BTEC Level 3 National in Engineering · BTEC Level 3 National in Engineering First teaching September 2016 Sample Marked Learner Work External Assessment Unit 1: Engineering Principles

BTEC Level 3 National in

Engineering

First teaching September 2016

Sample Marked Learner Work

External Assessment

Unit 1: Engineering Principles

In preparation for the first teaching from September 2016 and as a part of the on-going

support that we offer to our centres, we have been developing support materials to help

you better understand the application of Nationals BTEC Level 3 qualification.

What is Sample Marked Learner Work (SMLW)?

The following learner work has been prepared as guidance for centres and learners. It

can be used as a helpful tool when teaching and preparing for external units.

Each question explores two responses; one good response, followed by a poor response.

These responses demonstrate how marks can be both attained and lost.

The SMLW includes examples of real learners’ work, accompanied with examiner tips and

comments based on the responses of how learners performed.

Below displays the format this booklet follows. Each question will show a learner

response, followed by comments on the command verbs and the content of the question.

Tips may be offered where possible.

The appendix has attached a mark scheme showing all the possible responses that

perhaps were not explored in the SMLW, but can still be attained.

Tips offer helpful hints that the learner may find useful. For example:

Recommended length of the answer

Reference to the amount of marks awarded

General advice for the learner when answering questions

The red box comments on the command verbs used in the question. Command

typically means; to instruct or order for something to be done. Likewise, in

assessments, learners are required to answer questions, with the help of a

command verb which gives them a sense of direction when answering a

question.

This box may choose to highlight the command verb used and comments if the

learner has successfully done this, or not.

The green box comments on the content words and phrases. Content makes

reference to subject knowledge that originates from the specification. Learners

are required to use subject specific knowledge to answer the questions in order to gain maximum marks.

The comments may include:

Any key words/phrases used in the learner’s answer.

Why has the learner gained x amount of marks? And why/how have they

not gained any further marks?

Any suggestions/ ideas regarding the structure of the answer.

If the answer meets full marks- why it is a strong answer? What part of

the content has been mentioned to gain these marks?

You will need to refer to the appropriate specification alongside

these sample materials.

The link below will direct you to the SAMs that this SMLW refers to.

https://qualifications.pearson.com/content/dam/pdf/BTEC-

Nationals/Engineering/2016/specification-and-sample-

assessments/Sample-assessment-material-Unit-1-Engineering-

Principles.pdf

Question 1: Find, by factorisation, the time when the rocket reaches its highest

point. [Total marks for Q1- 2 marks]

The command word is find.

Good response: The learner has factorised the expression correctly. They have clearly

applied the rules of factorisation within their answer.

Good response: Two marks have been awarded for this answer.

The learner has gained one method mark for factorising the expression, and a second

mark for finding the correct answer for the time (1.5s). The learner has correctly

shown the two values for time (t) and has identified the positive value which is the

time taken to reach the highest point.

2

Poor response: The learner has factorised the expression correctly. They have

clearly applied the rules of factorisation within their response but have not found a

solution.

1

Poor response: One mark would have been awarded for this answer. The learner

has correctly factorised the expression to gain one method mark.

The learner has not however used this to find the correct value of ‘t’ therefore they

have not achieved the second mark.

There are lots of questions in this exam that require multiple

stages to be completed. It is important that learners show

these stages to gain method marks.

Learners should always read the question in full to check if

any specific techniques need to be demonstrated, as in this

question where factorisation had to be used.

Question 2: Find the co-ordinates where the grooves intersect.

[Total marks for Q2- 2 marks]

2

The command word is find.

Good response: The learner has shown their working in full to solve the simultaneous

equations to find the co-ordinates of where the grooves intersect.

Good response: Two marks have been awarded to the learner.

The learner has shown their working in full to solve the simultaneous equations to find

one variable (y) to achieve 1 mark, and has then substituted this back into the

equations to find the other variable (x) for a second mark.

0

Poor response: The learner has shown their working in full but has incorrectly

applied methods of solving simultaneous equations.

Poor response: Zero marks have been awarded to the learner.

The learner has shown their working in full to attempt to solve the simultaneous

equations to find the variables. Some stages have been completed correctly, for

example multiplying the equation -3y=1.5x-6 by 20, however they have made

arithmetic errors and have not found the correct value of y.

The value found has been substituted into equation 2, however further arithmetic errors

have been made with the negative numbers.

The learner would have gained one mark if the second variable had been correct in

relation to the first (incorrect) value of y.

Show working in full. The vast majority of questions have

method marks, or allow for errors to be carried forward.

This means learners can still get some marks even if

errors have been made at early stages.

When possible to do so learners should check their

answers for accuracy. Here the values obtained for ‘y’ and

‘x’ could be substituted back into one of the equations to

check that they have completed the calculation correctly.

If they don’t match then further work to correct one or

more errors should be completed.

Question 3: Calculate the area of the component represented by the shaded

sector. [Total marks for Q3- 2 marks]

2

The command word is calculate.

Good response: The learner has identified the correct information to use, and the

correct formula from the information booklet to calculate the area. The learner has used the correct mathematical operations in their calculations.

Good response: Two marks awarded. The learner has selected the correct formula

from the information booklet and has carried out the conversion of the angle from

degrees to radians. This gains one method mark. A second mark is awarded for the

correct area of the sector.

0

Poor response: Zero marks awarded. The learner has made a common error of not

converting the angular measurement from degrees to radians. They select the correct

formula from the information booklet this, however no credit is given. For this

question one method mark would have been awarded for the conversion to radians. A

second mark could then be awarded for the correct area of the sector.

Make sure learners are familiar with the units and

types of measurement that are used in formulae found

in the information booklet.

Encourage learners to show all stages of calculations

Question 4: Solve the equation to find the noise level of the third compressor.

Show evidence of the use of the laws of logarithms in your answer.


2

Command word is solve.

Good response: Learners are expected to find the answer to a problem, in this case

finding a missing value showing evidence of the use of the laws of logarithms.

Good response: Two marks awarded. The learner has correctly applied the laws of

logarithms at each stage of the solution. This gains one method mark. A second mark

is awarded for the correct value of the unknown variable ‘z’.

0

Poor response: Learners are expected to find the answer to a problem, in this case

finding a missing value showing evidence of the correct application of the laws of

logarithms.

