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## Design Of Equipment For Shipbuilding & Submersible Industry

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 F1 [kN] F2 [kN] F3 [kN/m] x1 [m] x2 [m] x3 [m] 25 20 4 3 9.5 3

Figure 1: Given data of task 1

RC x X2 = F1 x X1 + F2 x (X2 + X3) + F3 x (X2 + X3) x (X2 + X3)/2

RC x 9.5 = 25 x 3 + 20 x (9.5 + 3) + 4 x (9.5 + 3) x (9.5 + 3)/2

RC = 67.10 kN

It is known that the sum of total vertical force is zero.

Hence, ∑V = 0

[RC] + [RA] = 25 + 20 + 4 x (9.5 + 3)

67.10 + [RA] = 25 + 20 +50

RA = 27.9 kN

Hence the support reactions become,

RC = 67.10 kN & RA = 27.9 kN

### Task 2: Relative of mass of material under submerged condition

Here, it can be seen that one component having a mass of [A] kg was submerged in water completely. After submerging the component in water the mass of the component came out as [B] kg. The several results found from it are as follows.

Table 2: Given data of task 2

a) Mass of water displaced after submergence = [A] - [B] = 14.621 - 9.208 = 5.413

b) Taking the density of water at room temperature which is 998.2 Kg/m3

It is known that volume = mass/density

Hence, the volume of water displaced is = 5.413/998.2 = 5.42 x 10-3 m3

c) It is known that according to Archimedes' principle when a component is submerged in a liquid, the volume of the component is equal to the volume of water displaced by it.

Hence the volume of the component is 5.42 x 10-3 m3

d) It is known that volume = mass/density. The density of a material defines the mass of the material per unit volume of the material. Hence,

The Density of the component = Mass of the component / Volume of component = 14.621/5.42 x 10-3

The Density of the component = 2697.6 Kg/m3

e) It can be seen that the material that is used in building ships & in the submersible industry also having the average density to the determined figure is Aluminum. Its average density is 2739 Kg/m3

From this, it can be said that the component is Aluminum.

### Task 3: Strain of a sphere due to change in temperature

In this task, one sphere was heated. The different aspects of the change in the dimensions of the sphere are as follows.

 Material [A] B [mm] C [ºC-1] D [ºC] Aluminum 35 23.0 × 10-6 380

Table 3: Given data of task 3

Let the final radius of the sphere is r/

It is known that the linear expansion under a change in temperature is lαt.

Where

L is the initial length of the material

α = the coefficient of linear expansion of the material

t = the change in temperature

So, the equation becomes,

r/ = r + rαt = 35 + 35 x 23 x 10-6 x (380-20) = 35.289 mm.

The surface area of the sphere at 20ºC is = 4πr2 = 4π x (35)2 = 15393.8 mm2

After the change in temperature the new surface area becomes = 4[πr/2] = 4π x (35.289)2 = 15649.07 mm2

The volume of the sphere at 20ºC is 4/3[πr3] = 4/3 x π x [(35)3] = 179594.38 mm3

After the change in temperature, the new volume becomes = 4/3[πr/3] = 4/3 x π x [(35.289)3] = 184080.02 mm3

### Task 4: DC Network Analysis

 V1 [V] V2 [V] R1 [W] R2 [W] R3 [W] 9 6 4 5 12

Figure 4: Given data of task 4

Solution by Kirchhoff's law

Figure 4.1: Assumed network diagram of task 4

Applying Kirchhoff's “voltage law” to the loop AFEBA,

V1 – 4I1 – 5I2 – V2 = 0

9 – 4I1 – 5I2 – 6 = 0

4I1 + 5I2 = 3 eqn (i)

Applying Kirchhoff's “voltage law” to the loop BEDCB,

V2 - 5I2 – 12(I1 + I2) = 0

6 - 5I2 – 12I1 - 12I2 = 0

12I1 + 17I2 = 6 eqn (ii)

Doing solution of eqn (i) & (ii)

I1 = 2.625

I2 = -1.5

The sign of (I2) is negative which signifies that the actual direction of (I2) will be opposite to the assumed direction.

Hence,

Current through (R1) = (I1) = 2.625 amp

Current through (R2) = (I2) = 1.5 amp

Current through (R3) = (I1 + I2) = {(2.625) + (- 1.5)} = (2.625 - 1.5) = 1.125 amp

Solution by superposition theorem

Firstly, taking only [V1] as a source of power and replacing [V2] with internal resistance having a value equal to zero.

Figure 4.2: Assumed network after removal of 6 volt source of task 4

Total resistance = 4 + {(5 x 12)/(5+12)} = 7.529 Ω

Current passing through the 4Ω resistance (I1/) = (9/7.529) = 1.195 A

Current passing through the 5Ω resistance (I/) = 1.195 x {12/(12+5)} = 0.843 A

Current passing through the 12Ω resistance (I2/) = 1.195 x {5/(12+5)} = 0.351 A

Again, taking only [V2] as a source of power and replacing [V1] with internal resistance having a value equal to zero.

Total resistance = 12 + {(5 x 4)/(5+4)} = 14.222 Ω

Current passing through the 5Ω resistance (I//) = (6/14.222) = 0.421

Current passing through the 12Ω resistance (I2//) = 0.421 x {4/(12+4)} = 0.105 A

Current passing through the 4Ω resistance (I1//) = 0.421 x {12/(12+4)} = 0.315 A

The total current passing through each of the resistors will be the summation of the above.

Total current passing through the 4Ω resistance = (I1/) - (I1//) = 1.195 - 0.315 = 0.88 A (From F to E)

Total current passing through the 5Ω resistance = (I/) - (I//) = 0.843 - 0.421 = 0.422 A (From E to B)

Total current passing through the 12Ω resistance = (I2/) - (I2//) = 0.351 - 0.105 = 0.246 A (From D to C)

### Task 5: RLC Circuit Analysis

 1 A [V] B [kHz] R1 [W] L [mH] R2 [W] R3 [W] C [mF] 2 35 15 5 120 6 12 0.35

Figure 5: Given data of task 8

Total resistance R = 5+6+15 = 26 Ω

L = 0.12 H

XL = 2πfL = 2π x 1500 x 0.12 = 1130.4 Ω

XC = 1/2πfC = 1/(2π x 0.35 x 10-6 x 1500) = 303.1

X = 1130.4 - 303.1 = 827.3 Ω

The impedance of the circuit is,

Z = [{262 + 827.32}0.5] = 827.708 Ω

The current of the circuit is,

I = V/Z = 35/827.708 = 0.0422 A

The phase angle in between the “supplied voltage” & “circuit current” is ø = cos-1 (R/Z) = cos-1 (26/827.708) = 88.19º (Lagging)

Voltage in each of the resistors

XL2 = 2πfL = 2π x 1500 x 0.12 = 1130.4 Ω

Z2 = {(R2)2 + (XL2)2}0.5 = {62 + 1130.42}0.5 = 1130.41 Ω

V1 = IZ = 0.0422 x 827.708 = 34.929 V

V2 = IZ2 = 0.0422 x 1130.41 = 47.703 V

2] = cos-1 (R2/Z2) = cos-1 (6/1130.41) = 89.69º (Lagging)

[XC2] = 1/2πfC = 1/(2π x 1500 x 0.35 x 10-6) = 303.1 Ω

Z3 = {(R3)2 + (XC2)2}0.5 = {122 + 303.12}0.5 = 303.337 Ω

V3 = IZ3 = 0.0422 x 303.337 = 128.008 V

3] = cos-1 (R3/Z3) = cos-1 (12/303.337) = 89.69º (Leading)

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