3-DC Motor Starting

Power Converters and Drives, PCVD320B

Starting of of DC DC Motors starting conditions, conditions, the motor is not turning ( = 0), therefore the internal generated voltage E A = K   = 0V.

 At

 The

DC MOTORS: STARTING

only compon component ent to limit limit startin starting g current current is the armature resistance, which, in most DC motors is a very low value (approximately one ohm or less),

 Therefore,

a very high current flows in the armature circuit during starting condition.

ADEL ELGAMMAL

I a , Starting

ASSOCIATE PROFESSOR

Ra



V t Ra

Starting of DC motor

Starting a dc motor across the line. If a dc motor is sta started rted with full supply supply vol voltage tage across across its terminals, a very high current will flow, which may damage the motor due to heavy sparking at commutator and heating of the winding.

V t  E a

ADEL ELGAMMAL THE UNIVERSITY OF TRINIDAD AND TOBAGO UTT






• At stand-still, ( = 0),  E A = K   = 0 Volt

+

R a

I a  I

I a , Starting

Vt



– ADELELGAMMAL THE UNIVERSITY OF TRINIDAD AND TOBAGO UTT

I A



VT

 E A

R A



250  0 0.06

 4167 A

This current is over 20 times the motor’s rated full-load current and may sever ely damage the motor.

ADELELGAMMAL THE UNIVERSITY OF TRINIDAD AND TOBAGO UTT

1

Ra


DC motor problems on starting For instance, for a 50 hp, 250 V DC motor with armature resistance R A of 0.06  and a full-load current about 200 A, the starting current is

V t

Starting of DC motor Ther eref efor ore, e,  Th

it is ne nece cess ssar ary y to li limi mitt th thee current to a safe value during starting.

 For

normally designed machines, twice the rated current can be allowed to flow and for specially designed machines it can be 3.5 times.


Starter in DC Motor

Starting of DC motor I a

• We can limit Ia at start-up by:

o



V t

Ra

V  K  T d  K  t Ra

Ra Ia


V t

V  E a T d  K  t Ra

Motor

Operating Point

 E a

Ra

I st  

I st

V t

At Starting,  = 0, then E a = 0

Speed

1) Controlling V t using variable supply – e.g. using power electronics converter



V T st  K  t Ra

Current


Starting of DC motor (voltage reduction)

Starter in DC Motor • We can limit Ia at start-up by: 2) Adding external resistor  known as starter

I st





V t  Ia

Ra

–

Ist1 Ist2

 Ia


2

Vt R a

 R st

When Ea = 0

A solution to the problem of excessive starting current is to insert a starting resistor in series with the armature to limit the current until E A can build up to limit the armature current. ADEL ELGAMMAL THE UNIVERSITY OF TRINIDAD AND TOBAGO UTT

Starting DC Motor However, this resistor must be removed from the circuit as the motor speed is high since otherwise such resistor would cause losses and would decrease the motor’s torque-speed characteristic.  Starting resistors are usually incorporated into the motor design to limit starting current to 125 to 200 percent of full load current .  As motor accelerates and back emf rises, one section of the resistor is cut out at a time, either manually or automatically with the help of contactors, such that the current is kept within specified maximum and minimum values.



R st




Ia

+ Ea –

Vt

Vt1

Vt2

R a

+

Starting of DC motor (starting resistance)  



V t K 



Ra I a K 

R a R a + R

Ist1 Ist2

V t I st  Ra  Rst V Rst  t I st ADEL ELGAMMAL THE UNIVERSITY OF TRINIDAD AND TOBAGO UTT

 Ra

Ia

Example A dc separately excited motor has the following data: K  = V t = Ra = Ia =

Solution Rated torque = T d  Kø I a

 3 (5)  15 NM (Newton Meter)

3.0 Vs (Volt second)

600.0 V 2.0

Starting torque = T st  V t



5.0 A (armature current at full load)

Calculate the rated torque, starting torque and starting current at full voltage.

Kø Ra

 600.0 * 3.0 / 2.0  900.0 NM

Starting current = I st 

V t Ra

 300.0 A

starting torque is 60 times the rated torque starting current is also 60 times the rated current



Example

Solution

A dc separately excited motor has the following data: K  = V t = Ra = Ia =

3.0 Vs (Volt second)

600.0 V 2.0



5.0 A (armature current at full load)

Method 1: Redu ce the starting voltage

V st  I st Ra

 6 (5.0) (2.0)  60.0

Method 2: Increase the starting resistance

R  Ra How can you reduce the starting current to 6 times the rated current?


