Three-phase systems

alternator

In the previous subject we saw the systems of AC voltage using only one phase, and normally in electronics, he is the habitual thing, although in the systems where great power is used, the three-phase tension is used.

The clients of the electrical companies (usuary of electricity), that a certain consumption stops is commited to use the three-phase tension.

The three-phase Tension is generated in an alternator where we have three separated windings 120 and located in a static part (Stator). When by action of a revolving magnetic field located in the movable part (rotor), the coils receive that variation of field, an electrical, out of phase current 120 is generated with respect to the current of the neighboring winding.

Relation between the tension of line and phase.

He is not the same the tension that we have between one of the phases and the neutral one (phase tension) that the tension that there is between two tips of a generating winding (two phases). For it he is advisable to use a fasorial graph of tensones to obtain this relation.

relation between tension of line and phase
Figure 2

relation tension line and phase

The line tensions are taken between different lines. To we have named them them as Either12, Or23 and U31.

In the image of the left we represented a detail of the previous image. In her we see that the U12 (tension 1 with respect to 2) is equal to the U1 vector less U2.

Also it is possible to be seen by the vectors. If we added U2 and U12, we have the great vector of the right that goes from the N to the 1, that is the U1. Therefore:

U_1_2 + U_2 = U_1 \ Rightarrow {\ color {DarkOrange} {\ mathbf {U_1_2 = U_1 - U_2}}}

The rest of tension we see them in the drawing superior.

  • In the systems of well-balanced loads, the sum of the line current months will be equal to Zero.

In order to calculate the relation that we have between the tension of line and phase, we have represented two of the phases, with 120 an angular difference of .

In order to calculate the tension that there is between two phases, we added the tension of a respect neutral the other invested phase, respect the neutral one. of this form, we know the tension that there is between both.calculation of line tension

Note: It is possible to be demonstrated that, in the triangle is³celes that forms with sides U1, U2, U12, to the being the U1 sides = U2, and to being one of the angles (180 €“ 60 = 120), 30 the lateral attitudes are .

In the same image, we have removed one from the triangles that form when composing the U12. The part orange corresponds the line tension in the middle of. We can calculate with trigonometry since:

Sector orange (Half of U12) = green Sector (U1) * cos 30

This is:

\ tailcoat {U12} {2} = U1 * \ cos 30 = U1 * \ 2 3 tailcoats {\ sqrt {}} {} \ Rightarrow {\ color {Network} \ mathbf {U12 = U1 * \ sqrt {3}}}

The same we can say of the current of line and the one of phase. The current of A stage which it circulates around the load and current of line is called the one that circulates around the distribution cable.

Therefore, the tension of phase, VF, is the tension that appears between the terminals of a load and the tension of line, VL, is the tension difference that there is between the drivers of power supply.

2- Connection of three-phase loads

2,1 star Connection. In this case, we have in figure 2, a well-balanced star connection where one is fulfilled that, to the being all the equal loads, the absolute value of the intensity is the same in each branch and the value of the current that enters the neutral one is zero. If we calculated the power in one of the branches, we will have the power of all the system. The power comes given by the formulaCos P = U1 * I1 * \ \ varphi 1 cos + U2 * I2 * \ \ varphi 2 cos + U3 * I3 * \ \ varphi 3

As the loads are equal, the phase angle angle is the same and we will put it as Ï†. In addition, as the applied tensions are equal, U1 = U2 = U3 = Uf (Tension of phase). Therefore:  P = 3*Uf*IL*cosφ

If we want, we can put it as function of the line tension, since Vf = \ tailcoat {Ul} {\ sqrt {3}} \ Rightarrow P = 3* \ cos tailcoats {Ul} {\ sqrt {3}} *Il * \ \ varphi = \ mathbf {{\ color {DarkOrange} \ sqrt {3} * Ul * Il * \ cos \ varphi}}


Powers in load in triangle2,2 Connection in triangle
.

