Universal Motors

Generally, the electric motors are opearted either in DC Power or AC Power. But for some specific applications , it is desirable to introduce a motor that operates on either ac or dc supply. The word ‘Universal’ signifies that something which is compaible with versatile inputs. We have build small series motors upto ½ KW rating which operates on single phase ac supply as well as on dc . Such motors are called universal motors . An universal motor is a specifically designed series wound motor, that operates at approximately the same speed and output on either ac or dc voltage. In case of universal motor, the speed of rotation is slightly lesser when opearting in AC. Becuase, the reactance voltage drop is present on ac but not on dc. So, the motor speed is somewhat lower for same load in ac operation than dc . This takes place especially at high loads . Most universal motors are designed to operate at speeds exceeding 3500 rpm

Construction Of Universal Motor

Basically , an universal motor is similar to a dc series motor in construction. However it is constructed with few series field turns, laminated armature and field circuits, low reluctance magnetic path, increased armature conductors and commutator segments. The frames of universal motor are usually made up of alumunium , rolled steel and cast iron. The commutation on ac is much poorer than on dc , due to current induced in the short circuited armature coils .If we use wide brushes then the short circuited current is excessive and motor starting torque is reduced . Brushes used are high resistance carbon ones so as to aid commutation . Compensating windings which are used in most large rating motors to improve commutation are not used in universal motors . In this motor armature current is quite small which cannot cause any commutation problems .

Working principle of universal motor

Now let us discuss the operation of this kind of motor in brief . In series circuit, same amount of current flows throgh all components. Similarly in a series wound motor, the same current flows through field windings and armature winding both. In an universal motor, both windings connected in series with each other. When the motor is supplied from ac or dc supply, magnetic fields are developed around the armature winding and field winding. They  reacts on each other to produce an unidirectional torque forcefully. In some other words, the interaction in between series magnetic field and armature field causes to develop a torque and this torque leads to rotate the shaft. However a series motor which is specifically designed for dc operation suffers from following drawbacks when it is used on single phase ac supply :-

• Its efficiency is low due to hysteresis and eddy current losses .
• The p.f is low due to large reactance of the field and armature windings .
• The sparking at the brushes is excessive.

Characteristics Of Universal Motor

The torque – speed ( τ – N) characteristics of an universal motor is quite similar to that of a series wound dc motor . It has high starting torque at low speed and low starting torque at high speed. In small series motors losses are large at no-load to restrict the speed to a definite value. A centrifugal switch is placed on the motor shaft. The tension of the springs of the switch is adjusted so that the switch opens at a pre determined speed . An external resistor R is placed in series with armature to reduce the speed. When speed drops due to increase of load ,the switch contacts close thereby shorting the resistor R thus raising the speed . Universal motors are high speed , small size motors as compared to other motors of same output . Here the full load power factor is high (0.9).

Speed control of universal motor

Speed control of universal motor is best obtained by solid-state devices. Since the speed of these motors is not limited by the supply frequency and may be as high as 20,000 rpm , they are most suitable for applications requiring high speeds . The factors that determine the speed for any dc motor are the same as those for ac series or universal motors i.e flux and generated voltage . Generated voltage change is rarely employed in speed control method. Instead line voltage is varied .This has been accomplished by means of tapped resistor , rheostat in series with the line.

Another method is by using a tapped field , thereby reducing the flux and hence raising the speed . This can be achieved by any one of the methods that follow :

• By using field poles wound in various sections with wires of different size and bringing out the taps from each section .

• By using tapped nichrome wires coils wound over a single field pole. In this method torque decreases with increase in speed .

Applications Of Universal Motor

Universal motors finds its applications in various devices . These are :

• The very small power output rating universal motors, which usually does not exceed 5 to10 watts are employed in equipments such as sewing machines , fans , portable hand tools, hair dryers , motion picture projectors and electric shavers.

• The higher rating (5-500 W) universal motor are used in vacuum cleaners, food mixers, blenders, cameras and calculating machines .

• This type of machine is used in table fans , polishers , portable drills and other kitchen appliances .a

Open and Short Circuit Test on Transformer

Open Circuit Test:-The open-circuit test, or "no-load test", is one of the methods used in electrical engineering to determine the no-load impedance in the excitation branch of a transformer.


          

The secondary of the transformer is left open-circuited. A wattmeter is connected to the primary. An ammeter is connected in series with the primary winding. A voltmeter is optional since the applied voltage is the same as the voltmeter reading. Rated voltage is applied at primary.

