BASIC PRINCIPLES

Power In Single-Phase AC Circuit
Let instantaneous voltage be
v(t)=Vmcos(ωt+θv)
 Let instantaneous current be i(t)=imcos(ωt+θi)
 The instantaneous p(t) delivered to the load is
p(t)=v(t)i(t)=Vm imcos(ωt+θv) cos(ωt+θi)
 p(t) = |V||I| cosθ[1+cos2(ωt+θv) ] +
|V||I| sinθsin2(ωt+θv),
where | V |= Vm / 2 , | I |= I m / 2
 The first term is the energy flow into the circuit
 The second term is the energy borrowed and
returned by the circuit

Energy Conversion Lab
BASIC PRINCIPLES

Power In Single-Phase AC Circuit







p(t) = |V||I| cosθ[1+cos2(ωt+θv) ]
= |V||I| cosθ+ |V||I| cosθ cos2(ωt+θv) ]
The first term is the average power delivered to the
load
The second term is the sinusoidal variation with twice
of source frequency in the absorption of power by the
resistive load
power absorbed by the resistive component is active
power or real power
|V||I| is called apparent power in VA
P=|V||I| cosθ is real power in watts, cosθ is called
power factor
when current lags voltage, cosθ is lagging, and vice
versa
Energy Conversion Lab
BASIC PRINCIPLES

Power In Single-Phase AC Circuit





pX(t) = |V||I| sinθsin2(ωt+θv),
This term is the power oscillating into and out
of load of reactive element (capacitive or
inductive)
Q=|V||I| sinθ is the amplitude of reactive
power in “var” (voltage-ampere reactive)
For inductive load, current lagging voltage, θ
= (θv-θi) > 0, Q is positive
For capacitive load, current leading voltage, θ
= (θv-θi) < 0, Q is negative
Energy Conversion Lab
BASIC PRINCIPLES

The Instantaneous Power Has Some
Characteristics




For a pure resistor, cosθ=1, apparent power |V||I| =
real power
In a pure inductive or capacitive circuit, cosθ=0, no
energy change but the instantaneous power oscillate
between circuit and source
When pX(t) > 0, energy is stored in the magnetic field
with the inductive elements, when pX(t) < 0, energy is
extracted in the magnetic field with the inductive
elements
If load is pure capacitive, the current leads voltage by
90o, average power is 0, vise versa for pure inductive
load
Energy Conversion Lab
BASIC PRINCIPLES

Example 2.1(PSA-Saadat)


supply voltage v(t)=100cosωt, inductive load Z=1.25∠60o Ω
determine i(t), p(t), pR(t), px(t)
Energy Conversion Lab
COMPLEX POWER

The rms voltage phasor and rms current phasor


V = |V|∠θv and I = |I| ∠θi
The term VI* = |V||I| ∠(θv -θi) = |V||I| ∠ θ = |V||I| cosθ + j
|V||I| sinθ
V
θv
θ
θi

I
S
θ
Q
P
Define a complex power quantity




S = VI* = P + jQ, where S = P 2 + Q 2 , apparent power
The units of complex power S are kVA or MVA
apparent power S is used as a rating unit of power
equipment
apparent power S is used for an utility to supply power
to consumers
Energy Conversion Lab
COMPLEX POWER

Define a reactive power quantity Q

Q is (+) when phase angle θ is (+), i.e. load is inductive. Q is (-) when
load is capacitive
I
θ
θv

V
P
θ
Q (-)
S
Given a load impedance Z





θi
capacitive
V=ZI
S=VI*=ZII*= (R+jX)II* = (R+jX)|I|2
S=VI*=V(V/Z)*=V(V*/Z*)=(VV*)/Z*=|V|2/Z*
Z=|V|2/S*
Complex power balance


real power supplied by source = real power absorbed by load
see Example 2.2 (PSA-Saadat)
Energy Conversion Lab
POWER FACTOR CORRECTION AND
COMPLEX POWER FLOW

