Unit 3 Engineering Science Assignment Help - Higher National Certificate/Diploma in Engineering

Applications of the physical sciences to Engineering problems

Task 1.0

Analysis of scientific data

Observation
The stress strain analysis of a building under construction are taken. How much strain elongation each floor can withstand is the final outcome of this project. And based on this the safety of the persons in that building can be analysed earlier.

Data Collection
Various data are collected using the strain gauge device. The collected data is as shown below

Based on these values the strain is calculated
Results

The graphical representation is shown below

For Composite material

Task 2.0

Structural properties

The traditional thermal effects are:
Phase change, basically melting and boiling (phase transition temperatures).
Glass transition temperature.
Dimensional change, basically thermal expansion (in general, contraction if negative).
Elasto-plastic changes, due to thermal stresses.
Brittle/ductile transition temperature.
Chemical change, decomposition, oxidation, ignition.
Other physical changes as drying, segregation, outgassing, colour change, etc.
Thermal effects due to non-thermal causes: frictional heating, electrical heating, chemical heating, nuclear heating.

Plastic Material
Material Mis-selection 45%
and poor specification
Unsatisfactory design 20%
Poor processing 20%
Abuse and misuse 15%

Steel
Fatigue failure 30%
Corrosion 40%
Overload 15%
Fretting 10%
Creep 5%

Degradation may be a subjective term (e.g. hardening of ceramics by heat, or of metals by tempering, is degradation or enhancement of the material?)

Thermal degradation may be due to a mild sustained heating (giving rise to discolouration, increased creeping, thermal ageing) or to a wild heating as in fire exposure. Thermal ageing of materials is particularly important in the case of austenitic stainless steels, because chromium-carbide precipitation might occur at grain boundary, depleting the adjacent zones in chromium and hence reducing corrosion resistance.

The effect of fire on materials is a key issue in building (e.g. aluminium cannot be used as a structural building material because its collapse at high temperature). As for many of the thermal effects problems, first a purely thermal model is designed and solved, producing the input to the subsequent structural analysis; both models may be established and solved with the same tool (e.g. a finite element analyser for structures and heat transfer), or by independent tools (what may be cheaper and more versatile, at the expense of entangling the data interfacing).

Task 3.0
3.1 Consider the figure

3.11 Reaction forces at either end
R1 + R2 = 45kN
The moment about R1 is equal to R2 x 6 m = 45 kN x 4 m
R2 = (45 kN x 4) / 6 = 30 kN
R1 = 45 kN - R2 = 45 - 30 = 15 kN

3.12 If the actual weight is a UDL, the mass of the beam as 39kg/m and g = 9.81m/s2.
The total mass of the beam is 39 x 6 = 234 kg, which makes the weight of the beam
234 kg x 9.81 m/s2 = 2.2954 kN
The moment about R2 is equal to R1 x 6 m = 45 kN x 2 m + 2.29 kN x 3 m
R1 = 16.15 kN
R1 + R2 = 45kN + 2.29 kN
R2 = 31.15 kN

3.2 Applications of Buoyancy
Buoyancy explains the terminal velocity in skydiving

Consider the below figure the skydiver at terminal velocity.
Terminal Velocity

Weigth of the man displaced with terminal velocity almost same as the weight of man
80 Kg downward
The man is 80 kg(including his clothes and equipments). He is falling freely
Standard air density = 1.2kg/m3 .
Volume of air displaced by the skydiver = Velocity of Skydiver x Lower Surface Area of Skydiver
Lower Surface Area of Skydiver is calculated based on the Du Bois and Motseller formula.
Mass of Air Directly Displaced = Volume of Air Displaced x Air Density

Here in this experiment the man is 80 kg .From this table we can understand that the 80 kg man is having a terminal velocity of 66.7 m/s. A skydiver at terminal velocity to displace a mass of air equal to his mass each second.

Buoyancy in Boats
Consider the figure. The conclusion drawn from the figure is the mass of boat must be equal to the mass of water displaced. This can also be explained same as the terminal velocity of skydivers. Consider a boat in a river. The weight of the boat is exerting on water. Water will exert a force called the buoyant force to the boat. With this principle the boat does not sinks in water.

When a boat flows on water, according to the physics concepts the weight of the boat should be equal to the density of water. Boat will exert the buoyant force downwards. Both the pressure and force are due to gravity.
This can be explained by the equations
Weight = Buoyancy Force (Water Pressure)
Mass of Boat x Gravity = Mass of Water Displaced x Gravity
Mass of Boat = Mass of Water Displaced

3.3 Consider that a plastic plate is heated. First there is elastic convex deformation , and a plastic concave permanent deformation. This is due to material shrinkage due to plastic shrinkage formation.

Its main characteristics are:
Advantages: applicable to heavy plates without heavy equipment (no die, no press, no rolls), cheap (an oxyacetylene handheld torch is enough), and the bend gets thicker (more resistant) and not thinner as when rolling or hammering. The trend however is towards large automated heatingline equipment, with a heat source (preferably a laser) mounted on a gantry crane.
Handicaps: demands manual skill, high-temperature causes material degradation (grain growth, allotropic changes, species diffusion, surface reactions: carburation, nitruration, oxidation, combustion).
Materials: usually applied to thick mild-steel plates; the heating is below the transition temperature (995 K) to avoid hysteresis problems.
Geometrical parameters. The initial shape is a thick plate (D≈10..20 mm), and the location of lines, their quantity, breath (b≈10..20 mm) and depth (depending on the heating method), must be selected (up to now empirically).
Heating parameters. Heat-source input is a Gaussian profile with a global power of some 5 kW, and a size at the surface qI=q0exp(-r2/R2). The oxyacetylene torch is the cheapest source, but requires gas-flowrate-control and automated travelling (clearance distance, LT≈40 mm, and speed,
v≈10 mm/s) for constancy. Laser beam offers best control and inert atmosphere, but is more expensive. High-frequency induction depends a lot on the material, and its penetration is frequency-dependent. The welding arc is no good because the low penetration causes surface melting.
Cooling parameters. A water jet following the heat source at a distance LC≈100 mm is used, producing some hardening by quenching.

