SpecificationsModel for Electracycle all electric motorcycle - Spe.docx

SpecificationsModel for Electracycle all electric motorcycle -
SpecificationsBattery TypesNMCLMOLFPEnergy
Density0.5750.4950.4kWh/kgDensity150015001500kg/m3Cost1
40012001000$/kWhLife110010001200rechargesBike
Cost6000$Regen Brake FittednENTER Y / NRegen Brake
Mass0kgEnter 0 or 3Regen Brake Size0.001m3Brake Return
%17%Cycle mass80kgRider Mass80kgEnergy
Usage12%/sWheel Radius0.305mAverage
Speed30km/h8.3333333333m/sPin17.5kW17500WE48VNominal
Torque64.8NmEfficiency90%Internal Battery
Resistance50mΩMax Safe Temp50CHeat Trans Co-
Eff100W/m2/KAverage Ambient Temp40CN
Technical AnalysisModel for Electracycle all electric
motorcycle - Technical AnalysisNMCCapacityBattery MassBatt
SizeInit CostTot sizeTot MassBatt Life# Batt RqdCost BattTotal
CostAccelP to MaintBrk RetE TotBat LifeRangeTime to Av
VIQhArea rqdBatt areaFin areaSizekWhkgm3$m3kgYrsin 10
Yrs$$ms-2WWWhkmsAWm2m2m2-
444.47.650.00512,160.000.005167.6524.42.2733,181.8215,341.
821.2671397.1010.0001397.1012.83485.0336.57632.34052.2950
.1050.0590.045-
60610.430.00714,400.000.007170.4354.42.2733,181.8217,581.8
21.2471420.2900.0001420.2903.802114.0616.68532.87754.0450
.1080.0730.035-
757.513.040.00916,500.000.009173.0434.42.2733,181.8219,681
.821.2281442.0290.0001442.0294.681140.4276.78733.38055.71
20.1110.0850.027LMOCapacityMassSizeCostTot sizeMassBatt
Life# Batt RqdCost BattTotal CostAccelP to MaintBrk RetE
TotBat LifeRangeTime to Av VIQhAreaBatt areaFin
areaSizekWhkgm3$m3kgYearsin 10 Yrs$$ms-
2WWWhmsAWm2m2m2-
444.48.890.00611,280.000.006168.88942.53,000.0014,280.001.
2581407.4070.0001407.4072.81484.4116.62432.57953.0690.106
0.0650.041-

60612.120.00813,200.000.008172.12142.53,000.0016,200.001.2
341434.3430.0001434.3433.765112.9446.75133.20255.1200.110
0.0810.030-
757.515.150.01015,000.000.010175.15242.53,000.0018,000.001
.2131459.5960.0001459.5964.625138.7376.87033.78757.0780.1
140.0930.021LFPCapacityMassSizeCostTot sizeMassBatt Life#
Batt RqdCost BattTotal CostAccelP to MaintBrk RetE TotBat
LifeRangeTime to Av VIQhAreaBatt areaFin
areaSizekWhkgm3$m3kgYearsin 10 Yrs$$ms-
2WWWhmsAWm2m2m2-
444.411.000.00710,400.000.007171.0004.82.0832,083.3312,483
.331.2421425.0000.0001425.0002.77983.3686.70732.98654.404
0.1090.0750.033-
60615.000.01012,000.000.010175.0004.82.0832,083.3314,083.3
31.2141458.3330.0001458.3333.703111.0866.86433.75856.9790
.1140.0930.021-
757.518.750.01313,500.000.013178.7504.82.0832,083.3315,583
.331.1891489.5830.0001489.5834.531135.9447.01134.48159.44
70.1190.1080.011
GraphsModel for Electracycle all electric motorcycle -
GraphsInitial CostVolume of
componentsNMCLMOLFPNMCLMOLFP-
4412,160.0011,280.0010,400.00-440.00510.00590.0073-
6014,400.0013,200.0012,000.00-600.00700.00810.0100-
7516,500.0015,000.0013,500.00-
750.00870.01010.0125$m3Total
CostRangeNMCLMOLFPNMCLMOLFP-
4415,341.8214,280.0012,483.33-4485.0384.4183.37-
6017,581.8216,200.0014,083.33-60114.06112.94111.09-
7519,681.8218,000.0015,583.33-75140.43138.74135.94$kmFin
areaTime to Av Vm2sNMCLMOLFPNMCLMOLFP-
440.0450.0410.033-446.586.626.71-600.0350.0300.021-
606.696.756.86-750.0270.0210.011-756.796.877.01
Graphs
Total Cost NMC
Total Cost LMO

Total Cost LFP
Range NMC
Range LMO
Range LFP
Initial Cost NMC
Initial Cost LMO
Initial Cost LFP
Volume of components NMC
Volume of components LMO
Volume of components LFP
ENG1002 Design project Sem2 2013 - Client Brief Version 2.1
(9/9/2013) 1
© University of Southern Queensland
Client Brief Version 2.1
1. Project Outline (page 3 correction in blue)
Midas Gold Pty Ltd seeks submissions from suitably qualified
companies for the design an underground
passenger lift to service their new mine.
The lift is required to operate in a vertical shaft to 60m below
ground level. The lift is required to transport

300 miners at the change of shifts (150 up, 150 down) within a
period of 15 minutes. The steel cage is to be
suspended by multiple steel ropes from a winch drum positioned
10m above ground level. The winch is to
be used to raise and lower the cage. Figure 1 shows the
dimensional detail of the shaft, lift and winch
system.
Figure 1: Proposed lift equipment layout
Any company tendering a design must clearly specify the final
design parameters listed in bold for each of
the design sections below. Each design section requires a
technical analysis which must be summarised in
the final design proposal.
Ground
level
Passenger cage
W x W x 2.2m
Winch drum
Radius R
10m

60m
motion
Cage guides
and braking
surfaces
Brakes on cage
W
2.2m
W
NOT TO SCALE
= 0.25 m
2 ENG1002 – Introduction to Engineering and Spatial Science
Applications
1.1 Design Sections
The project has been divided into four distinct sections to
ensure clarity in the requirements and the
expected outcomes.
1. the passenger cage sizing (cage dimension W, number of

passengers, travel time, top speed of
lift, magnitude of acceleration and deceleration, quantities of
steel mesh required for the
cage)
2. the winch (model number, input power requirement, torque
delivered at drum)
3. the steel rope selection (rope type, total length of rope
required, number of ropes)
4. the budget and costs of all components of the system
All students please refer to the IMPORTANT NOTES on the
last page of this document.
1.2 Design Goals
The design goals for the project are to:
G1. maximise the rate at which workers can be transported
G2. do not exceed the budget ($40,000) for the project
G3. offer a minimum cost solution that meets the requirements
2. Specification of Requirements
2.1 Requirements
The following requirements must be met:
R1. The lift must be capable of moving 300 passengers (150 up,
150 down) the full travel of 60m
within a 15 minute period.

R2. The maximum acceleration experienced by the passengers is
not to exceed more than 3g’s, that
is – 1g due to gravity plus 2g’s due to the movement of the lift.
(1g = 9.81m/s
2
).
R3. The minimum acceleration experienced by the passengers is
not to be less than 0.5g, that is -
1g due to gravity plus -0.5g due to the movement of the lift.
2.2 Scope
The technical analysis and design work required for this project
only requires the selection of
components from those provided and the specification of values
for the parameters listed for each
Design Section. In particular, aspects of the project that are
outside the scope of the design include:
nstruction costs of the lift and lift shaft
counter weight
not specified in versions of this
brief

(9/9/2013) 3
2.3 Constraints
The following constraints apply:
set for the cost of
materials for the lift.
seconds to enter or to exit the lift.
ger in the lift is
0.25m
2
.
section of rope)
the steel ropes.

eel ropes to be used for the design
is 3.
lift is 6.
the drum is 3 full wraps around the
drum.
is 0.25 m.
2.4 Assumptions
The following simplifying assumptions have been made:
structure of the cage
rs is ignored
to the output power of the winch
by:
(Moment) Torque = Power * ω
(Moment) Torque (N.m) = Power (W) / ω (rad/s)

where ω (omega) is the rotational speed of the drum in rad/s
Assume ω is constant during the motion of the lift, calculated
from your chosen top speed of
the lift.
Notes to students:
You are expected to consider the forces on the lift cage to
include the weight of the
lift cage, the weight of the passengers and the force applied via
the cables generated
by the Torque of the winch. The largest magnitude of force will
be applied when the
lift accelerates upward at the bottom of the shaft.
4 ENG1002 – Introduction to Engineering and Spatial Science
Applications
3.0 Technical Information
Technical and cost information covering the components of the
project is provided in this section.
Table 1: Technical Information related to the Passenger Cage

