Subjects / Keywords:
Aerodynamics   ( lcsh )
Viscous flow   ( lcsh )
Genre:
federal government publication   ( marcgt )
technical report   ( marcgt )
non-fiction   ( marcgt )

Notes

Abstract:
The report records an experimental and theoretical investigation of the influence of high viscosity on the stagnation pressure of flow around cylinders and spheres. For the three-dimensional problem, a differential equation is set up which corresponds to the two-dimensional solution by Heimenz. Theoretical methods are developed for determining the stagnation pressure for the two- and three-dimensional cases of flow around cylinders and spheres, respectively. The results thus obtained are compared with experimental results obtained in an oil channel at Reynolds number up to approximately 100. Finally, a procedure to determine the velocity and the pressure variation, as well as the variation of the free-stream velocity gradient on the stagnation streamline, is shown and used for the practical case of Reynolds number 100.
Funding:
Sponsored by National Advisory Committee for Aeronautics
Statement of Responsibility:
by F. Homann.
General Note:
"Report date June 1952."
General Note:
"Translation of "Der einfluss grosser zähigkeit bei der strömung um den zylinder und um die kugel," ZAMM, vol. 16, no. 3, June 1936."

Record Information

Source Institution:
University of Florida
Rights Management:
All applicable rights reserved by the source institution and holding location.
Resource Identifier:
aleph - 003779458
oclc - 86175441
sobekcm - AA00006189_00001
System ID:
AA00006189:00001

Full Text
I'M-r i 3








'6&1 7r (

NATIONAL ADVISORY COMMITTEE FOR AERONAUTICS

TECHNICAL MEMORANDUM 1334


THE EFFECT OF HIGH VISCOSITY ON THE FLOW AROUND

A CYLINDER AND AROUND A SPHERE*L

By F. Homann


For the determination of the flow velocity one is accustomed to
measure the impact pressure, i.e., the pressure intensity in front of
an obstacle. In incompressible fluids the impact pressure is 7v2/2g
(7, kg/m3, is specific weight and v, m/sec, is velocity) if the
influence of viscosity can be neglected. Such an influence is appreci-
able, however, when the Reynolds number corresponding to impact tube
radius is under about 100, and must consequently be considered, if the
velocity determination is not to be faulty. The first investigations
of this influence are included in the work of Miss M. Barker2. In the
following pages, experiments will be reported which determine the inten-
sity of impact pressure on cylinders and spheres; furthermore a theory
of the phenomenon will be developed which is in good agreement with the
measurements.

The research apparatus consists of an oil circulation in which the
velocity of the oil can be varied from 0.5 centimeter per second to
30 centimeters per second with the help of a vane-type pump lying
entirely in the oil. A Russian bearing oil and a mixture of this with
fuel oil is used for the measurements. Figure 1 illustrates the test
setup. In this is indicated: P, the pump; a, turning vanes;
G, straightener; and V, the actual test section which possesses a
breadth of 0.148 meter, a depth of 0.15 meter calculated from the oil
surface, and a length of 0.74 meter. It was provided with wall ports A
in three different places. E is an entrance section for the pump;
D, a diffuser; the immersion heater T and the cooling coil K provide


"Der Einfluss grosser Zihigkeit bei der Strbmung um den Zylinder
und um die Kugel", ZAMM, vol. 16, no. 3, June 1936, p. 153-164.
1The suggestion of the present work, which was prepared in the
Kaiser Wilhelm Institute in G6ttingen, I obtained from Herr Professor
Dr. Prandtl, to whom in this place I express my most heartfelt thanks
for the energetic furthering of the work and the valuable suggestions
given me for its completion. Another work in the same field is published
in "Forshung a.d. Gebiet des Ingenieurwesens", 1936, vol. 7, no. 1.
M. Barker Proc. of Royal Society, 1922, Vol. A.-101, p. 435.







NACA TM 1334


temperature regulation. For impact pressure measurement a cylinder
which was provided with a port was built rigidly into the test section.
The diameters of the cylinders used were 1 centimeter, 1.377 centimeters,
1.953 centimeters, and 2'centimeters. The holes had a diameter of
0.1 centimeter and 0.2 centimeter. Two corrections one because of
wall effect, the other because of finite size of hole (which originated
with A. Thom and first had to be checked for the measuring range under
consideration here) were applied to the measurements, which are illus-
trated in figure 7. The solid curve represents the theory.

More precise information on the test setup and the measurement
technique are found in the work cited in footnote 1.

In the case of the measurement of static pressure on a sphere, a
sphere provided with a hole was affixed to a pitot tube, the sphere
having one time a diameter of 0.8 centimeter, the other time 1.6 centi-
meters. The execution of the measurements was in the same manner as in
the case of measurement of the Barker effect. The result is shown in
figure 8. The solid curve corresponds to the theory.

