You have learned the following equation for DC circuits:
P = VI
A more general equation for all circuits is this one:
p(t) = v(t) × i(t)
At any instant in time, this equation will be true. Nevertheless, engineers are usually more interested in average power than in instantaneous power.
P = I2R and P = V2/R are well known
equations for power consumption in DC circuits. Similar equations
apply for AC circuits:
P = Irms2R
P = Vrms2/R
P in the above two equations refers to average power. The rms subscript is usually left off so that the equations P = I2R and P = V2/R are understood to use rms values. This will be a general rule: When I and V are used in an AC circuit, they are meant to be rms values.
Similarly, an AC circuit can consume "reactive
power," given by these equations:
Q = I2X
Q = V2/X
X is the reactance, as in Z = R + jX. Unlike R, X can take on negative values. So, although P (real power) is always positive, Q (reactive power) may be positive or negative. The reactive power is not in the units of watts, because it is not real power. The units are VARs, which stands for volt-amps reactive. Real power represents the power available to do work. Reactive power can never do any work -- it's just a measure of energy storage in the circuit.
In some books, you will find Q replaced by Px for reactive power.
In a general AC circuit, the angles of the phasor voltage and current
will be different; e.g.
V = 120/32° V
I = 5/95° A
It will be useful to find the difference between these two
θ = θv - θi
θ is called the power factor angle.
In this specific case,
θ = 32° - 95° = -63°
Since the voltage and current angles are different, it should
not be surprising that the power P is not just the product of V
(the rms voltage) and I (the rms current). In fact, this is the
P = EIcosθ watts
In this equation, E is the same as V, the rms voltage. The cosθ part of the equation is referred to as the power factor. We usually work with the power factor rather than the angles themselves.
For the example above,
P = EIcosθ = (120)(5)cos(-63°) = 272 W
Power factor = .45
In manufacturers' literature, power factor is frequently expressed as a percentage. In the above case, the power factor would be given as 45%.
The power factor may be leading or lagging. Recall that cosθ = cos(-θ). How do we differentiate between the two? When the current lags the voltage, as it does in an inductor, we get a positive value for θ, and we say that the power factor is lagging. When the current leads the voltage, as it does in a capacitor, we get a negative value for θ, and we say that the power factor is leading. Sometimes power engineers speak of inductors absorbing magnetizing VARs and capacitors providing magnetizing VARs.
Here's a mnemonic for remembering the leading/lagging
relationships for inductors and capacitors:
ELI the ICE man.
The ELI part implies voltage leads current in an inductor. The ICE part implies current leads voltage in a capacitor.
If the load is primarily inductive, Q will be positive. Most motors have lagging power factors because they are made of wire and iron, just like inductors.
L and C modify instantaneous power but do not use any average power. Their presence can, nevertheless, change the amount of average (real) power consumed in a circuit.
1 Horsepower = 745.7 watts.
Motors and generators are not 100% efficient. In a motor, for example, the input electrical power is always reduced by windage, friction, and ohmic loss before it can appear as output mechanical power. If a motor has an efficiency of 89%, for example, and puts out 5 HP (mechanical), the input power (electrical) is 5 × 745.7/.89 = 4189 W.
Apparent power is defined as:
S = EI
E and I are rms values of voltage and current. S is called the apparent power because it is the power that would result if the power factor angle were zero. The units are volt-amps (not watts). We now have three sets of units: watts for real power, VARs for reactive power, and volt-amps for a combination of the two.
The relationship among P, Q, and S is best understood with the power triangle, shown below.
The relationships are just what you'd expect from a triangle:
S2 = P2 + Q2
P = Scosθ
Q = Ssinθ
This triangle clearly shows why apparent power (S) is not the same as real power. You will use this triangle repeatedly to solve power problems.
Real power in AC circuits is given by these equations:
P = EIcosθ watts
where cosθ is the power factor (pf).
Reactive power in AC circuits is given by these equations:
Q = EIsinθ VARs
X is positive for inductors and negative for capacitors.
kVA = kilo-volt-amps
By conservation of energy, total power consumed in a circuit is equal to the sum of the power consumed by each element. Also, power generated equals power consumed. Likewise, reactive power generated equals reactive power consumed.
