AC three phase electric power
1
Three phases AC motivations
▌ Whereas single-phase current requires two wires: the phase (potential V) and the neutral (potential zero), three-phase
current has four wires: three phases (potentials V1, V2 and V3) and a single neutral (Pot= 0).
▌ Three-phase lines are used to distribute large amounts of electrical power.
▌ Advantages:
Three-phase power lines reduce the current of the transmission lines by a factor of 3 for the same power P.
Three-phase allows twice as much power to be transmitted as single-phase lines with the same cable cross-section.
Three-phase allows the transport of power P with half the copper content of single-phase.
Three-phase allows the transport of voltages that are easily usable for driving forces.
The line losses of three-phase are half as small as those of single-phase for the same transmitted power and current density.
Three-phase also makes it possible to cancel fluctuating power.
2
Three phases AC power generator
In a 3 phase system there are three power wires, each 120⁰ out of phase with each other.
3
Three phases AC power generator
Airbus A320 .
4
Voltage associated to the wye configuration
Ph. 1:
𝑒1 𝑡 = 𝑉 2𝑐𝑜𝑠 𝜔𝑡
𝐼1
𝑒2 𝑡 = 𝑉 2𝑐𝑜𝑠 𝜔𝑡 +
𝑉1
𝐸1
𝐸3 Neutral:
𝐸2
𝐼2
𝐼3
Ph. 2:
Ph. 3:
Y Configuration
2𝜋
𝑒3 𝑡 = 𝑉 2𝑐𝑜𝑠 𝜔𝑡 +
3
𝑈12
Neutral wire (opt)
𝑈31
𝑉2
𝑉3
𝑈23
𝑉1 =
𝑉𝑒 𝑗0
=𝑉
𝑉2 =
4𝜋
𝑗3
𝑉𝑒
= 𝑎2𝑉1
𝑉3 =
2𝜋
𝑗3
𝑉𝑒
= 𝑎𝑉1
𝑈31
Imaginary axis
𝑈12
𝑉3
𝑉1
Real axis
𝑈12 = 𝑉1 − 𝑉2 = 𝑉 3𝑒
𝑈23 = 𝑉2 − 𝑉3 = 𝑉 3𝑒
𝑈31 = 𝑉3 − 𝑉1 = 𝑉
5
4𝜋
3
𝜋
𝑗6
𝜋
−𝑗 2
𝑉2
𝑈=𝑉 3
5𝜋
𝑗6
3𝑒
𝑈23
▌ We assume that The set (E1, E2, E3) is "direct balanced three-phase system". "Balanced" means that the amplitudes
are exactly the same and the phase shifts between the signals are 2ft/3. "Direct" indicates the sequence of phases.
Voltage associated to the wye configuration
Ph. 1:
𝐼1
𝑉1
𝐸1
𝑈12
Neutral wire (opt)
𝐸3 Neutral:
𝐸2
𝐼3
𝐼2
Ph. 2:
Ph. 3:
𝑈31
𝑉2
𝑉3
𝑈23
Y Configuration
𝑉1 =
6
𝑉𝑒 𝑗0
=𝑉
𝑉2 =
4𝜋
𝑗3
𝑉𝑒
𝑉3 =
2𝜋
𝑗3
𝑉𝑒
= 𝑎2𝑉1
= 𝑎𝑉1
𝑈12 = 𝑉1 − 𝑉2 = 𝑉 3𝑒
𝜋
𝑗6
𝑈23 = 𝑉2 − 𝑉3 = 𝑉 3𝑒
𝜋
−𝑗 2
𝑈31 = 𝑉3 − 𝑉1 = 𝑉 3𝑒
𝑗
5𝜋
6
Voltage associated to the wye configuration (Proof)
𝑉1 = 𝑉𝑒
𝑗0
=𝑉
𝑉2 =
4𝜋
𝑗3
𝑉𝑒
𝑈12 = 𝑉1 − 𝑉2 = 𝑉. 1 − 𝑒
𝑈12
7
4𝜋
𝑗3
𝑒
−
2𝜋
𝑗3
𝑒
𝑉3 =
1
= 𝑉. 1 − cos
2𝜋
𝑗3
𝑉𝑒
2𝜋
𝑗3
𝑒
= 𝑎𝑉1
𝑈31
Real axis
𝜋
3
1
𝑗6
+𝑗
= 𝑉. 3𝑒
2
2
= 𝑉. −𝑗 3 = 𝑉. 3 0 − 𝑗 = 𝑉. 3𝑒
𝑈12
𝑉3
𝑉1
4𝜋
4𝜋
2𝜋
2𝜋
= 𝑉. cos
+ 𝑗 sin
− cos
+ 𝑗 sin
3
3
3
3
− 1 = 𝑉. cos
Imaginary axis
4𝜋
4𝜋
+ 𝑗 sin
3
3
3
3
= 𝑉.
