1
A Brief Overview of Polarized MHO Characteristics in Steady-State
By
Luis G. Pérez
(Draft. For educational purposes only.)
TABLE I. INPUTS TO A CROSS-POLARIZED MHO RELAY
BASIC IMPLEMENTATION OF A POLARIZED
MHO DISTANCE RELAY SCHEME
A polarized MHO relay with circular
characteristic in the impedance plane can be
implemented using an angle comparator (see
Figure 1) whose phasor inputs are [1]:
𝑆̄1 = 𝐼̄ ⋅ 𝑍̄𝑟 − 𝑉̄
𝑆̄2 = 𝑉̄𝑃𝑂𝐿
(1)
(2)
COMP
TRIP
Comparator operates when:
𝑆̄1
−90∘ ≤ 𝑎𝑛𝑔𝑙𝑒 ( ) ≤ 90∘
𝑆̄2
Figure 1. Circular, cosine-type phase comparator
Input S1 is called “operating quantity”. Input S2 is
the polarizing -or reference- quantity. In a typical
distance relay scheme with six elements, voltage
V and current I are chosen according to Table 1
[1,2]. Voltage VPOL is a special characteristic of
the relay and its choice depends on a series of
factors [2,3]. There are different types of designs.
The design shown in Table 1 is known as “crosspolarized”. Complex number Zr is the relay
impedance setting (ohmic reach), which for the
first, instantaneous zone of the relay, is commonly
adjusted to under-reach the protected line’s
remote end.
The apparent impedance is defined, for each
element of Table I, as:
̄
𝑉
𝑍̄ = 𝐼̄ = 𝑅 + 𝑗𝑋
Luis Pérez
Last review: April 26, 2014
𝑉̄
̄𝑉𝑎 − 𝑉̄𝑏
𝑉̄𝑏 − 𝑉̄𝑐
𝑉̄𝑐 − 𝑉̄𝑎
𝑉̄𝑎
b-g
𝑉̄𝑏
c-g
𝑉̄𝑐
𝐼̄
̄𝐼𝑎 − 𝐼̄𝑏
𝐼̄𝑏 − 𝐼̄𝑐
𝐼̄𝑐 − 𝐼̄𝑎
𝐼̄𝑎 + 𝑘̄0
⋅ 𝐼̄𝑟𝑒𝑠
𝐼̄𝑏 + 𝑘̄0
⋅ 𝐼̄𝑟𝑒𝑠
𝐼̄𝑐 + 𝑘̄0
⋅ 𝐼̄𝑟𝑒𝑠
𝑉̄𝑃𝑂𝐿
̄𝑘𝑐 ⋅ 𝑉̄𝑐
𝑘̄𝑎 ⋅ 𝑉̄𝑎
𝑘̄𝑏 ⋅ 𝑉̄𝑏
𝑘̄𝑏𝑐 ⋅ (𝑉̄𝑏
− 𝑉̄𝑐 )
𝑘̄𝑐𝑎 ⋅ (𝑉̄𝑐
− 𝑉̄𝑎 )
𝑘̄𝑎𝑏 ⋅ (𝑉̄𝑎
− 𝑉̄𝑏 )
Note: 𝑘0 is a relay setting called residual compensation
factor. Ires is the residual current: 𝐼̄𝑟𝑒𝑠 = 𝐼̄𝑎 + 𝐼̄𝑏 + 𝐼̄𝑐 .
Values ka, kb, etc., are design constants.
S1
S2
Element
a-b
b-c
c-a
a-g
(3)
For a given fault, quantities V and I take a value
in steady state. The resulting apparent impedance
for each element, defined as in Eq. (3), can be
plotted in the complex Z-plane.
The understanding of this definition of apparent
impedance and the corresponding resulting
impedance plane is fundamental for the
understanding of distance relay analysis. For the
sake of stating this concept as clear as possible,
the apparent impedance defined for each element
is shown in Table II.
