Design of Electrical Systems (Solved Numerical Problems)

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Numerical

A 120 V DC shunt motor draws a current of 200A. The armature resistance is 0.02 ohms and the shunt field resistance is 30 ohms. Find back emf. If the lap wound armature has 90 slots with 4 conductors per slots, at what speed will the motor run when flux per pole is 0.04 Wb?​ (AMIE Summer 2023, 8 marks)

Solution

The back EMF (E_b) of a DC motor can be calculated using the formula:

E_b = V - I_aR_a 

Given:

V = 120 V

I_a = 200 A

R_a = 0.02 ohms

Substituting the values into the formula:

E_b = 120 − 200 × 0.02 = 120 − 4​ = 116 V

Now, let's calculate the speed (N) at which the motor will run using the given flux per pole (φ_p).
The formula to calculate the speed of a DC motor is:

N = 60×E_b/(P×φ_p)

Where P is the number of poles

= 2 (assuming it's a 2-pole motor)

φ_p = 0.04 Wb

Substituting the values into the formula:

N = 60×116/2×0.04 = 87000 RPM

Numerical

A 50 kW, 250 V shunt generator operates on full load at 1500 rpm. The armature has 6 poles and is lap wound with 200 turns. Find the induced emf and the flux/pole at full load. Given that the armature and field resistances are 0.01 and 125 Ω respectively. Neglect armature reaction. (AMIE Summer 2023, 8 marks)

Solution

I_L = (50 x 10³)/250 = 200 A

I_sh = V/R_sh = 250/125 = 2 A

For a shunt generator

I_a = I_L + I_sh = 202 A

Induced emf, E_g = V + I_aR_a

= 250 + 202 x 0.01 = 252.02 V

Z = 200 x 2 (As one turn = 2 conductors)

N = 1500 rpm

P = A = 6 (lap wound)

φ = E_g(60)A/PZN

= 252.02 x 60 x 6/(400 x 1500 x 6)

= 0.025205 Wb

= 25.205 mWb

Numerical

A 250 V, 25 kW, 4-pole DC generator has 328 wave connected armature conductors. When the machine is delivering full load, the brushes are given a lead of 7.2 electrical degrees. Calculate the cross magnetizing amp-turns/pole. (AMIE Summer 2023, 7 marks)

Solution

Given data:

Terminal Voltage V = 250 V

Output Power Pout = 25 kW

No. of Poles P = 4

No. of conductors Z = 328

θ_e = 7.2°

Wave connected A = 2

Cross-magnetising ampere-turns/pole. = ?

Load current supplied

I_a = (25 × 1000) / 250 = 100A

I = 100 / A = 100 / 2 = 50A (current/path)

\({\theta _m} = \frac{{2{\theta _e}}}{P} = \frac{{2x7.2}}{4} = {3.6^0}\)

Cross-magnetizing ampere-tums/pole

\(A{T_c}/Pole = ZI\left( {\frac{1}{{2p}} - \frac{{{\theta _m}}}{{360}}} \right)\)

\( = 328x50\left( {\frac{1}{{2x4}} - \frac{{3.6}}{{360}}} \right) = 1886\)

Numerical

A 220 V DC Shunt Motor runs at 500 rpm when the armature current is 50 A. Calculate the speed, if torque is doubled. Given the R = 0.2 ohm.(AMIE Summer 2023, 7 marks)

Solution

Performance eqn of DC shunt motors are
E_b = V_t - I_aR_a 

or, E_b = K_aφω

and τ = KaφI_a  

E_b1 = 220 - 50 × 0.2 = 210 V

Since the torque is double i.e., the armature current is doubled

i.e., I_a2 = 2 × I_a1 = 2 × 50 = 100 A

∴ E_b2 = 220 - 100 × 0.2 = 200 V

E_b2/E_b1 = N₂/N₁

⇒N₂ = N₁ x (E_b2/E_b1)

= 500 x (200/210)

= 476 rpm

Numerical

A single-phase transformer has 600 primary and 80 secondary turns. The mean length of the flux path in the ferromagnetic core is 1.6 m, the value of flux in the core for a magnetic field strength of 1.2 T is 425 AT/m, and the corresponding core loss is 1.5 W/kg at 50 Hz. The density of the core is 7400 kg/m3. If the maximum value of flux density is 1.2 T when the primary is connected to a 3300-V, 50-Hz supply, calculate (a) the cross-sectional area of the core, (b) the secondary voltage on no load, (c) the primary magnetizing current, and (d) the core loss. (AMIE Summer 2023, 8 marks)

Solution

(a) 3300=4.44 × 600 × 50 × Ø_m 

Ø_m = 0.0248 Wb

Cross-sectional area of the core 

= Ø_m/B_m = 0.0248/1.2

= 0.02067 m²

(b) Secondary voltage on load = 3300 × (80/600)= 440 V

(c) Primary magnetizing current = mmf/ primary turns

= (H × l)/600

= (425 × 1.6 )/ 600

= 1.133 A

Assuming sinusoidal current, the RMS value of the magnetizing current is

= 1.133/√2= 0.80 A

(d) Volume of the core = 1.6 × 0.02067 = 0.03307 m³ 

Mass of the core= 0.03307 × 7400 = 244.73 kg

Therefore, the core loss = 244.73 × 1.5= 367 W

Numerical

A 2400V/400V single-phase transformer takes a no-load current of 0.5 A and the core losses are 400 W. Determine the values of magnetising and core loss components of the no-load current. (AMIE Summer 2023, 7 marks)

Solution

V₁ = 2400 V

V₂ = 400 V

I₀ = 0.5 A

Core loss (i.e. iron loss) 

= 400 = V₁I₀cosφ₀

Hence, 400 = 2400 x 0.5 x cosφ₀

From this, cosφ₀ = 0.3333

Hence, φ₀ = 70.53⁰

The no-load diagram is shown below.

Magnetising component

I_M = I₀sinφ₀

= 0.5sin(70.53⁰) 

= 0.471 A

Core loss component

I_C = I₀cosφ₀

= 0.5cos(70.53⁰)

= 0.167 A

Numerical

A 440 V, 3-phase, 50 Hz, 4-pole, Y-connected induction motor has a full-load speed of 1425 r.p.m. The rotor has an impedance of (0 4 + j 4) ohm per phase and rotor/stator turn ratio is 0.8. Calculate (i) full-load torque (ii) rotor current and (Hi) full-load rotor Cu loss. (AMIE Summer 2023)

Solution

N_s = 120f/P = 120 x 50/4 

1500 rpm = 25 rps

s = (N_s - N)/N_s

= (1500 - 1425)/1500

= 0.05

E₁ = 440√3 = 254 V/phase

E₂ = KE₁ = 0.8 x 254 = 203.2 V

Full load torque

\({T_f} = \frac{3}{{2\pi {N_s}}}x\frac{{sE_s^2{R_2}}}{{R_2^2 + {{(s{X_2})}^2}}}\)

Putting values,

T_f = 78.87 N-m

Rotor current

\(I_2^ = \frac{{s{E_2}}}{{\sqrt {{{({R_2})}^2} + {{(s{X_2})}^2}} }} = \frac{{0.05x203.2}}{{\sqrt {{{0.4}^2} + {{(0.05x4)}^2}} }}\)

= 22.73 A

^{Total copper loss}

= 3(I₂’)²R₂ = 3 x 22.73² x 4

= 620 W

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