Let's take a coil with a ferromagnetic core and take out the ohmic resistance of the winding as a separate element as shown in Fig. 2.8.
Figure 2.8 - To the derivation of the transformer EMF formula
When an alternating voltage e c is turned on in the coil, according to the law of electromagnetic induction, an EMF of self-induction e L arises.
(2.8)
where ψ is the flux linkage,
W is the number of turns in the winding,
Ф is the main magnetic flux.
We neglect the scattering flux. The voltage applied to the coil and the induced EMF are balanced. According to the second Kirchhoff law for the input circuit, we can write:
e c + e L = i * R exchange, (2.9)
where R obm is the active resistance of the winding.
Since e L >> i * R exchange, we neglect the voltage drop across the ohmic resistance, then e c ≈ – . If the mains voltage is harmonic е с = E m cos ωt, then E m cos ωt = , whence . Let's find the magnetic flux. To do this, we take the indefinite integral of the right and left sides. We get
, (2.10)
but since we consider the magnetic circuit to be linear, only harmonic current flows in the circuit and there is no permanent magnet or constant component, then the integration constant c \u003d 0. Then the fraction in front of the harmonic factor is the amplitude of the magnetic flux, from which we express E m \u003d Ф m * W * ω. Its effective value is
Or we get
where s is the cross section of the magnetic circuit (core, steel).
Expression (2.11) is called the basic formula of the transformer EMF, which is valid only for harmonic voltage. Usually it is modified and the so-called shape factor is introduced, equal to the ratio of the effective value to the average:
. (2.12)
Let's find it for a harmonic signal, but we find the average value on the interval
Then the form factor is and the basic formula of the transformer EMF takes the final form:
(2.13)
If the signal is a meander, then the amplitude, effective and average values for half the period are equal to each other and its. You can find the form factor for other signals. The basic formula for transformer EMF will be valid.
Let's build a vector diagram of a coil with a ferromagnetic core. With a sinusoidal voltage at the coil terminals, its magnetic flux is also sinusoidal and lags behind the voltage in phase by an angle π / 2, as shown in Fig. 2.9a.
Figure 2.9 - Vector diagram of a coil with a ferromagnetic
core a) no loss; b) with losses
In a lossless coil, the magnetizing current - reactive current (I p) coincides in phase with the magnetic flux Ф m. If there are losses in the core (), then the angle is the angle of losses for remagnetization of the core. The active component of the current I a characterizes the losses in the magnetic circuit.
How is a transformer arranged?
(b, c) W x. W2 connects to the load.
U 1 i 1 F. This flow induces an emf e 1 and e 2 in the transformer windings:
EMF e 1 U 1, EMF e 2 creates tension U 2
· Step-down transformer - a transformer that reduces the voltage (K> 1).
What is the transformation ratio?
Transformation ratio - the ratio of the operating voltages at the ends of the primary and secondary windings with an open circuit of the secondary windings (idling of the transformer). K \u003d W 1 / W 2 \u003d e 1 / e 2.
For a transformer operating in idle mode, with sufficient accuracy for practice, we can assume that.
What do you know the nominal parameters of the transformer and what do they determine?
Rated power is the rated power of each of the transformer windings. Rated current, winding voltage. The external characteristic is the dependence of the voltage at the transformer terminals on the current flowing through the load connected to these terminals, i.e. dependence U2=f(I2) at U1=const. The load is determined by the load factor Kn=I2/I2nom ≈ I1/I1nom, efficiency - η = P2/P1
How to determine the rated currents of the transformer windings if the rated power of the transformer is known?
The rated power of a two-winding transformer is the rated power of each of the transformer windings.
Rated power equation: S H =U1 * I1 ≈ U2 * I2
I1 = S H / U1 ; I2 = S H /U2
What is called the external characteristic of the transformer and how to get it?
The external characteristic is the dependence of the voltage at the transformer terminals on the current flowing through the load connected to these terminals, i.e. dependence U 2 =f(I 2) at U 1 =const. When the load (current I 2) changes, the secondary voltage of the transformer changes. This is due to a change in the voltage drop across the resistance of the secondary winding I 2 " z 2 and a change in the EMF E 2 "=E 1 due to a change in the voltage drop across the resistance of the primary winding.
EMF and voltage equilibrium equations take the form:
Ù 1 = –È 1 + Ì 1 " z 1, Ù 2 "=È 2 – Ì 2" z 2 " (1)
The load value in transformers is determined by the load factor:
K n \u003d I 2 / I 2 nom ≈ I 1 / I 1 nom;
The nature of the load is the angle of the phase shift of the secondary voltage and current. In practice, the formula is often used
U 2 \u003d U 20 (1 - Δu / 100),
Δu \u003d K n (u ka cosφ 2 + u kr sinφ 2)
u ka \u003d 100% I 1nom (R 1 - R 2 ") / U 1nom
u ka \u003d 100% I 1nom (X 1 - X 2 ") / U 1nom
How to find the percentage change in the secondary voltage of a transformer for a given load?