3.4 mass = 60 kg
force = 150 N
Acceleration due to gravity using De Alembert's principle =
F=mg
g = F/m=150/60=2.5m/s2

3.5 Temperature of hot water = 70 0c
external diameter = 150 mm
overall diameter = 500 mm
surface temperature = 20 0C
thermal conductivity of the lagging = 0.09 W/mK.
Heat loss =UAΔT
ΔT=70-20=50
A=2πrL=2×3.14×75×500×10-6=0.23m2
Heatloss=0.09×0.23×50=1.03J

Task 4.0
4.11
To find the current through the 4Ω resistor

Remove the load 4Ω and open circuit that point and the voltage source is replaced by short circuit.
Calculate RTh

R_OC=10+ (8 x 6)/(8+6)=10+3.43=13.43 ?
V_OC=28 V x 8/(8+6)=16 V
I_L= V_OC/(R_OC+ R_L )=16/(13.43+4)= 16/17.43=0.917 A=917 mA

4.12 Consider the diagram we need to find the current I1, I2 and I3

If 4.5 V is acting alone

Reff = 1+ [0.5¦¦ 2 ]= 1.4Ω
I1 =4.5/1.4 = 3.2A
I2 = 3.2 × 0.5/2.5 = -0.64A
By KCL
I1+I2 =I3
I3 = 3.2-0.64 = 2.56A
If 8.5 V is acting alone

Reff = [1¦¦0.5]+2 =2.5Ω
I2 = 8.5/2.5 = 3.4A
I1 = 3.4× 0.5/1.5 = -1.13 A
I3 = I1+ I2 = 3.4-1.13 = 2.27A
Now the values for these three currents are
I1 3.2 -1.13 = 2.07A
I2 3.4 -0.64 = 2.76A
I3 2.56+2.27 = 4.83A

4.2 The spectrum of waveforms harmonics

A=10 V
square wave sawtooth wave triangular wave
Vn= 2A / nπ Vn= A / nπ Vn= 4A / (nπ)2
n = 1,2,3,7,5,6,7 n = 1,2,3,4,5,6,7 n = 1,2,3,4,5,6,7
V(1) 6.37 V(1) 3.18 V(1) 4.057 1
V(2) 3.18 V(2) 1.59 V(2) 1.014 2
V(3) 2.12 V(3) 1.06 V(3) 0.451 3
V(4) 1.59 V(4) 0.80 V(4) 0.254 4
V(5) 1.27 V(5) 0.64 V(5) 0.162 5
V(6) 1.06 V(6) 0.53 V(6) 0.113 6
V(7) 0.91 V(7) 0.45 V(7) 0.083 7

The spectrum of each waves are below


4.3 Consider the figure below

4.31 Inductance = 74 mH
Resistance = 28 ohms
capacitor = 50uF capacitor
connected to a 250 V,50 Hz supply

Current flowing Z = √(?[R?^2 )+(X_L ?-X?_C)2]
XL = 2πfL = 2× 3.14×50× 74 × 10-3 = 23.2
XC = 1/2πfC = 63
(XL-XC)2 = (23.2-63.7)2 = 1640.25

R2 = (28)2 = 784
Z = 49.24
I = 250/49.24 =5.077A

Voltage across the capacitor &

Voltage across the inductor
VC= I × XC
= I × 1/wC

= I ×1/2πfC
5.077 × 63 = 319.86V
VL = I × XL = I × wL
= 5.077 × 23.2 =117.78V
Phasor diagram

4.32 resistance = 25 Ω
inductance = 100 mH
capacitor across a 120 V =0.12 uf capacitor

Resonant frequency 1/(2π√LC) = 1/(2×3.14×√100×10-3×0.12×10-6) = 1.5KHz
Q factor (9128.6 x ?10?^(-1))/25 = 36.51

4.4
4.41 Consider the figure

Transformers works by electromagnetic induction. In core type transformers as in the figure we can understand that half of LV and HV winding are on two limbs. Consider that an voltage is applied to the primary turns let that be N1. As a result a current will flow I∅. An alternating flux will be induced in the core. A voltage emf will induce in primary causing an emf to be induced in secondary also.

4.42 Efficiency
Transformer efficiency = Output/Input = (Input-losses)/Input
= 1- Losses/Input
Input = 300KVA
Efficiency = 1 - (2)×103/300000
= 0.994 = 99.4%

4.5

4.51 Consider the circuit

R = 60 ohms, L = 318.4 mH, C= 15 uF, V= 200V, 50 Hz.
Current VC = 200V
Ic = 200 ×2×3.14×50× 15× 10-6 = 0.4A
ILR = 200/(60+2×3.14×50× 318.4× 10-3 ) = 1.25A
I = 0.4+1.25 =1.65A
Current in the capacitor Ic = 200 ×2×3.14×50× 15× 10-6 = 0.4A

Supply voltage and phase angle 200 and angle 0
Circuit impedance Z =√(?[R?^2 )+(X_L ?-1/X?_C)2] = 116.53
Phasor diagram

4.52 All are validated using simulations. Same results are obtained.