Voltage Regulation Of Synchronous Machines [Alternator] By Direct Loading Test

Voltage Regulation Of Synchronous Machines [Alternator] By Direct Loading Test

Today in this post we are going to learn what is Voltage regulation of synchronous machine and different methods to calculate Voltage regulation of synchronous machine.

Definition for Voltage regulation of a synchronous machine:

Voltage Regulation of synchronous machine is defined as the difference between terminal voltage at no load and terminal voltage at full load and excitation , speed must remain same.Voltage Regulation of synchronous machine is generally calculated in percentage of full load terminal voltage.

Objectives for calculating Voltage regulation of a synchronous machine:

1. Parallel operation of alternators is affected by the voltage regulation. By calculating voltage regulation of synchronous machine we can adjust the parallel operating machines to be in synchronism.

2. Calculating voltage regulation of a synchronous machine determines the type of automatic voltage control equipment required for resisting the voltage changes.

3.When the load is thrown off voltage rise must be known because with the rise in voltage the insulation must be able to withstand this rise.

So calculation of voltage regulation of synchronous machine has a great importance.

General expression for calculating Voltage regulation of synchronous machine:

Now let us derive general expression for calculating voltage regulation of a synchronous machine

Let E be the terminal voltage of the synchronous machine at no load. Now if the synchronous machine is given full load the terminal voltage will no longer be E because of the losses so let the terminal voltage now be V.

So general expression for Voltage regulation of a synchronous machine is given by

Voltage regulation% = (E - V / V) × 100

Methods for calculating voltage regulation of synchronous machine:

There are two types of methods for calculating voltage regulation of synchronous machine.

1. Direct load test method.

2. Indirect Method.

Indirect method of calculating voltage regulation of synchronous machine can be further classified into 3 types:



3.Zero power factor method or potier method.

Direct load test method for calculating voltage regulation of synchronous machine:

Now let's see how to calculate voltage regulation of synchronous machine by using direct load test method:

Circuit diagram for calculating Voltage regulation of synchronous machine by direct load test:

Circuit connections for calculating voltage regulation of synchronous machine by direct load test:

1.Firstly connections are to be made as given in the circuit diagram:

2. Armature which is star connected is connected to the three phase load with the help of TPST. TPST is a switch and it means triple pole single through. 

3. A rheostat is connected in series with the field winding. 

4. Field winding is excited by using D.C supply and flux is adjusted by adjusting the rheostat. Flux adjustment is nothing but adjust the current flow through field winding.

Procedure for calculating voltage regulation of synchronous machine by direct load test:

 1. Adjust the prime mover such that the alternator rotates at synchronous speed Ns.

 we know Eph α 𝞍 from emf equation

2. Now DC supply is given to the field winding and the current flow through field is adjusted so that the flux is adjusted such that the rated voltage is obtained at its terminals which can be seen on the voltmeter connected across the lines.

3. Now load is connected to alternator with the help of TPST switch.

4.The load is then increased such that the ammeter reads rated current. This is full load condition of alternator. Now as load is connected due to armature reaction there is loss in voltage so let the induced voltage be V. 

5.Now again adjust the rheostat of the field winding to get rated voltage at alternator terminals.

6.Now remove the load by opening TPST switch and the excitation , speed should not be changed it should be same as before removing the load.

7. As there is no load there is no armature reaction the induced emf is equal to terminal voltage which is E.  

Now we can calculate voltage regulation of synchronous machine by 

Voltage regulation% =( E - V / V) × 100 at a specific power factor.

Limitations for calculating voltage regulation of synchronous machine by using direct load method:

This method is applicable only for small capacity machines for larger capacity machines it is not economical because that much load cannot be given directly.

In this way we have calculated the voltage regulation of synchronous machine by direct load test method.

For larger capacity machines voltage regulation can be calculated by Indirect method.

In the next post we can see how to calculate voltage regulation of synchronous machine by Indirect method.

You can download PDF form of voltage regulation of synchronous machines here

Voltage Regulation of Transformer

Voltage Regulation Of Transformer

Hello everyone,
In this post we are going to discuss about Voltage regulation of a transformer.

