DC Machines: Important Solved Examples Problems

 

DC Machines - DC Generators and Motors

SOLVED EXAMPLES

 

EXAMPLE 1

Calculate the emf induced in the armature of a two pole DC generator whose armature has 280 conductors and is revolving at 1000 rpm. The flux per pole is 0.03 Wb.

Given Data: P=2, Z=280, N = 1000 rpm, ϕ=0.03 Wb ; A= P(Assumed).

SOLUTION:

Emf induced, E = ϕZN/60  × P/A Volts

 = ( 0.03×280×1000 / 60) × (2/2)

E = 140 V

 

EXAMPLE 2

Calculate the generated emf of a DC generator which has 4 poles, total number of conductors equal to 256 lap wound and running at 2000 rpm. The useful flux per pole is 0.2 Webers.

Given Data: P=4, Z=256, Lap wound, N=2000 rpm, ϕ=0.2 Wb.

SOLUTION:

Since lap wound, A = P = 4.

Emf induced, Eg = ϕZN/60  × P/A Volts

= ( 0.2×256×2000 / 60 ) × (4/4)

E_g = 1706.6 V

 

EXAMPLE 3

A 4 pole DC generator has a wave wound armature with 792 conductors. The flux per pole is 0.012 Wb. Determine the speed at which it should be run to generate 240 V on no load.

Given Data: P=4, wavewound, DC generator, Z=792, ϕ = 0.012 Wb, since wave wound, A = 2

SOLUTION:

Emf induced, Eg = ϕZN/60  × P/A Volts

240 = (0.012×792×N / 60) × (4/2)

N=758 rpm

 

EXAMPLE 4

Calculate the flux per pole required for a 4‒pole generator with 360 conductors generating 250 V at 1000 rpm. When the armature is (i) Lap connected (ii) Wave connected.

Given Data: P=4, Z=360, E_g = 250 V, N = 1000 rpm, ϕ = ?

SOLUTION:

(i) When the armature is lap connected

Emf induced, E_g = ϕZN/60  × P/A Volts

(A = P)

250= [ϕ×360×1000/60] × (4/4)

ϕ = 41.67 mWb

(ii) When the armature is wave connected

Emf induced, E_g = ϕZN/60  × P/A Volts

(Since wave connected, A = 2)

250= [ϕ×360×1000 / 60] × (4/2)

 ϕ = 20.83 mWb

 

EXAMPLE 5

An 8 pole DC shunt generator with 778 wave connected armature conductors and running at 500 rpm. Supplying a load of 12.5 Ω resistance at terminal voltage of 250 V. The armature resistance is 0.24 Ω and the field resistance is 250 Ω. Find the armature current, the induced emf, and flux per pole.

Given Data: P=8, DC shunt generator, Z=778, N=500 rpm, Wave wound, R_L = 12.5 Ω, V=250 V, R_a = 0.24 Ω, R_sh = 250 Ω

To find: I_a, E_g, ϕ = ?

SOLUTION:

Load current I_L = V / R_L = 250 / 12.5 = 20 A.

Shunt field current, I_sh = V / R_sh = 250/250 = 1 A.

Armature current, l_a = I_L+I_sh = 20+1

 I_a = 21 A

Induced emf, E_g = V+ l_aR_a

E_g = 250 + 21 × 0.24

E_g = 255 V

Emf induced, E_g = ϕZN/60  × P/A Volts

Flux, ϕ = [ E_g × 60 × A ] / ZNP

= [255 × 60 × 2] / [ 778 × 500 × 8 ]

 ϕ = 9.83 mWb

 

EXAMPLE 6

A shunt generator has an induced emf on open circuit of 127 V. When the machine is on load, the terminal Voltage is 120 V. Find the load current, if the field resistance and the armature resistance are 15 Ω and 0.02 Ω respectively, Ignore armature reaction.

Given Data: E=127 V, V= 120 V, R_sh = 15Ω, R_a = 0.02 Ω.

To find: I_L = ?

SOLUTION:

I_sh = V / R_sh = 120 / 15 = 8 A.