V

R 

V t I st

- Ra





V t I st

600.0 30.0

- 2.0  18 


DESIGN OF A MANUAL STARTER FOR A DC SHUNT MOTOR • A dc shunt motor has the following data: 240 V, 18 A, 1100 r/min, and armature-circuit resistance R a=0.33 . • The no-load saturation curve for the machine is given in the following Fig. • The motor is to be started by a manual starter, shown in next following Fig.; the field current is set at 5.2 A.

• No-load saturation curve for a 240-V, 1100-r/min dc generator.

• If the current during starting varies between 20 and 40 A, determine the starting resistors R 1, R 2, a n d R 3 needed in the manual starter. ADEL ELGAMMAL THE UNIVERSITY OF TRINIDAD AND TOBAGO UTT

3


I a (0 )

Manual starter for a shunt motor



I max



)  I min



)  I max



)



I min

I a (t 2



)

 I max

I a (t 2

I a (t 1 I a (t 1





t 1

t 1



t 2



t 2



t 3



t 3



E (t 3 

E (t 2 

E (t 1



)  E (t 3 )



)  E (t 2 )



)  E (t 1 )


DESIGN OF A MANUAL STARTER FOR A DC SHUNT MOTOR

1 . A t t = 0 ; I a = Imax , Ea = 0 V Calculate Armature Current at Standstill • At the first contact on the manual starter, 240 V dc is applied to the shunt field; the field rheostat has been set to allow I ƒ=5.2 A. • Thus, the armature current Ia=IL-Iƒ= 40 A-5.2 A=34.8 A. • The 40 A is the maximum permissible load current. ADEL ELGAMMAL THE UNIVERSITY OF TRINIDAD AND TOBAGO UTT

DESIGN OF A MANUAL STARTER FOR A DC SHUNT MOTOR 2 . A t t = t1- , Ia = Imin , Ea = ? ? V Calculate the Induced Armature Voltage, Ea, when the Machine Accelerates ((R1 + R2 +R3) still inserted in the armature) • As the machine accelerates, an induced armature voltage is created, causing the armature current to fall to Imin.


4


Calculate the Value of the Starter Resistance, (R1 + R2 +R3) • Use the equation: • (R1 + R2 +R3) =(Vt/Ia)-R a=(240 V)/(34.8 A)0.330 =6.57 .



• When the armature current drops to the minimum allowed during starting • Ia=IL- Iƒ=20 A-5.2 A=14.8 A, • the induced voltage is obtained from the equation: • Ea(t1-)=Ea(t1+)=Vt-Ia(R a+(R 1+R 2+R 3)) =240V-(14.8 A)(0.330 +6.57 )=137.9 V. ADELELGAMMAL THE UNIVERSITY OF TRINIDAD AND TOBAGO UTT


3. At t = t1+, R1 is removed from the circuit, Ia = Imax, Ea = 137.9 V Calculate the Starter Resistance, (R2+ R3)

• Use the equation: • (R2+ R3) = ((Vt-Ea) / Ia) - R a=(240 V137.9 V)/34.8 A-0.330 =2.60  .



4 . A t t = t2- , Ia = Imin , Ea = ? ? V Calculate the Induced Armature Voltage, Ea, when Machine Again Accelerates

• Ea(t2-) = Ea(t2+) = Vt-Ia(R a+(R2+ R3)) =240V- (14.8A)(0.330+2.60) =196.6 V.



5. At t = t2+, R2 is removed from the circuit, Ia = Imax, Ea = 196.6 V

6. Calculate the Three Starting Resistances, R1, R2, and R3

Calculate the Starter Resistance, (R3)

• From the preceding steps,

• Use the equation:

• R3= 0.917

,

• (R3) =(Vt-Ea)/Ia-R a=(240 V-196.6 V)/34.8 A-0.330 =0.917 .

• R2=1.683

, and


DC motor problems on starting In practice, a starting resistor is made up of a series of resistors that can be successively removed from the circuit as the motor speeds up. A shunt motor with an extra starting resistor that can be cut out of the circuit in segments by closing the 1A, 2A, and 3A contacts.

Therefore, two considerations are needed to be taken into account: Select the values and the number of resistor segments needed to limit the starting current to desired ranges; Design a control circuit shutting the resistor bypass contacts at the proper time to remove particular parts of the resistor from the circuit. ADELELGAMMAL THE UNIVERSITY OF TRINIDAD AND TOBAGO UTT

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• R1=3.97

.