If the loads are forming a triangle and whereas clause that are balanced, we have current them by all will be the same and, therefore, calculating one of them, we have the total power.

In the figure we have the three lines applied to the three well-balanced loads around which the I1 Intensities circulate, I2 and I3.

In knots 1, 2 and 3 the law of kirchhoff is fulfilled, by which, the currents that you enter are equal to the currents that leave, therefore:

Powers in configuration triangle

Knot 1 \ Rightarrow I1 = I12 - I31

Knot 2 \ Rightarrow I2 = I23 - I12

Knot 3 \ Rightarrow I3 = I31 - I23

When taking the vectors I12 and I31 to compose I1, we have the vectorial composition that is the left.

It is possible to be demonstrated as the previous case that the relation between I12 and I1 is:

I1 = \ sqrt {3} * I12 \ Rightarrow {\ color {DarkOrange} I_ 3 {Ls} = \ sqrt {} * I {_{f}}}

Considering the other two lines, the total power is:

Cos P = V_ {12} * I_ {12} * \ \ varphi _{12} cos + V_ {23} * I_ {23} * \ \ varphi _{23} cos + V_ {31} * I_ {31} * \ \ varphi _{31}

To the well-balanced being, we have them angles are equal

P = 3 cos V_L * I_L * \ varphi

. As we saw before, I_F = \ tailcoat 3 {I_L} {\ sqrt {}} \ Rightarrow 3 P = \ sqrt {cos} * V_L * I_L * \ \ varphi

For the Power it pretends and the reactive power, we have:

Q= \ sqrt {3} * V_L * I_L * \ without \ varphi

S = \ sqrt {3} * V_L * I_L

. That is:

For a well-balanced system, the powers are the same if the loads are connected in triangle that if we connected them in star

 

3. Compensation of the power factor.

The factor of power of a load obtains when calculating the cosine of the angle between the tension and the current in a load, that is to say, cosφ. That phase angle brings about load in the lines electrical that we must reduce them so that the electrical company does not penalize us in the invoice of the light.

triangle powers

If we want to reduce the power factor, we must put condensers next to the load. We will see how these capacities calculate.
The intention is to add to a Qc so that the Q pass to Q'.
When a current one happens through a condenser, this Q is generated, whose S-value:Z_c = X_C = \ tailcoat {1} {2* \ Pi *f*C}IC = \ tailcoat {V_C} {X_C} \ Rightarrow Q_C = 3 * 3 I_C*V_C= * \ tailcoat {V_C} {X_C} *V_C= \ tailcoat {3* (2 V_C^ {})}{\ tailcoat {1} {2* \ Pi *f*C}} = \ mathbf {3*2* \ Pi*f*V_C^ {2}}
Clearing the condenser, we have left:
C= \ tailcoat {Q_C} {3*2* \ Pi*f* (2 V_C) ^ {}} = {\ Blue color {} \ boldsymbol {C_ \ bigtriangleup}}
Where Vc are the terminal voltage of the condenser. Following the type of connection, we have the capacity for star and triangle.load-it stars
In the case of connection in triangle, tensions Vc and VL agree, and the capacity is the shown one in the previous case. If the connection is in star, the one of line is ˆš3 the phase tension, therefore:
{\ color {DarkRed} \ mathbf {C \ to bigstar}} = \ tailcoat {Q_C} {3*2* \ Pi*f* (\ tailcoat 3 {2 V_L} {\ sqrt {}}) ^ {}} = \ tailcoat {Q_C} {3*2* \ Pi*f* (\ tailcoat 2 {3 V_L^ {}} {})} = \ tailcoat {Q_C} {2* \ Pi*f* 2 {V_L} ^ {}} = {\ color {DarkRed} \ mathbf {C \ to bigstar}}
 We will see some examples how that factor can be fallen.


Example 1 

We have 3 single-phase motors of a consumption of 4000 W working to 220 volts and with a factor of power of 0,8. To calculate. We want to improve the system reducing facto of power to 0,95.
To calculate the capacity of the batteries of condensers (3 condensers, one by line) that we must connect in the load.