If the applied voltage is normal voltage then normal flux will be set up. Since iron loss is a function of applied voltage, normal iron loss will occur. Hence the iron loss is maximum at rated voltage. This maximum iron loss is measured using the wattmeter. Since the impedance of the series winding of the transformer is very small compared to that of the excitation branch, all of the input voltage is dropped across the excitation branch. Thus the wattmeter measures only the iron loss. This test only measures the combined iron losses consisting of the hysteresis loss and the eddy current loss. Although the hysteresis loss is less than the eddy current loss, it is not negligible. The two losses can be separated by driving the transformer from a variable frequency source since the hysteresis loss varies linearly with supply frequency and the eddy current loss varies with the square.

Since the secondary of the transformer is open, the primary draws only no-load current, which will have some copper loss. This no-load current is very small and because the copper loss in the primary is proportional to the square of this current, it is negligible. There is no copper loss in the secondary because there is no secondary current.

The current  is very small.

If  is the wattmeter reading then,

That equation can be rewritten as,

Thus,

Impedance[edit]

By using the above equations,  and  can be calculated as,

Thus,

or

Admittance[edit]

The admittance is the inverse of impedance. Therefore,

The conductance  can be calculated as,

Hence the susceptance,

or

Here,

 is the wattmeter reading

 is the applied rated voltage

 is the no-load current

 is the magnetizing component of no-load current

 is the core loss component of no-load current

 is the exciting impedance

 is the exciting admittance

Short Circuit Test on Transformer

The connection diagram for short circuit test on transformer is shown in the figure. A voltmeter, wattmeter, and an ammeter are connected in HV side of the transformer as shown. The voltage at rated frequency is applied to that HV side with the help of a variac of variable ratio auto transformer.

The LV side of the transformer is short circuited . Now with help of variac applied voltage is slowly increase until the ammeter gives reading equal to the rated current of the HV side. After reaching at rated current of HV side, all three instruments reading (Voltmeter, Ammeter and Watt-meter readings) are recorded. The ammeter reading gives the primary equivalent of full load current IL. As the voltage, applied for full load current in short circuit test on transformer, is quite small compared to rated primary voltage of the transformer, the core losses in transformer can be taken as negligible here.

Let’s, voltmeter reading is Vsc. The input power during test is indicated by watt-meter reading. As the transformer is short circuited, there is no output hence the input power here consists of copper losses in transformer. Since, the applied voltage Vsc is short circuit voltage in the transformer and hence it is quite small compared to rated voltage so core loss due to the small applied voltage can be neglected. Hence the wattmeter reading can be taken as equal to copper losses in transformer. Let us consider wattmeter reading is Psc.

Where Re is equivalent resistance of transformer.

If, Ze is equivalent impedance of transformer.


Therefore, if equivalent reactance of transformer is Xe

Operation of Transformers

Why Parallel Operation of Transformers is required ?

It is economical to installed numbers of smaller rated transformers in parallel than installing a bigger rated electrical power transformer. This has mainly the following advantages,

1) To maximize electrical power system efficiency: Generally electrical power transformer gives the maximum efficiency at full load. If we run numbers of transformers in parallel, we can switch on only those transformers which will give the total demand by running nearer to its full load rating for that time. When load increases we can switch no one by one other transformer connected in parallel to fulfill the total demand. In this way we can run the system with maximum efficiency.

2) To maximize electrical power system availability: If numbers of transformers run in parallel we can take shutdown any one of them for maintenance purpose. Other parallel transformers in system will serve the load without total interruption of power.

Condition for Parallel Operation of Transformer:

• For parallel connection of transformers, primary windings of the Transformers are connected to source bus-bars and secondary windings are connected to the load bus-bars.
• Various conditions that must be fulfilled for the successful parallel operation of transformers:

1.  Same voltage Ratio & Turns Ratio (both primary and secondary Voltage Rating is same).
2. Same Percentage Impedance and X/R ratio.
3. Identical Position of Tap changer.
4. Same KVA ratings.
5. Same Phase angle shift (vector group are same).
6. Same Frequency rating.

• Some of these conditions are convenient and some are mandatory.
• The convenient are: Same voltage Ratio & Turns Ratio, Same Percentage Impedance, Same KVA Rating, Same Position of Tap changer.
• The mandatory conditions are: Same Phase Angle Shift, Same Polarity, Same Phase Sequence and Same Frequency.
• When the convenient conditions are not met paralleled operation is possible but not optimal.

Losses in Transformer

As the electrical transformer is a static device, mechanical loss in transformer normally does not come into picture. We generally consider only electrical losses in transformer. Loss in any machine is broadly

defined as difference between input power and output power.
When input power is supplied to the primary of transformer, some portion of that power is used to compensate core losses in transformer i.e. Hysteresis loss in transformer and Eddy Current loss in transformer core and some portion of the input power is lost as I2R loss and dissipated as heat in the primary and secondary winding, as because these windings have some internal resistance in them. The first one is called core loss or iron loss in transformer and later is known as ohmic loss or copper loss in transformer. Another loss occurs in transformer, known as Stray Loss, due to Stray fluxes link with the mechanical structure and winding conductors.