Poor response: Zero marks awarded. The learner has correctly applied the laws of

logarithms for the first stage of solution, but has then not found the value of 22.

There are further errors in the application of the laws of logarithms, which prevents

the award of the method mark. A second mark is not awardable either as the

question asks for the value of ‘z’ and not ‘log z’.

Be careful to read the question and find out exactly

what is being asked for.

Make effective use of the information booklet.

Learners should simplify any equations where it is

possible to do enable calculations to be carried out

more effectively.

Question 5: Calculate the size of angle A. [Total marks for Q5- 2 marks]


Good response: The learner has identified from the information in the diagram that

this question requires the application of the cosine rule and has selected the correct

formula from the information booklet to calculate the unknown angle.

Good response: Two marks have been awarded for this answer. The learner has

correctly identified the cosine rule as being required for the calculation, and has

substituted the correct values for each variable to gain one method mark. The

correct value of the missing angle has been given for a second (accuracy) mark.

0

2

Poor response: The learner has not identified from the information in the diagram

that the application of the cosine rule is needed for this question. They have therefore

selected an inappropriate equation. This prevents them from being able to calculate

the unknown angle.

Poor response: zero marks have been awarded for this attempt. The learner has

incorrectly identified the sine rule as being required for the calculation, and although

they have substituted the correct values for each variable into part of the equation,

there is insufficient evidence to be able to award any marks.

Although no credit is given for copying equations from the

information booklet, it is good practice and shows learners

have an understanding of concepts.

Avoid just giving a numerical answer. If no working is

shown then no credit can be given for method.

Question 6: Calculate the magnitude and direction (from the horizontal) of the

resultant force for the system of coplanar forces shown in the diagram.


7


Good response: The learner has identified from the information in the diagram that

this question requires them to resolve forces vertically and horizontally through the

application of trigonometry. The correct formulae have been used from the

information booklet.

Good response: The learner has achieved full marks. They have gained one method

mark and one accuracy mark for calculating the vertical and horizontal components of

the resultant force. A further method mark is awarded for the application of

Pythagoras’ Theorem to find the magnitude of the resultant force (1 mark for

accuracy). The final mark is awarded for finding the value of the angle of the

resultant from the horizontal. The learner has some follow through variations for

rounding which is acceptable and full credit has been given.

Poor response: The learner has identified from the information in the diagram that

this question requires them to resolve forces vertically and horizontally through the

application of trigonometry. The correct formulae have been identified from the

information booklet.

Poor response: Three marks have been awarded for this response. The learner has

correctly applied trigonometric rules to determine the horizontal component of the

resultant force. This gains one method mark and one accuracy mark. A further method

mark has been awarded for the application of Pythagoras’ Theorem to find the value of

the resultant. Whilst the values used are incorrect, the learner has gained credit for

their knowledge and understanding of the processes required.

Arithmetic and process errors have been made in determining of the vertical

component of the resultant force, therefore no credit is given for these.

3

Learners should show their working in full to enable access

to method marks.

Sketches and diagrams are often helpful to support

calculations

Question 7: Calculate the hydrostatic thrust on the dam.



Good response: The learner has identified from the information presented in the

diagram and the written information the correct data to use. They have also selected

the correct formula to use from the information booklet.


mark for finding the height of the centre of pressure from the base of the dam. A

second method mark has been achieved for calculating the area of the vertical

surface. One accuracy mark has been awarded for inputting the correct values into

the given formula and calculating the correct answer.

3

Poor response: The learner has identified some the appropriate information from

the diagram and the written text, and has used this information appropriately in the

correct formula from the information booklet.

Poor response: Two marks have been awarded for this response. The learner has

correctly selected the appropriate formula from the information booklet. The learner

has identified the centre of pressure (5m) to gain one method mark, but has not

calculated the area of the vertical surface. The follow through rule has been applied,

therefore a further mark has been awarded for the force.

Learners must read written information in full as this will

often include vital data that needs to be included in

calculations.

Learners should show their answers to a sensible number of

decimal places, usually 2 or 3, or use standard form in their

responses.

2

Question 8: Describe the process of heat transfer through conduction and

convection. [Total marks for Q8- 4 marks]

The command word is describe.

Good response: This required the learners to show an understanding of the

processes and show how they transfer heat. The learner has successfully identified

each process and described how heat is transferred

Good response: Two marks have been awarded for each response. The learner

has correctly identified each process, with statements such as ‘this is the heat

energy through a solid’ for conduction. Each process is then described with further

details about how they work. Similar depth is provided in each response and full

marks have been awarded.

4

Poor response: This question required the learners to show an understanding of the

two processes and show how they transfer heat. The learner has successfully shown

that they understand the processes in question.

Poor response: One mark has been awarded for each response. The learner has

correctly identified each process, with statements such as ‘this is when hot air rises’

for convection. No further information is provided to describe how the process

works, therefore no further credit can be given.

For a describe question, learners must show a depth of

understanding of a concept in order to gain full marks.

Learners should avoid repeating the stem of the question as

this gains no credit.

2

Question 9: Calculate the reaction forces at points A and D.



Good response: The learner has identified from the information presented in the

diagram the correct processes that they need to use. They have selected the

appropriate equations and expressions from the information booklet and have applied

these with accuracy.


mark for taking moments around point A and an accuracy mark for the correct value

of reaction force at point D. They have applied the laws of static equilibrium to then

gain a further method an accuracy mark for finding the reaction force at point A. The

learner has presented answers to an appropriate degree of accuracy.

4

Poor response: The learner has identified some the appropriate information from

the diagram but has not considered the uniformly distributed load, therefore

calculations have been completed with incorrect values.

Poor response: One method mark has been awarded for the taking of moments,

however no further marks have been awarded. The learner has not considered the

overall load related to the uniformly distributed load and has therefore calculated an

incorrect support reaction at point D and an inaccurate overall loading for the beam.

This then results in the support reaction at point A also being incorrect

Learners must be familiar with each of the loading

conditions that are listed in the unit content.

Learners are encouraged to show working in full so as to

access method marks.

1

Question 10: Calculate the output flow velocity of the coolant.



Good response: The learner has interpreted the written information and the diagram

to be able to identify the formula needed to complete the calculation. Values have

been used with accuracy and the formula populated correctly.