Read here : Differences Between Core And Shell Type Transformers

What is meant my Voltage regulation of transformer?
Voltage regulation of a transformer may be defined as the difference between no load voltage of the secondary terminal of a transformer and full load voltage of the secondary terminal of that transformer at a certain power factor. Voltage regulation of a transformer is expressed in percentage of either no load secondary terminal voltage or full load secondary terminal voltage.

Read here : EMF Equation of Transformer & Voltage Transformation Ratio

Objective to calculate voltage regulation of transformer:

Calculating voltage regulation of transformer gives how much efficiently the transformer is resisting the voltage changes from no load to full load. If there is no change in value of secondary voltage from no load to full load then the transformer is ideal and has voltage regulation 0%. So the lower the value of voltage regulation the higher is the performance of the transformer. 

 Procedure to calculate voltage regulation of transformer:

Consider a transformer which is at no load which means the secondary of the transformer is open circuited. In this case the secondary voltage of the transformer and induced emf are same let it be E2 . Now full load is connected to the secondary of a transformer. In this case current I2 passes in the secondary which will lead to voltage drop and is given by I2Z2. Where Z2 is called secondary impedance of transformer. During this situation primary winding will draw equivalent full load current. Because of the voltage drop the secondary voltage cannot be E2 anymore so secondary induced emf will be V2.

Equivalent circuit for calculating voltage regulation of transformer:



Equation for calculating voltage regulation of a transformer:


Voltage regulation of transformer in percentage can be represented as:

Voltage regulation % = (E2-V2/V2)×100%. This is called regulation down. Power factor is specific.

Calculating voltage regulation of transformer for lagging power factor:

Now lets derive the expression for calculating voltage regulation of transformer for lagging power factor.

Phasor diagram:



here cos𝚹2 is lagging power factor

From diagram,

                       OC = OA + AB + BC
                       
                       OA = V2
                       
                       AB = AEcos𝚹=  I2R2cos𝚹2
          
                       BC=DEsin𝚹2=I2R2sin𝚹2

Angle between OC and OD is very less so OC is approximately equal to OD.

                 E2 = OC = OA + AB + BC.

                 E= OC = V2 + I2R2cos𝚹2 + I2R2sin𝚹2

Now voltage regulation of transformer at lagging power factor is,
Voltage regulation% = (E2-V2/V2) × 100%

Voltage regulation%=( I2R2cos𝚹2I2R2sin𝚹2V2 ) ×100%.

Calculating voltage regulation of transformer for leading power factor:

Now lets derive the expression for calculating voltage regulation of transformer for leading power factor.

Phasor diagram:




here cos𝚹is leading power factor
From diagram,

                         OC = OA + AB - BC

                          OA = V2
                       
                         AB = AEcos𝚹2 = I2R2cos𝚹2
          
                         BC = DEsin𝚹2 = I2R2sin𝚹2  

Angle between OC and OD is very less so OC is approximately equal to OD.

                         E2  = OC = OA +AB - BC.

                          E= OC = V2 + I2R2cos𝚹2 - I2R2sin𝚹2

Now voltage regulation of transformer at leading power factor is,
Voltage regulation% = (E2-V2/V2) × 100%

Voltage regulation%=(I2 R2cos𝚹2 - I2R2sin𝚹2 /V2 ) ×100%.

Thus we have learnt what is voltage regulation of transformer and derived expressions for voltage regulation of transformer for lagging and leading power factors. 
You can download this article of Voltage Regulation OF Transformer as a PDF here.

All Day Efficiency of Transformer/ Distribution Transformer All Day Efficiency

All Day Efficiency of Transformer/ Distribution Transformer All Day Efficiency

In our previous articles we have discussed transformer working,construction etc.Today we discuss one of the important parameter of distribution transformer I.e, "all day efficiency of a distribution transformer".We know for a transformer, the efficiency is defined as the ratio of output power to input power.

Transformer Efficiency= Output Power /Input Power 

The above equation is efficiency of any transformer. But for some special types of transformers such as distribution transformers power efficiency is not the true measure of the performance.For that purpose distribution transformer we calculate all day efficiency.Distribution transformer serve residential and commercial loads.


