losses and efficiency in transformer

Introduction

A transformer is one of the most vital electrical machines used in power systems. Its primary role is to step up or step down voltage levels, making power transfer more reliable and economical. Transformers ensure smooth and efficient power transmission from generating stations to end users, reducing waste and maintaining system stability.

The transformer’s performance greatly depends on its design, quality of materials, and operating conditions. Minimizing energy losses is crucial, as even small inefficiencies can result in significant power waste and financial loss when scaled to industrial levels. Understanding the working principles and efficiency parameters of transformers helps improve reliability and reduce costs.

Students, engineers, and industries alike depend on accurate transformer efficiency data. For students, it forms the foundation of learning in electrical engineering. For engineers, it is critical in designing, maintaining, and operating systems. For industries, higher efficiency directly translates into cost savings and sustainable energy use.

losses and efficiency in transformer

In simple terms, power losses in transformers are the portion of input energy that does not get converted into useful output. These losses appear in the form of heat, vibrations, or magnetic effects, and they reduce the overall efficiency of the device. Ideally, a transformer should transfer 100% of the input power to the output, but in practice, some energy loss in transformer systems is unavoidable.

This wasted energy is known as active power loss and is mainly of two types:

1. Iron-loss (Pi): This occurs in the magnetic core of the transformer. It is also referred to as Core Loss, No-load Loss, or Open-circuit Loss. It primarily arises due to hysteresis and eddy currents in the core material.
2. Copper-loss (Pcu): This takes place in the primary and secondary windings of the transformer, which are usually made of copper. It is also known as Ohmic Loss, and it is caused by the resistance of the winding conductors.

Thus, the Total Loss in a Transformer can be expressed as:

Total Loss = Iron Loss (Pi) + Copper Loss (Pcu)

The concept of real vs ideal transformer explains this better. An ideal transformer is a theoretical model with no losses, operating at 100% efficiency. On the other hand, a real transformer experiences transformer inefficiency due to winding resistance, core losses, leakage flux, and other practical limitations. Understanding these losses is essential for engineers and industries to design and operate more efficient systems.

Types of Transformer Losses

A transformer, though highly efficient, is never completely free from energy losses. These losses reduce overall performance and reliability, and understanding their nature helps engineers design better electrical systems. The main types of transformer losses are as follows:

1. Core (Iron) Losses

In electrical machines such as transformers, iron loss (core loss) is an important factor that reduces efficiency. This loss occurs in the magnetic core, which is continuously subjected to alternating magnetic flux. Iron loss can be divided into two main types:

1. Hysteresis Loss (Ph)
2. Eddy Current Loss (Pe)

Hysteresis Loss

Hysteresis loss is produced because the magnetic core undergoes repeated magnetization and demagnetization as the alternating flux (Φ) changes direction. Every cycle of reversal consumes energy in realigning the magnetic domains inside the core material, resulting in heat.

Minimization of Hysteresis Loss

To reduce this loss, the transformer core is made of highly permeable materials such as:

• CRGO (Cold Rolled Grain Oriented) silicon-steel alloy
• Amorphous alloy

These materials have a narrow hysteresis loop, which means less energy is wasted during each magnetization cycle.

Eddy Current Loss

When alternating flux passes through the core, it induces circulating currents (eddy currents) within the conductor material. These currents produce heat and wasteful energy loss.

Minimization of Eddy Current Loss

To minimize this, the transformer core is not a solid block but is constructed of thin insulated laminations (stamped sheets). This design restricts the path of eddy currents and reduces the overall loss.

Such a construction is known as a laminated core.

Mathematical Expression of Iron Loss

The total iron loss (Pi) is given by:

Pi = Ph + Pe

Hysteresis Loss Formula

Ph = Kh × Bmaxⁿ × f (W/kg)

Where:

• Kh = Hysteresis constant (depends on core material)
• Bmax = Peak flux density (Wb/m²)
• f = Frequency of flux reversal (Hz)
• n = Steinmetz constant (ranges between 1.5 to 2.5)

Eddy Current Loss Formula

Pe = Ke × Bmax² × f² × t² (W/kg)

Where:

• Ke = Constant depending on core resistivity and volume
• Bmax = Maximum flux density (Wb/m²)
• f = Supply frequency (Hz)
• t = Thickness of lamination (m)

Key Points

• Iron loss = Hysteresis loss + Eddy current loss
• Hysteresis loss depends on material properties and frequency.
• Eddy current loss depends on lamination thickness, flux density, and frequency.
• To reduce losses, cores are built with CRGO steel or amorphous alloys and designed with thin insulated laminations.