Apparent Power |S| > P if power factor (p.f.) < 1



Power factor




current supplied in p.f. < 1 will be larger than current supplied in
p.f. = 1
a larger current will cost more!
inductive load causes lagging current and result in a lagging
power factor (p.f. < 1) , Z=jX, I=V/(jX)=(V/X) ∠-90o (lagging)
capacitive load causes leading current and result in a leading
power factor (p.f. < 1), Z=-jX, I=V/(-jX)=(V/X) ∠+90o (leading)
resistive load has a unity power factor (p.f. = 1)
To keep p.f. ≈ 1, we need to




install capacitor banks through the network for inductive load
industry probably operate at low lagging p.f. since most of them
are motors (inductive load)
to correct low power factor, capacitor banks are connected in
parallel to the load
for p.f. correction, see Ex. 2.3 in pp.23, Ex. 2.4 in pp.24 (PSASaadat)
Energy Conversion Lab
COMPLEX POWER FLOW

Complex power flow (PSA-Saadat)


see Figure 2.9 in pp. 26
complex power S12
V1∠δ1
2
V1
V1 V2
S12 =
∠γ −
∠γ + δ1 − δ 2
Z
Z

real power at the sending end
2
V1
V1 V2
P12 =
cos γ −
cos(γ + δ1 − δ 2 )
Z
Z

reactive 2power at the sending end
Q12 =
V1
V V
sin γ − 1 2 sin(γ + δ1 − δ 2 )
Z
Z
Energy Conversion Lab
Z∠γ
V2∠δ2
COMPLEX POWER FLOW OBSERVATION

Observation on complex power flow

assume transmission lines have small resistance
compared to the reactance (Z=X∠90o, cosγ=0), P12
and Q12 becomes
V1 V2
P12 =
sin(δ1 − δ 2 )
X
V1∠δ1
V
Q12 = 1 (V1 − V2 cos(δ1 − δ 2 ))
X




Z∠γ
V2∠δ2
maximum power occurs at δ = 90o, where maximum
power transfer: Pmax=|V1||V2|/X
real power flow P12 could be changed by small
change in δ1 or δ2
reactive power Q12 is determined by terminal voltage
difference, (i.e., Q≒|V1|-|V2|)
line loss = S12 + S21
Energy Conversion Lab
(PSA-Saadat)
COMPLEX POWER FLOW OBSERVATION

Ex. 2.6 direction of power flow between two voltage
sources


voltage source 1 can change phase from ±30o in step of 5o
phase of voltage source 2 is constant
Energy Conversion Lab
PROJECT 2



Two voltage of sources V1=120∠-5 V rms and V2=100∠0 V
rms are connected by a short line of impedance Z=1+j7 Ω.
The figure is shown below
Write a Matlab program such that the voltage magnitude of
the source 1 is changed from 75% to 100% of the give in
steps of 1V. The voltage magnitude of source 2 and phase
angle are kept constant. Compute the complex power for
each source and the line loss. Tabulate the reactive powers
and plot Q1, Q2, and QL versus voltage magnitude |V1|.
From the results, show that the flow of reactive power along
the interconnection is determined by the magnitude
difference of the terminal voltages.
Z∠γ
V1∠δ1
Energy Conversion Lab
V2∠δ2
Transmission Line Modeling





Derivation of Terminal V,I relations
Transmission matrix
Transmission line circuit equivalent
Complex power transfer (short line)
Complex power transmission


Complex power transmission


short radial line
long or medium line
Power handling capability
SHORT LINE MODEL

Short line model assumption




length shorter than 80km/50 miles, voltage < 69kV
capacitance is ignored
line is represented with series RL circuits
Equivalent short line model is shown as below
IS
+
VS
-



IR
Z=R+jX
+
VR
SR
-
Phase voltage at sending end: VS=VR+ZIR
Phase current in the short line: IS=IR
For steady state transmission equations, sending ends V,I in terms
of receiving ends
VS  1 Z  VR 
VR 
 I  = 0 1   I  = T  I 
 R 
 R
 S 
1 Z
T = 
0 1 
SHORT LINE MODEL

For steady state transmission equations, receiving ends V,I in terms
of sending ends
V
VR   D - B  VS 
−1  S 
 I  =  − C A  I  = T  I 
 S 
 R 
 S

For short line (line length<80km/50mi)

Series RL circuit model, no shunt admittance
VS  1 Z  VR 
 I  = 0 1   I ,
 R 
 S 

Voltage regulation of the line
Percent VR =

Z = zl = ( r + jωL)l
VR ( NL ) − VR ( FL )
VR ( FL )
× 100
Transmission line efficiency
P
η = R ( 3φ )
PS ( 3φ )
MEDIUM LINE MODEL



