EEL305
DC Drives fed with 1- Φ converters:
1)A separately excited dc motor is fed from a 230V,50Hz source through a 1Φ half controlled converter.
Its field is fed through 1-Φ semi converter with zero degree firing angle delay. Ra=0.7Ω and motor
constant=0.5V.Sec/rad. For rated load torque of 15N-m at 1000rpm and for continuous ripple free
currents determine
a) firing angle delay of the armature converter
b) rms value of thyristor and freewheeling diode current
c) input power factor of the armature converter.
Sol:a)Motor constant=0.5 v-sec/rad = 0.5 Nm/A= KΦ
Armature current =Torque / KΦ = 15/0.5= 30 A
Motor back emf, Ea=KΦ*ωm=0.5*(2*3.14*1000) /60=52.36 V
For 1Φ half wave conveter feeding a dc motor,
Vt=(Vm(1+cosα))/π = Ea+Ia*Ra
Vt=√2*230*(1+cosα) / п=52.36+30*0.7=72.36 V
α =cos‾1[(73.36*2*3.14/ (230*√2) -1]=65.336º
Thus the firing angle of armature converter is 65.33°
b)Rms value of the thyristor current is
Thyristor current=Ia √[(п-α)/2п)]=30√[(180-65.336/360)]=16.931 A=Isr
Rms value of the free wheeling diode current is
Ifdr=Ia√[(п+α)/2п]=30√[(180+65.336) /360]=24.766A
C)Input power factor from the armature converter
Is=Vt*Ia/(Vs*Isr)=73.36*30/ (230*16.931)=0.5651 lag.
2)The speed of a 15hP,220V,1000rpm dc series motor is controlled using a 1Φ half controlled bridge
converter. The combined armature and field resistance is 0.2Ω. Assuming continuous and ripple free
motor current and speed of 1000 rpm and K=0.03Nm/A 2,determine a)motor current b)motor torque for a
firing angle α=30º. Supply AC voltage is 250V.
Sol:
for a DC series motor Vt=Ea+Ia(Ra+Rs)
Motor torque Te=K*Ia2
Ea=Ka*Φ*ωm=K*Ia*ωm
a)from the above ,Vt=Vo=Ea+Ia(Ra+Rs)
Vt=Vo=(Vm*(1+cosα))/п=Ea+Ia(Ra+Rs)=K*Ia*ωm+Ia(Ra+Rs)
= √2*250(1+cos30º)/п=0.03*Ia*(2п*1000/60)+0.2Ia
209.97=3.3416Ia
Motor armature current ,Ia=209.97/3.3416=62.84A
b)motor torque ,
Te=K*Ia*Ia=0.03(62.84)*(62.84)=118.466Nm.
3) The speed of a separately excited dc motor is controlled by a 1Φ fully controlled bridge converter
with field being also controlled by a full converter. The field current is set to the max possible value.
The supply voltage to armature & field converters is 220V, 50Hz. Ra=0.2Ω, Rf=150Ω and motor
voltage constant K’ =1.1 V.Sec/A.Rad. The Ia value corresponding to full-load is 25 A. The inductance
of armature and field are sufficient to make field and armature currents continuous and ripple free.
i)if delay angle of armature converter is α=45º Ia=25A determine
a)torque developed by motor b)speed and c)input power factor of drive
ii)if delay angle is changed to α=0º for same torque developed and determine motor speed
iii)for same load demand as in earlier condition find delay angle of the field converter, if speed
as to be increased to 1500 rpm
SOL:
Given V=220v Ia=25A Rf=150Ω
Ra=0.2Ω k=1.1(v.s) /A.rad
i)for first 2 cases asuume max field voltage for delay angle αf being 0º.
Vf=2√2V/п = 2√2*220/п=198 V
The field current If=Vf/Rf=198/150=1.32A
The motor constant value for this field current is
K= k’*If=1.1*1.32=1.452V.sec/rad
a) Torque developed =Td=K*Ia=1.452*25=3.63 Nm
ii)
iii)
b) Armature voltage for α=45º
Va=2√2*v*cosα/п=2*√2*cos45º/п=140V
Em=Va - Ia*Ra=140-(250*0.2)=135V
ω1=Eb1/K=135/1.452=93 rad/sec =888 rpm
c) Assume lossless converter total input power from supply is
Pi=Va*Ia+Vf*If=(140*25)+(198*1.32)=3.76kW
RMS value of input current is same as Ia i.e., 25 A.
The rms value of input current to field converter If=1.32A
Effective rms supply current Ir=√(Ia*Ia)+(If*If)=25.035A
Input kVA rating is VI=Vs*Is=220*25.035*10¯03=5.508 kva
Neglecting ripple in the current, Power factor =Power/VA=3.76/5.508=0.68(lag)
d)for same delay angle αf=0º,Vf=198v. For the same torque, Ia will be the same.
