## Purpose of the measurement

The measurement is carried out to determine the load-losses of the transformer and the impedanse voltage at rated frequency and rated current.

**separetely for each winding pair**(e.g., the pairs 1-2, 1-3 and 2-3 for a three-winding transformer), and furthermore on the principal and extreme tappings.

## Apparatus and measuring circuit

**On Figure 1 above (Circuit for the impedance and load-loss measurement) there are following figures:**

**G**– Supply generator_{1}**T**– Step-up transformer_{1}**T**– Transformer to be tested_{2}**T**– Current transformers_{3}**T**– Voltage transformers_{4}**P**– Wattmeters_{1}**P**– Ammeters (r.m.s. value)_{2}**P**– Voltmeters (r.m.s. value)_{3}**C**– Capacitor bank_{1}

The supply and measuring facilities are not described here. Current is generally supplied to the h.v. winding and the l.v. winding is short-circuited.

### Performance of the measurement

If the reactive power supplied by the **generator G _{1}** is not sufficient when measuring large transformers, a

**capacitor bank C**is used to compensate part of the inductive reactive power taken by the

_{1}**transformer T**.The voltage of the supply generator is raised until the current has attained the required value (25…100 % of the rated current according to the standard 4.1).

_{2}In order to increase the accuracy of readings will be taken at several current values near the required level. If a winding in the pair to be measured is equipped with an off-circuit or on-load tap-changer. the measurements are carried out on the principal and extreme tappings.

**It the transformer has more than two windings all winding pairs are measured separately.**

## Results

Corrections caused by the instrument transformers are made to the measured current, voltage and power values. The power value correction caused by the phase displacement is calculated as follows:

**Where:**

**P**= corrected power_{c}**P**= power read from the meters_{e}**δ**= phase displacement of the voltage transformer in minutes_{u}**δ**= phase displacement of the current transformer in minutes_{i}**ϕ**= phase angle between current and voltage in the measurement (ϕ is positive at inductive load)**K**= correction

The correction K obtained from **equation 4.1** is shown as a set of curves in **Figure 4.2**.

The corrections caused by the instrument transformers are made separately for each phase, because different phases may have different power factors and the phase displacements of the instrument transformers are generally different.

If the measuring current **I _{m}** deviates from the rated current

**I**, the power

_{N}**P**and the voltage

_{km}**U**at rated current are obtained by applying corrections to the values

_{km}**P**and

_{c}**U**relating to the measuring current.

_{c}**The corrections are made as follows:**

**The correction caused by the phase displacement of instrument transformers (Figure 2):**

**Where:**

**K**– correction in percent,**δ**– phase displacement in minutes_{u}– δ_{i}**cosδ**– power factor of the measurement.

**The sign of K is the same as that of δ _{u} – δ_{i}.**

Mean values are calculated of the values corrected to the rated current and the mean values are used in the following. According to the standards the measured value of the losses shall be corrected to a winding temperature of 75° C (80° C, if the oil circulation is forced and directed).

The transformer is at ambient temperature when the measurements are carried out. and the loss values are corrected to the **reference temperature 75° C** according to the standards as follows.

The d.c. losses **P _{Om }**at the measuring temperature

**ϑ**are calculated using the resistance values

_{m}**R**and

_{1m}**R**obtained in the resistance measurement (for windings 1 and 2 between line terminals):

_{2m}The additional losses Pamat the measuring temperature are:

Here **P _{km}** is the measured power, to which the corrections caused by the instrument transformer have been made, and which is corrected to the rated current according to equation (4.2).

The short-circuit impedance **Z _{km}** and resistance

**R**at the measureing temperature are:

_{km}**U**is the measured short-circuit voltage corrected according to Equation (4.3);_{km}**U**is the rated voltage and_{N}**S**is the rated power._{N}

The **short circuit reactance X _{k}** does not depend on the losses and X

_{k}is the same at the

**measuring temperature (ϑ**and the reference temperature (75 °C), hence:

_{m})When the losses are corrected to 75° C, it is assumed that d.c. losses vary directly with resistance and the additional losses inversely with resistance. The losses corrected to 75° C are obtained as follows:

**Where:**

**ϑs = 235° C** for Copper

**ϑs = 225° C** for Aluminium

Now the short circuit resistance **R _{kc}** and the short circuit impedance

**Z**at the reference temperature can be determined:

_{kc}## Results

The report indicates for each winding pair the **power S _{N}** and the following values corrected to 75° C and relating to the principal and extreme tappings.

- D.C. losses
**P**(PDC)_{Oc} - Additional losses
**P**(PA)_{ac} - Load losses
**P**(PK)_{kc} - Short circuit resistance
**R**(RK)_{kc} - Short circuit reaactance
**X**(XK)_{kc} - Short circuit impedance
**Z**(ZK)_{kc}

**Reference:** Testing of power transformers – ABB

Please help me calculate copper looses…I want to specify a high impedance value for transformers to reduce the fault level. I however established that this compromises/degrades the stability of the system/network and that there are high copper losses which may results into higher tariffs….please help me understand this.

Good Morning,

I like this site. it really help a lot.

I also need assistance. I have a very old transformer with a transformer nameplate invisible. I would like to know if is there any other way of calculating an estimated voltage impedance without using the standard transformer impedance measurement as indicated above using:

On Figure 1 above (Circuit for the impedance and load-loss measurement) there are following figures:

G1 – Supply generator

T1 – Step-up transformer

T2 – Transformer to be tested

T3 – Current transformers

T4 – Voltage transformers

P1 – Wattmeters

P2 – Ammeters (r.m.s. value)

P3 – Voltmeters (r.m.s. value)

C1 – Capacitor bank

Is there any other way to calculate this?

Hi i would like to call you in person in regsrds to testing of distribution transformers .

In the process of starting a manufacturing facility in Australia.

Thank you

Sir explain to movementing parallel operation of swithgear system

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I m nt a engineer nw, bt nw its too good.reading articles by you