DC motor problems on starting: Ex Example 5.6: A 100 hp, 250 V 350 A shunt DC motor with an armature resistance of 0.05 needs a starter circuit that will limit the max starting current to twice its rated value and which will switch out sections of resistor once the armature current decreases to its rated value. a. How many stages of starting resistance will be required to limit the current to the specified range? b. What must the value of each segment of the resistor to be? At what voltage should each stage of the starting resistance be cut out?




I a ( 0)





)  I min

I a (t 2 )  I min



)  I max

I a (t 2 )

I max I a (t 1 I a (t 1









t 1



t 1

 I max



t 2

DC motor problems on starting: Ex



t 3



t 2

t 3

The starting resistor must be selected such that the current flow at the start equals twice the rated current. As the motor speeds up, an internal voltage E A (which opposes the terminal voltage of the motor and, therefore, limits the current) is generated. When the current falls to the rated value, a section of the starting resistor needs to be taken out to increase the current twice. This process (of taking out sections of the starting resistor) repeats until the entire starting resistance is removed. At this point, the motor’s armature resistance will limit the current to safe values by itself. The original resistance in the starting circuit is 

E (t 3 

E (t 2 

E (t 1



)  E (t 3 )

 R1  R2  ...  RA 

Rtot

V T I max



)  E (t 2 ) After the stages 1 through i are shorted out, the total resistance left in the starting circuit is



)  E (t 1 )

Rtot ,i

 Ri1  ...  RA ADEL ELGAMMAL THE UNIVERSITY OF TRINIDAD AND TOBAGO UTT

DC motor starting: Ex The resistance R1 must be switched out of the circuit when the armature current falls to

I A,min



 E A,1

VT

Rtot

 I min  350 A

R A

After the resistance R1 is out of the circuit, the armature current must increase to

I A,max

VT



 E A,2

Rtot ,1

 I   Rtot ,n   min   I max 

Solving for n:

 VT  EA  I max Rtot ,1

Rtot 

V T I max

n

I min I max

Rtot



250 700

n

 I    min  Rtot  ADELELGAMMAL  I max  THE UNIVERSITY OF TRINIDAD AND TOBAGO UTT Rtot ,n

b. The armature circuit will contain the armature resistance R A and three starting resistors. At first, E A = 0, I A = 700 A, and the total resistance is 0.357 . The total resistance will be in the circuit until the current drops to 350 A. This occurs when

 VT  I A,min Rtot  250  350  0.357  125 V

At this time, the starting resistor R1 will be taken out making

Rtot ,1

 RA  R2  R3 

VT

 E A,1

I max



250  125 700

 VT  I A,min Rtot ,1  250  350  0.1786  187.5 V

At this time, the starting resistor R2 will be taken out leaving

Rtot ,2

 RA  R3 

VT

 E A,2

I max



250  187.5 700

 0.0893 


6

 0.357 

log  R A Rtot  log  I m in I ma x 



log  0.05 0.357  log  350 700 

 2.84  3

DC motor problems on starting: Ex This total resistance will be in the circuit until the current drops again to 350 A. This occurs when

E A,3

 VT  I A,min Rtot ,2  250  350  0.0893  218.75 V

At this time, the starting resistor R3 will be taken out leaving only R A in the circuit. The motor’s current at that moment will increase to

 0.1786 

This (new) total resistance will be in the circuit until the current drops again to 350 A. This occurs when

E A,2

log  I m in I ma x 


DC motor problems on starting: Ex

E A,1

Rtot

Notice that the number of stages n must be rounded up to the next integer.

The resistance left in the circuit is

Rtot ,1 

n

log  R A Rtot 

n

 I max  700 A

Since E A = K  , the quantity V T – E A must be constant when the resistance is switched out. Therefore

I min Rtot

DC motor starting: Ex The starting process is completed when Rtot,n is not greater than the internal armature resistance R A. At the boundary:

I A,3



VT

 E A,3 R A



250  218.75 0.05

 625 A

which is less than the allowed value. Therefore, the resistances are

 Rtot ,3  RA  0.0893  0.05  0.0393   Rtot ,2  R3  RA  0.1786  0.0393  0.05  0.0893  R1  Rtot ,1  R2  R3  RA  0.357  0.1786  0.0393 0.05  0.1786  R3

R2

The resistors R1, R2, and R3 are cut out when E A reaches 125 V, 187.5 V, and 218.75 V, respectively. ADEL ELGAMMAL THE UNIVERSITY OF TRINIDAD AND TOBAGO UTT

3-POINT STARTER

3-POINT STARTER

• A “3-point starter” is extensively used to start a D.C shunt motor. • It provides additional protective features such as over load protection and no volt protection. • The diagram of a 3-point starter connected to a shunt motor is shown in figure . • The starter is shown enclosed within the dotted rectangular box having three terminals marked as A, L and F for external connections. ADELELGAMMAL THE UNIVERSITY OF TRINIDAD AND TOBAGO UTT

3-POINT STARTER

3-POINT STARTER

• Terminal A is connected to one armature terminal Al of the motor. Terminal F is connected to one field terminal F1 of the motor and terminal L is connected to one supply terminal as shown. • F2 terminal of field coil is connected to A2 through an external variable field resistance and the common point connected to supply (-ve). • The external armatures resistances consist of several resistances connected in series and are shown in the form of an arc.

