Transformer efficiency is of great importance in the world of **electrical transformers**.

Through our study in this article, we will analyze and understand the importance of studying the efficiency of the transformer. I will discuss the subject in a practical and scientific in-depth to get the desired goal.

Before listing the **five facts** in detail, we must first clarify what the efficiency of the transformer is.

Primarily, any transformer on earth has input power and output power. We know that the **output power** is smaller than the **input power** due to the power losses within the transformer.

Therefore, the efficiency can be defined as the **ratio** between the output power and the input power, as is evident from the following equations:

we can represent efficiency as >>> ɳ = (output power / input power) * 100 % — >>> eq.1

That mean >>> ɳ = (output power / (output power +losses)) * 100 % — >>> eq.2

ɳ = (output power / (output power +iron losses+ copper losses)) * 100 % — >>> eq.3

And also may be >>> ɳ = ((input power-iron loss-copper loss)/ input power) * 100 % — >>> eq.4

ɳ = 1- ((iron loss +copper loss))/ input power) * 100 % — >>> eq.5

So, we can obtain the efficiency of the transformer from any equation above according to given data.

## Important Facts

Let us now turn to the important facts in turn …

### The first fact

Is there any transformer that has an efficiency of 100 %?

The efficiency of the ideal transformer** is 100%**. For those who do not know the ideal transformer, the input power is equal to the output power. However, this transformer is not available in reality. It is only a case for **scientific** and **theoretical studies**.

Looking at most of the topics that speak about the efficiency of the transformer does not address this fact.

### The second fact

Why do we analyze and study the efficiency of the transformer?

Of course, there are thousands, but millions of learners around the world chanting the transformer efficiency equations every day, but without a real understanding of what the goal of studying and analyzing of it.

In fact, we study and analyze the efficiency of transformers to determine the **performance** of the transformer, which is, of course, reflected on the **economic side**.

And all of us know that the **higher the efficiency** of the transformer the less power lost in the transformer and also reduce the costs of periodic maintenance of the transformer and also the cost of **energy consumption** in the transformer.

### The Third fact

Is loading the transformer have any effect on the transformer’s efficiency?

This means that it is possible that the efficiency is **variable** depending on the load of the transformer.

Many articles speak about the method of calculating the efficiency of the transformer but do not mention that the transformer is more than efficient and where the disagreement occurs.

##### And here comes the important question … Is the efficiency of the transformer fixed or variable?

Let me clarify it with the equations as it is the best evidence to answer this question.

Through the study of the losses inside the transformer, we find that there are two types of losses, namely **copper loss** and **iron loss**.

Suppose there is a transformer with a power of 1000 kVA and the iron losses are 1.2 kVA … In the case of non-loading (no-load condition), the value of the copper losses is zero.

- So, Compensation in equation no. 4 :

ɳ = ((input power-iron loss-copper loss)/ input power) * 100 % ————– >>> eq.4

ɳ = ((1000-1.2-zero)/ 1000) * 100 % = 99.88 %

- In the case of the load condition, copper losses have value suppose that is 9.2 KVA.

So,

ɳ = ((input power-iron loss-copper loss)/ input power) * 100 % ————– >>> eq.4

ɳ = ((1000-1.2-9.2)/ 1000) * 100 % = 98.96 %

Therefore, the higher the load, the greater the copper losses and the lower the efficiency.

And through this fact, we can deduce the maximum efficiency that the transformer can reach.

##### So, we can derive the equation of maximum efficiency of transformer:

Suppose that (X) is the fraction of the full load then compensation in equation no.3:

ɳ x= ((X*output power) / ((X*output power) +iron losses+ (X*copper losses)) * 100 % –>> eq.3

So that the copper losses vary according to the loading condition of the transformer (X fraction of the load).

Also, efficiency will be maximum if the denominator with respect to the variable copper losses is equated to zero.

We can conclude that we will obtain maximum efficiency if copper losses (Pc) are equal to iron losses (Pi).

