BTEC Level 3 National in Engineering · BTEC Level 3 National in Engineering First teaching...
Transcript of BTEC Level 3 National in Engineering · BTEC Level 3 National in Engineering First teaching...
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.
[Total marks for Q4- 2 marks]
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]
The command word is calculate.
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.
[Total marks for Q6- 7 marks]
7
The command word is calculate.
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.
[Total marks for Q7- 3 marks]
The command word is calculate.
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.
Good response: The learner has achieved full marks. They have gained one method
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.
[Total marks for Q9- 4 marks]
The command word is calculate.
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.
Good response: The learner has achieved full marks. They have gained one method
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.
[Total marks for Q10- 4 marks]
The command word is calculate.
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.
[Total marks for Q11- 7 marks]
The command word is calculate.
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.
Good response: The learner has achieved both marks. One mark has been awarded
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.
[Total marks for Q14- 2 marks]
The command word is calculate.
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.
Good response: The learner has achieved both marks. One mark has been awarded
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.
[Total marks for Q15- 2 marks]
The command word is calculate.
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.
Good response: The learner has achieved both marks. One mark has been awarded
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.
[Total marks for Q16- 3 marks]
The command word is calculate.
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]
The command word is calculate.
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]
The command word is calculate.
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]
The command word is calculate.
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]
The command word is calculate.
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]
The command word is calculate.
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]
The command word is calculate.
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]
The command word is calculate.
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]
The command word is calculate.
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)
Question number Working Answer Notes Mark
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)
Question number Working Answer Notes Mark
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
Question number Working Answer Notes Mark
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)
Question number Working Answer Notes Mark
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)
Questionnumber Working Answer Notes Mark
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
Question number Working Answer Notes Mark
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)
Question number Working Answer Notes Mark
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)
Question number Working Answer Notes Mark
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)
Questionnumber Working Answer Notes Mark
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)
Question number Working Answer Notes Mark
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)
Question number Working Answer Notes Mark
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
Accept final value that rounds to one dp.
Allow ft for rounding variations.
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
Question number Working Answer Notes Mark
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
Accept final value that rounds to one dp.
Allow ft for rounding variations.
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)
Question number Working Answer Notes Mark
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)
Question number Working Answer Notes Mark
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
Questionnumber Working Answer Notes Mark
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.
Allow ft for roundingvariations.
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)
Questionnumber Working Answer Notes Mark
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)
Questionnumber Working Answer Notes Mark
15 220100
= 2.2
7482.2
= 340 turns
340 turns M1 for transposingan equation andusing correct values.A1 for correct valueof turns. (2)
Question number Working Answer Notes Mark
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
Accept final value that rounds to one dp.
Allow ft for rounding variations.
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)
Question number Working Answer Notes Mark
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)
Question number Working Answer Notes Mark
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)
Question number Working Answer Notes Mark
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
Question number Working Answer Notes Mark
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.
Accept final value that rounds to one dp.
M1 for transposing formula and inserting correct values. A1 for each correct value of V1/V2/V3.
(3)
40
Questionnumber Working Answer Notes Mark
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)
Question number Answer Mark
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
Question number Working Answer Notes Mark
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.
Allow ft for rounding variations.
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)
Question number Working Answer Notes Mark
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
Questionnumber Working Answer Notes Mark
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.
Allow ft for roundingvariations.
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)
Questionnumber Working Answer Notes Mark
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
Question number Answer Mark
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)
Question number Working Answer Notes Mark
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
Question number Working Answer Notes Mark
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