| |>
|> | That may be true- it was routinely done in the area that I used to live
| in.
|> | Note that KVA demand metering was based on monthly peak where the meter
| had
|> | a thermal(?) lag so that it rode through starting transients and short
| peaks
|> | and represented what was important to the utility - temperature rises in
|> | equipment.
|>
|> Sounds like exactly the right way to measure for the component of costs
|> that relates to purchasing, installing, and maintaining transmission and
|> distribution facilities.
|>
|> So, given 2 customers that periodically have a 1000 kVAR demand, but run
|> regularly around 100 kW, where one of them has pf 0.97 and the other has
|> pf 0.66, how are they going to measure the added charges for the second?
|> A kVAR usage meter?
| ---
| The meters measured kVA (not kVAR) demand and the needle pushed another
| needle so that the peak kVA during the period is registered. Your two
| customers would have the same kVA demand charge as both require the utility
| to provide the same equipment capability but the first would have a higher
| kWH bill as he uses more energy. The effective cost per kWH would be lower
| for him than for the second person. Both are being charged for a high kVA
| (not necessarily kVAR) peak demand because of a fluctuating load and the
| second gets a whammy in this respect from poor pf. The KVA demand
| penalises high peak (over 15 to 30 minutes duration) to average loads as
| well as poor power factor.
The example I am giving is that both customers use the _same_ energy level
(an average of 100 kWH over the month) and both will run a peak kVA and kVAR
at the same high level (and hence the same demand reading).
-------------------
There seems to be some confusion here: 100kWh is, I assume the monthly
energy usage- if so then a peak kVA reading would last for less than 0.1
hours and a demand meter would register a negligable kVA for both loads. If
you are talking about 100kW average for the month (about 720 hours) then the
kWH would be 72000 kWH for both loads. If both loads indeed had a peak of
1000kVA with the pf's that you cited, the first load would have a peak
power of 970kW and the second a peak power of 660 kW but the same kVA peak.
--------------
But what the
difference is, when at times other than the peak, one customer has a lower
power factor despite the same energy. So that customer takes more power,
but then gives much of it back each half cycle, netting the same usage and
loss internally as the first customer, but imposing unmeasured (by that meter)
losses on the utility distribution. What I am saying is it can be measured
in terms of measuring both kW(H) and kVAR(H). Each of those figures would
have a rate. The kWH rate would be the usual energy rate. the kVARH rate
would be a secondary energy rate prportioned to the loss that would be
experienced sloshing reactive power around.
-----------
I have to disagree with you. The energy that is being shuffled around will
not appear on the kWH meter. That is true. The meter (and kW meters) measure
average power. The difference is that 100kW at 0.97 pf will have a kVA of
103 kVA while 100kW at 0.66 pf will have 151.+ kVA. Assuming the same
voltage and line resistances , the internal losses of the second customer
will be (151./103)^2 = 2.16 times that of the first customer. This loss WILL
be registered by the kWH meter and the customer pays for it.
Measuring kVARH will not give any useful information as the customer's
losses are already measured and this information will not be of use in
sizing of equipment or, as a matter of fact, determining the utility losses.
KVA demand will not determine utility losses either but will affect
equipment sizing and costs (a point that you agree with). Utility energy
losses will be determined by utilities measuring the difference between
their generated energy and the customers' usage.
------------
The first customer would have
a lower kVARH and the second customer would have a higher kVARH, while they
both have the same kWH and kVA peak demand.
| Take another case - steady 100KVA loads - first one at 0.97 pf lag and the
| second at 0.66 pf lag. Each pay $0.08 per KWH so that over the billing
| period (say 720 hours) he will pay $5587 for energy and (at $2/KVA) $2000
| for demand - total $7787 for $0.11 "average cost per KWH"
| The second pays $2000 for the demand charge and $3802 for energy - "average
| cost per KWH" becomes $0.13 per KWH. In this case it would be beneficial for
| the second user to install capacitors.
