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
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)
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