home » Style » Synchronous mode. Is it recommended to improve performance by using PRAGMA synchronous = OFF

Synchronous mode. Is it recommended to improve performance by using PRAGMA synchronous = OFF

I see that he got a big speed boost using PRAGMA synchronous=OFF.

I'm facing very slow sqlite update times (250ms) and I need to do a lot of updates from different threads.

I have many open database connections from different threads. Would it be better to just have a central DataBase class wrapping the DB with locks that all threads call and use PRAGMA synchronous=OFF to achieve this speed improvement?

Solution

PRAGMA synchronous only affects disk synchronization; that is. pause to ensure that data sent to the OS is written to disk. Moving the lock won't help with this.

Right now it feels like you're just guessing; You should do some profiling before you optimize. Where are your slow spots? Which queries are slow (use EXPLAIN QUERY PLAN)? Are you ANALYZING?

Also note that SQLite is not very concurrency friendly; only one connection can write to the database at a time. If you need high concurrency, consider a different database.

Change in reactive power. Synchronous compensator mode.

If all the conditions for turning on the generator for parallel operation are met, the armature current is zero, and the machine operates at idle speed. If the excitation current of the generator is increased after synchronization, then a current appears that lags behind 90 el. hail (Fig. 3.23, a). The machine will supply inductive current and reactive power to the network. If the excitation current of the generator is reduced, then a leading current appears (Fig. 3.23, b). The machine will supply capacitive current to the network and consume reactive power from the network.

A synchronous machine that does not carry a resistive load and is loaded with reactive current is called synchronous compensator.

Change in active power. Generator and engine mode.

In order for a machine switched on for parallel operation to generate active power and operate in generator mode, it is necessary to increase the mechanical torque on the shaft (Fig. 3.23c). In this case, a current arises that lags. The generator active power value is

If, on the contrary, you slow down the rotor of the machine, creating a mechanical load on its shaft, then the EMF will lag behind the angle, and the current - by the angle (Fig. 3.23, d). In this case, the active power will be equal, the machine will operate in engine mode, consuming active power from the network.

Asynchronous mode.

In this mode, the modem transmits data one byte at a time. At the beginning of each byte, synchronization bits are transmitted, and at the end of the byte, stop bits are transmitted. There is a certain pause between the transmission of two bytes. This mode works well on low-quality telephone lines because... if data corruption occurs, then a small amount of it (a small amount of data is transmitted in 1 second) and only a small number of bytes have to be repeated. However, the transfer speed is not high.

This mode is designed only for high-quality lines. The modem transmits not one byte at a time, but in clips (as with time multiplexing); those few bytes are transmitted half a row without pauses and intermediate start and stop bits. Start and stop bits are transmitted only at the beginning and end of the clip, and pauses are transmitted between clips. Due to such packet transfer of bytes, transmission is significantly accelerated, but with a bad line and numerous distortions, a large amount of data has to be retransmitted, which does not speed up, but on the contrary, slows down the transfer or makes it impossible at all.

Modems working only in asynchronous mode , usually support low data transfer rates - up to 1200 bps.

Modems working only in synchronous mode , can only be connected to a 4-wire termination. Synchronous modems use high-precision synchronization circuits to isolate the signal and are therefore usually significantly more expensive than asynchronous modems. In addition, synchronous operation places high demands on line quality. There are different standards for them:

· V.26 - transmission speed 2400 bps;

· V.27 - transmission speed 4800 bps;

· V.29 - transmission speed 9600 bps;

· V.32 ter - transmission speed 19,200 bps.

For a dedicated broadband channel 60-108 kHz, there are three standards:

· V.35 - transmission speed 48 Kbps;

· V.36 - transmission speed 48-72 Kbps;

· V.37 - transmission speed 96-168 Kbps.

Modems, working in asynchronous and synchronous modes, are the most versatile and frequently used devices. Most often, they can operate over both dedicated and switched channels, providing full-duplex operation. On dedicated channels they mainly support 2-wire termination and much less often 4-wire termination.

