CHAPTER10 : PRINCIPLES OF DIRECT CURRENT MACHINES

Example E01 : Pg 394

In [1]:
#  Example 10.1
#  Computation of (a) Induced emf (b) Frequency of the rectangular voltage 
#  wave in the armature winding
#  Page No. 394
#  Given data
E1=136.8;               #  Generated emf
P=6.;                    #  Number of poles
n=1180.;                 #  Operating speed of machine

#  (a) Induced emf 

E2=E1*0.75*2.;

#  (b) Frequency of the rectangular voltage wave in the armature winding

f=P*n*0.75/120.;

# Display result on command window
print"Induced emf =",E2,"V"
print"Frequency of the rectangular voltage wave =",f,"Hz"
Induced emf = 205.2 V
Frequency of the rectangular voltage wave = 44.25 Hz

Example E02 : Pg 399

In [2]:
#  Example 10.2
#  Computation of rheostat setting required to obtain an induced emf of 290 V
#  Page No. 399
#  Given data
Ebat=240.;               #  Induced emf
If=8.9;                 #  Field current
Rf=10.4;                #  Field resistance

#  Rheostat setting required to obtain an induced emf of 290 V

Rrheo=(Ebat/If)-Rf;

#  Display result on command window
print"Rheostat setting to obtain an induced emf of 290 V =",Rrheo
Rheostat setting to obtain an induced emf of 290 V = 16.5662921348

Example E03 : Pg 401

In [3]:
#  Example 10.3
#  Computation of no-load voltage if the voltage regulation is 2.3 percent
#  Page No. 401
#  Given data
Vrated=240.;               #  Rated voltage
VR=0.023;                 #  Voltage regulation


#  No-load voltage if the voltage regulation is 2.3 percent

Vnl=Vrated*(1.+VR);

#  Display result on command window
print"No-load voltage if the voltage regulation is 2.3 percent =",Vnl,"V"
No-load voltage if the voltage regulation is 2.3 percent = 245.52 V

Example E04 : Pg 405

In [4]:
#  Example 10.4
#  Computation of percentage reduction in field flux required to obtain a 
#  speed of 1650 r/min while drawing an armature current of 50.4 A.
#  Page No. 405
#  Given data
VT=240.;               #  Induced emf
R=95.2;               #  Shunt field resistance
IT=72.;                #  Total current
Ra=0.242;             #  Armature resistance
Ia2=50.4;             #  Armature current
n1=850.;               #  Rated speed of shunt motor
n2=1650.;              #  Speed of armature winding


#  Percentage reduction in field flux

If1=VT/R;             #  Field current
Ia1=IT-If1;           #  Armature current
Ea1=VT-Ia1*Ra;        #  Armature emf
Ea2=VT-Ia2*Ra;
phip2=(n1/n2)*(Ea2/Ea1);
PerRed=(phip2-1.)*100.;



#  Display result on command window
print"Percentage reduction in field flux =",PerRed,"Percent"
Percentage reduction in field flux = -47.4191394394 Percent

Example E05 : Pg 408

In [5]:
#  Example 10.5
#  Computation of no-load speed
#  Page No. 408
#  Given data
nrated=1750.;               #  Rated speed
SR=4.;                      #  Speed regulation
#  No-load speed
Snl=nrated*(1+SR/100);
#  Display result on command window
print"No-load speed =",Snl,"r/min"
No-load speed = 1820.0 r/min

Example E06 : Pg 418

In [6]:
#  Example 10.6
#  Computation of Induced emf
#  Page No. 418
#  Given data
P=25000.;               #  Power of the generator
VT=250.;                #  Rated voltade of the machine
Ra=0.1053;             #  Armature resistance
Rip=0.0306;            #  Resistance of interpolar winding
Rcw=0.0141;            #  Resistance of compensating windings
#  Induced emf
Ia=P/VT;               #  Armature current
Racir=Ra+Rip+Rcw;      #  Resistance of armature circuit
Ea=VT+Ia*Racir;        #  Induced emf
#  Display result on command window
print"Induced emf =",Ea,"V"
Induced emf = 265.0 V