Good response: The learner has achieved full marks. They have shown their

working in full, which is good practice, and have correctly calculated the initial and

final cross sectional areas. This is awarded one method and one accuracy mark. The

learner has then manipulated the formula from the information booklet to correctly

calculate the output velocity. This gains a further method and accuracy mark.

4

Poor response: The learner has identified some the appropriate information from the

diagram and has also selected the correct formula from the information booklet.

However the formula has not been applied appropriately.

Poor response: No marks have been awarded. The learner has not calculated the

areas of the pipe at either the input or the output. This means that the values

entered into the equation are incorrect and no further marks could be awarded

either. Had rounding errors been made in the calculating of areas, then follow

through marks could have been awarded, but this is not the case when the wrong

variables are used.

Learners must make sure they use the correct values in

formulae that are provided in the information booklet.

Learners should be familiar with the processes of

calculating surface areas for regular shapes

0

Question 11: Calculate the speed the motor must rotate (in rpm) to raise the

load if it travels up the inclined plane with a velocity of 0.8 m/s.



Good response: The learner has interpreted the written information and the diagram

to be able to identify the stages of calculation that need to be performed in order to

arrive at the solution. The correct formula have been identified and used from the

information booklet and applied with accuracy.

7

Good response: The learner has achieved full marks. They have resolved the weight

of the load into components along and normal to the inclined plane. This achieves two

marks. The learner has used the given coefficient of friction to calculate the frictional

force and then the total force parallel to the inclined plane for a further two marks.

This value has then been used to determine the power required to raise the load for a

further method mark. Finally one method mark and one accuracy mark have been

awarded for finding the rotational speed of the motor.

Full working is shown and the learner has gained full credit despite rounding some

values early in the calculation

Poor response: The learner has identified some the appropriate information from the

diagram but and has only applied this to a very limited extent. They have attempted

some parts of the question and have shown enough working for some credit to be

given.

Poor response: Two marks has been awarded for calculating the component of the

weight parallel to the inclined plane and the normal component.

The response shows minimal working, and credit can only be given for the correct

values shown.

Learners should attempt all questions. In many

questions on this paper, marks can be achieved by

providing partial solutions.

Learners should show their working clearly as not doing

so could prevent the award of method marks.

2

Question 12: Calculate the total resistance of the resistor network in the DC

circuit. [Total marks for Q12- 2 marks]

Good response: The learner has identified the formulae that are required to

calculate the total resistance in the network and has applied these correctly to find a

solution.

Good response: The learner has achieved both marks. One mark has been awarded

for finding the value of the resistance of the three parallel resistors; the learner has

shown good practice in presenting their working. They have then used the

appropriate formula to find the total resistance in the circuit, again showing their

calculation fully.

2

0

Poor response: The learner has used the information in the circuit diagram to

identify the values of the various resistors, however they have not selected or used

the correct formulae to determine the values in the parallel branches of the circuit.

Poor response: Zero marks have been awarded. The learner has not applied

appropriate DC circuit theory to carry out the calculations, and has incorrectly

calculated the value as being for a network of resistors in series. In questions where

there are only a limited number of stages, method marks are not awardable and

learners need to make sure they provide accurate values at all stages of a

calculation.

Learners should have an understanding of the

approaches for calculating resistance in both series

and parallel circuits.

Learners must be able to calculate total resistance in

networks that include series and parallel branches.

Question 13: Explain one factor affecting the force on a current carrying

conductor in a magnetic field. [Total marks for Q13- 2 marks]

The command word is explain

Good response: The command verb is explain and has a mark of 2 allocated which

implies that the learner should provide a response that has an initial statement that is

exemplified with a linked expansion point.


for identifying the magnetic field strength as a factor. This is then expanded upon

with an explanation that if the field strength is higher, then the force on the

conductor will be higher.

Poor response: The command verb is explain and has a mark of 2 allocated which

implies that the learner should provide a response that has an initial statement that is

exemplified with a linked expansion point as opposed to one or two word answers

Poor response: The learner has achieved one mark for correctly stating that

current will be a factor that impacts on the force on a conductor in a magnetic field.

To achieve the second mark, the learner would have needed to state that as the

current increases, the force on the conductor also increases.

Learners must provide a linked response in order to get both

marks for an explain question.

Statements must be related to each other to gain full marks,

for example stating current and field strength in this

question would only attract one mark.

2

1

Question 14: Calculate the current drawn by the motor.



Good response: The learner has identified the formulae that are required to

calculate the current drawn by the motor. They have interpreted the given data to

populate the formula correctly.


the transposition of the power equation and populating the aforementioned formula

with the correct values. A second mark is awarded for calculating the correct value of

the current.

Poor response: The learner has used the information in the question with a degree

of success and has identified the correct formulae to use. They have however not

considered the units of the values and have therefore made arithmetic errors.

Poor response: Zero marks have been awarded. The learner identified the correct

equation to use to calculate the current, however they have not converted the value of

power into Watts, and have therefore used an incorrect value (7.5 instead of 7500).

With single stage calculations such as this, learners will often get credit for the correct

answer only.

Learners should have an understanding of SI units and

multiples of them.

Learners must ensure that values are entered into equations

appropriately.

2

0

Question 15: Calculate the number of turns required in the secondary coil.



Good response: The learner has interpreted the given diagram and text to be able

to use and apply the correct formula to calculate the number of turns required.

Mathematical operations have been carried out with accuracy.


the transposition of the given equation in the information booklet to find the ratio

between the coils. A second mark is awarded for calculating the correct number of

turns on the secondary coil.

2

Poor response: The learner has interpreted the given diagram and text to be able to

use and apply the correct formula, however they have not populated the formula

correctly. Mathematical operations have been carried out with accuracy.

Poor response: The learner has achieved zero marks. The learner has identified the

correct formula, but has not populated it correctly, reversing the values. No marks

are awardable as a result, although the learner does have some understanding of the

concepts.

Learners should have an understanding of the application

of the formulae listed in the information booklet.

Learners should show their working in full, and enter

values into equations once they have been transposed.

0

Question 16: Calculate the current drawn from the supply.



Good response: The learner has used the data provided in the question to identify

the formulae that are required to calculate the current drawn from the supply. They

have interpreted the given data to populate the formula correctly.

