Electrical Machines - Basic vocational knowledge (Institut für Berufliche Entwicklung, 144 p.)
 8. Transformer
 8.2. Operational behaviour of a transformer
 8.2.1. Idling behaviour Idling features 8.2.2. Short-circuit behaviour 8.2.3. Loaded voltage behaviour 8.2.4. Efficiency

### 8.2.1. Idling behaviour Idling features

A transformer idles where mains voltage U1 remains applied to the primary side whilst no consumer is connected to the secondary side (Za) (Figures 125/126).

 Primary circuit U1 applies I0 flows (idling current) Secondary circuit Za = ¥ I2 = 0 U2 = U20

Idling current

The applied voltage U drives the idling current I0. This is needed to establish the magnetic field Iµ. This lags behind the voltage U1.

Figure 127 - Indicator image for idling operation

1 Iron loss current IFe

The phase position of the idling current I0 to voltage U1 can be determined according to the circuitry of Figure 128.

Figure 128 - Circuitry to determine idling losses

1 Rated voltage

The value of idling current I0 is between 2 and 5 per cent of idling current in big transformers and up to 15 per cent in smaller transformers.

The idling curve I = f (U1) in Figure 129 indicates that no-load current I0 increases proportionally to the input voltage U1. No-load current increases markedly over and beyond the input rated speed U1n. It can, moreover, even attain values greater than the rated current.

Figure 129 - Idling curve of a transformer I0 = f (U1)

Transformers shall not be driven by voltages greater than the rated voltage.

Idling losses (iron losses)

The active power derived from the circuit in Figure 128 can only be transformed into heat in the input winding and iron core as no current flows into the secondary winding during idling. The active power P0, which is converted into heat in the iron core, is made up of eddy current and hyteresis loss.

The following example shows that the iron losses almost always arise during idling.

Example:

The following idling values were measured in a transformer:

U1n = 220 V; I0 = 0.5 A; P0 = 40 W; R1 = 3.

What percentage of winding losses are contained in idling power?

Solution:

P0 = PVFe + PVW

PVW = 0.75 W

PVFe = P0 - PVW = 40 W - 0.75 W = 39.25 W

Thus, the power loss determined during idling is an iron loss.

Iron losses are determined during no-load operation and are independent of load.

### 8.2.2. Short-circuit behaviour

Short-circuit curves

Secondary current I2 increases if load resistance is decreased. Where Za = 0 the transformer has been short-circuited.

 Primary circuit U1 is applied IK flows Secondary circuit Za = 0 U2 = 0

Short-circuit voltage

The short-circuited transformer can be replaced by resistor Z1 which corresponds to the transformer internal resistor.

Figure 130 - Short-circuited transformer

1 Short-circuit current IK

Figure 131 depicts the commensurate duplicate circuit diagram.

Figure 131 - Duplicate circuit diagram for short circuit run

1 Ohmic winding resistance, 2 Scattered reactance (is made up of the scatter flow of the input and output coils), 3 Inner resistance of the transformer (impedance)

During a short-circuit attempt (Figure 132) the input voltage given a short-circuited output winding is increased until primary and secondary nominal currents flow. The voltage applied to the input side is then the short-circuit voltage UK.

Figure 132 - Circuitry to determine short-circuit losses

1 Short circuit voltage

The short-circuit voltage is the overall voltage decrease of a transformer during rated loading.

The relative short-circuit voltage UK in % is determined by the following equation:

The relative short-circuit voltage is, on average, 2 to 10% of input rated voltage (U1n) in mains transformers.

Short-circuit losses (winding losses)

In the short-circuit experiment (Figure 132) a power meter indicates short-circuit losses as the primary and secondary rated currents generate winding losses. The iron core is only slightly magnetised by the applied short-circuit voltage (UK U1).

The winding losses can be metered during the short-circuit experiment. They are dependent on the load current (PVW = I2 R).

In contrast to operational idling, during loading the secondary circuit is closed through an external resistance Za (Figure 126). Secondary current I2 flows. According to the energy conservation law the transformer must also take up commensurate primary power, thus a primary current I1 also flows.

 Primary circuit U1 is applied I1 > I0 Secondary circuit Za < ¥ I2 > 0 U2 ¹ U20

Voltage curve U2 = f (I2)

As the curve in Figure 133 shows, terminal voltage U2 decreases during loading.

Figure 133 - Voltage behaviour during loaded operation U2 = f (I)2

1 UK small, 2 UK big

Figure 134 depicts the duplicate circuit diagram for the loaded transformer.

Figure 134 - Duplicate circuit for the loaded transformer with a transformation ratio = 1:1

The duplicate circuit diagram corresponds to a transformer with a transformation ratio

As rated current flows the short-circuit voltage UK decreases at the internal transformer resistance Zi as a result of which the terminal voltage U2 declines by the power decrease of the short-circuit voltage UK.

Transformers with considerable short-circuit voltage UK have powerful internal resistors, that is to say pronounced voltage changes as load alters.

 UK = 2...10% minimal voltage losses voltage-rigid behaviour UK = 20...50% considerable voltage losses voltage-flexible behaviour

Example:

A 220/42 V transformer has a short-circuit voltage of 10%.

How great is the voltage change between idling and rated current loading?

Solution:

Given differing loads with ohmic, inductive or capacitive external resistance gives rise to the dependence of output voltage on load current as shown in Figure 135.

Figure 135 - Secondary terminal voltage depending on the degree and nature of loading

Given capacitive load, the output voltage may even be greater than no-load voltage.

The output voltage of a transformer depends on the

- degree of load current I2
- the magnitude of relative short-circuit voltage
- the nature of the load (ohmic, inductive or capacitive).

### 8.2.4. Efficiency

Efficiency represents the ratio of power output to power input.

Efficiency is determined by the separate loss indication.

Pin = Pout + PVFe + PVW

 Pout rated power of the transformer given cos j = 1 (resistive load) PVW metered winding losses in short-circuit trial PVFe metered iron losses in no-load trial

The following equation applies for determining efficiency for partially inductive or capacitive load:

Example:

Given:

Sn = 100 kVA
PVFe = 570 W
PVW = 2.1 kW
cos j = 0.85

Sought: h

Solution: