National Qualications AHSPECIMEN ONLY SQ10/AH/01 Engineering Science Date Not applicable Duration hours Total marks 60 Reference may be made to the Advanced Higher Engineering Science Data Booklet. SECTION 1 30 marks Attempt ALL questions SECTION 30 marks Attempt ALL questions 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. *SQ10AH01*
SECTION 1 30 marks MARKS Attempt ALL questions 1. A Wien-Bridge oscillator is intended to operate at a nominal frequency of 15 Hz. 4. 7 µf 70 Ω +V cc -V cc 4. 7 µf 70 Ω R 1 R 0 V (a) State the waveform produced by this oscillator. 1 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 R :R 1 affects the operation of the oscillator. page 0
. 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. MARKS The engineer draws up a table of tasks, precedents and times, shown below. Task Precedents Time (days) A Write program code None 5 B C Test program code on bench with sensors and motors Design and build control hardware to add to forklift A and E None 0 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 A node for an activity network is shown below. Earliest start time for task K Latest start time for task K K 4 6 1 Duration of task K (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. 3 1 [Turn over page 03
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. MARKS 3 Output lines from microcontroller 1 0 DAC Analogue voltage to 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 ±1 V DC Draw a complete circuit diagram for the DAC specified above, including all significant component values. 4 [Turn over page 04
4. A 5 m long mild-steel beam has a rectangular cross-section. It is designed for the maximum loading conditions shown.. 0 m 3. 0 knm -1. 0 knm -1. 0 m. 0 m MARKS R A R B An engineer used the free-body diagram shown below to find the maximum bending moment acting on this beam, which lies within the first. 0 m of the span, when measured from the left-hand end of the beam. 0 x <. 0 3. 0 knm -1 F S M b x R A (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, M b, measured in knm, is given by: 3 Mb = 5x x The mild-steel beam has a rectangular cross-section, with a width of 5 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. 5 mm (b) Calculate the required depth, d, of the beam. 4 d page 05
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. MARKS 114 mm t 9 mm 11 mm x x x x 114 mm 11 mm Replacement Beam Original Beam (a) Calculate the thickness, t, of the original rectangular beam. (b) State a design reason for making this replacement. 3 1 [Turn over page 06
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. MARKS 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. 005 Ω and 0. 45 Ω respectively. The core loss is. 75 kw. (b) Calculate the efficiency of the transformer. 3 page 07
SECTION 30 marks MARKS Attempt ALL questions 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 page 08 [Turn over
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. MARKS 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 CBASIC 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 55 (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 55 for this application mark: the duty cycle: between 0 (off) and 100 (fully on) (a) Using a high-level language appropriate for programming microcontrollers, write program code which will carry out the control sequence required. 4 page 09
7. (continued) The motors are driven by an L93D 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 L93D chip. (b) Copy and complete the diagram, showing all necessary connections to the microcontroller, the L93D IC and the motors. MARKS 3 Vs Vs 5 V 5 V 6 5 4 3 1 3 4 5 6 16 15 14 13 1 11 M A Microcontroller 7 8 10 9 M B L93D 0 V 0 V Extract from L93D Datasheet Enable 1 1 16 VSS Pin Description Input 1 Output 1 3 15 14 Input 4 Output 4 1 & 9 Connect HIGH to enable outputs & 7 Control inputs from microcontroller for Motor A GND 4 13 GND 10, 15 Control inputs from microcontroller for Motor A GND 5 1 GND 3, 6 Output connections for Motor A Output 6 11 Output 3 11, 14 Output connections for Motor B Input 7 10 Input 3 VS 8 9 Enable Pin-out diagram [Turn over page 10
7. (continued) MARKS (c) The footplate of the scooter is reinforced by an aluminium alloy beam. The freebody diagram for a test of the beam is shown below. 380 N 40 N 380 N 80 Nm -1 0.1 m 0.5 m 0.5 m 0.1 m 705 N 705 N (d) (i) Sketch the shear-force diagram, showing values at all significant points. (ii) Copy and complete the table below. 4 4 Distance from left-hand end (m) 0 0. 1 0. 35 Bending Moment (Nm) page 11
8. This diagram shows some detail of an intermediate shaft in the gearbox of a wind turbine. Gear 1 Gear MARKS 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. F 1 A F 1 F 0º F 0º B B,A 0 150 m Ø 0 600 m 0 50 m 0 150 m Ø 0 00 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 F 1 and F. 3 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. 4 [Turn over page 1
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. Sensor output voltage (V) 5. 0 4. 5 4. 0 3. 5 3. 0. 5. 0 1. 5 1. 0 0. 5 0. 0 Time (ms) Upper threshold Lower threshold The Schmitt trigger circuit is shown below. 5 V R 3 V in R R 1 V out 0 V page 13
8. (continued) MARKS The engineer selected a value of 10 kω for resistor R 3 and assumed that the op-amp output saturated to 3. 6 V and 0 V. (c) Calculate the required values of resistors R 1 and R 6 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] Acknowledgement of Copyright Question 7 Image from risteski goce/shutterstock.com page 14
AH National Qualifications SPECIMEN ONLY SQ10/AH/01 Engineering Science Marking Instructions These Marking Instructions have been provided to show how SQA would mark this Specimen Question Paper. 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 permissions@sqa.org.uk. 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.
