# What is Core Loss in transformer? With Calculation and Formula

Transformer is a static unit, and so there are no mechanical errors in it (like the lack of friction). The transformer consists only of loss of energy (iron and copper failure).
No machine is 100 % efficient worldwide, there are always some losses.

Due to variations in the magnetization of the core of the transformer and also the loss of copper due to the transformer winding strength, the physical phenomenon loss occurs. The description describes the different forms of losses.

Losses are like practical engineering components. There was an attempt to reduce the scale of these losses.

It is not an exception to the transformer. The perfect transformer performs without interruption. However, transformers practically have losses and there are two types.

1.)  Copper loss
2.)  Core loss

let us have a look at the two losses.

## COPPER LOSS:

Loss in copper defined as depletion in I*I*R is the lack of cover. The heat emitted from the wire while the transformer is working is known to be a loss.

The optimal winding of the transformer has zero power. Yet winding is virtually immune (although very small like copper), and as a result there is both a main and a secondary heat loss on the transformer side.

## CORE LOSS:

The core loss, as the name suggests, is the weakness in the heart. It’s also called loss of iron. Core loss can be differentiated by-

### Hysteresis loss –

This is characterized as energy loss when magnetizing and demagnetizing are done. We will look at alternating supply in order to grasp this. In the first positive loop, a charge passes in one direction and travels in the other direction for the negative time interval.

### Eddy Current loss –

The differing flow through the core of the transformer generates an emf. And with this emf, hot tubs are formed in the center like tiny currents. Their presence is such that they are against the existing availability. This may be reduced by splitting the core into fine laminated parts, increasing the region and thereby reducing the current.
A transformer core with perfect performance is 100 % effective but is approximately 96-97% in practical life.

In the transformer core losses are observed during the performance of the specified functions. They affect the initial amount of electricity dispensed by the device. What constitutes the transformer ‘s loss and efficiency will be debated later.

The transformer is a device of staticity. It is electrically powered. There are no movable sections in the architecture. Hence higher energy costs are excluded due to mechanical reasons. whereas its core is the main heart of function.

## Concept of Loss

Energy costs increase during off-hours during the operation of the power equipment. This is attributed to a spike in activated steel idle losses. A decrease in nominal load is observed in this situation, with an increase in the reactive form capacity. The energy loss in the transformer is calculated by relation to the active control. They show on the magnetic drive, on the unit’s windings and other parts.

A part of the power is supplied to the primary circuit during the installation operation. It is scattered all over the system. The incoming power to the load is then calculated at a lower point of the core.

There are two types of reasons that increase the energy consumption of equipment. They are influenced by various factors. They are divided into these types:

• Magnetic.
• Electric.

They should be understood in order to be able to reduce electrical losses in the core of the transformer.

## Magnetic losses

In the first case, the magnetic drive steel losses consist of eddy currents and hysteresis. They are directly proportional to the core mass and the magnetic induction thereof. That characteristic is affected by the iron itself, from which the magnetic drive is made. And the core is constructed of electric material. The plates are made very thin. A layer of insulation lies between them.

## Factors for Electrical Losses

The reduction in power in the windings can be determined when heated by the current. These costs account for 4-7 percent of the total energy consumed in networks. They rely on a number of factors.

• The electrical load of the system.
• The configuration of internal networks, their length and section size.
• Mode of operation.
• Weighted average system power factor.
• Location of compensation devices.
• Core losses in transformers are a variable. It is affected by the squared current in the circuits.

## Best Calculation Formula

The load factor in the method presented is determined by the formulation:

K = Ea / NM * OCh * cos π, where Ea is the amount of active power.

What losses occur during the load period in the transformer can be calculated according to the procedure established. The formulation is used for this:

P = XX * OCH * K2 * LF

## The calculation of three-winding transformers for cores

The latter approach is used to test double-winding transformer activities. For machines with three circuits, a variety of details need to be taken into account. They are indicated in the passport by the supplier.

The calculation includes the power rated for each circuit, as well as their loss of short circuit. In this case, it will be calculated according to the following formula:

E = ESN + ENN, where E is the actual amount of electricity passed through all the circuits; ESN-medium-voltage electric, ENN-low-voltage electricity.

## Here is a great example of calculation

To make it easier to understand the methodology presented you should use a specific example to consider the calculation. For example, the rise in energy consumption in a 630 kVA transformer needs to be determined. In a row, the source data is more conveniently viewed.

Based on the data obtained, a calculation can be made. The measurement result will be as follows:

K² = 4.3338

P = 0.38 kW * h

% loss is 0.001. Their total number is 0.492%

## How to measure the best performance?

The efficiency indicator is also set when calculating losses. It shows the active type of power ratio at both input and output. For a closed system this indicator is calculated with the following formula:

Efficiency = M1 / M2, where M1 and M2 are the transformer’s active core power, as determined by input and output circuits measurement.

The performance indicator is determined by the power factor (cosine of angle j squared) multiplying the installation ‘s rating capacity. Consideration is given in the above formula.

The efficiency indicator in transformers with 630 kVA, 1000 kVA and other powerful devices may be 0.98 or even 0.99. It shows just how effective the unit is. The higher the efficiency, the more energy is consumed economically. In this case, the energy consumption will be minimal while the equipment is operating.

Having considered the methodology for calculating the loss of transformer core, short circuit and idle, it is possible to determine the equipment ‘s efficiency and its efficiency. The method of calculation involves using a special calculator or carrying out a calculation in a special computer program.

Here is a youtube video from Electrical Tutorials about a great explanation of the losses in transformers.