Why is Iron Loss also Called Constant Loss?

In transformers, iron loss (core loss) is often referred to as a constant loss. Let’s understand why.

Relation with EMF Equation

From the emf equation of a transformer:

E ≈ 4.44 × f × Φ

Where:

• E = Induced emf (volts)
• f = Supply frequency (Hz)
• Φ = Maximum value of flux in the core (Wb)

Rearranging:

Φ ∝ E / f

This shows that the flux in the core is directly proportional to the ratio of input voltage (V1) to frequency (f).

Dependence of Iron Loss

From the iron loss equations:

• Hysteresis Loss (Ph) = Kh × Bmaxⁿ × f
• Eddy Current Loss (Pe) = Ke × Bmax² × f² × t²

It is clear that:

• Both losses depend on flux density (Bmax) and frequency (f).
• Neither equation contains winding currents (I1, I2, etc.), which means iron loss does not depend on the load current.

Why is Iron Loss Constant?

Since the input voltage (V1) and frequency (f) of the supply remain constant, the magnetic flux in the transformer core also remains constant from no-load to full-load conditions.

Thus:

• Hysteresis loss and eddy current loss remain the same regardless of load.
• Therefore, iron loss is independent of load current.

This is why iron loss is called a constant loss in transformers.

Experimental Verification

The value of iron loss can be determined experimentally using the Open-Circuit Test of a transformer. In this test:

• The secondary winding is kept open.
• The rated voltage is applied to the primary.
• The input power measured primarily represents the iron loss.

Key Takeaway

• Iron loss = Constant loss because it depends only on input voltage and frequency, not on load.
• It remains the same at no-load and full-load conditions.
• Verified practically by the Open-Circuit Test.

2. Copper Loss in Transformers

Copper loss is one of the most significant losses in a transformer and plays a crucial role in determining its efficiency. It is caused by the electrical resistance of the copper windings (primary and secondary). Since transformers carry alternating current, heat is generated in the conductors due to I²R loss, where I is the current flowing through the winding and R is the resistance of the winding.

1. Definition

Copper loss is defined as the active power loss in the windings of a transformer due to their inherent ohmic resistance. It is sometimes referred to as ohmic loss or winding loss.

2. Mathematical Expression

The total copper loss can be expressed as:

Pcu = I₁²R₁ + I₂²R₂ (Watts)

Where:

• I₁ = Current in the primary winding
• R₁ = Resistance of the primary winding
• I₂ = Current in the secondary winding
• R₂ = Resistance of the secondary winding

Alternative forms:

• Pcu = I₁²R₀₁ (primary side referred)
• Pcu = I₂²R₀₂ (secondary side referred)

This shows that copper loss is proportional to the square of the current.

3. Nature of Copper Loss

Copper loss is a variable loss because it depends directly on the load current:

• At Full Load → Pcu = Pcu(FL)
• At Half Load → Pcu = (½)² × Pcu(FL) = Pcu(FL)/4
• At One-Third Load → Pcu = (⅓)² × Pcu(FL) = Pcu(FL)/9

At No Load, copper loss is negligible because the no-load current (Io) is very small, and the secondary current (I₂) is absent.

4. Percentage Copper Loss

Percentage copper loss is used to compare copper losses with transformer output power:

% Copper Loss = (Copper Loss / Output Power) × 100

For primary side:
%R = (I₁²R₀₁ / V₁I₁) × 100%

For secondary side:
%R = (I₂²R₀₂ / V₂I₂) × 100%

Simplified:
% Copper Loss = (I₂R₂ / V₂) × 100%

5. Equivalent Resistance Determination

The equivalent resistance of transformer windings can be calculated as:

R₀₁ = (% Copper Loss × V₁) / (I₁ × 100)

R₀₂ = (% Copper Loss × V₂) / (I₂ × 100)

These values are important for transformer modeling and performance analysis.

6. Measurement of Copper Loss (Short-Circuit Test)

The most reliable method to measure copper loss experimentally is the Short-Circuit Test:

• The secondary winding is short-circuited.
• A low voltage is applied to the primary winding to circulate full-load current.
• The wattmeter reading under this condition gives the copper loss.

In this test, the applied voltage is very low, so the core flux is also very small. Hence, iron losses are negligible, and the wattmeter reading represents only copper losses.

7. Importance of Copper Loss

• Copper loss reduces transformer efficiency.
• It generates heat, which may affect insulation life and reliability.
• It is the major load-dependent loss in transformers.
• Reducing copper loss involves minimizing winding resistance by:
  • Using high-conductivity copper.
  • Increasing the cross-sectional area of windings.
  • Improving cooling systems.