TABLE II. APPARENT IMPEDANCE DEFINED FOR EACH
ELEMENT OF TABLE I.
Element
a-b
b-c
c-a
a-g
b-g
c-g
Apparent Impedance 𝑍̄
(𝑉̄𝑎 − 𝑉̄𝑏 )/(𝐼̄𝑎 − 𝐼̄𝑏 )
(𝑉̄𝑏 − 𝑉̄𝑐 )/(𝐼̄𝑏 − 𝐼̄𝑐 )
(𝑉̄𝑐 − 𝑉̄𝑎 )/(𝐼̄𝑐 − 𝐼̄𝑎 )
𝑉̄𝑎 /(𝐼̄𝑎 + 𝑘̄0 ⋅ 𝐼̄𝑟𝑒𝑠 )
𝑉̄𝑏 /(𝐼̄𝑏 + 𝑘̄0 ⋅ 𝐼̄𝑟𝑒𝑠 )
𝑉̄𝑐 /(𝐼̄𝑐 + 𝑘̄0 ⋅ 𝐼̄𝑟𝑒𝑠 )
2
It is important to recall that, in order to supply a
voltage for close-in three-phase faults, many
relays have a type of polarization voltage called
memory-polarizing voltage. For a close-in fault
(case for which voltages depress to near zero
value), this memory polarizing voltage expires
several cycles after the fault occurrence. In some
relays, the memory polarizing voltage is applied
in addition to the steady state polarizing voltage.
This document is not intended to analyze the
transient characteristic of memory-polarized
relays, but the characteristics of the relays in the
complex plane for steady state conditions without
considering the transient effect of the memory
polarizing voltage.
THE CHARACTERISTIC IN THE COMPLEX PLANE
It can be shown that, for a given relay setting Zr
and for a given ZP, Eq. (9) represents a circle in
the Z-complex plane, whose center, ZC, and
radius, r, are given by the following expressions:
1
𝑍̄𝐶 = 2 (𝑍̄𝑟 + 𝑍̄𝑃 )
1
𝑟 = |𝑍̄𝑟 − 𝑍̄𝑃 |
𝐼̄⋅𝑍̄ −𝑉̄
𝑃𝑂𝐿
) = ±90∘
(11)
2
The circle is shown in Figure 2. Point C is the
circle’s center. In terms of Z, the inequality stated
in Figure 1 becomes
𝑍̄ −𝑍̄
)
𝑍−𝑍̄𝑃
−90∘ ≤ 𝑎𝑛𝑔𝑙𝑒 ( ̄ 𝑟
≤ 90∘
(12)
This last equation represents the inside region of
the circle shown in Figure 2, including its border.
X
As defined in Figure 1, the operation threshold for
any of the six elements shown in Table I can be
expressed by the following equation:
𝑎𝑛𝑔𝑙𝑒 ( 𝑉̄ 𝑟
(10)
Q
Zr
(4)
C
The polarizing voltage, VPOL can always be
written in terms of V using the following
expression:
(5)
VPOL = V −V P
P
ZP
R
Where V and Vp depend on the particular case
being analyzed. With this, Eq. (4) becomes:
𝐼̄⋅𝑍̄ −𝑉̄
𝑟
𝑎𝑛𝑔𝑙𝑒 ( 𝑉̄−𝑉
) = ±90∘
̄
𝑃
(6)
Figure 2. Resulting circular characteristic of a MHO
polarized relay. R and X are defined in Equation (3)
Taking I as common factor:
Observe that, for a self-polarized relay:
𝑍̄ −𝑉̄/𝐼̄
𝑎𝑛𝑔𝑙𝑒 (𝑉̄/𝐼𝑟̄−𝑉̄ /𝐼̄)
𝑃
= ±90∘
(7)
Define a new complex number, with impedance
units (ohms), as:
𝑉̄
𝑍̄𝑃 = 𝑃
(8)
𝑉̄𝑃𝑂𝐿 = 𝑉̄ ⇒ 𝑉̄𝑃 = 0 ⇒ 𝑍̄𝑃 = 0
In this case, the circle passes through the origin
and the complex number 𝑍̄𝑟 − 𝑍̄𝑃 is a diameter of
the circle.