• The junctions of the resistances are brought out as terminals (called studs) and marked as 1,2,.. .12. Just beneath the resistances, a continuous copper strip also in the form of an arc is present.


• There is a handle which can be moved in the clockwise direction against the spring tension. The spring tension keeps the handle in the OFF position when no one attempts to move it.


3-POINT STARTER

3-POINT STARTER

• Now let us trace the circuit from terminal L (supply + ve). The wire from L passes through a small electro magnet called OLRC, and enters through the handle shown by dashed lines.

• On the other hand, the rectangular strip will make contact with the continuous arc copper strip.

• Near the end of the handle two copper strips are firmly connected with the wire. The furthest strip is shown circular shaped and the other strip is shown to be rectangular. • When the handle is moved to the right, the circular strip of the handle will make contacts with resistance terminals 1, 2 etc. progressively.


7


• The other end of this strip is brought as terminal F after going through an electromagnet coil (called NVRC). Terminal F is finally connected to motor field terminal Fl.


THE OPERATION OF THE STARTER • Initially the handle is in the OFF position. Neither armature nor the f ield of the motor gets supply. • Now the handle is moved to stud number 1. In this position armature and all the resistances in series gets connected to the supply. Field coil gets full supply as the rectangular strip makes contact with arc copper strip. • As the machine picks up speed handle is moved further r to stud number 2. In this position the external resistance in the armature circuit is less as the first resistance is left out. ADELELGAMMAL THE UNIVERSITY OF TRINIDAD AND TOBAGO UTT

THE OPERATION OF THE STARTER • The no volt release coil (NVRC) carries same current as that of the field coil. • In case supply voltage goes off, field coil current will decrease to zero. • Hence NVRC will be de-energised and will not be able to exert any force on the soft iron piece of the handle. • Restoring force of the spring will bring the handle back in the OFF position. ADELELGAMMAL THE UNIVERSITY OF TRINIDAD AND TOBAGO UTT

THE OPERATION OF THE STARTER • Now gap between the electromagnet and the soft iron piece is so adjusted that for I L Irated force of attraction will be sufficient to pull up iron piece. • This upward movement of the iron piece of OLRC is utilized to de-energize NVRC. To the iron a copper strip (A shaped in figure) is attached.


8

THE OPERATION OF THE STARTER • Field however, continues to get full voltage by virtue of the continuous arc strip. • Continuing in this way, all resistances will be left out when stud number 12 (ON) is reached. In this position, the electromagnet (NVRC) will attract the soft iron piece attached to the handle. • Even if the operator removes his hand from the handle, it will still remain in the ON position as spring restoring force will be balanced by the force of attraction between NVRC and the soft iron piece of the handle. ADEL ELGAMMAL THE UNIVERSITY OF TRINIDAD AND TOBAGO UTT

THE OPERATION OF THE STARTER • The starter also provides over load protection for the motor. The other electromagnet, OLRC overload release coil along with a soft iron piece kept under it, is used to achieve this. • The current flowing through OLRC is the line current IL drawn by the motor. As the motor is loaded, Ia hence IL increases. Therefore, IL is a measure of loading of the motor. • Suppose we want that the motor should not be over loaded beyond rated current. ADEL ELGAMMAL THE UNIVERSITY OF TRINIDAD AND TOBAGO UTT

THE OPERATION OF THE STARTER • During over loading condition, this copper strip will also move up and put a short circuit between two terminals B and C. • Carefully note that B and C are nothing but the two ends of the NVRC. In other words, when over load occurs a short circuit path is created across the NVRC. • Hence NVRC will not carry any current now and gets de-energised. The moment it gets de-energised, spring action will bring the handle in the OFF position thereby disconnecting the motor from the supply. ADEL ELGAMMAL THE UNIVERSITY OF TRINIDAD AND TOBAGO UTT

THE OPERATION OF THE STARTER • Three-point starter has one disadvantage. If we want to run the machine at higher speed (above rated speed) by field weakening (i.e., by reducing field current), the strength of NVRC magnet may become so weak that it will fail to hold the handle in the ON position and the spring action will bring it back in the OFF position. • Thus we find that a false disconnection of the motor takes place even when there is neither over load nor any sudden disruption of supply.