Quantity variable value or equation unit
Cage height h 2.2 m
Cage wall thickness w 0.02 m
Cage material – steel mesh 20% of surface area is solid
Density of steel ρ 7830 kg/m
3
Mass of a passenger (maximum) m 100 kg
Cost of steel mesh Cm 200 $/m
2
Table 2: Winch type, power rating and cost
Winch Type Input Power (kW) Cost Cw ($)
W50 50 12,250
W80 80 17,150
W100 100 24,000
W150 150 33,600
W200 200 47,000
Table 3: Steel rope type, diameter and cost (only Rope R-2 now

to be considered)
Steel Rope Type Diameter (mm) Cost Cr ($/m)
R-1 13 4
R-2 16 6
R-3 22 12
R-4 29 20
R-5 35 29
(9/9/2013) 5
Important note to students
The sections listed above are to be used to subdivide the
analysis and design process and
identify the sections you are to use for your Technical Analysis,
Presentation and Design
Proposal assessments, as detailed in the requirements of each
assessment.

IMPORTANT: This is a closed design problem where all
information required to
complete the technical analysis, calculations and evaluation of
possible solutions will be
available in the Client Brief, your text books or other provided
assignment material. The
problem presented is a simplified version of a real design
problem, so the fine details of the
components of the proposed system are ignored.
If you find yourself seeking information beyond that provided
in the Client Brief,
your text books or other assignment material then you are
probably over thinking the
problem. The three assessments using this problem are able to
be completed using just the
engineering fundamentals you are studying, supported by other
course material and tools
like the spreadsheet. There is no need to research commercial
equipment.
For the Technical Analysis assessment all students must
complete a technical analysis
and prepare a short technical report on Design Section 1 (only)
of the project. Your

memorandum to a (pretend) colleague is to request a technical
analysis and short
technical report on either section 2 or 3.
For the Presentation assessment each student will select a
design section of the project
(not section 1) on which to complete a technical analysis and
prepare a short oral
presentation. [This can be the same as the section identified in
your memo.] You are to
present a summarised technical analysis of that section of the
design and how it
depends-on / influences any other section of the design. The
presentation is to be
prepared and delivered as if to other colleagues in your
company who are working with you
on the larger project.
For the Design Proposal assessment students are expected to
complete the technical
analysis for the whole project, model the design on a
spreadsheet, evaluate some
alternatives within the design and select a specific design
solution to recommend in their
report. The recommendation must clearly specify all of the
parameters listed in the design

sections in bold, as they define each section of the design.
Students should note there is more than one correct answer to
this problem, as several
possible solutions will meet the requirements of the design.
Furthermore - a technical analysis of a single design section
ALONE is unlikely to
identify a set of design parameters that results in the final
project design, as the
sections are somewhat dependent on each other. Hence when
you complete a technical
analysis on a single section of the design you are not looking
for a specific ‘answer’ to
that section.
Your analysis should show the relationships between the
quantities within a section
and possibly with those in other sections of the design, to help
you understand the
problem. This analysis may enable you to eliminate some of the
possible choices of
equipment on offer (when it is evident it cannot do the job) or
you may be able to reduce the
range of values for some variables over which you expect they
will to contribute to finding

a viable solution.
---------------------------------------------------------------------------
---------------------
To : Mark Senott , (Design Engineer)
From : Daeej Ali , (Project Engineer)
Date : 12 Sep. 2013
Subject : Design a lift
---------------------------------------------------------------------------
---------------------
Mark
We want a design for an underground passenger lift to service
their
new mine. It’s supposed to take 300 passengers (150 up , 150
down).
I need from you the design for the winch part with the
calculations

and what power we need for that. It is necessary to provide me
with
the whole information about it ,model number, input power
requirement, torque delivered at drum. The winch efficiency is
70%.
I have completed the design for the cage sizing and I want you
to see
it that you might have a better idea about what we want
exactly.In
addition I want you to get which steel ropes should we use with
and
also with the information … the length, the number, the
calculations
and everything that’s about it. That should be designed with a
safety
factor = 8 and The maximum number of steel ropes that can be
fitted
to the lift is 6. The minimum length of steel rope that must
remain
wound on the drum is 3 full wraps around the drum.
Our budget to design this project is limited with an about
$40000 so

you should but that in your eye while getting what we need. We
will
see how much will it cost to do it with the final results that we
will get
later. I need all this to be provided and totally completed in the
20
th
of
Sep. so try to finifh it ASAP.
Regards
Ali
Mobile : 0468717944
Copy : John Terry , (Chief Engineer)
Technical Analysis (Lift Design)
By Daeej Ali
September 2013
#Introduction
This report is for designing an an underground passenger lift to

