OBJECTIVE
a) To observe the flow of real and reactive power in three-phase circuits.
b) To interpret the meaning of positive, negative, real and reactive power.
INTRODUCTION
Three-phase electric power is a common method of alternating current electric
power generation, transmission, and distribution. It is a type of polyphase system and is
the most common method used by electrical grids worldwide to transfer power. It is also
used to power large motors and other heavy loads.
A three-wire three-phase circuit is usually more economical than an equivalent twowire single-phase circuit at the same line to ground voltage because it uses less conductor
material to transmit a given amount of electrical power. There are three types of power,
real, reactive and apparent power.
Real Power (P):
Alternative words used for Real Power (Actual Power, True Power, Watt-full Power,
Useful Power, Real Power, and Active Power)
In a DC Circuit, power supply to the DC load is simply the product of Voltage across the
load and Current flowing through it i.e., P = V I. because in DC Circuits, there is no concept
of phase angle between current and voltage. In other words, there is no Power factor in DC
Circuits.
But the situation is Sinusoidal or AC Circuits is more complex because of phase difference
between Current and Voltage. Therefore, average value of power (Real Power) is P = VI
Cosθ is in fact supplied to the load.
In AC circuits, when circuit is pure resistive, then the same formula used for power as used
in DC as P = V I.
Real Power formulas:
P = √3VL IL cos θ
P = 3VP IP cos θ
P = 3I2 R
1
Reactive Power (Q):
Also known as (Use-less Power, Watt less Power)
The powers that continuously bounce back and forth between source and load is known as
reactive Power (Q)
Power merely absorbed and returned in load due to its reactive properties is referred to as
reactive power
The unit of Active or Real power is Watt where 1W = 1V x 1 A.
Reactive power represent that the energy is first stored and then released in the form of
magnetic field or electrostatic field in case of inductor and capacitor respectively.
Reactive power is given by Q = V I Sinθ which can be positive (+ve) for inductive, negative
(-Ve) for capacitive load.
The unit of reactive power is Volt-Ampere reactive. I.e. VAR where 1 VAR = 1V x 1A.
In more simple words, in Inductor or Capacitor, how much magnetic or electric field made
by 1A x 1V is called the unit of reactive power.
Reactive power formulas:
Q = √3VL IL sin θ
Q = 3VP IP sin θ
Q = 3I2 XL
Apparent Power (S):
The product of voltage and current if and only if the phase angle differences between
current and voltage are ignored.
Total power in an AC circuit, both dissipated and absorbed/returned is referred to as
apparent power
The combination of reactive power and true power is called apparent power
In an AC circuit, the product of the r.m.s voltage and the r.m.s current is called apparent
power.
2
It is the product of Voltage and Current without phase angle
The unit of Apparent power (S) VA i.e. 1VA = 1V x 1A.
When the circuit is pure resistive, then apparent power is equal to real or true power, but
in inductive or capacitive circuit, (when Reactance's exist) then apparent power is greater
than real or true power.
Apparent power formulas:
S = 3VP IP
S = √3VL IL
S = P + jQ
S = √P 2 + Q2
MATERIALS AND EQUIPMENT
1.
2.
3.
4.
5.
6.
7.
Power supply module EMS 8821
Resistance Module EMS 8311
Inductance Module EMS 8321
AC ammeter Module EMS 8425
AC voltmeter Module EMS 8425
3-phase watt-varmeter Module EMS 8446
Wound rotor IM Module EMS 8231
EXPERIMENTAL PROCEDURE
1. Connect the circuit shown in figure 1
Three phase connection Fig 1
3
2. Change the load as the table 1 shows: resistor, inductor, capacitor, resistorcapacitor, resistor-inductor, inductor-capacitor, and three phase induction motor.
3. Take the measurements and record the readings in table 1.
4. Do the calculation for each load and compare between the readings and your
calculations.
SAFETY PRECAUTIONS
You must wear uniform and safety shoes, and Do not switch power ON in lab until
the instructor has checked the connection.
DATA AND RESULTS
I (A)
P (W)
Q (VAR)
Load
E
(V)
Measured
Calculated
Measured
Calculated
Measured
Calculated
Resistor
208
0.44
0.4
142
144.21
0
Inductor
208
0.46
0.4
15
0
Capacitor 208
0.45
0.4
5
208
0.65
0.6
208
0.67
208
208
ResistorCapacitor
ResistorInductor
InductorCapacitor
Three
Phase IM
S (VA)
E1 √𝟑
0
144.21
144.21
147
144.1
144.1
144.1
0
-147
-144.1
144.1
144.1
147
145
-145
-145
216.16
205.06
0.6
165
160
152
150
216.16
219.32
0.01
0
12
0
5
0
0
0
0.75
0.7
252.18
252.18
103.43
230
Table 1
4
DISCUSSION OF RESULTS
We can see that the resistive load absorbed the real power but not the reactive power. While
the inductive and capacitive load absorbed the reactive power but not the real power. Also,
we can see that there is some real power in inductive and capacitive load because the
inductor and capacitor are not pure.
CONCLUSION
In conclusion from this experiment we find out that Resistor absorbs the real power.
Inductor absorbs the reactive power. Capacitor absorbs the reactive power. Also, we learn
about the power triangle by knowing the relationship between real, reactive and apparent
power
TEST YOUR KNOWLEDGE
1. An electrical load Z is connected to the terminals of a 120-volt AC source. Show the
direction of real and reactive power flow if Z is composed of a) a resistance, b) an
inductance c) a capacitance d) a resistance and inductance e) a resistance and
capacitance f) a single-phase motor. See Fig 2.
Fig 2
ANS:
Load
Real Power
Reactive Power
Resistor
Source → Resistor
0
Inductor
0
Source → Inductor
Capacitor
0
Capacitor → Inductor
Resistor-Inductor
Source → Resistor
Source → Inductor
Inductor-Capacitor
Source → Resistor
Capacitor → Source
Motor
Source → Motor
Source → Motor
5
2. Calculate the real and reactive power which is delivered by the single-phase source in
the two single-phase circuits shown in Fig 3.
Fig 3
ANS:
Z=
I=
V 120
=
= 2.4A∟ − 36.9
Z
50
P = VI cos θ = 120 ⋅ 2.4 cos(36.9) = 230.3 W
Q = VI sin θ = 120 ⋅ 2.4 sin(36.9) = 172.92 VAR
6
3. A three-phase source having a line-to-line voltage of 69kV supplies a wye-connected
resistive load having an impedance of 100 ohms per phase. Calculate the real power
delivered.
ANS:
VP =
IP =
VL
√3
=
69k
√3
= 39.83kV
VP 39.83k
=
= 398.37A
R
100
P = √3VL IL cos θ = √3 ⋅ 69k ⋅ 398.37 cos(0) = 47.6MW
4. Explain what is meant by the statement that an inductor absorbs reactive power while
a capacitor supplies reactive power.
ANS:
7
5. A three-phase power line, shown schematically in Fig 4. delivers real and reactive
power as given in Table 2. Calculate the real and reactive power absorbed by the line.
Fig 4
ANS:
kW1
Kvar1
kW2
Kvar2
LINE
kW
LINE
kvar
+100
+10
+95
+5
+5
+5
+100
+10
+95
-10
+5
+20
+100
-10
+95
-25
+5
+15
-100
+10
-105
+5
+5
+5
6. A three-phase line operating at a line-to-line voltage E supplies power to a wyeconnected load whose impedance is Z ohms per phase. Show that the total apparent
power P is given by the equation.
E2
P=
Z
ANS:
8