In order to arrive at a clearer picture of the viscous flow around
a cylinder or a sphere, the case of viscous flow against a plate was
next calculated. The differential equation appearing in the two-
dimensional case has already been solved by Hiemenz3 and will be sketched
once more for the sake of a better understanding of the final form. The
solution will than be used in the flow around the cylinder.

After this, the three-dimensional flow against a plate in a fluid
jet will be treated, to be used on the sphere.


VISCOUS FLOW IN THE VICINITY OF A STAGNATION POINT

(TWO-DIMENSIONAL CASE)


Potential-flow theory gives for the velocity components in the
neighborhood of a stagnation point, for the case of flow perpendicular
to a plane wall (fig. 2):


U = -ax V = ay



3Hiemenz Dissertation, Dingl. Polyt. Journal, V.326 (1911),
No. 21-26.






NACA TM 1334 3


The pressure is found from the Bernoulli equation to be

S- p (U2 + V2) pa22 2)
Po 2 2


With consideration of viscosity, Hiemenz3 makes the formulation:


u = -f(x) v = yf'(x) (1)


po p a2 (x) (2)


The continuity equation is fulfilled; the boundary conditions read:


for x = 0 (that is, at the wall): u = v = O, f = f' = O

for x = m: v = V, f' = a


In equation (2) if po signifies the pressure at the stagnation point,
then F(o) = 0. From the equations of motion

6u u 1 vp .2 vfu 62u
u u+ v + W+
ax ay P ax a2 y2


u-+v + +
6x dy P 6y Yx2 y 2]


one obtains as determining equations for f(x) and F(x)


ff' =- F' Vf" (3)


f,2 ff" = a2 + Vf'' (4)

with the boundary conditions above.









Equation (4) has already been integrated by Heimenz.
make the coefficients equal to unity, he sets


f(x) = AO(k)


NACA TM 1334


In order to


S= ax


From comparison of the coefficients, it follows that

A = va a = f


With this, equation (4) becomes


0,2 0", = 1 + 0"'


The new boundary conditions read


for 0 = 0: 0 = = 0

for w = m: 0' = 1


The behavior of 0 and the first two derivatives is shown in figure 4.

We need the pressure difference between the stagnation point and
the pressure for x = a% For x = m:


0' = 1


f = a


0 = 0.647


From integration of equation (3), F is determined to be


a2 1
3- F = Vf' +1
2 2


If one now forms (P p) minus (p p'), as given by the Bernoulli
equation, one obtains:

(P0 P) (Po' P') a (F() y2) (f2 y }2f2)






NACA TM 1334


For the stagnation streamline, for which y = 0:


(Po- P) (P' P')= p2 F P 2
2 2


If one puts in for F the previously obtained value, there results


(Po p = (2vff + f 2 2 f2)= pvfm


For x = mo, f = a; therefore, one obtains as a final formula:


(6a)


(Po P) (Po' P') = pva


VISCOUS FLOW AT A STAGNATION POINT

ROTATIONALLYY SYMMETRIC CASE)


For the solution of the differential equation arising, all the
expressions, such as the equations of motion, the velocity components,
etc., were reduced to cylindrical coordinates.

If z, r and p are the coordinates (fig. 3), then corresponding
to the two-dimensional case, there will apply:


v = -f(z)


Vr = (z)


pa2 2
P p F Z) r2)


The continuity equation is again fulfilled; v- = O, since we are dealing
with a rotationally symmetric process. These expressions stem from the
frictionless problem of a fluid jet against a plate, where


Vr = ar
r


Vz = -2az


p p a2 (4z2 r2)
o 2






6 NACA TM 1334


The quantity 2az in the frictionless case is replaced by f(Z) in
the viscous case. In the case at hand the equations of motion read:


avr
r




aVz
Vr +


vr
vz
VZ Z



v z
Z z


-_ 8+ Vr
P 6r r2


1 6r
r 6r


r2


+z2 r)
oz2


(9)


=1 -
P 6z


6z2V


1
S
r


St 2
+-r
r z2 /


Substituting equations (7) and (8) in equation (9) gives:


1 f,2 ff"' = 2a2 + vf'''
2


2
ff" a2
2


(10)



(11)


F' Vf"''


The boundary conditions read

for z = 0: f = f' = 0

for z = o: f' = 2a


If one finds
from equation


f from equation (10), one can therewith determine
(11). One next substitutes the transformation


f(z) = Ao(()


t = az


(12)


into equation (10), in order to make the two coefficients equal to unity.
This yields:


a 2A2~'2 a2A2"' = 2a2 + 3AO'''
2

From equating the coefficients:
1 2 2 2 3
aA = 2a = Va3A
2






NACA TM 1334


A = 2v- a = (13)


From equation (10) with equations (12) and equations (13) there results
the final differential equation:


0"' + 20" 0'2 + 1 = 0 (14)


with the boundary conditions:
I

for = 0: 0 = = 0

for t = o: = 1


The differential equation (14), just as Hiemenz', is no longer
elementarily integrable. Its solution was obtained, accordingly,
through a power series development from zero:


