ah - sqa.org.uk · reference may be made to the advanced higher engineering science data booklet....

*SQ10AH01*©

NationalQualicationsSPECIMEN ONLY AH

Total marks — 60

Reference may be made to the Advanced Higher Engineering Science Data Booklet.

SECTION 1 — 30 marks

Attempt ALL questions

SECTION 2 — 30 marks


Write your answers clearly in the answer booklet provided. In the answer booklet you must clearly identify the question number you are attempting.

Show all working and units where appropriate.

Numerical answers should include units, and be rounded to an appropriate number of significant figures.

Use blue or black ink.

Before leaving the examination room you must give your answer booklet to the Invigilator; if you do not, you may lose all the marks for this paper.

SQ10/AH/01 Engineering Science

Date — Not applicable

Duration — 2 hours

MARKSSECTION 1 — 30 marks


1. A Wien-Bridge oscillator is intended to operate at a nominal frequency of 125 Hz.

4.7 µF

4.7 µF

0 V

270 Ω R1

R2

+Vcc

-Vcc

270 Ω

(a) State the waveform produced by this oscillator.

The tolerance of the components used in the circuit is 5%.

(b) Calculate the minimum frequency at which the circuit output will oscillate.

(c) Describe how the resistor ratio R2:R1 affects the operation of the oscillator.

1

2

2

MARKS 2. An engineer, working on a project to automate the movement of a forklift truck, uses

critical path analysis to help plan the production of a prototype.

The engineer draws up a table of tasks, precedents and times, shown below.

Task Precedents Time (days)

A Write program code None 5

B Test program code on bench with sensors and motors A and E 2

C Design and build control hardware to add to forklift None 20

D Assemble all parts B, C 3

E Design and test interface circuit for microcontroller, motors and sensors None 10

F Test fully assembled automated forklift D 2

A node for an activity network is shown below.

Earliest start time for task K Duration of task K

Latest start time for task K

24

26

K

1

(a) Complete the activity network for the project, using the above information.

(b) State the order of tasks on the critical path.

(c) Describe two steps the engineer could take to ensure that, if a critical task is delayed, then the overall project is still completed on time.

[Turn over

3

1

2

MARKS 3. The microcontroller in an air-conditioning system accurately controls the speed of a

motor-driven fan. A 4-bit digital-to-analogue converter (DAC) allows the motor speed to be varied. A block diagram of the DAC is shown below.

DAC

3

2

1

0

Output linesfrom microcontroller

Analogue voltageto motor

The DAC has the following specification:

• It is constructed from operational amplifiers and appropriate resistors.

• The outputs from the microcontroller are each 0 V or 5 V

• The maximum output from the DAC should be 10 V

• The least significant bit (lsb) is line 0

• The system power supply is ±12 V DC

Draw a complete circuit diagram for the DAC specified above, including all significant component values.

[Turn over

4

MARKS 4. A 5 m long mild-steel beam has a

rectangular cross-section.

It is designed for the maximum loading conditions shown.

An engineer used the free-body diagram shown below to find the maximum bending moment acting on this beam, which lies within the first 2.0 m of the span, when measured from the left-hand end of the beam.

3.0 kNm-1

RA

Mb

FS

x

0 ≤ x < 2.0

(a) By reference to the two diagrams above, show that, at a distance x metres from the left-hand end of the beam, the bending moment, Mb, measured in kNm, is given by:

23= 5

2−bM x x

The mild-steel beam has a rectangular cross-section, with a width of 25 mm. The beam is designed such that the maximum stress in the beam is 50% of the material’s yield stress for a maximum bending moment of 5 kNm.

(b) Calculate the required depth, d, of the beam.

2.0 m

3.0 kNm-1 2.0 kNm-1

2.0 m 2.0 m

RA RB

2

d

25 mm

4

MARKS 5. A rectangular beam used to support a module on an oil-rig is replaced by an I-beam

of the same material, length, depth and stiffness. Details of the beam cross-sections are shown below.

114 mm

ReplacementBeam

OriginalBeam

t

x x x x

9 mm

11

mm

114

mm

11

mm

(a) Calculate the thickness, t, of the original rectangular beam.

(b) State a design reason for making this replacement.

[Turn over

3

1

MARKS 6. Step-up transformers are a necessary part of the National Grid, which enables

electricity transmission.

(a) Explain the major benefit of including a step-up transformer at a power station for the transmission of electricity through the National Grid.

A single-phase step-up transformer has an input power rating of 100 kVA and increases the 415 V AC input voltage to 11000 V AC at its output. The primary and secondary windings of the transformer have resistances of 0.0025 Ω and 0.45 Ω respectively. The core loss is 2.75 kW.

(b) Calculate the efficiency of the transformer.

2

3

MARKSSECTION 2 — 30 marks


7. A self-balancing electric scooter has two motors which control the movement and balance of the machine.

Any movement causing the scooter to tilt forwards or backwards is sensed using a triple-axis electronic gyroscope.

The signal from the gyroscope is processed by a microcontroller which then controls the speed and direction of the two motors, using pulse width modulation.

During the development of the scooter’s control, a section of a test program was written to meet the following specification:

• The tilt angle is checked.

