Problems & Solutions on Electric Drives

DC Motor Drives Example 1

A 220V, 200 A, 800 rpm dc separately excited motor has an armature resistance of 0.06Ω. The motor armature is fed from a variable voltage source with an internal resistance of 0.04Ω. Calculate internal voltage of the variable voltage source (VVS) when the motor is operating in: (i)

Motoring at 50% of the rated motor torque and 400 rpm

(ii)

Regenerative braking at 80% of the rated motor torque and 600 rpm.

Solution

We know that the motor torque,Tm is proportional to the armature current, Ia. The motor speed, N is proportional to the back emf, E.

Rint Ra Ia + Vs Ea -

(i) The motor armature current at 50% of the rated motor torque is I a = 0.5 × 200 = 100A

The back emf at the rated speed and the rated torque is E = V a − I a × R a = 220 − 200 × 0.06 = 208V

The back emf at 400 rpm is

E=

N2 400 ×E= × 208 = 104V 800 N1

Internal voltage of VVS is Vint = E + I a ( R a + Rint) = 104 + 100(0.06 + 0.04) = 114V

(ii) The motor armature current at 80% of the rated motor torque is

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I a 2 = 0.8 × 200 = 160A

The back emf at the rated speed and the rated torque is E 2 = V a − I a 2 × R a = 220 − 200 × 0.06 = 208V

The back emf at 600 rpm is

E2 =

N2 600 × E1 = × 208 = 156V 800 N1

Internal voltage of VVS is Vint = E 2 + I a 2 ( R a + Rint) = 156 − 160(0.06 + 0.04) = 140V

Example 2 A 200V, 875 rpm, 150A separately excited dc motor has an armature resistance of 0.06 Ω. It is fed from a single-phase fully controlled rectifier with an ac source voltage of 220V, 50 Hz. Assuming continuous conduction, calculate (i)

Firing angle for rated motor torque and 750 rpm.

(ii)

Firing angle for rated motor torque and –500 rpm

(iii)

Motor speed for α = 160° and rated torque.

Solution

E1 = 200 − 150 × 0.06 = 191V

At rated operation:

I 1 = 150 A N 1 = 875 rpm (i) E at 750 rpm, E 2 =

750 ×191V = 163.7V 875

Because of the motor operated at the rated torque I a1 = I a 2 = 150 A

and V a 2 = E 2 + I a 2 × R a = 163.7 + 150 × 0.06 = 172.7V

Now

2

2Vm

π

cos(α ) = 172.7

2 × 220 × 2

cos(α ) = 172.7

π cos(α ) = 0.872

(ii) E at -500 rpm, E 2 =

α = 29.3 o

or

− 500 × 191V = −109V 875

Because of the motor operated at the rated torque

I a1 = I a 2 = 150 A and V a 2 = E 2 + I a 2 × R a = −109 + 150 × 0.06 = −100V

Now 2Vm

π

cos(α ) = −100

2 × 220 × 2

cos(α ) = −100

π cos(α ) = −0.5

or

α = 120 o

(iii) At α =160°

Va =

2V m

π

cos(α ) =

Since

V a = E + I a × Ra

Or

E = −195V

2 × 220 × 2

π

cos(160) = −186V

− 186 = E + 150 × 0.06

Speed =

− 195 × 875 = −893.2 rpm 191

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Example 3

A 230V, 960 rpm and 200A separately excited dc motor has an armature resistance of 0.02 Ω. The motor is fed from a chopper, which provides both motoring and braking operations. The source has a voltage of 230V. Assuming continuous conduction: (i)

Calculate duty ratio of chopper for motoring operation at rated torque and 350 rpm.

(ii)

Calculate duty ratio of chopper for braking operation at rated torque and 350 rpm.

(iii)

Calculate the speed of the motor for motoring operation if the duty ratio of 0.6 and the armature current of 150 A.

(iv)

If maximum duty ratio of chopper is limited to 0.95 and maximum permissible motor current is twice the rated, calculate maximum permissible motor speed obtainable without field weakening and power fed to the source.

(v)

If motor field is also controlled in (iii), calculate field current as a fraction of its rated value for a speed of 1200 rpm.