So,

ɳ max= (output power) / (output power +Pi+ Pc)) * 100 %

Pc = Pi at maximum efficiency so,

ɳ max= (output power) / (output power +2 Pi)) * 100 %

Also copper losses (Pc) = x2Pc (where Pc is the full load copper losses)

For maximum efficiency — >>> Pi = X2 * Pc

X = √(Pi/Pc)

so

ɳ max = X * full load KVA

ɳ max =√(Pi/Pc)* full load KVA ———- >>>>>>>>> final equation at maximum condition.

### The fourth fact

Does the power factor of the load effect on the efficiency of the transformer?

Again, I will clarify in numbers as is the best evidence to prove the facts:

- Suppose there is a transformer with a power of 100 kVA and the iron losses has 0.2 kVA and the current in the secondary coil is 8 amperes with a lag power factor of 0.8, then :

Copper losses = V2 * I2 * Cos ɸ = 220 * 8 * 0.8 = 1.41 KVA

ɳ = ((input power-iron loss-copper loss)/ input power) * 100 % ————– >>> eq.4

ɳ = ((100-0.2-1.41)/ 100) * 100 % = 98.39 %

- For the same transformer and the same load but the current in the secondary coil is 8 amperes with a lag power factor of 0.6, then:

(Pc) Copper losses = V2 * I2 * Cos ɸ = 220 * 8 * 0.6 = 1.06 KVA

ɳ = ((input power-iron loss-copper loss)/ input power) * 100 % ————– >>> eq.4

ɳ = ((100-0.2-1.06)/ 100) * 100 % = 98.74 %

Therefore, the **higher the load** power factor, the **less the efficiency** when the load is constant.

### The fifth fact

Is the efficiency of the transformer calculated and analyzed only momentarily?

After identifying the above facts we can understand that the transformer efficiency is **variable** all the time, but will the efficiency be analyzed and recorded every hour or every day or every month?

As we explained earlier that this efficiency has many factors, and most importantly, the download and since the loading is not fixed, we have to conduct a comprehensive study **every hour** to produce a daily report to analyze the performance of the transformer.

To ask ourselves a question; how is the transformer efficiency calculated throughout the day … The answer is very simple and lies in the following **equations**:

ɳ = ((input power-iron loss-copper loss)/ input power) * 100 %

Note that: iron loss is fixed all day and Copper loss changes according to loading condition

ɳ (for 24 hour) = ((input power -(iron loss)-((∑copper loss for 24 hours / 24))/ input power) * 100 %

## 10 Secrets about Losses in Transformer

Previously we talked about the efficiency of the transformer and how this efficiency affects the lifetime of the transformer. Here, we will discuss more on losses in the transformer and how to reduce them.

The losses of any machine in a simple scientific way are the amount of **difference** between the **input power** to the machine and the **output power** coming out of it.

But we will not let things go this way and as we would go deeper into the transformer to see what exactly happens to consume this power within the transformer and the factors affecting it.

### The first secret

We previously understood the ideal transformer and its characteristics and it has become clear to us that the ideal transformer is found only in the **world of theories** and has no practical presence on the ground.

So, the efficiency of the ideal transformer is 100%, so the losses in the ideal transformer are 0%.

That means the **input power** is equal to the **output power** and this is theoretically only.

### The second secret

Most of the scientific sites that deal with this subject have classified the losses in transformer into two parts, namely** iron loss**, and **copper loss**, but the secret here is that the losses within the transformer are divided into 6 categories.

- Hysteresis losses.
- Eddy current losses.
- Dielectric losses.
- Copper losses.
- Stray losses.
- Conductor eddy current losses.

Now I will reveal the set of secrets for each type of loss in the transformer in detail.

### The third Secret

The third secret about **No-load losses** in details are below:

When we disconnect the loads from the transformer while still connecting it to the source, the transformer continues to draw power from the source. We consider this power as a loss because there is **no output**.

But this power has a **key benefit** that is making the transformer energized and ready for service.

This power is withdrawn when the transformer is not loaded; it also remains a lost power while loading the transformer with the same value.

