End effect in magnetohydrodynamic ducts with nonconducting partition

Lavrentyev, I. V.

English

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https://ntrs.nasa.gov/search.jsp?R=19690003819 2020-03-12T08:08:30+00:00Z
t	 14w
International Atomir energy t_¢ency
SYMPOSIUM ON THE PRODUCTION OF ELECTRICAL ENERGY BY MEANS OF MHD-GENERATORS
Warsaw, 24-30 July 1968
I
N 69 - 13 150  SM-107/137
END EFFECT IN MHD DUCT$ WITH NON-CONDUCTING PARTITIONS
by
I. V. Lavrentyev
D.V. Efremov Research Institute for Electrophysical Equipment,
Leningrad, USSR
As has been shown previously [ref. #1, 2, 3], the introduction into the
end zones of an MHD duct of non-conducting partitions whose plane is
parallel to the induction vectors of the magnetic field and the veloci_cy
of the fluid improves its electrical characteristics (increasing the
efficiency and the output voltage and raising the power to be extractedl.
For the purpose of qualitative evaluation of the effect of introducing
the partition on the integral characteristics of a duct, two problems
were solved whose results are also presented in the report.
1. Let us examine a flat duct of constant cross section with insulating
walls and two electrodes MN and KL (figure l.a).
A (n - 1) and a (m - 1) partition is introduced respectively into the
duct on the left and on the right of the electrodes. It is assumed
that the partitions are non-conducting, infinitely thin, and one end
approaches infinity (points Ak and Ci in figure l.a). We shall assume
that the electrical conductivity of the medium ^ is constant. The
distribution of velocity is given in the form V = (V, 0, 0) (V = const),
and the magnetic field is expressed by the relationship
B = (0, 0, -B(x))	 (1.1)
B (x) =BoB (x), D(x) -►
 0	 when x -► -
Then at small magnetic Reynolds numbers for determining the electrical
characteristics of the duct in question, it is necessary to solve the
following system of equations and boundary conditions [ref. #4]:
,jx	 - Qa , i — -aay + VB(x), V2(P - 0	 (1.2)
(P _ ?1	 at electrode MN
9 _ 92	 at electrode KL
VB(x) at the insulating walls and partitions.
Here j is the density of the current, 9 is the electrical potential,
91 and ?2 aro potential values at the electrodes, We shall introduce,
into the band with slots along the partition corresponding to the exterior
	
1	 ^ - C) 3
t' 7c- ^ f i n 7 C- 7 f'.'l .,^ .. I S/ J N L /^ ^ 4 A J	 9 6^ ^^^ io r^ 9 7^^L'
of the duct, an analytic function of complex variable z = x + iy
F= y+i4i
	 (1.3)
where ^ is a function of the current that satisfies the ratios:
(1.4)
ax	 ay., ay ax
Let us further represent the indicated band of plane z as the upper half-
plane of plane t = T + iv with..the aid of the conform conversion [ref.#5]:
z = f(t) = n hk ln(t - a k ) - M ki In (t - ci ) + C	 (1.5)
7T	 Trk=1	 i=1
where the designations hk and ki are clear from figure l.a, and ak and
ci are point coordinates on the T axis, corresponding to points Ak and Ci
of plane z. The points on the T axis with the coordinates bj and d8
which are found from equation
hk 
-I	
^Z 
=0
k=1
-ak i=1 t-ci
correspond to the ends of the partitions (points Bj and DS of plane z).
The parameters ak, b • , ci, ds, and C are defined from the values of
Re f(d^j) and Re f(ds^ known from the assignment of the duct geometry
and its position relative to the selected system of coordinates. Three
parameters (ak, bj, ci, ds) are arbitrarily selected. If one of the
parameters ak or ci is equal to -, the formulas remain in force if the
corresponding components are selected If one of the parameters b • or
ds is equal to -, then the formulas dt- not change. Without disrupting
the generalities, it is possible to consider that the points (-1, -k,
k, 1) (figure l.b)(k < 1) will correspond to the points of the electrodes
on the T axis.
Problem (1.2) can be solved by various methods. The first method in-
cludes the fact that we initially represent the upper half-plane by
a rectangle, and then we solve the Laplace equation for the potential
under the condition that the value of the potential is given on two
sides of the rectangle, and the value of its normal derivative is on
the order of two. The solution of the problem is obtained then in the
form of a trigonometric series. Thus, in the paper [ref. #61, the
problem was solved for a duct without partitions. It should be noted
that this method of solution is suitable only in the case when there
is a total of 2 electrodes. The second method is included in the use
of the Keldysh-Sedov formula for this problem. Actually, problem (1.2)
is reduced to finding the following analytic function in the upper half-
plane
F1 (t) = F[z (t) ] = (p i + i4j	 (1.7)
whose real part is assigned in the sectors ITIe(k,1) and whose imaginary
part in the remaining section of the T axis. Since F 1 (t) is limited on
the T axis and it is always possible to define it so that it disappears
at infinity (t	 then the solution is written in the following manner[ref. #71:
2
-k	 1
fR(T)(T-t)F1 (t)° R2 	 1 IR( T)TT-) 	 +
-1	 k	 (1.8)
k	 -1
R(t)
fR(T)
l ( -r)d -r	 r V(T)d -r	 rl/(T)d-r
Tr 
	
