# Chapter 6: Direct-Current Motors¶

## Example 6.1, Page 360¶

In :
import math

#Variable declaration
P_o=10*746;#output power (in Watts)
V_s=220;
P_rot=1040;#rotational loss (in Watts)
R_a=0.75;#armature resistance (in ohms)
R_s=0.25;#series winding resistance (in ohms)
N_m= 1200;#(in rpm)

#Calculations&Results
P_d=P_o+P_rot;
def root(a,b,c):
return ((-b)-math.sqrt((b**2)-(4*a*c)))/(2*a);

I_a=root(1,-220,8500);
print '(a) armature current at rated load (in Amperes)=%.f'%I_a
P_in=V_s*I_a;
print '(b) Efficiency at full load (%%)=%.2f'%((P_o/P_in)*100)
N_s=150/I_a;
print '(c) no. of turns per pole=%.f'%N_s
I_an=16.67;
E_an=V_s-(I_an*(R_a+R_s));
N_mn=(E_an*N_m)/90;
print '(d) new speed of motor (in rpm)=%.f'%(int (N_mn))
T_dn=(E_an*I_an)/283.9;
print 'driving torque (in Newton-meter)=%.2f'%T_dn

(a) armature current at rated load (in Amperes)=50
(b) Efficiency at full load (%)=67.82
(c) no. of turns per pole=3
(d) new speed of motor (in rpm)=2711
driving torque (in Newton-meter)=11.94


## Example 6.2, Page 365¶

In :
import math

#Variable declaration
Vt = 240.  #V
Il = 3.5   #A
Ra = 0.5   #armature resistance,ohms
Rsh = 160. #shunt field winding resistance,ohms

#Calculations
I_fnl = Vt/Rsh  #shunt-field current,A
I_anl = Il - I_fnl  #armature current,A
E_anl = Vt-I_anl*0.4    #back emf,V
P_dnl = E_anl*I_anl #power developed,W
P_cunl = I_anl**2*0.4+Vt*I_fnl  #toal copper loss,W
P_innl = Vt*Il    #power input,W

If = 24   #A
N = 2400  #rpm
P_infl = Vt*If   #power input,W
I_afl = If-I_fnl #armature current,A
E_afl = Vt-0.4*I_afl  #back emf,V
P_dfl = E_afl*I_afl   #power developed,W
P_ofl = P_dfl-P_dnl   #power output,W
P_cufl = Vt*I_fnl+I_afl**2*0.4  #total copper loss,Q

n = P_ofl/P_infl*100
T_dfl = P_dfl/w_mfl   #N-m
T_sfl = P_ofl/w_mfl   #N-m
N_mnl = N*E_anl/E_afl  #rpm
SR = (N_mnl-N)/N*100

#Results
print "(a)Motor efficiency = %.2f %%"%n
print "(b)Torque developed = %.2f N-m\n   Torque available = %.2f N-m"%(T_dfl,T_sfl)
print "(c)No load speed = %d rpm"%N_mnl
print "(d)Speed regulation = %.2f %%"%SR

(a)Motor efficiency = 81.93 %
(b)Torque developed = 20.68 N-m
Torque available = 18.78 N-m
(c)No load speed = 2485 rpm
(d)Speed regulation = 3.55 %


## Example 6.3, Page 370¶

In :
import math

#Variable declaration
print '(a) For Cumulative compound motor'
V=240;#in volts(Refer to exa:6.2)
R_a=0.4;#armature resistance (Refer to exa:6.2)
T=20.68;#torque (Refer to exa:6.2)
R_x=0.1;#in ohms
I_a=22.5;#armature current of shunt motor (Refer to exa:6.2)

#Calculations&Results
I_ac=I_a/(1+0.125);#armature current of cummulative compound motor
E_ac=V-(I_ac*(R_a+R_x));
P_dc=E_ac*I_ac;
print 'Power developed (in Watts)=%.f'%P_dc
N_mc=(P_dc*60)/(T*2*math.pi);
print 'speed (in rpm)=%.f'%int(N_mc)
print '(b) For differential compound motor'
I_ad=I_a/(1-0.125);#armature current of cummulative compound motor
print 'Power developed (in Watts)=%.3f'%P_dd
N_md=(P_dd*60)/(T*2*math.pi);
print 'speed (in rpm)=%.f'%(int(N_md))

(a) For Cumulative compound motor
Power developed (in Watts)=4600
speed (in rpm)=2124
(b) For differential compound motor
Power developed (in Watts)=5840.816
speed (in rpm)=2697


## Example 6.4, Page 374¶

In :
import math

#Variable declaration
V=120.  #V
R_a=0.4;#armature resistance (in ohms)
R_sh=160.;#shunt field winding resistance
I_fL=14.75;#current drawn at full load (in Amperes)
I_nL=2.;#current drawn at no load (in Amperes)
R_x=3.6;#external resistance

#Calculations&Results
I_f=V/R_sh;#feild current
E_a_n=V-(I_afL*(R_a+R_x));#new back emf
P_d_n=E_a_n*I_afL;#new power developed
N_m_n=math.ceil((E_a_n/E_afL)*N_mfL);
print 'After insertion of external resistance in the armature ckt'
print '(a) motor speed (in rpm)=%.1f'%N_m_n
P_rot_n=(N_m_n/N_mnL)*P_dnL;
P_o_n=P_d_n-P_rot_n;
P_x=(I_afL**2)*R_x;
print '(b) power loss in external resistance (in Watts)=%.3f'%P_x
Eff=P_x/P_in_fL;
print '(c) efficiency (%%)=%.2f'%(Eff*100)

