Download speed control of dc shunt motor

SPEED CONTROL OF DC SHUNT MOTOR
AIM
To Control the speed of the DC Shunt motor by armature control (varying
armature voltage with field current kept constant) and field control (varying field
current with armature voltage kept constant) methods
THEORY
The voltage equation of a DC shunt motor is given by
Va = Eb + IaRa Volts
Eb =
∅ZN P
Volts
60 A
From the above relation speed of DC Shunt Motor is given by relation
N = (Va - IaRa)/KØ
Where
N
=
speed (rpm)
Va
=
Armature voltage (volts)
Ia
=
armature current (amps)
Ra
=
resistance of the armature (ohms)
K
=
motor design constant
φ
=
strength of the field flux (weber)
Z
=
Total number of armature conductors
P
=
Number of field Poles
A
=
Number of parallel paths
If
=
Field Current (amps)
From the relation of speed above, the speed of DC shunt motor can be varied by
varying
1. Armature resistance Ra
2. The Field Flux Ø
PROCEDURE
Field Control Method
Make the Connection as per circuit diagram shown below
1. The armature rheostat should be kept at the maximum position so that there
should be maximum resistance in series with armature.
2. The field rheostat should be at minimum position.
3. Switch on the supply by closing DC On/Off Isolator and DC MCB
4. Adjust the armature voltage to some fixed value.
5. Change the speed by varying the field rheostat in steps and note down the
corresponding values of speed and field current.
6. Repeat the steps 5 and 6 for different armature voltages
7. Plot field current V/s Speed graph.
Tabulation
S.No.
Armature Voltage Va =____ (V)
Field Current
Speed
If(A)
N (rpm)
1.
2.
3.
4.
5.
Model Graph
Armature Voltage Va =____ (V)
Field Current
If(A)
Speed
N (rpm)
Armature Control Method
1. The armature rheostat should be kept at the maximum position so that there
should be maximum resistance in series with armature
2. The field rheostat should be at minimum position.
3. Switch on the supply by closing DC On/Off Isolator and DC MCB
4. Adjust the field current to some fixed value and change the speed in steps by
varying armature rheostat.
5. Note down the corresponding values of armature voltage, armature current and
speed.
6. Repeat step 5 for different values of field current.
7. Plot Back Emf v/s Speed.
Tabulation
Field Current If =____ (A)
S.No. Armature Armature
Voltage
Current
Va (V)
Ia (A)
1.
2.
3.
4.
5.
Model Graph
Back
Emf
Eb (V)
Speed
N
(rpm)
Field Current If =____ (A)
Armature
Voltage
Va (V)
Armature Back
Current
Emf
Ia (A)
Eb (V)
Speed
N
(rpm)
SWINBURNES TEST
AIM
To pre-determine the percentage efficiency of the given DC Machine
whenworking as a generator and as a motor.
THEORY
In Swinburne’s method the d.c. machine (generator or motor) is run asa motor
at no-load and losses of the machine are determined. Once the losses of the machine
are known, its efficiency at any desired load can be determined in advance. It may be
noted that this method is applicable to those machines in which flux is practically
constant at all loads. E.g., Shunt and Compound machines.
As a motor,
Motor input power Pi(M) = VLoILo (watts)
No load armature currentIao= (ILo- Ish) (amps)
Armature Copper Loss Wcu = I2ao Ra(watts)
Constant losses Wc =VLoILo – I2ao Ra (watts)
Gross power developed by armaturePo(M) = VLoIao-I2ao Ra
= (VLo -Iao Ra)Iao
Po(M) = EbIao
Percentage efficiency, η(M) = (Po(M) / Pi(M)) *100
Where,
VLo
=
Motor input voltage (volts)
ILo
=
No-Load current (Amps)
Iao
=
No load armature current (Amps)
Ish
=
Motor Shunt field current (Amps)
Ra
=
Armature resistance (Ohms)
As a generator,
Generator output Po(G) = VIL (Watts)
Armature copper lossWcu = I2a Ra(Watts)
Armature current Ia = IL + Ish (Amps)
Total losses Wt = Wc+ I2a Ra (Watts)
Generator input Pi(G) = Po(G)+ Wt(Watts)
Percentage efficiency, η(G) = (Po(G) / Pi(G)) *100
= (VIL) / (VIL+ Wc+ (IL + Ish)2 Ra) * 100
Procedure
1. The armature rheostat should be kept at the maximum position so that there
should be maximum resistance in series with armature.