• After the tilt angle is checked, the variable ANGLE holds a value between 0 and 100

• If ANGLE < 50, then the variable ERROR = 50 − ANGLE and the motors are driven forwards.

• If ANGLE >= 50, then the variable ERROR = ANGLE − 50 and the motors are driven backwards.

• ERROR should be multiplied by a gain factor so that, when ERROR is a maximum, it holds the value required for the motors to be fully on.ERROR is used as the value of duty cycle for a PWM command detailed below, in order to control the speed of the motors.

• The sequence should repeat continuously until the stop switch is pressed.

Pin connections

Input Pin number Output

6 Left motor A (forward)

5 Left motor A (backward)

4 Right motor B (forward)

3 Right motor B (backward)

Stop button 2

[Turn over

MARKS 7. (continued)

Notes

• Assume ANGLE and ERROR have been defined as integer variables, which can take values in the range 0–65535.

• For a test program written in a C-based language, the tilt angle is checked by calling the function getAngle().

• For a test program written in a form of BASIC, the tilt angle is checked by calling the sub-procedure get_angle.

The following information is provided for reference. Assume that both commands are available for the output pins identified in the “Pin Connections” table above.

Arduino C BASIC

Command

analogWrite()

Description

Writes a PWM signal to a pin until the next call to analogueWrite() or digitalWrite() on the same pin.

Syntax

analogWrite(pin,value)

Parameters

pin: a variable/constant which is the pin to write to

value: the duty cycle: between 0 (off) and 255 (fully on)

Command

pwmout

Description

Writes a continuous PWM signal to a pin until another pwmout command is sent to the pin.

Syntax

pwmout pin,period,mark

Parameters

pin: a variable/constant which is the pin to write to

period: set to 255 for this application

mark: the duty cycle: between 0 (off) and 1020 (fully on)

(a) Using a high-level language appropriate for programming microcontrollers, write program code which will carry out the control sequence required. 4


The motors are driven by an L293D Push-Pull 4-channel driver integrated circuit. An incomplete wiring diagram is given below, with a pin-out diagram and some accompanying notes on interfacing a microcontroller with the L293D chip.

(b) Copy and complete the diagram, showing all necessary connections to the microcontroller, the L293D IC and the motors.

M

M

1

2

3

4

5

6

7

8

6

5

4

3

16

15

14

13

12

11

10

9

A

BMicrocontroller

L293D

0 V 0 V

5 VVs Vs 5 V

1

2

3

4

5

6

7

8

Enable 1

Input 1

Output 1

GND

GND

Output 2

Input 2

VS

16

15

14

13

12

11

10

9

VSS

Input 4

Output 4

GND

GND

Output 3

Input 3

Enable 2

Pin Description

1 & 9 Connect HIGH to enable outputs

2 & 7 Control inputs from microcontroller for Motor A

10, 15 Control inputs from microcontroller for Motor A

3, 6 Output connections for Motor A

11, 14 Output connections for Motor B

Extract from L293D Datasheet

Pin-out diagram

[Turn over

3


(c) The footplate of the scooter is reinforced by an aluminium alloy beam. The free-body diagram for a test of the beam is shown below.

0.25 m 0.25 m 0.1 m0.1 m

820 Nm-1

380 N

705 N 705 N

240 N 380 N

(d) (i) Sketch the shear-force diagram, showing values at all significant points.

(ii) Copy and complete the table below.

Distance from left-hand end (m) 0 0.1 0.35

Bending Moment (Nm)

4

4

MARKS 8. This diagram shows some detail

of an intermediate shaft in the gearbox of a wind turbine.

The intermediate shaft of the gearbox is fitted with two spur gears. The forces acting on each gear when the shaft is being driven at constant speed are shown in the diagrams below.

F2

F2

F1

F1

A

0∙150 m

0∙250 m

0∙150 m

20º

20º

BB,A

Ø 0∙600 m

Ø 0∙200 m

Assume that there are no energy losses in the gearbox. When the input power is 1.85 MW, the intermediate shaft rotates at 450 rev min-1.

(a) Calculate forces F1 and F2.

Roller bearings are located at the points A and B shown in the previous diagram.

(b) Calculate the magnitude and direction of the reaction at bearing B.

[Turn over

Gear 2

Gear 1

3

4

8. (continued)

During testing, the speed of the gearbox output shaft is monitored by means of a toothed wheel and magnetic sensor, as shown.

An electronic engineer uses a Schmitt trigger to create a digital signal from the analogue signal produced by the magnetic sensor.

A graph of the signal from the sensor is shown below. Selected threshold voltages for the Schmitt trigger circuit are also shown.

Upper threshold

Lower threshold

Time (ms)

Sens

or o

utpu

t vo

ltag

e (V

)

5.04.54.03.53.02.52.01.51.00.50.0

The Schmitt trigger circuit is shown below.

5 V

Vin Vout

0 V

R3

R2

R1


The engineer selected a value of 10 kΩ for resistor R3 and assumed that the op-amp output saturated to 3.6 V and 0 V.

(c) Calculate the required values of resistors R1 and R2

It may seem unnecessary to design efficient drive trains, generators and transformers for wind turbines when they are used to produce energy from a renewable resource.