In this case energy usage is definitely different. ---
Right
----
| Another case- first one 100KW steady at 1.0 pf and second 100KW steady at
| 0.66pf lag
| First pays $200+5760=$5960 Second pays $303+7760 =$8063 for the same actual
| energy used.
|
| Third case: both at 100KW average- first as above and the second with a peak
| of 1000kVA- both at unity pf. First pays $5960, second pays $7760
|
| A demand charge based on KVA peaks which exist long enough for equipment to
| reach a steady state operating temperature reflects both the effect of high
| load peaks as well as poor power factor.
Now go with BOTH having a peak of 1000kVA. Both demand meters will read the
same thing (1000kVA peak since reset). It could be a one day long peak. So
they would both have the same demand charge. The rest of the time both use
power such that over the whole month (720 hours) the energy used, including
that used during the one day peak, totals 72000 kWH (the equivalent of 100kW
over 720 hours). So their energy readings will be the same, too. But the
difference is that during the times other than peak, they have different pf.
What I am saying is that kVA demand peak and kWH total energy can still have
exactly the same readings even though the kVAH would differ.
--------
That is true.
---------------
The customer
with the lower pf is, of course, putting more load on the distribution due
to more kVAR, but this load isn't contributing to the peak demand since that
one bad day set the peak level. Still, it is overall heating of equipment,
and more importantly, is energy wasted in distribution that does not get
measured by the kWH reading (I'm not talking about what little loss there
is in the customer's own wires, which would be measured by the kWH meter);
I'm talking about the extra current drawn in by the kVAR, not see by the
kWH meter at the customer, but has some partial loss in the distribution
coming in.
---------
OK, noting that the customer is paying for the additional loss on his side
of the meter due to poor pf.
--------------
And that is an energy loss (in addition to anything else).
Maybe we don't need both kW and kVA peak demand readings, but I do think
we need to have both kWH and kVA(R)H readings to measure the cost of the
customer (not just what they benefit from).
-------
The problem is that with both kVA(R)H and kWh readings you still do not have
a handle on the utility's losses. Only if the loads were constant over the
metering period would you get such information and even then it would be
only after some "D'Bougerre" factoring.
Consider a load at 1.0 pf which is at 10kW for 9 hours and 100kW for 1 hour.
in the 10 hour period the energy usage would be 190kWH and kVAH would be
190kVAH (kVARh =0). Note that the current would be (at 1000V) 10A for 9
hours and 100A for the remaining hour. For a fixed resistance of 0.1 ohms
equivalent the loss would be 10 watts for 9 hours and 1000 watts for the
remaining hour. The watthour meter will show 190 +0.09 +1.0 =191.09 kWH and
the customer would be billed accordingly.
At a load of 19 kW for 10 hours the kWH reading would be 190.1 kWH
including losses. The kVAH and kVARH values will be the same in both cases.
The kVA peak would only be 19+kVA.
Now consider a load which is 190kW for 1 hour-and 0 for 9 hours. same kVAH
and kVARH but the metered power is now 190+(190)^2*0.1=193.6 kWH The kVA
peak would be 194- kVA.
In all cases the customer is paying for his own losses. This is true at
any power factor. The increase in utility losses is NOT reflected in kVAH
or kVARH readings which integrate losses. kVA demand in the 3 cases is 100,
19 and 190 respectively. with losses of 1.1, 0.1 and 3.6 kWH respectively.
The same sort of situation would occur at any pf. The customer's energy
charge picks up the customer's loss but neither kVAH or kVARH readings have
any relationship to the utility or customer energy losses that are incurred
and serve no purpose. In fact kVA demand, in this set of examples, is a
better measure- but it is, admittedly, not a good measure. It is a good
measure for equipment costs and sizing (and kVA(R)H is not useful in this
respect).
In fact, the utility measures its generation and the loads and from these
gets a handle on the losses which then is a factor in the rates that are
set. (i.e. generate 100,000,000MWH at $30/MWH and sell 90,000,000 MWH
Cost of power delivered is 10/9*$30=$33.3/MWH + operating cost, interest
and depreciation, taxes, and return to the investor and the customer cost
will be well above $33.3/MWH or 3.3 cents/KWH- say 10cents/kWH?
Please do not trust my math- it is affected by vino. In addition, I think
that we are headed in the same direction, eventually. Much easier one on one
over a beer.
--
Don Kelly
[email protected]
remove the urine to answerhttp://ka9wgn.ham.org/ |
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