# 5 Facts That Nobody Told You about Transformer efficiency

July 28, 2017 ## 5 Facts That Nobody Told You about Transformer efficiency

There is no doubt that this important subject, which is 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 so 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 efficiency of transformer from any equation above according to given data. Let us now turn to the important facts in turn … The first fact:

#### Is this any transformer 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 on the reality. It is only a case for scientific and theoretical studies.

And looking at most of the topics that speak about the efficiency of the transformer does not address the presentation of this important fact.

The second fact:

### Why 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 it is.

In fact, we study and analyze the efficiency of any 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 reduced 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 has an effect on the transformer’s efficiency?

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

We understand from the many articles available on this subject the method of calculating the efficiency of the transformer, but it was not mentioned that the transformer 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 that matter with the equations 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 case 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 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 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:

##### Is the power factor of the load effect on the efficiency of the transformer?

Some students are also problematic here … Will the power factor of the load affect small or large on the efficiency of the transformer?

I will clarify the matter in numbers 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, most important of 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.

##### All day transformer efficiency To ask ourselves a question is how to calculate the transformer efficiency throughout the day … The answer is very simple 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 %