MEO CLASS 4 KOCHI MMD & CLASS 2 EXAM PREPARATIONS

Recently asked questions in Kochi mmd and Class 2 Numerical solutions

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12 June 2024

212.MET December 2022 Q.9(b)

June 12, 2024 Posted by AK No comments

 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.

             If you are noticing some error in problems kindly comment below.Thanks

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

211.MET December 2022 Q.8(b)

June 12, 2024 Posted by AK No comments

 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


          If you are noticing some error in problems kindly comment below.Thanks

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

210.MET December 2022 Q7(b)

June 12, 2024 Posted by AK No comments

  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.


 If you are noticing some error in problems kindly comment below.Thanks

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

209.MET December 2022 Q.6(b)

June 12, 2024 Posted by AK No comments

 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

 IF YOU ARE NOTICING ANY ERROR KINDLY COMMENT BELOW

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

209.NA November 2022 Q.10(b)

June 12, 2024 Posted by AK No comments

 A watertight bulkhead is 8m high and is supported by vertical stiffeners 700mm apart, connected at the tank top by brackets having 10 rivets 20mm diameter. The bulkhead is flooded to its top edge with sea water. Determine: (a) Shearing force at top of stiffeners, (b) Shear stress in the rivets, (c) Position of zero shear.










11 June 2024

208.NA November 2022 Q.9(b)

June 11, 2024 Posted by AK No comments

 A vessel, when floating at a draught of 3.6 m has a displacement of 8172 tonne, KB 1.91 m and LCB 0.15 m aft of midships. From the following information, calculate the displacement, KB and position of the LCB for the vessel when floating at a draught of 1.2rn









207.NA November 2022 Q.8(b)

June 11, 2024 Posted by AK No comments

  The ½ ordinates of a water plane at 15m intervals, commencing from aft, are 1, 7, 10.5, 11, 11, 10.5, 8, 4 and 0 m. Calculate:

i. TPC

ii. Distance of the centre of flotation from midships

iii. Second moment of area of the water plane about a transverse axis through the centre of flotation.

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Solution

½ ordinate

SM

Product of area

lever

Product for 1st moment

lever

Product for 2nd moment

1

1

1

+4

+4

+4

+16

7

4

28

+3

+84

+3

+252

10.5

2

21

+2

+42

+2

+84

11

4

44

+1

+44

+1

+44

11

2

22

0

0

0

0

10.5

4

42

-1

-42

-1

+42

8

2

16

-2

-32

-2

+64

4

4

16

-3

-48

-3

+144

0

1

0

-4

0

-4

0

 

 

∑m = 190

 

∑m1 = +52

 

∑m2 = +646

 

i) Common interval h = 15m

Area of water plane Aw = 2 x ((h/3) x product of area ) --------- 2 given because 1/2 ordinates are given

                                       = 2 x (15/3) x 190

                                       = 1900 m3

TPC =  (Aw x density of water ) / 100

        = (1900 x 1.025) / 100

        = 19.475


ii)Distance of the centre of flotation from midships = (h x product of first moment) / product of area

                                                                               X  = (15 x +52) / 190

                                                                                X = 4.11 m towards aft (As value is +ve)


iii) Second moment about midship Im = (2/3) x h3 x Product of 2nd moment

                                                              = (2/3) x 153 x 646

                                                              = 1453500 m4

 

Second moment of area about centroid = Im - (Aw x X2)

                                                               = 1453500 - (1900 x 4.112)

                                                               = 1421405 m4

206.NA November 2022 Q.7(b)

June 11, 2024 Posted by AK No comments

 A 6m model of a ship has a wetted surface area of 7m2 and when towed in fresh water at 3knots has a total resistance of 35 N. Calculate the effective power of the ship, 120 m long, at its corresponding speed. n=1.825. f from formula SCF=1.15





205.NA November 2022 Q.6(b)

June 11, 2024 Posted by AK No comments

 A ship 150m long and 8.5m draught has a rudder whose area is one sixtieth of the middle-line plane and diameter of stock 320mm. Calculate the maximum speed at which the vessel may travel if the maximum allowable stress is 70 MN/m? the centre of stock 0.9m from the centre of effort and the maximum rudder angle is 35 degrees.




208.MET November 2022 Q.10(b)

June 11, 2024 Posted by AK No comments

 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