  • If the power factor is 0.8, it means that the angle obtains when doing arc cos of 0.8, giving a value of 36.87 .
  • As the power loss is 4000 W, this is going to be the value P, that is the product of V * I * cosφ, but also we can obtain the values of the apparent power of the form S = P/cosφ =>  S1 = \ tailcoat cos {P} {\ varphi} = \ 4000 tailcoats {} {0,8} = 5000 GOES
  • The reactive power is: Q1 = P1 * sen \ varphi = 5000 * sen 36.87 = 3000 GOES

Now we happened to a factor of power of 0,95. In this case, we have a 18.19 arc angle 0.95 = . The real power has not changed and, therefore we used the same P to calculate the new S2 and Q2:

  • S_ {2} = \ 4000 tailcoats {} {0,95} = 4210.05 GOES
  • The Reactive power is Q_2 = S_2 * sen \ varphi _2 = 4210.5 GOES * sen 18.19 = 1314.1 BAR
  • So that we have that reactive power, it is necessary to add to the condenser of capacity C, that contributes the differences of reactive powers, this is: Q_c = 3000 - 1314.4 = \ boldsymbol {\ mathbf {} 1685.6 BAR}
  • On the other hand, the power is the product of an Intensity by a tension, or, in the form of the tension:Q_c = V * I = V^ * \ tailcoat {V} {Z_c} = \ tailcoat {V^ {2}} {Z_c} = \ tailcoat {V^ {2}} {\ tailcoat {1} {2* \ pi *f*C}} \ Rightarrow Q_c = {V^ {2}} * 2* \ pi *f*C
  • If we cleared the value of the condenser in this development, we have:Q_c = V * I = V^ * \ tailcoat {V} {Z_c} = \ tailcoat {V^ {2}} {Z_c} = \ tailcoat {V^ {2}} {\ tailcoat {1} {2* \ pi *f*C}} \ Rightarrow Q_c = {V^ {2}} * 2* \ pi *f*C

{\ color {DarkGreen} \ boldsymbol {C = \ tailcoat {1} {{V^ {2}} * 2* \ pi *f*Q_c} \ rightarrow C= \ 110.85 2 1685.6 tailcoats {} {2* \ pi*50*220^ {}} = \ mu F}}

This it is the value that we must put in each line (in this case between neutral line and) to compensate the power factor.

Example 2. In the circuit of the figure, to calculate the powers consumed by all the loads. problem-Power-three-phasea) The power is a product of the effective Current and the effective Tension. We see that the tension that gives the problem us is between line and line, reason why must find out the phase tension, dividing the tension of 380 between the root of 3, which gives about 220 volts.

b) We are going to pass to polar form the load:

|Z| = \ sqrt {10^ {^ {2}} +10^ {^ {2}}} = 14,14

\ varphi = arc tag \ 10 10 tailcoats {} {} = 45

c) In order to calculate the tension of the green load as 220 ˆ 0 and we have:I_v = \ 15.55 14.14 220 tailcoats {\ angle 0} {\ angle 45‚} = \ angle -45‚

 where IV are the current by the green load (current of line that in star is the same that the phase current). The rest of the current months In (current load orange) and Ir, will be the same only that desasadas 120 and the other advanced 120, this are: I_r = 15.55 \ angle (45 +120) = 15.55 \ angle 165

I_v = 15.55 \ angle (45 - 120) = 15.55 \ angle -75

d) The power is:\ mathbf {P = I^ {2} * R = 15,55^ {2} * 10 \ cong 2418 W \ Rightarrow Power (total) = 7254 W}

3421 IT GOES

e) Powers I pretend and reactive. We can make our triangle of powers with the following values:
  1. The power pretends S is the product of the Tension by the Intensity, giving a value of 220 Xs 15,55 = 3421 GOES
  2. The reactive power is obtained when multiplying S by the sine of Ï†. We have then 3421 GOES * sen Ï† = 3421 GOES * sen 45 = 2419 BAR
  3. For the active power (to calculated before) we have Pa = 3421 * cos Ï† = 2419 W

