212.MET December 2022 Q.9(b)

 Three conductors fitted side by side in the stator of a salient-pole alternator. Each generates maximum voltage of 200V (sinusoidal). The angle subtended at the centre of the stator between adjacent conductors is 20 electrical degrees. If the three conductors are connected in series, find

(i) the r.m.s. value of the effective voltage and
(ii) the ‘breadth factor’ Using the theory that is the basis of this problem, give one reason why three-phase current has been introduced.

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Given

E1 = E2 = E3 =200V
Angle 1 = angle 2 = angle 3 = 20 deg
Max possible EMF = 600 (CONNECTED IN SERIES )

To find

i) Rms value
ii) Breadth factor

Solution

Resolving the phasor into horizontal and vertical components

Horizontal component = E1 Cos 0 + E2 cos 20 + E3 Cos 40
                                     = 200 cos 0 + 200 cos 20 + 200 cos 40
                                     = 541.14 V

Vertical component  = E1 sin 0 + E2 sin 20 + E3 sin 40
                                 = 200 sin 0 + 200 sin 20 + 200 sin 40
                                 = 196.96 V

Resultant EMF = Square root of (541.142 + 196.962)
                               = 576 V

i) RMS value = 0.707 x Max EMF
                      = 0.707 x 576
                      =  407.23 V
ii) Breadth factor = Resultant EMF / Max possible EMF
                             = 576 / (200+200+200)
                             = 0.96

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211.MET December 2022 Q.8(b)

 A three phase induction motor is wound for four poles and is supplied from a 50 Hz system. Calculate. i. The synchronous speed; ii. The speed of the rotor when the slip is 4 per cent; iii. The motor frequency when the speed of the rotor is 600 r.p.m

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Given 

No of poles P = 4
Frequency f = 50 Hz
Slip s = 4% = 0.04

To Find

 i. The synchronous speed
ii. The speed of the rotor when the slip is 4 per cent
iii. The motor frequency when the speed of the rotor is 600 r.p.m

Solution

i) Synchronous speed Ns = 120f / P
                                         = (120 x 50) / 4
                                         = 1500 rpm

                                slip s = (Ns - N) / Ns

ii) Rotor speed N = Ns (1 - s)
                             = 1500 (1 - 0.04)
                             = 1440 rpm

iii) Motor frequency when N = 600 rpm
                                    Slip s  =(Ns - N) / Ns
                                               = (1500 - 600) / 1500
                                               = 0.6

                   Motor frequency = slip x f
                                               = 0.6 x 50
                                               = 30 Hz

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210.MET December 2022 Q7(b)

  A d.c. motor takes an armature current of 110 A at 480 V. The resistance of the armature circuit is 0.2𝛀. The machine has six 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.

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Given

Ia = 110 V
V = 480 V
Ra = 0.2 ohm
No of poles , P = 6
No of conductors Z = 864
Lap connected, No of parallel path  A = P = 6
Flux per pole ø = 0.05 Wb

To find

i) Speed, N
ii) Gross torque , T

Solution

Terminal EMF , E = V - (Ia x Ra)
                              = 480 - (110 x 0.2)
                              = 458 V

i)                        E = ( ø x Z x N / 60) X (P / A)
                     
                       458 = (0.05 x 864 x N /60 ) x (6 / 6)

                        N   = 636 rpm

ii)    Gross torque  = 0.159 x ø x Z x Ia x (P / A) 

                          T  =  0.159 x 0.05 x 864 x 110 x (6 / 6)   

                         T   =  756.3 Nm

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209.MET December 2022 Q.6(b)

 A diode whose internal resistance is 20 Ω is to supply power to 1000 Ω load from 110 V rms source. Calculate i) peak load current ii) DC load current iii) AC load current

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Solution

i) We know that Vmax = Vrms x √2

                                     = 110 x √2

                                     = 155.56 V

We know that I = V / R

Here total resistance R = 20 + 1000 = 1020 Ω

So Peak load current  I = 155.56 / 1020 

                                     = 0.152 A

 

ii) DC load current = Peak load current / π

                                = 0.152 / 3.14

                                = 0.048 A

iii)  AC load current = Peak load current / 2

                                 = 0.152 / 2

                                 = 0.076 A

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208.MET November 2022 Q.10(b)

 A diode valve, having the following characteristic is connected in series with a resistor of 10,000 ohm to a 240 V d.c. supply. If a resistor of 40,000 ohms is connected between the anode and cathode, determine the current through the diode.

