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BTEC Level 3 Engineering: Unit 1 - Engineering Principles (Past Paper with Solutions)
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Some examples from this set of practice questions
1.
Identify the term that describes the relationship between the change in length of a material and its original length. (1 Mark) A - Stress B - Strength C - Strain D - Strip
Answer: Correct Answer C - Strain
2.
State one factor that affects the Young’s modulus of a material (1 Mark) Materials can be exposed to a range of environments and in-service conditions.
Answer: Correct Answers *Impurities in the material *Temperature of the material
3.
Which of the following is the correct answer Identify the unit of measure for pressure. A - Celcius B - Coulomb C - Litre D - Pascal
Answer: Correct Answer D - Pascal
Content preview
PearsonBTEC Level 3 National
in
Engineering
Unit 1: Engineering Principles
Sample
Assessment
Materials (SAMs)
For use with:
• Extended Certificate, Foundation Diploma, Diploma and Extended
Diploma in Engineering
• Diploma and Extended Diploma in Electrical and Electronic Engineering
• Diploma and Extended Diploma in Mechanical Engineering
• Diploma and Extended Diploma in Computer Engineering
• Diploma and Extended Diploma in Manufacturing Engineering
• Diploma and Extended Diploma in Aeronautical Engineering
First teaching from September 2016 Issue 2
,Edexcel, BTEC and LCCI qualifications
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All the material in this publication is copyright
© Pearson Education Limited 2018
, Pearson BTEC Level 3
Engineering
Sample assessment material for first teaching Paper Reference
September 2016 XXXXX/XX
Information Booklet of Formulae and
Constants Insert
You do not need any other materials.
Instructions
• You will need the information in this booklet to answer most questions.
• Read the information carefully.
• You must not write you answers in this booklet.
• your answers given in the question paper will be marked.
Only
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*S50749A*
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©2015 Pearson Education Ltd.
1/1/1/1/1/1
Pearson BTEC Level 3 Nationals in Engineering – Unit 1 – Sample Assessment Materials 1
Issue 2 – February 2018 © Pearson Education Limited 2018
,Formulae and Constants
Maths
Rules of Indices
am x an = a (m+n)
am ÷ an = a(m-n)
(am)n = amn
Rules of Logarithms
log AB = log A + logB
A
log = logA – logB
B
logAx = xlogA
Trigonometric rules
b C
a
A B
c
Sine rule
a b c sinA sinB sinC
= or =
sinA sinB sinC a b c
Cosine rule
a2 = b2+c2 – 2bc cos A
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, Volume and area of regular shapes
s
θ
r
length of an arc of a circle s = rθ (where θ is expressed in radians)
area of a sector of a circle A = ½ r2θ (where θ is expressed in radians)
r
h
volume of a cylinder v = π r2 h
total surface area of a cylinder TSA = 2 π rh + 2 π r2
r
volume of sphere v = 4/3 π r3
surface area of a sphere SA = 4 π r2
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, l h
r
volume of a cone v = 1⁄3 π r2 h
curved surface area of cone CSA = π r l
Quadratic Formula
To solve ax2 + bx + c = 0 , a ≠ 0
x = –b± √(b –4ac)
2
2a
Physical constants
Acceleration due to gravity g = 9.81m/s2
Permittivity of free space ε0 = 8.85×10−12 F/m
Permeability of free space µ0 = 4π×10−7 H/m
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, Equations of linear motion with uniform acceleration
v = final velocity, u = initial velocity, a = acceleration, t = time and s = distance
v = u + at
s = ut + ½ at2
v2 = u2 + 2as
s = ½(u + v)t
Stress and strain
Direct stress σ = F/A
Direct strain ε = ΔL/L
Shear stress τ = F/A
Shear strain γ = a/b
Young’s Modulus (modulus of elasticity) E = σ/ε
Modulus of rigidity G =τ/γ
Work, Power, Energy and Forces
Force F = ma
Components of forces Fx = Fcosθ, Fy = Fsinθ
where θ is measured from the horizontal
Mechanical work W = Fs
Mechanical power P = Fv, P = W/t
P
Mechanical Efficiency Efficiency (η) = Pinout
Force to overcome limiting friction F = µN
(where N is the normal force)
Gravitational potential energy PE = mgh
Kinetic energy KE = ½ mv2
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,Angular parameters
Centripetal acceleration a=ω²r or a= v²/r
Power P = Tω
Rotational Inertia I = k m r2
The inertial constant for a solid
cylinder (flywheel) k = ½ and for a thin
walled hollow cylinder k ≈ 1 (along the
axis of rotation).
Rotational Kinetic energy KE = ½ Iω2
Angular frequency ω =2πf
1
Frequency f=
time period
360 θ(radians)
Radians to degrees conversion θ(degrees) =
2π
where 2π radians = 360°
2 π θ(degrees)
Degrees to radians conversion θ(radians) =
360
Fluid Principles
Continuity of volumetric flow A1v1 = A2v2
Continuity of mass flow ρA1v1 = ρA2v2
Hydrostatic thrust on an immersed plane surface F = ρgAx
Density ρ = m/V
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, Static and DC Electricity theory
q
Current/electron flow I=
t
Coulomb’s law (q1q2)
F=
(4πε0 r2)
pl
Resistance R=
A
Resistance: temperature coefficient ∆R
= α∆T
R0
V
Ohm’s Law DC circuits I=
R
Total for resistors in series R =R +R +R…
T 1 2 3
1 1 1 1 …
Total for resistors in parallel = + +
RT R1 R2 R3
V2
Power P = IV, P = I2 R, P =
R
P
Electrical Efficiency Efficiency (η) = Pout
in
Kirchoff’s Current Law I = I1 + I2 + I3…
Kirchoff’s Voltage Law V = V1 + V2 +V3… or ∑PD = ∑IR
Capacitance
Electric Field Strength F V for uniform electric fields
E= or E =
q d
Capacitance C = εA
d
Time constant τ = RC
Charge stored Q = CV
Energy stored in a Capacitor W = 1 CV2
2
1 1 1 1 …
Capacitors in series c = c +c +c
T 1 2 3
Capacitors in parallel C =C +C +C …
T 1 2 3
Voltage decay on Capacitor discharge vc = Ve (–t⁄τ)
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, Magnetism and Electromagnetism
ф
Magnetic Flux Density B=
A
Magnetomotive Force Fm = NI
NI
Magnetic Field Strength or Magnetising force H=
l
B
Permeability = μ0 μr
H
Reluctance S = Fm
ф
dф dl
Induced EMF E = Blv, E = –N = –L
dt dt
1 LI2
Energy stored in an inductor W=
2
ф
Inductance of a coil L= N
I
V1 N
Transformer equation = 1
V2 N2
Single Phase Alternating Current Theory
Time Period T= 1
f
1
Capacitive reactance Xc =
2πfC
Inductive reactance XL = 2πfL
V
Ohm’s Law AC circuits I= (when voltage and current are
Z
in phase)
peak voltage
Root mean square voltage r.m.s.voltage =
√2
Total impedance of an inductor in series Z = √ X 2+ R2
L
with a resistance
Total impedance of a capacitor in series Z = √ X 2+ R2
c
with a resistance
2
Waveform average value Average value = × maximum value
π
r.m.s.value
Form factor of a waveform Form factor =
average value
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