0 = ao + al + a 2+ + agn (15)


By the method of undetermined coefficients, ai can be determined.
Since, however, one boundary condition lies at infinity, one coefficient
remains undetermined; and in fact it turns out to be a2. From the
recursion formulas


= ao + al1 + a 22 + a33 +


= al + 2a2E + 3a3 2 +


0" = 2a2 + 2 x 3a 3 + 3 x 4a2 + .


result as coefficients:






NACA TM 1334

ao = al = 0

a2 = for the present undetermined

a3 = -0.166667

a4 = 0

a5 = 0

a6 = 0.555556 x 10-2 a2

a7 = -0.396825 x 10-3
ag =0

a9 = -0.440917 x 10-3 a22

al0 = 0.793651 x 10-4 a2

all = -0.360750 x 10-5

a12 = 0.374111 X 10-4 a 3

al3 = -0.114597 X 10-4 a22

a14 = 0.115735 x 10-5 a2
ai5 = -0.301482 x 10-5 a4 0.385784 x 10-7

a16 = 0.134896 x 10-5 a23
a17 = -0.211005 X 10-6 a22

a18 = 0.224141 x 10-6 a25 + 0.157758 x 10-7 a2

a19 = -0.135546 x 10-6 a24 0.415153 x 10-9

a20 = 0.316633 x 10-7 a23

a21 = -0.152798 x 10-7 a26 0.295658 x 10-8 a22

a22 = 0.119505 x 10-7 a25 + 0.199390 x 10-9 a2
a23 = -0.371665 x 10-8 a24 0.433457 x 10-11

a24 = 0.956242 x 10-9 a27 + 0.554360 x 10-9 a23

a25 = -0.943031 x 10-9 a26 0.462914 x 10-11 a22






NACA TM 1334 9

In order now to be able to determine a2, a second series develop-
ment from infinity was set up, which was adjusted to the boundary condi-
tion for 0 at infinity. To this end one sets


0 = 0 + 1 (16)


in which 0i corresponds to a small quantity, which one can neglect in
the following expressions when it appears squared. 0o is the solution
for 5 = m.


0' = 0o + 0 = 1 + '


since for e = m:


00' = 0' = 1 0" = 0" "'' = 1'" (16a)

The boundary condition reads


for = c: 1' = 0

Furthermore, 0o= "


The integration constant is omitted, since in the following calculation
it comes in again automatically.

If one substitutes the above values into equation (10), one obtains

1"' + 2(o + 0101) (Oo2 + 20o'0'1 ,+ + 1 = 0 (17)


Or if one neglects the squared terms in 0:


01''' + 2l"'' 201' = 0 (18)






10 NACA 'M 1334


To solve this differential equation one sets


Q = 1


With this, equation (18) gives


0" + 2W0' 20 = 0 (19)


A special, not identically vanishing solution of equation (19) is:


Ol= E


If 02 is an additional solution, then



01 2 1 ID = e a


_2
t02 2 = e 2


02- 2 1e- 2 = 0


This equation is directly solvable. Its solution is


O 1e-n TI
02 = E f J ,



The general solution of equation (19) is then:


= + = + 1 e2
S= Clz + C2"2 = CiE + C2 21 e-2 dT]
go T1






NACA TM 1334


Since for t = i' = 0 = 0, then C1 = 0. Therefore


S' = D = C2 J
'J


01=)


Se-'2 dT
q2


= C2 e-2 2
Ic


0 da


1 = 3 + C2 e-2 2q'


I


e-u du dq


C32
= C C2 e-' d'q
iJ 0


- 2C2f


The double integral becomes, according to Blasius ,


dq e-u2 du


= 2C2 2


Se- 2 d9 e"12 dTi


With this, equation (20) becomes:


i C3
C02


-1 e-2
2


2 d
e'- d


2


e-2 di 2 f
O o


2 PR
e-2 d'i = 2
0


2
e- d2] J/

(22)


d = 2
FU fo


2
e- dq 1


4Blasius Dissertation, Z.F.M.u. Phys. 1908, p. 1.


and since


e-'2


d f 2
JC


2C2 f TI


(20)


+ -


Now:



2f
0


(21)


e-2 d' = 2
O o


.-2






12 NACA TM 1334


If one substitutes equation (22) in equation (21), one can calculate
pointwise, since


2 '
Ff


e-72 dq


is tabulated. Therefore


1 C3 = e-2
C2 2


C2


= -e-2


C2


e- 2 d j -


+ t 7


- t {A f e-T2
[n o


dT 1.


= e-2 1d
0- Jfo


Herewith 0, 0', and 0" are determined for the development at
infinity, as a comparison with equations (16) and equation (16a) shows.