Armature Voltage drop, I_a R_a = 127‒120 =7 V.

Armature current, I_a = Armature drop / R_a = 7/0.02

 I_a = 350 A

Load current, I_L = I_a ‒ I_sh

I_L=350‒8

I_L= 342 A

 

EXAMPLE 7

A DC shunt generator supplies a load of 10 KW at 250 V. Calculate the induced emf if the armature resistance is 0.5 Ω and shunt field resistance is 100 Ω.

Given Data: D.C shunt generator, output power = 10 KW, V=250 V, R_a = 0.5 Ω, R_sh = 100 Ω.

To find: E_g = ?

SOLUTION:

I_L = output power / V = 10×10³ / 250 = 40 A

I_sh = V / R_sh = 250 / 100 = 2.5 A

∴. I_a = I_L+I_sh = 40+2.5 = 42.5 A

∴ E_g = V+I_aR_a = 250+ (42.5 × 0.5)

E_g = 271.25 V

 

EXAMPLE 8

A 4 pole lap wound DC shunt generator has a useful flux/pole of 0.6 Wb. The armature winding consists of 200 turns, each turn having resistance of 0.003 Ω, calculate the terminal voltage when running at 1000 rpm, if armature current is 45 A.

Given Data: P=4, DC shunt generator, Lap wound, A = P. ϕ=0.6 Wb, No. of turns = 200, R_T=0.003 Ω.

To Find: V=? when N = 1000 rpm, I_a = 45 A

SOLUTION

Z = 2× No. of turns

= 2×200=400

Emf induced, E_g = ϕZN/60  × P/A Volts

= [0.6×400×1000 / 60] × [4/4]

E_g = 4000 Volt

Turns per parallel path, T_p = T_a /A = 200 / 4 = 50

Resistance per parallel path, R_p = T_p × R_Ț

= 50 × 0.003

R_p = 0.15 Ω

Armature resistance, R_a = R_p/A = 0.15 / 4

R_a = 0.0375 Ω

V = E_g ‒ I_aR_a

= 4000 ‒ (45 × 0.0375)

V=3998 V

 

EXAMPLE 9

The resistance of the field circuit of DC shunt generator is 200 Ω. When the output if the generator is 100 KW, the terminal voltage is 500 V and the generated emf 525 V, calculate (i) the armature resistance and (ii) the value of the generator emf when the output is 60 KW if the terminal voltage is then 520 V.

Given Data: DC shunt generator R_sh=200Ω, output = 100 KW, V=500 V; Eg = 525 V.

To Find: R_a=?, E_g = ?, when output = 60 KW at V = 520 V.

SOLUTION:

(i) Armature drop, I_aR_a = E_g ‒ V

= 525‒500=25 V

 I_aR_a = 25 V

Load current, I_L = output / voltage = 100 × 10³ / 500 = 200 A

I_sh = V / R_sh =  500/200 = 2.5 A

I_sh = 2.5 A

∴ Armature current, I_a = I_L + I_sh = 200 +2.5.

= 202.5 A

∴ Armature Resistance, R_a = Armature drop / l_a

R_a = 25 / 202.5 = 0.123 Ω

R_a = 0.123 Ω

(ii) Output power = 60 × 10³ W

Terminal voltage V = 520 V

∴ Load current I_L = output / voltage = 60×10³ / 520

= 115.38 A

Armature current, I_a = I_L+I_sh

= 115.38+2.5

= 117.88 A

Generated emf, E_g = V + I_aR_a

= 520+ (117.88 × 0.12)

E_g = 534.15 V

 

EXAMPLE 10

A DC series generator delivers a load of 20 kW at 400 V. Its armature and series field resistance are 0.3 Ω and 0.2 Ω respectively. Calculate the generated emf and the armature current. Allow 1.0 V per brush contact drop.

Given Data: DC series generator, output power = 20 kW, V=400 V, R_a = 0.3 Ω and

R_se=0.2 Ω; V_b = 2×1 = 2V.

To Find: Eg=?, I_a= ?