A number of V series standards have been developed for asynchronous-synchronous modems:

· V.22 - transmission speed up to 1200 bps;

· V.22 bis - transmission speed up to 2400 bps;

· V.26 ter - transmission speed up to 2400 bps;

· V.32 - transmission speed up to 9600 bps;

· V.32 bis - transmission speed 14,400 bps;

· V.34 - transmission speed up to 28.8 Kbps; - selection of speed and other parameters depending on the quality of the line

· V.34+ - transmission speed up to 33.6 Kbps. – improved encoding method, works better on noisy lines

At high speeds, V.32-V.34+ modems actually always use synchronous mode in the communication channel.

1. Occurrence and general characteristics of asynchronous modes

In normal steady state, synchronous machines switched on for parallel operation operate synchronously. The synchronous mode is characterized by the fact that the EMF of all electrical machines have the same frequency and, therefore, their vectors rotate at the same angular speed (Figure 1a). Violation of stability leads to the fact that the machines stop working synchronously. In this case, the EMF of synchronous machines that have fallen out of synchronism rotate relative to the EMF of machines operating synchronously (1b).

Rice. 1. a – normal steady state, b – asynchronous mode

A violation of stability can occur as a result of loss of excitation (1), a sharp disturbance (2), or a small disturbance of an overloaded system (3).

Figure 2. Loss of excitation, sudden disturbance, violation of static stability

In a complex system, there may be cases when an asynchronous movement that occurs in one part of the system can lead to a generator or group of generators falling out of synchronism. This is case 4.

Rice. Station 1 falls out of synchronism and the subsequent swaying of station 2 causes it to fall out of synchronism.

In asynchronous mode, the generator acquires additional properties of an asynchronous machine, since currents appear in the rotor due to the presence of slip. Therefore, in a simplified manner, the power of a synchronous generator in asynchronous mode can be represented by two components: synchronous P s and asynchronous P ac. Accordingly, the turbine is counteracted by the synchronous torque M c and the asynchronous torque M ac . But if no voltage is applied to the excitation winding, then the synchronous power will be zero, i.e. There will be only asynchronous power.

For the considered cases of instability - in 1 case the generator will produce only asynchronous power, and in 2,3,4 - both synchronous and asynchronous.

The asynchronous power of the generator, by analogy with an asynchronous machine, can be found by the expression:

where r 2 ∑, x s ∑ are the parameters of the equivalent circuit of an asynchronous machine, taking into account the external resistance of the system.

As slip increases, asynchronous power increases.

Slip is the difference in angular speeds of rotation or electrical frequencies is called slip:

ω s =ω 0 - ω 1

where ω 0 ω 1 are the angular speeds of rotation of the EMF of electrical machines (If there is one machine, then ω 0 is the speed of the system voltage vector, ω 1 is the EMF of the generator).

Asynchronous modes are characterized by periodic changes in the angle between the EMF from 0 to 360, changes (swings) in voltage, current, active and reactive power. Since such changes can be very significant, asynchronous operation in an EPS is not a normal mode and is unacceptable for a long time.

When δ changes, the power of a synchronous generator changes over time approximately according to a sinusoidal law. For large swings, in contrast to asynchronous motion, a dip in the dependence P=f(t) is characteristic, which appears when δ passes through 90. Asynchronous motion is characterized by a periodic change in the sign of synchronous power.

Rice. Towards the definition of asynchronous motion

Let us consider the transition of the generator to an asynchronous operating mode due to a violation of dynamic stability.

Fig. Transition to asynchronous mode of a synchronous generator: power characteristics in normal and asynchronous mode (curves 1,2); change in slip and asynchronous torque (curves 3,4)

Let one of the power lines suddenly turn off and then turn on again. In this case, a transition occurs from characteristic 1 to characteristic 2 and back. But the switching angle δ on is so large that the acceleration area f abcd exceeds the largest possible deceleration area f def . The angle δ exceeds the critical value δ cr. An accelerating torque begins to act on the rotor, leading to a further increase in angle δ.