(T-t) + R (T) (T-t) + R (T) (T- t)
-k	 -0	 1
In fulfilling the conditions of solubility:
(1.9)
f
kTi -1dT + (P T
J-1 dT + 
z	
k(T)TJ -1dT 
+-
f*(T)T`^- 1dT +	 (-r) -r -1dT = 0
R T	 f R T	 j	 R T)	 R(•c)	 f	 R(T)
-1	 k	 -k	 -CO	 1
where j = 1, 2; R(t) _ (tT )(t+k)(t-k)(t- ), and the values of the roots
are considered positive when T > 1. Condition (1.9), when j = 2, defines
91 and 92 at a given difference of potentials at the electrodes 91=Q2=U
and when ,j = 1 yields the expression for the total current at the elec-
trodes. The last condition (j = 1) makes it possible to easily find
one of the basic values that determines the electrical characteristics
of the duct, i.e., the dimensionless integral resistance of the duct Q-1.
Actually, assuming ^ = 0 in the sectors k k (1, -) and	 coast in
the sector Tc(-k, k) from (1.9) when j = 1, we find
0 = K(k')12K(k)	 (1.10)
where K(k) is the total elliptic integral of the first kind, k' = vr1_-_k_7.
The parameter k (see figure l.b) is defined by the formula of the con-
form conversion of the internal duct to the upper half-plane v > 0 and
is only a function of the duct geometry. Let us note that formula
(1.10) is justified for a duct of arbitrary form having 2 electrodes,
it is necessary only that its exterior be represented as an upper half-
plane v > 0; then the points corresponding to the ends of the electrodes
will go over to points T1, T2r T 3 , and T 4 of the T axis, and then by
means of an additional linear conversion of them, it is always possible
to transfer them to points (-1, -k, k, 1).
2. Let us now examine the case of the distribution of partitions of
practical interest. The geometry of the duct in question is clear in
figure 2.a. In order to simplify the solution, we shall consider that
the electrodes are semi-infinite, i.e., we shall examine the left half
of the real duct as having sufficiently long symmetrically distributed
electrodes. In practice, the formulas obtained below can be used when
c > 1, where c = 2a/H, 2a is the length of the electrodes, H is the
width of the duct. Actually, as was shown [ref. #6] for a duct without
partitions, formulas that take into consideration the influence of the
end effects on the electrical eharacteriFtics of the duct when e > 1
in practice do not differ from the corresponding formulas obtained in
the assumption that the electrodes are :.nfinitely long. In our case
here, the introduction of partitions ir.co the duct can only reduce the
3
scale of the zone of inhomogeneity of the electrical field at the
input or output of the duct, therefore, we shall consider that all
formulas that will be obtained below are of course justified when
c > 1. Thus, the problem is reduced to the solution of the system of
equations of (1.2) with the following boundary conditions:
0	 when	 J x J -►
a
(P - 
0 when x = = ^, (- U12 when y= H, u< x < -y (2.1)
a9 - H wh en x = ^, 9 _ -U/2 when Y=O, u < x <00y
a9 - VB(x) on the insulating walls and partitions.
ay
B (x) - 0 when x -► -^
An example of this problem can be conveniently demonstrated by yet another
method of solution.
Let us introduce the analytic function
	