(d) Explain why optimising the efficiency of these components remains a necessary design consideration.

[END OF SPECIMEN QUESTION PAPER]

6

2

Acknowledgement of CopyrightQuestion 7 Image from risteski goce/Shutterstock.com

SQ10/AH/01 Engineering Science

Marking Instructions

The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis. If it is to be used for any other purpose, written permission must be obtained from SQA’s Marketing team on [email protected].

Where the publication includes materials from sources other than SQA (ie secondary copyright), this material should only be reproduced for the purposes of examination or assessment. If it needs to be reproduced for any other purpose it is the user’s responsibility to obtain the necessary copyright clearance.

These Marking Instructions have been provided to show how SQA would mark this Specimen Question Paper.

©

AH NationalQualificationsSPECIMEN ONLY

page two

General Marking Principles for Advanced Higher Engineering Science

This information is provided to help you understand the general principles you must apply when marking candidate responses to questions in this Paper. These principles must be read in conjunction with the Detailed Marking Instructions, which identify the key features required in candidate responses.

(a) Marks for each candidate response must always be assigned in line with these General Marking Principles and the Detailed Marking Instructions for this assessment.

(b) Marking should always be positive. This means that, for each candidate response, marks are accumulated for the demonstration of relevant skills, knowledge and understanding: they are not deducted from a maximum on the basis of errors or omissions.

(c) Where a candidate makes an error at an early stage in a multi-stage calculation, credit should normally be given for correct follow-on working in subsequent stages, unless the error significantly reduces the complexity of the remaining stages. The same principle should be applied in questions which require several stages of non-mathematical reasoning.

(d) All units of measurement will be presented in a consistent way, using negative indices where required (eg m s-1). Candidates may respond using this format, or solidus format (m/s), or words (metres per second), or any combination of these (eg metres/second).

(e) Answers to numerical questions should normally be rounded to an appropriate number of significant figures. However, the mark can be awarded for answers which have up to two figures more or one figure less than the expected answer.

(f) Unless a numerical question specifically requires evidence of working to be shown, full marks should be awarded for a correct final answer (including unit) on its own.

(g) A mark can be awarded when a candidate writes down the relevant formula and substitutes correct values into the formula. No mark should be awarded for simply writing down a formula, without any values.

(h) Credit should be given where a labelled diagram or sketch conveys clearly and correctly the response required by the question.

(i) Marks should be awarded, regardless of spelling, as long as the meaning is unambiguous.

(j) Candidates may answer programming questions in any appropriate programming language. Marks should be awarded, regardless of minor syntax errors, as long as the intention of the coding is clear.

(k) Where a question asks the candidate to “explain”, marks should only be awarded where the candidate goes beyond a description, for example by giving a reason, or relating cause to effect, or providing a relationship between two aspects.

(l) Where separate space is provided for rough working and a final answer, marks should normally only be awarded for the final answer, and all rough working ignored.

page

thr

ee

Que

stio

n Ex

pect

ed r

espo

nse

Max

m

ark

Add

itio

nal i

nfor

mat

ion

1 a

Si

ne w

ave

1

1 b

=1.05×270

=1.05×4.70

×10

=1

2××1.05×

270×1.05×

4.70×10

=114

Hz(3)

2 1

mar

k —

rec

ogni

sing

tha

t m

inim

um f

requ

ency

req

uire

s m

axim

um

com

pone

nt v

alue

s 1

mar

k —

cal

cula

ting

low

er lim

it o

f fr

eque

ncy

corr

ectl

y __

____

____

____

____

____

____

____

____

____

____

____

____

____

___

Eg p

ossi

ble

part

ially

corr

ect

resp

onse

:

=125 1.05×1.05

=113Hz(3

) Tot

al o

f 1

mar

k, c

andi

date

has

not

che

cked

tha

t th

e co

mpo

nent

va

lues

act

ually

prod

uce

125

Hz.

Eg

pos

sibl

e pa

rtia

lly

corr

ect

resp

onse

:

=1

2××0.95×

270×0.95×

4.70×10

=139Hz(3

) Tot

al o

f 1

mar

k, c

andi

date

has

app

lied

tol

eran

ce t

o co

mpo

nent

va

lues

inc

orre

ctly

, w

orki

ng o

ut t

heir

min

imum

pos

sibl

e va

lues

.

Firs

t m

ark

not

awar

ded,

but

sec

ond

mar

k is

ava

ilab

le a

s a

corr

ect

calc

ulat

ion,

alt

houg

h ba

sed

on inc

orre

ct v

alue

s.

page

fou

r

Que

stio

n Ex

pect

ed r

espo

nse

Max

m

ark

Add

itio

nal i

nfor

mat

ion

1 c

<2: W

hen

the

osci

llat

or is

turn

ed o

n, t

he

osci

llat

ion

will co

ntin

uous

ly d

ecre

ase

in

ampl

itud

e.

=2: Whe

n th

e os

cillat

or is

turn

ed o

n, t

he

osci

llat

ion

will gr

ow a

nd q

uick

ly s

ettl

e at

a

cons

tant

-am

plit

ude.