The load on distribution transformers vary considerably during the period of the day. For most period of the day these transformers are working at 30 to 40 % of full load only or even less than that. But the primary of such transformers is energised at its rated voltage for 24 hours, to provide continuous supply to the consumer.

The core loss which depends on voltage, takes place continuously for all the loads. But copper loss depends on the load condition. For no load, copper loss is negligibly small while on full load it is at its rated value. Hence power efficiency can not give the measure of true efficiency of distribution transformers. in such transformers, the energy output is calculated in kilo watt hour (kWh). Then ratio of total energy output to total energy input (output + losses) is calculated. Such ratio is called energy efficiency or All Day Efficiency of a transformer.

Based on this efficiency, the performance of various distribution transformers is compared. All day efficiency is defined as,


While calculating energies, all energies can be expressed in watt hour (Wh) instead of kilo watt hour (kWh). Such distribution transformers are designed to have very low core losses. This is achieved by limiting the core flux density to lower value by using a relative higher core cross-section i.e. larger iron to copper weight ratio.

The maximum efficiency in distribution transformers occurs at about 60-70 % of the full load. So by proper designing, high energy efficiencies can be achieved for distribution transformers. Numerical problems on all day efficiency of a transformer will be posted soon.

Related:

maximum efficiency of distribution
transformer all day efficiency of transformer pdf 
efficiency of transformer formula
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all day efficiency of transformer examples
distribution transformer efficiency in 24hrs

No Load Test on Induction Motor / Procedure & Application

No Load Test on Induction Motor

There are three types of tests to be performed for checking the performance of an induction motor

1.No load test or open circuit test.
2.Blocked rotor test or short circuit test or locked rotor test.
3.Stator winding resistance test.
Now let's see each test in detail:

No load test:

Main objective of conducting the no load test is:

1. To find the constant losses of the induction machine.
2. To divide constant losses into iron losses and mechanical losses.
3. To find shunt branch elements of the equivalent circuit.
4. To find no load power factor of the induction machine.

Condition to conduct no load test:

1.To conduct the no load test we need to apply rated voltage and rated frequency to stator under no load condition and also the air gap flux is to be maintained constant.
2. If the power supply given to the induction machine is constant then one watt meter is enough in the circuit if not two watt meter's are required.

Circuit diagram to conduct no load test:



Procedure to conduct no load test:

1. Connections are to be done as per the circuit diagram given above.
2. load connected to the motor is to be removed.
3. Now we need to supply rated voltage at rated frequency to the stator and also maintain constant air gap flux.
4. Input power can be measured with the help of two watt meters connected in the circuit.
5. Ammeter and voltmeter connected in the circuit gives no load voltage and current.
6. As said there is no load given to the motor the total input power given to the motor is equal to the sum of constant iron loss, mechanical losses like friction and windage losses and copper loss but it is in small amount.
7. Copper losses at rotor side can be neglected in no load test because the slip is small at no load.

Experiment values table:  

               P1
            P2
           ILINE
           VLINE









Equivalent circuit:

at no load as slip is very less we can take slip approximately equal to zero.
R2'( 1/s -1) ≈ ∞  since no load I2r' = 0
so the equivalent circuit now gets modified as follows:

                        
And the current flowing through xm is Iu  and R0 is I2.

Calculations:

V = Line to line voltage.
I = Average value of line currents flowing throw ammeters.
WNL  = P1  + Pwhich is input power or three phase power.
All the readings we get from the circuit are line values.

With star connected stator winding:

VNLperphase = V/√3 here v is the voltmeter reading in the circuit.
I0 perphase =  I here I is the ammeter reading in the circuit.
cosᴓ0 =  WNL / 3×VNLperphase×I0perphase
I1perphase = I0perphase × cosᴓ0.
Iuperphase = I0perphase × sinᴓ0.
R0 = VNLperphase / I2perphase.
Xm = VNLperphase / Iu perphase.

With delta connected stator winding:

VNL  = V = vlotlmeter reading of the circuit.
I0 perphase = I/√3 I is the ammeter reading.
cosᴓ0 =  WNL / 3×VNLperphase×I0perphase.
I1perphase = I0perphase × cosᴓ0.
Iuperphase = I0perphase × sinᴓ0.
R0 = VNLperphase / I2perphase.
Xm = VNLperphase / Iu perphase.

