WAZIPOINT Engineering Science & Technology: The Efficiency of a Transformer

Thursday, November 2, 2023

The Efficiency of a Transformer

In our other article All-day Efficiency of Transformer, we have learned what is all-day efficiency and how to calculate it. In this article, we want to know the efficiency of a transformer and the calculation formula.

Efficiency Curve of a Transformer
Fig: Transformer Efficiency Curve

What Exactly is Meant by Transformer Efficiency?

A transformer's ordinary or commercial efficiency is defined as the ratio of output power to input power.

Efficiency = η = Output / Input

Efficiency = η= Output / (Output + Losses) 

As Input = Output + Losses

If we go into more detail about the transformer losses, just remember the previous articles about the transformer loss calculation.

Transformer Copper Loss the,

WC = I12 x R1   or I2x R2

Transformer Iron Loss,

WI = Hysteresis Loss + Eddy Current Loss = W= WH + WE

Suppose to the primary side of the transformer,

Primary Input = P1 = Vx I1 Cosθ1

Efficiency = η = Output / Input

Efficiency = η = (Input – Losses) / input 

(As Output = Input – Losses)

Efficiency = η = (Input – Copper losses – Iron Losses)/Input

Efficiency = η = (P– W– WI) / P1

Efficiency = η = (V1 x I1 Cosθ– I12  x R– WI) / V1 x I1 Cosθ1

Efficiency = η = 1- (I12 x R/ V1I1 Cosθ1) – (W/ Vx ICosθ1)


Efficiency = η = 1- (Ix R/ VCosθ1) – (WI / Vx I1 Cosθ1)

Differentiating both sides w.r.t I1.

Dη / dI1 = 0 – ( R/ V1 Cosθ1) + (W/V1 x I12 Cosθ1)

Dη / dI1= – ( R/ V1 Cosθ1) + (W/ Vx I12 Cosθ1)

For maximum efficiency, the value of (Dη / dI1) should be minimum i.e.

Dη / dI1 = 0

Thus, the above equation can be rewritten as:

R/ (V1 Cosθ1) = (W/V1 x I12 Cosθ1)

Or, WI = I12 x R1or       I2x R2

Iron Loss = Copper Loss

Thus, the transformer will give the maximum efficiency when its copper loss is equal to iron loss:

I= √ (W/ R2)

The value of output current (I2) is the factor that makes it possible to equal the value of copper loss and iron loss (i.e. copper loss = iron loss).

The range of transformer efficiency

The transformer's efficiency is generally in the range of 95 – 99 %. The efficiency can be as high as 99.7% for great power transformers with very low waste. Ideally, the efficiency of the Transformer should be hand 100%, but practically it is not possible due to various types of losses such as core losses or ohmic loss.

A case study on the efficiency of transformers.

A 500 KVA transformer has 2500 watts iron loss, and 7500 watts copper loss at full load. The power factor is 0.8 lagging. Calculate transformer efficiency at full load,
maximum efficiency of the transformer,
output KVA corresponding to maximum efficiency,
transformer efficiency at half load.

Solution: Transformer rating = 500 KVA
Transformer output power = 500,000 x 0.8 = 400,000 watts

Iron losses (Pi) = 2500 W
Full load copper loss (Pcu) = 7500 W

The efficiency of Transformer When Operating at Maximum Capacity

= [(output power)/(output power + Pi +Pcu)] x 100
= [(400,000)/(400,000 + 2500 + 7500)] x 100
= 97.56% (Ans)

Maximum Efficiency of Transformer

The maximum efficiency of a transformer refers to the highest level of energy efficiency that can be achieved by the device. This efficiency is typically expressed as a percentage and is calculated by dividing the output power by the input power. A higher efficiency rating means that less energy is wasted as heat and more is used to power the device, resulting in a more environmentally friendly and cost-effective operation.