SOLUTION

 I_L = output power  / V

= 20×10³ / 400

= 50 A

 I_a= I_L = 50 A

 E_g = V + I_a(R_a+R_se) + V_brush = 400 + 50(0.3+0.2) + 2

E_g = 400+25+2

 E_g = 427 V

 

EXAMPLE 11

A short shunt compound generator has armature, shunt and series field resistance of 0.5, 100 and 0.3 Ohms respectively and supplies a load of 10 KW at 230 V. Calculate the emf generated in the armature.

Given Data: R_a=0.52Ω, R_sh=100Ω, R_se = 0.32Ω, output power = 10 KW, V_L = 230 V

To Find: Eg = ?

SOLUTION:

 I_L = Output power / V_L = 10×10³ / 230 = 43.47 A

Voltage across armature, V_a = V_L + I_LR_se

V_a = 230 + (43.47 × 0.3)

V_a = 243 V

I_sh = V_a /  R_sh = 243/100 = 2.43 A

I_a=I_L+I_sh

= 43.47 + 2.43

I_a = 45.9 A

E_g = V + I_aR_a + I_LR_se

= 230 + (45.9 × 0.5) + (43.47 × 0.3)

= 230 + 22.95 + 13.04

E_g = 265.99 V

 

EXAMPLE 12

A long shunt compound generator delivers a load current of 19 A at 230 Volts. The resistance of the armature, series field and shunt field are 0.04 Ω, 0.02 Ω and 100 Ω respectively. Calculate the emf induced in the armature. Allow a brush contact drop of 1 V per brush.

Given Data: I_L = 19 A, V = 230 V, R_a = 0.04 Ω, R_se = 0.02 Ω, R_sh = 100 Ω V_brush= 2×1 = 2V

To Find: E_g = ?

SOLUTION:

 I_sh = V / R_sh = 230 / 100 = 2.3 A

I_a = I_L + I_sh =19 +2.3 = 21.3 A

I_a = I_se = 21.3 A

E_g = V+ I_aR_a + I_seR_se + V_brush

 = 230 + (21.3 × 0.04) + (21.3 × 0.02) + 2

 E_g  = 233.27 V

 

EXAMPLE 13

A 230 V DC shunt motor takes 12A at full load. The resistance of the armature and shunt field windings are 0.5 Ω and 230 Ω respectively. Determine the back emf on full load.

Given Data: V=230 V, DC shunt motor, I_L = 12 A, R_a = 0.52 Ω, R_sh = 230 Ω

To Find: E_b = ?

SOLUTION:

I_sh = V/ R_sh  = 230 / 230 = 1 A

I_a = I_L‒I_sh

= 12‒1 = 11 A

E_b=V‒I_aR_a = 230‒(11×0.5)

 E_b = 224.5 V

 

EXAMPLE 14

A 4 pole, 230 V DC shunt motor has 720 wave connected conductors in its armature. The full load armature current is 19 A, and the flux per pole is 0.03 Wb. The armature resistance is 0.2Ω, and the contact drop is 1 Volt per brush. Calculate the full load speed and power developed by the motor.

Given Data: P=4, DC shunt motor, V=230 V, Z=720, wave wound, A=2, I_a= 19 A, ϕ = 0.03 Wb, R_a = 0.2 Ω, V_brush = 2 × 1 = 2,

To Find: N=?, P=?

SOLUTION:

V = E_b + I_aR_a + Brush Drop

E_b = V ‒ l_aR_a ‒ Brush drop

= 230 ‒ (19 × 0.2) ‒ 2

 E_b = 224.2 Volts

We know back emf, E_b =  ϕZN/60  × P/A Volts

 N = E_b/ϕ (60A / PZ) rpm

= ( 224.2 × 0.03 ) × ( 60×2 / 4×720 )

 N = 311 pm

Power developed in the armature

P_m= E_bI_a

= 224.2 × 19

P_m = 4260 W

 

EXAMPLE 15

A 4 pole DC motor has a wave wound armature with 556 conductors. Determine the torque developed when the armature current is 80 A and the flux per pole is 0.037 Wb.

