Testing, Efficiency & Regulation

 

25.1 Goals of the lesson

 

In the previous lesson we have seen how to draw equivalent circuit showing magnetizing reactance (X_m), resistance (R_cl), representing core loss, equivalent winding resistance (r_e) and equivalent leakage reactance (x_e). The equivalent circuit will be of little help to us unless we know the parameter values. In this lesson we first describe two basic simple laboratory tests namely (i) open circuit test and (ii) short circuit test from which the values of the equivalent circuit parameters can be computed. Once the equivalent circuit is completely known, we can predict the performance of the transformer at various loadings. Efficiency and regulation are two important quantities which are next defined and expressions for them derived and importance highlighted. A number of objective type questions and problems are given at the end of the lesson which when solved will make the understanding of the lesson clearer.

Key Words: O.C. test, S.C test, efficiency, regulation.

After going through this section students will be able to answer the following questions.

 

• Which parameters are obtained from O.C test?

 

• Which parameters are obtained from S.C test?

 

• What percentage of rated voltage is needed to be applied to carry out O.C test?

 

• What percentage of rated voltage is needed to be applied to carry out S.C test?

 

• From which side of a large transformer, would you like to carry out O.C test?

 

• From which side of a large transformer, would you like to carry out S.C test?

 

• How to calculate efficiency of the transformer at a given load and power factor?

 

• Under what condition does the transformer operate at maximum efficiency?

 

• What is regulation and its importance?

 

• How to estimate regulation at a given load and power factor?

 

• What is the difference between efficiency and all day efficiency?

 

25.2 Determination of equivalent circuit parameters

After developing the equivalent circuit representation, it is natural to ask, how to know equivalent circuit the parameter values. Theoretically from the detailed design data it is possible to estimate various parameters shown in the equivalent circuit. In practice, two basic tests namely the open circuit test and the short circuit test are performed to determine the equivalent circuit parameters.

25.2.1 Qualifying parameters with suffixes LV & HV

For a given transformer of rating say, 10 kVA, 200 V / 100 V, 50 Hz, one should not be under the impression that 200 V (HV) side will always be the primary (as because this value appears

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first in order in the voltage specification) and 100 V (LV) side will always be secondary. Thus, for a given transformer either of the HV and LV sides may be used as primary or secondary as decided by the user to suit his/her goals in practice. Usually suffixes 1 and 2 are used for expressing quantities in terms of primary and secondary respectively - there is nothing wrong in it so long one keeps track clearly which side is being used as primary. However, there are situation, such as carrying out O.C & S.C tests (discussed in the next section), where naming parameters with suffixes HV and LV become imperative to avoid mix up or confusion. Thus, it will be useful to qualify the parameter values using the suffixes HV and LV (such as r_{e HV}, r_{e LV} etc. instead of r_e1, r_e2). Therefore, it is recommended to use suffixes as LV, HV instead of 1 and 2 while describing quantities (like voltage V_HV, V_LV and currents I_HV, I_LV) or parameters (resistances r_HV, r_LV and reactances x_HV, x_LV) in such cases.

25.2.2 Open Circuit Test

To carry out open circuit test it is the LV side of the transformer where rated voltage at rated frequency is applied and HV side is left opened as shown in the circuit diagram 25.1. The voltmeter, ammeter and the wattmeter readings are taken and suppose they are V₀, I₀ and W₀ respectively. During this test, rated flux is produced in the core and the current drawn is the no-load current which is quite small about 2 to 5% of the rated current. Therefore low range ammeter and wattmeter current coil should be selected. Strictly speaking the wattmeter will record the core loss as well as the LV winding copper loss. But the winding copper loss is very small compared to the core loss as the flux in the core is rated. In fact this approximation is built-in in the approximate equivalent circuit of the transformer referred to primary side which is LV side in this case.

The approximate equivalent circuit and the corresponding phasor diagrams are shown in figures 25.2 (a) and (b) under no load condition. Open Circuit Test HV side LV side Autotransformer 1-phase A.C supply V A WML S Figure 25.1: Circuit diagram for O.C test V0 (a) Equivalent circuit under O.C testI0 Xm(LV) Rcl(LV) I0 Open circuit V0 I0 Im Icl θ0 θ0 (b) Corresponding phasor diagramFigure 25.2: Equivalent circuit & phasor diagram during O.C test Version 2 EE IIT, Kharagpur

Below we shall show how from the readings of the meters the parallel branch impedance namely R_cl(LV) and X_m(LV) can be calculated.

