XYLENE POWER LTD.

DAILY kVA AND kVAh

By C. Rhodes

INTRODUCTION:
Cumulative daily kVAh metering can be applied to generators and loads of all sizes for fair allocation of transmission/distribution costs. With cumulative daily kVAh metering a customer can be sometimes a load and sometimes a generator without any changes in the metering arrangement.

This web page reviews basic electrical engineering concepts that can be found in many electrical engineering textbooks and then introduces the concepts and features of daily kVA and daily kVAh metering. The daily kVAh cumulated over a month is a good measure for allocating transmission/distribution costs in a transmission/distribution system involving distributed generation.

WYE CONNECTION:
A wye electricity service presents three AC phases with 120 degree (2 Pi / 3 radians) phase separations. The voltage reference point is the junction of the wye, which is normally close to or at ground potential. The phase voltages are measured with respect to this junction. If the individual phases are denoted by a, b, c, the instantaneous phase voltages are Vai, Vbi, Vci and the instantaneous phase currents are Iai, Ibi, Ici. The instantaneous power Pai fed to load phase a is:
Pai = (Vai X Iai).
If the voltage source is the electricity grid, to a good approximation the voltage is sinusoidal. Hence:
Vai = Vaio sin(WT)
where:
Vaio = peak voltage on phase a
W = 2 Pi F
Pi = 3.1415928
F = 60 Hz
T = time in seconds relative to an arbitrary initial time
sin = sine

The RMS voltage Va is defined as:
Va = [Integral from T1 to T2 of (Vai^2 dT)/ (T2 - T1)]^0.5

The RMS current Ia is defined as:
Ia = [Integral from T1 to T2 of (Iai^2 dT)/ (T2 - T1)]^0.5

For a sinusoidal voltage Vai the cycle period is (2 Pi / W) giving:
Va = [Integral from T1 to T1 + (2 Pi / W) of [Vaio sin(WT)]^2 dT / (2 Pi / W)]^0.5
= Vaio / (2)^0.5

For a linear but reactive load the corresponding instantaneous current is given by:
Iai = Iaio sin(WT - Phi)
where:
Iaio = the peak instantaneous current
Phia = phase angle in radians between the Vai and Iai wave forms

For a sinusoidal current Iai the cycle period is (2 Pi / W) giving:
Ia = [Integral from T1 to T1 + (2 Pi / W) of [Iaio sin(WT - Phi)]^2 dT / (2 Pi / W)]^0.5
= Iaio / (2)^0.5

The instantaneous power on phase a is defined as:
Pai = Vai Iai
It can be shown that for sinusoidal voltages and currents the average power on phase a is:
Pa = Va Ia cos(Phia)
In this expresion:
cos(Phia) = Power Factor

For a 3 phase wye fed system the total instantaneous power P is defined as:
P = Pai + Pbi +Pci
= (Vai X Iai) + (Vbi X Ibi) + (Vci X Ici)
where:
Vbi = instantaneous voltage on phase b
Ibi = instantaneous current on phase b
Vci = instantaneous voltage on phase c
Ici = instantaneous current on phase c

If the voltages and currents are sinusoidal it can be shown that the total average power Pt on all three phases is:
Pt = Va Ia cos(Phia) + Vb Ib cos(Phib) + Vc Ic cos(Phic)
where:
Vb = RMS voltage on phase b
Ib = RMS current on phase b
Phib = phase angle between Vbi and Ibi
Vc = RMS voltage on phase c
Ic = RMS current on phase c
Phic = phase angle between Vci and Ici

For the special case of balanced sinusoidal voltages and currents:
Va = Vb = Vc = V
and
Ia = Ib = Ic = I
and
Phia = Phib = Phic = Phi
Thus:
Pt = 3 V I cos(Phi)

Note that even in a perfectly balanced system the total power on each phase is less than or equal to the product of the RMS voltage multiplied by the RMS current due to the factor cos(Phi). In general cos(Phi) is less than or equal to unity.