Copper Loss in Transformer

Copper loss is I2R loss, in primary side it is I12R1 and in secondary side it is I22R2loss, where I1 & I2 are primary & secondary current of transformer and R1 & R2are resistances of primary & secondary winding. As the both primary & secondary currents depend upon load of transformer, so copper loss in transformer vary with load.

Core Losses in Transformer

Hysteresis loss and eddy current loss, both depend upon magnetic properties of the materials used to construct the core of transformer and its design. So these losses in transformer are fixed and do not depend upon the load current. So core losses in transformer which is alternatively known as iron loss in transformer and can be considered as constant for all range of load.

Hysteresis loss in transformer is denoted as,

Eddy Current loss in transformer is denoted as,

Where, Kh = Hysteresis Constant.

Ke = Eddy Current Constant.

Kf = form Constant.

Copper loss can simply be denoted as,

IL2R2′ + Stray loss

Where, IL = I2 = load of transformer, and R2′ is the resistance of transformer referred to secondary.

Now we will discuss Hysteresis loss and Eddy Current loss in little bit more details for better understanding the topic of losses in transformer

Hysteresis Loss in Transformer

Hysteresis loss in transformer can be explained in different ways. We will discuss two of them, one is physical explanation other is mathematical explanation.

Physical Explanation of Hysteresis Loss

The magnetic core of transformer is made of ′Cold Rolled Grain Oriented Silicon Steel′. Steel is very good ferromagnetic material. This kind of materials are very sensitive to be magnetized. That means whenever magnetic flux passes through,it will behave like magnet. Ferromagnetic substances have numbers of domains in their structure. Domain are very small region in the material structure, where all the dipoles are paralleled to same direction. In other words, the domains are like small permanent magnet situated randomly in the structure of substance. These domains are arranged inside the material structure in such a random manner, that net resultant magnetic field of the said material is zero. Whenever external magnetic field or mmf is is applied to that substance, these randomly directed domains are arranged themselves in parallel to the axis of applied mmf. After removing this external mmf, maximum numbers of domains again come to random positions, but some few of them still remain in their changed position. Because of these unchanged domains the substance becomes slightly magnetized permanently. This magnetism is called " Spontaneous Magnetism". To neutralize this magnetism some opposite mmf is required to be applied. The magneto motive force or mmf applied in the transformer core is alternating. For every cycle, due to this domain reversal there will be extra work done. For this reason, there will be a consumption of electrical energy which is known as Hysteresis loss of transformer.

Mathematical Explanation of Hysteresis Loss in Transformer

Determination of Hysteresis Loss

Consider a ring of ferromagnetic specimen of circumference L meter, cross - sectional area a m2 and N turns of insulated wire as shown in the picture beside,

Let us consider, the electric current flowing through the coil is I amp,

Magnetizing force,

Let, the flux density at this instant is B,

Therefore, total flux through the ring, Φ = BXa   Wb

As the electric current flowing through the solenoid is alternating, the flux produced in the iron ring is also alternating in nature, so the emf (e′) induced will be expressed as,

According to Lenz,s law this induced emf will oppose the flow of electric current, therefore, in order to maintain the current I in the coil, the source must supply an equal and opposite emf. Hence applied emf ,

Energy consumed in short time dt, during which the flux density has changed,

Thus, total work done or energy consumed during one complete cycle of magnetism,

Now aL is the volume of the ring and H.dB is the area of elementary strip of B - H curve shown in the figure above,

= total area enclosed by Hysteresis Loop.

Therefore, Energy consumed per cycle = volume of the ring X area of hysteresis loop.

In the case of transformer, this ring can be considered as magnetic core of transformer. Hence this work done is nothing but electrical energy loss in transformer core and this is known as hysteresis loss in transformer.

What is Eddy Current Loss ?

In transformer we supply alternating current in the primary, this alternating current produces alternating magnetizing flux in the core and as this flux links with secondary winding there will be induced voltage in secondary, resulting current to flow through the load connected with it. Some of the alternating fluxes of transformer may also link with other conducting parts like steel core or iron body of transformer etc. As alternating flux links with these parts of transformer, there would be an locally induced emf. Due to these emfs there would be currents which will circulate locally at that parts of the transformer. These circulating current will not contribute in output of the transformer and dissipated as heat. This type of energy loss is called eddy current loss of transformer. This was a broad and simple explanation of eddy current loss. The detail explanation of this loss is not in the scope of discussion in that chapter.