In both developments a2, C2, and C3 appear as unknowns. If
one now combines both solutions at the point t = to, and determines
that the value of the function and the first two derivatives of the
series development at zero are equal to the corresponding values that
one obtains from the development at infinity, then three determining


-(









NACA TM 1334


H 0
0 u
a +
a )

E- +













0
0 0
--
(U I









a,






a F
+
: CM












0
0 I


a)
0
cu-







0





Cd
Cu1 +
*H I





0, I
'H 0


'II



rd


I





OJ
0

0








C-)
I
0



a
0





N
0












I
o





















0
up















od
ru
*M




O





r-i
C
01
II
r-i
+U


r-d



a)
I


01
0













II
(,1




Cd
I'
CQ
x

CM

+




-I

*n

01
rLO


crd

CU
+
*
*I


1-
a0
4,





rd
-d





a, 0






c
rj
0=



"a
0 0
aL






D a
Cd


o a)

0 m


- 0












0 +
Ur-






a
4C



o
-c








O i
0
H U


Ul








In

U 3
: .
0







OO






NACA TM 1334


Thereby a2 is determined accurately
coefficients of the power series this


a2

a3

a6

a7


ag

a10

all

12

al 3
a13

al4

a15


to at least five places.
yields:


= 0.658619

= -0.166667

= 0.365900 x 10-2

= -0.396825 x 10-3

= -0.191261 x 10-3

= 0.522714 x 10-4

= -0.360750 x 10-5

= 0.106882 x 10-4

= -0.497098 x 10-5

= 0.762253 x 10-6

= -0.605859 x 10-6


For the


al6

a17

a18

a19

a20

a21

a22

a23

a24

a25


a = a1 = a = a = a = 0


The values for ', and are shown more accurately how-
ever, in Table I. In this case, 0' is calculated accurately to two
decimal places, 0' to two, and 0 to three. With this the differ-
ential equation (14) is solved.

From integration of equation (ll), one obtains


a F = Vf + f2 =2av( + 2) (23)
2 2


= 0.385391 x 10-6

= -0.958673 x 10-7

= 0.381678 x 10-7

= -0.259200 x 10-7

= 0.904605 x 10-8

= -0.252966 x 10-8

= 0.161233 x 10-8

= -0.703675 x 10-9

= 0.209783 x 10-9

= -0.970520 x 10-10






NACA 1 1334


As in the plane case, one uses again the pressure difference
between the stagnation pressure and the pressure for z = c. For
equation (23) is equal to zero; for E =


0' = 1


f' = 2a


If one now forms again (po P)
the Bernoulli equation, one obtains:


0 = t 0.557611


minus (po' P'), as given by


(P P) (P' P') = P( Vf' + I f"2 f.2) = pVf'

As a final formula one obtains


(P P)- (Po' P') = 2pva


STAGNATION PRESSURE ON A CYLINDER


For the stagnation streamline the
tion gives


Navier-Stokes differential equa-


du 1 dP 82u
u-- +1 v= -- +
ax P 6x ax2


(25)


Figure 12 shows the variation of u on this streamline. The different
behavior of u at the stagnation point from potential flow is explained
by the influence of viscosity. If one integrates between the boundary R
and -, one gets


UR2 2 -
2 2 + (P


R
P)= v 6 + .
/ OX 10


fR w2u
a y2


Figure 13 shows that


R
v-u =0
ax m


t = 0,


(24)









likewise uR = 0, and we want to identify PR with po
ceding calculation. One obtains, therefore


(P P)- 2


NACA TM 1334


of the pre-


2- dx
8y2


(26)


r u dx
" oy2
00 6


is calculated approximately in that for u the value corresponding to
the potential flow is put in. The contribution of the boundary layer
to the integral is, in the case of not too small Reynolds number, small
in comparison. As potential function $ of the flow around the cylinder,
one obtains

0 = Uo x + 2


and with it:


82u

By2


2R2
r4


8x2R2
r6
r6


S8y2R2
r6


48x2y2R2)
r8


For y = 0, therefore, along the stagnation streamline:


S2
ay


6UoR2
x4


R 2
0oo oy


dx = 2pUo
dx -
R


Herewith equation (26) gives


S- ) 2
(p p) = 2


2pvUo
R


(27a)


- -Uo






NACA TM 1334 17


or

P p 4
p = 1 +- (27)
pUo2/2 Re


If one substitutes 7 = pg, then formula (27) reads

Po p 4
o 1 + -- (27b)
Uo2/2g Re


where g = 9.81 meters per second2

In order to be able to accomplish a comparison of test results with
theory, the "displacement thickness" (see Tolmien: Hdb. d Experimental-
physik, v. 4, 1st part, p. 262, "Grenzschichttheorie (Boundary Layer
Theory)") on the cylinder must yet be considered in the calculation.
Solution of the differential equation (5) yields (fig. 5):

E* = ax* = nb* = 0.647

where 5* is the displacement thickness. Therefore


0.647 = 5*" (28)


If one compares the flow in the region nearest the stagnation point for
the cylinder and for the flow against a plate, one obtains from equa-
tions (27a) and (6a)

2pVUo 2Uo
pva 2v a
R

If one substitutes this value in equation (28), one obtains for the
displacement thickness


0.647 5 0 = *Re
RV R


R V/R-e


The dependence of 5*/R on Re is shown in figure 6.