𝐼̄
Considering Eq. (3) and this last definition, Eq.
(7) becomes:
𝑍̄ −𝑍̄
𝑟
𝑎𝑛𝑔𝑙𝑒 (𝑍̄−𝑍
) = ±90∘
̄
𝑃
Luis Pérez
Last review: April 26, 2014
(9)
FAULT ANALYSIS
In order to examine the relay performance, the
behavior of the six elements for different kinds of
3
faults can be analyzed in the complex plane
shown in Figure 2. The impedance represented in
that plane is the one defined in Equation (3) and
Table II. In other words, the apparent impedance
is different for each element for a given fault. Let
us study the behavior of element b-c for the case
of a b-c fault. For this element, the comparator’s
inputs are:
𝑆̄1 = 𝐼̄𝑏𝑐 ⋅ 𝑍̄𝑟 − 𝑉̄𝑏𝑐
𝑆̄2 = 𝑘̄𝑎 ⋅ 𝑉̄𝑎
(13)
(14)
Thus,
𝑉̄𝑃 = 𝑉̄𝑏𝑐 − 𝑘̄𝑎 ⋅ 𝑉̄𝑎
(15)
and
̄
̄
̄
𝑉
𝑘 ⋅𝑉
𝑍̄𝑃 = 𝐼̄ 𝑏𝑐 − 𝐼𝑎̄ 𝑎
𝑏𝑐
𝑏𝑐
(17)
𝐹1
(21)
Luis Pérez
Last review: April 26, 2014
(22)
For this design, Eq. (21) becomes
̄
̄
̄
𝛽 ⋅𝑚⋅𝑍𝐿1 +2𝑍1
𝑍̄𝑃 = 𝑚 ⋅ 𝑍̄𝐿1 − 𝛽̄+2𝐶
̄
(23)
or
̄
̄
𝛽
2𝑍1
𝑍̄𝑃 = (1 − 𝛽̄+2𝐶̄ ) ⋅ 𝑚 ⋅ 𝑍̄𝐿1 − 𝛽̄+2𝐶
̄
1
1
(24)
𝑍̄𝑃 = −𝑍̄𝑆1
𝑘̄𝑎 ⋅ 𝑅𝑓
𝛽̄ + 2𝐶̄1 (−𝑗√3)(𝛽̄ + 2𝐶̄1 )
𝑘̄𝑎 ⋅ (𝛽̄ ⋅ 𝑚 ⋅ 𝑍̄𝐿1 + 2𝑍̄1 )
−−
(−𝑗√3)(𝛽̄ + 2𝐶̄1 )
−
𝑘̄𝑎 = −𝑗√3
(20)
Substituting Eqs. (17)-(19) in Eq. (16):
𝑅𝑓
In order to suppress the dependency with the fault
resistance, the relay can be designed with:
(19)
This is the ratio of the pre-fault current to the
positive sequence fault current. It is an indicator
of how significant the pre-fault line current is with
respect to the severity of the fault.
𝑍̄𝑃 = 𝑚 ⋅ 𝑍̄𝐿1 +
DETERMINATION OF THE DESIGN CONSTANT:
THE FAULT-DEPENDENT CHARACTERISTIC
The dependence with the load and the fault
location is evident in this equation. In the extreme
case of not having pre-fault load current (β=0),
with the notation of the Appendix:
In these expressions:
𝐼̄
𝛽̄ = 𝐼̄𝑃𝐹
This means that the radius and the center of the
circle shown in Figure 2 both change according to
those factors.