As soon as the rotor speed begins to differ from synchronous speed, slip s appears, increasing with increasing speed difference. The occurrence of slip causes the appearance of asynchronous power Pac, which increases with increasing slip.

As the rotor speed increases, turbine power regulators begin to operate, reducing Pt.

At a certain slip value s ∞ the turbine power will be balanced by the average asynchronous power. P t = P ac (M t = M ac). This condition determines the beginning of the steady-state asynchronous mode (stroke).

If voltage is applied to the excitation winding, then in addition to the asynchronous torque and turbine torque balancing each other, the synchronous torque M s will also act on the generator-turbine shaft. Synchronous power has a pulsating nature, its average value is zero.

Fig. Changing synchronous torque and slip in asynchronous mode

Fluctuations in synchronous power cause periodic changes in the rotor speed in asynchronous mode and, consequently, slip pulsations. The slip varies from s max to s min.

In the transition process that occurs when stability is disrupted, three stages can be distinguished: 1) loss of synchronism; synchronous swings 2) transition to asynchronous mode 3) steady-state asynchronous mode.

It is assumed that the operation of a synchronous machine occurs at balanced voltages at the terminals of the stator winding; a flowing symmetrical system of currents creates a magnetic field rotating at a speed of co = 2l/. The rotor and the magnetic field from the field winding rotate at the same speed, the current 1 f in which does not depend on the load.

Under the considered conditions of projections of the image vector U s on the axis d And q will be constant values independent of time, i.e., equivalent to constant voltages. The same remark applies to projections from other image vectors (Fig. 5.4). Therefore, the equations for the considered steady state will be obtained from (5.14), assuming p = 0.

Rice. 5.4.A- for the engine; o - for generator

Taking into account the comments made characterizing the regime under consideration, we obtain the equations of motion from (5.14), assuming p = 0 (constant currents):

Here xd=co x L d , x q=co- synchronous inductive resistance of the armature winding along the longitudinal and transverse axes; x ad = with L ad- inductive resistance of mutual induction between the armature and excitation windings located along the longitudinal axis; x ad I f = E f- voltage drop equal to EMF,

created by the excitation current in the armature winding. It is also called excitation emf and no-load emf.

Load angle. The role of load angle 0 in the study of synchronous machines is similar to the role of slip in the analysis of asynchronous machines. In the theory of synchronous machines, this angle is taken to be the angle between the resulting voltage vector U s and vector E f,

coinciding with the positive direction of the transverse axis. His choice will become clearer if we consider the ideal idle speed of the machine, assuming that there are no losses in it, that is, it neither consumes nor produces power. In the mode under consideration, the resulting voltage vector and l. must coincide with the transverse axis (0 = 0). In this case, the flux vector? l will be directed along the longitudinal axis, i.e. will it coincide in direction with the flux vector of the excitation field? f. The latter is easy

establish if we remember that when R s= 0 (no losses), the voltage equation for the stator winding has the form U, = = j v . The torque developed in this case will be equal to zero, which characterizes the idle speed of the machine.

Let us note that the accepted starting point for the load angle corresponds to the stable equilibrium position that the rotor will tend to occupy after the removal of external disturbances.

Thus, a change in the load angle is associated with a change in the position of the rotor relative to the running resultant field of the machine, which is determined by the applied voltage. If, due to an external mechanical force, the rotor is ahead of it, then the resulting electromagnetic torque keeps the latter at synchronous speed. This operating mode corresponds to the generator mode. If the traveling field drags the rotor along with it, overcoming external resistance forces, and keeps it at synchronous speed, then the operating mode under consideration corresponds to the motor mode. For the first case, generator, the load angle is considered positive (0 > 0); for the second, motor, - negative (0

However, one cannot help but notice that the proposed reading of the load angle differs from the generally accepted system, in which the positive reading is made counterclockwise. And this circumstance becomes especially noticeable if we consider the load angle as the angle between two coordinate systems - the coordinate system d, q and synchronous, in which it is customary to combine the axis of real numbers with the voltage vector (see Fig. 5.4). With such a reading, angle 0 is the angle between the imaginary (q) and the real axis of two coordinate systems, and not the axes of the same name.