Eb2=198-(25*0.2)=193V
ω2=Eb2/K=133 rad/sec = 1270 rpm
e) ω3=1500rpm = 157.1 rad/sec
Eb = 193 V = Knew * ω3 = k’*If3*157.1
If3=193/ (1.1*157.1)=1.117 A
Vf3=If3*Rf=1.117*150=167.6 V
Cosαf=162.6/198=0.846
Hence, delay angle αf=32.2º
4)The speed of a 20hP ,210V ,1000rpm series motor is controlled by a single phase a)semi converter
b)full converter. The combined armature and field circuit resistance is 0.25Ω. Motor constants are
Kaf=0.03 Nm/ A2 and Kres=0.075 v-s/rad. Supply voltage is 230V. Assuming continuous and ripple free
motor current, determine the following for a α=30º and N=1000 rpm. a)Motor torque b)Ia c)PF
sol: semi converter controlled dc drive:
N=1000rpm=1000*2п/60=104.72rad/sec; Ia = ((Vm/π)(1+cosα) - Eb) /(Ra+Rf)
Eb = (Kres*ω + 0.03 * Ia *ω) = 7.8525 + 3.141 Ia
(193 -7.8525 – 3.141 Ia) / 0.25 = Ia Therefore Ia = 54.5 A
We have T= Kres Ia + Kaf*Ia*Ia =93.195 Nm
Supply PF = 193 x 54.5 / 230 x 54.5 sqrt (150°/180°) = 0.91
Full converter controlled dc series motor:
Ia = 50.58 A T = 78 Nm PF =0.78
so PF is better in semi converter system.
5) A 220V ,1000 rpm, 60A separately excited dc motor has Ra of 0.1Ω. It is fed from a single phase full
converter with an ac source voltage of 230V,50Hz. Assuming continuous conduction compute
a) Firing Angle for rated motor torque at 600 rpm
b) Firing Angle for rated motor torque at -500 rpm
c) Motor speed for α=150º and half-rated torque.
Sol: Under rated operating conditions of motor, 1000 rpm =104.72 rad/sec
Vt= Ea+IaRa=K*ωm + IaRa.
220=K*104.72 + 60*0.1
K= 2.044 N-m/A
a) For rated motor torque,Ia=60Amp
 Vo=Vt= K*ωm+IaRa where ωm =62.8 rad/sec
 Vt = 2.044*62.8 + 60*0.1 = 134.43 V = (2Vm/π) Cosα
 α= cos‾1(134.43* п) / (2√2*230) = 49.56º
b) At (-500)rpm or -52.35 rad/sec
2√2*(230/п)*cosα = 2.044*(-52.35) + 60*0.1=-101.024 V
 α= cos‾1(-101.024 * п)/ (2√2*230) = 119.2º
c) At half-rated torque, Ia =(1/2)*rated current= 30A
d) ((2√2*230)/п)*cos150° = 2.044 *ωm + (30*0.1)
-179.30=2.044 *ωm+ 3
ωm=-182.3/2.044=-89.188rad\sec = 851.7 rpm
DC Drives fed with 3 Φ converters
6) The speed of a separately excited dc motor is controlled by means of a 3Φ semi converter from a 3Φ,
415V, 50Hz supply. The motor constants are L=10mH, Ra=0.9Ω, KΦ=1.5 V.S/rad. Calculate the speed
of this motor at a torque of 50N-m when the converter is fired at 45º. Neglect losses in the converter.
SOLUTION: Motor torque, Te=KΦ*Ia=50N-m
Motor armature current, Ia=50/1.5 = 33.33 A
The equation for the semi-converter motor combination is
3Vm(1+cos α)/2п = Ea+Ia*Ra
3√2*415(1+cos45°) / 2π=1.5*ωm + 33.33*0.9
ωm =(478.3-30) /1.5=298.867rad/s
hence motor speed N=298.867*60/ (2п)=2854 rpm.
7)The speed of a 10kW, 1000 rpm, 560V, 20 A separately excited dc motor is controlled by 3Φ full
converter. Input to the converter is 3Φ, 415V, 50 Hz. Ra = 1Ω. When the motor is delivering rated
torque determine a) firing angle of converter and power factor at rated speed b) firing angle of converter
and power factor at 10% rated speed c) active, reactive powers drawn from system at rated speed d)
active, reactive powers drawn from system at 10% rated speed e) ratio of reactive powers drawn from
system at 10% rated speed and at rated speed. Neglect system losses and effect of commutation angle.
SOL: Rated back emf = 560-20*1 = 540 V
Back emf constant KΦ = 540/104.72 V.S/rad
a) For 3Φ fully controlled converter, we have
Va=3Vm*cos α/ п = 1.35 * 415 * cosα
=3√2*415*cos α/п= Eb rated + IaRa = 560 V
α =0º
Since effect of commutation angle is neglected PF cos Φ=cos α= 1
b) at 10% rated speed Eb = 54 V Therfore, Va = Eb + IaRa = 74 V
74=1.35*415*cos α
α=82.4°
PF at this firing angle =Cos 82.4° = 0.1322
c) Active power at rated speed and rated load Vd Id = 560 x 20 =11.2 kW
Reactive power at rated speed, rated load = 0 as PF is unity
d) Active power at 10% speed P= VdId =74 * 20 = 1480 W
Reactive power =P*tanα =11.092 kVAR
e) Ratio of reactive powers drawn from the system at rated speed to 10% rated speed is zero.