IA (milliamperes) 0  5.5  13  22  32    42    52   59   63
VA(Volts)             0   25   50  75  100  125 150 175 200

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207.MET November 2022 Q.9(b)

  An 18.65 KW,4-pole,50HZ, 3 phase induction motor has friction and windage losses of 2.5 percent of the output. The full load slip is 4% compute for full load (a) the rotor Cu loss (b) the rotor input (c) the shaft torque (d) the gross electromagnetic torque.

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Given

P = 18650 W

F = 50 Hz

Poles P = 4

Friction and windage loss = 2.5 % = 18650 X 0.025 = 466 KW

Slip s = 4 % = 0.04

To find

     a)      Rotor copper loss

     b)      Rotor input

     c)       Shaft torque

     d)      Gross electromagnetic torque

Solution

Losses = 466 KW

So total output PT = 18650 + 466 = 19116 KW

      a)      Rotor copper loss / Rotor total output = s / (1-s)

      Rotor copper loss / 19116 = 0.04 / (1 – 0.04)

                   Rotor copper loss = 796.6 W

      b)      Rotor input = Rotor total output + Rotor copper loss

                          = 19116 + 796.6

         Rotor input = 19912.6 W

      c)        Ns = 120 f / P

             = (120 x 50) / 4

             = 1500 rpm

Speed N = ( 1 – slip ) x Ns

               = (1 – 0.04) X 1500

               = 1440 rpm

Shaft torque TSH = (9.55 x P) / N

                            = (9.55 x 18650) / 1440

                            = 123.7 Nm

       d)      Gross torque TTOTAL  = (9.55 x PT) / N

                                            = (9.55 x 19116) / 1440

                                            = 126.8 Nm

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206.MET November 2022 Q.8(b)

 A 20 kVA , 2000/220 V single-phase transformer has a primary resistance of 2.1 ohms and a secondary resistance of 0.026 ohms. The corresponding leakage reactance are 2.5 and 0.03 ohms. Estimate the regulation at full load under power factor condition of a) Unity b) 0.5 (lagging) and c) 0.5 (leading)

Given

kVA = 20

V1 = 2000

V2 = 220

R1 =2.1

R2 = 0.026

X1 = 2.5

X2 = 0.03

To find

Estimate the regulation at full load under power factor condition of 

a) Unity 

b) 0.5 (lagging)  

c) 0.5 (leading)

Solution

We know that turn ratio = N1/N2 = V1/V2

So Turn ratio = 2000/220 = 9.09

Full load secondary current I2 = VA/V

                                                 = 20000 / 220

                                                 = 90.9 A

Equivalent resistance, R2' = R2 + R1 x (N2/N1)2

                                          = 0.026 + 2.1 x (1/9.09)2 

                                          = 0.026 + 0.0255

                                          = 0.0515 ohm

Equivalent reactance, X2' = X2 + X1 x (N2/N1)2

                                          = 0.03 + 2.5 x (1/9.09)2 

                                          = 0.03 + 0.0302

                                          = 0.0602 ohm

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205.MET November 2022 Q.7(b)

Determine the line current taken by a 440V, three-phase, star-connected motor having an output of 45kW at 0.88(lagging) power factor and an efficiency of 93 per cent

 Apparent Power (S): Apparent power is the product of voltage and current in amperes reactive (A). It represents the total electrical energy flowing in a circuit.

S = P × (1 + (PF)^2)

Where: P = Real Power (kW) = 45,000 W PF = Power Factor = 0.88 (lagging)

S = 45,000 W × (1 + (0.88)^2) = 63,761.2 W

Real Power (P): Real power is the actual useful power being consumed by the motor. It is equal to the product of apparent power and power factor.

P = S × PF P = 63,761.2 W × 0.88 = 54,937.2 W or 54.9 kW

Reactive Power (Q): Reactive power represents the electrical energy that flows back and forth between the supply source and load due to capacitive or inductive effects in the circuit. It can be calculated as follows:

Q = S × (1 - PF) Q = 63,761.2 W × (1 - 0.88) = 7,824.2 W or 7.8 kVar

Line Current: Line current can be calculated using the following formula: Iline = S / VL^2 or Iline = P / (VL^2 × PF) where VL is line voltage in volts: VL for this problem is given as 440V.**

Iline_star_connected = S / VL^2 or Iline_star_connected = P / (VL^2 × PF) Iline_star_connected = 63,761.2 W / (440V)^2 or Iline_star_connected = 54,937.2 W / (440V)^2×(0.88) Iline_star_connected≈195 A or≈195 Amperes for star-connected motors.