To bring the reading of the load angle to the reading accepted in circuit theory, we take the load angle as the angle (5 - the angle between the real axes of two systems, measured from the axis d. As can be seen from the vector diagrams in Fig. 5.4, it can be represented through the known angle 0:

with the significant difference that its countdown will be reversed. Now in the motor mode, in which the voltage vector leads the transverse axis, the angle will be positive 0 > 0, while in the generator mode this angle will be negative 0

Components of stress along the longitudinal u d and transverse and h axes will be determined as a result as

and the resulting voltage vector corresponding to them

Rice. 5.5. To determine the longitudinal and transverse components of currents and voltage (Rs. =0)

Thanks to the introduced angle p, the counting of pathological angles in both coordinate systems will be the same, and the convenience of this approach can be observed when determining the longitudinal and transverse components of the current through its active and reactive components in a synchronously rotating coordinate system. Let us carry out this operation for the motor mode, which corresponds to the vector diagram in Fig. 5.4, when the machine operates with a leading power factor coscp>0, then, taking into account (5.17), we obtain

where are the active and reactive components of the current

The given connections between the currents of both coordinate systems will be valid both for the generator mode and for cases when the power factor is negative - in all these cases it is enough to change the sign in front of the angle 0 and the angle φ to the opposite.

Power and electromagnetic torque of a synchronous machine.

In Fig. Figure 5.5 shows a vector diagram constructed in accordance with equations (5.16).

From the diagram it is easy to establish connections between real quantities and their components in the axes d, q:

Power at machine clamps P s we'll get it, come on Ud And U q (5.21):

As you can see, minus the losses due to the armature resistance, the remaining power is transmitted through the air gap. This part of the power represents electromagnetic power R sh, connecting the electrical and mechanical parts of the machine:

Where Ef q represents the algebraic sum of the emf, of which the excitation emf is created by the excitation current E f = x ad I f and EMF,

created by the longitudinal composition of the armature current Id and the difference in inductance along the longitudinal and transverse axes caused by the salient polarity of the machine I d (x d - x q). Both components are created

magnetic fluxes directed along the longitudinal axis.

The relationship between electromagnetic power and load angle 0 can be found if we substitute currents in the power expression Id And 1 (/9 found from solving equations (5.16), in which the stress components must be expressed according to (5.18). Considering that synchronous machines are usually operated at high power with high efficiency, the resistance of the armature winding can be neglected without a noticeable error, i.e. R s= 0. Then, taking into account (5.18)

And, accordingly, we get

The first power term in (5.25) is the result of the interaction of the stator and rotor currents. Unlike an asynchronous machine, the latter is set by an external direct current source.

The nature of the second term of power will be discussed in detail below; here we will only point out that it is associated with the interaction of the stator (armature) current with the spatial harmonic of the magnetic field caused by the salient polarity of the rotor, and created, we emphasize, by the same current. Synchronous machines, the operation of which is based on this phenomenon, are called synchronous reluctance machines.

The electromagnetic torque of the machine will be obtained directly from (5.16) or through the value of the electromagnetic power

In Fig. Figure 5.6 shows the dependence of power, presented as the sum of its components, on the load angle.

Positive load angle values correspond to engine mode, negative values correspond to generator mode. Moreover, in full agreement with circuit theory, in the first case the power is positive (consumed from the network), in the second case it is negative (supplied to the network).

Rice. 5.6. Power of a synchronous salient pole machine as a function of load angle at constant excitation current (R, = 0)

The angular characteristic determines the potential capabilities of the machine. It shows what maximum power the machine can develop with a given excitation current. If external forces turn out to be greater than the maximum values determined by the points A, then in the case of the motor mode the machine will stop, in the generator mode, on the contrary, the speed of the machine will increase due to the supplied mechanical power, it goes into overdrive. In both cases they speak of falling out of synchronicity.

