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Transformer Efficiency, All day Efficiency & Condition for Maximum Efficiency

Transformer Efficiency, All day Efficiency & Condition for maximum Efficiency

Transformer Efficiency, All day Efficiency & Condition for maximum Efficiency
Transformer Efficiency, All day Efficiency & Condition for Maximum Efficiency

Efficiency of Transformer

Transformer efficiency may be defined as the ratio between Output and Input.
Efficiency = Output/Input
On specified Power factor and load, the Transformer efficiency can be found by dividing its output on Input (Similar to other Electrical Machines i.e. motors, generators etc). But the values of both input and Output should be same in unites (i.e. in Watts, kilowatts, megawatts etc)
But note that a transformer has very high efficiency because the losses occur in transformer is very low. Since the Input and Output almost equal, therefore measurement of input and output is not possible practically. The best way to find the transformer efficiency is that, first determine the losses in transformer and then calculate the transformer efficiency with the help of these losses.
Efficiency = η= Output / Input
Efficiency = η= Output / (Output + Losses) …….… (As Input = Output +Losses)
Efficiency = η= Output / (Output +Cupper Losses + Iron Losses)
You may also find the Efficiency by the following formula
Efficiency = η= Output / Input
Efficiency = η = (Input – Losses) / Input …….… (As Output = Input – Losses)
Taking LCM
Efficiency = η = 1 – (Losses /Input)
As we know that the rating of Transform is expressed in kVA not in kW. But the efficiency doesn’t depend on VA i.e. it would be expressed in Power Watts (kW) not in kVA. Although, the Losses are directly proportional to VA (Volt-Amperes), thus, efficiency depends on Power factor on every kind of VA load. And the efficiency would be maximum on unity (1) Power factor.
 
Good to Know
  • We can also find Transformer Efficiency by determining:
  • Core losses by open circuit test or No Load test, and
  • Copper losses by short circuit test.

Condition for Maximum Efficiency of Transformer

We know that,
Copper Loss = WC = I12. R1or I22R2
Iron Loss=Wi = Hysteresis Loss + Eddy Current Loss
WI = WH + WE
Suppose to Primary Side…
Primary Input = P1 = V1I1 Cosθ1
Efficiency = η = Output / Input
Efficiency = η = (Input – Losses) / input ….. (As Output = Input – Losses)
Efficiency = η = (Input – Copper losses – Iron Losses)/Input
Efficiency = η = (P1 WC – WI) / P1
Efficiency = η = (V1 I1 Cosθ1 I12. R1 WI)/ V1 I1 Cosθ1
Taking LCM
Efficiency = η = 1- (I12. R1 /V1I1 Cosθ1) –(WI/ V1 I1Cosθ1)
Or
Efficiency = η = 1- (I1. R1 /V1Cosθ1) – (WI/ V1 I1 Cosθ1)
Differentiate both sides with respect to I1
Dη/ dI1 = 0 – ( R1 /V1 Cosθ1) + (WI/V1 I12 Cosθ1)
Dη/ dI1= – ( R1 /V1 Cosθ1) + (WI/V1 I12 Cosθ1)
For Maximum Efficiency, the value of (Dη/ dI1) should be Minimum i.e.
Dη/ dI1 = 0
The above Equation can be written as
R1 / (V1 Cosθ1) = (WI/V1 I12 Cosθ1)
Or
WI = I12. R1      or       I22R2
Iron Loss = Copper Loss
The value of Output current (I2) on which Maximum efficiency can be gained
I2 = √ (WI/ R2)
The value of Output current (I2) is the actor who equals the value of Copper Loss and Iron Loss (i.e. Copper Loss = Iron Loss)
Doing so, the maximum efficiency can be gained. Therefore, with proper designing, maximum efficiency can be attained at any desired load i.e. Copper loss and Iron Loss can be equaled.
Good to Know
·        Efficiency is usually less than 1 and it is often expressed as a percent (%).
·        Ideal Transformer is 100% efficient i.e. the efficiency of ideal transformer is 1.
·        Practical Transformers efficiencies are generally quite high in compression to other electrical machines and electronics devices (i.e. Motors, Generators etc) on the ordure of 90 t0 98%.

All Day Efficiency of Transformer

As we know that the commercial or typical efficiency of a transformer is the ratio of Output and Input in watts
Efficiency = Output (in Watts)/Input (in Watts)
But there are number of transformers whose performance can’t be monitored according the above general efficiency.
Those distribution transformers which supply electrical energy to lighting and other general circuits, their primary energize for 24 hours, but the secondary windings does not energize all the time. In other words, Secondary windings only energize at the night time when they supply electrical energy to lighting circuits. I.e. secondary windings supply eclectic power for very small load or no load for maximum time in 24 hours. It means that core loss occurs for 24 hours regularly but copper loss occurs only when transformer is on loaded.
Therefore it realizes the necessity to design a transformer in which the core loss should be low. As copper loss depends on load, therefore, they should be neglected. In this type of transformers, we can track their performance only by all day efficiency. All day efficiency may be also called “Operational efficiency”. On the base of usable energy, we estimate the all day efficiency for a specific time (During the 24 hours =one day). And we can find it by the following formula
All Day Efficiency = Output (in kWh)/Input (in kWh)
To understand about the all day efficiency, we must know about the load cycle i.e. how much load is connected, and for how much time (in 24 hours).

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2 comments

  1. If we connect the transformer in the circuit.
    Then circuit follow the ohm’s law or not??????

  2. Good matter

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