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) (g) 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. 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) (j) Marks should be awarded, regardless of spelling, as long as the meaning is unambiguous. 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 two
Question Expected response 1 a Sine wave 1 1 b Max mark Additional information 1 mark recognising that minimum frequency requires maximum component values 1 mark calculating lower limit of frequency correctly Eg possible partially correct response: = 15 1.05 1.05 = 113 Hz (3 ) = 1.05 70 =1.05 4.70 10 Total of 1 mark, candidate has not checked that the component values actually produce 15 Hz. = 1 1.05 70 1.05 4.70 10 = 114 Hz (3) Eg possible partially correct response: = 1 0.95 70 0.95 4.70 10 = 139 Hz (3) Total of 1 mark, candidate has applied tolerance to component values incorrectly, working out their minimum possible values. First mark not awarded, but second mark is available as a correct calculation, although based on incorrect values. page three
Question Expected response 1 c <: When the oscillator is turned on, the oscillation will continuously decrease in amplitude. =: When the oscillator is turned on, the oscillation will grow and quickly settle at a constant-amplitude. >: When the oscillator is turned on, the oscillation will grow continuously, and the output will begin to saturate after a few cycles. Max mark Additional information Total of marks: correct description for all three conditions with critical resistor ratio clearly indicated. Total of 1 mark: correct descriptions for two of three conditions with the critical resistor ratio mentioned, or correct descriptions for all three conditions but no critical value for the resistor ratio. Total of 0 marks: only one condition mentioned, or two conditions but no critical value for the resistor ratio. If the ratio is too small, then the oscillation will die out and if it is too large the oscillation will grow until it saturates, so the resistor ratio has a critical value 1 mark page four
Question Expected response a 0 0 0 0 0 Max mark 0 13 5 5 10 18 h 0 0 3 3 5 8 0 3 5 Additional information 3 1 mark precedence of tasks is correct 1 mark earliest start time for each task is correct There may be follow-through error from first mark, if the sequence is not correct but there is branching. 1 mark latest start time for each task is correct There may be follow through from the first and second marks for the final mark if the sequence is not correct but there is branching and if the finish time has been miscalculated. Note that times for tasks shown on arrows are not necessary. b C,D,F 1 There may be follow-through error from question (a), if the sequence is not correct but there is branching, and if the pathway that takes the longest time to complete is selected. 1 mark may be awarded for a correct critical path based on the candidate s answer to (a), applying general marking instruction (c). c Bring in more staff to work on a task to catch up. Where there is slack/float in other tasks, staff can be reassigned to a critical task to help catch up. Hire/buy new equipment to speed up a task. Authorise staff overtime to catch up on the delay. If the delay is due to a shortage of parts, alternative suppliers can be found. 0 0 0 Any two relevant points page five
Question Expected response Max mark Additional information 3 4 1 mark circuit diagram: 4-input summing amplifier 0 3 1 0 + + = + + + = 1 + 1 + 1 + 1 1 mark inverting amplifier gain is either unity, or is correct for first-stage gain = 10, =5, =, = 4, = 8 10 5 = 15 1 mark calculations correct for summing amplifier feedback and input resistors 1 mark inputting resistor ratios on first stage correctly for lsb to msb and in kω range = = 10, h = 15 = 800 h = 400, = 00, = 100 = 107 page six
Question Expected response Possible alternative response R-R 0 + = 1 + 1 4 + 1 8 + 1 16 10 5 = 15 16 + Max mark Additional information 1 mark circuit diagram (does not have to include inverting amplifier) 1 mark inverting amplifier gain is either unity, or is correct for first-stage gain 1 mark calculations correct for first stage feedback and input resistors 1 mark R-R resistor ratio on first stage correct and in kω range.(* is marked R) page seven