The quantity (Va Ia + Vb Ib + Vc Ic) is the normally measured kVA and is referred to herein as the single cycle kVA. For sinusoidal voltages and currents the single cycle kVA equals the power Pt in kW if:
Phia = Phib = Phic = 0

DELTA CONNECTION:
Recall that the instantaneous power P is given by:
P = Iai Vai + Ibi Vbi + Ici Vci
For a three phase delta connected customer, at any instant in time the instantaneous phase currents Iai, Ibi, Ici conform to:
Ici = -Ibi - Iai
Substituting into the three phase formula for P gives:
P = Iai (Vai - Vci) + Ibi (Vbi - Vci)
The corresponding expression for kVA is: kVA = Ia (Va - Vc) + Ib (Vb - Vc)
where:
(Va - Vc) is the RMS value of (Vai - Vci)
and
(Vb - Vc) is the RMS value of (Vbi - Vci)

Hence the fair share of the distribution costs for a delta connected load during a particular time interval is proportional to:
[(Ia(Va - Vc) + Ib (Vb - Vc)]

CONSTANT TORQUE:
An important property of a balanced 3 phase system is that:
P = Pt = constant
Hence the torque on a three phase generator driving a balanced load is constant. Similarly the torque exerted by a three phase motor is nearly constant.

The results to this point are available from numerous basic electrical engineering text books. The following results are not readily available elsewhere:

DAILY kVA:
Recall that the definition of single cycle kVA involved the calculation of RMS voltage Va and the calculation of RMS current Ia over a single cycle period (1 second / 60). The concept of daily kVA is to calculate kVA over 24 hours instead of over (1 second / 60). Using modern microprocessor instrumentation it is straight forward to calculate daily RMS voltage and current values where the calculation period instead of one cycle period is:
24 hours = 86,400 seconds = 5,184,000 X (1 second / 60)

The daily RMS voltage Vad is given by:
Vad = [Integral from T1 to T2 of (Vai^2 dT')/ (T2 - T1)]^0.5
where T2 - T1 = 86,400 seconds. In order to evaluate this integral numerically dT' is chosen to be 1 second / 3600, which is sufficient for sampling the 30th harmonic of 60 Hz.

The expression for Vad can be simplified using the substitution:
Va^2 = [Integral from T to T + dT of (Vai^2 dT')/ dT)]
where dT = 1 second / 60
Then:
Vad = [Integral from T1 to T2 of (Va^2 dT)/ (T2 - T1)]^0.5
where T2 - T1 = 86,400 seconds. This integral is easier for the layman to understand. Furthermore, provided that Va is accurate this integration result is almost independent of minor variations in dT. Hence this formula is easy to numerically evaluate.

Similarly the daily RMS current Iad is given by:
Iad = [Integral from T1 to T2 of (Iai^2 dT')/ (T2 - T1)]^0.5
where T2 - T1 = 86,400 seconds. In order to evaluate this integral numerically dT' is chosen to be 1 second / 3600, which is sufficient for sampling the 30th harmonic of 60 Hz.

The expression for Iad can be simplified using the substitution:
Ia^2 = [Integral from T to T + dT of (Iai^2 dT')/ dT)]
where dT = 1 second / 60
Then:
Iad = [Integral from T1 to T2 of (Ia^2 dT)/ (T2 - T1)]^0.5
where T2 - T1 = 86,400 seconds. This integral is easier for the layman to understand. Furthermore, provided that Ia is accurate this integration result is almost independent of minor variations in dT. Hence this formula is easy to numerically evaluate.

For the special case of:
Vaio = constant
Iaio = constant
then:
Va Ia = Vad Iad
and for the special case of:
Vbio = constant
Ibio = constant
then:
Vb Ib = Vbd Ibd
and for the special case of:
Vcio = constant
Icio = constant
then:
Vc Ic = Vcd Icd
Thus if the single cycle RMS voltages and currents are both constant throughout a 24 hour period the daily kVA equals the single cycle kVA. However, if the single cycle RMS voltages or the single cycle RMS currents are not constant throughout the 24 hour period then the daily kVA is not equal to the single cycle kVA. The daily kVA creates the desired metering function which causes flat loads to have lower connection costs than peaky loads that draw the same amount of energy per day.

Consider a practical example. Suppose a 1 kW resistive load operates for 24 hours. The energy consumed is 24 kWh. At an RMS voltage of 120 volts the RMS current is:
1000 W / 120 v = 8.333 A
The single cycle kVA value is:
.120 kV X 8.333 A = 1.0 kVA
Since the single cycle RMS voltage and the single cycle RMS current are constant, in this case:
daily kVA = single cycle kVA = 1.0 kVA.