Good response: The learner has achieved three marks. One mark has been

awarded the calculating the reactance on the circuit. Despite working not being

shown, the learner has gained credit for calculating the impedance. Finally, an

accuracy mark has been awarded for the correct value of current. One decimal place is an appropriate level of accuracy for this question.

Poor response: The learner has interpreted the information in the question with

limited success. As a result they have not applied the correct formulae and have

arrived at an incorrect solution.

3

0

Learners should have an understanding of each of the

concepts listed in the unit content and how these relate to

each other.

Learners should show their working in full as this allows for

access to all of the marks available for a question.

Poor response: Zero marks have been awarded. The learner has failed to identify

that there is an impedance in the circuit and has simply applied the formula for

inductance in a coil. This was a common error by learners. As learners progress

through the paper, the demand of questions increases and often they will need to

synthesise knowledge of a number of concepts to arrive at an answer.

Question 17a: Calculate the total equivalent capacitance of the three series

capacitors. [Total marks for Q17a- 1 mark]


Good response: The learner has shown an understanding of capacitor theory in

order to be able to select and populate the correct formula to calculate the total

equivalent capacitance of the network.

Good response: The learner has shown their working in full. Although the answer

is provided to 9 decimal places, the value is correct. Learners should however be

encouraged to limit the number of decimal places given in answers

Poor response: The learner has interpreted the information in the question with

limited success and has shown an understanding of capacitor theory. However an

inaccurate calculation has been performed.

1

0

Poor response: Zero marks have been awarded. The learner has identified the

correct formula to use, but has made arithmetic errors in the addition of the

fractions. As such no credit can be given.

Learners should have an understanding of each of the

concepts listed in the unit content and how these relate

to each other.

Learners should apply formulae accurately and check

their calculations for accuracy.

Question 17b: Calculate the charge stored on each capacitor.

[Total marks for Q17b- 1 mark]


Good response: The learner has interpreted the data correctly, and has shown an

understanding of capacitor theory to select and use the correct formula to answer the

question.

Good response: The learner has demonstrated that they understand that the same

charge is stored on each capacitor when in series. As a result they have identified and

applied the correct formula to arrive at the correct solution.

Poor response: The learner has interpreted the information in the question but has

not applied capacitor theory correctly. The correct formula has however been used,

although this gains no credit

Poor response: Zero marks have been awarded. The learner has calculated the

charge based on individual capacitors, and has not demonstrated that they

understand that the charge is the same for all of the capacitors in series. This was a

common error amongst learners

Learners should be familiar with concepts relating to

capacitors in series and parallel.

Learners should be encouraged to write out formulae that

they use to ensure that they populate them with the

correct values.

1

0

Question 17c: Calculate the total energy stored.

[Total marks for Q17c- 1 mark]


Good response: The learner has interpreted the data correctly, and has used

appropriate values to populate the formula. Working has been shown, which is good

practice and allows for answers to be verified.

Good response: The learner has demonstrated that they understand capacitance

concepts. They have shown that they are able to perform calculations with accuracy.

There is a second solution provided, however the learner has been marked positively

as the correct answer is given on the answer line.

Poor response: The learner has interpreted the information in the question to select

the correct formula from the information booklet. However, the formula has not been

populated accurately and an incorrect answer generated.

Poor response: Zero marks have been awarded. The learner has not include the

supply voltage in their calculations, therefore the answer is incorrect. No marks are

awarded for the selection of the formula.

Learners should be encouraged to write down the variables

that they are going to enter into given formulae.

Learners should populate formulae systematically to ensure

that no variables are omitted.

1

0

Question 17d: Calculate the voltage on each capacitor.

[Total marks for Q17d- 1 mark]


Good response: The learner has shown an understanding of capacitor theory in

order to be able to select and populate the correct formula to calculate the voltage on

each capacitor. The correct formula has been manipulated to make voltage the

subject of the formula.

Good response: The learner has not shown their working in full, rather in abridged

form at the top of the response. The answers given are correct, and the learner has

used the appropriate values determined in previous part questions to produce

solutions.

3

0

Poor response: The learner has interpreted the information in the question to select

the correct formula from the information booklet. The formula has been transposed

correctly, however incorrect values have been used to populate the formula.

Poor response: Zero marks have been awarded. The learner has used incorrect

values of the charged stored in the capacitors, and has, as a result produced

inaccurate values of voltage. The values of voltage given are unrealistic given the

total supply voltage is 240V and the sum of the individual voltages should be

approximately 240V.

Learners should be encouraged to check that the

answers that they produce are realistic and

reasonable for the data given.

Learners should show working clearly.

Question 18: Draw a phasor diagram to represent V1 + V2 and find the resultant

phasor. [Total marks for Q18- 3 marks]

The command word is draw.

Good response: Learners are expected to make a graphical representation of given

data that is accurate and shows results clearly.

Learners make a graphic representation of data by hand (as in a diagram). For example, ‘Draw a diagram to represent…’

Learners make a graphic representation of data by hand (as in a diagram).

For example, ‘Draw a diagram to represent…’ se

Good response: The learner has achieved three marks. One mark has been

awarded for plotting the two phasors in proportion to each other, and at the correct

angle. A second mark has been awarded for plotting the resultant phasor. The final

mark is achieved for the angle and magnitude of the phasor being within the correct

range of values.

3

1

The command word is draw

Poor response: The learner has interpreted the information and has sketched the

phasors for the two voltage waveforms. It is expected for draw questions that

learners will use appropriate drawing instruments to ensure diagrams are accurate.

Poor response: One mark has been awarded. The learner has sketched the two

phasors, and they are approximately in proportion. This has allowed one mark to be

awarded. The angle is not accurate, neither is the resultant phasor. To achieve full

marks the learner would need to present an accurate drawing using a ruler and

protractor.

Learners should be encouraged to use drawing

instruments to produce accurate diagrams when

answering questions that require them to draw.

Learners should select scales that allow information to be

shown clearly.

Question 19a: Calculate the time constant for the discharge of the capacitor

through the variable resistor. [Total marks for Q19a- 5 marks]


Good response: The learner has used the data provided in the question to identify

the processes that are required to calculate the time constant. They have interpreted

the given data to populate the formula correctly and then apply the laws of

logarithms correctly.