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204.MET November 2022 Q.6(b)

 The low-voltage release of an a.c. motor-starter consists of a solenoid into which an iron plunger is drawn against a spring. The resistance of the solenoid is 35 ohm. When connected to a 220 V, 50 Hz, a.c. 2supply the current taken is at first 2A, and when the plunger is drawn into the “full-in” position the current falls to 0.7 A. Calculate the inductance of the solenoid for both positions of the plunger, and the maximum value of flux-linkages in weber-turns for the “full-in” position of the plunger.

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Given

R = 35 ohm

V = 220 V

F = 50 Hz

Iinitial = 2 A

Ifinal = 0.7 A

To find

1. Inductance of solenoid at initial position

2. Inductance of solenoid at final(plunger in) position

3. Maximum value of flux linkage in weber turns

Solution

Initial Impedence Zinitial  = V/ Iinitial

                                         = 220 / 2

                                         = 110 ohm

Impedence 2 = Reactance2 + Resistance2

Reactance Xinitial  = √ (Zinitial2 – Resistance2)

                               = √ (1102 - 352)

                               = 104.25 ohm

            X = 2 x Pi x F x L

   104.25 = 2 x 3.14 x 50 x Linitial

1. Linitial = 104.225 / (2 x 3.14 x 50)

               = 0.332 Henry

When plunger is in Impedence Zfinal  =  V / Ifinal

                                                             = 220 / 0.7

                                                             = 314.28 ohm

Reactance Xfinal = √ (Zfinal2 – Resistance
                            = √ (314.282 - 352)

                            = 312.3 ohm

          X = 2 x Pi x F x L

   312.3  = 2 x 3.14 x 50 x Lfinal

2. Lfinal = 312.3 / (2 x 3.14 x 50)

              = 0.995 Henry

3. Maximum flux linkage = inductance x peak value of current

                                       = 0.995 x √ 2 x 0.7

                                       =  0.985 Weber Turns

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203.MET October 2022 Q.10(b)

 A 440V shunt motor takes an armature current of 30 A at 700 rev/min. The armature resistance is 0.7 ohm. If the flux is suddenly reduced 20 per cent, to what value will the armature current rise momentarily? Assuming unchanged resisting torque to motion, what will be the new steady values of speed and armature current? Sketch graphs showing armature current and speed as functions of time during the transition from initial to final, steady state conditions.

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Given

V = 400 V
Ia = 30 A
N1 = 700 
Ra = 0.7 Ω
Flux reduced to 20 %. Therefore Φ2 = 0.8 Φ1

To find

1.Momentarily increase in armature current
2.Speed & armature current
3.Graph

Solution

As we know that the voltage equation of motor is 
V = Eb + Ia Ra
440 = Eb + (30 x 0.7)
 Eb = 440 - 21
      = 419 V

We know that Eb ∝ Φ N

Consider emf Eb1 and flux Φ1 as initial values and emf Eb2 and flux Φ2 as after 20% flux reduction.
Taking rpm as constant.
Therefore Eb ∝ Φ 

Eb1 / Eb2 = Φ1 / Φ2
419 / Eb2 = Φ1 / 0.8 Φ1
         Eb2 = 335.2 V

1. Again V = Eb2 + Ia2 Ra
           440 = 335.2 + Ia2 x 0.7
   0.7 x Ia2 = 104.8
   
    Momentarily increase in armature current Ia2 = 149.71 A 

2. As we know that torque T ∝ Φ Ia
   
    It is given that after flux reduction torque remains constant.
    
    Therefore T1 = T2
              Φ1 Ia1 =  Φ2 Ia2
            Φ1 x 30 = 0.8  Φ1 x Ia2
                    Ia2 = 30 / 0.8

    Armature current Ia2 = 37.5 A

    V = Ebnew + Ia2new  Ra
    440 = Ebnew + (37.5 x 0.7)
    Ebnew = 413.75 V

    We know that Eb ∝ Φ N

     Eb1 / Ebnew = ( Φ1 / Φ2 ) x ( N1 / N2 )
     419 / 413.75 = (Φ1 / 0.8 Φ1) x (700 / N2)
           1.012 N2 = 1.25 x 700
                     N2 = 875 / 1.012

          Speed N2 = 864.6 rpm

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