When length>80km (50miles), line charging current becomes
appreciable and shunt capacitance must be considered
To consider the shunt capacitance, half of the shunt capacitor may
be lumped at each end of line
The referred shunt circuit is called the nominal π model
IS
VS


R
Y/2
jX
IL Y/2
IR
VR
Y is the total shunt admittance of line with length l: Y=(g+jωC)*l
Derivation of the shunt circuit matrix




From KCL, I = I + Y V
,
L
R
R
2
from KVL the sending end voltage: VS = VR + ZI L
We have VS in terms of receiving end variables:
 ZY 
VS = 1 +
VR + ZI R
2 

MEDIUM LINE MODEL

Derivation of the shunt circuit matrix


Y
VS
2
The sending end current:
We have IS in terms of receiving end variables:
IS = IL +
 ZY 
 ZY 
I S = Y 1 +
VR + 1 +
IR
4 
2 



For medium line, nominal π circuit model
 ZY

1
+
Z
 VR 
VS  
2
  , Z = zl ,
I  = 
 S  Y (1 + ZY ) 1 + ZY   I R 

4
2 

Y = yl
In general, the medium line matrix, ABCD have
complex constants and symmetrical with A=D,
and AD-BC=1
DERIVATION OF V,I RELATIONS

From Kirchhoff’s voltage law, taking limit as Δx -> 0
V ( x + ∆x) = V ( x) + z∆xI ( x)
V ( x + ∆x) − V ( x)
or
= zI ( x)
∆x

dV ( x)
= zI ( x)
dx
From Kirchhoff’s current law, taking limit as Δx -> 0
I ( x + ∆x) = I ( x) + y∆xV ( x)
I ( x + ∆x) − I ( x)
or
= yV ( x)
∆x
Energy Conversion
Lab
dI ( x)
= yV ( x)
dx
DERIVATION OF V,I RELATIONS

Second order V equation is
d 2V
= yzV = γ 2V
2
dx
d 2I
= yzI = γ 2 I
2
dx

General solution is:
VS = VR cosh γl + Zc I R sinh γl
IS = I R cosh γl +
where
VR
sinh γl
Zc
is characteristic impedance
γ = zy is the propagation constant



Zc =
z/ y
γ=α+jβ, α is attenuation constant, β is phase constant
For a desired load voltage or complex power is specified, we can
use this equation to calculate supply side variables to satisfy its
loads
Under normal operation, a power system is driven by the
demand of loads
TRANSMISSION MATRIX

Arrange V,I equations in matrix form
V ( x ) = cosh γx VR + Zc sinh γx I R
I ( x) =

The transmission parameters A,B,C,D form the
transmission matrix T
B = Zc sinh γx
where A = cosh γx,
A B
T =

C D 

1
sinh γx VR + cosh γx I R
Zc
C=
Therefore,
1
sinh γx, D = cosh γx
Zc
VR 
V ( x )  A B  VR 
=
=
T
I 
 I ( x )  C D   I 

 
 R 
 R


As before, A=D and AD-BC=1
If we want to calculate V,I from sending end to
receiving end. Just simply use T-1
LUMP-CIRCUIT EQUIVALENT

The π equivalent circuit for transmission
line
I1
IR
Z’
V1

Y’/2
Y’/2
VR
We end up with the solution
Z 'Y '
) VR + ( Z' ) I R
2
Z 'Y '
Z 'Y '
I = Y ' (1 +
) VR + (1 +
)IR
4
2
V = (1 +
 Z 'Y '

+
Z'
1
 VR 
V  
2
 
I  = 
  Y ' (1 + Z ' Y ' ) 1 + Z ' Y '   I R 

4
2 
LONG LINE LUMP-CIRCUIT




For the l meter transmission line
sinh γl
Z ' = Z C sinh γl = ( zl )
,
γl
Y' Y tanh(γl / 2) yl tanh(γl / 2)
=
=
γl / 2
γl / 2
2 2
2
where γ = α + jβ = zy
As you can see, the equivalent circuit is affected
by “z”,”y”, and the length of the transmission line “l”
For example, z=r+jwL=0.169+j0.789, y=j5.38e-6
this means if the length is not too long, the effect
of z,y could be neglected
REVIEW FOR SIMPLIFIED MODELS

