Download Analsysis of CT and Clamp-on Response to Motor Starting Waveform

Analsysis of CT and Clamp-on Response to Motor Starting Waveform
electricpete
Overview:
DC Hall effect probes are not readily available for measuring in the range necessary to measure
75 hp 460vac motor starting current transients which may reach 1200 – 1300A true peak.
Therefore, our plant has used AC current probes such as AEMC SR651. The response of an ac
current probe is similar to the response of a CT. Specifically, it acts as a high-pass filter. This
may reduce the indicated magnitude of initial transient peak below the actual.
CT Model
1:1
L
+
|
v(t)
|
-
i2(t)
+
|
v(t)
|
-
Ideal 1:1
transformrer
im(t)
CT
magnetizing
current
Input
Current
Output
Voltage
For clamp-on
R
Measuring
Device
i1(t)
Figure 1 - CT Model
Where:
i1(t) = Actual current (input)
i2(t) = Indicated current (output)
L – represents CT magnetizing inductance
R – represents measuring device resistance
Turns ratio is assumed 1 for simplicity
Leakage reactance is neglected (primary leakage reactance has no effect and secondary leakage
reactance is typically much lower than measurement resistance).
v(t) = voltage accross magnetizing branch = voltage accross measuring resistance (these are the
same)
For simplicity, the circuit is redrawn eliminating the ideal transformer as follows:
i2(t)
i1(t)
im(t)
L
+
|
v(t)
|
-
R
Since the same voltage is seen accross L and R, we can write:
voltage accross inductor = voltage accross resistor
d
L* [im(t)] = R * i2(t)
dt
We can substitute into the above equation im(t) = i1(t) - i2(t) (based on Kirchoff's current law),
which gives:
d
L* [i1(t)-i2(t)] = R * i2(t)
dt
Taking the Laplace transform of both sides:
L * s*[I1(s) – I2(s)] = R*I2(s)
where I1(s) and I2(s) are laplace transforms of i1(t) and i2(t), and we have assumed that all
currents are 0 at time 0-.
Multiply out the LHS:
s L I1(s) – s L I2(s) = R*I2(s)
Add to each side: s L I2(s)
s L I1(s) = R*I2(s) + s L I2(s) = [R + sL ] I2(s)
Solve for I2(s) in terms of I1(s)
I2(s) = I1(s) *
sL
sL  R
Introduce transfer function H(s) = I2(s) / I1(s)
sL
H(s) = I2(s) / I1(s) =
sL  R
Divide numerator and denominator by L
s
H(s) =
R
s
L
Define measurement Time constant meas = L/R
Define measurement decay parameter meas = 1/meas = R/L
s
H(s) =
s
1
=
 meas
s
s   meas
Obeservations about this transfer function: The transfer function has a "zero" at s=0 and a "pole"
at s = -meas. Therefore, for high sinusoidal frequencies (relative to meas), we expect that the
pole and the zero will cancel out the the CT is very accurate. However, for low frequency
content (relative to meas), the zero begins to dominate and the low frequency content does not
show up in the output (indicated current). The CT is a high-pass filter which blocks lowfrequency content. The half-power frequency is f = meas/(2*pi) = 1/(2*pimeas). (At this halfpower frequency the only 70.7% of signal is transmitted.).