Solution

At rated operation V a1 = E1 + I a × R a ⇒ E1 = V a − I a × R a = 230 − ( 200)(0.02) = 226V

(i) Back emf at 350 rpm, E =

350 × 226 = 82.4V 960

Motor terminal voltage V a = E + I a R a = 82.4 + (200)(0.02) = 86.4 Duty ratio D =

86.4 = 0.376 230

(ii) V a = E − I a R a = 82.4 − ( 200)(0.02) = 78.4V Duty ratio D =

78.4 = 0.34 230

(iii) V a = DV s = 0.6 × 230 = 138V E = V a − I a R a = 138 − (150)(0.02) = 135V

Speed =

135 × 960 = 573.45 rpm 226 4

(iv) Maximum available armature voltage, V a = DmaxV s = 0.95 × 230 = 218.5V E = V a + 2 I a R a = 218.5 + ( 2)(200)(0.02) = 226.5V

Maximum permissible motor speed =

226.5 × 960 = 962 rpm 226

Assuming lossless chopper, power fed into the source V a I a = 218.5 × 400 = 87.4kW

Example 4

The speed of a separately excited dc motor is controlled by a single-phase full-wave converter. The field circuit is also controlled by a full converter and the field is set the maximum possible value. The ac supply voltage to the armature and field converter is one phase, 440 V, 60 Hz. The armature resistance is Ra = 0.25 Ω, the field circuit resistance is Rf = 175 Ω, and the motor voltage constant is Kv = 1.4 V/A rad/s. The armature current corresponding to the load demand is Ia = 45 A. The viscous friction and no-load losses are negligible. The inductances of the armature and field circuits are sufficient to make the armature and field currents continuous and ripple free. If the delay angle of the armature converter is αa = 60° and armature current is Ia = 45 A, determine: i)

The electromagnetic torque, Te, developed by the motor.

ii)

The speed, ω

iii)

The input PF of the drive

Solution

Vs = 440V, Vm = √2×440 = 622.25 V, Ra = 0.25 Ω, Rf = 175 Ω, αa = 60° and Kv = 1.4 V/A rad/s. i)

The maximum field voltage (and current) would be obtained for a delay angle of αf =

0° and Vf =

2V m

π

cos(α f ) =

2 × 622.5

π

= 396.14 V

The field current is If =

Vf Rf

=

396.14 = 2.26 A 175

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The developed torque is Te = TL = K v I f I a = 1.4 × 2.26 × 45 = 142.4 N The armature voltage is Va =

2V m

π

cos(α a ) =

2 × 622.5

π

cos 60 o = 198.07 V

The back emf is E a = V a − I a R a = 198.07 − 45 × 0.25 = 186.82 V

ii)

The speed is

ω=

Ea 186.82 = = 59.05 rad/s or 564 rpm K v I f 1.4 × 2.26

iii) Assuming lossless converter, the total input power from the supply is

The input current of the armature converter for a highly inductive load is shown below. It rms value is Isa =Ia = 45 A. The same goes to input current of field converter. The rms value of input current of field converter is Isf = If = 2.26 A. The effective rms supply current can be found from 2 I s = I sa + I sf2 = 45 2 + 2.26 2 = 45.06 A

and the input VA rating, VI = VsIs = 440x45.06 = 19 826.4. Neglecting the ripples, the input power factor is approximately PF =

Pi 9808.4 = = 0.495 lagging VI 19826.4

If

Ia

Is Isf

Isa

Ra La

Isf Rf Lf M

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ia Ia ωt Ia

isa

π

αa

π+ αa

ωt

-Ia Example 5 If the polarity of the motor back emf in Example 4 is reversed by reversing the polarity of the field current, determine: The delay angle of the armature circuit converter, αa, to maintain the armature

i)

current constant at the same value of Ia = 45 A. The power fed back to the supply due to regenerative braking of the motor.

ii)

Solution i)

From part ii) of Example 4, the back emf at the time of polarity reversal is Eg =

186.82 V and after polarity reversal Eg = -186.82 V. V a = E g + I a R a = −186.82 + 45 × 0.25 = −175.57 V

and

2Vm

π

cos(α a ) = −175.57 V

− 175.57 × π 2 × 622.25 α a = 116.31o cos(α a ) =

ii)

The power fed back to the supply is Pa = V a I a = 175.57 × 45 = 7900.7 W

AC Motor Drives

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Example 1 A Y-connected squirrel-cage induction motor has following ratings and parameters:

400V, 50 Hz, 4-pole, 1370 rpm, Rs = 2Ω, Rr’ = 3Ω , Xs =Xr’ = 3.5Ω.

