Solve Transformer Problems
Solve and Answer Transformer Problems and Questions
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 R1 = 0.8 ohms and secondary resistance of R2 = 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.
Ipri = Rated kVA x 1000/Rated primary voltage = 500 x 1000/13,800 = 36.2 Amp
Copper losspri = (Ipri)2x R1 = 36.22 x 0.8 = 1,048 watts
Isec = Rated kVA x 1000/Rated secondary voltage = 500 x 1000/4,160 = 120.2 Amp
Copper losssec = (Isec)2 x R2 = 120.22 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:
Energyoutput = (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
Energyinput = Output + total loses loss
Total copper and iron loses = (1.626kw x 3hrs x 12) + (1.626kw x 5hrs x 0.752) + (1.626kw x 7hrs x 0.252) + (3kwiron 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%
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; Epri/Esec 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.
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%
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 31/2 Cosø.
Note: The 31/2 is equivalent to square root of 3.