5

Good response: The learner has achieved five marks. One mark has been awarded

the populating the formula with the correct values. A further mark has been awarded

for using natural logs in the process. The application of natural logarithms allows the

learner to be awarded a further mark. The learner has correctly noted that ln’e’ = 1

which is awarded one mark. The final mark is achieved for the solution which is

correct based on the values used throughout the calculation.

Poor response: The learner has used the data provided in the question to identify

the processes that are required to calculate the time constant. They have interpreted

the given data to populate the formula to an extent, although this has been done

towards the end of the process which has potentially disadvantaged them.

Poor response: The learner has achieved zero marks. Whilst there is some evidence

of the application of the laws of logarithms, the learner would have benefited from

substituting values into the formula at an earlier stage of the process. As a result,

laws of logarithms have not been employed correctly and errors made.

Learners should be encouraged to populate formulae

with values at an early stage of calculations.

Learners should show working clearly to ensure

method marks can be awarded.

0

Question 19b: Calculate the resistance setting of the variable resistor.

[Total marks for Q19b- 2 marks]


Good response: The learner has used the data provided in the question to select

and use the correct formula from the information booklet and has rearranged this

after populating with values. The learner has performed mathematical skills

accurately.

Good response: Two marks have been awarded. The learner has achieved one

method mark for the transposition of the formula and populating it with the correct

values derived from part (a) and the information in the question. Had the learner

provided an incorrect value for the time period in part (a) but then performed an

accurate calculation with that value in part (b), then full credit would have been

awarded.

The second mark is awarded for the correct solution that has been given with

appropriate units.

Poor response: The learner has used the data provided in the question to select

and use the correct formula from the information booklet and has rearranged this

before population with values. The learner has performed mathematical skills

accurately.

Poor response: One mark has been awarded. The learner has achieved one method

mark for the transposition of the formula and populating it with the correct values

derived from part (a) and the information in the question. Had the learner then copied

the correct values through to the next stage of the calculation then they would have

achieved two marks.

2

1

Learners should take care when transferring values from one

part of a calculation to another. It is easy to change a value

by omitting factors of 10.

It is good practice to populate formulae at an early stage of

calculations.

Question 20a: Explain how energy loss processes in both the mechanical and

electrical equipment affects the efficiency of the system.

[Total marks for Q20a- 4 marks]

The command word is explain

Good response: The command verb is explain and has a mark of 4 allocated which

implies that the learner should provide two linked responses that have an initial

statement that is exemplified with an expansion point. For this question two distinct

reasons for energy loss are required, one related to mechanical equipment and one

related to electrical equipment.

Good response: The learner has noted that energy will be lost from the generator

due to friction in the form of noise. This is a linked response and therefore is awarded

two marks. The response continues to state that heat will also be lost due to electric

currents and is awarded a further two marks. Specific forms of energy loss are

suitable and appropriate responses.

4

Poor response: The command verb is explain and has a mark of 4 allocated which

implies that the learner should provide two responses that have an initial statement

that is exemplified with a linked expansion point. For this question two distinct

reasons for energy loss are required, one related to mechanical equipment and one

related to electrical equipment. It would not be accepted to repeat the same

statements for the two types of energy loss.

Poor response: The learner has been awarded one mark for stating that energy will

be lost through sound and heat for the mechanical part of the system. The same

types of energy loss are given for the electrical part of the system. There are no

explanations of how these losses occur, therefore no further marks are awardable.

Learners should make sure that if a question asks them to

consider two factors that both are addressed in the answer.

Credit will not be given for repeated responses.

1

Question 20b: Calculate the efficiency with which the heat engine provides

mechanical work to the generator. [Total marks for Q20b- 7 marks]


Good response: The learner has used the data provided in the initial question stem

to identify the processes and calculations that need to be performed in order to

answer this part of the synoptic question. Synoptic questions draw strands from

both the electrical and mechanical aspects of the unit, with both needing to be

considered in a calculation.

Good response: The learner has achieved full marks for this part of the question.

One method mark has been awarded for multiplying the mass flow rate by the

energy content; a further mark has been awarded for the accurate value for power.

A further method mark is awarded for the process of finding out the angular

velocity, with an accuracy mark awarded for the correct value of omega.

One mark is awarded for calculating the output power.

Finally one method mark and one accuracy mark have been awarded for calculating

the accuracy of the system.

7

2

Poor response: The learner has used the data provided in the initial question stem

to identify some of the processes and calculations that need to be performed in order

to answer this part of the synoptic question. Synoptic questions draw strands from

both the electrical and mechanical aspects of the unit, with both needing to be

considered in a calculation.

Poor response: The learner has made an attempt at answering some parts of the

question and has been rewarded appropriately.

No credit is given for the input power, as there is no consideration of the relationship

between the mass flow and energy content.

Two marks (1 method, 1 accuracy) are however awarded for calculating the output

power, with a further mark awarded for calculating omega (although the value is not

given)

A final method mark is awarded for the relationship between efficiency, input and

output power.

Learners should be encouraged to attempt synoptic

questions as there may be some parts of the question that

can be answered.

Working should be shown in full so that method marks can

be awarded.

Question 20c: Calculate the overall system efficiency.

[Total marks for Q20c- 3 marks]


Good response: The learner has used the data provided in the initial question stem,

and their answers for previous sections, to identify the processes and calculations

that need to be performed to answer this final part of the synoptic question.

Good response: The learner has been awarded three marks, and the ‘follow

through’ rule has been applied for the incorrect input power being used.

The learner has shown their working in full and has achieved one mark for calculating

the correct input power from the given data. A further mark is awarded for the

relationship between input and output power. The third mark is awarded for the

value for efficiency being correct (based on the values used)

Poor response: The learner has used the data provided in the initial question stem

to provide a partial response to the question. Whilst the response is limited to only

one aspect of the question, this does allow some credit to be given.

Poor response: The learner has been awarded one mark for the calculation of the

input power. No further work has been submitted, but the learner has demonstrated

good practice by providing a partial answer to part of the question that they could

complete.

3

1

Learners should avoid leaving answers blank, there could

be some part of a question that can be answered and

marks could be awarded.

Learners should be encouraged to check their answers and

make sure that the values given in responses are realistic

*S50749A02424*24

(c) Calculate the overall system efficiency.