For maximum efficiency, Copper loss (Pc) = Iron losses (Pi) = 2500 W
= [(output power)/(output power + Pi +Pc)] x 100
Therefore, maximum efficiency = [(400,000)/(400,000 + 2500 + 2500)] x 100
= 98.76% (Ans),

Output KVA Corresponding to Maximum Efficiency

= full load KVA x √(Pi/Pc)
= 500 x √(2500/7500)
= 500 x √0.333 = 166.5 KVA (Ans)

Efficiency can vary between different types of transformers and their load conditions. For example, power transformers used in electrical substations are designed for high efficiency under continuous full-load operation, while distribution transformers might have varying loads and need to maintain reasonable efficiency over a wider load range. Therefore, understanding and optimizing transformer efficiency is essential in the design and operation of electrical power systems.

You may know the details about the electrical transformer from the following articles:

  1. Working Principle of Transformer;
  2. Transformer Construction;
  3. Core-type Transformers;
  4. Shell-type Transformers;
  5. Elementary Theory of an Ideal Transformer;
  6. E.M.F. Equation of Transformer;
  7. Voltage Transformation Ratio;
  8. Transformer with losses but no Magnetic Leakage;
  9. Transformer on No-load;
  10. Transformer on Load;
  11. Transformer with Winding Resistance but no Magnetic Leakage;
  12. Equivalent Resistance;
  13. Magnetic Leakage;
  14. Transformer with Resistance and Leakage Reactance;
  15. Simplified Diagram;
  16. Total Approximate Voltage Drop in Transformer;
  17. Exact Voltage Drop;
  18. Equivalent Circuit Transformer Tests;
  19. Open-circuit or No-load Test;
  20. Separation of Core Losses;
  21. Short-Circuit or Impedance Test;
  22. Why Transformer Rating in KVA?;
  23. Regulation of a Transformer;
  24. Percentage Resistance, Reactance, and Impedance;
  25. Kapp Regulation Diagram;
  26. Sumpner or Back-to-back-Test;
  27. The efficiency of a Transformer;
  28. Condition for Maximum Efficiency;
  29. Variation of Efficiency with Power Factor;
  30. All-day Efficiency;
  31. Auto-transformer;
  32. Conversion of 2-Winding Transformer into Auto-transformer;
  33. Parallel Operation of Single-phase Transformers;
  34. Questions and Answers on Transformers;
  35. Three-phase Transformers;
  36. Three-phase Transformer Connections;
  37. Star/Star or Y/Y Connection;
  38. Delta-Delta or ∆/∆ Connection;
  39. Wye/Delta or Y/ Connection;
  40. Delta/Wye or ∆/Y Connection;
  41. Open-Delta or V-V Connection;
  42. Power Supplied by V-V Bank;
  43. Scott Connection or T-T Connection;
  44. Three-phase to Two-Phase Conversion and vice-versa;
  45. Parallel Operation of 3-phase Transformers;
  46. Instrument Transformers;
  47. Current Transformers;
  48. Potential or Voltage Transformers.

1 comment:

  1. The efficiency of a transformer is a measure of how well it converts electrical power from the primary (input) side to the secondary (output) side. In an ideal world, a transformer would be 100% efficient, converting all input power into output power without any losses. However, real-world transformers have inherent losses, which are divided into two main categories: copper losses (I²R Losses) and iron losses (Hysteresis and Eddy Current Losses). Copper losses occur due to the resistance of wire windings, generating heat when current flows through them. These losses can be minimized by using larger wire sizes and high-conductivity materials. Iron losses are associated with the core of the transformer, caused by the repeated reverse of the magnetic field in the core and circulating currents induced in the core due to the changing magnetic field. The overall efficiency of a transformer is the ratio of output power to input power, taking into account both copper and iron losses. Maintaining high efficiency is critical in power distribution and transmission applications to minimize energy waste. Efficiency can vary depending on the transformer's design, material quality, and load conditions. Proper maintenance and load management are also essential for ensuring transformers operate at their highest efficiency.abogado de lesiones personales playa de virginia


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