+𝑗
= 𝑉. 3
2
2
1
3
1
3
= 𝑉. − − 𝑗
− − +𝑗
2
2
2
2
𝑈31 = 𝑉3 − 𝑉1 = 𝑉.
𝑈31
4𝜋
3
1
3
= 𝑉. 1 − − − 𝑗
2
2
𝑈23 = 𝑉2 − 𝑉3 = 𝑉.
𝑈23
𝑗
=
𝑎2𝑉
𝜋
−𝑗 2
2𝜋
2𝜋
+ 𝑗 sin
−1
3
3
5𝜋
1
3
3
3
3
1
𝑗6
= 𝑉. − + 𝑗
− 1 = 𝑉. − + 𝑗
= 𝑉. 3 −
+𝑗
= 𝑉. 3𝑒
2
2
2
2
2
2
𝑉2
𝑈=𝑉 3
𝑈23
Voltage associated to the wye configuration (Remark)
Let z1 and z2 two complex number the product z1.z2 gives a complex number such that : 𝑧 = 𝑧1 𝑧2 and Arg z = Arg z1 + Arg z2
𝑗
2𝜋
3
Thus by multiplying a complex number by 𝑎 = 𝑒 the module does’nt change and the argument increase of
corresponds to a rotation in counterclockwise of 120°. By this way we can verify the following properties.
𝑉2 =
= 𝑎2𝑉1
𝑉3 =
2𝜋
𝑗3
𝑉𝑒
= 𝑎𝑉1
𝑈31
2𝜋
3
𝑉3
𝑈12
𝑉1
Real axis
𝑉1 +𝑉2 +𝑉3 = 𝑉1 + 𝑎2𝑉1 + 𝑎𝑉1 = 𝑉1 1 + 𝑎 + 𝑎2 = 0
𝑉2
𝜋
𝑗6
3𝑒
𝜋
−𝑗 2
𝑈12 = 𝑉1 − 𝑉2 = 𝑉
𝑈23 = 𝑉2 − 𝑉3 = 𝑉 3𝑒
= 𝑎2 𝑈12
𝑈31 = 𝑉3 − 𝑉1 = 𝑉
= 𝑎 𝑈12
5𝜋
𝑗6
3𝑒
𝑈12 +𝑈23 +𝑈31 = 𝑈12 + 𝑎2𝑈12 + 𝑎𝑈12 = 𝑈12 1 + 𝑎 + 𝑎2 = 0
8
which
Imaginary axis
𝑉1 = 𝑉𝑒 𝑗0 = 𝑉
4𝜋
𝑗
𝑉𝑒 3
2𝜋
3
𝑈=𝑉 3
𝑈23
Examples
Ph. 1: R
Generator
Y Configuration
𝑉1
Receptor
D Configuration 𝐽
12
𝐼1
Z=R+j L ;
Z
𝐼3
𝐼2
Ph. 2: S
Ph. 3: T
𝐽23
f = 50 Hz, V= 230 V,
Z
Z
𝑉2
𝑉3
R=15 , and L = 27,6 mH.
𝑈31
𝑈12
𝐽31
V1 is phase origin.
𝑈23
The problem will be dealt with using complex notations.
1 - Give the complex representation of the magnitudes of simple Vi and compound Uij tensions.
2 - Deduce the effective values and the phases in relation to the origin.
3 - Calculate the complex intensities of the currents flowing in the load and in the lines.
4 - Determine the active and reactive power in each branch.
5 - Deduce the total active and reactive power consumed by the three-phase load.
9
Examples
10
Examples
11
Examples
Ph. 1: R
Generator
Y Configuration
𝑉1
Receptor
Y Configuration
𝐼1
Z=R+j L ;
Z
𝑈12
𝑉1
Z
Z
𝐼2
𝐼3
Ph. 2: S
Ph. 3: T
𝑈31
𝑉3
𝑉2
𝑉3
𝑉2
𝐼1
𝐼2
𝐼3
R=10 , and L = 10 .
f = 50 Hz, V= 220 V,
V1 is phase origin.
𝑈23
The problem will be dealt with using complex notations.
1 - Give the complex representation of the magnitudes of simple Vi and compound Uij tensions.
2 - Deduce the effective values and the phases in relation to the origin.
3 - Calculate the complex intensities of the currents flowing in the load and in the lines.
4 - Determine the active and reactive power in each branch.
5 - Deduce the total active and reactive power consumed by the three-phase load.
12
Examples
13
Examples
14
A simple way to draw a cosinus function
Divide x axis in 14 blocs and 12 blocs correspond to 2 pi, thus each step is equal to pi/6 =30°.
Remind you that cos(60°) =0.5; cos(30°)=0.86; cos(theta+pi) =- cos (theta). That’a all.