Now suppose that the same 24 kWh of energy is drawn by a 3 kW resistive load that operates for 8 hours of the 24 hour period. The energy consumed is:
3 kW X 8 h = 24 kWh
The RMS voltage is constant at 120 volts.
During the 8 hour load on period the single cycle RMS current is:
3 X 8.33 A = 25 A
The daily RMS current is given by:
{[(16 h X (0 A)^2) + (8 h X (25 A)^2)] / 24 h}^0.5
= {(25 A)^2 / 3}^0.5
= {(8.333 A)^2 X 3}^0.5
= 8.333A X (3)^0.5
Hence the daily kVA value is:
120 V X 8.333 A X (3)^0.5
= 1 kVA X (3)^0.5
= 1.732 kVA

Notice that concentrating the energy into an 8 hour period instead of spreading it out over a 24 hour period had the effect of increasing the daily kVA from 1.0 kVA to 1.732 kVA. By comparison a single cycle peak kVA recording meter would have registered 3 kVA. However, that 3 kVA measurement would have been the same regardless if the 3 kVA peak was hit once or 20 times a month.

DAILY kVAh:
We can now introduce a unit of:
daily kVAh = 24 h X (measured daily kVA)
Note that if the load is resistive and constant 24 hours per day, then:
measured daily kVAh = measured kWh in a 24 hour period
However, if the load is time varying:
measured daily kVAh > measured kWh
The cumulative daily kVAh is a good proportional indicator of the grid connection costs for both loads and generators.

FEATURES OF CUMULATIVE DAILY kVAh METERING:
1. Cumulative daily kVAh meters can be applied to generators and loads of all sizes for fair allocation of transmission/distribution costs.
2. Cumulative daily kVAh meters cumulate so that a building that has multiple load peaks in a month is charged more for its grid connection than a building that has only a single load peak during the month.
3. The use of cumulative daily kVAh simplifies meter reading and administration issues. Bills and real estate transactions can easily be settled to the nearest day.
4. Cumulative daily kVAh metering mitigates the cost effect of load swings that occur only one or two days per month but captures the cost effect of load swings that occur almost every day.
5. Use of daily kVAh cumulated monthly instead of peak kVA would have the effect of slightly shifting the rate burden from low load factor to high load factor customers, which would encourage more energy conservation.
6. All the existing advantages of single cycle kVA metering that encourage high power factor and low harmonic content are retained.
7. Like peak kVA metering, cumulative daily kVAh metering encourages high load factor.
8. Unlike peak kVA metering, cumulative daily kVAh metering allows brief equipment shutdowns for maintenance purposes without an undue cost penalty to the building owner.

GENERAL BENEFITS OF CUMULATIVE DAILY kVAh METERING:
1. The cumulative daily kVAh electricity meter calculates the RMS values over a 24 hour time interval to encourage efficient utilization of distribution and transmission.
2. If the rate of power transfer to or from the grid is not uniform over the 24 hour RMS calculation period use of cumulative daily kVAh metering will cause the generator/load to be charged more per kWh for grid access.
3. Use of cumulative daily kVAh metering would encourage wind generators to build electro-chemical energy storage on the generator site to reduce variations in net power output.
4. Cumulative daily kVAh metering is fair to behind the meter co-generation because it mitigates the cost effect of short generator shutdowns for maintenance or repair.
5. Cumulative daily kVAh metering is believed to be fairly applicable to generators and loads of all sizes ranging from small apartment suites to the largest nuclear generation stations.
6. If a customer presents a constant resistive impedance to the grid, then the calculated daily kVA is the same as the average power in kW sensed by an induction type kWh meter.
7. If a customer presents a reactive impedance or harmonic distortion to the grid then the calculated daily kVA is greater than the average power in kW, causing that customer to be allocated a larger fraction of the distribution costs.
8. If a customer presents a resistive impedance to the grid that varies over the 24 hour RMS calculation period, that customer will be charged more for distribution per kWh consumed than is a customer that consumes the same amount of energy at a constant rate.
9. The contemplated metering technology for measuring daily kVAh is readily able to resolve voltage and current harmonics up to the 30th harmonic of the power line frequency. Generally power transformers effectively filter out higher frequency harmonics.
10. The daily kVAh for each 24 hour period is calculated, cumulated and displayed for distribution / transmission cost billing purposes. As part of this calculation the single cycle RMS voltages Va, Vb, Vc and the single cycle RMS currents Ia, Ib, Ic are calculated and are available for display.
11. The use of daily kVAh metering would encourage installation of behind the meter energy storage and and behind the meter electricity generation to minimize swings in the power transfer rate to and from the grid.
12. The use of daily kVAh metering allows LDCs to fairly recover their costs from net metering customers.

This web page last updated February 11, 2008.