Transformer Efficiency Calculations

In summary, the conversation is about calculating the efficiency of a 500W transformer that reduces 230 AC voltage to 24V. The calculations involve considering the resistances of the primary and secondary windings as well as the losses in the transformer core. The final efficiency is found to be 99.07% by taking into account the total input power and total losses. There is a discussion about the discrepancy between the expected efficiency and the actual efficiency, and the potential impact on the current and losses in the primary winding.
  • #1
MathsDude69
26
0
Hey. I am currently doing a bit of revision for an exam and I am struggling on the following question:

Calculate the efficiency of a 500W transformer, which reduces the 230 AC voltage to 24 V. For the calculations, assume the resistances of the primary and secondary windings as 0.05 Ohm and 0.01 Ohm, respectively. The losses in the transformer core Pc = 70mW.

I understand the whole voltage/current/power/impedance ratio stuff, but can't seem to find any information on calculating efficiencies using the resistances or the transformer core losses. Any Suggestions?
 
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  • #2
Hey. I've found a bit more information regarding copper losses in transformers. It states:

Copper Loss = Ip2Rp + Is2Rs

I would think given the question that the asnwer would be:

P = VI ∴ I = P/V

For the primary:
500/230 = 2.27 Amps

For the secondary:
500/24 = 20.83 Amps

Copper loss for primary:
2.272 x 0.05 = 0.258 W

Copper loss for secondary:
20.832 x 0.01 = 4.34 W

Total Copper Loss:
0.258 + 4.34 = 4.598 W

Total loss including core loss:

4.598 + 0.07 = 4.668 W

Total efficiency =
(500-4.668)/500 = 0.9907 = 99.07% efficient.

The problem with this is if we were only getting an efficiency of 99.07% the secondary wouldn't be developing 20.83 Amps! Does anyone know how to actually solve this question?
 
  • #3
@ Maths:
I skimmed through this pretty fast but I would say that you should not have 500 watts for both the primary and secondary. With any efficiency of less than 100% you will naturally get you a lower secondary current than 20.83.
 
  • #4
You are nearly there except for the last bit.

The 500W refers to the output not the input for transformers so

input power = output power + total losses = 500 + 4.668 = 504.668.

efficiency = 1 - losses/input power = 1 - 4.668/504.668 = 99.07% expressed as a percentage.
 
  • #5
See this is where I am getting confused. If the input power is 504.668W instead of 500W wouldn't that make the current and thus the losses higher on the primary winding??
 
  • #6
More power is lost, but more is put in too. The percentage of loss should be the same.
 

What is the purpose of calculating transformer efficiency?

The purpose of calculating transformer efficiency is to measure how well a transformer is able to convert electrical energy from one circuit to another. It allows engineers to evaluate the performance and effectiveness of a transformer and determine if it is operating at its optimal level.

How is transformer efficiency calculated?

Transformer efficiency is calculated by dividing the output power by the input power and multiplying by 100. The output power is the power delivered to the load, while the input power is the power drawn from the source. The result is expressed as a percentage, with higher percentages indicating a more efficient transformer.

What factors affect transformer efficiency?

There are several factors that can affect transformer efficiency, including the design and construction of the transformer, the quality of the materials used, the condition of the transformer, and the operating temperature. Additionally, using the transformer at higher voltages and lower currents can improve efficiency.

Why is high transformer efficiency important?

High transformer efficiency is important because it results in less energy loss during power transmission and distribution. This not only reduces energy costs, but also helps to minimize environmental impact and promote sustainability. Additionally, efficient transformers are less likely to overheat, which can lead to equipment failures and power outages.

Is there a standard efficiency level for transformers?

Yes, there is a standard efficiency level for transformers established by the U.S. Department of Energy. This standard, known as the Federal Energy Management Program (FEMP) efficiency requirement, sets a minimum efficiency level for transformers used in federal facilities. Additionally, many countries have their own efficiency standards for transformers to promote energy efficiency and conservation.

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