NACA TM 1334


Now since in the
radius R, the actual
tion (27b) is altered


test results Re is formed from the cylinder
effective radius is therefore (R + 8*), and equa-
to:


Po 4v
yUo2g 1 + +
7Uo2/2g Uo(R + 6*)


With this one obtains
cylinder


as a final rule for the stagnation pressure on the


P p 4
7Uo 1 +
7Uo2/2g Re t 0.457+e


(29)


In figure 7 the solid curve again gives the theory, which agrees very
well with the practice.


STAGNATION PRESSURE ON A SPHERE


Corresponding to a cylinder, for the stagnation streamline of a
sphere


8u 1 6p
U -+ --
6x P ax


a2u
-- +
ox2


va
( 6y


If one integrates again over x from o to R:


pU 2
Po P) = 2
2


+ pv /
Sco Vy2


2 d
68z2


(30)


since


HR
Lm


udx = 0,
6x2


UR = O


The integral on the right side of equation (30) one again solves by






NACA TM 1334


substituting for u the value for the potential flow.
function of this flow is


0 = -Uo


For potential flow it is further true that


x2
3x


The potential


62u
3+ 0
2
az


82u
3y2


Therefore


R( 2
Jo\Ia


r u d.x
R ax2


X IR


If one substitutes in the last formula the value for
one obtains


f, aRf
2^
= oby


6u/3x given by 0,


,+ udx = 3Uo
2 R
z /


Herewith equation (30) becomes

2
pUo 3pVUo
p -p +
0 2 R


(31a)


(31)


Po P 6+
PUo2/2 Re


or with 7 = pg


(31b)


Po P 6
2, = 1 + R-
YU 2 29g


or


z2






20 NACA TM 1334


If here one also puts the displacement thickness into the calculation,
one gets (since in the rotationally symmetric case E* = 0.5576):


0.5576 = B7i/v

Comparison of equation (24) and equation (31a) yields

a 3 0.5576- D= 5*
2R V 2VR
(32)
6* 0.455
R %fRe


It appears that the displacement thickness for a sphere and a cylinder
are equal within I percent, although the displacement thickness in the
case of plane flow against a plate is different from the corresponding
three-dimensional flow.

If one considers the displacement thickness in equation (31b), one
obtains as a final stagnation pressure formula for a sphere

p p
0 = + 6(33)
7Uo2/2g Re + 0.455J-Fc


The solid curve in figure 8 corresponds to the theory; the agreement
with test results is again satisfactory.

From the final stagnation pressure formula the dependence of the
numerical factor c on Re can be determined, if osne sets

Po P C
po-p
7Uo2/2g Re

For the sphere there results

6 Re
Re + 0.(34)
Re + 0.455V1ie






NACA TM 1334


From Stokes' calculation one obtains for small Re: C = 3. In
figure 9 is drawn log Re as abscissa, e as ordinate. In the region
from about Re = 0.1 to Re = 1 the course of e is essentially dif-
ferent, since Stokes' law describes an approximation for very small
Reynolds number and the above law is an approximation for large Reynolds
number.

For the cylinder one obtains in the same fashion

e= 4Re (35)
Re + 0.457vRe


According to Lamb5, for small Re, for which the validity of the formula
extends to about Re = 0.5:


Re Re (36)
1.309 ZnRe


In figure 10 is again shown the dependence of e on log Re. Within
the accuracy of measurement the test results here also confirm equa-
tion (29).

With the help of the flow against a plate it is now also possible
to establish approximately the course of u, 6u/6x, and from this p,
on the stagnation streamline. A single curve was assumed in which,
inside the displacement thickness 6*, the magnitudes as given by the
flow against a plate were used. From the displacement thickness on,
which had a value of 0.0455 in the foregoing case for Re = 100, the
potential flow was calculated. To explain the transition from viscous
to potential flow, I would like to go through the calculation of u as
an example. The solution of the viscous problem ul/uo has as asymptote
the tangent to the curve u2/uo, which was determined from potential
theory, at the point 5* = 0.0455 centimeter. In figure 11 this tangent
is labelled t. The difference k between the asymptote t and ul
at the point xo gives the deviation of viscous flow from potential
flow at this point. Therefore to the value ul at the point xo was
added the proper k. With the help of this procedure one obtains point-
wise the transition from ul to u2.



5Lamb "Hydrodynamics" (2nd Edition 1931; German Edition by
E. Helly, p. 696, par. 343).