3. Stray Losses in Transformers

Stray losses are relatively small but unavoidable losses in a transformer. They occur because of stray flux, which is the portion of magnetic flux that does not follow the designed path in the transformer core. Instead, this leakage flux links with nearby metallic parts such as the transformer tank, clamps, or structural supports. When this happens, small currents are induced in these metallic parts, leading to leakage flux losses.

Although these are minor compared to iron and copper losses, they are still categorized as additional transformer losses and must be considered in the design stage. Proper shielding, magnetic circuit design, and use of non-magnetic materials in structural parts can help minimize stray losses.

4. Dielectric Losses in Transformers

Dielectric losses occur within the insulating materials of a transformer. When subjected to high alternating electric fields, the insulation experiences dielectric heating, which is the conversion of electrical energy into heat.

These losses become more significant in high-voltage transformers, where electric stress is high. Prolonged dielectric stress can lead to insulation breakdown, reducing the safety, reliability, and lifespan of the transformer.

To reduce dielectric losses, high-quality insulating materials with low loss factors are used, along with proper design of insulation thickness and cooling arrangements.

Efficiency of a Transformer (η)

The efficiency (η) of a transformer is defined as the ratio of active output power to active input power. It represents how effectively a transformer converts input electrical power into useful output power. In practice, some power is lost in the form of iron losses (Pi) and copper losses (Pcu). Therefore, efficiency is always less than 100%.

It is also known as power efficiency or commercial efficiency because it is based on active power (in watts) and not on apparent power (in volt-amperes). Since efficiency depends on the load power factor, it reaches its maximum when the power factor = unity (cosθ = 1).

1. Formula for Efficiency

The general expression for efficiency is:

%η = (Output Power / Input Power) × 100

Or equivalently,

%η = (S cosθ) / (S cosθ + Losses) × 100

Where,

• S = Apparent power (V₂ × I₂)
• cosθ = Load power factor
• Losses = Pi + Pcu (Iron and Copper Losses)

Another form:

%η = Output Power / (Output Power + Pi + Pcu) × 100

Or,

%η = (Input Power – (Pi + Pcu)) / Input Power × 100

2. Factors Affecting Efficiency

1. Iron Loss (Pi): Constant loss, independent of load.
2. Copper Loss (Pcu): Variable loss, proportional to I² (load current squared).
3. Load Power Factor (cosθ): Efficiency improves as the load power factor approaches unity.

Maximum Efficiency (ηmax) in a Transformer

The efficiency of a transformer varies with load and power factor (cos θ). At no-load, efficiency is zero because there is no output power, but iron losses still occur. As the load increases, efficiency improves because output power increases compared to fixed losses. However, beyond a certain load point, copper losses (which increase with load current) become dominant, reducing efficiency.

Thus, there exists an optimum load at which transformer efficiency is maximum.

General Efficiency Equation

From the efficiency formula:

η = (Input power – Pi – Pcu) / Input power

Where:

• Pi = iron (core) loss (constant with load)
• Pcu = copper (I²R) loss (varies with load)

Mathematical Derivation

Considering input power = V₁ I₁ cos θ₁, we have:

η = (V₁ I₁ cos θ₁ – Pᵢ – I₁² R₀₁) / (V₁ I₁ cos θ₁)

Simplifying:

η = 1 – (Pᵢ / V₁ I₁ cos θ₁) – (I₁² R₀₁ / V₁ I₁ cos θ₁)

For maximum efficiency, the derivative concerning load current (I₁) is zero:

dη / dI₁ = 0

This leads to the condition:

Pᵢ = I₁² R₀₁

Thus, maximum efficiency occurs when iron loss = copper loss.

Expression for Maximum Efficiency

The maximum efficiency (ηmax) occurs when:

Pᵢ = Pcu

The corresponding load kVA at which maximum efficiency occurs is:

Sₘₐₓ η = SFL × √(Pᵢ / Pcu(FL))

Where:

• SFL = full-load kVA rating of transformer
• Pᵢ = iron (core) loss
• Pcu(FL) = full-load copper loss

Key Points:

1. Iron losses are constant, independent of load.
2. Copper losses increase with load (I²R).
3. Maximum efficiency is obtained when:
   Iron Loss = Copper Loss
4. For distribution transformers, efficiency is designed to be maximum near 50–70% of full load (as they usually operate below full load).
5. For power transformers, maximum efficiency is designed to occur near full load, as they are usually operated closer to rated capacity.