2. The field rheostat should be at minimum position.
3. Switch on the supply by closing DC On/Off Isolator and DC MCB.
4. Bring the motor to its rated speed by adjusting motor field rheostat.
5. Note down the corresponding motor input voltage (V), No-Load current (ILO)
and field current (Ish).
Tabulation
S.No.
Input
Voltage
V (Volts)
No load
Current
ILO (Amps)
Field
Current
Ish (Amps)
No Load
Armature
Current
Iao (Amps)
Constant Loss
Wc (Watts)
Calculation
As a motor
S.No.
Assumed
Load
Current
IL(Amps)
Armature
Current
Iao (Amps)
Armature
Copper
Loss
Wcu
(watts)
Motor
input
power
Pi(M)
(watts)
Motor
output
power
Po(M)
(watts)
Efficiency
η(M) (%)
Armature
Current
Ia (Amps)
Armature
Copper
Loss
Wcu
(watts)
Generator
output
Po(G)
(watts)
Generator
input Pi(G)
(watts)
Efficiency
η(G) (%)
1.
2.
3.
4.
As a Generator
S.No.
Assumed
Load
Current
IL(Amps)
1.
2.
3.
4.
Model Graph
LOAD TEST ON SINGLE PHASE TRANSFORMER
AIM
To find out the efficiency of the given single phase transformer by conducting
direct load test in it.
THEORY
The transformer is a device which transfers energy from one electrical circuit to
another electrical circuit through magnetic field as coupling medium. In this process it
does not change the frequency of voltage or current. It works on the basic principle of
electromagnetic induction (mutually induced EMF). Being a static device it has a very
high efficiency as compared to rotating machine of same rating as the losses are less.
When primary winding of transformer is energized with source of voltage V pan
EMF Esis induced across the secondary winding and it is also equal to secondary
terminal voltage Vs till there is no load across secondary winding. As soon as load is
applied across the secondary winding the terminal voltage is decreased from Es to Vs.
Percentage Efficiency Ƞ = [Output Power (Ws) / Input Power (Wp)]*100
Vp
- Primary Voltage (Volts)
Vs
- Secondary Voltage (Volts)
Ip
-Primary Current (Ampere)
Is
- Secondary Current (Ampere)
Wp
-Power consumed by primary winding (Watts)
Ws
-Power consumed by secondary winding (Watts)
Ƞ
- Percentage Efficiency (%)
PROCEDURE
1. Make the connections as per the circuit diagram.
2. Keep the switch on secondary side open so that load is zero to measure no load
voltage. Also keep knob of auto transformer at zero output voltage position.
3. Now increase the voltage through auto transformer until voltage in voltmeter
V2 reads rated value of secondary winding & read no load voltage E2.
4. Switch onresistive load such that secondary winding current be approximately
10% of the rated current of secondary side.
5. Take the corresponding readings from WattmetersWs&Wp,Voltmeters Vs& Vp
and AmmetersIs& Ip
6. Increase the load current in steps of 10% of the rated value by switching on few
more loads and take the readings of the Wattmeter, Ammeter & Voltmeter till it
reaches 120-125% of rated value.
7. Reduce the load to zero by switching of the loads one-by-one.
8. Switch off the Supply.
TABULATION
S.No. Primary Primary Primary Secondary Secondary Secondary Efficiency
Voltage Current Power
Voltage
Current
Power
(%)
Vp (V) Ip (A) Wp (W)
Vs (V)
Is (A)
Ws (w)
1.
2.
3.
4.
5.