The value of the load angle corresponding to the maximum torque (power) can be found from the known condition s/M, m/s/0 = 0. As a result of differentiation, we obtain

We find the solution to the resulting equation by representing cos 20.„ as

then the last equation is transformed to the form from which we find

The ratio of the maximum torque to the nominal torque characterizes the overload capacity of the machine. She usually sucks

i.e., the working value of the load angle is approximately equal to 0 1УМ » 30°.

Comparing the expressions of the electromagnetic moments of synchronous and asynchronous machines, one cannot help but highlight their fundamental difference. The reason for this situation is due to the fact that the operation of an asynchronous machine is based on the interaction of the running field of the machine with the current induced by it in the rotor winding. Or, in other words, the current in the rotor winding arises as a result of a transformer connection with the stator winding. As a consequence, the inductive parameters that determine the currents, torque and other parameters of the machine are transient inductances, which, as was shown in the previous chapter, are much less than the self-inductances of the stator or rotor circuits, amounting to approximately double the inductance from stray fields.

For synchronous machines we see a different picture. At synchronous speed of rotation of the rotor, there is no transformer connection between the windings, since the rotor and the traveling magnetic field of the armature rotate at the same speed, i.e., they are relatively stationary. Therefore, the behavior of a synchronous machine is determined by the own inductances of the stator windings, which are many times greater than the transient inductances. Their values are inversely proportional to the size of the air gap. Consequently, if you build a synchronous machine on the basis of an asynchronous machine, it will develop significantly less power. In this regard, it becomes clear why in synchronous machines the air gap is much larger than in asynchronous ones. Accordingly, the size and weight of the excitation winding increase, which turns out to be much more expensive than the short-circuited rotor winding of the latter.

Another way to increase the overload capacity and the associated stable operation of the machine is achieved by using high-speed regulators that provide a rapid change in the excitation current when the load fluctuates. In this case, the air gap can be made significantly smaller; the machine operates in nominal mode at a lower overload capacity, at load angles close to the critical angle.

Vector diagrams of a synchronous machine and determination of the excitation emf. For the case when the voltage is specified Us, excitation emf E/ and load angle 0, current components Ij, I, are found according to equations (5.24) (Rs = 0):

For the case under consideration, vector diagrams of the motor and generator modes are presented in Fig. 5.8, clearly reflecting equations (5.16). They allow you to calculate all indicators characterizing the operation of the machine: active and reactive power

current and power factor electromagnetic torque

In the diagrams shown E/= l t x ad correspond to voltage drops across inductive resistances. The Oshes are ahead of the corresponding currents by a distance of 2/2.

In practice, the calculation problem is often posed differently: the total power is given S, power factor coscp, terminal voltage U s and it is required to find the excitation emf that provides the specified output indicators. The task posed essentially boils down to determining the load angle 0. The value of Esh can most easily be found from the graphical construction shown in Fig. 5.7. It was carried out based on the contrary, assuming that the EMF E f excitation and the desired angle 0 between the coordinate systems are known.

Rice. 5.

As can be seen from the construction, right triangles OBC And tkn similar. It follows that the vector

determining the position of the transverse axis, is found through known quantities - current I s given Thus, we can recommend the following procedure for determining the load angle:

1) we plot (arbitrarily) on the complex plane the voltage vector C„ equal to the segment OK;
2) from the end of the voltage vector (point To) draw a perpendicular ahh to the direction of the current and plot the vector on it PC = = ~jXqh
3) through points and and ABOUT draw a straight line that defines the transverse axis With]. The longitudinal axis will be obtained as a perpendicular restored to the axis ts from the origin, and its direction should lag behind the transverse axis c.

At a known angle 0, the currents / and / are easily found, with the help of which the desired value of the excitation emf is determined Er

The given graphical method for searching for electromotive force E/ is a classic solution; it is given in all textbooks and monographs on electrical machines. Its author is, apparently, A.A. Gorev", one of the creators of the theory of synchronous machines based on transformation d, c. A graphical solution based on vector diagrams allows you to avoid analytical transformations, the cumbersomeness of which is due to the difference in parameters along the longitudinal and transverse axes - xj f X h.