Question Expected response 4 a 0 m 0 m 1 0 m 3 0 knm -1 0 knm -1 + RA =0 RB 5 3 4 1 1 =0 =5 3 0 knm -1 + RA, =0 3 =0 = 3 Max mark Additional information 1 mark finding the value for, the reaction at the left-hand end of the beam 1 mark using first principles to find an expression for the bending moment, in terms of the reaction at the left-hand end of the beam and the 3 knm -1 UDL page eight
Question Expected response 4 b = 0 =, = 1, = = 1 = 6 5 10 110 = 5 6 =99 Max mark Additional information 4 1 mark finding the value for from data booklet and establishing the maximum working stress 1 mark finding expressions for and 1 mark rearranging the beam bending equation and substituting to leave as unknown 1 mark calculating the value for depth of beam The final mark will involve checks on units for substituted values of breadth, yield stress and bending moment =5 = 50% = 110 =5 10 = 5 10 page nine
Question Expected response 5 a For the I-beam: = 1 1 = 114 9 = 5.5, = 114 11 = 9 = 114 114 1 5.5 9 1 = 14074668 681350 = 7.61 10 For the rectangular beam: = 114 1 = 7.61 10 =58.8 Possible alternative response, using parallel-axis theorem: = + h = 1 + h = 114 11 = 9, h = 114 11 =51.5 1 + Max mark Additional information 3 1 mark knowing how to calculate for I-beam, using either subtraction or parallel axis theorem 1 mark calculating for I-beam 1 mark calculating the necessary value of rectangular beam breadth to equate values of second moment of area h h page ten
Question Expected response 5 b The cross-sectional area of the beam is reduced. The amount of material in the beam is reduced. The weight of the beam is reduced. There is an increase in surface area on the top and bottom surfaces of the beam (providing more flexibility for potential component attachments to the beam). 6 a Using a transformer steps up voltage and steps down current without significant loss of power. Transmission of low currents, rather than high currents, reduces power losses along transmission lines. ( = ) Max mark Additional information 1 Any relevant point which is a design consideration. A response might combine all four, eg The cross-sectional area of the beam is reduced, so the amount of material in the beam is reduced, hence reducing its weight and possibly its cost. (1 mark) However, The beam may have corroded because of the wet, salty environment, and/or The beam may have been overloaded during operation are both reasons why the beam might have to be replaced, but neither response answers the question (design reason required), so (0 marks) 1 mark describing what the transformer at the power station would do, either increasing the voltage or reducing the current (or both) 1 mark explanation for stepping up voltage and stepping down current page eleven
Question Expected response 6 b = = 11 10 = 100 10 100 10 415.75 10 0.45 0.005 0.45 +11 10 97105 = 0 =8.8 Ƞ= = 8.8 11 10 100 10 =0.97 Max mark Additional information 3 1 mark explicit or implicit relationship between power input, power output and losses 1 mark calculating the output current using solution of a quadratic* 1 mark calculation of efficiency *see data booklet page twelve
Question Expected response 7 a Arduino C solution void loop() { do { value = digitalread(1); getangle(); if (ANGLE < 50) { } ERROR = 50 - ANGLE; DIRSET = true; else if (ANGLE > 50) { } ERROR = ANGLE - 50; DIRSET = false; ERROR = ERROR * 55/50; if (DIRSET == true) { analogwrite(5,error) analogwrite(3,error) } else if (DIRSET == false) { analogwrite(6,error) analogwrite(4,error) } } while (value==low); } PBasic solution main: if VAL=1 then main gosub get_angle if ANGLE<50 then ERROR=50-ANGLE DIRSET=1 else ERROR=ANGLE-50 DIRSET=0 endif ERROR=ERROR*103/ 50 if DIRSET=1 then low B.6 low B.4 pwmout B.5,55,ERROR pwmout B.3,55,ERROR else pwmout B.6,55,ERROR pwmout B.4,55,ERROR low B.5 low B.3 endif Max mark 4 Additional information 1 mark generating an error value correctly from the getangle result ANGLE 1 mark scaling the error (applying the required gain factor) 1 mark using If..Else If structure to control direction of movement Note that if(angle<=50) / if ANGLE<=50 then would work instead of solution shown, but Angle =50 must be accounted for so motor control lines are all off. 1 mark using command structure for motors (two lines on as PWM signal; two lines off) goto main page thirteen
