ANALYSIS OF PERFORMANCES IN ELECTRIC AUTOMOBILES (II)

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ANALYSIS OF PERFORMANCES IN ELECTRIC AUTOMOBILES
(II)
PROF. EMILIO LARRODÉ University of Zaragoza
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ANALYSIS OF PERFORMANCES IN ELECTRIC AUTOMOBILES
1. TYRES 2. MECHANICAL RESISTANCE 3. POWER 4.
TRANSMISSION 5. CHARACTERISTIC CURVES 6.
ADHERENCE 7. ENERGETIC DESIGN 8. WORKING CYCLES
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6 ADHERENCE (FRICTION)In this chapter we are
going to analyze the four most extreme situations
in which the vehicle adherence to the road is
usually involved. - maximum acceleration -
maximum slope - maximum speed for a given
radius during turning - roll over verification
Preliminary assumptions and notation- 50 of
total vehicle weight over each axe. - vehicle
mass m- vehicle weight P- approximate
position of gravity center cdg- approximate
height of gravity center hcdg- wheel base d-
distance from front axe to cdg a- distance
from rear axe to cdg b- adherence coefficient
(friction) ma- rolling coefficient m

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• 6.1 MAXIMUM ACCELERATIONS STUDY
• Maximum accelerations and decelerations of the
  vehicle occurs during starting and braking.
• 6.1.1 Braking.
• Maximum force during braking is calculated in
  friction and dynamic conditions.
• Fmax P ma m g ma (1.50)
• Fmax m g deceleration max (1.51)
• Making equal two previous equations we obtain the
  value of maximum deceleration g deceleration
  max g ma (1.52)
• Once max. deceleration is obtained, we can obtain
  the braking distance for a given initial speed.
  To calculate braking distance (D) from vi to vf
  (vf 0) 1st we calculate braking time (t). In
  order to calculate the maximum braking distance
  we assume initial speed is the vehicle maximum
  speed (vi vmax).
• (1.53)
• (1.54)

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• 6.1.2 Maximum acceleration.
• To calculate maximum acceleration by friction
  conditions, we solve the forces system that can
  be seen in next figure
• where N1 normal reaction in front axe wheels
• N2 normal reaction in rear axe wheels
• F1 Traction force in front axe wheels
• F2 Traction force in rear axe wheels
• R1 Rolling resistance in front axe wheels
• R2 Rolling resistance in rear axe wheels

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• - 4 wheels drive case.
• By doing summatory of forces in the displacement
  axe (x) and in the normal axe (y), we obtain the
  following equations.
• P N1 N2 (1.55)
• m g F1 F2 - R1 - R2 (1.56)
• Moreover, it is verified that
• F1 N1 ma (1.57)
• F2 N2 ma (1.58)
• R1 N1 m (1.59)
• R2 N2 m (1.60)
• We obtain the following equation
• (N1 N2) ma - (N1 N2) m m g (1.61)
• substituting in the vehicle weight equation, we
  have

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• - Front wheels drive.
• In this case
• (1.65) (1.66)
• Fmax N1 ma (1.67)
• where m has been calculated by the S.A.E.
  formula.
• Time needed to reach determined speed by the
  vehicle (v)
• (1.68)
• However, this time is not real because that
  acceleration can not be maintained during all
  time, but only at the beginning at v 0 Km/h.
  If the calculated acceleration here is higher
  than the maximum acceleration capability obtained
  during performances analysis, our calculated
  vehicle will not skip. The maximum acceleration
  force will be then
• Fmax m g (1.69)

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• 6.2 MAXIMUM SLOPE
• In this case we do not consider
• the aerodynamic resistance. Then,
• we solve the following system
• of forces in the next figure
• We calculate the total resistance
• of the vehicle as well as
• the total traction force as
• RT R1 R2 P sin a (1.70)
• FT F1 F2 (1.71)
• As we are calculating maximum slope and speed
  will be very slow, we assume a very low value of
  acceleration (F m g (with m g 0 for
  maximum slope)), that is to say, we assume we
  are in total traction, then RT FT.

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• Solving the previous system equation
• F1 F2 R1 R2 P sin a (1.72)
• As
• F1 F2 P ma (1.73)
• R1 R2 P cosa m (1.74)
• Then P ma P cosa m P sina (1.75)
• ma m cosa sina (1.76)
• Taking into account and
  substituting in the previous equation, we obtain
  
• cos4a (1 - m2)2 4 m2 cos2a 2(1 - m2) (ma2
  - 1) - 4 m2 (ma2 - 1)2 0 (1.77)
• Solving this equation , obtaining ma depending on
  the type of soil and m from the SAE formula, we
  obtain the value of a. Then, the maximum value of
  the slope expressed in will be
• nmax 100 sina (1.78)

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• 6. 3 MAXIMUM VELOCITY IN TURNING
• To calculate maximum velocity in turning we
  compose centrifugal force with total resistance
  and then make equal to adherence or friction
  force. We assume that slope is 0. Using
  following figure
• (1.79)
• where Fc centrifugal force
• RT total resistance
• Fad adherence or friction force

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• Taking into account that
• RT Rr Ra Rp (1.80)
• Rr P m (1.81)
• Ra (1.82)
• Rp 0 (for this case)
• (1.83)
• Fad P ma (1.84)
• where r curve radius (m)
• v speed in m/s
• Because is 4 driving wheels, is independent which
  axe is. Substituting values
• (1.85)

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• 6.4 ROLL OVER WHILE TURNING
• To evaluate the possibility of roll over while
  turning we will solve the forces system which
  appears in turning. Looking to the figure
• where q vehicle inclination angle
• Ld front way
• Lt rear way
• Fd force done by the frontal suspension
  system.
• Ft force done by the rear suspension system.