1
𝑉̄𝑏𝑐 = −𝑗√3 ⋅ [𝑚 ⋅ 𝑍𝐿1 ⋅ (𝛽̄ + 2 ⋅ 𝐶̄1 ) + 𝑅𝑓 ] ⋅ 𝐼̄𝐹1
(18)
𝑉̄𝑎 = (𝛽̄ ⋅ 𝑚 ⋅ 𝑍̄𝐿1 + 2 ⋅ 𝑍̄1 + 𝑅𝑓 ) ⋅ 𝐼̄𝐹1
a) The pre-fault load ()
b) The fault location (m)
c) The fault resistance Rf
(16)
For this fault, the voltages and currents can be
found using the sequence network connection
shown in the Appendix. The results of interest are:
𝐼̄𝑏𝑐 = −𝑗√3 ⋅ (𝛽̄ + 2 ⋅ 𝐶̄1 ) ⋅ 𝐼̄𝐹1
Notice from Eq. (21) that ZP depends on:
(25)
NUMERICAL EXAMPLE
The statements demonstrated in the previous
paragraphs can be illustrated with some fault
examples on the simple system shown in Figure
3. Table III shows the five simulations performed
on the system. Figure 4 represents the b-c
polarized MHO element characteristic (ZP as in
Equation (24)) for two different fault locations
and a certain fixed pre-fault condition. Figure 5
shows the characteristic of the same element for a
fixed fault location and for two different pre-fault
conditions. Figure 6 compares the characteristic
for a forward fault and the characteristic obtained
4
for a reverse fault. Notice that point Q is fixed for
all conditions, regardless of the ZP value.
Cases 1 and 2
12
10
TABLE III. FAULT CASES. (ANGLE SR) IS THE ACTUAL LOADFLOW ANGLE BETWEEN THE VOLTAGES VS AND VR AT BUSES
S AND R, RESPECTIVELY.
1
2
3
4
5
SR
(deg.)
m
(p. u.)
Ipf
(A)
Re{S}
(MW)
8
8
0
10
0
0.67
0.10
0.67
0.67
(-0.3)
755
755
0
942.5
0
138.4
138.4
0
166.1
0
X in primary ohms
Case
Q
8
6
4
2
0
-2
-4
-10
Note: SR is the angle that the voltage at bus S leads the
voltage at bus R.
𝑍̄𝐶
𝑟
1
4.413
+7.439j
1.717
+1.085j
4.613
+7.887j
4.356
+7.339j
-2.79
-1.193j
-1.66
-5.51j
-1.25
-5.76j
-0.8
-6.0j
-1.85
-5.38j
0.58
+5.67j
0.131
+2.042j
0.333
+1.92j
0.56
+1.8j
0.036
+2.108j
1.252
+7.633j
7.766
7.842
7.918
10
S
I
Line 1
Forward
Fault
2.078
2
0
-4
-10
-5
Positive sequence impedances of the 115 kV Line and
Equivalent System Data:
𝑍̄𝐿1 (Line 1) = 0.8 + 4𝑗 Ω
𝑍̄𝐿1 (Line 2) = 2.4 + 12 Ω
𝑍̄𝐻1 = 𝑍̄𝐻2 = 2𝑗 𝛺
𝑍̄𝑅1 = 𝑍̄𝑅2 = 3𝑗 𝛺
𝑉𝑆 = 1.0𝑝. 𝑢. ; 𝑉𝑅 = 0.99 𝑝. 𝑢.
Figure 3. Example system
5
10
Cases 3 and 5
12
R
Line 2
0
R in primary ohms
Figure 5. Cases 3 and 4(---)
10
Relay
Luis Pérez
Last review: April 26, 2014
4
-2
Q
8
X in primary ohms
Reverse
Fault
6
7.724
Notes: a) 𝑍̄ is the apparent impedance as defined in Eq. (3)
and Table II, using RF =2 ohms. It is represented with the
symbol “*” in Figures 4 to 6. The point represented with a
“o” indicates the corresponding 𝑍̄ for a bolted fault at the
same location. b) All values are in primary ohms.