The percentage change in the secondary voltage ∆U 2% at a variable load is determined as follows: , where are the secondary voltages at no load and at a given load, respectively.
What do you know about the equivalent circuits of a transformer and how are their parameters determined?
T-shaped transformer equivalent circuit:
How is a transformer arranged?
A transformer is a static electromagnetic device designed to convert, by means of a magnetic flux, AC electrical energy of one voltage into AC electrical energy of another voltage at a constant frequency.
The electromagnetic circuit of the transformer (a) and the conventional graphic symbols of the transformer (b, c) shown in Fig.1. Two windings are located on a closed magnetic circuit made of sheets of electrical steel. Primary winding with number of turns W x connected to a source of electrical energy with voltage U . Secondary winding with number of turns W2 connects to the load.
What determines the EMF of the transformer windings and what is their purpose?
Under the influence of the supplied alternating voltage U 1 current flows in the primary winding i 1 and there is a changing magnetic flux F. This flow induces an emf e 1 and e 2 in the transformer windings:
EMF e 1 balances the bulk of the source voltage U 1, EMF e 2 creates tension U 2 at the output terminals of the transformer.
3. In what cases is the transformer called step-up and in what case - step-down?
· Step-down transformer - a transformer that reduces the voltage (K> 1).
step-up transformer - a transformer that increases the voltage (K<1).
Let's take a coil with a ferromagnetic core and take out the ohmic resistance of the winding as a separate element as shown in Figure 1.
Figure 1. Inductor with a ferromagnetic core
When an alternating voltage e c is applied in the coil, according to the law of electromagnetic induction, an EMF of self-induction e L arises.
(1) where ψ - flux linkage, W- the number of turns in the winding, F is the main magnetic flux.We neglect the scattering flux. The voltage applied to the coil and the induced EMF are balanced. According to the second Kirchhoff law for the input circuit, we can write:
e c + e L = i × R exchange, (2)where R obm - active resistance of the winding.
Because the e L >> i × R exchange, then we neglect the voltage drop across the ohmic resistance, then e c ≈ −e L. If the mains voltage is harmonic, e c = E m cosω t, then:
(3)Let's find the magnetic flux from this formula. To do this, we transfer the number of turns in the winding to the left side, and the magnetic flux Ф to the right:
(4)Now take the indefinite integral of the right and left sides:
(5)Since we consider the magnetic circuit to be linear, then only harmonic current flows in the circuit and there is no permanent magnet or a constant component of the magnetic flux, then the integration constant c \u003d 0. Then the fraction in front of the sine is the amplitude of the magnetic flux
(6)whence we express the amplitude of the input EMF
E m = F m × W × ω (7)Its effective value is
(8) (9)Expression (9) is called the basic formula of the transformer EMF, which is valid only for harmonic voltage. With a non-harmonic voltage, it is modified and the so-called shape factor is introduced, equal to the ratio of the effective value to the average:
(10)Find the shape factor for a harmonic signal, while the average value is found in the interval from 0 to π/2
(11)Then the form factor is and the basic formula of the transformer EMF takes the final form:
(12)If the signal is a sequence of rectangular pulses of the same duration (meander), then the amplitude, effective and average values for half the period are equal to each other and its k f = 1. You can find the form factor for other signals. The basic formula for transformer EMF will be valid.
Let's build a vector diagram of a coil with a ferromagnetic core. With a sinusoidal voltage at the coil terminals, its magnetic flux is also sinusoidal and lags the voltage in phase by an angle π / 2, as shown in Figure 2.
WORKSHOP
FOR ELECTRIC MACHINES
AND DEVICES
Tutorial
For full-time and part-time students
in the field of instrument making and optotechnics
as a teaching aid for students of higher education
institutions studying in the specialty 200101 (190100)
"Instrument making"
Kazan 2005
UDC 621.375+621.316.5
BBC 31.261+31.264
Prokhorov S.G., Khusnutdinov R.A. Workshop on electrical machines
and devices: Textbook: For full-time and part-time students. Kazan: Kazan Publishing House. state tech. un-ta, 2005. 90 p.
ISBN 5-7579-0806-8
It is intended for conducting practical classes and performing independent work in the discipline "Electrical Machines and Apparatus" in the direction of training a certified specialist 653700 - "Instrument Engineering".
The manual may be useful for students studying disciplines
"Electrical engineering", "Electromechanical equipment in instrument making",
"Electrical machines in instrumentation", as well as students of all
engineering specialties, including the electrical profile.
Tab. Il. Bibliography: 11 titles.
