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.

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

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

Read More

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

Read More

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.

Read More

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.

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

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

Read More

204.NA October 2022 Q.9(b)

 A ship 85 m long displaces 8100 tonne when floating in seawater at draughts of 5.25 m forward & 5.55 m aft. TPC 9.0, GML 96 m, LCF 2 m aft of midships. It is decided to introduce water ballast to completely submerge the propeller & a draught aft of 5.85 m is required. A ballast tank 33 m aft of midships is available. Find the least amount of water required & the final draught forward.

IF YOU ARE NOTICING ANY ERROR KINDLY COMMENT BELOW

 

 Given

L = 85 m

Δ = 8100 t

df = 5.25 m

da = 5.55 m

TPC = 9

GML = 96m

LCF = 2 m

 

To find

i) Amount of water required to make draught aft 5.85 m

ii) Final Forward draught

 

Solution

 We know that moment to change trim one cm, MCT1 cm = (Δ x GML) / (100 x L)

                                                                                             = (8100 x 96) / (100 x 85)

                                                                                             = 91.48 t m

 

We know that trimming moment = mass moved x distance moved

In question it is given that ballast tank is 33 m aft of midship and LCF is 2 m aft of midship.

i) So distance at which water has to be taken = 33 - 2 = 31 m

So trimming moment = m x 31 

Change in trim = trimming moment / MCT1 cm

                         = (31 x m) / 91.48

                      t  =  0.339 x m -----------------------------------------------(1)

We know that bodily sinkage = mass added / TPC

                                                = m / 9 

                                                = 0.111 x m -------------------------------(2)

Distance from LCF to aft = (85/2) - 2 (As 85/2 will be mid ship. then 2 m from mid to LCF)

                                  WF = 40.5 m

Change in draught aft = (t / L) x WF

                                    = {(0.339 x m) / 85} x 40.5

                                    = 0.161 x m ------------------------------------------(3)

So new draught aft = old draught + bodily sinkage + change in draught

5.85 = 5.55 + 0.111 m + 0.161 m

5.85 = 5.55 + 0.272 m

0.272 m = 0.3

m = 1.10 = 110 t (Bodily sinkage and change in trim are in cm)

So amount of water required = 110 t

ii) So Distance from LCF to fwd = (85/2) + 2 (As 85/2 will be mid ship. then 2 m from mid to LCF)

                                                FL = 44.5 m

Change in draught forward = - (t / L) x FL

                                            = -{(0.339 x m) / 85} x 44.5

                                            = - {(0.339 x 110) / 85} x 44.5

                                            = -  19.52 cm

So new draught fwd = old draught + bodily sinkage + change in draught

                                 = 5.25 + (0.111 x 110) - 0.195

                                 = 5.25 + 0.122-0.195

                                 = 5.177 m

Read More

203.NA October 2022 Q.8(b)

 A ship 160m long and 8700 tonne displacement floats at a waterline with

Station            AP        ½         1         2         3         4         5         6         7         71/2      FP

½ ordinate       0        2.4       5.0      7.3      7.9       8.0      8.0       7.7     5.5        2.8      0m

While floating at this waterline, the ship develops a list of 100 due to instability. Calculate the

negative metacentric height when the vessel is upright in this condition.

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

Given

L = 160 m
Δ = 8700 t
List = 10 deg

To find

1.Negative metacentric height

Solution

½ Ordinates	½ Ordinate3	SM	Product
0	0	1/2	-
2.4	13.82	2	27.64
5.0	125.00	3/2	187.5
7.3	389.02	4	1556.08
7.9	493.04	2	986.08
8.0	512.00	4	2048.00
8.0	512.00	2	1024.00
7.7	456.53	4	1826.12
5.5	166.38	3/2	249.57
2.8	21.95	2	43.9
0	0	1/2	-
		TOTAL	7948.9