Power factor:

Based on watt meter readings we can find power factor approximately
If  P1 = P2  then power factor (lagging) = Unity.
if  P1 = 2Pthen power factor(lagging) = 0.866.
if  P2= 0 , P1 = total 3 phase power then power factor(lagging) = 0.5
if P1 = - P2 then power factor(lagging) =  Zero.

Finding out constant losses:

At no load input power supplied WNL = iron losses of stator + mechanical losses + stator copper losses.
1.Here the core losses of rotor are not considered as there is no load losses at rotor side are negligible.
2.And as there is no load mechanical power output is zero.
3.For induction machine mechanical losses are almost constant irrespective of load because of less speed variations.
Therefore iron loss of stator + mechanical losses = Constant losses.
WNL = Constant losses + stator copper loss at no load .
constant losses = WNL - stator copper losses at no load.
constant losses =  WNL - 3I02R1.
where R1 is the per phase resistance of stator winding.
R1 can be measured by using bridges.

Dividing constant losses into Iron losses and Mechanical losses:

1. We need to conduct the no lo load test using variable voltage at constant frequency.
2. By changing the applied voltage note down the readings of watt meter.
3. Now find constant losses using, constant losses = WNL - 3I02R1.

Calculating iron loss:

As frequency is constant  iron losses α V12.
V1 is the varying voltage.

Calculating mechanical loss:

1.As we said that speed is almost constant in induction motor at constant frequency the mechanical losses are constant.
2. By adding load to the induction motor and decreasing the voltage at constant frequency the speed of induction motor falls to maintain load torque constant so mechanical losses will decrease. If applied voltage becomes zero iron losses will become zero.
Now draw the graph and from graph we can calculate the values of mechanical loss and iron loss.
Test has to be conducted from Vrated to Vmin.

In this way we can find out Iron losses and mechanical losses.
Now  all the objectives of No load test are fulfilled and hence the performance of machine can be known.

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Unijunction Transistor (UJT) With Operation & Applications

What Is Unijunction Transistor (UJT) ?

The unijunction transistor, abbreviated UJT, is a three-terminal, single-junction device. The basic UJT and its variations are essentially latching switches whose operation is similar to the four-layer diode, the most significant difference being that the UJT’s switching voltage can be easily varied by the circuit designer. Like the four-layer diode, the UJT is always operated as a switch and finds most frequent applications in oscillators, timing circuits and SCR/TRIAC trigger circuits.

Unijunction Transistor (UJT) Operation

A typical UJT structure, pictured in figure, consists of a lightly doped, N-type silicon bar provided with ohmic contacts at each end. The two end connections are called base-l , designated B, and base-2, B2. A small, heavily doped P-region is alloyed into one side of the bar closer to 82. This P-region is the UJT emitter E, and forms a P-N junction with the bar.
An interbase resistance, RBB, exists between B1 and B2. It is typically between 4 kΩ and 10kΩ, and can easily be measured with an ohmeter with the emitter open. RBB is essentially the resistance of the N-type bar. This interbase resistance can be broken up into two resistances, the resistance from B1 to emitter, called RB1 and resistance from B2 to emitter, called RB2. Since the emitter is closer to B2, the value of RB1 is greater than RB2 (typically 4.2 kΩ versus 2.8 kΩ).

The operation of the UJT can better be explained with the aid of an equivalent circuit.The UJT’s circuit symbol and its equivalent circuit are shown in below. The diode represents the P-N junction between the emitter and the base-bar (point x). The arrow through RB1, indicates that it is variable since during nonnal operation it may typically range from 4 kΩ down to 10 Ω.

The essence of UJT operation can be stated as follows:

(a) When the emitter diode is reverse biased, only a very small emitter current flows. Under this condition, RB1 is at its normal high-value (typically 4 kΩ). This is the UJT’s “off" state.
(b) When the emitter diode becomes forward biased, RB1 drops to a very low value (reason to be explained later) so that the total resistance between E and BI becomes very low, allowing emitter current to flow readily. This is the “on” state.