NACA TM 1334


au/6x was determined correspondingly; the pressure p was found
from the equations of motion to be, in the case of the sphere:


Po P Ux2 1 bux 6 (37)
1+ (37)
Uo2/2g 400 200 6x Uo(R + x)4


Instead of 6 in the last term of the preceding equation (37), in
the case of the cylinder one gets the factor 4. Figures 12, 13, and 14
are the results; by way of comparison the corresponding curves for the
cylinder and the sphere are shown on one sheet. The curves are true,
as already said, for Re = 100, in which R = 0.01 meter; Uo = 1 meter;
v = 0.0001 kilogram x second per meter2 was assumed.

Naturally the last curves give only an approximation, which can be
made essentially better through a second approximation; yet this task
in the framework of the foregoing work would lead too far.


SUMMARY


In the foregoing work the stagnation pressure increase on cylinders
and spheres brought about through the influence of large viscosity, was
reported on.

For the three-dimensional problem, hence the flow around a sphere,
a differential equation was set up which corresponded to that of Heimenz,
who had already solved the two-dimensional case. The solution was
ascertained likewise through an approximate method. The solutions for
the two- and for the three-dimensional case were used for the flow
around the cylinder and sphere respectively; the formulas so obtained
for the stagnation pressure increase stood in good agreement with the
reported test results. Finally, a procedure to determine the velocity
and pressure variation, as well as the variation of 6u/ox on the
stagnation streamline was shown and used on the practical case of
Re = 100.


Translated by D. C. Ipsen
University of California
Berkeley, California






NACA TM 1334


TABLE I


0 0 0 1.3172 1.4 0.8546 0.9476 0.1697
.1 .0064 .1267 1.2172 1.5 .9502 .9635 .1301
.2 .0250 .2434 1.1173 1.6 1.0472 .9762 .0895
.3 .0548 .3502 1.0181 1.7 1.1453 .9863 .0622
.4 .0974 .4471 .9200 1.8 1.2424 .9905 .o418
.5 .1439 .5343 .8235 1.9 1.3436 .9935 .0276
.6 .2012 .6129 .7298 2.0 1.4430 .9962 .0180
.7 .2659 .6804 .6400 2.1 1.5413 .9979 .0115
.8 .3370 .7400 .5548 2.2 1.6409 .9986 .0073
.9 .4137 .7847 .4742 2.3 1.7416 .9991 .0042
1.0 .4951 .8352 .4015 2.4 1.8417 .9995 .0027
1.1 .5805 .8712 .3351 2.5 1.9420 .9997 .0016
1.2 .6653 .9025 .2760 2.6 2.0423 .9999 .0009
1.3 .7608 .9247 .2241






NACA TM 1334


Figure 1.- Test tunnel.


Figure 2.- Streamline picture for flow against a plate (two-dimensional).


Figure 3.





NACA TM 1334


0 0.5 I I5 2.0 25

Figure 4.- 0, 0', 0 ". The curves drawn out illustrate the two-dimensional
solution, those not drawn out the three-dimensional.







L/
0.647
Figure 5.


R


Figure 6.- Momentum thickness
c*


on a cylinder = 0.457 and on a
R -. d


0 U.45D
sphere = .
R







NACA TM 1334


o Test Point


S P -P
20


4
Re + 0.457'C


Figure 7.- Stagnation pressure on cylinder.


o Test Point
PO -P ____
YUL Re + 0.455 Nf
29


Po-P
2u
29


40
Re


Figure 8.- Stagnation pressure on sphere.


PO-P

29







NACA TM 1334


o Test Point

Curve I: e = 3 (stokes solution)
Curve II: f 6Re
Re +*o0.455 V
a

8---- --
6 f o




2

3 -2 -1 0 I 2


g- Log Re


Figure 9.- e for sphere.


o Test Point

Curve I : E x Re +4 Re (Oseens' solution)
1.309 in Re
Curve I: 'Re o.45 -
Ike + 0.457q~


---Log Re


Figure 10.- e for cylinder.


-







NACA TM 1334


Figure 11.- Illustration of interpolation for transition from viscous to
potential flow.









U
0






0 0.2 0.4 0.6 0 1.0 1.2 IA
X
R

Figure 12.- Velocity variation on stagnation streamline; cylinder: Curve I;
sphere: Curve IL Re = 100.






NACA TM 1334


u 150
dx


Figure 13.-


3U
Variation of x on stagnation streamline; cylinder: Curve I;
sphere: Curve II. Re = 100.


PO-P
Yru
2g


Figure 14.- Pressure variation on stagnation streamline; cylinder: Curve I;
sphere: Curve II. Re = 100.


NACA-LI gle 8-11-52 1000












u = C








E o 0 Q -,;
EcN uwE F


a o C i0 Q '0


- Nfl U.* in


E



,' LEa a 14U
MF-
cgs i'- s
cn~
Sz .E" ES
U D- P
0


O U.<= <
rn Z m w



u,. CZ

-~zL.- L...
n > U t~ i
2 E < L .n-
r. mr Z t




u ;2 C O



u NCE
L40 -Lo J _*
z -
>^r S3 m.* C


Lcb5M> L ^B
pj 2u0 -N EM -
,a ~ls


cV,
c0 l 'C 0 'Oi O
r -cr
C- DCc 10.cu




E. E wl 'a* uCLCUC)Z
m E on c



2 a *'. Ln u .