>2: Whe

n th

e os

cillat

or is

turn

ed o

n, t

he

osci

llat

ion

will gr

ow c

onti

nuou

sly,

and

the

ou

tput

will be

gin

to s

atur

ate

afte

r a

few

cy

cles

.

2 Tot

al o

f 2

mar

ks:

cor

rect

des

crip

tion

for

all t

hree

con

diti

ons

wit

h cr

itic

al r

esis

tor

rati

o cl

earl

y in

dica

ted.

Tot

al o

f 1

mar

k: co

rrec

t de

scri

ptio

ns f

or t

wo

of t

hree

con

diti

ons

wit

h th

e cr

itic

al r

esis

tor

rati

o m

enti

oned

, or

cor

rect

des

crip

tion

s fo

r al

l th

ree

cond

itio

ns b

ut n

o cr

itic

al v

alue

for

the

res

isto

r ra

tio.

Tot

al o

f 0

mar

ks:

onl

y on

e co

ndit

ion

men

tion

ed,

or t

wo

cond

itio

ns

but

no c

riti

cal va

lue

for

the

resi

stor

rat

io.

“I

f th

e ra

tio

is t

oo s

mal

l, t

hen

the

osci

llat

ion

wil

l di

e ou

t an

d if

it

is

too

larg

e th

e os

cill

atio

n w

ill

grow

unt

il i

t sa

tura

tes,

so

the

resi

stor

rat

io h

as a

cri

tica

l va

lue”

— 1

mar

k

page

fiv

e

Que

stio

n Ex

pect

ed r

espo

nse

Max

m

ark

Add

itio

nal i

nfor

mat

ion

2 a

3 1

mar

k —

pre

cede

nce

of t

asks

is

corr

ect

1 m

ark

— e

arlies

t st

art

tim

e fo

r ea

ch t

ask

is c

orre

ct

The

re m

ay b

e fo

llow

-thr

ough

err

or f

rom

fir

st m

ark,

if

the

sequ

ence

is

not

cor

rect

but

the

re is

bran

chin

g.

1 m

ark

— lat

est

star

t ti

me

for

each

tas

k is

cor

rect

The

re m

ay b

e fo

llow

thr

ough

fro

m t

he f

irst

and

sec

ond

mar

ks f

or

the

fina

l m

ark

if t

he s

eque

nce

is n

ot c

orre

ct b

ut t

here

is

bran

chin

g an

d if

the

fin

ish

tim

e ha

s be

en m

isca

lcul

ated

.

Not

e th

at t

imes

for

tas

ks s

how

n on

arr

ows

are

not

nece

ssar

y.

2 b

C,D

,F

1

The

re m

ay b

e fo

llow

-thr

ough

err

or f

rom

que

stio

n 2(

a),

if t

he

sequ

ence

is

not

corr

ect

but

ther

e is

bra

nchi

ng,

and

if t

he p

athw

ay

that

tak

es t

he lon

gest

tim

e to

com

plet

e is

sel

ecte

d.

1 m

ark

may

be

awar

ded

for

a co

rrec

t cr

itic

al p

ath

base

d on

the

ca

ndid

ate’

s an

swer

to

2(a)

, ap

plyi

ng g

ener

al m

arki

ng ins

truc

tion

(c

).

2 c

Brin

g in

mor

e st

aff

to w

ork

on a

tas

k to

cat

ch u

p.

Whe

re t

here

is

slac

k/fl

oat

in o

ther

tas

ks,

staf

f ca

n be

rea

ssig

ned

to a

cri

tica

l ta

sk t

o he

lp c

atch

up.

Hir

e/bu

y ne

w e

quip

men

t to

spe

ed u

p a

task

.

Aut

hori

se s

taff

ove

rtim

e to

cat

ch u

p on

the

del

ay.

If

the

del

ay is

due

to a

sho

rtag

e of

par

ts,

alte

rnat

ive

supp

lier

s ca

n be

fou

nd.

2 Any

tw

o re

leva

nt p

oint

s

0 80 135 18

0 0

20 2023 23

00 0

ℎ

0 0

5 10

203

22

25 25

page

six

Que

stio

n Ex

pect

ed r

espo

nse

Max

m

ark

Add

itio

nal i

nfor

mat

ion

3

4 1

mar

k —

cir

cuit

dia

gram

: 4-

inpu

t su

mm

ing

ampl

ifie

r

=−

×+++

×−

=−×

1 +1+1 +

1 ×−

=10,=

5,= 2,

= 4,= 8

10 5=

×15×

==

10,ℎ

=2 15

=8

00ℎ

=400,=2

00,=100

=107

1 m

ark

— inv

erti

ng a

mpl

ifie

r ga

in is

eith

er u

nity

, or

is

corr

ect

for

firs

t-st

age

gain

1

mar

k —

cal

cula

tion

s co

rrec

t fo

r su

mm

ing

ampl

ifie

r fe

edba

ck a

nd

inpu

t re

sist

ors

1 m

ark

— inp

utti

ng r

esis

tor

rati

os o

n fi

rst

stag

e co

rrec

tly

for

lsb

to

msb

and

in

k Ω r

ange

1

− +

− + 0 2 3 0

page

sev

en

Que

stio

n Ex

pect

ed r

espo

nse

Max

m

ark

Add

itio

nal i

nfor

mat

ion

Po

ssib

le a

lter

nati

ve r

espo

nse

R-2

R

1 m

ark

— c

ircu

it d

iagr

am (

does

not

hav

e to

inc

lude

inv

erti

ng

ampl

ifie

r)