Example E07 : Pg 418

In [7]:
#  Example 10.7
#  Computation of cemf
#  Page No. 418
#  Given data
Rf=408.5;               #  Field resistance 
VT=500.;                 #  Rated voltade of the machine
IT=51.0;                #  Total current
Ra=0.602;               #  Armature resistance
Ripcw=0.201;            #  Resistance of interpolar winding and compensating windings

#  Induced emf
If=VT/Rf;             #  Current
Ia=IT-If;             #  Armature current
Racir=Ra+Ripcw;       #  Resistance of armature circuit
Ea=VT-Ia*Racir;      


#  Display result on command window
print"Induced emf =",Ea,"V"
Induced emf = 460.029864137 V

Example E08 : Pg 420

In [8]:
#  Example 10.8
#  Computation of new armature current
#  Page No. 420
#  Given data
Rf=120.;                #  Resistance of inserted resistor
VT=240.;                #  Rated voltade of the machine
IT=91.;                 #  Total current
Racir=0.221;           #  Armature sircuit resistance
n2=634.;                #  New speed after resistor was inserted
n1=850.;                #  Rated speed OF THE MACHINE
Rx=2.14;               #  Resistance inserted in series witH armature

#  New armature current

If=VT/Rf;           #  Resistor current
Ia1=IT-If;          #  Armature current
Ia2=(VT-(n2/n1)*(VT-Ia1*Racir))/(Racir+Rx);


#  Display result on command window
print"New armature current =",Ia2,"A"
New armature current = 32.0453177866 A

Example E09 : Pg 421

In [9]:
#  Example 10.9
#  Computation of (a) Steady state armature current if a rheostat in the 
#  shunt field circuit reduces flux in air gap to 75% of its rated value 
#  (b) Steady state speed for the conditions in (a)
#  Page No. 421
#  Given data
Rf=160.;               #  Field resistance
VT=240.;               #  Rated voltade of the machine
IT=37.5;              #  Total current
Ra=0.213;             #  Armature resistance
Rip=0.092;            #  Resistance of interpolar winding
Rcw=0.065;            #  Resistance of compensating windings
n1=2500.;              #  Rated speed of the machine


#  (a) At rated conditions

If=VT/Rf;               #  Field current
Ia1=IT-If;              #  Armature current
Ia2=Ia1*0.50*1./0.75;

#  (b) steady state speed for the above mentioned conditions

Racir=Ra+Rip+Rcw;

n2=n1*(VT-(Ia2*(1.+Racir)))/0.75*(1./(VT-(Ia1*Racir)));


#  Display result on command window

print"Steady state armature current =",Ia2,"A"
print"Steady state speed =",n2,"r/min"
Steady state armature current = 24.0 A
Steady state speed = 3045.70319393 r/min

Example E10 : Pg 427

In [10]:
#  Example 10.10
#  Computation of (a) Mechanical power developed (b) Torque developed 
#  (c) Shaft torque
#  Page No.427
#  Given data
T=40.;                  #  Hp rating of motor
Rf=95.3;               #  Field resistance
VT=240.;                #  Rated voltade of the machine
IT=140.;                #  Total current
Racir=0.0873;          #  Armature circuit resistance
n=2500.;                #  Rated speed of the machine


#  (a) The mechanical power developed

If=VT/Rf;               #  Field winding current
Ia1=IT-If;              #  Armature current
Ea=VT-Ia1*Racir;        #  Armature emf
Pmech=Ea*Ia1;           #  Mechanical power
Pmechhp=Ea*Ia1/746.;

#  (b) Torque developed

TD=7.04*Ea*Ia1/n;

#  (c) Shaft torque

Tshaft=T*5252./n;

