# Solve Transformer Problems

## Solve and Answer Transformer Problems and Questions

###### Question-1: Transformer all day efficiency

Transformers are on operation 24/7 whether they are partially or heavily loaded for a particular length of time. It is, therefore, for this consideration that a transformer must be evaluated for its 24-hour efficiency which is the ratio of energy output and energy input for a 24 hour period:

**A 500KVA, 1ph, 13.8/4.160kV, 60hz transformer has a primary resistance of R _{1} = 0.8 ohms and secondary resistance of R_{2} = 0.04 ohms. The iron loss is 3,000 watts. Calculate the copper loss and the full load efficiency when the transformer’s daily load is 3hrs @ full load, 5hrs @ 3/4 load and 7hrs @ 1/4 load.**

Solution:

*I _{pri} = Rated kVA x 1000/Rated primary voltage = 500 x 1000/13,800 = 36.2 Amp*

*Copper loss _{pri} = (I_{pri})^{2}x R_{1} = 36.2^{2} x 0.8 = 1,048 watts *

*I _{sec} = Rated kVA x 1000/Rated secondary voltage = 500 x 1000/4,160 = 120.2 Amp*

*Copper loss _{sec} = (I_{sec})^{2} x R_{2} = 120.2^{2} x 0.04 = 578 watts*

*Total full load copper loss = 1,048 + 578 = 1,626 watts*

Since the input is equal to the output + the total losses, the all day efficiency is:

Output x 100/Output + losses; with the copper loses varying directly as the square of the fractional load.

Assuming a 100% power factor, the output energy is:

Energy_{output} = (500 x 3hrs) + (500 x 3/4 x 5hrs) + ( 500 x 1/4 x 7hrs) = 2, 750 kw-hr

Thus, the Energy input = energy output + loses

Energy_{input} = Output + total loses loss

Total copper and iron loses = (1.626kw x 3hrs x 1^{2}) + (1.626kw x 5hrs x 0.75^{2}) + (1.626kw x 7hrs x 0.25^{2}) + (3kw_{iron loss} x 24hrs)] = [4.878 + 6.40 + 0.711 + 72 = 84kw-hr.

All day efficiency = (2,750/2,750 + 84)100% = 97%

Typical efficiency calculation is: (500/500 + 3+ 1.048 + 0.578)100% = 99.3%

###### Question-2: Parallel operation of transformers with different ratings

**Explain how a two single phase transformers of different ratings can be connected in parallel on both the primary and secondary for them to share a common load in proportion to their kva ratings.**

Answer: The following should be strictly observed:

1. Ensure that the transformers’ reactance and resistance ratio are the same.

2. The transformers must have the same transformation ratios; E_{pri}/E_{sec} i.e., for both the primary and secondary voltage.

3. The transformers must have the same %regulation, i.e., the reduction in the voltage level of both transformers in the primary and secondary are the same when they are loaded.

###### Question-3: Utilizing a 3ph transformer for 1ph service

** For a particular service, a single-phase transformer is needed but what’s available is one three-phase transformer. If the available transformer have the same voltage ratio and frequency, (1) how can you configure the three-phase transformer for a single-phase service for the maximum single-phase output, and (2), what is the percentage of the three-phase kva can be utilized?**

Answer: The requirement of this set up must utilize the maximum power from the 3ph transformer to which only 2/3 of the transformer’s KVA rating is possible:

1. Parallel the two windings of the transformer – say Phase-A and Phase-B – taking care that polarity markings are observed; parallel their corresponding secondaries in the same manner.

2. The maximum load that can be utilized for 1ph service is 2/3 x 100% = 66.67%

###### Question-4: Why transformers are rated in volt-amperes or kilovolt-amperes:

Explain why transformers are rated in volt-ampere (va) or kilovolt-ampere (kva)? Justify and prove your answer.

Answer: The heat equivalent of current Idriven into the circuit by the voltage* E *is determined by *Joule’s Law: Q = 0.24EI x time. * Thus it is advantageous and practical to rate a transformer in volt-amperes rather than in watts because the crucial relationship of the primary and secondary volt-ampere are equal when the core and copper losses amounting to about 1 to 3% is neglected. Since the power factor of a transformer is unity or close to unity, the volt-ampere rating provides the engineer with the ease to calculate the output and input expeditiously.

*On the other hand, if transformers are rated in watts or kilowatts, the rating would not be depictive of its actual performance. Consider the power conveyed by a three-phase system: the Power = EI x 3*^{1/2 }*Cosø.
*

*Note: The 3*^{1/2}* is equivalent to square root of 3.*