TD - Mono and Triphasic - FERRAH
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ScribdTranslationsTD - Mono and Triphasic - FERRAH
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ScribdTranslationsExercise 01: single-phase system
Give the expression:
the active power consumed by the resistor
of the reactive power consumed by the coil
Deduce the expression:
of the apparent power of the circuit
of the power factor of the circuit
Digital application: Given R = 10 Ω, L = 200 mH, f = 50 Hz and I = 3.6 A.
Calculate U and the phase shift of u with respect to i.
Corrected:
Give the expression:
the active power consumed by the resistance PR = RI²
from the reactive power consumed by the coil QL = LωI²
Deduce the expression:
of the apparent power of the circuit
Boucherot's Theorem: S= 2+ √
( )2 .I2
the power factor of the circuit k = cos ϕ = =
√ 2+( ) 2
Calculate U and the phase shift of u with respect to i
U = = √ 2+ ( )2 .I = 229 V
cos ϕ = 0.157 so ϕ = 81° (inductive circuit)
Exercise 02: single-phase system
A single-phase electrical installation 230 V / 50 Hz includes:
ten bulbs of 75 W each;
an electric heater of 1.875 kW;
three identical electric motors each absorbing a power of 1.5
kW with a power factor of 0.80.
These different devices function simultaneously.
What is the active power consumed by the light bulbs?
2- What is the reactive power consumed by a motor?
3- What are the active and reactive powers consumed by
the installation? (it is assumed that the bulbs and the radiator are
purely resistant
4- What is its power factor?
5- What is the effective current intensity in the line cable? We
add a capacitor in parallel with the installation.
6- What should be the capacity of the capacitor to raise the factor of
power at 0.93?
7- What is the interest?
Corrected
What is the active power consumed by the light bulbs?
10×75 = 750 W
2- What is the reactive power consumed by a motor?
Power factor = cos ϕ = 0.80 hence tan ϕ = 0.75
Qm= Pmtan ϕ = 1500×0.75 = + 1125 vars (Q > 0 because a motor is inductive).
3- What are the active and reactive powers consumed by
the installation?
P = 750 + 1875 + 3×1500 = 7,125 kW
Q = 0 + 0 + 3×1125 = +3,375 kvar
4- What is its power factor?
Apparent power of the installation: S = (7.125² + 3.375²)1/27,884 kVA
Power factor: cos ϕ = 7.125/7.884 = 0.904
5- What is the effective current intensity in the power line cable?
I = S/U = 7884/230 = 34.3 amperes
A capacitor is added in parallel with the installation.
6- What should be the capacitor's capacity to improve the factor of
power at 0.93?
A capacitor does not consume active power, so the installation
always consumes P’= P = 7.125 kW.
Power factor = cos ϕ' = 0.93
from where tan ϕ' = 0.4
Q' = P' tanϕ' = 7.125×0.4 = + 2.85 kvar
The capacitor consumes reactive power:
QC = Q' - Q = 2850 - 3375 = -525 vars
(QC < 0 : a capacitor is a reactive power generator).
QC = -U²Cω where C = 32 µF
What is the interest?
The capacitor allows the installation to consume less power.
reactive for the same active power.
The apparent power is therefore lower, as is the line current:
S' = (P'² + Q'²)1/2 = (7.125² + 2.85²)1/2 = 7.674 kVA (instead of 7.884 kVA)
I’ = S’/U = 7674/230 = 33.4 A (instead of 34.3 A without capacitors).
The line current being less important, the voltage drops and the
losses due to Joule effect in distribution lines are reduced
Exercise 03: equivalent electrical diagram of a single-phase transformer
empty
230 V
1- Calculate the active power PRconsumed by the resistance.
2- Calculate the reactive power QLconsumed by the coil.
3- Use Boucherot's theorem to calculate the apparent power S
of the circuit.
4- Deduce Ieffand the power factor of the circuit.
5- What is the phase shift of v with respect to i?
Corrected
1- Calculate the active power PRconsumed by the resistance.
Joule's Law: PR= Veff² / R = 230² / 1600 = 33 W (this is electrical power)
is degraded in thermal form: this results in heating of the
magnetic circuit of the transformer
2- Calculate the reactive power QLconsumed by the coil.