1
0.86
0.5
- 0.5
- 0.86
-1
15
𝜋
6
2𝜋
3
𝜋
2
2𝜋
Three phase electrical power
Imaginary axis
▌ Three-phase balanced/wye-coupled receivers
𝑈31
𝑉3
𝜑
𝑈12
𝐼3
𝑉1
Real axis
𝐼2
𝜑
𝜑
𝑉2
𝐼1
𝑈23
𝑃𝑇 =𝑃1 + 𝑃2 + 𝑃3 = 𝑉1 . 𝐼1 . cos𝜑1 + 𝑉2 . 𝐼2 . cos𝜑2 +𝑉3 . 𝐼3 . cos𝜑3 = 3𝑉. 𝐼. cos𝜑 = 3𝑈. 𝐼. cos𝜑
𝑄𝑇 =𝑄1 + 𝑄2 + 𝑄3 = 𝑉1 . 𝐼1 . sin𝜑1 + 𝑉2 . 𝐼2 . sin𝜑2 +𝑉3 . 𝐼3 . sin𝜑3 = 3𝑉. 𝐼. sin𝜑 = 3𝑈. 𝐼. sin𝜑
𝑆𝑇 =𝑃1 + 𝑃2 + 𝑃3 + 𝑗 𝑄1 + 𝑄2 + 𝑄3 = 3𝑉. 𝐼. cos𝜑 + 𝑗 3𝑉. 𝐼. sin𝜑
16
Here 𝜑 > 0, corresponding to the case of an impedance equivalent composed by resistor and inductor.
The load received active power P which is dissipated by Joule’s effect in the resistor and received reactive
power Q in the inductor. 𝜑 < 0 correspond the case of capacitor absorbing a negative reactive power.
Three phase electrical power
▌ Three-phase balanced/D-coupled receivers
𝑉3
𝑈31
𝐼=𝐽 3
𝐽31
𝐼3
𝐽12
𝐽23
𝐼2
𝑈12
Real axis
𝑉1
𝐼1
𝑉2
Imaginary axis
𝑈23
𝑃𝑇 =𝑃1 + 𝑃2 + 𝑃3 = 𝑈12 . 𝐽12 . cos𝜑1 + 𝑈23 . 𝐽23 . cos𝜑2 +𝑈31 . 𝐽31 . cos𝜑3 = 3𝑈. 𝐽. cos𝜑 = 3𝑈. 𝐼. cos𝜑 = 3𝑉. 𝐼. cos𝜑
𝑄𝑇 =𝑄1 + 𝑄2 + 𝑄3 = 𝑈12 . 𝐽12 . sin𝜑1 + 𝑈23 . 𝐽23 . sin𝜑2 +𝑈31 . 𝐽31 . sin𝜑3 = 3𝑈. 𝐽. sin𝜑 = 3𝑈. 𝐼. sin𝜑 = 3𝑉. 𝐼. sin𝜑
𝑆𝑇 =𝑃1 + 𝑃2 + 𝑃3 + 𝑗 𝑄1 + 𝑄2 + 𝑄3 = 3𝑉. 𝐼. cos𝜑 + 𝑗 3𝑉. 𝐼. sin𝜑
17
Here 𝜑 > 0, corresponding to the case of an impedance equivalent composed by resistor and inductor.
The load received active power P which is dissipated by Joule’s effect in the resistor and received reactive
power Q in the inductor. 𝜑 < 0 correspond the case of capacitor absorbing a negative reactive power.
Power measured with a wattmeter
▌ A wattmeter is a measuring instrument for alternative current given the value of 𝑹𝒆 𝑉. 𝐼 ∗ .
A
𝐼
Imaginary axis
𝐖
V
𝑉
18
𝑉. 𝐼 ∗
𝑊 = 𝑅𝑒
𝑊 = 𝑉. 𝐼. cos𝜑
𝑉
𝑍
𝜃 = −𝜑
𝐼
Real axis
2 wattmeter method
▌ This method of measurement gives the total active power and the total reactive power in a balanced state.
▌ In unbalanced mode and without neutral current (no neutral cable) it gives the total active power.
▌ This method does not allow the active and reactive power to be measured if the neutral current is not zero.