f (z) = p + iq = ay + iax	 (2.2)
Then problem (2.1) is reduced to finding this function that satisfies
the boundary conditions: q = 0 on the electrodes and p = VB(x) on the
insulating walls and partitions. In order to solve the formulated
problem, we shall represent conformally the band JxJ < -, 0 < y < H
with slots along the partitions as the upper half-plane of plane
t = T + iv (figure 2.b) by means of the following formula (ref. #51:
	
z - k Tr 
In Tn (t)	 (2.3)
where
Tn(t)= 2 I(t + ^) n
 + (t - ^)n]
is Chebyshev's polynomial.
Here points (Tk, 0) and (Tm, 0) of plane t will correspond to points
Dk and Bm of plane z, where
Tk — cos ek, ek 
— 2n + n^
^ Tn(Tk) — 0, k - 1, ..., n
(2.4)
TM — cos em, em	
mn^ dTndTm 
m 0, m = 1, ..., n-1
The correspondence of the remaining points is clear from figure 2.a,b,
and the value a is found from the equation: 1
	
Tn (a) - exp I
	 g J	 (2.5)
The analytic function f(z) from (2.2) in the region of the complex
variable t will change into function
4
f l (t) = p i + iq
	 (2.6)
satisfying the boundary conditions:
p = VB (T) at A_A + Tc (-a, a)
q = 0 at CA_ and A+C , a < T <
	
(2.7)
f i (Tk) = 0 and f l (-) = U/H
Thus, we have come to a mixed boundary value problem for the harmonic
function f l , while f l has the properties of a polar type at points
corresponding to the ends of the partitions. We shall first plot the
function f l o corresponding to theuniform problem (p = 0 at A_A+). It
is easy to see that
n-i
f^ o	 HgU^ + 11 ( t ) ^Yo +m = 1tYT r^	 (2.8)
where
g M- /t-a3 -t -+a ,	 1I (t)= t-a t+a
In order to obtain a solution of the initial problem, it is sufficient
to add some special solution of the non-uniform problem (2.7) to flo.
As this special solution one can take the function [ref. #71:
a	 _
f ll a VB°g(T)B(T)dT	 (2.4)
2 Tr t	 T - t
-a
then, finally, we shall obtain the following expression for f l solving
the initial problem:
T) B(T)dT	 U	 1	 n-1
f' (t)	
VB o	 g(
- zTrg (1) j T - t	 + Hg t + ^^ Yo +m^1ta	 , I	 (2.10)-	 m1
The constants Yo and Ym are found from the condition f l (Tk) = 0, i.e.,
they satisfy the system of equations:
n-1
YO + I Ym -- - (J - k) (a + TO + F k	 (2.11)
m=1 Tn - Tm
wnere Yo - Yo/B o V, Ym - Ym/B 0 V k - U/HBoV is the coefficient of load
Tr	 I
J	 oB	 (8) d8, Fk = a co^ 8k o B (0) ctg (s-0)do	 (2.12)Tr
cos $ - T/a, cos 9k - T k /a, b(0) - B [x ($) ]
From (2.10) with the use of (2.11) and (2.12) we obtain the following
expression for the current extracted from the electrodes:
5
	I - Q g (cG, - kO + $1)
	
(2.13)
where	 = BoVH is the Faraday e.m.f.,
a+u
	
G 1
	a-1 I B(x)dx	 (2.14)
P
=c+ 2 In 2 + 2u-2 Ina
Tr	 H	 7T
a
s l ° n42 f arc sin (a) B(T)d 1nTn(T)	 (2.15)
0
Then the electric power taken from the electrodes will be equal to:
P1 - k9I
	
(2.16)
The electromagnetic power developed in the duct is defined in the fol-
lowing manner:
u+X	 H	 X
	 (2.17)
r	 z
P 2	 2 I dx f dy (j x B)V = 2 H^ f B(x)dx - G &2	  + fS VB(x)g(x)dx,
00	 0	 -00
where IS designates the integration along the insulating walls and both
sides of the partitions. Omitting the intermediate calculations, we
shall present the final expression for the power developed
P 2 = c& 2 [c(G 2 - kG l ) - kal + R2].
	