All-Day Efficiency (Energy Efficiency) of a Transformer

Transformers used in distribution systems (for residential and commercial supply) are energized 24 hours a day. The secondary load, however, varies throughout the day — sometimes operating at no-load, at times at light-load, and occasionally near full-load.

Hence, the efficiency of such transformers cannot be judged by the usual power efficiency formula, since it only considers input and output power at one instant. Instead, All-Day Efficiency (also called Energy Efficiency) is used, which is based on total energy delivered in 24 hours.

Definition

All-day efficiency is defined as:

ηₐₗₗ₋dₐy = (Total output energy in watt-hours over 24 hours) / (Total input energy in watt-hours over 24 hours)

Mathematically:

ηₐₗₗ₋dₐy = (Output energy in Wh) / (Input energy in Wh) (for 24 hours)

Key Characteristics

1. Iron Loss (Pᵢ):
   • Occurs continuously (even at no-load).
   • Independent of load, since it depends only on supply voltage.
   • Present throughout the entire 24-hour period.
2. Copper Loss (Pcu):
   • Depends on load current (∝ I²).
   • Varies throughout the day depending on loading conditions.
   • Zero at no-load, maximum at full-load.

Thus, in distribution transformers:

• Iron loss dominates because the transformer is energized continuously.
• Copper loss fluctuates with load variations.

Expression

Let:

• Pᵢ = Iron loss (constant)
• Pcu(t) = Copper loss at time t (variable)
• Eout = Total output energy over 24 hours (Wh)
• Ein = Total input energy over 24 hours (Wh)

Then:

ηₐₗₗ₋dₐy = Eout / Ein

And:

Ein = Eout + (Pᵢ × 24) + ∑(Copper Loss × time interval)

Step-by-Step Calculation

1. Divide the 24-hour cycle into load periods (e.g., 6 hrs no-load, 10 hrs half-load, 8 hrs full-load).
2. Compute output energy for each period:
   • Output Energy = Load (kW) × Time (h).
3. Compute copper loss for each period:
   • Copper Loss ∝ (Load/Full load)² × Full-load Copper Loss × Time.
4. Add iron loss × 24 hours.
5. Substitute into the formula:

ηₐₗₗ₋dₐy = Total Output Energy / (Total Output Energy + Total Iron Loss Energy + Total Copper Loss Energy)

Example

Suppose a 100 kVA transformer has:

• Iron loss = 1 kW
• Full-load copper loss = 2 kW
• Daily load cycle:
  • 12 hours at 50 kW load
  • 6 hours at 80 kW load
  • 6 hours at no-load

Step 1: Output Energy

• 12 hrs @ 50 kW = 600 kWh
• 6 hrs @ 80 kW = 480 kWh
• 6 hrs @ 0 kW = 0
• Total Output = 1080 kWh

Step 2: Copper Loss Energy

• Full-load copper loss = 2 kW
• At 50 kW (50/100)² × 2 = 0.5 kW → 12 hrs = 6 kWh
• At 80 kW (80/100)² × 2 = 1.28 kW → 6 hrs = 7.68 kWh
• At 0 kW → 0
• Total Copper Loss = 13.68 kWh

Step 3: Iron Loss Energy

• 1 kW × 24 hrs = 24 kWh

Step 4: Input Energy

• Ein = 1080 + 24 + 13.68 = 1117.68 kWh

Step 5: All-Day Efficiency

• ηₐₗₗ₋dₐy = 1080 / 1117.68 = 96.61%

Key Points

1. Power efficiency is instantaneous (depends on load at that moment).
2. All-day efficiency considers energy over 24 hours.
3. Distribution transformers are designed for high all-day efficiency (since they run at light load most of the time).
4. Power transformers (used in generating stations) are designed for maximum power efficiency near full load.

Methods to Reduce Transformer Losses

Transformer losses mainly consist of iron (core) losses and copper (winding) losses. Reducing these losses improves the efficiency and reliability of transformers, making them more energy-efficient and cost-effective in the long run.

1. Use of Better Core Materials

• CRGO (Cold Rolled Grain Oriented) Steel: This type of silicon steel is widely used in modern transformers because it reduces hysteresis loss and improves efficiency.
• Amorphous Core Materials: Amorphous alloys have very low hysteresis loss compared to conventional silicon steel. They significantly reduce no-load (iron) losses and are commonly used in energy-efficient distribution transformers.

Benefit: Minimizes core losses, leading to improved performance and reduced energy wastage.