MODEL GRAPH
+++
OPEN CIRCUIT AND SHORT CIRCUIT TEST ON SINGLE PHASE
TRANSFORMER
AIM
To obtain the equivalent circuit parameters of a single phase transformer by
conducting open circuit and short circuit test in it.
THEORY
OPEN CIRCUIT TEST
The shunt branch parameters can be determined by performing this test. Since,
the core loss and themagnetizing current depend on applied voltage, this test is
performed by applying the rated voltageto one of the windings keeping the other
winding open (generally HV winding is kept open and ratedvoltage is applied to LV
winding). Since,the secondary terminals are open (no load is connected across the
secondary), current drawn from thesource is called as no load current. Under no-load
condition the power input to thetransformer is equal to the sum of losses in the primary
winding resistance Rand coreloss. Since, no load current is very small, the loss in
winding resistance is neglected. Hence, on no loadthe power drawn from the source is
dissipated as heat in the core.
SHORT CIRCUIT TEST
Suppose the input voltage is reduced to a small fraction of ratedvalue and
secondary terminals are short-circuited. A current will circulate in the secondary
winding.Since a small fraction of rated voltage is applied to the primary winding, the
flux isgenerated in the core and hencethe core loss is very small. Hence, the power
input on short circuit is dissipated as heat in the winding.In this test, the LV terminals
of thetransformer are short circuited. The primary voltage is gradually applied till the
rated current flows inthe winding. Since, the applied voltage is very small (may be of
the order of 5-8%).
FORMULA
OPEN CIRCUIT TEST (LV Side)
1. Iron or core loss, Wi=Woc
2. No load power input power, Woc = Voc* Ioc* cos Φ oc
3. No load power factor,Cos Φ oc = Woc/(Voc*Ioc)
4. Working component of no load current,Iw = IoccosΦoc
5. Magnetizing component of no load current,Iµ = Ioc sin Φoc
6. No load Resistance R0=Voc/Iw
7. No load Reactance X0=Voc/ Iµ
Voc
= Open circuit voltage (Volts)
Ioc
= Open circuit current (Amps)
Woc
= Open circuit power (Watts)
SHORT CIRCUIT TEST (HV Side)
1. Equivalent impedance of winding referred to HV sideZ02=Vsc/Isc
2.
Equivalent resistance of winding referred to HV side, R02 =Wsc/ Isc
3. Equivalent Reactance of winding referred to HV side, X02= √( Z202– R202 )
4. Transformation Ratio, K= V1 / V2
5. Equivalent resistance of winding referred to LV side, R01= R02 / K2
6. Equivalent reactance of winding referred to LV side, X01= X02 / K2
V1
= LV side or Primary side voltage (Volts)
V2
= HV side or secondary side voltage(Volts)
TABULATION
OPEN CIRCUIT TEST
S.No.
No load Voltage
(Voc)
No load Current
(Ioc)
No load Power
(Woc)
SHORT CIRCUIT TEST
S.No.
Short Circuit
Voltage (Vsc)
Short Circuit Current
(Isc)
Short Circuit Power
(Wsc)
PROCEDURE
OPEN CIRCUIT TEST
1. Make the connections as per the circuit diagram.
2. Switch on the AC breaker and isolator.
3. Adjust single phase auto transformer till the voltage connected across the
primary winding reads rated primary voltage.
4. Note down the corresponding open circuit voltage (Voc), no load current (Ioc)
and no-load power (Woc) indicated by the respective voltmeter, ammeter and
wattmeter.
5. Reduce the auto transformer voltage to zero.
6. Switch off the AC breaker and isolator.
SHORT CIRCUIT TEST
1. Make the connections as per the circuit diagram.
2. Switch on the AC breaker and isolator.
3. Adjust single phase auto transformer till the ammeter reads rated load
current.
4. Note down the corresponding short circuit voltage (Vsc), short circuit
current (Isc), and short circuit power (Wsc) indicated by the respective
voltmeter, ammeter and wattmeter.
5. Reduce the auto transformer voltage to zero.
6. Switch off the AC breaker and isolator.
EQUIVALENT CIRCUIT (referred to primary)