Advice: ALWAYS DRAW A WELL-LABELED DIAGRAM SHOWING THE POWER TRIANGLES(S).
In the example below, we consider the case of an AC generator feeding a motor through a transmission line.
One of the first things to notice is that the motor is not specified in terms of its impedance, but rather in the way it uses power. This is a 20 kW motor. This means that it uses 20 kW of real power. It has a power factor of 0.6. We could use this to calculate θ, but we will see that finding the angle is not necessary.
The task is to find the motor current, then find the voltage, power, and power factor of the generator.
We begin with a power triangle:
Here's how we calculate the triangle values:
S = P/pf = 20×103/.6 = 33.3 kVA
Q = √S2 - P2
Recalling that S = EI, we can find I, the motor current:
I = S/E = 33.3 × 103/120 = 278 A
The motor current is the same as the transmission line
current, so we can make the following calculations for losses in
the transmission line:
P = I2R = 3.86 kW
Q = I2X = 23.19 kVAR
Now we will use conservation of energy to determine the net
power loss in the motor/transmission line combination:
PT = Pmotor + Pline = 20 kW + 3.86 kW = 23.86 kW
QT = Qmotor + Qline = 26.7 kVAR + 23.19 kVAR = 48.89 kVAR
With these two values, we can now draw another power triangle for the whole circuit.
S is found from S2 = P2 + Q2.
The power factor is just P/S:
pf = P/S = 23.86/55.3 = .431 lagging
We know the power factor is lagging, because the load is inductive.
Finally, we'll find the generator voltage:
Eg = S/I = 55.3×103/278 = 199 volts
In the above example, a 22 kVAR capacitor is placed across the motor to partially correct the pf. Find the new values of Eg, pf, and P necessary for the generator.
Before beginning this solution, notice that the capacitor was specified in kVAR, not in microfarads. This is common practice in industrial applications. Catalogs of industrial capacitors will specify the kVAR and voltage rating of the capacitor. Recall that Q is positive for inductors and negative for capacitors. Nevertheless, the capacitor is specified as 22 kVAR and not -22 kVAR. This is because the engineer using industrial catalogs is expected to know that Q is negative for capacitors, and therefore using the minus sign is superfluous. You, the student, will also be expected to know this. In the diagram below, the actual (negative) value is used. This may not always be the case with other diagrams.
From Example 1, we know that for the motor,
Q = +26.7 kVAR
Therefore the total Q for the motor and capacitor together is,
Q = 26.7 - 22 = 4.7 kVAR
This results in the following power triangle:
Notice that the power of the motor did not change as a result
of adding the capacitor. Now, we'll calculate the current in the
I = S/E = 20.54×103/120 = 171 amps
Before going on, let's note that the line current (171 A) is less than the motor current (278 A). This occurs because the capacitor has corrected the power factor to be more nearly unity.
For the transmission line:
P = I2R = 1.47 kW
Q = I2X = 8.79 kVAR
Now we apply conservation of energy to get the totals:
PT = 20 kW + 1.47 kW = 21.5 kW
QT = 4.7 + 8.79 = 13.5 kVAR
This allows us to draw the composite power triangle.
pf = P/S = 21.5/25.4 = 0.85 lagging
Eg = S/I = 25.4×103/171 = 149 volts
Notice that addition of the capacitor has allowed us to reduce the generator voltage. This is a direct result of improving the power factor.
Before going on, you should complete Tutorial 12 on AC power.
It's simple for residential users of electric power. They pay about 10 cents per kW-h. Note that this is an energy charge, not a power charge. In a few markets, residential users may be charged at a different rate dependent on the time of day; electricity may be cheaper at night.
Commercial users, such as banks and shops, may pay a lower energy charge (perhaps 4 cents/kW-h) than residential users, but they may also pay a base service charge plus a maximum load charge. They may also be required to keep the power factor above some specified minimum.