Motor is controlled by a voltage source inverter at constant V/f ratio. Assuming motor speed torque curves to be parallel straight lines in the region of interest, calculate: (i)

Speed for a frequency of 30 Hz and 80% of full-load torque.

(ii)

Frequency for a speed of 1000 rpm and full-load torque.

(iii)

Torque for a frequency of 40 Hz and speed of 1100 rpm.

Solution (i)

At 50 Hz, drop in speed from no load to full-load torque = 1500-1370 = 130 rpm Drop in speed from no load to 80 % of full load = (130)(0.8)= 104 rpm Synchronous speed at 30 Hz =

120 f 120 × 30 = 900 rpm = p 4

Therefore motor speed at 30 Hz = 900-104= 796 rpm

(ii)

Drop in speed from no load to full load torque = 130 rpm

Synchronous speed Ns =1000+130 = 1130 rpm

f =

(iii)

p × N s 4 ×1130 = = 37.67 Hz 120 120

At 40 Hz synchronous speed Ns =

120 f 120 × 40 = 1200 rpm = p 4

Drop in speed from no load to 1100 rpm = 1200-1100= 100 rpm

Torque =

100 TFL = 0.769 TFL 130

where TF is the full load torque. At full-load s =

1500 − 1370 = 0.08667 1500

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T=

3

ωs

V ×

2

Rr' s

2

 R'   R s + r  + ( X s + X r' ) 2  s  

= 38.06 Nm

Hence torque = 0.769 TFL = (0.769)(38.06)=29.267 Nm

Example 2 For regenerative braking of inverter-fed induction motor drive of example 1, determine approximate values of: (i)

Speed for the frequency of 30 Hz and 80 % of full load torque

(ii)

Frequency for a speed of 1000 rpm and full load torque

(iii)

Torque for a frequency of 40 Hz and speed of 1300 rpm

Solution (i)

Increase in speed from no-load to full load torque = 130 rpm Increase in speed from no load to 0.8 of full load torque = (0.8)(30)=104 rpm Synchronous speed at 30 Hz =

30 ×1500 = 900 rpm 50

Machine speed at 30 Hz = 900 + 104 = 1004 rpm.

(ii)

Synchronous speed = 1000-130 rpm = 870 rpm f =

(iii)

p × N s 4 × 870 = = 29 Hz 120 120

At 40 Hz synchronous speed

Ns =

120 f 120 × 40 = 1200 rpm = p 4

Increase in speed from no-load speed = 1300 – 1200 rpm = 100 rpm Motor torque =

− 100 × T FL = −29.277 Nm 130

Example 3 A 4-pole, 3-phase, 50 Hz, 1420 rpm induction machine has a rated torque of 40 Nm. It is used to drive a load with the speed-torque characteristics given by TL = Kω2, such that the speed

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equals the rated value at the rated torque. If a constant V/f control method is used, find the synchronous frequency of the motor at 0. 25 and 0.75 rated torque.

Solution

At rated torque and speed:

Ns = 1500 rpm , Nr = 1420 rpm, Nsl = 80 rpm, TL = Tm = 40 Nm, f = 50 Hz. At 0.25 rated torque:

TL = Tm = (0.25)(40) = 10 Nm N sl =

Nr =

T Trated

× N sl ( rated ) =

T Trated

10 × 80 = 20 rpm 40

× N r ( rated ) =

10 × 1420 = 710 rpm 40

N s = N r + N sl = 710 + 20 = 730 rpm

At 0.75 rated torque:

TL = Tm = (0.75)(40) = 30 Nm N sl =

Nr =

T Trated

× N sl ( rated ) =

T Trated

30 × 80 = 60 rpm 40

× N r ( rated ) =

30 ×1420 = 1229.76 rpm 40

N s = N r + N sl = 1229.76 + 60 = 1289.76 rpm

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-NDM-13 Oktober 2005

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