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 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 the ease to calculate the output and input expeditiously.
On the other hand, if transformers are rated in watts or kilo-watts, 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.