We can divide these losses into three main types: A lot of sites classified no-load losses as hysteresis losses and eddy current losses only … but no-load losses are divided into **3 types of losses**:

a. Hysteresis Losses.

b. Eddy’s current Losses.

c. Dielectric losses.

### The fourth secret

The fourth secret about **hysteresis losses** in details:

This loss is part of the core losses and is also a type of no-load losses, that is, whether the transformer is connected to a load or not.

The idea of this type of loss is simply is that “*every time the magnetic field is reflected, a fraction of the power is lost as a result of hysteresis*”

So, what’s the meaning of hysteresis? This word means the tendency of **magnetism** to stay in its old state.

And the fact that a part of the magnetism remains in the iron core is “**residual flux**”. Whiles a lost part of the magnetic power within the iron core is called “**hysteresis losses**”. Their effect is represented in the circuit equivalent to the real transformer in the form of a winding with an **impedance Xm**.

When passing the sinusoidal current in the primary winding it produces the **magnetic flux**. This magnetic flux is quite similar to the voltage that created and it has the same sinusoidal curve.

##### Sinusoidal curve

In the first half of the sinusoidal cycle, this magnetic flux causes magnetization of the **magnetic material** in the core, which we can consider being internally composed of a group of domains that are aligned in one direction because of the magnetic field.

When the direction of the magnetic field is **reversed** in the second half of the second cycle of the sinusoidal curve, the domains inside the core must change direction and the poles must align parallel in the new direction of the field.

Thus we can consider hysteresis loss as the **lost power** during the friction of the particles of matter and the movement of its poles in each cycle of the magnetic field.

##### Hysteresis losses

In fact, the energy consumed in the row of poles of the magnetic material comes from the input power. Of course does not transfer this energy to the secondary side, which does not move to the load and then considers it **lost energy**.

##### Calculation of Hysteresis losses

We conclude from this that the **stronger** the magnetic field and the **higher** the frequency, the higher the energy consumed in the row, and the change of the direction of the electrodes inside the material.

We can calculate the value of hysteresis losses from the following equation:

Kh: constant depends on the material and its quality

Bmax: peak value of the flux density.

F: frequency of the supply.

V: volume of the core

We can calculate the value of hysteresis loss in another way, through the area within the loop known as hysteresis loop as in the following figure:

So, the larger the area within the loop, the greater the hysteresis losses.

From the previous figure:

It is clear that we can reduce hysteresis losses by improving the quality of the material used in the core.

### The fifth secret

The fifth secret in details about **eddy current losses**:

We all know that magnetic materials such as iron also have the ability to connect the electrical current and not only to pass the magnetic flux. When the flux lines cut off the wires of copper winding, they generate a great e.m.f under the **law of Faraday**.

When they cut off these winding, they also cut off Iron core which generates an electric current called **eddy current.**

Of course, these eddy currents are not desirable. They do not reach the load and cause the iron core to warm unnecessarily. Usually, eddy currents cause about **50%** of the iron core loss.

This type of loss depends mainly on the type of magnetic material, frequency, and magnetic field density.

We can calculate according to the following formula:

Such that :

Ke is constant depends on the type and thickness of the material

##### Reduce Eddy Current Losses

The value of the thickness of the **magnetic material** has a strong effect on the Eddy current loss. So, the less thickness the greater the electrical resistance and the less the current passing through the iron core.

Therefore all the iron cores of all electrical transformers should be in the form of **lamination** separated from each other and compressed together.

It involves transformer lamination core1 and transformer lamination core2.

In terms of magnetism, the lamination set gives the area of the cube of the iron core ability to withstand the passage of the flux.

In the electric sense, it is **high resistance** to the small area of the section in each slice. Usually does not exceed the thickness of the single lamination is 0.35 mm. There is only a small current and this can be the way to reduce the eddy current.

### The sixth secret

The seventh secret in detail is about **Dielectric losses**:

The dielectric losses are classified under **no-load losses** … they are present during loading and at the no-load condition of the transformer. The insulation materials used to isolate the conductors from each other inside the transformer causes a kind of capacitors known as the **stray capacitors**.