H
Q
8
X in primary ohms
𝑍̄𝑃
5
10
Cases 3 and 4
𝑍̄
4
5
12
Case
3
0
R in primary ohms
Figure 4. Cases 1 and 2 (----)
TABLE IV. RESULTS
2
-5
6
4
2
0
-2
-4
-10
-5
0
R in primary ohms
5
10
Figure 6. Cases 3 and 5(---)
The fact that the characteristic depends on the
fault conditions implies that engineers cannot plot
a fixed characteristic for all faults, as in the case
5
of a self-polarized relay. Thus, when presenting
the results of a test on a relay polarized MHO
element, the fault and load conditions for which
the presented characteristic is valid must be
clearly stated. There are methods to represent a
fixed characteristic of polarized relays. This can
be achieved re-defining the apparent impedance
(and the corresponding Z-plane) such that the
resulting circle is fixed for all faults conditions
[4].
the pre-fault load conditions. This dependency is
observable even for those cases where the relay
has been designed to suppress the dependency
with the fault resistance.
Engineers must be careful when the characteristic
of a polarized MHO element is presented. The
fault and load conditions must be clearly specified
in each case.
REFERENCES
THE OTHER ELEMENTS IN TABLE I AND SOME
COMMENTS ABOUT DIFFERENT POLARIZING
SCHEMES
The same classic procedure used to analyze the bc element can be used to analyze and choose the
design constants for the other five elements
shown in Table I [1-5]. This is left as an exercise
to the reader.
The analysis technique can also be applied to any
type of voltage-polarized relay. The goal is to find
ZP for the particular polarization voltage being
used. In general:
̄
̄
̄
𝑉
𝑉−𝑉
𝑍̄𝑃 = 𝐼̄𝑃 = 𝐼̄𝑃𝑂𝐿
(26)
VPOL can be any of the known polarizing voltages
used by different manufacturers. For example, for
a b-c element, VPOL can be a combination of
positive sequence voltage Va [5] and a sample of
this voltage a half a cycle before (dual positivesequence and memory polarization scheme). In
this case ZP changes in time because part of the
polarizing voltage changes (expires). However,
the steady-state characteristic would remain as
defined by the steady-state value of the positive
sequence voltage.
CONCLUSION
For a plot in an apparent impedance plane, as
traditionally defined, the size and position of the
steady-state circular characteristic of a polarized
MHO element depends on the fault location and
Luis Pérez
Last review: April 26, 2014
[1] A. R. Van C. Warrington, Protective Relays: Their
Theory and Practice. Vols. One and Two. John
Wiley & Sons, N. Y., 1969.
[2] Wedepohl, L. M. “Polarised Mho Distance
Relay,” Proceedings of IEE, Vol. 112, No 3,
March 1965, pp. 525-535.
[3] Cook, V. Analysis of Distance Protection, R.S.P.
and J.W.& Sons, Inc. Great Britain, 1985.
[4] Martilla, R. J., "Directional Characteristics of
Distance Relay Elements,” (Parts I and II), IEEE
Transactions on Power Apparatus and Systems,
Vol. PAS-100, No. 1, 1981, pp. 96-113.
[5] Roberts, J.; Guzman, A., Schweitzer, E. O.
“Z=V/I does not Make a Distance Relay,”
Proceedings of WPRC, 1993. SEL paper 6022.pdf.