3 marks

Answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

END OF EXAM TOTAL FOR SECTION C = 14 MARKSTOTAL FOR PAPER = 80 MARKS

Total for Question 20 = 14 marks

Unit 1: Engineering Principles – sample mark scheme

General marking guidance All learners must receive the same treatment. Examiners must mark the first

candidate in exactly the same way as they mark the last.

Mark schemes should be applied positively. Candidates must be rewarded forwhat they have shown they can do, rather than be penalised for omissions.

Examiners should mark according to the mark scheme, not according to theirperception of where the grade boundaries may lie.

All marks on the mark scheme should be used appropriately.

All the marks on the mark scheme are designed to be awarded. Examinersshould always award full marks if deserved, i.e. if the answer matches the markscheme. Examiners should also be prepared to award zero marks if thecandidate’s response is not worthy of credit according to the mark scheme.

Where some judgement is required, mark schemes will provide the principlesby which marks will be awarded and exemplification may be limited.

When examiners are in doubt regarding the application of the mark scheme toa candidate’s response, the team leader must be consulted.

Crossed-out work should be marked UNLESS the candidate has replaced it withan alternative response.

Types of Mark and Abbreviations

This mark scheme uses the following types of marks:

M marks – method marks are awarded for ‘knowing a method and attemptingto apply it’, unless otherwise indicated

A marks – accuracy marks can be awarded only if the relevant method (M)marks have been earned

B marks – unconditional accuracy marks (independent of M marks)

marks should not be subdivided.

Abbreviations:

ft – follow through

cao – correct answer only

cso – correct solution only, there must be no errors in this part of the questionto obtain this mark

isw – ignore subsequent working

awrt – answers which round to

SC – special case

oe – or equivalent (and appropriate)

dp – decimal places

sf – significant figures

Types of Mark and Abbreviations

This mark scheme uses the following types of marks:

M marks – method marks are awarded for ‘knowing a method and attemptingto apply it’, unless otherwise indicated

A marks – accuracy marks can be awarded only if the relevant method (M) marks have been earned

B marks – unconditional accuracy marks (independent of M marks)

marks should not be subdivided.

Abbreviations:

ft – follow through

cao – correct answer only

cso – correct solution only, there must be no errors in this part of the questionto obtain this mark

isw – ignore subsequent working

awrt – answers which round to

SC – special case

oe – or equivalent (and appropriate)

dp – decimal places

sf – significant figures

Section A – Applied Mathematics

Question number Working Answer Notes Mark

1 (2t − 3)(t + 4) 2t = 3

t = 32

t = 1.5s

t = 1.5s or 1.5 cao

M1 for correct factorisation as shown in working (can be reversed).

B1 for correct positive value of t = 1.5s Do not penalise if t = -4 is also given. (2)


2 (1) 20y = 30x + 9

(2) −3y = 1.5x − 6 multiply (2) by 20

20y = 30x + 9

−60y = 30x – 120 subtract

80y = 129 y = 1.6125

Substitute into (1) 32.5 = 30x + 9 x = 0.775 Or any similar/appropriate method giving correct answers.

y = 1.6125 x = 0.775 or (0.775, 1.6125)

Accept final values that round to two dp. Allow ft for rounding variations. Allow ft if second variable is correct in relation to the first incorrect answer.

A1 for correct value of x.

A1 for correct value of y.

(2)


3 48180

= 0.864 rad

2

2rA

= 25 × 25 × 0.8642

= 261.8 mm2

Or

A = 261.8 mm2

Or 83 13

Accept final values that round to one dp. Allow ft for rounding variations.

M1 for conversion to radians or developing a formula using the ratio of angle to area. A1 for the correct area of the sector.

(2)

33

A = × 252 × 48360

A = 261.8 mm2


4 log24 = logz + 2log2 log24 = logz + log22

log24 = logz + log4 logz = log24 – log 4

logz = log 244

logz = log6 z = 6

z = 6 cao

M1 for correct application of laws of logarithms.

A1 for correct value of z.

(2)


5 cos(A) =

2 2 2

2b c a

bc

cos(A) = 2 2 230 42 602 30 42

cos(A) = −0.37 A = cos−1(−0.37) A = 111.80°

A = 111.80

Accept answers wrt 112

M1 for manipulating the cosine rule formula.

A1 for 111.8 or answers wrt 112

(2)

34

A = × 252 × 48360

A = 261.8 mm2

Questionnumber Working Answer Notes Mark

4 log24 = logz + 2log2log24 = logz + log22

log24 = logz + log4logz = log24 – log 4

logz = log 244

logz = log6z = 6

z = 6cao

M1 for correctapplication of laws of logarithms.

A1 for correct valueof z.

(2)


5cos(A) =

2 2 2

2b c a

bc

cos(A) = 2 2 230 42 602 30 42

cos(A) = −0.37A = cos−1(−0.37)A = 111.80°

A = 111.80

Accept answers wrt 112

M1 for manipulatingthe cosine rule formula.

A1 for 111.8 or answers wrt 112

(2)

Section B – Electrical, Electronic and Mechanical Principles


6 10 N + 7sin25 = 5sin45 + RV 10 + 2.96 = 3.54 + RV

RV = 12.96 – 3.54 = 9.42 N

7cos25 = 5cos45 + RH 6.34 = 3.54 + RH

RH = 6.34 – 3.54 = 2.80 N

FR = 2 2H vR R

FR = 2 22.80 9.42 =

7.84 88.74 = 96.58

FR = 9.83 N

tan = 2.809.42

= 0.297

=tan-1 (0.297) = 16.55 o

= 73.45 o (from the horizontal)

FR = 9.83 N = 16.55 o

Do not accept graphical solutions.

Accept final values that round to one dp.

Allow ft for rounding variations

M1 for correct process for RV.

A1 for correct value of RV.

M1 for correct process for RH.

A1 for correct value of RH.

M1 for process of finding resultant using Pythagoras. A1 for the correct value of FR.

A1 for the correct value of .

(7)


7 F = gAx

A = 10 × 5 = 50

x = 102

= 5

F = 1000 × 9.81 × 50 × 5

F = 2 452 500 N or 2453 kN (ft) Or 2450 kN

F = 2 452 500 N Allow ft.

Only accept 2453 or 2450 if correct units given (kN).

Do not penalise if centre of pressure is

calculated as 13

height or similar.

M1 for calculation of x. M1 for calculation of A. A1 for correct answer.