1 m

ark

— inv

erti

ng a

mpl

ifie

r ga

in is

eith

er u

nity

, or

is

corr

ect

for

firs

t-st

age

gain

1 m

ark

— c

alcu

lati

ons

corr

ect

for

firs

t st

age

feed

back

and

inp

ut

resi

stor

s

1 m

ark

— R

-2R r

esis

tor

rati

o on

fir

st s

tage

cor

rect

and

in

k Ω

rang

e.(*

is

mar

ked

2R)

+

− +

2 2

2 2

−

0 ∗ 2

=−×

1 2+1 4+1 8+1 16×−

10 5=×15 16×

page

eig

ht

Que

stio

n Ex

pect

ed r

espo

nse

Max

m

ark

Add

itio

nal i

nfor

mat

ion

4 a

=0 ×5−3

×2×4−2×1

×1 2=0 =5

,

=0

×−3× 2−

=0 =−3 2

2 1

mar

k —

fin

ding

the

val

ue f

or

, th

e re

acti

on a

t th

e le

ft-h

and

end

of t

he b

eam

1

mar

k —

usi

ng f

irst

pri

ncip

les

to f

ind

an e

xpre

ssio

n fo

r th

e be

ndin

g m

omen

t, in

term

s of

the

rea

ctio

n at

the

lef

t-ha

nd e

nd o

f th

e be

am a

nd t

he 3

kN

m-1 U

DL

3·0

kNm

-1

R

A

+

RA

2·0

m

3·0

kNm

-1

RB

1·0

m

2·0

kNm

-1

2·0

m

+

page

nin

e

Que

stio

n Ex

pect

ed r

espo

nse

Max

m

ark

Add

itio

nal i

nfor

mat

ion

4 b

=220

=,= 12

,= 2

= 122=

6

5×10 110=

25 6

=99

4 1

mar

k —

fin

ding

the

val

ue f

or

fro

m d

ata

book

let

and

esta

blis

hing

the

max

imum

wor

king

str

ess

1 m

ark

— f

indi

ng e

xpre

ssio

ns f

or

and

1 m

ark

— r

earr

angi

ng t

he b

eam

ben

ding

equ

atio

n an

d su

bsti

tuti

ng

to lea

ve

as unkn

own

1 m

ark

— c

alcu

lati

ng t

he v

alue

for

dep

th o

f be

am

The

fin

al m

ark

will in

volv

e ch

ecks

on

unit

s fo

r su

bsti

tute

d va

lues

of

bre

adth

, yi

eld

stre

ss a

nd b

endi

ng m

omen

t

=25

=50%

=110

=5×10

=5×10

page

ten

Que

stio

n Ex

pect

ed r

espo

nse

Max

m

ark

Add

itio

nal i

nfor

mat

ion

5 a

Fo

r th

e I-

beam

:

= 12−2×

12

=114−9 2=

52.5,=114

−2×11=92

=114×114 12

−2×52.5×9

2 12

=1407

4668−681352

0

=7.261×10

Fo

r th

e re

ctan

gula

r be

am:

=×114 12=

7.261×10

=58.8

Po

ssib

le a

lter

nati

ve r

espo

nse

, us

ing

para

llel

-axi

s th

eore

m:

=

+×ℎ

= 12+2×

12+×ℎ

=114−2×1

1=92,ℎ

=114 2−11 2=51

.5

3 1

mar

k —

kno

win

g ho

w t

o ca

lcul

ate

for

I-be

am,

usin

g ei

ther

su

btra

ctio

n or

par

alle

l ax

is t

heor

em

1 m

ark

— c

alcu

lati

ng

for

I-be

am

1 m

ark

— c

alcu

lati

ng t

he n

eces

sary

val

ue o

f re

ctan

gula

r be

am

brea

dth

to e

quat

e va

lues

of

seco

nd m

omen

t of

are

a

ℎ ℎ

page

ele

ven

Que

stio

n Ex

pect

ed r

espo

nse

Max

m

ark

Add

itio

nal i

nfor

mat

ion

5 b

•

The

cro

ss-s

ecti

onal

are

a of

the

bea

m is

redu

ced.

•

The

am

ount

of

mat

eria

l in

the

bea

m is

redu

ced.

•

The

wei

ght

of t

he b

eam

is

redu

ced.

•

The

re is

an inc

reas

e in

sur

face

are

a on

the

top

and

bo

ttom

sur

face

s of

the

bea

m (

prov

idin

g m

ore

flex

ibilit

y fo

r po

tent

ial co

mpo

nent

att

achm

ents

to

the

beam

).

1 Any

rel

evan

t po

int

whi

ch is

a d

esig

n co

nsid

erat

ion.