(2.18)
where	 X
 f
u
G 2 = ^' 1 
	
B2 (x)dx
u
2
 - H
u	 Tr	 Tr	
(2-^]	
rl(2.19)
f B2(x)dx - 
n 2 Tr 3
 j d9 I B 	 dlnTnl2^
J j 2 1nTn20	
8 
	 t	 o 	 l
No-ce that the special integral with Hilbert's nucleus enters into the
expression for $2- It is possible to calculate this integral for the
aarbitrary function B(x) by expanding the function
f(8) = B(8) d1nTn(0/2)
2	 d8
into the Fourier series
00
	
f(e) - I bm sin m8	 (2.20)
M=1
where
6
Tr
bm 	J B (0/2) si.n r;is dlnTn (s/2)	 (2.21)
0
Substituting (2.20) into (2.19) and integrating, having used the
relationship [ref. #51:
2 Tr
lI sin m8 1 ctg 
012$ d$1 — 27 cos m8,
b
we shall obtain
u
S2	 H j B 2 (x)dx — n2 IT I A m ,.	 (2.22)
m=1
where	 Tr 1 2
Am = 
J 
B(8) sin 2m8 d1nTn(8)
0
When n = 1 (duct without partitions) expressions (2.15) and (2.22) for
a1 and s2 coincide with the corresponding expressions obtained in a
previous paper [ref. # 6], and when n = 2,those obtained in [ref. #81
If the partitions approach right up to the electrode zone (u = 0), then
$1 = Sl o/n 	and	 $2 — $2o/n	 (2.23)
where Q10 and a 20 are the corresponding [ appropriate] coefficients for
the subduct (width H/n).
Let us examine the practically important case fo the distribution of the
magnetic field
1 when 0 < x < U
	
B(x) _	 (2.24)
0 when	 x < 0
then ( 2.15) and ( 2.19) will assume the form:
0/
01 i 2u + nn2 J In Tn(o)do,	 (2.15')
arc cos
1/a
0 In Tn (8) do
a 2 — 
u	
arc sin 
a 
+ ( 1 —- 2^ arc sins)$1 - n 4	 ram	
1 J 
a2cos28-
arccos
1/a
n	 F
— n2^  J In Tn (8) k^1a cos 8 kcos sk d8
arc cos
1/a
Figures 3 and 4 show the dependence of $1 and $2 on u/H at various values
of n calculated according to formulas (2.15 1 ) and ( 2.19 1 ). From formulas
(2.19')
7
(2.15 1 ) and (2.19 1 ), it is possible to oL-tain an expression for the
Joulean losses Q in the case where all walls of the duct are insulating,
by going over to the limit when u/H -►
 -
  (u	 here
Q
	
(2.25)
where i
	
4	 1n2Tn 1 /x) dx 4 n-1	 ' 1n 2 Tn (1 /x) d,S - lim (Bz-^i) - 	
_ x
	
+^	 I Yml J	
-T x
U/H- w	 m=1	 °	 m
n-1	 (2.26)
1 Yml/ Tk - Tm	 - sin Tk arccth1Tk1-'
m=1
As can be easily confirmed, when n = 1 and n = 2 (2.26) coincides with
the corresponding expressions obtained in the papers [ref. ; 9`, 101,
the only difference being that the value $ from (2.26) is twice as
large as in [ref. #9, 10], because we take into account the total
losses at the input and output of the magnetic field. Figure 5 shows
the dependence of S on the number of partitions n calculated according
to formula (2.26).
3. All the above examined problems pertain to the case wher. the parti-
tions have infinite length. In actual constructions the partitions can
not be infinitely long, moreovc;r, in order to reduce the hydraulic
resistance in the duct they are rationally made as short as possible.
If several partitions of finite ?ength are introduced into a duct,
then it is not possible to obtain an analytic solution of the correspond-
ing problems by virtue of the greater mathematical difficulties connected
with the fact that the region corresponding to the interior of the duct
becomes multiply connected. If then one pair of partitions of finite
length is introduced into the duct symmetrically with respect.to  the
electrodes, then using the conditions of symmetry, it is possible to
so!-re the problem for a single connected region. We shall present some
results cf the solution of such a problem. The geometry of the duct is
clear from figure 6.a, the electrodes are considered semi-infinite, the
assumed tolerances are the same as at point 1.
Thus, it i:i necessary to solve system (1.2) with the following boundary
conditions (figure 6.a):
a - °, ay - 0 when x - -
	