#  Display result on command window
print"Mechanical power developed=",Pmech,"%0.0f W ",Pmech
print"Mechanical power developed=",Pmechhp,"hp ",Pmechhp
print"Torque developed =",TD,"lb-ft "
print"Shaft torque =",Tshaft,"lb-ft "
Mechanical power developed= 31345.5180615 %0.0f W  31345.5180615
Mechanical power developed= 42.0181207258 hp  42.0181207258
Torque developed = 88.2689788611 lb-ft 
Shaft torque = 84.032 lb-ft 

Example E11 : Pg 430

In [11]:
#  Example 10.11
#  Determine (a) Electrical losses (b) Rotational losses (c) Efficiency
#  Page No. 430
#  Given data
T=124.;                    #  Hp rating of motor
Rf=32.0;                  #  Field resistance
VT=240.;                   #  Rated voltade of the machine
IT=420.;                   #  Total current
Ra=0.00872;               #  Armature resistance
RipRcw=0.0038;            #  Resistance of interpolar winding and compensating windings
Pout=92504.;
Vb=2.0;                   #  Rated speed of the machine
Racir=Ra+RipRcw;

#  (a) Electrical losses 

If=VT/Rf;                  #  Field current
Ia=IT-If;                  #  Armature current
Pf=If**2.*Rf;                #  Field power
Paipcw=Ia**2.*(Ra+RipRcw);
Pb=Vb*Ia;                  #  Brush loss power
Plosses=Pf+Paipcw+Pb;      #  Total power loss

#  (b) Rotational losses

Ea=VT-(Ia*Racir)-Vb;         #  Armature emf      
Pmech=Ea*Ia;               #  Mechanical power
Pshaft=T*746.;              #  Shaft power 
Protational=Pmech-Pshaft;

#  (c) Ffficiency

eeta=Pout/(VT*IT)*100.;

#  Display result on command window

print"Electrical losses =",Plosses,"W"
print"Rotational losses =",Protational,"W"
print"Efficiency =",eeta,"Percent"
Electrical losses = 4755.35625 W
Rotational losses = 3540.64375 W
Efficiency = 91.7698412698 Percent

Example E12 : Pg 433

In [12]:
#  Example 10.12
#  Determine (a) Rated torque (b) Armature current at locked rotor if no
#  starting resistance is used (c) External resistance required in the armature
#  circuit that would limit the current and develop 200 percent rated torque
#  when starting (d) Assuming the system voltage drops to 215V, determine the 
#  locked rotor torque using the external resistor in (c)
#  Page No. 433
#  Given data
n=1750.;                   #  Rotor speed
P=15.;                     #  Hp rating of motor
VT=230.;                   #  Rated voltade of the machine
Ea=0;
Racir=0.280;              #  Armature circuit loss
Rf=137.;                   #  Field resistance
ItRated=56.2;             #  Total current drawn
VT1=215.;                  #  Rated voltage after drop

#  (a) Rated torque
Trated=P*5252./n;

#  (b) Armature current
Ia=(VT-Ea)/Racir; 

#  (c) External resistance required
If=VT/Rf;                      #  Field current
IaRated=ItRated-If;            #  Rated armature current

Ia2=IaRated*2.;                 #  Armature current for 200% rated torque

Rx=((VT-Ea)/Ia2)-Racir;        #  External resistance required

#  (d) Locked rotor torque 
If215=VT1/Rf;                  #  Field current at 215V
Ia215=(VT1-Ea)/(Racir+Rx);     #  Armature current at 215V
TD2=Trated*( (If215*Ia215) / (If*IaRated) );

#  Display result on command window

print"Rated torque =",Trated,"lb-ft "
print"Armature current =",Ia,"A"
print"Armature current for 200 percent rated torque =",Ia2," %0.1f A "
print"External resistance required =",Rx," %0.2f Ohm "
print"Locked rotor torque =",TD2,"lb-ft "
Rated torque = 45.0171428571 lb-ft 
Armature current = 821.428571429 A
Armature current for 200 percent rated torque = 109.042335766  %0.1f A 
External resistance required = 1.82927249846  %0.2f Ohm 
Locked rotor torque = 78.6736267891 lb-ft