C OMPOUND DC M OTOR

Lecture # 9

C ^OMPOUND DC M ^OTOR

C ^OMPOUND DC M ^OTOR

o Compound motor combines the effect of both a shunt and series field.

o Can be wired in either a long shunt or a short shunt configuration.

o Under no-load or light load conditions, the series field current will be very

small and the shunt field will have its full excitation.



Long Shunt Short Shunt

D IFFERENTIAL C ^OMPOUND DC M ^OTOR

o Differential Compound: the series and shunt field windings are connected in

such a way that their fields oppose each other.

o where �_f represents the shunt field flux and �_s represents the series field flux.

o As the load increases, both I_a and �_s will increase, causing the term (�_f - �_s) decreases, the increase in current predominates over the decrease in flux.

o Thus, the induced torque (T) will increase and the motor speed will increase too, and the motor may reach an unsafe uncontrollable speed or

"run away" very quickly. (undesirable and dangerous).

o The differential compound motor maintains even better constant speed

within its load limit than the shunt motor. But it has very poor starting torque and is unable to handle serious overloads.

 The differential compound motor has very few practical applications.

T = k_a ϕ_f − ϕ_s I_a Speed = ω ≈ V_t k_a ϕ_f − ϕ_s

C ^UMULATIVE C ^OMPOUND DC M ^OTOR

o Cumulative Compound: the shunt and series windings are connected so that

their respective fields aid each other.

o As the load increases, the mmf of the series increases but the mmf of the shunt field remains constant. The total mmf is greater under load that at no-load.

o An increase in load causes I_a to increase, producing a larger torque (T) which is greater than the shunt and the net flux will increase as well.

o The motor speed falls with increasing load and the speed drop from no-load to full-load.

o The cumulative compound motor has somewhat higher starting torque than a shunt motor of the same rating.

 Compounding improves the torque characteristic and degrades the speed characteristic.

T = k_a ϕ_f + ϕ_s I_a Speed = ω ≈ V_t k_a ϕ_f + ϕ_s

C ^OMPOUND DC M ^OTOR

Armature current

Differential compound Cumulative compound

Torqu e

Armature current

Differential compound

Cumulative compound

Speed

DIFFERENTIAL COMPOUND

CUMULATIVE COMPOUND

T = k_a ϕ_f + ϕ_s I_a

Speed = ω ≈ V_t k_a ϕ_f + ϕ_s

T = k_a ϕ_f − ϕ_s I_a

Speed = ω ≈ V_t k_a ϕ_f − ϕ_s

C OMPARISON OF DC M ^OTORS

Armature current

Differential compound

Torqu e

Series Cumulative compound Shunt

Speed

Cumulative compound

Armature current

Shunt Series

Differential compound

S ^TARTING DC M ^OTORS

If we examine the 120V/10A shunt DC motor shown in the figure and calculate the currents at starting, the need for a special device becomes apparent.

At starting the speed ( is zero, the back EMF (E_g) is zero. Thus,

 This not only might damage the source, but the motor brushes would most likely burn up. In addition, the motor’s armature winding which is rated at 10A, would burn.

I_f  ^Vt  ¹²⁰  2 A R_f

60

Ra 1

V_t  E_g

Ia _ start ^  120 

0 

120A

starting rated

IL _ rated 10 I_{L _ start}  I_{a _ start}  I_f  122 A

I_{L _ start} 122

%Increase  100  100  1220%  I

 12  I

S ^TARTING DC M ^OTORS

If we examine the 120V/10A shunt DC motor shown in the figure and calculate the currents at starting, the need for a special device becomes apparent.

At starting the speed ( is zero, the back EMF (E_g) is zero. Thus,

I =f V^t 120

R_f = 60 = 2A

I_{L_start} = I_{a_start} + I_f = 122A I_{a_start} =

V_t − E_g

R_a = 120 − 0

1 = 120A

I_{L_start} 122

%Increase = × 100 = × 100 = 1220% ⇒ I

≅ 12 × I

starting rated

I_{L_rated} 10

 This not only might damage the source, but the motor brushes would most likely burn up. In addition, the motor’s armature winding which is rated at 10A, would burn.