6
APPENDIX
ANALYSIS OF A B-C FAULT
Current divider factor:
𝑍̄𝑦1
𝑥1 +𝑍̄𝑦1
𝐶̄1 = 𝑍̄
Forward
Fault
(A4)
Thevenin voltage at the fault point (open circuit
voltage at the fault point before the fault occurs):
𝐸̄𝑇𝐻 = 𝐶̄1 ⋅ 𝐸̄𝑆 + (1 − 𝐶̄1 ) ⋅ 𝐸̄𝑅
Relay
(A5)
Positive sequence fault current:
a) One-line diagram
̄
𝐸
𝐼̄𝐹1 = 2𝑍̄ 𝑇𝐻
+𝑅
1
ES
ER
Z S1
Z R1
m  Z L1
(A6)
Pre-fault current through the line:
Z S1
V1
I1
𝑓
I2
(1 − m)  Z L1
m  Z L1
I F1
̄
̄
𝐸 −𝐸
𝐼̄𝑃𝐹 = 𝑍̄ 𝑆 +𝑍̄𝑅 = 𝑍̄
Z R1
𝑥1
V2
𝑦1
𝐸̄𝑆 −𝐸̄𝑅
𝑆1 +𝑍̄𝐿1 +𝑍̄𝑅1
(A7)
Positive and negative sequence currents measured
by the relay:
(1 − m)  Z L1
Rf
𝐼̄1 = 𝐼̄𝑃𝐹 + 𝐶̄1 ⋅ 𝐼̄𝐹1
(A8)
𝐼̄2 = −𝐶̄1 ⋅ 𝐼̄𝐹1
(A9)
b) Sequence diagram
Figure A1. Sequence network connection
Positive and negative
measured by the relay:
sequence
voltages
V1 = m  Z L1  I1 + ( Z1 + R f )  I F 1
The usual notation [1] is employed.
= m  Z L1  ( I PF / I F 1 + C1 )  I F 1 + ( Z1 + R f )  I F 1
Sending-side total impedance:
𝑍̄𝑥1 = 𝑚 ⋅ 𝑍̄𝐿1 + 𝑍̄𝑆1
(A1)
𝑉̄2 = [−𝑚 ⋅ 𝑍𝐿1 ⋅ 𝐶̄1 + 𝑍̄1 ] ⋅ 𝐼̄𝐹1
Receiving-side total impedance:
𝑍̄𝑦1 = (1 − 𝑚) ⋅ 𝑍̄𝐿1 + 𝑍̄𝑅1
𝑉̄1 = [𝑚 ⋅ 𝑍𝐿1 ⋅ (𝛽̄ + 𝐶̄1 ) + (𝑍̄1 + 𝑅𝑓 )] ⋅ 𝐼̄𝐹1
(A10)
(A2)
Where
𝐼̄
𝛽̄ = 𝐼̄𝑃𝐹
Parallel equivalent impedance for positive (and
negative) sequence:
̄
̄
𝑍𝑥1 ⋅𝑍𝑦1
𝑍̄1 = 𝑍̄ +𝑍̄
𝑥1
Luis Pérez
Last review: April 26, 2014
𝑦1
(A3)
(A11)
𝐹1
Phase currents:
(A12)
7
𝐼̄𝑏𝑐 = 𝐼̄𝑏 − 𝐼̄𝑐 = −𝑗√3 ⋅ (𝐼̄1 − 𝐼̄2 )
𝐼̄𝑏𝑐 = −𝑗√3 ⋅ (𝛽̄ + 2𝐶̄1 ) ⋅ 𝐼̄𝐹1
(A13)
Phase voltages:
𝑉̄𝑎 = 𝑉̄1 + 𝑉̄2 = (𝛽̄ ⋅ 𝑚 ⋅ 𝑍̄𝐿1 + 2𝑍̄1 + 𝑅𝑓 ) ⋅ 𝐼̄𝐹1
(A14)
𝑉̄𝑏𝑐 = 𝑉̄𝑏 − 𝑉̄𝑐 = −𝑗√3 ⋅ (𝑉̄1 − 𝑉̄2 )
𝑉̄𝑏𝑐 = −𝑗√3 ⋅ [𝑚 ⋅ 𝑍̄𝐿1 ⋅ (𝛽̄ + 2𝐶̄1 ) + 𝑅𝑓 ] ⋅ 𝐼̄𝐹1
(A15)
Luis Pérez
Last review: April 26, 2014