2. Improved Winding Techniques

• Use of Low-Resistance Conductors: High-quality copper or aluminum with larger cross-sectional area is used to reduce I²R copper losses.
• Better Winding Arrangement: Optimized placement and design of windings reduce leakage reactance and enhance efficiency.
• Foil or Strip Windings: Used in high-current, low-voltage windings to minimize eddy current losses.

Benefit: Reduces copper (I²R) losses and improves thermal performance.

3. Proper Maintenance

• Regular Oil Testing and Replacement: Ensures proper insulation and cooling, reducing overheating and additional copper losses.
• Cleaning and Inspection: Prevents dust and moisture accumulation, which may increase stray losses.
• Cooling System Maintenance: Ensures that the transformer operates within safe temperature limits, minimizing additional resistive losses.

Benefit: Extends transformer lifespan and keeps losses within designed limits.

Practical Applications of Transformer Efficiency

Transformer efficiency plays a critical role in ensuring reliable and cost-effective operation in various sectors. By reducing losses and maximizing output, efficient transformers directly contribute to energy savings and system stability.

1. Power Transmission Systems

Efficient transformers are vital in power transmission systems where electricity is transferred over long distances. High-efficiency transformers minimize transmission losses, ensuring that more of the generated power reaches end users. This not only improves efficient power distribution but also reduces the overall cost of electricity.

• Step-up transformers increase voltage for long-distance transmission, reducing current and losses.
• Step-down transformers deliver electricity at safe, usable voltage levels to homes and industries.

2. Industrial Applications

In industrial power systems, transformers supply energy to heavy machinery, motors, and equipment. High efficiency ensures stable power supply and reduces energy bills.

• Supports continuous operation of factories with minimal downtime.
• Reduces heat generation and improves reliability of industrial systems.
• Energy-efficient transformers lower operating costs and enhance productivity.

3. Smart Grids

Modern smart grids rely on efficient transformers to balance load demand, integrate renewable energy, and ensure grid reliability.

• Efficient transformers help in real-time load management.
• Enable better integration of solar, wind, and other renewable energy sources.
• Reduce stress on the grid during peak demand periods.

Key Benefits of Efficient Transformers

1. Lower operating costs due to reduced energy losses.
2. Improved grid reliability and stability.
3. Enhanced sustainability through reduced carbon emissions.
4. Long-term savings for utilities, industries, and consumers.

Challenges and Solutions in Transformer Efficiency

1. Heat Dissipation
   • Challenge: Transformers generate heat due to copper losses (I²R losses) and core losses. Excessive heat reduces insulation life and transformer efficiency.
   • Solution: Use advanced cooling techniques (oil cooling, forced air cooling, or liquid-immersed systems) and improved ventilation designs to manage overheating in transformers.
2. High Initial Cost of Efficient Designs
   • Challenge: Energy-efficient transformers with amorphous cores, CRGO steel, or advanced insulation materials are more expensive initially.
   • Solution: Consider the life-cycle cost instead of just the purchase cost, since efficient designs save more energy and reduce operating costs over time. Many governments also provide incentives for installing energy-efficient transformers.
3. Maintenance Issues
   • Challenge: Lack of regular inspection can lead to problems such as insulation breakdown, oil leakage, or winding degradation, affecting long-term efficiency.
   • Solution: Implement preventive maintenance schedules, oil testing, thermal scanning, and condition monitoring systems to detect issues early and improve transformer reliability.

Future Trends in Transformer Efficiency

1. Smart Monitoring Systems
   Integration of sensors, AI, and IoT for real-time efficiency tracking. This enables predictive maintenance and reduces unexpected failures.
2. IoT-Enabled Efficiency Tracking
   Digital transformers can communicate performance data to operators, allowing optimization of load distribution and reduction of unnecessary losses.
3. Eco-Friendly Transformer Designs
   Use of biodegradable insulating oils, amorphous core materials, and recyclable components to promote sustainable electrical machines and support green energy systems.
4. Smart Transformers in Grids
   Essential for smart grids, they help improve grid reliability, allow integration of renewable energy sources, and ensure stable voltage regulation.

Conclusion

Transformer losses—iron losses, copper losses, stray losses, and dielectric losses—directly affect efficiency and reliability. By focusing on efficiency optimization, industries and utilities can achieve significant transformer energy savings.

• High-efficiency transformers reduce operational costs, extend equipment lifespan, and support reliable power supply.
• With advancements in smart monitoring and eco-friendly designs, the future of transformers is moving toward greener, smarter, and more sustainable solutions.