Question Expected response Max mark Additional information 7 b 3 1 mark all power connections to Vss and 0 V and enable pin connections to 5 V 1 mark microcontroller connections to L93D 1 mark connections to motors 7 c i 4 1 mark (0 1, 705 ) 1 mark (0.1, 705 ) (0.6, 705 ) (0.1, 35 ) (0.6, 35 ) (0 1, 35 ) 1 mark (0.35, 10 ) (0.35, 10 ) (0 35, 10 ) (0.7, 0 ) 1 mark vertical, horizontal and sloping lines between points and units must be shown in some way, either like the solution given or on sketched axes. (0, 0 ) (0 35, 10 ) (0 6, 35 ) (0 6, 705 ) page fourteen
Question Expected response 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) 80( 0 1) ( 0 1) h = 0 35, = 705 0 35 380 0 5 80 0 5 = 16 8 a = No losses: =, = = 450 60 cos 0 0 =1 85 10 = 418 = cos 0 = cos 0 0 6 0 = 418 = 139 Max mark Additional information 4 1 mark stating the value of bending moment when =0 1 mark calculating the bending moment when =0 1 1 mark calculating the bending moment produced by point loads when =0 35 1 mark calculating the bending moment produced by UDL when =0 35 3 1 mark expressing torque as a product of the horizontal (tangential) component of the gear contact force and half the PCD 1 mark calculating one of the forces from the power and shaft speed given 1 mark calculating the other force by recognising that input and output torques are equal and opposite page fifteen
Question Expected response 8 b For equilibrium in the yz plane sin 0 sin 0 0 15 0 5 0 15 Taking moments about A, for equilibrium: sin 0 0.15+ sin 0 0 4 0 55=0 139 sin 0 0 15 + 418 sin 0 0 4 0 55=0 = 117 For equilibrium in xz plane cos 0 cos 0 0 15 0 5 0 15 Taking moments about A, for equilibrium: cos 0 0 15+ cos 0 0 4 0 55=0 139 cos 0 0 15 + 418 cos 0 0 4 0 55 =0 = 50 117 kn 50 = 50 + 117 = 76, =tan 117 50 =5 1 Max mark 4 page sixteen Additional information sin 0 cos 0 A sin 0 y x cos 0 B z 1 mark using moment equilibrium in two perpendicular planes (evidence of normal and tangential force components) marks establishing equations of moment equilibrium correctly in each plane 1 mark calculating the magnitude and direction of reaction at bearing
Question Expected response 8 c 5 h Max mark Additional information 6 1 mark Kirchhoff s Current Law applied at node =0 + =0 0 + 5 =0 1 mark Ohm s Law applied for each unknown current 1 mark identifying the link between two required threshold voltages and two required states 1 mark substituting the values twice to find two expressions for unknown resistances 0 h =3 6, =3 1 mark solving the first unknown value 1 mark calculating the second unknown value 3 6 3 3 0 + 5 3 10 =0 1 + 8 = 1 0.4 1 8 10 1 + 8 = 9 0 1 h =0, =1 0 1 1 0 + 5 1 10 =0 1 + 1 = 1 1 3 8 10 1 + 1 = 19 60 1 & : 9 = 46 60 =11 7 : 1 + 1 11.7 = 19 60 =4 3 page seventeen
Question Expected response 8 d The efficiency of an electromechanical system is a measure of how much of the energy supplied to the system is usefully converted. If energy is not being converted to a useful output, then it may be being converted to other forms of unwanted energy, damaging to the components in the system. OR If design is inefficient in electromechanical components, unwanted energy conversion is regularly manifested as heat of components which causes wear or more rapid failure in operation. Max mark Additional information 1 mark stating the meaning of low efficiency in an electromechanical machine, in terms of potential heating and damage/wear as the machine is operating Heat shortens the working life of components. Equipment would have to be serviced more frequently. A greater number of turbines would therefore be required to cover individual outages and this would be more expensive in terms of materials and workforce. OR Components have to be manufactured from raw materials, which generally have to be mined, and replacement can be expensive. Sustainable design must take account of material resources and inefficient electromechanical designs will use these at a greater rate over a fixed working lifespan. 1 mark linking the potential for higher risks of wear/failure to higher running costs and adverse effects on sustainability Other possible responses: If the design of the wind-turbine was inefficient, then more turbines would be required to produce a given energy output, so more materials would be used in their construction and they would have a larger footprint. (Total of 1 mark) Efficient machinery will reduce initial capital investment requirements and reduce the operational time for the investor to run into profit (1 mark), because more energy will be generated by each machine constructed and the machines are likely to require less maintenance (1 mark). [END OF SPECIMEN MARKING INSTRUCTIONS] page eighteen