Q,, CL 0, -
3~ CLC 0.L5C1 C





IL 0~ 9.
.C % U a 0
o Y' Cc 0L O E L 'A
E g .ic

0.'L CuE lCPt Lb

'aLrL CL.O Cu
cL

H-u1,... CO.01C'
-JSL C^SS?
2tt~gai ^i.
gc. El.9i



LC a s5 L '


S' *i-53 ; *
^~~ (L s'^ '"^-


c' ,TI 3 0 L' ,3
i- cnr c-E o^ ^c -


I----












it




:Jm m






-g o
I I' o





Sa V >






. 3 9 i C- Z
a.- ^ ^ a ," N w -* N*
E C 47 C S a go


ca x

-0lU L B B U < w.0 E 0


U -c <1 M S z N


0 bD, ,



r" 1. 0 -
aZ u CU0
5.O-o'0-'.o
-6 0 a

0-~
(U JA kQ0 < *





|S2 R bw

U0 ( Q D
R- N m3""20
u R wod-&.|4
C^%[-raZ a? ^
z z E. 0!


'0 faU~0
wu I ) w

^ -2S '. 4






0 1 0 L cU ~ ~
tu .s|a a) M H z


vm~







5.0~00
T Lo 0 c '

~jgg^ ^Ssg



0.Dg|0,0 5.a



k 0. L 5 f B

0 cu 0. (D 45
a n^ o-t' ai or
gpp z3 '
C|.s a)s
tsS '^gl^-g0
W w be W w
e~lssilgE.


( 0) .-. C' C- qc O
'0.

eS :a gM3 ES m.E
to o o,
5 ~ 0 640.0o)~yag
S.00.


*U 0)E
0 0 cc CU 10 'U)
5. 0 0 0)U' 0 a) 4E'
0. 05 a. t E
5 -001 0.0L '.4 v m
f0 ok m P gw -W-.t 2



.-4 A'C 'Ui U) L6 .4


g Q< -0 E





ca
Q -. 0'Ea--












0 1 C A.
z zr P < .4 e :0


A~J


I7


-40'01 0.m0 LU0.
r q o r 4)m r
0 bb- > X

CU
;, a M t :z 0 )

o 0 -0 14 = U .=a
DO M.-~
0..290.0 S.2 41 --
o-"60 t' sSa^'3
0 CLB. =, 3
o o F-r Z! h > >>

01
0, E.l-0

E- = 3 j E 0 *S ok E -
02 9 co c S



2- eo w o 'a'
oIF. qS5
LU 0 0 0 .0

t:sc PIst- ~


rL i
aji h. u '" i~ i
025.00; U 0^
uoo
0 2 -
VU bo01 WU z 0
ar. e 2


uI. L -1
0 M- 01 -


0'01.


u S .l~g3-3, <
S o
S. E no
0. 020m ; i< Z 9

Im L z NC P


-?? C'S C-


wu a
01..s. 0 .l
.0
c cc
>a k u
w- 0- w- fn C- M acn t -<
LU |.-|-| ->






4 g-igillvi 4 w
LU ~" 0~S (ULU4 00)




h bCarai.<6Sa Cisi a ,Si 5~$~NE
-. *. 0
0.i cUco UL^ io i-


s


Q 0.'-V


0z 0 m
20 o



0 o.u 0 glo
Sa a





w 0 1.. N M1




to$1 = a, r 00
zSW~W S z I
s;41~ a F.o *


0 .
0 t 0 CU w i r.
5.>, CI w 0CUCA 0
g5 mS~s-s-s
0 uiP


^IIE IEp1
LCU Sv ~
02 W'- 0.'eB -g
c '. c, E (. 0
01r 001
B G k
> %ti
tj.' 00 1 0..
0.v P 5 ; = 5;!



0) .-S 5..g
6 -L W~' (D= L0 0.0

ca

8" |KS 0.S-.00
C'0 (D o CU 0
Q~ CU -c-. 01
6 .06 v


'0 -' LU
w. 0.20L

0 mr0 L a vO)* BS

t1ubf60 ) 0 0 U-.0 0
I S n E-45
*e^dml: $~g
?SiiS-5ESS


r



























Cu









Cu'u
t






u 0nC


0G 0 :. c





ElE
o0

07 o ca

(A 0 4'l
ed 0
o 14 J 11 a





















Cuu















14 -


0 -













I 0
W. A
I KE 0.4 tS a

4 CT CW B w

!: 0.
0 La a)"? '1 ''*



























0 ,

4 1
g S. M


U (UiflG0






-" -S N P
cu-
wu ~y r cr% ^
i Ui Ai -, Ei
E *


5- ia M E e-i
2 o. o E Ui z a C
= I % 0CQ < w
C m C -- 5 '