For steady state transmission equations, sending ends
V,I in terms of receiving ends
VS   A B  VR 
VR 
 I  = C D   I  = T  I 
 R 
 R
 S 

A B
T =

C
D


For steady state transmission equations, receiving
ends V,I in terms of sending ends
V
VR   D - B  VS 
−1  S 
 I  =  − C A  I  = T  I 
 S 
 R 
 S
REVIEW FOR SHORT, MEDIUM LINE MODELS




If |rl|<<1,
Z ' = ( zl )
sinh rl
Y' Y tanh( rl / 2) Y yl
= Z = zl ,
=
= =
rl
2 2
rl / 2
2 2
For short line (l<50mi)
Series RL circuit model
no shunt admittance
VS  1 Z  VR 
 I  = 0 1   I ,
 R 
 S 
Z = zl
For medium line (50<l<150 mi, approximately)
Nominal π circuit model, shunt admittance becomes
more apparent
 ZY

1
+
Z
 VR 
VS  
2
  , Z = zl ,
I  = 
ZY
ZY
IR 

 S  Y (1 +

) 1+

4
2 
Y = yl
LONG LINE MODEL

For long line (l>150 mi, approximately)
Use π equivalent circuit, contains trigonometric term
and Z,Y parameters
Equivalent π circuit model
 Z 'Y '

+
1
Z
'
 VR 
VS  
2
  ,
I  = 
 S  Y ' (1 + Z ' Y ' ) 1 + Z ' Y '   I R 

4
2 
sinh γl
sinh γl
=Z
Z' = zl
,
γl
γl
Y' yl tanh(γl / 2) Y tanh(γl / 2)
=
=
2
2
2
γl / 2
γl / 2
Complex Power Transmission

From steady-state V,I equations
VS   A B  VR  1 Z  VR 
 I  = C D   I  = 0 1   I 
 R  
 R 
 S 


The complex power transfer delivered to the
receiving end can be computed from
For lossless line
− S RS = −( PR + jQR ) = −VR I R*
 V − AVR 
= −VR  S

B


*
S SR = PS + jQS = VS I S* = VS (CVR + DI R )*
Complex Power Transmission

Complex power transfer from sending end to
receiving end
Assume Z = R + jωL
VS = VS e jθ S , VR = VR e jθ R
Z = Z e j∠Z , θSR ≡ θ S − θ R

Calculate the complex power transfer from sending
end
S SR = VS I S* = VS (
VS − VR *
)
Z
2
VS
VS VR j∠Z jθ SR
j∠Z
e
e e
=
−
Z
Z
Complex Power Transmission

Complex power transfer from receiving end to
sending end
Assume Z = R + jωL
VS = VS e jθ S , VR = VR e jθ R
Z = Z e j∠Z , θSR ≡ θ S − θ R

Calculate the complex power transfer from receiving
end
(
)
− S RS = VR − I R* = VR (
=−
VR
Z
2
e
j∠Z
+
VS − VR *
)
Z
VR VS
Z
e j∠Z e − jθ SR
Power Circle Analysis




Sending end circle and receiving end circle
S SR = CS − Be jθ SR , - S RS = CR + Be − jθ SR
Circles do not intersect if |VS|≠|VR|
θSR increase will increase the complex power
transfer until ultimate limit
2
For the transmission line
VS
∠Z ,
CS =
2
Z
VS
VS VR j∠Z jθ
j∠Z
S SR =
− S RS = −
Z
VR
Z
e
−
2
e
j∠Z
+
Z
VR VS
Z
e
e
SR
2
VR
∠Z ,
CR = −
Z
e j∠Z e − jθ SR
VS VR
B=
Z
COMPLEX POWER CIRCLE
2
S SR
VS
VS VR j∠Z jθ SR
j∠Z
=
−
e
e e
Z
Z
sending end circle
Q
2
− S RS = −
VR
Z
2
e j∠Z +
VR VS
Z
VS
∠( Z )
Z
e j∠Z e − jθ SR
δ
∠δ = θ S − θ R = θ SR
P
2
V
− R ∠( Z )
Z
receiving end circle
δ
VS VR j∠Z
radius =
e
Z
COMPLEX POWER FLOW

For lossless line, the real and reactive power
transfer from sending end to receiving end