(3)

35

Question number Answer Mark

8 Award up to 2 marks for describing the process of conduction/convection and how it transfers heat. 1 mark for identifying the process in each case and 1 mark for explaining how it works.

Conduction In conduction, electrons gain kinetic energy (1) and their vibrations

cause heat to transfer through the material (1). In conduction, energy is transferred through physical contact (1) with

heat being transferred as atoms bump into each other (1).

Convection In convection, heated particles are replaced by cooler particles (1) and

this movement creates circulation of heat throughout the liquid/gas(1).

In convection, the addition of heat make a gas/liquid less dense causingit to rise (1) this is replaced by cooler gas/liquid in a continuousprocess (1).

Accept any other reasonable response. (4)


9 Taking moments about A: (2 × 45) + (4 × 70) + (35 × 3.5 × 7) + (20 × 7) = 6 × RD

90 + 280 + 857.5 + 140 = 1367.5 = 6 × RD

Calculating RD

RD = 1367.56

= 227.9 N

vertical equilibrium RA + RD = 45 + 70 + (35 × 7)

+ 20 RA + 227.9 = 380 RA = 152.1 N

RD = 227.9 N

RA = 152.1 N Accept final values that round to whole numbers.

Allow ft for rounding variations.

M1 for taking moments around A or D. A1 for value of RD. M1 for total reaction forces = total load

A1 for RA

(4)

36

Questionnumber Answer Mark

8 Award up to 2 marks for describing the process of conduction/convectionand how it transfers heat.1 mark for identifying the process in each case and 1 mark for explaininghow it works.

Conduction In conduction, electrons gain kinetic energy (1) and their vibrations

cause heat to transfer through the material (1). In conduction, energy is transferred through physical contact (1) with

heat being transferred as atoms bump into each other (1).

Convection In convection, heated particles are replaced by cooler particles (1) and

this movement creates circulation of heat throughout the liquid/gas (1).

In convection, the addition of heat make a gas/liquid less dense causingit to rise (1) this is replaced by cooler gas/liquid in a continuous process (1).

Accept any other reasonable response. (4)


9 Taking moments about A:(2 × 45) + (4 × 70) +(35 × 3.5 × 7) + (20 × 7)= 6 × RD

90 + 280 + 857.5 + 140 = 1367.5 = 6 × RD

Calculating RD

RD = 1367.56

= 227.9 N

vertical equilibriumRA + RD = 45 + 70 + (35 × 7)

+ 20RA + 227.9 = 380RA = 152.1 N

RD = 227.9 N

RA = 152.1 NAccept finalvalues thatround to wholenumbers.

Allow ft for roundingvariations.

M1 for takingmoments around Aor D.A1 for value of RD.

M1 for total reactionforces = total load

A1 for RA

(4)


10 A1 =

20.054

= 1.96 × 10-3 m2

A2 = 20.04

4 = 1.26 × 10-3 m2

A1V1 = A2V2 V2 = 1 1

2

A VA

V2 = 3

3

1.96 10 61.26 10

V2 = 9.38 m/s (ft)

V2 = 9.38 m/s

Accept final value that rounds to one dp.

Allow ft for rounding variations

M1 mark for entering the correct values into A1 and A2. A1 mark for correct values of A1 and A2. M1 for manipulation of formula and entering the correct values. A1 mark for calculating V2.

(4)


11 Fi = mgsin30 = 8 × 9.81 × sin30 = 39.24 N

N = mgcos30 = 8 × 9.81 × cos30 = 67.97 N

Frictional resistance = µN = 0.4 × 67.97 = 27.19 N

F = Fi + N = 39.24 + 27.19 = 66.43 N

Power to angular velocity connection

Power = Fv = 66.43 × 0.8 = 53.14 W

Power =

angular speed = P

= 53.1412

= 4.43 rad/s

Rotational speed = × 602

= 4.43 × 602

= 42.30 rpm

Rotational speed = 42.30 rpm



M1 mark for entering the correct values into Fi and Fn. A1 mark for correct values of Fi and Fn. A1 mark for calculating frictional resistance. A1 for determining the total force down the slope (ft error allowed). M1 for the calculating the power required. M1 for determining the angular speed of the motor. A1 for correct conversion into rpm.

(7)

37


12 1 1 1 112 2.2 10R

= 0.639

R = 1.6k or 1k6

R = 1k + 2k2 + 1k6 R = 1k + 2.2k + 1.6k R = 4.8k (ft)

R = 4.8k Or 4k8 Or 4800



A1 resistance in the parallel resistors.A1 for correct value of total resistance.

(2)

Question number Answer Marks

13 Award 1 mark for a correct identification and a further mark for expansion. The current flowing in the conductor (1)/increasing the current will

increase the force on the conductor (1). The magnetic field strength (1)/increasing the field strength will

increase the force on the conductor (1). The length of the conductor within the field (1), increasing the length

of the conductor in the magnetic field will increase the force (1). Accept any other reasonable response. (2)


14 P = IV to I = P

V7500240

I =31.25 A

31.25 A

cao

M1 for transposing equation and using correct values of P and V.

A1 for correct value of I. (2)


15 220100

= 2.2

7482.2

= 340 turns

340 turns M1 for transposing an equation and using correct values. A1 for correct value of turns. (2)

38


12 1 1 1 112 2.2 10R

= 0.639

R = 1.6k or 1k6

R = 1k + 2k2 + 1k6R = 1k + 2.2k + 1.6kR = 4.8k (ft)

R = 4.8kOr 4k8Or 4800

Accept finalvalue thatrounds to one dp.


A1 resistance in theparallel resistors.A1 for correct valueof total resistance.

(2)

Questionnumber Answer Marks

13 Award 1 mark for a correct identification and a further mark for expansion. The current flowing in the conductor (1)/increasing the current will

increase the force on the conductor (1). The magnetic field strength (1)/increasing the field strength will

increase the force on the conductor (1). The length of the conductor within the field (1), increasing the length

of the conductor in the magnetic field will increase the force (1).Accept any other reasonable response. (2)


14P = IV to I = P

V7500240

I =31.25 A

31.25A

cao

M1 for transposingequation and usingcorrect values of Pand V.