Nevertheless, we can offer a simple analytical way to achieve our goals. Looking at the vector diagrams in Fig. 5.7, you can see that the load angle is determined by the vector E f = On, the module of which, as is easy to establish, is equal to

This is more clearly visible from the vector diagram in Fig. 5.8. Thus, the task was reduced to finding the vector F. /(/ = On,

which was previously introduced in the search for electromagnetic power (5.25a).

Rice. 5.8.

Gorev A.A. Transient processes in synchronous machines. - M.-L.: SEI, 1950.-552 p. ill.

We will carry out the search based on stress equations (5.16). To introduce the indicated value into them, it is enough to add and subtract to the right side of the equation for Ud term l d x q , then by

The resulting system of equations corresponds to a non-salient pole machine in which the inductive parameters along the longitudinal and transverse axes are equal X (/ . Multiplying the first equation by - 1 and the second

on -j and adding them up we get

iti

From the resulting equation, “written in the rotor coordinate system, we move on to a synchronous coordinate system, in which the axis of real numbers, as can be seen from the diagrams above, is determined by the voltage vector. Representing the variables through the real and imaginary components in the new coordinate system, the above equation will be written as

from here, equating the imaginary and real parts on the left and right sides, we find

and, thus, for the desired load angle 0 and EMF Efq the following expressions will be obtained:

Here the currents are determined by the given load mode, and for the leading power factor they will be positive, for the lagging power factor they will be negative, regardless of the operating mode.

excitation emf Ef can be found using the found expressions using the formula

where is the current Id is determined using (5.20). The resulting expression for the excitation emf makes it possible to find the corresponding excitation current from the no-load characteristic E f = E f (I f).

Current diagram and equivalent equivalent circuit. The role of current diagrams and the corresponding equivalent circuits will become clear if we consider that, obtained on the basis of voltage equations, in addition to current, we can find all energy quantities, such as power, torque, power factor, efficiency. This kind of expectation will be justified if the diagram represents the dependence on the parameter characterizing the load in the form of a circle, as was the case in the analysis of asynchronous machines. For synchronous salient-pole machines, the change in the current vector from the load angle is more complex - it is an ellipse in the rotor axes, which degenerates into Pascal's snail in the synchronous (or natural) coordinate system. The construction of such dependencies is a cumbersome task, the solution of which is of purely theoretical interest. Nevertheless, the problem can be significantly simplified if we introduce into the analysis a conditional emf along the transverse axis F f 4(5.23), with the help of which

equation (5.28) was obtained:

from where the vector E was found using the given currents fq. It's obvious that

on its basis it is possible to solve the inverse problem, when for a given Ef q you need to find the current

As can be seen from the above equation, the current vector describes a circle with a center determined by the vector U s z~", the radius of which is Efq(R j+ x q j. Actual value of excitation emf E f for each load angle is found through the found

active and reactive current components (5.23).

We will show how, based on the voltage equation, in addition to the current, we can find energy relationships that characterize the steady-state operating mode. To do this, multiply the left and right sides of equation (5.28) 0.5 m s, where, recall, I represents the conjugate current complex

and as a result we get

Here the left side represents the power at the machine terminals P s, on the right - the first term determines the electrical losses in the stator winding p el1, the second - the electromagnetic power. Their expanded expressions are given below:

It is not difficult to establish that in the special case when R s = 0, you get an expression for electromagnetic power identical to that found above under the specified condition.

Based on equation (5.28), it is possible to obtain a stress equation taking into account steel losses, which are determined by the total flux linkage of the stator winding, as can be observed from the following equation:

This equation corresponds to the equivalent circuit shown in Fig. 5.9, and the corresponding system of equations along the longitudinal and transverse axes:

Rice. 5.9.

Here 0" is the angle between the vector J a and vector E lying

on the transverse axis. It also follows that for a fixed value Ej- with angle change

load 0" the end of the current vector describes a circle, the radius of which E /s/ / x c.