DC motor starting circuits Several different schemes can be used to short contacts and cut out the sections of a starting resistor. Some devices commonly used in motor-control circuits are Fuses: protects against short circuits

Spring-type push button switches

Time delay relay similar to ordinary relay except for having adjustable time delay.


Overload: a heater coil and normally closed contacts ADEL ELGAMMAL THE UNIVERSITY OF TRINIDAD AND TOBAGO UTT

DC motor starting circuits

DC motor starting circuits

A common DC motor starting circuit:

Another type of motor starter:

A series of time delay relays shut contacts removing each section of the starting resistor at approximately correct times.

A series of relays sense the value of armature voltage E A and cut out the starting resistors as it riches certain values.

Notice that the relay 1TD is energized at the same time as the motor starts – contacts of 1TD will shut a part of the starting resistor after some time. At the same instance, relay 2TD is energized and so on… Observe also 4 fuses protecting different parts of the circuit and the overload in series with the armature winding.


Relay: a main coil and a number of contacts

This starter type is more robust to different loads. FL is the field loss relay: if the field is lost for any reason, power to the M relay will be turned off. Armature current in a DC motor during starting.


Problem # 1

Chapter's Assignment ADEL ELGAMMAL THE UNIVERSITY OF TRINIDAD AND TOBAGO UTT

9

• A 10 kW , 100 V , 1000 rpm dc machine has Ra=0.1 ohm and is connected to a 100 V dc supply. a) Determine the starting current if no starting resistance is used in the armature circuit b) Determine the value of the starting resistance if the starting current is limited to twice the rated current c) This dc machine is to run as a motor, using starter box. Determine the values of resistance required in the starter box such that the armature current Ia is constraint within 100% to 200% of its rated value during start-up. ADEL ELGAMMAL THE UNIVERSITY OF TRINIDAD AND TOBAGO UTT

Problem # 2

Problem # 3

• Calculate the require resistance for a four-step starter to limit the starting current of a DC shunt motor to 150% 0f rate current. Assume that all four steps have an equal resistance value. The motor has rating as 25 Hp, 240 Volts, 860 rpm. And armature resistance is 0.08 Ohms, the overall efficiency 90%. Determine at each speed of the starter resistance must be take into the starting circuit to maintained the rate value during start up period. Assume that the field current is negligible compared to rate armature current.

• A 10 hp (7.46kW) 200 V shunt motor has full-load efficiency of 85%. The armature has a resistance of 0.25. Calculate the value of the starting resistance necessary to limit the starting current to 1.5 times the full-load current at the moment of first switching on. The shunt current may be neglected. Find also the back emf of the motor, when the current has fallen to its full load value, assuming that the whole of the starting resistance is still in circuit.


Problem # 4 •

A 25-kW, 230-V shunt motor has an armature resistance of 0.064 Ω and a field-circuit resistance of 95 Ω . The motor delivers rated output power at rated voltage when its armature current is 122 A. When the motor is operating at rated voltage, the speed is observed to be 1150 r/min when the machine is loaded such that the armature current is 69.5 A.

a) Calculate the rated-load speed of this motor. In order to protect both the motor and the dc supply under starting conditions, an external resistance will be connected in series with the armature winding (with the field winding remaining directly across the 230-V supply). The resistance will then be automatically adjusted in steps so that the armature current does not exceed 200 percent of rated current. The step size will be determined such that, until all the external resistance is switched out, the armature current will not be permitted to drop below rated value. In other words, the machine is to start with 200 percent of rated armature current and as soon as the current falls to rated value, sufficient series resistance is to be cut out to restore the current to 200 percent. This process will be repeated until all of the series resistance has been eliminated. b) Find the maximum value of the series resistance.


Problem # 5 • A dc shunt motor drives a centrifugal pump at a speed of 1000 rpm when the terminal voltage and line currents are 200 V and 50 A, respectively. The armature and field resistances are 0.1 Ω and 100 Ω, respectively. Design a starting resistance for a maximum starting current of 120 A in the armature circuit.

c) How much resistance should be cut out at each step in the starting operation and at what speed should each step change occur? ADEL ELGAMMAL THE UNIVERSITY OF TRINIDAD AND TOBAGO UTT

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3-DC Motor Starting

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