QL= +Veff² / (Lω) = +134.7 vars
3- Use the Boucherot theorem to calculate the apparent power S
of the circuit.
The circuit consumes active power: P = PR+ PL= PR(the reel does not
does not consume active power.
The circuit consumes reactive power: Q = QR+ QL =L(the resistance does not
does not consume reactive power
4- Deduce Ieffand the power factor of the circuit.
Power factor: k = PR/ S = 0.238
5- What is the phase shift of v with respect to i?
cos φ = 0.238 hence φ = +76.2°
The phase shift is positive (the voltage leads the current) because the
the circuit is inductive (Q>0).
It will be noted that a no-load transformer consumes a lot of
reactive power: it is strongly inductive.
Exercise 04: Single-phase load
We consider the single-phase load represented in the figure, placed under
a sinusoidal voltage with an effective value of V = 230 V and a frequency of 50 Hz.
1) Calculate the effective value of the current I1circulating in the resistance R1.
2) Calculate the effective value of the current I2circulating in the resistance R2.
3) Calculate the effective value of the current I absorbed by the entire system.
circuit.
4) Calculate the value of active power P, reactive power Q and apparent power S
related to this circuit.
5) Deduce the value of the power factor of this load.
Corrected
Exercise 05: three-phase system
Consider a balanced three-phase receiver made up of three radiators R = 100 Ω.
This receiver is powered by a three-phase network of 230 V / 400 V at 50 Hz.
1- Calculate the effective value I of the line current and the active power P
consumed when the receiver coupling is in a star configuration.
2- Take up the question with a triangle coupling.
3- Conclude.
Corrected
1- When the receiver coupling is in star.
Voltage across a radiator: V = 230 V (voltage between phase and neutral).
The current in a radiator is also the line current: I
Ohm's Law: I = V/R = 2.3 A
The three-phase receiver consumes 3RI² = 1.6 kW (Joule's Law).
2- With a triangular coupling.
Voltage across a radiator: U = 400 V (voltage between phases).
The current in a radiator is the phase current: J.
Ohm's Law: J = U/R = 4.0 A
Hence the line current: I = J√3 = 6.9 A
Joule's Law: 3RJ² = RI² = 4.8 kW
3- Conclude.
In triangle coupling, the line current is three times higher than with a
star coupling.
The same applies to active power: in triangle, the device provides three
times more heat than in a star.
Exercise 06: Three-phase network with balanced and unbalanced load
A three-phase network (U = 400 V between phases, 50 Hz) supplies a
resistive receiver (star connection without neutral)
R = 50 Ω
Calculate the effective values of the line currents I1, I2, and I3.
Calculate the active power P consumed by the three resistors.
A short circuit occurs on phase 3:
Calculate the effective values of the line currents I1 and I2.
3- Phase 3 is cut: Calculate the effective values of the currents
line I1, I2and I3.
Corrected
1. Calculate the effective values of the line currents I1, I2, and I3.
I24.62 A
I34.62 A
Calculate the active power consumed by the three resistors:
P = 3VI cos φ = 3 × 400/3×4.62
√ ×1= 3200 W
A short circuit occurs on phase 3:
Calculate the effective values of the line currents I1and I2.
I1= U/R = 400/50 = 8 A
I2= 8 A
3- Phase 3 is cut off:
Calculate the effective values of the line currents I1I2, and I3.
400
I1= = = 4A
2 x 50
I2= 4 A
I3= 0 A
Exercise 07: three-phase system
On a network (230 V / 400 V, 50 Hz) without neutral, three are connected in a star.
identical capacitive receivers with resistance R = 20 Ω in series with a
Capacitance C = 20 µF.
1- Determine the complex impedance of each receiver. Calculate its
module and its argument.
2- Determine the effective value of the currents in line, as well as their
phase shift relative to simple voltages.
3- Calculate the active and reactive power consumed by the receiver
three-phase, as well as the apparent power.
Corrected
1- Determine the complex impedance of each receiver. Calculate its
module and its argument.
2- Determine the effective value of the currents in line, as well as their
phase shift with respect to simple tensions.
3- Calculate the active and reactive powers consumed by the receiver
three-phase, as well as the apparent power.