Test: Transformers - EmSAT Achieve MCQ

• The energy lost due to hysteresis per cycle is 6 J, but to find the power loss,
  we need to consider the frequency (50 Hz) at which the transformer operates.
• Here's how to calculate the power loss due to hysteresis:
• Cycles per Second:
  • The transformer operates at 50 Hz, which means it completes 50 cycles in one second.
• Energy Loss per Second:
• We know the energy lost per cycle is 6 J.
  To find the total energy loss per second, we multiply the energy per cycle by the number of cycles per second:
• Energy loss per second (Wh) = Energy per cycle (J) * Cycles per second
• Energy loss per second (Wh) = 6 J * 50 cycles/s
• Conversion to Watts:
• The unit for energy loss per cycle is Joules (J),
  and the unit for power loss is Watts (W).
  We need to convert Joules to Watt-seconds (Ws) before calculating the power loss in Watts.
  There are 1 Joule per Watt-second (1 J/Ws).
  Energy loss per second (Ws) = Energy loss per second (Wh) * (1 Ws/J)
  Energy loss per second (Ws) = 6 J * 50 cycles/s * (1 Ws/J)
• Power Loss:
• Power is the rate of energy loss per unit time. Since we have the energy loss per second in Watt-seconds (Ws),
  we can directly define it as the power loss due to hysteresis.
• Therefore, the power loss due to hysteresis in the transformer is 300 Watts (W).

Given:

kVA rating (kva) = 20
Iron loss (iron_loss) = 300 W
Copper loss (copper_loss) = 250 W
Calculations:

Total Losses:

total_loss = iron_loss + copper_loss
total_loss = 300 W + 250 W
total_loss = 550 W
Apparent Power Output (Sout):

We are considering full load, so Sout is equal to the kVA rating converted to VA.
Sout_va = kva * 1000 (conversion factor from kVA to VA)
Sout_va = 20 kVA * 1000 VA/kVA
Sout_va = 20000 VA
Output Power (Pout):

At unity power factor (unity PF), all the apparent power is transferred as real power (Pout).
Pout_w = Sout_va (since PF = 1)
Pout_w = 20000 VA
Efficiency:

efficiency = (Pout_w / (Pout_w + total_loss)) * 100
efficiency = (20000 W / (20000 W + 550 W)) * 100
Result:

efficiency ≈ 97.32%
Therefore, the efficiency of the 20 kVA transformer at full load and unity power factor is approximately 97.32%.

• Eddy current losses in a transformer are a form of power loss due to the induction of circulating currents in the core material when it is subjected to alternating magnetic fields.
• These losses contribute to the overall inefficiencies in a transformer and generate unwanted heat.

However, there are several strategies to minimize eddy current losses in transformers:

• Laminated Cores:
  • The core of the transformer is made from thin sheets of electrical steel, laminated and insulated from each other by a thin layer of insulating material. This construction increases the electrical resistance to the flow of eddy currents perpendicular to the laminations, thereby reducing the magnitude of these currents and the resulting losses.
  • The thinner the laminations, the lower the eddy current losses, but this comes with increased manufacturing costs.
• High-quality core material:
• Using high-grade silicon steel, or electrical steel, with specific properties that reduce hysteretic and eddy current losses.
  • Silicon steel improves the electrical resistivity of the core material, further reducing the eddy currents.
    • Amorphous steel, though more expensive, offers even lower loss characteristics and is used in high-efficiency transformers.
• Optimal Operating Frequency:
  • Eddy current losses increase with the square of the frequency. While this is a limitation in applications with a fixed supply frequency (e.g., 50 Hz or 60 Hz mains power), in applications where the frequency is controllable (such as in some types of electronic transformers), operating at the most efficient frequency can reduce these losses.
• Reduced Core Flux Density:
  • Operating the transformer at a lower magnetic flux density can also reduce eddy current losses, as these losses are proportional to the square of the flux density. However, reducing flux density might require a larger core or a different core material to maintain the same level of performance, which could increase costs and physical size.
• Improved Core Design:
  • The design of the transformer core can be optimized to reduce pathways for eddy currents.
  • This includes both the shape of the core and the arrangement of the laminations. Techniques such as step-lap and cross-lap designs in the lamination stacks can help reduce eddy currents by aligning the magnetic domains more effectively.
• Use of grain-oriented steel laminations:
  • The use of grain-oriented (GO) silicon steel, where the magnetic properties are optimized in one direction, can further reduce losses if the grain orientation is aligned with the direction of the magnetic flux.

• The efficiency of two identical transformers under load conditions can be determined by performing a back-to-back test.
• In this test, the two transformers are connected in parallel, and the primary windings of both transformers are connected to the same source.
• The secondary windings are then connected to a load.
• The efficiency of two identical transformers under load conditions can be measured and compared using a method known as the back-to-back test, also referred to as the Sumpner's test.
• This method is particularly effective because it allows the assessment of the transformer efficiency under load conditions without the necessity to supply the full load power from the external circuit, thus saving energy and reducing the requirements for high power sources for testing.
  

• Advantages:
  • Energy-efficient, as most of the load power is circulated internally.
  • Reduced requirements for a high-capacity power source for testing under full-load conditions.
  • Allows assessment of efficiency under simulated full-load conditions without the need for actual full-power dissipation.
• Limitations:
  • It mainly provides information on steady-state efficiency and losses, without insights into transient behaviors or other dynamic characteristics.
  • Accurate for similar types of transformers but might not fully replicate all operational conditions.​​​​

Voltage regulation for the transformer is given by the ratio of change in secondary terminal voltage from no load to full load to no load secondary voltage.

Voltage regulation =E₂−V₂ / E₂

It can also be expressed as,

Regulation =I₂R₀₂cosϕ₂±I₂X₀₂sinϕ₂ / E₂

+ sign is used for lagging loads

• Short circuit test performs on the high-voltage (HV) side of the transformer where the low-voltage (LV) side or the secondary is short-circuited.
• A wattmeter is connected to the high voltage side. An ammeter is connected in series with the high voltage side

When the load is not connected to the electric machine, the machine draws a low value of current to keep the machine active or running

• The current which is taken by the machine is called no-load current
• Current is drawn by the core of the machine 
• The equivalent circuit is represented by a high value of resistance in parallel.
• The power factor of the machine will be very low.

E = 111 V, f = 50 Hz, N = 200

111 = 4.44 × 50 × 200 × ϕ_m

ϕ_m = 111 / (4.44 × 50 × 200)

= 2.5 mWb

• Magnetizing inrush current in transformer is the current which is drawn by a transformer at the time of energizing it.
• To have minimum inrush current in the transformer the switch-on instant should be at maximum input voltage
• This inrush current is transient in nature and exists for few milliseconds.
• The inrush current may be up to 10 times higher than the normal rated current of the transformer.
• Although the magnitude of inrush current is high it generally does not create any permanent fault in the transformer as it exists for a very small time.
• But still, inrush current in a power transformer is a problem, because it interferes with the operation of circuits as they have been designed to function.
• Some effects of high inrush current include nuisance fuse or breaker interruptions, as well as arcing and failure of primary circuit components, such as switches. Another side effect of high inrush is the injection of noise.

• As in the step-up transformer, the number of coils in the secondary coil is more than that in the primary coil. So the electrical potential increases.
• A step-up transformer converts low voltage at high current into a high voltage at low current. So option 4 is correct.