Therefore due to decrease in speed the reactive power increased tremendously.
8) A 80 kW, 440V, 800 rpm dc motor is controlled by 3Φ, 6 pulse thyristor converter to run at
600 rpm delivering 75% rated torque. Back emf at rated speed is 410 V. Determine trigger angle
of the converter. Input to the converter is 3Φ ,415,50 Hz,AC.
SOL: given Eb1=410 V at 800 rpm KΦ = 4.89 V.Sec/rad
Eb2 at 600 rpm = 410*600/800=307.5V
Rated torque = 80000 * 60 / (2π*800) = 954.9 Nm
Corresponding Ia = Torque /KΦ =195.28 A
410 = 440- Ia*Ra
Ra=0.1536 Ω
Now terminal voltage of dc motor at 600 rpm and 75% rated torque
Eb2+Ia*Ra=307.5+(0.75*195.28*0.1536 )=330 V = 3Vmcos α/п = 1.35 V cosα
α =53.9 º
9) A 100kW, 500V, 2860 rpm separately excited dc motor is energized from 400V, 50 Hz, 3Φ
full converter. The voltage drop in conducting thyristor is 2V. The dc motor parameters are
Ra=0.1 Ω, Km=1.6 v-s/rad, La=8mH; Rated Ia=210A. Noload armature current=10% of
rated current. Armature current is continuous and ripple free. Find a) noload speed at firing angle
of 30 º b) firing angle for a speed of 2000 rpm and rated Ia. determine also the supply power
factor. c) Find the speed regulation for the firing angle obtained in part(b)
SOL:a) The motor terminal voltage at 30° firing angle
Va = 1.35Vcos 30=467.64V
Also Va = Eb + Ia*Ra + 2 Therefore Eb = 467.64 – 2 -21*0.1
No load motor speed=Eb/Km =289.71 Rad/sec = 2766.55 rpm.
b) At rated armature current and at 2000rpm firing angle ?
Va= Km*ωm+IaRa+2 = 1.6*209.44 + 210*0.1 + 2 = 358.1 V = 1.35 V cosα
α= 48.46º
RMS value of source current per phase , Isr = Ia√(2/3)=210*√(2/3)=171.46A
Supply power factor =Va*Ia/ (√3*Vs*Isr)=358.1*210/ (√3*400*171.46)=0.633(lag)
c) Full-load speed at the above α = 2000 rpm
No-load current = 21 A.
No-load Eb = 358.1 – 2 – 21.*0.1 =354.1 V
No-load speed = No-load Eb / Km = 221.3 rad/sec = 2113 rpm
Speed regulation is 113 / 2000 = 5.65% .
10) The speed of a 20 hp, 300 V, 900 rpm separately excited dc motor is controlled by a 3 Φ full
converter. The field circuit is also controlled by a 3 Φ full converter. The ac input to both the
converters is 3 Φ star connected, 208V, 60Hz supply. Ra=0.25Ω, Rf=145 Ω and the motor
voltage constant is K=1.2 V.Sec/(A.rad). Assume ripple free armature and field currents. . If the
field converter delivers maximum field current, at a torque value of Td=116N-m the motor runs
at 900 rpm. a) If the field current is set to maximum value with Td being 116 Nm, what will be
the motor speed with armature converter firing angle being zero. b) For the same load demand
and same firing angle of the armature converter as in part (a), determine the delay angle of the
field converter if speed has to be increased to 1800 rpm.
SOL: Given: Ra=0.25 Ω, Rf=145 Ω, K=1.2V.sec/Arad, VL-L=208V.
3-phase full-converter output = 1.35 V cosα
a)Td = 116N-m and ω=900п/30=94.25 rad/s at max field current.
Vf=1.35 * 208* cos 0°=280.8 V
If=280.8/145=1.936A
Armature current at this torque and If, Ia=Td/ (K*If)=116/ (1.2*1.936)=49.93A
Back emf at 900 rpm speed Eb=Kv*If*ω=1.2*1.936*94.25=218.96 V
At this torque and 900 rpm speed armature voltage
Va=Eb + Ia*Ra=218.96+49.93*0.25=231.44 V
Va=231.44=1.35 *208* Cosαa from which armature converter firing angle can be deduced.
At αa = 0° and Td = 116 Nm, back emf Eb’=280.8 - 49.93*0.25 = 268.22 V
And the speed ω=Eb’/ (K*If)=268.22/ (1.2*1.936)=115.45rad/s or 1102.5 rpm
b) ω=1800п/30=188.5 rad/s
Eb= 268.22V (as in part (a))=1.2*188.5*If or If=1.186A
Vf=1.186*145=171.97v
Vf=171.97=1.35*400*cosαf
The delay angle αf = 52.2º