They are capacitors that are not visible to the eye and are not held by a hand but do the same work as the real capacitor and cause some kind of **loss of energy. **

In fact, we can represent the ideal capacitor by **capacitance** only without resistance … where there is no loss in power, it is charged by the first half of the sine wave and then discharge in the second half. And then charged and discharged and so on …

And therefore does not lose anything of the active power so the angle between the voltage and the current **90 degrees**. But this is only **theoretically possible** … There is no capacitor, in fact, only capacitance, but always with small resistance.

##### The severity of dielectric loss

So, the angle between voltage and current is less than 90 degrees by a small angle called **delta (δ)**.

Tan δ is a standard measure to give an indication of the magnitude of the capacitor spacing as an ideal capacitor. The smaller the angle, the closer the capacitor is to the **ideal capacitor**.

Thus, dielectric losses are proportional to the other with tan δ, and this amount called “**dissipation factor**” … and we can calculate the dielectric losses from the following equation:

Such that:

Pd: Dielectric losses in Watt.

F: applied frequency in HZ.

C: Capacitance in Farad.

V: Operating Voltage in r.m.s

Tan δ: dissipation factor.

Although dielectric loss is a small fraction of the losses in the transformer, it is one of the **most dangerous** types of losses in the transformer because temperature strongly affects **tan δ**. “The higher the temperature the more tan δ”.

### The seventh secret

The seventh secret in details about **Load losses**:

These set of load losses appears only during the loading of the transformer as a result of the passage of the load current through the **winding**. Therefore it consists mainly of **copper losses** in the resistance of winding, either in the primary or secondary winding.

The importance of calculating load losses is a fundamental element when **estimating the size** of the transformer. The heats resulting from the passage of the current in the winding raises the temperature to dangerous degrees.

So it is necessary to work to reduce these losses, which is often by reducing the **value of resistance** winding.

We can calculate load losses through the **short circuit test** of the transformer.

Load losses divided into three main types:

a. Copper losses.

b. Stray losses.

c. Leakage flux losses.

### The eight secret

The eighth secret about **Copper losses** in details:

These losses are the **first type of load loss** in the sense that it does not appear as an effective value until after loading the transformer. The more the transformer loads, the more power losses are high.

It is known that the copper winding (primary and secondary) has a certain resistance and therefore the passage of a current which **causes loss of power** is calculated from the following equation:

P= I² * R

For accuracy, the loss in the copper winding is at **no-load condition** because of the passage of no-load current. The no-load current is the current that passes through the **primary winding** only at the rated voltage with the fact that the other winding open and often within 1% to 2% of full load.

##### Effect of temperature on copper losses

As the temperature rises, the value of winding resistance increases in contrast to the eddy current loss, which decreases with increasing temperature.

Statistically, each **1 ° C increase** can cause a rise in **losses of 0.4%**.

It is known that the value of the resistance is calculated from the law (R = ρ *L /A). But this law is absolutely true in the case of DC Current only if the current AC is considered a kind of approximation acceptable and not accurate. Because the AC tends to pass at the ends of the connector away from its center it causes a skin effect.

##### skin effect

The concentration of the current in the sides makes the area of the actual section of the connector less than the engineering area of the conductor and then Rac increases the value of Rdc and this is the skin effect.

Therefore, the copper losses increase more if the current of AC and of course shows this effect clearly whenever the area of the section of the conductor is larger … We can neglect this effect In small conductors.

### The ninth secret

The ninth secret about **stray losses** has the following details:

These stray losses produce as a result of the **leakage flux** … We know that the flux that generates when the current passes in a winding is not completely coupled to the other winding but there is a missing part.

We can express this missing part of the flux as leakage flux. This leakage flux may cut off the **external iron** parts of the transformer and create the **eddy current**.

### The tenth secret

The tenth secret about **conductor eddy current** losses in details:

This eddy current causes the hotness of these **non-current metal parts**. This is a type of energy loss that shows an effect on large transformers only.

This type of loss is generated in the **copper conductors** of the winding due to leakage flux but has small values and its effect is weak … its value is calculated by **experimentation** and **measurement** and not by equations.