The following article will guide you about how to determine transformer efficiency.
Meaning of Transformer Efficiency:
The rated capacity of a transformer is defined as the product of rated voltage and full-load (rated) current on the output side. The power output depends upon the power factor of the load.
The efficiency (η) of a transformer, like that of any other apparatus, is defined as the ratio of useful power output to the input power, the two being measured in same units (either in watts or kilowatts).
Now power output = V2 I2 cos ɸ
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Where V2 is the secondary terminal voltage on load, I2 is the secondary current at load and cos ɸ is the power factor of the load.
Iron loss, Pi = Hysteresis loss + eddy current loss
Copper loss = I12 R1 + I22 R2 = I12 R01 = I22 R01
Determination of Transformer Efficiency:
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The ordinary transformer has a very high efficiency (in the range of 96—99%). Hence the transformer efficiency cannot be determined with high precision by direct measurement of output and input, since the losses are of the order of only 1-4 %. The difference between the readings of output and input instruments is then so small that an instrument error as low as 0.5 % would cause an error of the order of 15 % in the losses.
Further, it is inconvenient and costly to have the necessary loading devices of the correct current and voltage ratings and power factor to load the transformer. There is also wastage of large amount of power (equal to that of power output + losses) and no information is available from such a test about the proportion of copper and iron losses.
The best and accurate method of determining of efficiency of a transformer would be to compute losses from open-circuit and short-circuit tests and determine efficiency as follows:
Iron loss, P1 = W0 or P0, determined from open-circuit test
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Copper loss at full load, Pc = Wc or Ps, determined from short-circuit test
Copper loss at a load x times full load = I22 R02 = x2 Pc
Where x is the ratio of load current I2 to secondary full-load secondary current.
In Eq. (10.45) the effect of instrument readings is confined to losses only so that overall efficiency obtained from it is far more accurate than that obtained by direct loading. Another great advantage of this method is that it is not necessary for the transformer to be loaded to its full-load rating during testing, and the kW rating of the test plant need be equal to the value of the individual transformer losses.
Efficiency versus Load:
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It has been pointed out that with constant voltage the mutual flux of the transformer is practically constant from no-load to full load (maximum variation is from 1 to 3 %). The core or iron loss is, therefore, considered constant regardless of load. Copper loss varies as the square of the load current or kVA output. The variation in copper loss with the increase in load current (or kVA) is shown in Fig. 10.25.
The efficiency vs load curve as deduced from these is also shown in the Fig. From the efficiency-load curve shown in Fig. 10.25 it is obvious that the efficiency is very high even at light load, as low as 10 % of rated load. The efficiency is practically constant from about 20% rated load to about 20% overload.
At light loads the efficiency is poor because of constant iron loss whereas at high loads the efficiency falls off due to increase in copper loss as the square of load. From Fig. 10.25 it is also obvious that the transformer efficiency is maximum at the point of intersection of copper loss and iron loss curves i.e. when copper loss equals iron loss.
Condition for Maximum Efficiency:
Where P is equal to the full load (rated output in volt-amperers or kVA) × load power factor (cos ɸ), Pi is the total iron loss, Pc is the full-load copper loss and x is the fraction of full load-kVA at which efficiency is maximum.
Differentiating both sides of Eq. (10.46) we get:
Output current corresponding to maximum efficiency is determined as below:
Copper loss at given load at which efficiency is maximum = I22 R02
And since for maximum efficiency to occur it is necessary that copper loss equals Pi
So I22 R02 = Pi
Power transformers employed for bulk power transmission are operated continuously near about full load and are, therefore, designed to have maximum efficiency at full load. On the other hand, the distribution transformers, which supply load varying over the day through a wide range, are designed to have maximum efficiency at about three-fourths the full load.
It is important to appreciate that, since copper loss depends on current and the iron loss depends on voltage, the total loss in the transformer depends on the volt-ampere product, and not on the phase angle between voltage and current, i.e., is independent of the load power factor. The transformers are, therefore, rated in kilo-volt amperes (KVA) and not in kilowatts.
Efficiency versus Power Factor:
Transformer efficiency is given as:
The variations of efficiency with power factor at different loadings for a typical transformer are illustrated in Fig. 10.26.
All-Day Efficiency (or Energy Efficiency):
The transformer efficiency discussed so far is the ordinary, also called commercial efficiency, which is defined as the ratio of power output to power input.
There are certain types of transformers whose performance cannot be judged by ordinary or commercial efficiency. For instance, distribution transformers are energized for 24 hours, but they deliver very light loads for major portion of the day.
Thus iron or core loss occur for the whole day but the copper loss occurs only when the transformer is loaded. The performance of such a transformer must be judged by its all-day efficiency, also called the energy efficiency or operational efficiency which is computed on the basis of energy consumed during the whole day (24 hours).
The all-day efficiency is defined as the ratio of energy (kWh) output over 24 hours to the energy input over the same period.
Since the distribution transformer does not supply the rated load for the whole day so the all-day efficiency of such a transformer will be less than ordinary or commercial efficiency.
For determination of all-day efficiency of a transformer, it is necessary, of course, to know how the load varies from hour to hour during the day.
Higher energy efficiencies are achieved by designing distribution transformers to yield maximum commercial efficiency at less than full load (usually 50-75 percent of full load). This is achieved by restricting the core flux density to the lower values by using a relatively larger x-section. Thus the ratio of iron loss to copper loss is reduced.
Example:
A 100 kVA transformer has its maximum efficiency of 98% at full load and unity pf. During the day it is loaded as follows:
12 hrs — 20 kW at a pf of 0.5 lag
6 hrs — 50 kW at a pf of 0.9 lag
6 hrs — 75 kW at a pf of 0.8 lag
Calculate the “all day efficiency” of the transformer:
Solution:
For maximum efficiency:
