Experimental Set-up of Direct Loading Test on Transformer

In this test, the secondary winding is connected directly across the load for which regulation and efficiency are to be obtained. The experimental setup for this is shown in Fig 1.

Figure 1: Experimental setup for Direct Loading Test on Transformer.

A variac is connected across the single-phase ac supply in order to keep the primary voltage constant to its rated value. Ammeters A₁ and A₂ are used to measure the primary and secondary currents respectively. Two wattmeters W₁ and W₂ are used to measure the active powers on the primary and secondary sides respectively. The voltmeters V₁ and V₂ are connected to measure the primary voltage and load voltage respectively. The load for which regulation and efficiency are to be obtain is connected across the secondary as shown in Fig. 1.

Procedure and Observations for Efficiency Measurement:

Connect the desired load across the secondary as shown in Fig. 1. We obtain the values of efficiency and regulation at this load current. Switch on the single-phase AC supply connected across the variac and adjust the variac to get the rated primary voltage. This is measured by voltmeter V₁. Measure the primary and load active powers by means of wattmeter W₁ and W₂ respectively, and calculate the percent efficiency as follows,

\[\%\eta =\frac{\text{Output power delivered to load}}{\text{Input power to the primary}}\times 100\]

\[\%\eta =\frac{{{W}_{2}}}{{{W}_{1}}}\times 100\]

Note that,

\[{{W}_{1}}={{V}_{1}}{{I}_{1}}\cos {{\phi }_{1}}\] And \[{{W}_{2}}={{V}_{2}}{{I}_{2}}\cos {{\phi }_{2}}\]

Procedure and Observations for the Regulation Measurement:

Do not connect any load. This is the no load condition I₂ = 0 and V₂ = E₂. Switch on the AC power supply and adjust the variac to obtain the rated primary voltage V₁. Measure the voltage V₂. This voltage is called as the no load voltage. V₂ = E₂ under the no load condition.  Switch off the ac supply and connect the desired load. Turn on the power supply again and measure V₂. This voltage is the full load secondary voltage. Calculate the regulation as,

\[\%\text{Regulation = }\frac{{{V}_{2NL}}-{{V}_{2FL}}}{{{V}_{2NL}}}\times 100\]

Also

\[\%\text{Regulation =}\frac{{{E}_{2}}-{{V}_{2}}}{{{E}_{2}}}\times 100\]

Regulation characteristics:

Using this table we can plot the regulation characteristics of the transformer. To do so, we have to change the load from 0 to full load, measure the corresponding values of load current I₂ and load voltage V₂. Then plot the load regulation characteristics of load current (I₂) versus the load voltage V₂.

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1. The losses are zero (No iron loss, no copper loss).
2. The primary and secondary winding resistances are zero.
3. The leakage flux is zero. Therefore all the flux produced by the primary winding is coupled to the secondary.
4. A small current is required to develop flux inside the core. This happens because the permeability of the core is very large.
5. The external voltage applied to the primary, V₁ is same as the primary induced voltage E_l. This is because the primary winding resistance is zero and so there is no voltage drop across it.

\[{{E}_{1}}={{V}_{1}}\]

7. Similarly the voltage induced in the secondary winding (E₂) will be equal to the load voltage V₂, because the secondary resistance is zero.

\[{{E}_{2}}={{V}_{2}}\]

8. The transformation ratio for an ideal transformer is given by

\[k=\frac{{{E}_{2}}}{{{E}_{1}}}=\frac{{{V}_{2}}}{{{V}_{1}}}\]

9. Efficiency of an ideal transformer is 100%. This is because there are no losses taking place.
10. The voltage regulation is That means the secondary voltage will remain constant irrespective of the load current.

Ideal Transformer on No Load of Ideal Transformer

Figure 1: Ideal Transformer.

An ideal transformer is as shown in Fig. 1(a). An ideal transformer is the one which has no power losses. In order to have a transformer with zero loss, the following conditions should be satisfied.

Conditions for an ideal or loss free transformer:

1. The transformer primary & secondary winding resistance should be zero.
2. The hysteresis loss and eddy current loss should be zero.
3. There should not be any leakage flux.

Operation and Phasor Diagram of an Ideal Transformer

An ac voltage V₁ is applied across the primary winding of the transformer (see Figure 1(a)). As the load on the transformer is zero i.e. R_L ideally the primary current I₁ = 0. But practically a small current called the magnetizing current I_m flows through the primary winding. The magnetizing current is used by the transformer for magnetizing the transformer core. As the primary winding is assumed to be purely reactive (R = 0) the magnetizing current lags behind the primary induced voltage by 90º as shown in Fig. 1(b). Due to the sinusoidal magnetizing current, a sinusoidal varying magnetic flux is produced in the iron core. This flux is in phase with I_m as shown in Fig. 1(b). Due to this varying flux, emfs are induced in the primary (self-induced voltage) E_l and secondary (mutually induced voltage) E₂ respectively.

\[{{E}_{1}}=-{{V}_{1}}\text{ and  }{{E}_{2}}={{V}_{2}}\]

The magnitudes of E₁ and E₂ are proportional to the number of tums N₁ and N₂ respectively. The secondary induced voltage E₂ will also oppose V₁. So E₂ also appears 180º out of phase with V₁. The magnitude of E₂ however is dependent on the turns ratio N₂/N₁. E₁ and E₂ appear in phase with each other and 180º out of phase with V₁.

Power input on no load:

The input power to the ideal transformer on no load is given by,

\[{{P}_{0}}={{V}_{1}}{{I}_{m}}\cos {{\phi }_{0}}\]

Where,

\[\begin{align}
& {{\phi }_{0}}=\text{ Angle between }{{\text{V}}_{1}}\text{ and }{{\text{I}}_{m}}\text{ which is 90}{}^\circ \\
& \\
\end{align}\]

Thus the input power to an ideal transformer on no load is zero. The output power is zero and there are no losses taking place in the ideal transformer.

Ideal Transformer on Load

Figure 2: Ideal Transformer on load.

When some load is connected between the secondary terminals of the transformer it is said that the transformer is loaded or it is on load. Due to the load on the secondary, a finite secondary current (I₂) starts flowing. If the load is (R + L) type then I₂ will lag behind V₂ by an angle as shown in Fig. 2(b). As per the Lenz’s law, the secondary current I₂ will oppose the cause producing it. Hence it opposes the magnetic flux. This is called as demagnetizing effect of I₂. Due to demagnetizing, the flux is weakened and it reduces the amount of self-induced voltage E_l. Due to reduction in E_l, the difference between V₁ and E_l will increase and the additional primary current I’₂ called as load component starts flowing as shown in Fig. 2(b).

\[{{{{I}’}}_{2}}={{I}_{2}}\times \frac{{{N}_{2}}}{{{N}_{1}}}=K{{I}_{2}}\]

The current I’₂ = KI₂ and it is 180º out of phase with the current I₂. The net primary current I₁ is the phasor sum of I’₂ and I_m as shown in Fig. 2(b).

\[{{\overline{I}_{2}}}={{{\overline{{I}’}}_{m}}}+{{I}_{m}}\]

Thus due to load on secondary side, the primary current of the transformer increases to supply the additional power to the load. The angle between V₁ and I₁ is ϕ₁ as shown in Fig. 2(b).

Why does primary current increase when the load current is increased ?

When the transformer is loaded, the load current I₂ will start flowing. Due to increase in load current I₂ the secondary ampere tums N₂I₂ will increase. This increased secondary mmf (N₂I₂) will increase the flux set up by the secondary current. This flux opposes the main flux ϕ₁ set up in the core by the current flowing through the primary winding. Hence the secondary mmf N₂I₂ is called as the demagnetizing ampere. Due to reduction in the main flux ϕ₁, the induced emf in the primary winding E₁ will also reduce. Hence the difference between V₁ and E_l will increase and the primary current will Increase and the primary power factor is ϕ₁.

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Figure1: Losses in a Transformer.