Common interval h = L/ no of equidistant ordinates

There are 8 equidistant sapces

h = 160 / 8

   = 20 m

Second moment of area about centre line I = (h / 9) x Total products

Since half ordinates are given.For full ordinates

I =  2 x (h/9) x Total products

  = 2 x (20/9) x 7948.9

  = 35328.44 m4

Distance from B to M, BM = (I / Δ) x 1.025

                                            = (35328.44 / 8700) x 1.025 

                                            = 4.162 m

At angle of loll tan θ = √(-2 GM/BM)

Squaring both sides tan2 θ = - 2 GM/ BM

                                    GM = - (tan2θ x BM) / 2

                                            = - 0.17632 x 4.162 / 2

Negative metacentric height = - 0.0646 m  

Read More

202.NA October 2022 Q.7(b)

 A ship 120m long displaces 10500 tonne and has a wetted surface area of 3000m2 . At 15 knots the shaft power is 4100KW, propulsive coefficient 0.6 and 55% of the thrust is available to overcome frictional resistance; calculate the shaft power required for a similar ship 140m long at the corresponding speed. ∫ = 0.42 and n = 1.825

Read More

201.NA October 2022 Q.6(b)

 An oil tanker 160m long and 22m beam floats at a draught of 9m in seawater. Cw is 0.865. The midships section is in the form of a rectangle with 1.2m radius at the bilges. A midships tank 10.5m long has twin longitudinal bulkheads and contains oil of 1.4m3/t to a depth of 11.5m. The tank is holed to the sea for the whole of its transverse section. Find the new draught

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

Given

L = 160 m
B = 22 m
d = 9 m
Cw = 0.865
For the tank
Lt = 10.5 m
dt = 11.5 m
r = 1.2 m

 To find

1.New draught

Solution

Cw = Aw /Lx B
Area of water plane, Aw = Cw x L x B
                                        = 0.865 x 160 x 22
                                        = 3045.8 m2

It is given that the tank is holed.
Therefore area lost = Lt x B
                               = 10.5 x 22
                               = 231 m2

Intact water plane area = Aw - Area lost
                                     = 3045.8 - 231
                                     = 2814.8 m2

We need to find out area of space where oil is occupied.
So first we can consider the bilge part. Both bilge part together form a half circle.(Section (3))
Therefore Area of circle = π x r2

Since it is a half circle Area = (1/2) x π x r2
                                             = (1/2) x 3.14 x 1.22

              Area of section (3) = 2.26 m2 -------------------------------------------(1)

Consider section (2)
Bredth = 22 - (1.2 + 1.2)
            = 19.6 m
depth = 1.2 m
So area of section (2) = 19.6 x 1.2
                                   = 23.52 m2--------------------------------------------------(2)

Consider section (1)
Breadth = 22 m
Depth = 11.5 - 1.2
           = 10.3 m2
So area of section (1) = 22 x 10.3
                                   = 226.6 m2---------------------------------------------------(3)

Total area of oil = (1) + (2) + (3)
                          = 226.6 + 23.52 + 2.26
                          = 252.38 m2

Area of immersion = Total area of oil - (Breadth x Depth of non immersion)
                               = 252.38 - (22 x (11.5 - 9))
                               = 197.38 m2

As we know that density of oil = 1.4 m3/t
Density = Volume  / mass (Because the unit is given as m3/t)

Total mass of oil = Volume of oil / Density
                            = Area x Lt / Density
                            = 252.38 x 10.5 / 1.4
                            = 1892.85 t

So the compartment is holed we can assume that buoyancy is lost.

Mass of buoyancy lost = Area of immersion x Lt x SW density
                                     = 197.38 x 10.5 x 1.025
                                     = 2124.30 t

Net loss in buoyancy = Mass of buoyancy lost - Total mass of oil
                                   = 2124.30 - 1892.85
                                   = 231.55 t

Equivalent volume comparing to SW = Mass / density
                                                             = 231.55 / 1.025
                                                             = 225.9 m3

Increase in draught = Volume lost in buoyancy / Area of intact water plane
                                = 225.9 / 2814.8
                                = 0.0802 m

Increase in draught = 9 + 0.0802
                                = 9.08 m

Read More