aq
ax — ° ►
 ^ —
	
when x —
Y
a- VB(x) when- - < x < 0, y - 0
	
- u2 < x < Ui y - 6	 (3.1)
- 0 wheny - 6, - - < x < -P2, -PI < x <
9 - -U/2wheny - 0, 0 < x < m.
Introducing the analytical function f(a)-- + ti^ - p + iq and represent-
ay
8
the band IxI	 < -,	 0 < y < 6 of plane z by means of formula t = exn(nz/d)
on the upper half-plane of plane t = T + iv(figure 6.b) we come to the
following problem:	 to find the function f l (t)	 = f[z(t)] = p 1 + iq that
satisfies the boundary conditions
p l	 = VB(T)
	 at	 B 1 A 1	 and O1D1
q1	 - 0	 at	 C 1 B 1,	 A 1 0 1,	 and D1C1.
Having applied the Keldysh-Sedov formula [ref. #71, we shall obtain the
following expression for the function f1:
-m	 1
fl (t) U	 + h(t)	 +	 VB0 jB(T)9'(T) Ct T +
J
_
(T)B(T)9d (3.2)
dg	 t	 Tr 2g 	 t  T— t T- t
where
/t+m) t- ) h(t) -
	
Yo + Y1tg(t)
	 3 	 t(t+R)	
t+m) (t+k) t t-1)
m = exp ( - Tr U2 / 6 ),	 k = exp ( - 1r U1/ 6 ),	 v = U2 - U1
The roots are considered positive when T > 1. From the boundary con-
ditions of (3.1) it follows that Y = 0. The constant Y 1 is defined
form the condition that the increase of the potential along the partition
is equal to 0.
Formulas for R1, S2, and 0 obtained from (3.2) at an arbitrary position
of the partition are very cumbersome, therefore, they are not presented
in the report. The basic conclusion which can be drawn from the analysis
of the obtained solution includes the fact that the values of the
coefficients $1, $2, and 0 quite slowly (-6/v) approach their correspond-
ing values for the semi-infi,iite partition. For example, when u 1 = 0
and v/d > 2,
( 12 In 2	 4^)J -1	 (3.3)
l	 Tr	 -3-j
where p. is the integral conductivity of the duct with a semi-infinite
partition.
A similar solution of the problem when 1j1 = 0 where the magnetic field
is absent beyond the electrodes is presented in the paper [ref. #11).
OFg
References
1. Sutton, G.W., Vista Astronautics, 3, 53, 1960.
2. Vatazhin, A.B., Nyemkuva, N.G., PMTF, 2, 40, 1964.
3. Lavrent'y r^v, I.V., Magnetic HydroTy—namics (Magnitnaya
Cidrodinamika), 3, 153, 1967.
4. Vatazhin, A.B., Regiser, S.A., PMM, 3, 548, 1962.
5. Volkovyskii, L.I., Lunts, G.L., Aramanovich, I.G.,A Col-
lection of Problems on the Theory of Functions of Complex
Variables (Sbornik zadach po teorii funktsii kompleksnogo
peremennogo), Fizmatgiz (Translator's note: Publishing
House for Material in Physics), 1960.
6. Sutton, G.W., Hurwitz, H. Jr., and Poritsky, H., Commun.
and Electron., 58, 687, 1962.
7. Muskhelishvili, N.I., Singular Integral Equations (Sin-
gulyarnyye integral'nyye uravneniya) Fizmatgiz, 1962.
8. Lavrent'yev, I.V., Magnetic Hydrodynamics, 4, 89, 1967.
9. Vatazhin, A.B., PMTF, 5, 59, 1962.
10. Vatazhin, A.B., PMTF, 4, 122, 1964.
11. Lavrent'yev, I.V., Magnetic Hydrodynamics, 1, 1968.
Figures
End Effects in a MHD Duct with Nonconducting Partitions.
Fig.l. Schematic of a duct and conforming reflection.
Fig.2. Schematic of a duct and conforming reflection.
Fig.3. Dependence of the coefficient of S, on u/H.
Fig.4. Dependence of the coefficient of a 2 on	 u/f.'.
Fig.5. Dependence of the coefficient of a on the number of partitions.
Fig.6. Schematic of a duct with conforming reflection.
ti


End effect in MHD ducts with non-conducting partitions

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