Sunday, September 18, 2011

Transformer: three phase

 Introduction:
Three phase transformers are used throughout industry to change values of three phase voltage and current. Since three phase power is the most common way in which power is produced, transmitted, an used, an understanding of how three phase transformer connections are made is essential. In this section it will discuss different types of three phase transformers connections, and present examples of how values of voltage and current for these   connections are computed.      
Three Phase Transformer Construction:
A three phase transformer is constructed by winding three single phase transformers on a single core. These transformers are put into an enclosure which is then filled with dielectric oil. The dielectric oil performs several functions. Since it is a dielectric, a nonconductor of electricity, it provides electrical insulation between the windings and the case. It is also used to help provide cooling and to prevent the formation of moisture, which can deteriorate the winding insulation.
Three-Phase Transformer Connections:
There are only 4 possible transformer combinations:
  1. Delta to Delta - use: industrial applications
  2. Delta to Wye - use : most common; commercial and industrial
  3. Wye to Delta - use : high voltage transmissions
  4. Wye to Wye - use : rare, don't use causes harmonics and balancing problems.
Three-phase transformers are connected in delta or wye configurations. A wye-delta transformer has its primary winding connected in a wye and its secondary winding connected in a delta (see figure 1-1). A delta-wye transformer has its primary winding connected in delta and its secondary winding connected in a wye (see figure 1-2).


Figure 1-1: Wye-Delta connection


Figure 1-2: Delta-Wye connection
Delta Conections:
A delta system is a good short-distance distribution system. It is used for neighborhood and small commercial loads close to the supplying substation. Only one voltage is available between any two wires in a delta system. The delta system can be illustrated by a simple triangle. A wire from each point of the triangle would represent a three-phase, three-wire delta system. The voltage would be the same between any two wires (see figure 1-3).


Figure 1-3:
Wye Connections:
In a wye system the voltage between any two wires will always give the same amount of voltage on a three phase system. However, the voltage between any one of the phase conductors (X1, X2, X3) and the neutral (X0) will be less than the power conductors. For example, if the voltage between the power conductors of any two phases of a three wire system is 208v, then the voltage from any phase conductor to ground will be 120v. This is due to the square root of three phase power. In a wye system, the voltage between any two power conductors will always be 1.732 (which is the square root of 3) times the voltage between the neutral and any one of the power phase conductors. The phase-to-ground voltage can be found by dividing the phase-to-phase voltage by 1.732 (see figure 1-4).


Figure 1-4:
Connecting Single-Phase Transformers into a Three-Phase Bank:
If three phase transformation is need and a three phase transformer of the proper size and turns ratio is not available, three single phase transformers can be connected to form a three phase bank. When three single phase transformers are used to make a three phase transformer bank, their primary and secondary windings are connected in a wye or delta connection. The three transformer windings in figure 1-5 are labeled H1 and the other end is labeled H2. One end of each secondary lead is labeled X1 and the other end is labeled X2.


Figure 1-5:
Figure 1-6 shows three single phase transformers labeled A, B, and C. The primary leads of each transformer are labeled H1 and H2 and the secondary leads are labeled X1 and X2. The schematic diagram of figure 1-5 will be used to connect the three single phase transformers into a three phase wye-delta connection as shown in figure 1-7.


Figure 1-6:


Figure 1-7:
The primary winding will be tied into a wye connection first. The schematic in figure 1-5 shows, that the H2 leads of the three primary windings are connected together, and the H1 lead of each winding is open for connection to the incoming power line. Notice in figure 1-7 that the H2 leads of the primary windings are connected together, and the H1 lead of each winding has been connected to the incoming primary power line.
Figure 1-5 shows that the X1 lead of the transformer A is connected to the X2 lead of transformer c. Notice that this same connection has been made in figure 1-7. The X1 lead of transformer B is connected to X2, lead of transformer A, and the X1 lead of transformer B is connected to X2 lead of transformer A, and the X1 lead of transformer C is connected to X2 lead of transformer B. The load is connected to the points of the delta connection.
Open Delta Connection:
The open delta transformer connection can be made with only two transformers instead of three (figure 1-8). This connection is often used when the amount of three phase power needed is not excessive, such as a small business. It should be noted that the output power of an open delta connection is only 87% of the rated power of the two transformers. For example, assume two transformers, each having a capacity of 25 kVA, are connected in an open delta connection. The total output power of this connection is 43.5 kVA (50 kVA x 0.87 = 43.5 kVA).


Figure 1-8: Open Delta Connection
Another figure given for this calculation is 58%. This percentage assumes a closed delta bank containing 3 transformers. If three 25 kVA transformers were connected to form a closed delta connection, the total output would be 75 kVA (3 x 25 = 75 kVA). If one of these transformers were removed and the transformer bank operated as an open delta connection, the output power would be reduced to 58% of its original capacity of 75 kVA. The output capacity of the open delta bank is 43.5 kVA (75 kVA x .58% = 43.5 kVA).
The voltage and current values of an open delta connection are computed in the same manner as a standard delta-delta connection when three transformers are employed. The voltage and current rules for a delta connection must be used when determining line and phase values of voltage current.
Closing a Delta:
When closing a delta system, connections should be checked for proper polarity before making the final connection and applying power. If the phase winding of one transformer is reversed, an extremely high current will flow when power is applied. Proper phasing can be checked with a voltmeter at delta opening. If power is applied to the transformer bank before the delta connection is closed, the voltmeter should indicate 0 volts. If one phase winding has been reversed, however, the voltmeter will indicate double the amount of voltage.
It should be noted that a voltmeter is a high impedance device. It is not unusual for a voltmeter to indicate some amount of voltage before the delta is closed, especially if the primary has been connected as a wye and the secondary as a delta. When this is the case, the voltmeter will generally indicate close to the normal output voltage if the connection is correct and double the output voltage if the connection is incorrect.
Overcurrent Protection for the Primary:
Electrical Code Article 450-3(b) states that each transformer 600 volts, nominal or less, shall be protected by an individual overcurrent device on the primary side, rated or set at not more than 125% of the rated primary current of the transformer. Where the primary current of a transformer is 9 amps or more and 125% of this current does not correspond to a standard rating of a fuse or nonadjustable circuit breaker, the next higher standard rating shall be permitted. Where the primary current is less than 9 amps, an overcurrent device rated or set at not more than 167% of the primary current shall be permitted. Where the primary current is less than 2 amps, an overcurrent device rated or set at not more than 300% shall be permitted.