Approach for Calculations With Transfer Function:
The transfer function enables us to estimate the indicated current i2(t) for any input current i1(t).
The approach would be as follows:
I1(s) = L {i1(t)} (where L indicated Laplace transform )
I2(s) = I1(s) * H(s)
i2(t) = Linv {I2(s)} (where Linv indicates inverse Laplace transform )
Step Response of CT:
Use the above approach to find the response of the CT to a step increase in DC current.
i1(t) = step function (1 for t>0, 0 for t<0)
I1(s) =
L {i1(t)}
= 1/s
1
s
1
I2(s) = I1(s) * H(s) = *
=
s
s   meas
s   meas
t
i2(t) = Linv {I2(s)} = e
t * meas
e
 meas
Assuming a value meas = 0.1 sec, the step response would be as shown below:
Step Response
-=
actual
current
(input)
-=
indicated
current
(output)
Expressing  in different forms:
For a fixed power system frequency, the decay characteristics of a given system are given in a
number of different ways in different references. For example, they can be expressed as: power
factor, time constant (), decay parameter ( = 1/), Halving time (T1/2), and X/R ratio. To
aid comparison of these different terms which all express the same thing, the following graph is
provided (based on 60hz frequency):
100
XoverR
10
Tau (msec)
Thalf (msec)
1
0
0.1
0.2
0.3
0.4
0.5
0.6
p.f.
Applicability of CT model To Clamp-ons
The clamp-on may contain internal circuitry for processing the output waveform which are not
well known. It is believed that in most cases the circuit is simply a CT with shunt resistor to
provide voltage output as shown in Figure 1.
However, as shown in Figure 2 , measurement of the response of a similar (smaller) AEMC
clamp-on probe to a step change in dc current (during Viper MOV testing) showed the same
form of response as above (step increase, followed by exponential decay) with a time constant
estimated at 0.04 seconds.
Switch Position
(sensed via
another
Contact on
same sw)
Indicated
Current
Through
switch
Closed
Open
T1/2 = 0.04 sec
meas = T1/2 / ln(2) = 0.058 sec
0.04 sec
transition
Actual
Current
flowing
Actual Current is 0
Indicated current decaying due to probe time constant
Figure 2 - Measured Response Of Small Clampon to Step Current Change
If we assume that the clamp-on is a linear time-invariant (LTI) system, then its step response is
unique. All LTI systems which have the same step response are the same and will respond the
same way to any input. Therefore we can apply the CT model to predict the behavior of the
clamp-on as long as the clamp-on has a similar step response to the one shown above.
Motor Starting Waveform:
The current waveform for worst-case closing angle (closing when associated phase to neutral
voltage is 0) is as follows:
 t * RL Motor