Given Data: P=4, Wave wound, Z= 556, I_a = 80 A, ϕ = 0.037 Wb, Since wave wound, A=2

To find: T=?

SOLUTION:

Torque developed, T = 0.159ϕZI_a(P/A)  N‒m

= 0.159 × 0.037 × 556 × 80 × 4/2

T = 523.35 Nm

 

EXAMPLE 16

A 4 pole DC motor takes an armature current of 50 A. The armature has lap connected 480 conductors. The flux per pole is 20 mWb. Calculate the gross torque developed by the motor.

Given Data: P=4, I_a = 50 A, Lap wound, Z = 480, ϕ = 20 mWb, Since Lap Wound A=P=4

To Find: T = ?

SOLUTION:

Gross torque developed, T = 0.159ϕZI_a(P/A) Nm

 = 0.159 × 20 × 10^‒3 × 480 × 50 × 4/4

 T = 76.32 Nm

 

EXAMPLE 17

A 250 V, 4 pole wave wound DC series motor has 782 conductors on its armature. It has armature and series field resistance of 0.75 ohm. The motor takes a current of 40 A. Determine its speed and gross torque developed if it has a flux per pole of 25 mWb.

Given Data: V = 250 V, P = 4, wave wound, Z=782, R_a and R_se = 0.75 ohm, I_a = 40A, ϕ = 25 mWb, since wave wound, A = 2

To Find: N, T= ?

SOLUTION:

Torque developed, T = 0.159ϕZI_a(P/A) N‒m

= 0.159 × 25 × 10^‒3 × 782 × 40 × 4/2

T=248.67 Nm

Back emf, E_b = V ‒ I_a(R_a + R_se)

E_b = 250 ‒ 40(0.75) = 220 V

E_b × I_a = 2πΝΤ / 60

N = E_b×I_a×60 / 2πΤ

N = (220×40×60) / (2×π×248.67)

N=338 rpm

 

EXAMPLE 18

Four pole, 500 V, DC shunt motor has 700 wave connected conductors on its armature. The full load armature current is 60 A and flux per pole is 30 mWb. Calculate the full load speed if the rotor armature resistance is 0.2Ω and the brush drop is 1 volt per brush.

Given Data: P=4, V=500 V; DC shunt motor, Z=700, wave wound, I_a=60A, ϕ=30 mWb, R_a=0.2 Ohms, V_b = 1 Volt per brush = 1×2 = 2V, A = 2

To Find: N=?

SOLUTION:

 E_b = V ‒ I_aR_a = 500 ‒ (60 × 0.2) = 488 V

We know,

E_b =  ϕZN/60  × P/A Volts

488 = ( 30×10^‒3×700×N /60 ) × (4/3)

N = 488×60×2 / 30×10^‒3×700×4 = 697 rpm

 N = 697 rpm

 

EXAMPLE 19

A DC motor takes an armature current of 110 A at 480 V. The armature current resistance is 0.2 ohm. The machine has 6 poles and the armature is lap connected with 864 conductors The flux per pole is 0.05 Wb. Calculate (i) the speed (ii) the gross torque developed by the armature.

Given Data: I_a=110 A, V=480 V, R_a = 0.2 Ohm, P=6, Lap connected, Z=864, ϕ=0.05 Wb, Since Lap wound, A = P = 6.

To Find: N, T=?

SOLUTION:

Torque developer, T = 0.159ϕZl_a(P/A)  N‒m

= 0.159 × 0.05 × 864 × 110 × 6/6

 T=755.56 Nm

Back emf, E_b=V‒I_aR_a

E_b=480‒(110×0.2)

= 458 V

E_bI_a= 2πΝΤ / 60

N = E_b×I_a×60 / 2πΤ

 = 458×110×60 / 2π×755.56

N=637 rpm

 

EXAMPLE 2.20

A 220 V DC motor has an armature resistance of 0.5Ω. The full load armature current is 20 A. Find the induced emf.