Definitions:


Let x be the instantaneous phase of a signal of frequency f at time t:
x=2\pi ft\,\!
Using this, the waveforms for the three phases are
V_{L1}=V_P\sin \left(x\right)\,\!
V_{L2}=V_P\sin \left(x-\frac{2}{3} \pi\right)
V_{L3}=V_P\sin \left(x-\frac{4}{3} \pi\right)
where VP is the peak voltage and the voltages on L1, L2 and L3 are measured relative to the neutral.Balanced loads
Generally, in electric power systems, the loads are distributed as evenly as is practical between the phases. It is usual practice to discuss a balanced system first and then describe the effects of unbalanced systems as deviations from the elementary case.

 Star connected systems with neutral

This refers to a system with a resistive load R between each phase and neutral
Constant power transfer
An important property of three-phase power is that the power available to a resistive load, \scriptstyle P \,=\, V I \,=\, \frac{1}{R}V^2, is constant at all times.
\begin{align} P_{Li} &= \frac{V_{Li}^{2}}{R}\\ P_{TOT} &= \sum_i P_{Li} \end{align}
To simplify the mathematics, we define a nondimensionalized power for intermediate calculations, \scriptstyle p \,=\, \frac{1}{V_P^2}P_{TOT} R
p=\sin^{2} x+\sin^{2} \left(x-\frac{2}{3} \pi\right)+\sin^{2} \left(x-\frac{4}{3} \pi\right)=\frac{3}{2}
Hence (substituting back):
P_{TOT}=\frac{3 V_P^2}{2R}
since we have eliminated x we can see that the total power does not vary with time. This is essential for keeping large generators and motors running smoothly. Actually, the load need not be resistive for achieving a constant instantaneous power since, as long as it is balanced or the same for all phases, it may be written as
Z=|Z|e^{j\varphi}
so that the peak current is
I_P=\frac{V_P}{|Z|}
for all phases and the instantaneous currents are
I_{L1}=I_P\sin\left(x-\varphi\right)
I_{L2}=I_P\sin\left(x-\frac{2}{3}\pi-\varphi\right)
I_{L3}=I_P\sin\left(x-\frac{4}{3}\pi-\varphi\right)
Now the instantaneous powers in the phases are
P_{L1}=V_{L1}I_{L1}=V_P I_P\sin\left(x\right)\sin\left(x-\varphi\right)
P_{L2}=V_{L2}I_{L2}=V_P I_P\sin\left(x-\frac{2}{3}\pi\right)\sin\left(x-\frac{2}{3}\pi-\varphi\right)
P_{L3}=V_{L3}I_{L3}=V_P I_P\sin\left(x-\frac{4}{3}\pi\right)\sin\left(x-\frac{4}{3}\pi-\varphi\right)
P_{L1}=\frac{V_P I_P}{2}\left[\cos\varphi-\cos\left(2x-\varphi\right)\right]
P_{L2}=\frac{V_P I_P}{2}\left[\cos\varphi-\cos\left(2x-\frac{4}{3}\pi-\varphi\right)\right]
P_{L3}=\frac{V_P I_P}{2}\left[\cos\varphi-\cos\left(2x-\frac{8}{3}\pi-\varphi\right)\right]
which add up for a total instantaneous power
P_{TOT}=\frac{V_P I_P}{2}\left\{3\cos\varphi-\left[\cos\left(2x-\varphi\right)+\cos\left(2x-\frac{4}{3}\pi-\varphi\right)+\cos\left(2x-\frac{8}{3}\pi-\varphi\right)\right]\right\}
Since the three terms enclosed in square brackets are a three-phase system, they add up to zero and the total power becomes
P_{TOT}=\frac{3V_P I_P}{2}\cos\varphi
or
P_{TOT}=\frac{3V_P^2}{2|Z|}\cos\varphi

 No neutral current

For the case of equal loads on each of three phases, no net current flows in the neutral. The neutral current is the sum of the phase current.
\begin{align} I_{L1} &= \frac{V_{L1}}{R},\; I_{L2}=\frac{V_{L2}}{R},\; I_{L3}=\frac{V_{L3}}{R}\\ I_{N} &= I_{L1} + I_{L2} + I_{L3} \end{align}
We define a non-dimensionalized current, i=\frac{I_{N}R}{V_P}:
i=\sin (x)+\sin \left(x-\frac{2}{3} \pi\right)+\sin \left(x-\frac{4}{3} \pi\right)= 0
Hence also \scriptstyle I_N \,=\, 0\,\!
Since we have shown that the neutral current is zero we can see that removing the neutral core will have no effect on the circuit, provided the system is balanced. Such connections are generally used only when the load on the three phases is part of the same piece of equipment (for example a three-phase motor), as otherwise switching loads and slight imbalances would cause large voltage fluctuations.
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