Industrial customers pay the lowest energy charge of all, perhaps 3 cents/kW-h or less. Instead they may pay in many other ways:
Base Service Charge: This is a basic monthly fee of perhaps $300.
Demand Charge: This is a fee based on the maximum amount of power (not energy) that is used during the month. The power a customer uses varies from time to time during the day and from day to day. The power company monitors this power use and notes the peak power used during the billing month. For example, the industrial customer may have had a maximum power usage of 300 kW; at the rate of (say) $8/kW, the customer would pay $300 × 8 = $2400 as a demand charge. Usually, the demand charge varies throughout the year, with the highest demand charge occurring during the summer months.
Power Factor: The power company rewards industrial customers who have good power factors (PF near 1), and punishes those who have poor power factors (PF << 1). This carrot/stick approach is usually applied as an adjustment to the demand charge. For example, a customer might have a lower demand charge rate if his power factor is above 80% (PF = .80), and a higher demand charge if it's below 80%.
We learned earlier that we could maximize the power consumed in a resistive load if we set the load resistance equal to the Thevenin resistance of the circuit the load was connected to, (i.e. RL = RTH). For complex circuits, the result is similar:
ZL = ZTH*
The asterisk (*) indicates complex conjugate. To find the complex conjugate of a complex number, change the sign of the imaginary part (if in rectangular form) or change the sign of the angle (if in polar form). Below are two examples.
(5 + j6)* = (5 - j6)
20/-35° * = 20/35°
Consider the circuit below.
For ZL to consume the most power, it must have an impedance of (8 - j7)* = 8 + j7 Ω.
Now let's calculate the power consumed in ZL. Note that the total impedance seen by the voltage source is (8 - j7) + (8 + j7) = 16 Ω . The impedance seen by the voltage source is totally real; the imaginary parts cancel out. We can now calculate the current:
I = 100/25° /16 = 6.25/25° A
To calculate the power consumed in the load, we just want the magnitude of I (i.e. 6.25 A). Recall that phasor currents are based on the maximum value of the current, not the rms value. To calculate power, we need the rms value:
Irms = Im/√2 = 6.25/√2 = 4.42 A
We then calculate the power in the load as follows:
P = Irms2RL = (4.42)2 × 8 = 156.25 W
Note that we use only the real part of the load impedance in this calculation. We do not use the imaginary part of ZL as this would give us reactive power.
Before going on to the homework, you should complete Tutorial 12A on maximum power transfer in complex circuits.
1. In Example 1, what kVAR value of capacitance would be needed to exactly correct the power factor of the load (the motor and capacitor together) to unity? What must the voltage of the generator be so that the motor voltage (with the capacitor in place) is 120 V?
2. Find the real and reactive generator power. The motor puts out 3 HP and has an efficiency of 89.5%.
3. Two machines and a lighting load are connected in parallel to a 220 V, 60 Hz line. The first machine uses 40 kW of power at 0.81 power factor lagging. The second machine uses 27 kVA at 0.93 power factor leading. The lighting load consumes 13 kW. What is the current in the line feeding the three loads? What kVAR value of capacitance connected across the line is necessary to obtain a power factor of 0.98 lagging? With this capacitor installed, what will be the current flowing in the line feeding the four loads?
4. E = 100/15°, R = 70 Ω, Z1 = j330 Ω, Z2 = -j275 Ω. Find the impedance ZL that will consume the most power in ZL. How much power is consumed in ZL?
BONUS (No Partial credit) It is said that a friend once gave Norbert Wiener this problem. Dr. Wiener is the famous mathematical physicist who invented cybernetics, and his friend wanted to see if Dr. Wiener thought like a physicist or like a mathematician. A garbage truck starts out 5 miles from the city landfill. The truck travels at 10 mph. A fly on the garbage truck takes off and heads for the dump at 20 mph. When the fly gets to the dump, it turns around and heads back to the truck. Arriving at the truck, the fly again turns around and heads for the dump. The fly repeats this procedure over and over again until the truck arrives at the dump. How far does the fly go (the total distance, so the answer is not "5 miles")?