Example 3 problem several three-phase loads

An industry has three-phase feeding of 380 V and 50 Hertz
with the following more important loads:
1) An three-phase motor of 20 CB, yield 80% and factor of power 0,82.
2) illumination of lamps leds with a consumption of 3500 W (we despised the small inductive load of the feeders)
3) A set of variable and well-balanced loads, connected in triangle j has a 40 impedance of 60 + „¦
To calculate:
a) In each one of the loads, the current, each one of the powers and their factor of power.
b) The total current next to each one of the powers that the installation has. To obtain its factor of power.
c) The capacity of the condensers connected in star to be the installation with a factor of power of 0,95.

Solution:

First that we have to do it is that, to the being loads with inductive components, the currents are going to take a certain delay with respect to the tension, except for the resistiva load. We take one from phases S, as home point. That phase has a tension of 220 volts with angle of zero degrees. Starting off of that point, we will calculate the currents that leave that phase (IS1, IS2 and IS3) THAT WILL GIVE RISE To is.

1€ Load 1

In this load, we have a motor of an three-phase motor of 20 CB, yield 80% and factor of power 0,82. First it is to obtain the electrical consumption that it has, since it leaves from the energy is lost by heat, frictions, etc.P (useful) = 20 CB * \ tailcoat {736W} {CB} = 14720 W

 The active power that consumes is:

\ eta = \ tailcoat {Power - used} {given Power} \ Power-given Rightarrow = \ tailcoat {Power - used} {\ eta} = \ tailcoat {14.72 KW} {0,8} = 18.4 KW

The angle between the tension and the current gives the power factor us:\ varphi = arc cos 0.82 = 34,92

In a triangle rectangle, if we know two angles and a side, we can know the rest. We will see:

diagram of powers

 

 

 

 

 

We can obtain Q if we know Ï† and S, so that: Q = S * sen Ï† and P = cos S * Ï†. When dividing the two expressions we have:

\ tailcoats {Q} {P} = \ tailcoat {S *Sen \ varphi} {S * Cos \ varphi} = tang \ varphi \ Rightarrow \ mathbf {Q = P * tang \ varphi}

Therefore, the reactive power has a value of: Q = 18400 * tang 34.92 = BAR

The power pretends we can calculate it applying pit¡goras:

S_1 = \ sqrt {{P_1} ^2 + {Q_1} ^2} = \ sqrt {{18400} ^2 + {12845.56} ^2} = \ mathbf {22440.33 GOES}

The phase current is:

I_1_L = \ tailcoat {P1} {\ sqrt {3 cos} *V_L * \ varphi} = \ 34.09 0.82 3 18400 tailcoats {} {\ sqrt {} *380 *} = A

As the current goes in delay in this inductive system, we have:

\ overrightarrow {I_1_S} = I_1_L \ angle - \ varphi 1 = 34.09 \ angle -34,92 = \ boldsymbol {(27.95 - 19,51j) To}

2€ Load 2

We are going to consider as pure resistance the loads leds. Therefore, the power of 3500 has a factor of power of 1, the reactive power is zero and the apparent power agrees with the resistiva power.

when connected being in triangle, the tension of phase and line are the same and the line current is ˆš3 * IF, therefore:

I_2_L = \ cos tailcoats {P_2} {\ sqrt {3} *380 * \ varphi _3} = \ 3 3500 tailcoats {} {\ sqrt {} *380*1} = 5.37 To \ rightarrow \ overrightarrow {I_2_S} = 5.37 + 0 j A

3€ Load 3  

We have 3 loads connected in triangle, that receive a phase tension just as the tension of line of 380 volts. The impedance is of 60 + 40 j „¦ that happened to polar form:\ overrightarrow {Z_3} = 60 + 40 j \ Rightarrow Z = \ sqrt {60^ {2} +40^ {2}} = 72,11; \ varphi _3 = arctag \ 33.69 60 40 tailcoats {} {} = \ Rightarrow \ overrightarrow {Z_3} = 72.11 \ angle 33,69