B=jX, θA=0, θB=90o, A=cos βl
PSR ( 3φ ) = − PRS ( 3φ ) =
QSR ( 3φ ) =


VS ( L− L ) VR ( L− L )
VS ( L− L ) VR ( L− L )
X
X
cos δ −
sin δ
VR ( L− L )
X
2
cos βl
Power is transferred proportional to power angle δ,
as load increases, δ increase
For lossless line, max power occurs when δ=90o,
but for the adequate margin of stability, δ is
between 30o to 45o
Power Circle Analysis
 The circles do not intersect of |VS|≠ |VR|
 As θ12 increases, the active power sent and received
increases.
o
 The maximum limit of active power sent is θ12 =180 –
angle(Z)
 The maximum limit of active power received is θ12 =angle(Z)
Q
sending end circle
2
VS
∠( Z )
Z
δ
∠Z
Pmax(sent)
∠δ = θ S − θ R = θ SR
P
2
V
− R ∠( Z )
Z
receiving end circle
δ
Pmax(received)
Ploss = PS − PR = 43.2315 MW
Qloss = Qs − QR =104.6636 MVar
NCKU EE Energy Conversion Lab. 92781
Power Circle Analysis
 too much θSR increase will force synchronous
machine running out of synchronism
 How can we increase the ultimate
transmission capability? Decrease X or
increase |V|
 Decrease X: line design (space, material)
 Increase |V|: sometimes not feasible
 For lossless line |VS| ≈ |VR|,|Z| ≈90o, θSR
<10o , PSR coupled with θSR , QSR coupled with
|VS|-|VR|
POWER TRANSMISSION CAPABILITY

Power handling capability is limited by



thermal loading: Sthermal=3VφratedIthermal
stability limit: load angle δ
In practice, when generator and transformer are
connected to transmission line, the combined reactance
will result in a larger δ for a given load
Power Handling Capability of Line
 Thermal limit: line temperature to prevent line sag,
irreversible stretching and insulation
 Bundling for greater spacing






Current limit: Thermal limit can affect current carrying
Voltage limit: Conductor size, spacing and insulation
MVA limit: MW and MVAr, typically a 345kV line has a
thermal rating of 1600MVA
θSR limit: synchronism max limit between
40-50o, which is about 65 to 75% of the ultimate
transmission capability
Power handling capability is limited by thermal rather
than stability for short lines, vice versa for long lines
Project analysis (project 2-1)
Distributed Parameter Model

The distributed parameter model states that


∂v
i ( x ) − i ( x + ∆x ) = (Gv + C ) ∆x
∂t
∂i
v ( x ) − v ( x + ∆x ) = ( Ri + L ) ∆x
∂t
I

Let Δx -> 0, the above equation

−
∂i
∂v
= Gv + C
∂x
∂t
−
∂v
∂i
= Ri + L
∂x
∂t

Laplace
transform
dI
= −(G + Cs )V = −YV
dx
dV
= −(R + Ls )I = − ZI
dx
Distributed Parameter Model

When the line is lossless, R and G are zero

characteristic impedance
∂v
L
=
∂i
C

propagation speed of v and I along the line
∂x
=
∂t

1
CL
Take the time derivative of the left equations


dI
= −(G + Cs )V = −YV
dx
d 2I
2
=
(
G
+
sC
)(
R
+
sL
)
I
=
γ
I
2
dx
dV
= −(R + Ls )I = − ZI
dx
d 2V
2
=
(
R
+
sL
)(
G
+
sC
)
V
=
γ
V
2
dx
Distributed Parameter Model

The solution of the second order differential equation

Ib


Vf
If
Propagation constant
γ=

V ( x, s ) = Z c A2e − rx − Z c A1e rx


 
I ( x, s ) = A1e rx + A2 e − rx
 
characteristic impedance
R C
+ s LC
2 L
Voltage component
 Vf=IfZc, Vb=-IbZc
 V(x,s)= Vf+ Vb=Zc(If –Ib)
For terminal impedance Zd

Vb
V (d , s) Z c ( I f − I b )
Z d ( s) =
=
I (d , s)
I f + Ib
Zc =
( R + sL)
(G + sC )
Distributed Parameter Model

Define current reflection coefficient


ρi=Ib/If
from: Z ( s) = V (d , s) = Z c ( I f − I b ) =
d
I (d , s )