A r

espo

nse

mig

ht c

ombi

ne a

ll f

our,

eg

“The

cro

ss-s

ecti

onal

are

a of

the

bea

m i

s re

duce

d, s

o th

e am

ount

of

mat

eria

l in

the

bea

m i

s re

duce

d, h

ence

red

ucin

g it

s w

eigh

t an

d po

ssib

ly i

ts c

ost.

” (1

mar

k)

How

ever

, “T

he b

eam

may

hav

e co

rrod

ed b

ecau

se o

f th

e w

et,

salt

y en

viro

nmen

t”,

and/

or

“The

bea

m m

ay h

ave

been

ove

rloa

ded

duri

ng o

pera

tion

” ar

e bo

th r

easo

ns w

hy t

he b

eam

mig

ht h

ave

to b

e re

plac

ed,

but

neit

her

resp

onse

ans

wer

s th

e qu

esti

on (

desi

gn r

easo

n re

quir

ed),

so

(0

mar

ks)

6 a

Usi

ng a

tra

nsfo

rmer

ste

ps u

p vo

ltag

e an

d st

eps

dow

n cu

rren

t w

itho

ut s

igni

fica

nt los

s of

pow

er.

Tra

nsm

issi

on o

f lo

w c

urre

nts,

rat

her

than

hig

h cu

rren

ts,

redu

ces

pow

er los

ses

alon

g tr

ansm

issi

on

lines

. ( =

)

2 1

mar

k —

des

crib

ing

wha

t th

e tr

ansf

orm

er a

t th

e po

wer

sta

tion

w

ould

do,

eit

her

incr

easi

ng t

he v

olta

ge o

r re

duci

ng t

he

curr

ent

(or

both

) 1

mar

k —

exp

lana

tion

for

ste

ppin

g up

vol

tage

and

ste

ppin

g do

wn

curr

ent

page

tw

elve

Que

stio

n Ex

pect

ed r

espo

nse

Max

m

ark

Add

itio

nal i

nfor

mat

ion

6 b

=

−

−

−

=−×−

−

11×1

0×=100

×10−100×

10 415×0.00

25−2.75

×10 −0.45×

0.45×+11×

10×−9710

5=0

=8.82

Ƞ=×

=8.82×11×

10100×

10

=0.97

3 1

mar

k —

exp

lici

t or

im

plic

it r

elat

ions

hip

betw

een

pow

er inp

ut,

pow

er o

utpu

t an

d lo

sses

1

mar

k —

cal

cula

ting

the

out

put

curr

ent

usin

g so

luti

on o

f a

quad

rati

c*

1 m

ark

— c

alcu

lati

on o

f ef

fici

ency

*s

ee d

ata

book

let

page

thi

rtee

n

Que

stio

n Ex

pect

ed r

espo

nse

Max

m

ark

Add

itio

nal i

nfor

mat

ion

7 a

A

rdui

no C

sol

utio

n

void

loop

()

do

va

lue

= di

gita

lRea

d(1)

; ge

tAng

le();

if

(AN

GLE

< 5

0)

ER

RO

R =

50

- AN

GLE

; D

IRS

ET

= tru

e;

else

if (A

NG

LE >

50)

E

RR

OR

= A

NG

LE -

50;

DIR

SE

T =

fals

e;

ER

RO

R =

ER

RO

R *

255

/50;

if

(DIR

SE

T ==

true

)

ana

logW

rite(

5,E

RR

OR

)

ana

logW

rite(

3,E

RR

OR

) el

se if

(DIR

SE

T ==

fals

e)

a

nalo

gWrit

e(6,

ER

RO

R)

a

nalo

gWrit

e(4,

ER

RO

R)

w

hile

(val

ue==

LOW

);

PBas

ic s

olut

ion

mai

n:

if V

AL=

1 th

en m

ain

gosu

b g

et_a

ngle

if

AN

GLE

<50

then

E

RR

OR

=50-

AN

GLE

D

IRS

ET=

1 el

se

ER

RO

R=A

NG

LE-5

0 D

IRS

ET=

0 en

dif

ERR

OR

=ER

RO

R*1

023/

50

if D

IRS

ET=

1 th

en

low

B.6

lo

w B

.4

pwm

out

B.5

,255

,ER

RO

R

pwm

out

B.3

,255

,ER

RO

R

else

pw

mou

t B

.6,2

55,E

RR

OR

pw

mou

t B

.4,2

55,E

RR

OR

lo

w B

.5

low

B.3

endi

f

goto

mai

n

4 1

mar

k —

gen

erat

ing

an e

rror

val

ue c

orre

ctly

fro

m t

he g

etA

ngle

re

sult

AN

GLE

1 m

ark

— s

caling

the

err

or (

appl

ying

the

req

uire

d ga

in f

acto

r)

1 m

ark

— u

sing

If…

..El

se If

……

str

uctu

re t

o co

ntro

l di

rect

ion

of

mov

emen

t

Not

e th

at “

if(A

NG

LE<=

50)”

/ “i

f AN

GLE

<=50

the

n” w

ould

wor

k in

stea

d of

sol

utio

n sh

own,

but

Ang

le =

50 m

ust

be a

ccou

nted

for

so

mot

or c

ontr

ol lin

es a

re a

ll o

ff.