Circuit Operation of Uni junction Transistor (UJT) 

The UJT is normally operated with both B2 and E biased positive relative to B1 as shown in below figure. B1 is always the UJT reference terminal and all voltages are measured relative to B1. The VBB source is generally fixed and provides a constant voltage from B2 to B1. The VEE source is generally a variable voltage and is considered the input to the circuit. Very often, VEE is not a source but a voltage across a capacitor.
We will analyze the UJT circuit operation with the aid of the UJT equivalent circuit, shown inside the dotted lines in Fig.(a). We will also utilize the UJT emitter-base-1 VE-IE curve shown in Fig.(b). The curve represents the variation of emitter current  IE, with emitter-base-1 voltage, VE, at a constant B2-B1 voltage. The important points on the curve are labelled, and typical values are given in parentheses.

The “Off ” state If we neglect the diode fora moment, we can see in Fig.(a) that RB1 and RB2 form a voltage divider that produces a voltage Vx, from point x relative to ground.
Where η (the greek letter “eta") is the internal UJT voltage divider ratio RB1/RBB and is called the intrinsic stand of ratio.

Values of η typically range from 0.5 to 0.8 but are relatively constant for a given UJT.

The voltage at point x is the voltage on the N-side of the P-N junction. The VEE source is applied to the emitter which is the P-side. Thus, the emitter diode will be reverse-biased as long as VEE is less than Vx This is the “off” state and is shown on the VE-IE curve as being a very low current region. In the “off" state, then, we can say that the UJT has a very high resistance between E and B1, and IE is usually a negligible reverse leakage current. With no IE, the drop across RE is zero and the emitter voltage, VE, equals the source-voltage.

The UJT "off " state, as shown on the VE-IE curve, actually extends to the point where the emitter voltage exceeds Vx by the diode threshold voltage, VD, which is needed to produce forward current through the diode. The emitter voltage and this point, P, is called the peak-point voltage, VP, and is given by

VP= Vx + VD= ηVBB+ VD 

where VD is typically 0.5 V. For example, if η = 0.65 and VBB= 20V, then VP=13.5 V. Clearly, VP will vary as VBB varies.

The “On " state As VEE increases, the UJT stays “off” until VE approaches the peak-point value VP, then things begin to happen. As VE approaches Vp, the P-N junction becomes forward biased and begins to conduct in the opposite direction.
Note on the VE-IE curve that IE becomes positive near the peak point P. When VE exactly equals VP, the emitter current equals IP, the peak-point current. At this point, holes from the heavily doped emitter are injected into the N-type bar, specially into the B1 region. The bar, which is lightly doped, offers very little chance for these holes to recombine. As such, the lower half of the bar becomes replete with additional current carriers (holes) and its resistance RB1, is drastically reduced. The decrease in RB1 causes Vx to drop. This drop in turn causes the diode to become more forward biased, and IE increases even further. The larger IE injects more holes into B1 further reducing RB1, and so on. When this regenerative or snowballing process ends, RB1. has dropped to a very small value (2-25 Ω) and IE can become very large, limited mainly by external resistance RE.

The UJT operation has switched to the low-voltage. high-current region of its VE- IE curve. The slope of this “on” region is very steep, indicating a low resistance. In this region, the emitter voltage VE, will be relatively small, typically 2V, and remains fairly constant as IE is increased up to its maximum rated value,IE(sat). Thus, once the UJT is “on,” increasing VEE will serve to increase IE while VE remains around 2V.

Turning “Off” the UJT Once it is “on,” the UJT‘s emitter current depends mainly on VEE and RE. As VEE decreases, IE will decrease along the “on” portion of the VE - IE curve. When IE decreases to point V, the valley point, the emitter current is equal to IV, the valley current, which is essentially the holding current needed to keep the UJT “on”. When 15 is decreased below IV, the UJT turns “off" and its operation rapidly switches back to the “off” region of its VE - IE curve,where IE = 0 and VE - VEE. The valley current is the counterpart of the holding current in PNPN devices, and generally ranges between 1 and 10 mA.

Applications of UJT:

Unijunction transistors are used extensively in oscillator, pulse and voltage sensing circuits. Some of the important applications of UJT are discussed below :
(i) UJT relaxation oscillator.
(ii) Overvoltage detector.