A1 for correct valueof I. (2)


15 220100

= 2.2

7482.2

= 340 turns

340 turns M1 for transposingan equation andusing correct values.A1 for correct valueof turns. (2)


16 Reactance = 2fL

= 2 × 50 × 0.4

= 125.66 Ω

Total impedance: 2 2

Lz X R = 2 2125.66 5

= 125.8 Ω

240125.8

rmsrms

VI

Z = 1.91 A

1.91 A



A1 for identifying the presence of impedance and using the formula correctly to calculate the reactance.

A1 for using Pythagoras to calculate the resultant impedance. A1 for the correct value of I. (3)


17(a) 1c

= 11c

+ 12c

+ 13c

= 1.714 μF

or awrt 1.7 μF

awrt 1.71 μF Accept final value that rounds to one dp. Allow ft for rounding variations.

A1 Answers which round to 1.71 μF or responses in standard form are acceptable.

(1)


17(b) Q = CV = 1.714 × 10-6 × 240

=411 μC (the same charge is stored on each capacitor)

Q = 411 μC Allow ft from 17a. Allow ft for rounding variations.

A1 for correct value of Q.

(1)


17(c) E = 0.5CV2 = 0.5 × 1.714 × 10-6

× 2402 = 0.049 J

E = 0.049 J Ft

Accept final value that rounds to two dp.

A1 for correct value of E.

(1)

39


17(d) V1 11

QC

= V1 = 6

6

411 103 10

= 137 V

V2 22

QC

= V2 = 6

6

411 106 10

= 68.5 V

V3 33

QC

= V3 = 6

6

411 1012 10

= 34.25 V

V1 = 137 V V2 = 68.5 V V3 = 34.25 V Ft.


M1 for transposing formula and inserting correct values. A1 for each correct value of V1/V2/V3.

(3)

40


17(d) V111

QC

= V1 = 6

6

411 103 10

= 137 V

V222

QC

= V2 =6

6

411 106 10

= 68.5 V

V333

QC

= V3 =6

6

411 1012 10

= 34.25 V

V1 = 137 VV2 = 68.5 VV3 = 34.25 VFt.

Accept finalvalue thatrounds to one dp.

M1 for transposingformula andinserting correctvalues.A1 for each correct value of V1/V2/V3.

(3)


18 Correct phasor for V1 Correct phasor for V2 Correct phasor for Vr = V1 + V2 = 290 V including phase angle correct 18–22 degrees.

Notes M1 Take account of the scale chosen by the candidate. V2 should be twice the length of V1. 30° angle should be drawn accurately using a protractor. M1 VT should be drawn and measured as accurately as possible. A1 magnitude awrt 290 at angle below horizontal 18–22 degrees. (3)

41


19(a) 3 = 12

t

e

loge3 = loge12 + t logee

loge3 − loge12 = − t

= 1.39

t = 7.19

= 1.39

t

= 7.19

Accept final value that rounds to two dp.


M1 for substituting the correct voltages into the correct parts of the formula. M1 for selecting use of natural logs to the original equation. M1 for applying laws of logs to the equation. M1 for substituting logee with 1 A1 for correct answer. Answers in the form ‘ln’ are also acceptable.

(5)


19(b) = RC so R =

C = 6

7.19470 10

R = 15.34k Ω or 15 297.87 Ω

R = 15.34k Ω or = 15 297.87 Ω

ft

Accept final value that rounds to two dp (if using a prefix).

M1 for correct transposition of formula and substitution of values. A1 for correct value of R or equivalent.

(2)

42


19(a)3 = 12

t

e

loge3 = loge12 + t logee

loge3 − loge12 = − t

= 1.39

t = 7.19

= 1.39

t

= 7.19

Accept finalvalue thatrounds to two dp.


M1 for substitutingthe correct voltages into the correctparts of theformula.M1 for selecting useof natural logs to the original equation.M1 for applyinglaws of logs to the equation.M1 for substitutinglogee with 1A1 for correct answer.Answers in the form ‘ln’ are alsoacceptable.

(5)


19(b) = RC so R =

C = 6

7.19470 10

R = 15.34k Ω or 15 297.87 Ω

R = 15.34k Ωor= 15 297.87 Ω

ft

Accept finalvalue thatrounds to two dp (if using aprefix).

M1 for correcttransposition of formula andsubstitution of values.A1 for correct valueof R or equivalent.

(2)

Section C – Synoptic Question


20(a) Award 1 mark for identification in mechanical system and a further mark for justifying how it affects the mechanical system. Award 1 mark for identification in electrical system and a further mark for justifying how it affects the electrical system.

Mechanical Energy losses/conversion to unwanted forms of energy in the heat

engine (1) due to friction/poor combustion/unwanted heat transfer which direct energy away from the desired output (1).

Electrical Energy losses in generator of subsequent transformer equipment (1)

due to friction/electrical resistance/eddy current/hysteresis processes (1).

Accept any other appropriate explanation. (4)


20(b) 0.003 × 40exp6 = 120 kW Or 3 × 10-3 × 40 × 106 = 120 kW

Output power = T

= 1500 rpm × 60N = 157 rad/s

= 157 rad/s

Output power = 255 × 157 = 40 kW

Efficiency of heat engine = (output power/input power) × 100%

= 40120

× 100= 33.4%

0.334 or 33.4%

120 kW or 120 000 W

CAO

Accept final value that rounds to 62.8 kW or 62 800 W to nearest thousand.

ft Accept final value that rounds to two dp (0.33) or nearest whole number for per cent answer.

M1 for recognising the need to multiply energy content by mass flow rate. M1 for correct method to calculate ω. A1 for the correct value of ω. A1 for correct value of input power. A1 for correct value of output power.

M1 for recognising the relationship between input and output. A1 for correct efficiency value given (ft acceptable).

(7)

43


20(c) Power out of generator

P = IV = 62.5 × 400 = 25 kW

Overall efficiency = 25120

kW =

0.21 or 21%

0.21 or 21%

ft Accept final value that rounds to 25 kW or 250 00W to nearest thousand.

Allow ft for rounding variations Accept final value that rounds to two dp (0.21) or nearest whole number for per cent answer.

A1 for correct value of power. M1 for recognising the relationship between power input and output (even if 40 kW is used instead of 120 kW).

A1 for correct value given (ft).

(3)

44

BTEC Level 3 National in Engineering · BTEC Level 3 National in Engineering First teaching September 2016 Sample Marked Learner Work External Assessment Unit 1: Engineering Principles

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