Design Of Equipment For Shipbuilding & Submersible Industry Assignment Sample

Design Of Equipment For Shipbuilding & Submersible Industry

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Task 1: Beam Analysis

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

Taking moments about A

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

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

R_C = 67.10 kN

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

Hence, ∑V = 0

[R_C] + [R_A] = 25 + 20 + 4 x (9.5 + 3)

67.10 + [R_A] = 25 + 20 +50

R_A = 27.9 kN

Hence the support reactions become,

R_C = 67.10 kN & R_A = 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

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

X_C = 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

X_L2 = 2πfL = 2π x 1500 x 0.12 = 1130.4 Ω

Z₂ = {(R₂)² + (X_L2)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

[Ø₂] = cos^-1 (R₂/Z₂) = cos^-1 (6/1130.41) = 89.69º (Lagging)

[X_C2] = 1/2πfC = 1/(2π x 1500 x 0.35 x 10^-6) = 303.1 Ω

Z₃ = {(R₃)² + (X_C2)²}^0.5 = {12² + 303.1²}^0.5 = 303.337 Ω

V₃ = IZ₃ = 0.0422 x 303.337 = 128.008 V

[Ø₃] = cos^-1 (R₃/Z₃) = cos^-1 (12/303.337) = 89.69º (Leading)

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