As shown in Fig. 1, the total loss in a transformer can be divided into two types namely the copper loss and the iron loss. The iron loss is further classified into two types namely the hysteresis loss and eddy current loss.

Copper Loss (P_cu) in a Transformer

The total power loss taking place in the windings resistance of a transformers is known as the copper loss.

Copper loss = Primary copper loss + secondary copper loss

The copper loss is denoted by P_cu.

\[{{P}_{cu}}=I_{1}^{2}{{R}_{1}}+I_{2}^{2}{{R}_{2}}\]

Where R₁ and R₂ are resistances of primary and secondary respectively.

Where,

\[I_{1}^{2}{{R}_{1}}\text{ = Primary copper loss}\]

and

\[I_{2}^{2}{{R}_{2}}\text{ = Secondary copper loss}\]

The copper loss should be kept as low as possible to increase the efficiency of the transformer. To reduce the copper loss, it is essential to reduce the resistances R₁ and R₂ of the primary and secondary windings. Copper losses are also called as variable losses as they are dependent on the square of load current. The relation between copper loss at full load and that at half load is as follows:

\[{{P}_{cu(HL)}}={{\left( \frac{1}{2} \right)}^{2}}{{P}_{cu(FL)}}=\frac{{{P}_{cu(FL)}}}{4}\]

Where,

\[{{\text{P}}_{cu(FL)}}=\text{ Copper loss at full load and}\]

\[{{\text{P}}_{cu(HL)}}\text{ = Copper loss at half load}\]

Iron Loss (P_i) in a Transformer

Iron loss P_i is the power loss taking place in the iron core of the transformer. It is equal to the sum of two components called hysteresis loss and eddy current loss.

P_i = Hysteresis Loss + Eddy current loss

Hysteresis Losses in a Transformer

The hysteresis loss taking place in a magnetic material. The area enclosed by the hysteresis loop of a material represents the hysteresis loss. Hence special magnetic materials should be used in order to reduce the hysteresis loss. Materials such as silicon steel has hysteresis loops with very small area. Hence such materials are preferred for the construction of core. Commercially such steel is called as Lohys, means low hysteresis materials. Mathematically the hysteresis loss is given by:

\[\text{Hysteresis loss = }{{K}_{H}}.B_{m}^{1.67}fV\text{ watt}\]

Where:

K_H – Hysteresis constant,

B_m – Maximum flux density

f – Frequency and

V – Volume of the core

Thus the hysteresis loss is frequency dependent As we increase the frequency of operation, the hysteresis loss increase proportionally.

Eddy current losses in a Transformer

Due to the time varying flux, there is some induced emf in the transformer core. This induced emf causes some currents to flow though the core body. These currents are known as the eddy currents. The core is made of steel and has some finite resistance. Hence due to the flow of eddy currents, heat will be produced. The power loss due to the eddy currents is given by :

Eddy current loss = (Eddy current)² × r

Where,

r – Resistance of the core.

The eddy current losses are minimized by using the laminated core. The core is manufactured as a stack of laminations rather than a solid iron core. These laminations are insulated from each other by means of a varnish coating on all the laminations. Hence each lamination acts as a separate core with a small cross-sectional area, providing a large resistance to the flow of eddy currents. Mathematically the eddy current loss is given by,

\[\text{Eddy current loss = }{{K}_{E}}B_{m}^{2}{{f}^{2}}{{T}^{2}}\]

K_E – Eddy current constant,

T – core

Thus the eddy current loss also is frequency dependent. It is directly proportional to the square of operating frequency. It can be reduced by using the laminated core for transformer.

Hence the iron loss P_i of the total loss is dependent on the frequency but the copper loss P_cu is constant inspective of frequency. The iron loss is denoted by P_i. It is the sum of hysteresis and eddy current loss. Iron loss is a constant loss and does not depend on the level of load.

Total Loss in a Transformer

The total power loss taking place in a transformer can be obtained by adding the copper loss and iron losses together.

Total loss = Copper loss + Iron loss

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