SOLUTION:

Induced emf, E_b=V‒I_aR_a

= 220 ‒ (20 × 0.5)

E_b = 210 volts

 

EXAMPLE 21

A four pole generator having wave wound armature winding has 51 slots, each slot containing 20 conductors. What will be the voltage generated in the machine when driven

at 1500 rpm. Assuming the flux per pole to be 7.0 mWb.

SOLUTION:

Number of conductors, (Z) = 51 × 20 = 10.20

P=4

N=1500 rpm

ϕ = 7.0 mwb = 7 × 10^‒3 Wb

E_g =  ϕZN/60  × P/A Volts

= (7×10^‒3×1020×1500 / 60) × (4/2)

 E_g = 357 V

 

EXAMPLE 22

A dc shunt generator supplies a load of 10 kW at 220 V through feeders of resistance 0.1 Ω. The resistance of armature and shunt field windings is 0.05 Ω and 100 Ω respectively Calculate the terminal voltage.

Given Data: Output power = 10 kW, V=220 V, R_feeder = 0.1 Ω, R_a=0.05 Ω, R_sh= 100 Ω

To Find: V=?

SOLUTION:

I_L = output power  / V

= 10×10³ / 200

= 45.45 A

I_sh = V / R_sh = 220/100 = 2.2 A

I_a = I_L + I_sh = 45.45 + 2.2 = 47.65A

E_g= V + I_aR_a + I_LR_feeder

=220+(47.65 × 0.05) + (45.45 × 0.1)

=220 +2.38 +4.545

 E_g = 226.925 V

 

EXAMPLE 23

A 200 V dc shunt motor takes a total current of 100 A and runs at 750 rpm. The resistance of the armature winding and of shunt field winding is 0.1 and 40 Ω respectively. Find the torque developed by the armature.

Given Data: I_L= 100 A, N=750 rpm, R_a=0.1 Ω, R_sh = 40 Ω

SOLUTION

T_a = 9.55E_bI_a / N    N‒m

 I_a = I_L‒I_sh

I_sh = V / R_sh = 220/40 = 5 A

I_a=100‒5=95 A

E_b = V ‒ I_aR_a =200‒95 × 0.1= 190.5 V

T_a = (9.55 × 190.5 × 95) / 750

T_a = 230.44 N‒m

 

EXAMPLE 24

A DC motor connected to a 460 V supply has an armature resistance of 0.15 ohms. Calculate

1. the value of back emf when the armature current is 120 A

2. the value of armature current when the back emf is 447 V

Given Data:

V= 460 V, R_a = 0.15 ohms, I_a = 120 A, E_b = 447 V

Solution

1. The value of back

E_b=V ‒ I_aR_a = 460‒120 × 0.15

= 460‒18

E_b = 442 V

2. The value of armature current

E_b = V ‒ I_aR_a

447 = 460 ‒ I_a×0.15

I_a×0.15=460‒447

I_a = 460‒447 / 0.15

 I_a = 86.67 A

 

EXAMPLE 25

A dc, shunt generator supplies a load of 7.5 kW at 200 V. Calculate the induced emf if armature resistance is 0.6 Ω and field resistance is 80 Ω.

Solution

Output power, P_out= VI_L

⇒ I_L = P_out / V = 7.5×10³ / 200

⇒ I_L = 37.5 A

I_sh=V/R_sh=200/80=2.5A

I_a=I_L+I_sh= 37.5+2.5= 40 A

Induced emf, E_g = V + I_aR_a

=200+ (40 × 0.6)

E_g =  224 V

 

EXAMPLE 26

A 8 pole, lap wound armature rotated at 350 rpm is required to generate 260 V. The useful flux / Pole is 0.05 wb. If the armature has 120 slots, calculate the number of conductors per slot.

Solution

E_b =  ϕZN/60  × P/A Volts

Conductors / Slot, Z = ( E_b60 / ϕN ) × (AP)

= (260×60 / 0.05×350) × (8/8)

⇒ Z=891.43

⇒ Z = 891 Conductors.