- NOn 'U ino _


E


: 3 Ei i
Cu NE a -0 c





M2 ZD 0"Il
~Z x itCu

E"; '.%^w w.r


E 2 bE l
0 a
z OCu F ui.
a. V :3 0c"

L 0 tn C.
bLC bl,*C
U V3QS i
U )S..3"


w < T3 L1
LL. 11 a
SSS~s-s',


CC;
ci mo wa"^a 0
cu m 0 c cu C4 r_
o = b- > S
S.. M Er 00 d cc aj ai




GO w EU '0"C;
L.. a, r- O
" 0 W 0 Cm

c a- F C c dm



Cu C i 0 D" 'L ai~~.~C
0j a. 0 E o 0
'' = E 16 0,5.2 Qc i:- 0,- u

z 0 mLn it 3, 0 w r_
w U E -'a c -?,S

5CLC CuCO m1


xcu 1 r- 0 -. -a
0 5 "b 0
o Zn.2 w -X, -0
rn T a a, s a -

Sa C u0 CUo '. W
wu E 6 Cu '2 a, Vj 2
F- = f iSE L ow 5- E


-.~ N .


3=. a M Ms
,Q-- UC'D 0 --
m C <' 0i C .
0 uS Cu CJ N 2 --a
~Ln
E oC0aC
o .- = o .. -' .

U 0 3 ( C C : C : U C
0 o w-i 0 < ,
m 7- N. 430CuC


C
0







0 z u ;8~




N cu *bf,
E; CM E m 3
U E
SZfc .0 E :3
Lu o ra -
-1 'o :Bl'

u a n =L -i
<,~ n 1, '03 u

L; 0 E n 61 1
w (D<> C "'n -
z z P .


' ,


CMu
n CA O r ,*C
2U,0 m '0
C 0 i c U w 0.C

m rE i r

Cu 0,~ E L

.C 0C G .U
Cq CLu .
wu.OE Q.0C'0 Cu
CC **S0.* |a


C CuE..
Gu z '- = Cu
E- -5 0j 0u S u o-

0 O M. C m
Q5 C w 2T 0 .u 0 9


gM_ u .- '.0,^ >


Cu Cu l M -M
k C 'n Q0 3.2 ,i| w. 0 S M
0U* w 1- 3 >air -0
If3, E^ "5 -



W Go b SA ( 6 u.'oi


-- N' t '

'00
a-: -~.i! N" C C<




g g g -= <
0E~7 0 0 w
E 1

^ no nw w -w w

M < = S a ^ <


CK v i 3 1 ; v i : ; Z ? S
o U h
r^ zMIv n h^


0 ;, X 0 Ei-4t!
&'L E O h-



0 IV 0 wQr "t>
oL.J 0p


F E-a4 0 E- I




_%jS to<* C -
2o 20













CU 0 Ci
ZgH v "kB


M^ ." N N* i*o
C' Ci '2i 5 C


0 Cu 0 0)Cuu~
wuw Cu .'mo


> ~ E-1-4
*a4 ca 4) IC
w< 0 o S (D HS ''^\
v i l rnl sl




-4 cz ic L6 ia -a


2

III^




.- -,o.-
~O0 ha q u
.4S S 9 3



'-0 4) (D










,l0 "" C', o 'C




< 1 n1 ,
-0-
0 u. Br



-N CD b, g"
ft w V;,
Zaj 100 s
C)S l.v'
'r 0 1 b B C: Oll-l


2: W[ C 3 ri0 -
z P< -,!


m
ro 'a 0 c S 1 o-







0) 'j; = 0
S(D 0C w C0
Cu.03 C.-.C can."



- 0 C C 0 > u "




20 0 m C ,
(L) 3 1 u *'= %0. .


'p C .0 Cm2~E
ul 4) 0 0 it5 M 0cc



'0 w:S 2 ;

w0. n -0



> Cu cu a, (U
0 0 FO
o ^l2.2: C^-


g" Cut- U iCu w .C" 1 Q
.i0 > Cu i<-S L. 0 ~u-~C
0-i j.Cu ^ l ^ ii
































01 0







CU
>
w .1 C
*S U C










V 6 cr 1-0
L .3



Os
i- a



N
c 5 VU 0
a0 CuO


'E 2o o- E
0 B S0








Z Ct.,>W


Mi


a-




C1 =i 6


0 Ed



'3 o 5 -
C:

- c a L




B.S g ''
=aS !m


0t

C: P=:0 m


I










1-0
a
U ? C >
b% Q .fl( -










a'~, = .N
CD L. 5 -


0) M
CD c i
- 'a Cut1










- 5- 0 Cu

ag.l
5- Cu >^ -oi


0).. Cu 2 a 3
c *o r0>M







U1NIVrbl I UP I- LUHIUA


3 1262 08105 808 2