Zc (
I f + Ib
The following relations hold

Z d 1 − ρi
=
Z c 1 + ρi
or
Zc − Zd
ρi =
Zc + Zd
If
If
If
If
−
+
Ib
)
If
Ib
If
= Zc
(1 − ρ i )
(1 + ρ i )
Distributed Parameter Model

Define voltage reflection coefficient




ρv=Vb/Vf
ρv =
Vb − I b Z C
Z − Zd
=
=− c
Vf
I f ZC
Zc + Zd
ρi = - ρv
The reflection coefficients of some common terminations
are:


Open-circuit
Zd = ∞
ρv = 1
ρ i = −1
Short-circuit
Zd = 0
ρ v = −1
ρi = 1
Zd = Zc
ρv = 0
ρi = 0
Matched termination
Wave Propagation Characteristics

The solution of the second order differential equation

I ( x, s ) = A1e rx + A2 e − rx
 
Ib
I (0, s ) ρ i e −2 rd
A1 =
1 + ρ i e −2 rd



Vf
e
2 rd
ρi
)
A2 =
I (0, s )
1 + ρ i e −2 rd
Let A1 -> 0, A2 = I(0,s) for d->∞ (infinite line)
-γx=I
 I(x,s)=I(0,s)e
f
Neglecting G, the propagation constant reduced to


If
I (0, s ) = A1 + A2 = A1 (1 +

V ( x, s ) = Z c A2e − rx − Z c A1e rx


 
γ=
R C
+ s LC
2 L
Current wave:
I ( x, s ) = I (0, s )e
−(
R C
+ s LC ) x
2 L
Vb
Distributed Parameter Model

Transient wave propagation characteristics, for
∞ long line
i ( x, t ) = e
−(
R C
)x
2 L
atten. Factor



i (0, t − LC x )u(t − LC x )
delay
The time-domain response of the forwarding
voltage
v ( x, t ) = Z C i ( x, t )
V,I reflection coefficients, ρv, ρi
Simulation of a single-phase line
The Bewley Lattice Diagram



V is applied at t=0 to a line of length d, characteristic impedance
ZC that is terminated with an impedance Zd at the other end
Assume the source impedance at the sending end is zero, i.e., ρvs
= -1
The bewley lattice diagram is as follows
Simulation of A Single-phase Line


Source circuit
Load circuit
di
e − vS = RS iS + LS S
dt
iR =
iS =
1
( vR − RLiR )dt
LL ∫

State variable: vbs, vfR
vS = Z C iS + 2vbs
vR = 2v fR − Z C iR
forward, backward wave relations:

sending end and receiving end relations

v fs = vS − vbs
v fR ( x, t ) = e
−(
R C
)x
2 L
1
( e − vS − RS iS )dt
∫
LS
vbR = vR − v fR
v fs u(t − LC x )
vbs ( x, t ) = e
−(
R C
)x
2 L
vbR u(t − LC x )
Distributed Parameter Model

Line model
v fR ( x, t ) = e
vbs ( x, t ) = e
−(
−(
R C
)x
2 L
R C
)x
2 L
v fs u(t − LC x )
vbR u(t − LC x )
Distributed Parameter Model

Single phase line simulation
Project 3

A transmission system with a supply source at
sending end in rated voltage 500 kV, 60Hz.
The 1000km single phase line has the following
electrical parameters



R = 0.15 Ω/km, L=2.96 mH/km C=0.017 µF/km
with 1000 km long line.
The source electrical parameters are: RS=50 Ω,
LS=0.1H
Now the transmission line is connected to a
single phase RL load of RL=5.4kΩ, LL=10.743H
Project 3

Operating status


Observe and plot the following
1)
2)
3)

show the trace of e, VS, iS, VR, iR, from t=0 through t=0.3 in
one figure with different subplots
show the traces of Vfs, ibs, VfR, ibR from t=0 through t=0.3
what are the ρv and ρi at the receiving end from the results
of 2) you obtain? could you verify it by Eq.(3.132) and
Eq.(3.134)?
Operating status


the voltage source “ e = 500 2 / 3 cos ωet ” is operated at
rated value to supply the load
when there is a step wave of VS=500u(t) volt injected into
the sending end of line, use the same line and load
Observe and plot the following
1)
2)
show the traces of VX and iX at the point of x=600km
show the Bewley Lattice diagram of VX