Theory of Transformer

Theory of Transformer On No-load, and Having No Winding Resistance and No Leakage Reactance of Transformer

Let us consider one electrical transformer with only core losses. That means the it has only core losses but no copper lose and no leakage reactance of transformer. When an alternating source is applied in the primary, the source will supply the electric current for magnetizing the core of transformer. But this current is not the actual magnetizing current, little bit greater than actual magnetizing current. Actually total electric current supplied from the source has two components one is magnetizing current which is merely utilized for magnetizing the core and other component of the source current, is consumed for compensating the core losses in transformer. Because of this core loss component, the source current in transformer on no-load condition, supplied from the source as source current is not exactly at 90o lags of supply voltage but it lags behind an angle θ is less than 90o.

If total electric current supplied from source is Io, it will have one component in phase with supply voltage V1 and this component of the current Iw is core loss component. This component is taken in phase with source voltage, because it is associated with active or working losses in transformer. Other component of the source current is denoted as Iμ.

This component produces the alternating magnetic flux in the core, so it is watt-less means it is reactive part of the transformer source current.

Hence Iμ will be in quadrature with V1 and in phase with alternating flux Φ.

Hence, total primary current in transformer on no-load condition can be represented as

Now you have seen how simple to explain the theory of transformer in no-load.

Theory of Transformer On Load But Having No Winding Resistance and Leakage Reactance

Now we will examine the behavior of above said transformer on load, that means load is connected to the secondary terminals. Consider, transformer having core loss but no copper loss and leakage reactance. Whenever load is connected to the secondary winding, load current will start to flow through the load as well as secondary winding. This load current solely depends upon the characteristics of the load and also upon secondary voltage of the transformer. This current is called secondary current or load current, here it is denoted as I2. As I2 is flowing through the secondary, a self mmf in secondary winding will be produced. Here it is N2I2, where, N2 is the number of turns of the secondary winding of transformer.

This mmf or magneto motive force in the secondary winding produces flux φ2. This φ2 will oppose the main magnetizing flux and momentarily weakens the main flux and tries to reduce primary self induced emf E1. If E1 falls down bellow the primary source voltage V1, there will be an extra current flows from source to primary winding. This extra primary current I2′ produces extra flux φ′ in the core which will neutralized the secondary counter flux φ2. Hence the main magnetizing flux of core, Φ remain unchanged irrespective of load.

So total current, this transformer draws from source can be divided into two components, first one is utilized for magnetizing the core and compensate the core loss i.e. Io. It is, no-load component of the primary current. Second one is utilized for compensating the counter flux of the secondary winding. It is known as load component of the primary current.
Hence total no load primary current I1 of a transformer having no winding resistance and leakage reactance can be represented as follows

Where θ2 is the angle between Secondary Voltage and Secondary Current of transformer.
Now we will proceed one further step toward more practical aspect of a transformer.

Theory of Transformer On Load, With Resistive Winding, But No Leakage Reactance

Now, consider the winding resistance of transformer but no leakage reactance. So far we have discussed about the transformer which has ideal windings means winding with no resistance and leakage reactance, but now we will consider one transformer which has internal resistance in the winding but no leakage reactance. As the windings are resistive, there would be a voltage drop in the windings.

We have proved earlier that total primary current from the source on load is I1. The voltage drop in the primary winding with resistance, R1 is R1I1. Obviously induced emf across primary winding E1, is not exactly equal to source voltage V1. E1 is less than V1 by voltage drop I1R1.

Again in the case of secondary, the voltage induced across the secondary winding, E2 does not totally appear across the load since it also drops by an amount I2R2, where R2 is the secondary winding resistance and I2 is secondary current or load current.

Similarly voltage equation of the secondary side of the transformer will be

Theory of Transformer On Load, With Resistance As Well As Leakage Reactance in Transformer Windings

Now we will consider the condition , when there is leakage reactance of transformer as well as winding resistance of transformer.

Let leakage reactances of primary and secondary windings of the transformer are X1 and X2 respectively.

Hence total impedance of primary and secondary winding with resistance R1 and R2 respectively, can be represented as,

We have already established the voltage equation of a transformer on load, with only resistances in the windings, where voltage drops in the windings occur only due to resistive voltage drop. But when we consider leakage reactances of transformer windings, voltage drop occurs in the winding not only because of resistance, it is because of impedance of transformer windings. Hence, actual voltage equation of a transformer can easily be determined by just replacing resistances R1 & R2 in the previously established voltage equations by Z1 and Z2.

Therefore, the voltage equations are,
 & 

Resistance drops are in the direction of current vector but reactive drop will be in perpendicular to the current vector as shown in the abovevector diagram of transformer.