service their
new mine by Midas Gold Pty Ltd seeks submissions.
The lift is required to operate in a vertical shaft to 60m below
ground level. The
lift is required to transport 300 miners at the change of shifts
(150 up, 150
down) within a period of 15 minutes. The steel cage is to be
suspended by
multiple steel ropes from a winch drum positioned 10m above
ground level. The
winch is to be used to raise and lower the cage. Figure 1 shows
the dimensional
detail of the shaft, lift and winch system.
#Cage Size
To get the cage size we need to calculate a few things that
would help to design
it to be suitable. We need to have the mass and the weight of the
miners. Also

the cage floor area and the other parts of the cage area that is
the steel mesh.
Mass of the miners = Mass of a passenger (maximum) * number
of passengers
Mass of the miners =100*150 =15000 Kg
Weight of the miners = mass of the miners * g
Weight of the miners = 15000 * 9.81 = 147.15 KN
Then we can get the area of the cage floor by the equation
(w*w)
W*W = 0.25*150 = 37.5 m^2
W^2 = 37.5
W = √37.5 = 6.124 m
Cage Area = 2w^2 + (4*2.2*w)
Cage Area =2w^2 + 8.8 * w m^2
Cage Area = (2*6.124^2) + (8.8*6.124) = 128.898 m^2
Cage Volume = Cage Area * Thickness
Cage Volume = 128.898 * 0.02 = 2.578 m^3

ρ (Density of the cage) = Mass of the cage / Volume of the
cage
Mass of the cage = ρ (Density of the cage) * V (Volume of the
cage)
Mass of the cage = 7830 * 2.578 = 20185.74 Kg
Weight of the cage = mass of the cage * g
Weight of the cage = 20185.74 * 9.81 = 198.022 KM
T = m (g + a) ,,, when it’s up
Tmax = m (g+2g)
Tmax = m * 3g
M = 15000 (mass of the miners) + 20185.74 (mass of the cage)
= 35185.74 Kg
Tmax = 35185.74 * 3 * 9.81 = 1035.516 KN

T = m (g - a) ,,, when it’s down
Tmin = m (g – 0.5 g)
Tmin = m * 0.5 g
Tmin = 35185.74 * 0.5 * 9.82 = 172.586 KN
#SAFETY FACTOR = 8
Design for 1035.516 * 8 = 8284.128 KN
δ = F / A ,,,, F = δ * A
F=700*10^6 * π/4 * 0.061^2
F= 140.743 KN
-The force that can be carried by one cable.
The Winch
Efficiency = P(out) / P(in) X 100 %
70/100 = P(out) / P(in)
τ (Torque) = P(out) / w
τ = T(max) * r

τ = (1035516.328 * 8 ) * 0.25
τ = 2071032.656 Nm
P(out) = τ x w
P(out) = 2071032.656 x 20 = 41.421KW
P(in) = P(out) x 100 / 70
P(in) = 41.421x 100 / 70 = 59.173 KW
The cost of the Winch
our power input is about 60 KW , so we can find in the table
that
we can use W80 which costs $17,150
The cost of the steel ropes
#Number of ropes need = 6
R-2 cost is $6 / m . We need to use 10 meters for one rope. And
the
minimum length of the steel rope that must remain wound on
the

drum is 3 full wraps around the drum. Also we can remember
that the
radius of the drum is 0.25 m so we can figure out this.
10 + 2 π r * 3
10 + 2 π * 0.25 * 3 = 14.712 m (one rope)
And for 6 ropes
14.712 X 6 = 88.272 m
So the cost = 88.272 x 6 = $529.632
The cost of the steel cage
Total surface area of the mesh = w^2 + (2.2 x w) x 4
Total surface area of the mesh = w^2 + 8.8 x w
Total surface area of the mesh = (6.124)^2+(8.8 x
6.124)=91.395 m^2
Then from the table above we can see that it costs $200 for 1
m^2

So for the cage it will cost $200 x 91.395 = $18279
# in conclusion
The goals of this design are to maximise the rate at which
workers can be transported. In addition we don’t exceed the
budget of about $40000 that we can see with this project
design we just spent less than $36000 by offering a minimum
cost solutions that meets the requirements that we need to do
this project.
Daeej Ali
U1054594
ENG 1002 Sep. 2013
ClientBrief_V2.1memoReport

SpecificationsModel for Electracycle all electric motorcycle - Spe.docx

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