A complete equivalent circuit is obtained by adding armature winding resistance to the above circuit R s and resistance /‘ m,

Rice. 5.10.

taking into account losses in steel. The result will be an equivalent circuit shown in Fig. 5.10. Here's the inclusion

In this way, it is possible to take into account losses in steel caused by both the main magnetic flux and stray fields.

However, the practical value of the given circuit is small, since it includes the 0" angle, and not the actual 0 angle. Another inconvenience is associated with the determination of the current I sa. To eliminate this drawback, we will perform transformations based on the following considerations. Current/m

increases the current at the machine terminals. It is in phase with the voltage U and, therefore, to the currents Id And I q found from the solution

equations (5.32)

it is necessary to add currents due to losses in the steel. As a result we get

Here 0" is the angle between the vector Ua and transverse axis q. Solving the given equations for

To go to mains voltage U s, it is necessary to take into account the voltage drop across the armature resistance R s , for which we use the equations

substituting into which the expressions for Uad And U ic, we get
Where

Let us move from the rotor coordinate system to a synchronous one, as was done earlier, in which the real axis is aligned with the voltage vector, and take into account that

where positive angle values correspond to the motor mode, we obtain

where do we find the current?

Rice. 5.11.

The obtained result corresponds to the equivalent circuit shown in Fig. 5.11.

As is easy to establish, for a given E /q with a change in load, characterized by an angle of 0, the current hodograph again represents a circle whose radius is R:

with center defined by vector T = U Jz r Its construction

Rice. 5.12.

shown in Fig. 5.12. The point of the current circle corresponding to the angle 0 = 0 is determined by the position of the radius of the vector R drawn from the center ABOUT x, as shown in Fig. 5.13. Note that in the absence of losses (Rs = 0, = °°) the center of the circle lies on the axis of imaginary numbers.

Using a current diagram, all quantities can be found, I characterize! i still established operating mode. In this case, in the future, the pie chart will represent the effective values of current, voltage and EMF:

Let us recall the basic quantities characterizing the operation of a synchronous machine.

1. Electrical power at terminals

is determined by the active component of the current, which is equal to the length of the perpendicular E X Q X , omitted from the end of the current vector OE x to the abscissa axis. Thus, the segment E X Q X , measured on a current scale and multiplied by m[U], gives power

The power will be positive for motor mode and negative for generator mode. The apse axis c (of imaginary numbers), to which the length of the perpendicular is measured E X Q X is called the power input line.

Rice. 5.11

2. Electromagnetic power R em proportional to the length of the perpendicular E X N X, lowered onto the electromagnetic power line bb x:

which passes through the points of the current circle at which R y= 0. Such points can be found based on equations (5.34). Multiplying each equation accordingly by the currents /d, 1(/k adding the left and right sides, we obtain the expression for electromagnetic power

which contains information about the required angles. To do this, let us substitute the currents Id And 1 „, found from the same system of equations.

As a result, we obtain the following equation for power, expressing the voltage components through the load angle (5.18):

Where

Thus, from the condition R em = 0 you get the following equation to find the required load angles:

from which there are points of stable a and unstable a 2 equilibrium in which R w = 0.

Note that for the motor mode, electromagnetic power will be transmitted to the rotor of the machine as part of the power supplied to the stator winding terminals. For generator mode, it represents part of the power transferred to the stator from the mechanical power supplied to the rotor.

3. The power factor cos(p, as can be seen from Fig. 5.13, can be found by directly measuring the angle ep or using the construction shown in the figure: draw a circle with a radius equal to one and a center at the origin ABOUT. Projection of a point h x on the ordinate axis gives the desired value of cos(p.
4. Efficiency. For motor mode, the output power is R um, so it's equal

For the point E x we have q = E X N X /E X Q. Line segment a x q = E X Q X -E X N X determines electrical losses and steel losses. The given efficiency value does not take into account mechanical and additional losses.