Reviewers: Department of Electric Drive and Automation of Industrial Installations and Technological Complexes (Kazan State Power Engineering University); professor, cand. Phys.-Math. Sciences, Associate Professor V.A. Kirsanov (Kazan branch of the Chelyabinsk Tank Institute)
ISBN 5-7579-0806-8 © Kazan Publishing House. state tech. university, 2005
© Prokhorov S.G., Khusnutdinov R.A.,
The proposed tests in the discipline "Electrical Machines and Apparatus" are intended for practical training and independent work. The tests are compiled in the sections "Transformers", "Asynchronous machines", " Synchronous machines”, “DC Collector Machines”, “Electrical Apparatuses”. Answers in the form of a table are given at the end of the manual.
TRANSFORMERS
1. Why are air gaps in a transformer kept to a minimum?
1) To increase the mechanical strength of the core.
3) To reduce the magnetic noise of the transformer.
4) To increase the mass of the core.
2. Why is the transformer core made of electrical steel?
1) To reduce the no-load current.
2) To reduce the magnetizing component of the no-load current
3) To reduce the active component of the no-load current.
4) To improve corrosion resistance.
3. Why are the core plates of the transformer pulled together with studs?
1) To increase mechanical strength.
2) For attaching the transformer to the object.
3) To reduce moisture inside the core.
4) To reduce magnetic noise.
4. Why is the transformer core made of electrically insulated plates of electrical steel?
1) To reduce the mass of the core.
2) To increase the electrical strength of the core.
3) To reduce eddy currents.
4) To simplify the design of the transformer.
5. How are the beginnings of the primary winding of a three-phase transformer indicated?
1) a, b, c 2) x, y, z 3) A, B, C 4) X, Y, Z
6. How are the primary and secondary windings of a three-phase transformer connected if the transformer has 11 groups (Y - star, Δ - triangle)?
1) Y/Δ 2) Δ/Y 3) Y/Y 4) Δ/Δ
7. How do the magnetic core and the winding of a conventional transformer differ in mass from an autotransformer if the transformation ratios are the same To=1.95? The power and rated voltages of the devices are the same.
1) Do not differ.
2) The masses of the magnetic circuit and the winding of the autotransformer are less than the masses
magnetic circuit and windings of a conventional transformer, respectively.
3) The mass of the magnetic circuit of an autotransformer is less than the mass of the magnetic circuit of a conventional transformer, and the masses of the windings are equal.
4) The masses of the magnetic circuit and windings of a conventional transformer are less than those of the corresponding values of an autotransformer.
5) The mass of the winding of an autotransformer is less than the mass of the windings of a conventional transformer, and the masses of the magnetic circuits are equal.
8. On what law of electrical engineering is the principle of operation of a transformer based?
1) On the law of electromagnetic forces.
2) On Ohm's law.
3) On the law of electromagnetic induction.
4) On the first law of Kirchhoff.
5) On the second law of Kirchhoff.
9. What will happen to the transformer if it is connected to a DC voltage network of the same magnitude?
1) Nothing will happen.
2) May burn out.
3) The main magnetic flux will decrease.
4) The leakage magnetic flux of the primary winding will decrease.
10. What does a transformer convert?
1) The magnitude of the current.
2) The magnitude of the voltage.
3) Frequency.
4) Current and voltage values.
11. How is electrical energy transferred from the primary winding of an autotransformer to the secondary?
1) Electrically.
2) Electromagnetically.
3) Electrical and electromagnetic way.
4) As in a conventional transformer.
12. What magnetic flux in a transformer is a carrier of electrical energy?
1) Magnetic leakage flux of the primary winding.
2) Magnetic leakage flux of the secondary winding.
3) The magnetic flux of the secondary winding.
4) Magnetic flux of the core.
13. What influences the EMF of self-induction of the primary winding of the transformer?
1) Increases the active resistance of the primary winding.
2) Reduces the active resistance of the primary winding.
3) Reduces the current of the primary winding of the transformer.
4) Increases the current of the secondary winding of the transformer.
5) Increases the current of the primary winding of the transformer.
14. What influences the EMF of self-induction of the secondary winding of the transformer?
1) Increases the active resistance of the secondary winding.
2) Reduces the active resistance of the secondary winding.
3) Reduces the current of the secondary winding of the transformer.
4) Increases the current of the primary winding of the transformer.
5) Reduces the inductive resistance of the secondary winding
transformer.
15. What is the role of the EMF of the mutual inductance of the secondary winding of the transformer?
1) Is the source of EMF for the secondary circuit.
2) Reduces the primary current.
3) Reduces the secondary current.
4) Increases the magnetic flux of the transformer.
16. Choose the formula for the law of electromagnetic induction:
Choose the correct spelling of the effective value of the EMF of the secondary winding of the transformer.
18. How are the magnitudes of the short circuit voltage U 1k and nominal U 1n in medium power transformers?
1) U 1k ≈ 0.05. U 1n 2) U 1k ≈ 0.5. U 1n 3) U 1k ≈ 0.6. U 1n
4) U 1k ≈ 0.75. U 1n 5) U 1k ≈ U 1n
19. What parameters of the T-shaped equivalent circuit of the transformer are determined from the experience of idling?
1) r 0 , r 1 2) X 0 , r 1 3) r' 2 , X' 2