From this angle we obtain the factor of power of load 2, like fdp3 = cosφ3 = 0,83

The current by that cable I3F comes to apply the law of ohm, that stops the module of the current is:

I_3_F = \ 5.27 72.11 380 tailcoats {} {} = A

In a triangle connection:9.13 3 I_3_L = \ sqrt {} *5,27 = A. As we have inductive loads again and we know the phase angle that there is of 33,69. This does not give that the current is:

\ overrightarrow {I_3_S} = 9.13 I_3_L \ angle -33,69 = \ angle -33,69; a = 9.13 * cos ones 33.69 = 7,60; b = 9.13 * sen 33.69 = 4.21 \ Rightarrow \ overrightarrow {I_3_S} = 7.6 - 4.21 j

The powers we calculated them in each load, considering the current that happens through each:P_2 = 3 R_2 * (2 I_2_F) ^ {} = 3*60 * (5,27) ^ {2} = 4999.12 W

Q_2 = 3 X_2 * (2 I_2_F) ^ {} = 3*40 * (5,27) ^ {2} = 3332.78 WARS_2 = 3 Z_2 * (2 I_2_F) ^ {} = 3*72,11 * (5,27) ^ {2} = 6008.11 GOES

This data can also be removed by trigonometry.

We obtain the factor of power by the relation of the triangle of powers so that:P_3 = S_3 * cos \ varphi _3 \ rightarrow fdp_3 = \ tailcoat {P_3} {S_3} = \ 6008.11 4999.12 tailcoats {} {} = 0,832

. The phase angle angle is the arc whose cosine is 0.832 = 33,68

4€ Calculation of the current by cable S

Since we have passed to bin³mica the intensities of each cable, we will add the real and imaginary part of each, therefore we have:\ overrightarrow {I_S} = \ overrightarrow {I_S_1} + \ overrightarrow {I_S_2} + \ overrightarrow {I_S_3} = (27.95 - 19,51j) + (5,37) + (7.6 - 4.21 j) A = (40.87 - 23.72 j) To =47.25 \ angle -30,13

4€ Calculation of the total powers

Since we did with correintes, the active and reactive powers are added and soon to apply pit¡ginas to obtain S

P_T = P_1 + P_2 + P_3 = 18400 + 3500 + 4999.12 = 26899.12 W

Q_T = Q_1 + Q_2 + Q_3 = 12845.56 +0+3332.78 = 16178.34 BAR

and the Power pretends is:

2 2 S_T = \ sqrt {26899^ {} +16178,34^ {}} = 31389.4 GO

The factor of power of the set is the quotient between the active power and the apparent one, therefore:fdp = \ 31389.4 26899.12 tailcoats {} {} =0.857

4€ Calculation of the condensers to reduce the Factor of Power

We must apply a series of condensers so that new fdp is 0.95. Therefore, the previous and later situation is the following, considering that the active power does not change and that the respective angles for those factors of power are\ varphi _a = arc cos 0.857 = 31,024; \ varphi _d = arc cos 0.95 = 18,19

Being Ï†a the angle before putting the condensers and Ï†d the angle after the condensers

:Q_a_n_t_e_s = P * tg \ varphi _a; Q_d = P * tg \ varphi _d

The difference of the Q is due to the condensers. reducing both expressions it gives us:Q_C = P * (tg_a - tg_b) = 26899.12 * (tg31,024 - tg 18.19) = 26899.12 * (0.6014 - 0.3285) = 7338.35 BAR

That Qc is the one that it must offer the condensers, that the case of connection in triangle stops, is:

C_ \ bigtriangleup = \ tailcoat {Qc} {2* \ Pi *f* (2 V_L) ^ {}} = \ 7338.35 tailcoats {} {2* \ Pi *50* (380) ^ {2}} = \ boldsymbol {161 \ mu F}