Calculate no load power factor cos θ₀ = 000WVI

Hence θ₀ is known, calculate sin θ₀

Calculate magnetizing current I_m = I₀ sin θ₀

Calculate core loss component of current I_cl = I₀ cos θ₀

Magnetising branch reactance X_m(LV) = 0mVI

Resistance representing core loss R_cl(LV) = 0clVI

We can also calculate X_m(HV) and R_cl(HV) as follows:

X_m(HV) = ()2mLVXa

R_cl(HV) = ()2clLVRa

Where, a = LVHVNN the turns ratio

If we want to draw the equivalent circuit referred to LV side then R_cl(LV) and X_m(LV) are to be used. On the other hand if we are interested for the equivalent circuit referred to HV side, R_cl(HV) and X_m(HV) are to be used.

25.2.3 Short circuit test

Short circuit test is generally carried out by energizing the HV side with LV side shorted. Voltage applied is such that the rated current flows in the windings. The circuit diagram is shown in the figure 25.3. Here also voltmeter, ammeter and the wattmeter readings are noted corresponding to the rated current of the windings.

Suppose the readings are V_sc, I_sc and W_sc. It should be noted that voltage required to be applied for rated short circuit current is quite small (typically about 5%). Therefore flux level in Short Circuit Test HV side LV side Autotransformer 1-phase A.C supply V A WML S Figure 25.3: Circuit diagram during S.C test

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the core of the transformer will be also very small. Hence core loss is negligibly small compared to the winding copper losses as rated current now flows in the windings. Magnetizing current too, will be pretty small. In other words, under the condition of the experiment, the parallel branch impedance comprising of R_cl(HV) and X_m(LV) can be considered to be absent. The equivalent circuit and the corresponding phasor diagram during circuit test are shown in figures 25.4 (a) and (b).

(a) Equivalent circuit under S.C test

Therefore from the test data series equivalent impedance namely r_e(HV) and x_e(HV) can easily be computed as follows:

Equivalent resistance ref. to HV side r_e(HV) = 2scscWI

Equivalent impedance ref. to HV side z_e(HV) = scscVI

Equivalent leakage reactance ref. to HV side x_e(HV) = ()()22eHVeHVz-r

We can also calculate r_e(LV) and x_e(LV) as follows:

r_e(LV) = a²r_e(HV)

x_e(LV) = a²x_e(HV)

where, a = LVHVNN the turns ratio

Once again, remember if you are drawing equivalent circuit referred to LV side, use parameter values with suffixes LV, while for equivalent circuit referred to HV side parameter values with suffixes HV must be used.

25.3 Efficiency of transformer

In a practical transformer we have seen mainly two types of major losses namely core and copper losses occur. These losses are wasted as heat and temperature of the transformer rises. Therefore output power of the transformer will be always less than the input power drawn by the primary from the source and efficiency is defined as VSC

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η = Output power in KWOutput power in Kw + Losses

= ()Output power in KW 25.1Output power in Kw + Core loss+Copper loss

We have seen that from no load to the full load condition the core loss, P_core remains practically constant since the level of flux remains practically same. On the other hand we know that the winding currents depend upon the degree of loading and copper loss directly depends upon the square of the current and not a constant from no load to full load condition. We shall write a general expression for efficiency for the transformer operating at x per unit loading and delivering power to a known power factor load. Let,

KVA rating of the transformer be = S

Per unit degree of loading be = x

Transformer is delivering = x S KVA

Power factor of the load be = cos θ

Output power in KW = xS cos θ

Let copper loss at full load (i.e., x = 1) = P_cu

Therefore copper loss at x per unit loading = x² P_cu

Constant core loss = P_core (25.2)

(25.3)

Therefore efficiency of the transformer for general loading will become:

2corecuxS cos θη=xS cos θ + P+xP

If the power factor of the load (i.e., cos θ) is kept constant and degree of loading of the transformer is varied we get the efficiency Vs degree of loading curve as shown in the figure 25.5. For a given load power factor, transformer can operate at maximum efficiency at some unique value of loading i.e., x. To find out the condition for maximum efficiency, the above equation for η can be differentiated with respect to x and the result is set to 0. Alternatively, the right hand side of the above equation can be simplified to, by dividing the numerator and the denominator by x. the expression for η then becomes:

corePcuxS cos θη=S cos θ++x P

For efficiency to be maximum,ddx(Denominator) is set to zero and we get,

or corecuPdS cos θ++x Pdxx⎛⎞⎜⎟⎝⎠ = 0

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or 2corecuP+Px− = 0

or x² P_cu = P_core

The loading for maximum efficiency, x = corecuPP

Thus we see that for a given power factor, transformer will operate at maximum efficiency when it is loaded to corecuPPS KVA. For transformers intended to be used continuously at its rated KVA, should be designed such that maximum efficiency occurs at x = 1. Power transformers fall under this category. However for transformers whose load widely varies over time, it is not desirable to have maximum efficiency at x = 1. Distribution transformers fall under this category and the typical value of x for maximum efficiency for such transformers may between 0.75 to 0.8. Figure 25.5 show a family of efficiency Vs. degree of loading curves with power factor as parameter. It can be seen that for any given power factor, maximum efficiency occurs at a loading of x =corecuPP. Efficiencies η_max1, η_max2 and η_max3 are respectively the maximum efficiencies corresponding to power factors of unity, 0.8 and 0.7 respectively. It can easily be shown that for a given load (i.e., fixed x), if power factor is allowed to vary then maximum efficiency occurs at unity power factor. Combining the above observations we can say that the efficiency is obtained when the loading of the transformer is x = corecuPP and load power factor is unity. Transformer being a static device its efficiency is very high of the order of 98% to even 99%.

25.3.1 All day efficiency

In the earlier section we have seen that the efficiency of the transformer is dependent upon the degree of loading and the load power factor. The knowledge of this efficiency is useful provided the load and its power factor remains constant throughout.

For example take the case of a distribution transformer. The transformers which are used to supply LT consumers (residential, office complex etc.) are called distribution transformers. For obvious reasons, the load on such transformers vary widely over a day or 24 hours. Some times the transformer may be practically under no load condition (say at mid night) or may be over loaded during peak evening hours. Therefore it is not fare to judge efficiency of the Power factor = 1Power factor = 0.8Power factor = 0.7Degree of loading x corecuPx=PEfficiency ηmax 1 ηmax 2 ηmax 3 Figure 25.5: Efficiency VS degree of loading curves.

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transformer calculated at a particular load with a fixed power factor. All day efficiency, alternatively called energy efficiency is calculated for such transformers to judge how efficient are they. To estimate the efficiency the whole day (24 hours) is broken up into several time blocks over which the load remains constant. The idea is to calculate total amount of energy output in KWH and total amount of energy input in KWH over a complete day and then take the ratio of these two to get the energy efficiency or all day efficiency of the transformer. Energy or All day efficiency of a transformer is defined as:

η_{all day} = Energy output in KWH in 24 hoursEnergy input in KWH in 24 hours

= Energy output in KWH in 24 hoursOutput in KWH in 24 hours + Energy loss in 24 hours

= Output in KWH in 24 hoursOutput in KWH in 24 hours + Loss incore in 24 hours+Loss in the Winding in 24 hours

= coreEnergy output in KWH in 24 hoursEnergy output in KWH in 24 hours +24 +Energy loss (cu) in the winding in 24 hoursP

With primary energized all the time, constant P_core loss will always be present no matter what is the degree of loading. However copper loss will have different values for different time blocks as it depends upon the degree of loadings. As pointed out earlier, if P_cu is the full load copper loss corresponding to x = 1, copper loss at any arbitrary loading x will be x² P_cu. It is better to make the following table and then calculate η_{all day}.

Time blocks KVA Loading Degree of loading x P.F of load KWH output KWH cu loss
T1 hours	S1	x1 = S1/S	cos θ1	S1 cos θ1T1	21xPcu T1
T2 hours	S2	x2 = S2/S	cos θ2	S2 cos θ2T2	22xPcu T2
…	…	…	…	…	…
Tn hours	Sn	xn = Sn/S	cos θn	Sn cos θnTn	2nxPcu Tn