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Power Quality for Beginners

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Fig 1: From V = AR, power VA becomes AR x A = A²R watts.

The higher the resistance, the less the current and vice versa. The current A = V/R which means that V = A x R. The power consumed is calculated by multiplying voltage by current (volts x amperes = watts), or VA = W.

With alternating current (AC) things are not as simple. However, with resistive loads, such as incandescent lamps, toasters, or cooking oven heaters, whether the current is AC or DC makes no difference. (Fig. 2).

Fig 2: Note that V and A pass through o to 360 degrees fo the AC wave form at the same time, i.e., in phase (zero degree difference).

Thus, the AC current that flows will be in phase with the applied voltage, and the power consumed (as in DC) will still be VA = W.

Without going too deep into mathematics, this can be described as follows: The cosine of zero = 1. If the cosine is zero, as for a resistive load such as an incandescent lamp, VA always equals watts. (VA x cos 0° = VA x 1 = watts) Cos 0° is known as the power factor (PF) here. For a resistor, the PF is 1.0 which is the highest - and best -value possible for the power factor.

However, other properties of an electric load, such as inductive reactance (XL) and capacitive reactance (XC), make things a bit more complex. When AC voltage is applied to a load that is not a resistor, but more electromagnetic in nature, things become different. For example, in a conventional fluorescent ballast, the current does not start at the same instant as the voltage is applied. The core of the ballast has to become magnetized before it will allow the current to pass. This delay is called a phase shift of the current in relation to the voltage (Fig. 3).

Fig 3: With AC operation of an inductor, the current A starts after the voltage V. The current lags by 90°.

When the load is a capacitor instead of an inductor, the exact opposite happens: The current flows first in order to build up the voltage across the capacitor (Fig. 4).

Fig 4: With AC operation of a capacitor, the current A starts ahead of the voltage V. The current leads by 90°.

For both inductive and capacitive loads, when the voltage is at its maximum, the current is zero. There will thus be no power dissipated in the load, despite the presence of both a voltage and current! Inductive and capacitive loads store energy in one part of the AC cycle and return it in another part. This means that voltage and current never interact, being always out of phase by 90°. Mathematically, cos 90 = 0, which means that their power factor is zero.

ELECTROMAGNETIC CFLs: PHASE SHIFT
A conventional (electromagnetic) compact fluorescent lamp (CFL) consists of a folded fluorescent tube with a standard ballast inside. The tube dissipates some power. The ballast has some resistance but is more inductive to control the current (Fig. 5). Due to this combination of resistance and inductance, the lamp current is neither in phase nor lagging by 90°. It typically lags somewhere between 55 and 60 degrees (Fig. 5).

Fig 5: In the electromagnetic CFL, the combination of resistance and inductance makes the lamp current lag somewhere between 55 and 60 degrees.

This means that the power factor of the conventional CFL would be between 0.5 (cos 60°) and 0.57 (cos 55°).

However, it should be noted that for cosine of angle (theta ) = PF, it is necessary that both the voltage and current are in the form of sine waves. The simplified formula power factor = cosine of angle does not apply if the current flows in a different wave shape or in pulses, e.g., during only part of the voltage cycle.

ELECTRONICS: PEAKY WAVE FORM
This is normally the case with electronically-ballasted CFLs. Here the lamp is controlled by a built-in, light-weight electronic circuit. The fluorescent tube is fed via a switch-mode power converter/inverter: The alternating current is rectified into high-voltage direct current, stored in a reservoir capacitor and then fed to a DC/AC inverter oscillating around 45kHz. Because of the electronics and the high frequency, the efficacy of the lamp exceeds that of conventional CFLs, and there is virtually no flicker. This kind of circuit is known to have one important characteristic; i.e., the current is drawn in the form of pulses. (Fig. 6)

Fig 6: In the electronic CFL the current is drawn in pulses.

Where is the power factor problem then? The current is not a sine wave anymore, and it has a distorted wave form. A high proportion of such a current and its components are not at the fundamental frequency of 50Hz; instead they are at various odd harmonics: three-times the frequency, five-times, seven-times and so on (Fig. 7).

Fig 7: Various odd harmonics of the 50Hz frequency.

In other words, if all the harmonic currents are added or subtracted with reference to the fundamental frequency, a "peaky" current similar to "A" in fig. 6 will be the result. These odd frequency currents make no contribution to the real power consumed by the lamp, but do add to the overall current. The effect is therefore very much like the conventional CFL itself: a higher current and a lower PF (usually between 0.5 and 0.6). Also, since the lamp current in an electronic CFL is only in pulses, it cannot be corrected with a capacitor.