For generator efficiency mode, the output power is the power at the stator terminals, therefore

At each point on the circumference of the current, the actual electromotive force Ef, created by excitation current If, let's find out how

Ej=E Jq+ I d (x d -x q), where is the current Id calculated according to (6.21) through the active 1 A and reactive I R current components. Using the current diagram, the problem of finding the excitation emf is easily solved Ef, providing a given load mode. To do this, it is enough to plot the current vector Ij on the complex plane at an angle Ef q, and then the desired value Ef. In Fig. 5.14 shows a solution for the case when losses in the machine can be neglected. In this case, the center of the circle O lies on the abscissa axis, the lines of supplied and electromagnetic power coincide with the axis of imaginary numbers.

Rice. 5.14. To determine the required value Ef

Active for this case I A and reactive I R current components can be represented through circle parameters

and based on the expression from here

11a, based on the above diagram, it is easy to construct different characteristics of a synchronous machine, showing the nature of the change in the armature current from the excitation current

calculated at constant power on the shaft for an engine or at the terminals for a generator. In this case, the connection between the current Ef q / X q and excitation current I f can be found from the given relations.

Operation of a synchronous machine in cross-firing mode.

In connection with the development of autonomous sources based on semiconductor elements with high technical and economic indicators, providing a wide range of supply frequencies, synchronous motors are increasingly used in adjustable drives, successfully competing with direct current drives. As is known, DC motors have control characteristics, which are the standard for constructing adjustable AC drives. The nature of this position is associated with ensuring orthogonality between the axes of the magnetic fields from the stator and rotor currents (armature and excitation), which is achieved using a commutator-brush assembly. As a result, the developed electromagnetic torque depends only on the instantaneous currents of the current values. However, this node is the most vulnerable, weakest point of DC motors due to switching phenomena and sliding contacts that require constant attention. The use of AC motors allows us to get rid of this disadvantage, while the orthogonality of the fields is achieved by hardware, for example, as is done in the so-called frequency-current drive.

For synchronous motors in this mode of operation between the resulting vectors of the stator current 1 5 and excitation I f

the angle l/2 electric radians is provided, as shown in Fig. 5.15. In this mode, the longitudinal component of the armature current l d - 0, which gave reason to call the regime under consideration the transverse field regime. Although under the created conditions the picture of the formation of an electromagnetic torque turns out to be similar to a direct current machine, however, a synchronous motor will be significantly inferior in terms of energy indicators due to its low power factor.

Rice. 5.15. Vector diagram for the transverse field case Id= O

Let us show this using the example of a synchronous non-salient pole machine, neglecting electrical losses

From the condition that the longitudinal component of the current is equal to zero I d = O we find based on equation (5.3) the excitation emf E f and stator current 1 (/ :

The resulting current value can be found from the current diagram in Fig. 5.16, taking into account that the angle A at the vertex of the triangle is a right angle. This easy to establish if you remember that the segment OO x = U s /x a, and the radius of the current circle E f /x a =

= (U s sin Q)/x a.

Thus, it can be observed that the current diagram shown corresponds to the transverse field mode, with the triangle OBO" is an isosceles rectangle. Since the angles φ and 0 are equal, the result obtained means that operation in the transverse field mode occurs at a lagging power factor equal to

Rice. 5.16. Pie chart point for l d = 0, l s = 1 H, 0 =

which will lead to twice the losses per unit of power compared to a DC motor, the picture of torque generation in which is realized in our case. Moreover, the engine, despite the presence of an excitation winding, is a consumer of reactive power.

The conclusion drawn also applies to synchronous salient-pole machines. Although in this case, for each load angle, the radius of the current circle is determined by the resulting EMF along the transverse axis E fq , as can be observed from Fig. 5.13, for mode

transverse field, it will be equal to the excitation emf, since in the considered mode the longitudinal component of the current is zero. Note also that for the same reason there will be no reactive component of the electromagnetic moment, caused by the difference in magnetic conductivity along the longitudinal and transverse axes.

To obtain high energy levels, allowing you to obtain cos

Views