𝐖𝟏
V
𝑈12
𝐼2
𝑈31
𝑈23
𝐼3
A
𝐖2
19
𝐼1 +𝐼2 +𝐼3 = 0
A
V
three-phase load
𝐼1
𝑊1 = 𝑅𝑒 𝑈12. 𝐼 1∗
𝑊2 = 𝑅𝑒 𝑈32. 𝐼 3∗
𝑃 = 𝑊1 + 𝑊2
𝑄 = 3 𝑊2 − 𝑊1
2 wattmeter method: 𝑷 = 𝑾𝟏 + 𝑾𝟐 (Proof)
𝐼1 +𝐼2 +𝐼3 = 0
𝐼2 = −𝐼1 − 𝐼3
𝑊1 = 𝑅𝑒 𝑈12. 𝐼 1∗
𝑊2 = 𝑅𝑒 𝑈32. 𝐼 3∗
𝑊1 + 𝑊2 = 𝑅𝑒 𝑈12. 𝐼 1∗ + 𝑅𝑒 𝑈32. 𝐼 3∗
𝑊1 + 𝑊2 = 𝑅𝑒
𝑉 1 − 𝑉 2 . 𝐼 1∗ + 𝑅𝑒
𝑉 3 − 𝑉 2 . 𝐼 3∗
𝑊1 + 𝑊2 = 𝑅𝑒 𝑉 1. 𝐼 1∗ + 𝑉 2. −𝐼 1∗ − 𝐼 3∗ + 𝑉 3. 𝐼 3∗ using the relation 𝐼2 = −𝐼1 − 𝐼3 we obtains
𝑊1 + 𝑊2 = 𝑅𝑒 𝑉 1. 𝐼 1∗ + 𝑉 2. 𝐼 1∗ + 𝑉 3. 𝐼 3∗ = 𝑅𝑒 𝑆𝑇 = 𝑃𝑇== 𝑊1 + 𝑊2
Remark:
To reduce the cost it is possible to use only one
monophasic meter by using a wattmeter switch
20
2 wattmeter method: 𝑸 =
𝟑 𝑾𝟐 − 𝑾𝟏 (Proof)
𝑈32 = 𝑉 3 3𝑒
𝑊1 = 𝑅𝑒 𝑈12. 𝐼 1∗
−𝑗
𝜋
6
𝑈12 = 𝑉 1 3𝑒
𝜑 𝐼3
∗
𝑊2 = 𝑅𝑒 𝑈32. 𝐼 3
𝑊2 − 𝑊1 = 𝑅𝑒 𝑈32. 𝐼 3∗ − 𝑅𝑒 𝑈12. 𝐼 1∗
𝑊2 − 𝑊1 = 𝑅𝑒
𝑉 3 3𝑒
𝑊2 − 𝑊1 = 𝑅𝑒
2𝜋
𝑗3
𝑒
𝑉3
𝑊2 − 𝑊1 = 𝑅𝑒 𝑉3
−𝑗
𝜋
6
3𝑒
∗
. 𝐼3
𝜋
−𝑗 6
.
𝜋
𝑗 𝜑3 − 6
3. 𝐼3 . 𝑒
− 𝑅𝑒
𝑉 1 3𝑒
∗
2𝜋
𝑗 3 −𝜑3
𝐼3 𝑒
𝑗
𝜋
6
𝐼2
𝜑
. 𝐼 1∗
− 𝑅𝑒
− 𝑅𝑒 𝑉1 3. 𝐼1 𝑒
𝑉1
𝑒0
3𝑒
𝜋
𝑗6
. 𝐼1 𝑒 −𝑗𝜑1
∗
𝜋
𝑗𝜑1 + 6
Assuming that the load is balanced then 𝑰𝟏 = 𝑰𝟐 = 𝑰𝟑 = 𝑰 and 𝝋𝟏 = 𝝋𝟐 = 𝝋𝟑 = 𝝋, thus
𝜋
𝜋
𝑊2 − 𝑊1 = 3. 𝑉3. 𝐼3 cos 𝜑 −
− 3. 𝑉1. 𝐼1 cos 𝜑1 +
6
6
𝜋
𝜋
𝜋
𝜋
𝑊2 − 𝑊1 = 3. 𝑉. 𝐼 cos𝜑cos + sin𝜑sin − cos𝜑cos + sin𝜑sin
6
6
6
6
3. 𝑉. 𝐼sin𝜑
𝑄
𝑊2 − 𝑊1 = 3. 𝑉. 𝐼sin𝜑 =
=
𝑄 = 3 𝑊2 − 𝑊1
3
3
21
𝐼1𝜑
𝑗
𝜋
6
Example
Z=R+j L ;
R=15 , and L = 27,6 mH.
f = 50 Hz, V= 230 V,
V1 is phase origin.
We want to carry out a power measurement using the two-watt meter method.
1 - Explain why this method can be used in this case.
2 - Make a diagram explaining the connection of the two wattmeters on the lines, taking the
phase 2 as the origin of the voltages.
3 - Determine the indications given by the two wattmeters by calculating them.
4 - Deduct the active and reactive power from the results of the previous section.
5 - Compare these results with the values of these powers calculated directly.
22
6 - We want to reduce the power factor of the ka charge to 0.9 backwards using identical
capacitors connected in star configuration. Calculate the capacitance of these capacitors.
Examples
23
Examples
24