After insertion of external resistance in the armature ckt
(a) motor speed (in rpm)=1343.0
(b) power loss in external resistance (in Watts)=705.600
(c) efficiency (%)=39.86


## Example 6.5, Page 377¶

In :
import math

#Variable declaration
R_x=80;#external resistance
#Refer to Exa 6.4
R_sh=160;#shunt resistance
V=120.;#in volts
N_m=2400;#speed of motor
P_rot=143;#rotational losses

#Calculations&Results
I_fn=V/(R_x+R_sh);#new field-winding current
c=math.sqrt(I_f/I_fn);#ratio of new flux to old flux
R_a=0.4;#armature resistance
I_a=14;#armature resistance
I_an=I_a*c;
E_an=V-(I_an*R_a);
N_mn=c*(E_an/E_a)*N_m;
print '(a) new motor speed (in rpm)=%.f'%(int(N_mn))
P_x=(I_fn**2)*R_x;
print '(b) Power loss in external resistance (in Watts)=%.f'%P_x
P_in=V*(I_fn+I_an);
P_dn=E_an*I_an;
P_o=P_dn-P_rot;
Eff=P_o/P_in;
print '(c) Efficiency (%%)=%.1f'%(Eff*100)

(a) new motor speed (in rpm)=2907
(b) Power loss in external resistance (in Watts)=20
(c) Efficiency (%)=84.9


## Example 6.6, Page 380¶

In :
import math

#Variable declaration
V_s=120;#in Volts
R_fe=30;#resistance of feild winding
I_a=50;#armature current (in Amperes)
R_ag=0.2;#armature resistance of generator (in ohms)
R_am=0.3;#armature resistance of motor (in ohms)
N_m1=2000;
N_m2=715;
T=30;#torque (in Newton-meter)

#Calculations
w_m=(N_m1*2*math.pi)/60;
P_d=T*w_m;#power developed
E_am=P_d/I_a;#back emf of motor
E_amn=E_am*N_m2/N_m1;#new back emf
V_t=E_am+(I_a*R_am);
V_tn=E_amn+(I_a*R_am);
E_ag=V_t+(I_a*R_ag);#induced emf of generator
E_agn=V_tn+(I_a*R_ag);#new induced emf of generator
I_f=1.75;#Refer to magnetization curve
I_fn=0.4;#Refer to magnetization curve
R_f=V_s/I_f;
R_fn=V_s/I_fn;
R_x=R_f-R_fe;
R_xn=R_fn-R_fe;

#Result
print '(a)   external resistance (in ohms)= %.2f'%R_x
print '(b)   external resistance (in ohms)= %.f'%R_xn

(a)   external resistance (in ohms)= 38.57
(b)   external resistance (in ohms)= 270


## Example 6.7, Page 383¶

In :
import math

#Variable declaration
V_s=120;#in volts
N_m=2400;#speed of motor (in rpm)
I_in=7;#input current (in Amperes)
L=0.5;#arm length (in meter)
F_d=4.57;#deflection force (in Newton)
W=0.03;#weight (in Newton)

#Calculations&Results
F=F_d-W;
T_s=F*L;
print 'shaft torque of motor (in Newton-meter)=%.2f'%T_s
w_m=(2*math.pi*N_m)/60;
P_o=T_s*w_m;
P_in=V_s*I_in;
Eff=P_o/P_in;
print 'Efficiency of motor (%%)=%.1f'%(Eff*100)

shaft torque of motor (in Newton-meter)=2.27
Efficiency of motor (%)=67.9


## Example 6.8, Page 385¶

In :
import math

#Variable declaration
P_o=5*746;#power output (in Watts)
N_m=1200;#speed of motor (in rpm)
L=0.4;#arm length (in meter)

#Calculations
w_m=(2*math.pi*N_m)/60;
T_s=P_o/w_m;
F=T_s/L;#force reading on the scale (in Newton)

#Result
print 'Reading on the scale (in Kg)=%.3f'%(F/9.81)

Reading on the scale (in Kg)=7.564


## Example 6.9, Page 391¶

In :
#Variable declaration
V_s=400;#voltage applied
R_f=200;#resistance of field winding
I_L=30;#in Amperes

#Calculations&Results
I_f=V_s/R_f;
R_a=1;#armature resistance (in ohms)
I_a=I_L-I_f;
E_a=V_s-(I_a*R_a);#back emf (in Volts)
V_t=E_a+V_s;#total voltage in armature ckt
I_t=1.5*I_a;
R=(V_t/I_t)-R_a;
print '(1) external resistance (in ohms)=%.2f'%R
K_3=(E_a*V_s)/((R+R_a)*w_m);
K_4=((E_a/w_m)**2)/(R+R_a);
T_b=K_3+(w_m*K_4);
print '(2a) breaking torque at the instant of plugging (in Newton-meter)=%.2f'%T_b
print '(2b) breaking torque when speed of motor approaches zero (in Newton-meter)=%.2f'%K_3

(1) external resistance (in ohms)=17.38
(2a) breaking torque at the instant of plugging (in Newton-meter)=156.24
(2b) breaking torque when speed of motor approaches zero (in Newton-meter)=80.95