PF AND THE UTILITY
The power factor of a conventional 18-W CFL can be calculated using the formula:

P = VA cos theta (= VA PF)
V = 230 V, A = 0.15 Amp, and PF = 0.52

Therefore the power consumed =
230 x 0.15 x 0.52 = 17.94 watts.

We can also re-write:

PF	.	active power	.	.
___	=	______________	=	0.52
VA	.	apparent power	.	.

Theoretically, if the PF could be as high as 1.0, then the apparent power (the power generating capacity needed by the electric utility) needs only to match the active power. But a low PF of 0.52 means that the volt-amps (VA) has to be nearly twice as great as the watts. In other words, there is twice as much current to be handled for this power. Electric generators are rated based on their current capacity in kVA (kilovolt-amperes) or MVA (megavolt-amperes). However, power plants are often rated in kilowatts or megawatts. This is because loads were once purely resistive (the first electric generating companies were lighting companies that installed incandescent lamps on their customers premises!), and VA and W were almost the same for such early loads.

The generating capacity (in kVA or MVA) of an electric utility must be able to meet the maximum demand on the system at any given time. It is only when generating capacity is utilized by customers to the maximum extent, by keeping the current in phase with the voltage (i.e., with a high PF), that the utility can utilize its own capacity best and also get maximum revenue from its customers.

However, in the real world, there is never such a thing as loads with a consistently high power factor, since the load is made up of a mixture of machines and appliances, some with high PF and others with very low ones. The utility must therefore maintain more capacity in VA just to supply the reactive currents, which do not produce watts. In a simplified way this can be described using the water pump analogy in Fig. 8.

Fig 8.

In the analogy the water pump (or power generator) and the water pipe (the transmission line) are designed for a capacity of 100 liters/min. If the customer's tanks can be filled at 100 liters/min, then the utilization is 100%. But if the customer takes water only in spurts, e.g. 100 liters for one minute, tap closed for one minute, and so on, the effective usage rate is only 50 liters/min. The customer will only pay for that actual amount, while the utility must maintain twice that capacity unnecessarily; i.e. the utility must maintain a capacity to deliver 100 liters per minute.

TRANSMISSION LOSSES
There is also the problem of transmission losses in the electricity system (analogous to friction in the water pipe example) and harmonics (analogous to turbulence in the water pipe), which are additional burdens. Turbulence slows the flow of water through the pipe. In a similar way, harmonics bring about a lower PF, which reduces the transmission system's ability to convey (active) power.

An electric utility can therefore require that a customer must have a power factor of 0.8 or 0.85 to minimize the reactive (unproductive) current which does not contribute to this power consumption. (In practice, this is only done with large customers).

Let us now take a schematic example of a distribution situation where the load happens to be one thousand units of 100-W incandescent bulbs. The transmission line in our example has its own resistance of R ohms. The current A will cause a loss of A2R watts in the transmission line, apart from the load itself (Fig. 9).

Fig 9.

GLS load of 1000 x 100 W = 100 kW
PF = 1.00
Demand = 100 kVA

.	.	100 000 VA	.	.
Current A	=	__________	=	400 Amp
.	.	250 V	.	.

Transmission losses:
A2R = 400 x 400 R watts = 160 R kW

(Actual losses in kW in the example will depend on the value of R, which is determined by the material used in the line, as well as the cable thickness and length. It is not uncommon for a good city network to sustain distribution losses of 4-5%. Thus, in our example, transmission and distribution losses could be 4-5kW in addition to the load of 100 kW. However, in a rural application in a developing country, the T&D losses may be as high as 15-20%.)

Let's now see what happens if all incandescents are replaced by conventional CFLs (Fig. 10).

Fig 10.

If we assume that T&D losses in the incandescent example were 4-5 kW, then the T&D losses in the conventional CFL example would be 1-1.25 kW, a 75% reduction in T&D losses.

So, when incandescents are replaced with conventional CFL's, even with a low PF, the following occurs:

• The line current is reduced by half.
  
  
• The kVA demand is reduced by half, which means that half the capacity has been saved.
  
  
• The new kW load is only 25% of that existing in the GLS situation.
  
  
• The customer will save 75% of the energy, since customer energy use is dependent on kW (active load) and duration.
  
  
• The savings in electricity production at the power plant will also be ~75%.

• They reduce the PF of an electric system. The implications for the low PF of electronic CFLs is very much the same as for conventional CFLs, but depending on the amount of harmonic distortion and the load status of the distribution grid, these currents can cause additional losses.
  
  
• They can interfere with other electronic and signaling equipment (so-called mains pollution).

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