C ^OMPOUND DC M ^OTOR

a) The back EMF

I_f = V_t/R_f =120/100 = 1.2 A I_a = I_L – I_f = 15-1.2 = 13.8A

E_g = V_t – I_a (R_s+R_a) = 120-13.8(0.2+0.8)=106.2 V b) The output power rating

P_o = P_in – P_Losses

I² R_a  I² R_s ^  P_st

PLosses  Pcu  Pst  I² Rf a a

 f



P _Losses1.2² 100 13.8² ^{0.8  0.2} 65  399.44 W

P_i  V_t I_L  12015  1800 W P_cu P_st

P_dev

P_i P_o

PLosses  I² R  I² ^{R R} ^{  P} Example 5-8:

A long shunt cumulatively compound motor is run at full load. It is rated 120 V, 15 A, 140 rad/s. If R_f =100 , R_a=0.8 , R_s=0.2 , and the stray losses are 65 W, find;

Sol.

f f a a s st

  P_o  1800  399.44  1400.54 W  1400

C ^OMPOUND DC M ^OTOR

T_o  ^Po  ¹⁴⁰⁰  10 N.m.

 140

e) The developed torque

P_dev = P_o + P_st = 1400+65=1465W

or P_dev = E_g I_a= 106.213.8=1465 W

T_dev  ^Pdev  ¹⁴⁶⁵  10.46 N.m.

Example 5-8:

A long shunt cumulatively compound motor is run at full load. It is rated 120 V, 15 A, 140 rad/s. If R_f =100 , R_a=0.8 , R_s=0.2 , and the stray losses are 65 W, find;

Sol. (cont.)

c) The efficiency

%  ^Po 100  ¹⁴⁰⁰ 100  78%

P_i 1800

d) The torque rating

 140

C ^OMPOUND DC M ^OTOR

L s f f a a

P_cu  I² R  I² R  I²

R  ²⁴^{2 0.05 } ^1.194^{2 200 } ^22.81^{2 0.4}

 522 W

b) P_st = P_i – P_cu – P_o ; P_i = V_t I_L

= (240) 24 –522 – 5000

P_o 5000

c) %

 x 100  x 100 

86.8%

P_i 240

d) E_g = V24_t – I_L R_s – I_a R_a

= 240 – (24) (0.05) – (22.81) (0.4) Example 5-9:

A short shunt compound motor is 5kW, 240V, and draws a line current of 24A. If R_f =200 , R_a=0.4 , and R_s=0.05 , find at rated load: a) the total copper

losses,

b) the stray power losses, c) the efficiency, d) the back EMF.

Sol.

a) V_f = V_t – I_L R_s = 240 – 24 (0.05) = 238.8 V I_f = V_f /R_f = 238.8/200 = 1.194 A

I_a = I_L – I_f = 24-1.194 = 22.81A

= 238 W = 229.68 V

S ^HUNT DC M ^OTOR

b) T_o  ^Po  ⁵⁵⁹⁵  66.4 N.m

   84.2 rad / s (or 804.4 rpm)

2 4 217.2 540 4 3010^3

Eg  ^{Z P}   2 a E_g

   and a=P

2 a Z P

 Example 5-10:

A 4-pole, 220-V shunt motor has 540 lap-wound conductor. It takes 32 A from the supply mains and develops output power of 5.595 kW. The field winding takes 1 A. The armature resistance is 0.09 Ω and the flux per pole is 30 mWb. Calculate:

a) the speed and b) the shaft torque in N.m.

Sol.

a) I_a = 32  1 = 31 A

E_g = V_t – I_a  R_a = 220 – 310.09 = 217.2 V

 84.2