1 m

ark

— u

sing

com

man

d st

ruct

ure

for

mot

ors

(tw

o line

s on

as

PW

M s

igna

l; t

wo

line

s of

f)

page

fou

rtee

n

Que

stio

n Ex

pect

ed r

espo

nse

Max

m

ark

Add

itio

nal i

nfor

mat

ion

7 b

3

1 m

ark

— a

ll p

ower

con

nect

ions

to

Vss a

nd 0

V a

nd e

nabl

e pi

n co

nnec

tion

s to

5 V

1 m

ark

— m

icro

cont

roller

con

nect

ions

to

L293

D

1 m

ark

— c

onne

ctio

ns t

o m

otor

s

7 c

i

4 1

mar

k —

1

mar

k —

1

mar

k —

1

mar

k —

ver

tica

l, h

oriz

onta

l an

d sl

opin

g line

s be

twee

n po

ints

and

un

its

mus

t be

sho

wn

in s

ome

way

, ei

ther

lik

e th

e so

luti

on

give

n or

on

sket

ched

axe

s.

(0·1,705

) (0·1

,325) (0·35

,120)

(0·35,−12

0) (0·6,−325

) (0·6

,−705)

(0.7,0)

(0,0)

( 0.1,705

) (0.6,−705

) ( 0.1

,325) (0.6

,−325)

( 0.35,120

) (0.35,−12

0)

page

fif

teen

Que

stio

n Ex

pect

ed r

espo

nse

Max

m

ark

Add

itio

nal i

nfor

mat

ion

7 c

ii 0≤≤

0·1

=705

ℎ=

0,=705

×0=0

ℎ=

0·1,=705

×0.1=70·5

0·1≤

≤0·35

=705−380(

−0·1)−820

(−0·1)(

−0·1) 2

ℎ=

0·35, =705×

0·35−380×

0·25−820×

0·25 2=126

4 1

mar

k —

sta

ting

the

val

ue o

f be

ndin

g m

omen

t w

hen

=0 1

mar

k —

cal

cula

ting

the

ben

ding

mom

ent

whe

n =0·1

1

mar

k —

cal

cula

ting

the

ben

ding

mom

ent

prod

uced

by

poin

t lo

ads

whe

n =0·3

5 1

mar

k —

cal

cula

ting

the

ben

ding

mom

ent

prod

uced

by

UD

L

whe

n =0·3

5

8 a

2

=2

No

loss

es:

=,

=

= 2×4

50 60×cos20°

×0·2 2=1·85×1

0

=418

=

cos20°×=

cos20°×

×0·6 2=418×0

·2 2

=139

3 1

mar

k —

exp

ress

ing

torq

ue a

s a

prod

uct

of t

he h

oriz

onta

l (tan

gential)

com

pone

nt o

f th

e ge

ar c

onta

ct f

orce

and

ha

lf t

he P

CD

1 m

ark

— c

alcu

lati

ng o

ne o

f th

e fo

rces

fro

m t

he p

ower

and

sha

ft

spee

d gi

ven

1 m

ark

— c

alcu

lati

ng t

he o

ther

for

ce b

y re

cogn

isin

g th

at inp

ut a

nd

outp

ut t

orqu

es a

re e

qual

and

opp

osit

e

page

six

teen

Que

stio

n Ex

pect

ed r

espo

nse

Max

m

ark

Add

itio

nal i

nfor

mat

ion

8 b

Fo

r eq

uilibr

ium

in

the

yz p

lane

Tak

ing

mom

ents

abo

ut A

, fo

r eq

uilibr

ium

:

sin20°×0.15

+sin20° ×0

·4−×0·55

=0 139sin

20° ×0·15+4

18sin20° ×0·

4−×0·55

=0 =117

Fo

r eq

uilibr

ium

in

xz p

lane

Tak

ing

mom

ents

abo

ut A

, fo

r eq

uilibr

ium

:

−cos20° ×

0·15+cos20°

×0·4−×

0·55=0

−139cos20° ×

0·15+418co

s20° ×0·4−

×0·55=0

=250

= √25

0+117=2

76,

=tan117 250=

25·1°

4

1 m

ark

— u

sing

mom

ent

equi

libr

ium

in

two

perp

endi

cula

r pl

anes

(e

vide

nce

of n

orm

al a

nd t

ange

ntia

l for

ce c

ompo

nent

s)

