See Also: Peak Alert, Peak Demand
On Friday, July 28, 1995 Glen Johnson of Howell-Oregon called and said he was
worried that they were going to set a peak. AECI had called a peak alert and
in the early morning hours the load was strong, but by 11:00 a.m. it had slowed
its growth - typical for a Friday in the summer. Howell-Oregon’s load, on the other hand, was running approximately 800 kW higher than it had
on the July 12 and July 13 high peak days.
What worried Glen the most was that the A-phase Voltage at West Plains #1 was 127 volts while B- and C-phase were in the 124.5 to
125 volt range. The reason he was worried about high voltage was that it would
drive up his demand. The last thing Glen wanted was for high voltage to cause
him to set a new peak. I told him I would make some calculations and get back
with him.
From Glen and Sho-Me records I came up with the following information:
Metering CT Ratio: 80 Phase A Voltage: 127 Phase B Voltage: 125
Phase C Voltage: 125
STEP ONE: Calculate The Load On Phase A.
As a general rule there is an inverse relationship between voltage and Power on a circuit, the higher the voltage tries to run, all other things being
equal, the lower the load on that circuit. Since Phase A’s voltage is approximately 1.5 percent higher than Phase B and C, I
arbitrarily assumed it had 30 percent of the load and Phase B and C each had 35 percent.
With 30 percent of the load, Phase A had approximately 1,860 kW.
STEP TWO: Calculate The Power Factor
P.F. Angle = TAN (VARS/ Watts)
= ATAN (1.6/6.2)
= 14.47
Power Factor = COS(PF Angle) = COS(14.47
STEP THREE: Calculate The A-Phase Current
The total power on a three-phase circuit can be calculated in two basic ways.
One, calculate the power on each phase and sum the three using:
P(PHASE)= V(L-G) X I(PHASE) X PF (assuming a Y-circuit)
The second way is to assume the circuits are balanced, which means all phase
voltages and currents are equal, and use the following:
P(3-Phase) = 1.732 X V(L-L) X I(PHASE) X PF (assuming a Y-circuit)
In this situation we are only interested in Phase-A so I used the first
equation.
P(PHASE) = V(L-G) X I(PHASE) X PF
The Phase-A voltage equals 127 times 60 (the PT ratio). Rearrange as follows
to calculate the current:
I(Phase) = P(PHASE)/(V(L-G) X PF)
= 1,860 kW/(7.62 kV X .9683)
= 252.1 amps (see Ampere )
STEP FOUR: Calculate The Magnitude Of The Phase-A Impedance
The actual impedance is 30.23 at an angle of 14.47 Degrees. Since this is an inductive (see Inductance) load the angle of the impedance is positive.
STEP FIVE: Lower The Voltage From 127 To 125. Calculate The Current Flow
I = V/Z = (125X60)/30.23 = 248.10
We assume the impedance characteristics of the load out there doesn’t change (more on this later).
STEP SIX: Calculate The Amount of Power On This Circuit.
P(PHASE) = V(L-G) X I(PHASE) X PF
= (125X60) (248.1) (0.9683)
= 1,802 kW
STEP SEVEN: Calculate The Savings From Dropping The Voltage
Saving = 1,860 kW - 1,802 kW
= 58 kW or 3.12 percent.
Since Glen was worried about 800 kW, the 58 kW was not going to break him. Also, Motors will do their best to produce their rated horsepower or KVA. That means the
3.12 percent figure in step seven is not accurate. For example, assume a
single-phase five horsepower motor that is ninety percent efficient, operates at
ninety percent power factor, and 127 volts. How much power will it consume?
STEP ONE: Calculate The Current
Amps = (HP X 746)/(Volts X Efficiency X PF)
= (5 X 746)/(127 X 0.9 X 0.9)
= 36.26
STEP TWO: Calculate The Amount of Power The Motor Draws
P(PHASE) = V X I(PHASE) X PF
= 127 X 36.26 X 0.9
= 4,144.4 W
Now drop the voltage down to 125 and see how much less power the motor uses.
STEP THREE: Calculate The New Current
Amps = (HP X 746)/(Volts X Efficiency X PF)
= (5 X 746)/(125 X 0.9 X 0.9)
= 36.84
STEP FOUR: Calculate The Amount of Power The Motor Draws
P(PHASE) = V X I(PHASE) X PF
= 125 X 36.84 X 0.9
= 4,144.4 W
So the motor uses as much power at 127 volts as it does at 125. Lower the
voltage and the motor draws more amps. Lower the voltage enough and the motor
will burn itself up as it draws more current than its windings are rated for.
What does all this mean? A utility can lower its peak demand and Energy usage by lowering its voltage. But it only works on purely resistive loads
like Light bulbs, strip heaters, etc. On a circuit containing primarily inductive type
loads there would be little difference. With respect to Howell-Oregon I would
suspect they could save 2 to 2.5 percent by lowering their circuit voltage from
127 to 125. What is the downside? When you lower your voltage the power may just barely be dripping out to
those customers on the end of the line. Voltage that is too high will burn out
light bulbs (pretty cheap), while voltage that is too low will burn out
air-conditioning (see Air Conditioner) Compressor motors and refrigerators (pretty expensive - see Refrigeration).
Approx. MW Load: 6.2 Approx. MVAR Load: 1.6 PT Ratio: 60
) = 96.83%