i(t) = Imax * e Motor  cos( w * t ) for t>0 (0 for t<0)


where
Imax = peak symmetrical amplitude of LRC
w = radian frequency = 2*pi*f = 2*pi*60hz
Rmotor, Lmotor – characteristics of motor with slight influence from the power system.
For simplicity, we will set Imax = 1, and define another time constant
motor = Lmotor/Rmotor
Then we can write:
  t

i(t) = e Motor  cos( w * t )


Equation 1
For motor starting, various references give motor starting power factor in the range 0.1 – 0.25.
For purposes of this analysis, a motor starting p.f. = 0.13 was chosen. This corresponds to
T1/2=14 msec which is the value estimated from one of the motor starting traces. This
corresponds to motor = 0.020 sec
Using this value of motor in n Equation 1, the motor starting waveform would look as follows
(peak-to-zero symmetrical component chosen as 1.0 for simplicity):
Figure 3 - Computed Motor Starting Waveform
Solution of Indicated Waveform for motor starting:
We proceed using the transfer function as above.
  t

i1(t) = e Motor  cos( w * t ) = et * Motor  cos( w * t )




I1(s) =
L {i1(t)}
1
s Motor

s
s  w2
2
I2(s) = I1(s) * H(s)
 1
s
s 
I2(s) = 
*
 2
2
 s Motor s  w  s   meas
This can be simplified to:
( Tau_mot
I2 ( s ) :=
:
w
2
 s ) s Tau_meas
2
2
( s Tau_mot  1 ) ( s  w ) ( 1  s Tau_meas
)
i2(t) = Linv {I2(s)}
To calculate the inverse Laplace transform, we could expand the above expression by partial
fractions. However, for ease of computaation, a computer solution was performed which gave
the following inverse Laplace transform:
i2(t) = (-Tau_meas/(Tau_mot-Tau_meas)*exp(-t/Tau_mot)w^2*Tau_meas^2/(1+w^2*Tau_meas^2)*cos(w*t)+w*Tau_meas/(1+w^2*Tau_meas^2)*sin(w
*t)+Tau_meas/(1+w^2*Tau_meas^2)/(Tau_mot-Tau_meas)*exp(t/Tau_meas)+Tau_meas^2/(1+w^2*Tau_meas^2)/(Tau_mot-Tau_meas)*exp(t/Tau_meas)*w^2*Tau_mot)
Indicated Motor Starting Waveform for various values of meas
The solution computed above was applied to several values of meas, starting far above Motor and
decreasing below Motor (in all cases Motor remains 0.020)
Results as follows:
Motor Starting Current for
meas = 0.200 sec
- = actual
current
(input)
-=
indicated
current
(output)
Motor Starting Current for
meas = 0.110 sec
- = actual
current
(input)
-=
indicated
current
(output)
Motor Starting Current for
meas = 0.050 sec
- = actual
current
(input)
-=
indicated
current
(output)
Motor Starting Current for
meas = 0.025 sec
- = actual
current
(input)
-=
indicated
current
(output)
Motor Starting Current for
meas = 0.015 sec
- = actual
current
(input)
-=
indicated
current
(output)
Motor Starting Current for
meas = 0.010 sec
- = actual
current
(input)
-=
indicated
current
(output)
Motor Starting Current for
meas = 0.005 sec
- = actual
current
(input)
-=
indicated
current
(output)
Commentary/Conclusions from above graphs:
Three types of features / errors can be noted from the above graphs as summarized in Figure 4.
As we proceed from the first graph (meas = 0.200 sec) to the last graph (meas = 0.005 sec), each
of these errors tends to become more pronounced.
Error Type 3 (steady state lower) – not easily visible until meas = 0.005
Error Type 1 (peak lower)
Error Type 2
(oscillating offset)
Figure 4 - Errors due to measurement time constant
Error Type 1 - Indicated current peak is lower than actual current peak. The difference increases
as meas decreases. Although not shown above, the difference also increases as motor increases.
Additionally, although not shown here, some difference will be present for any suddenly-applied
sin waveform even when no dc offset is present.
Error Type 2 - The average value (dc component) oscillates below and above zero instead of
simply decaying monotonically to 0 as we expect from the R/L circuit. We see similar behavior
in some of our traces, although that oscillation continues over a longer period than I would
expect if it was simple result of two time constants (one for motor and one for measurement).
DC offset positive
DC offset negative
Figure 5 - Actual data resembles type 2 error (3rd run 9/4/08)
Error Type 3 - Steady state current magnitude is less than actual. The very last graph (meas =
0.005 sec) shows this behavior. The probe manufacturer would not allow this much lag since it
would not properly capture the amplitude for 60hz. We can be sure that this last plot represents
an extreme case, i.e. that our meas is higher than this (higher than 0.005 seconds).
Probe Specifications:
The AEMC probe spec sheet is available here:
http://www.aemc.com/products/pdf/2113.45.pdf
Two specifications of interest:
"Frequency Range: 30hz - 50 khz"
"Load Impedance: 100 kOhm min"
The fact that a minimum frequency range (30hz) is specified is consistent with our model. The
type 3 error due to simple high-pass filtering worsens as frequency decreases relative to meas.
The graphs above show worsening type 3 error as we decrease meas = increase meas. We expect
similar worsening would occur as we decrease frequency relative to meas
The fact that we have a minimum [external] load resistance specified is consistent with the
circuit model we have envisioned in Figure 1. Any resistance in parallel with the calibration
shunt resistor will change the calibration. Also worth noting that it will change the
measurement time constant. Requiring a very high external resistance limilts these effects. I
don't know what type of load resistance are presented by the Viper or the Yokogawa
instruments.
Possible Additional Followups:
Investigate alternate current probes for future starts.
Measure Step response of existing probes (could be a challenge to create a dc step current at
appreciable fraction of the probe range)
.