2 m

arks

— e

stab

lish

ing

equa

tion

s of

mom

ent

equi

libr

ium

cor

rect

ly

in e

ach

plan

e 1

mar

k —

cal

cula

ting

the

mag

nitu

de a

nd d

irec

tion

of

reac

tion

at

bear

ing

0·15 0·15

0·25 sin20°

sin20°

cos20°

cos20°

0·15 0·15

0·25

y A

B

x z

cos20°

sin20°

sin20°

cos20°

250

117

kN

page

sev

ente

en

5

0

Que

stio

n Ex

pect

ed r

espo

nse

Max

m

ark

Add

itio

nal i

nfor

mat

ion

8 c

ℎ

=0 −+=0

− −

−0 +5− =

0

ℎ =

3·6,=3·2

3·6−3·2 −

3·2−0 +5−

3·2 10=0

−1 +8 =1 0.4×1

·8 10−1

+8 =9 20

1 ℎ

=0,

=1·2

0−1·

2 −1·2−0 +

5−1·2 10=

0

1 +1=1 1·2×

3·8 101 +

1 =19 602

1

&2:9 =4

6 602=11

·7

2:1 +

1 11.7=19 60

2=4·3

6 1

mar

k —

Kir

chho

ff’s

Cur

rent

Law

app

lied

at

node

1

mar

k —

Ohm

’s L

aw a

pplied

for

eac

h un

know

n cu

rren

t 1

mar

k —

ide

ntif

ying

the

lin

k be

twee

n tw

o re

quir

ed t

hres

hold

vo

ltag

es a

nd t

wo

requ

ired

sta

tes

1 m

ark

— s

ubst

itut

ing

the

valu

es t

wic

e to

fin

d tw

o ex

pres

sion

s

for

unkn

own

resi

stan

ces

1 m

ark

— s

olvi

ng t

he f

irst

unk

now

n va

lue

1 m

ark

— c

alcu

lati

ng t

he s

econ

d un

know

n va

lue

page

eig

htee

n

Que

stio

n Ex

pect

ed r

espo

nse

Max

m

ark

Add

itio

nal i

nfor

mat

ion

8 d

The

eff

icie

ncy

of a

n el

ectr

omec

hani

cal sy

stem

is

a m

easu

re o

f ho

w m

uch

of t

he e

nerg

y su

pplied

to

the

syst

em is

usef

ully

con

vert

ed.

If e

nerg

y is

not

bei

ng

conv

erte

d to

a u

sefu

l ou

tput

, th

en it

may

be

bein

g co

nver

ted

to o

ther

for

ms

of u

nwan

ted

ener

gy,

dam

agin

g to

the

com

pone

nts

in t

he s

yste

m.

OR

If d

esig

n is

ine

ffic

ient

in

elec

trom

echa

nica

l co

mpo

nent

s, u

nwan

ted

ener

gy c

onve

rsio

n is

reg

ular

ly

man

ifes

ted

as h

eat

of c

ompo

nent

s w

hich

cau

ses

wea

r or

mor

e ra

pid

failu

re in

oper

atio

n.

Hea

t sh

orte

ns t

he w

orki

ng lif

e of

com

pone

nts.

Eq

uipm

ent

wou

ld h

ave

to b

e se

rvic

ed m

ore

freq

uent

ly.

A g

reat

er n

umbe

r of

tur

bine

s w

ould

the

refo

re b

e re

quir

ed t

o co

ver

indi

vidu

al o

utag

es a

nd t

his

wou

ld b

e m

ore

expe

nsiv

e in

ter

ms

of m

ater

ials

and

wor

kfor

ce.

OR

Com

pone

nts

have

to

be m

anuf

actu

red

from

raw

m

ater

ials

, w

hich

gen

eral

ly h

ave

to b

e m

ined

, an

d re

plac

emen

t ca

n be

exp

ensi

ve.

Sus

tain

able

des

ign

mus

t ta

ke a

ccou

nt o

f m

ater

ial re

sour

ces

and

inef

fici

ent

elec

trom

echa

nica

l de

sign

s w

ill us

e th

ese

at a

gre

ater

ra

te o

ver

a fi

xed

wor

king

lif

espa

n.

2 1

mar

k —

sta

ting

the

mea

ning

of

low

eff

icie

ncy

in a

n el

ectr

omec

hani

cal m

achi

ne,

in t

erm

s of

pot

enti

al h

eati

ng

and

dam

age/

wea

r as

the

mac

hine

is

oper

atin

g 1

mar

k —

lin

king

the

pot

enti

al f

or h

ighe

r ri

sks

of w

ear/

failur

e to

hi

gher

run

ning

cos

ts a

nd a

dver

se e

ffec

ts o

n su

stai

nabi

lity

Oth

er p

ossi

ble

resp

onse

s:

If t

he d

esig

n of

the

win

d-tu

rbin

e w

as ine

ffic

ient

, th

en m

ore

turb

ines

wou

ld b

e re

quir

ed t

o pr

oduc

e a

give

n en

ergy

out

put,

so

mor

e m

ater

ials

wou

ld b

e us

ed in

thei

r co

nstr

ucti

on a

nd t

hey

wou

ld

have

a lar

ger

foot

prin

t. (

Tot

al o

f 1

mar

k)

Effi

cien

t m

achi

nery

will re

duce

ini

tial

cap

ital

inv

estm

ent

requ

irem

ents

and

red

uce

the

oper

atio

nal ti

me

for

the

inve

stor

to

run

into

pro

fit

(1 m

ark)

, be

caus

e m

ore

ener

gy w

ill be

gen

erat

ed b

y ea

ch m

achi

ne c

onst

ruct

ed a

nd t

he m

achi

nes

are

like

ly t

o re

quir

e le

ss m

aint

enan

ce (

1 m

ark)

.

[E

ND

OF

SPEC

IMEN

MA

RKIN

G IN

STRU

CTIO

NS]

ah - sqa.org.uk · reference may be made to the advanced higher engineering science data booklet....

Documents