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• Moreover we define
• td front way / 2 Ld / 2 (1.86)
• tt rear way / 2 Lt / 2 (1.87)
• Taking moments respect point O we obtain
• Ft tt Fd td Ft tt Fd td Fc
  hcdg (1.88)
• 2 Ft tt 2 Fd td Fc hcdg (1.89)
• Ft y Fd represent forces done by the suspension
  system of each frontal and rear suspensions,
  which are proportional to the displacements of
  suspension system, and then we can put as
• Ft tt q kt (1.90)
• Fd td q kd (1.91)
• where kt rear suspension stiffness
• kd frontal suspension stiffness
• Establishing the roll over condition (a pair of
  wheels do not touch the floor)
• Ft tt Fd td Fc hcdg (1.92)
• and substituting we obtain
• tt2 q kt td2 q kd m v2/r
  hcdg (1.93)

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• Then, we can obtain the vehicle inclination angle
  q
• (1.94)
• Establishing the forces equilibrium
• Fd Ft P m g (1.95)
• q tt kt td kd m g (1.96)
• Then
• (1.97)
• and substituting in the equation obtained for the
  angle q, we allow us to obtain the velocity at
  which roll over take place.
• (1.98)

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• 7 ENERGETIC DESIGN
• Assuming we have a vehicle with the following
  dimensions, and principal characteristics are
• Weight P (Kg)
• Speed v (m/s)
• Frontal surface S (m2)
• Aerodynamic coefficient Cx
• Slope n ()
• Rolling coefficient m
• Transmission efficiency hT
• 1. First, we calculate total resistance to
  movement (RT), which is the addition of rolling
  resistance (Rr), aerodynamic resistance (Ra) and
  resistance to overcome slopes (Rp).
• RT Rr Ra Rp (1.99)
• Rr m P (1.100)
• Ra 1/16 Cx S v2 (1.101)
• Rp P n / 100 (1.102)

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• 2. Once total resistance is obtained, we can
  calculate the power needed for a given speed v
• (1.103)
• 3. In order to determine the electric power that
  we need from the electric motor (Pb), it is
  necessary to check that our electric motor is
  able to give us the needed power to move the
  vehicle. This consumed power is calculated
  applying the efficiency of the electric motor
  (hE).
• f(P) (1.104)
• 4. Once we know the power of our electric motor,
  it is possible to determine the energy given by
  the batteries (Ebt), taking into account the
  quantity of time that this power is used.
• Ebt Pb t f(P,t) (1.105)

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• 5. Range of the vehicle (A) at a given speed can
  be calculated as a function of time that we use
  the power of the electric motor (t) as
• (1.106)
• Then the equation of the energy given by the
  battery system can be expressed as
• f(P,A) (1.107)
• 6. Analyzing the working of the battery system,
  we need to consider other efficiencies that are
  related to the charge (hc) and discharge (hd)
  process.
• If we add those efficiencies to the calculation
  of the given energy by the battery system, we
  will obtain, the total energy that our battery
  must to have (Ebt0)
• f(P,A) (1.108)

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• 7. Once we know the value of the needed energy to
  impulse the vehicle, we can calculate the weight
  of our battery (Pbt), knowing the the value of
  the battery energy capacity (density) (dE), which
  value will depend on the type of battery used.
• f(P,A) (1.109)
• 8. If we consider the possibility to have a deep
  discharge of the battery (because a high
  extension of range), we need to make an over
  dimensioning of the capacity of the battery by
  using an efficiency ratio (hdp) which used to be
  between 80 and 90 of the total capacity needed.
  Then we have
• f(P,A) (1.110)
• 9. Now, we can define the specific range concept,
  (Ae), which give us the vehicle range only as a
  function of the vehicle weight. Is defined as
  number of Km driven per battery kilogram, and can
  be expressed as
• f(P) (1.111)

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• 10. To evaluate the vehicle range (A), and
  considering the vehicle weight without batteries
  we can raw the following curves
• Specific range (Km/Kg) Specific range
  (Km/Kg)
• vs. Vehicle weight (Kg) vs. Battery
  weight (Kg)

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• Range (Km) Range (Km)
• vs. Battery weight (Kg) vs. Total vehicle
  weight (Kg)

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• 8 WORKING CYCLES
• For comparison purposes, we need study how our
  vehicle behave in different conditions, that is
  the reason because several kind of working cycles
  have been defined. These cycles represents the
  pattern for comparison.
• So, we have urban cycles, where we include
  accelerations, braking, and more frequent stops,
  as well as a more irregular velocity, against the
  road cycles, where we include a more regular way
  of driving, and with a less number of
  accelerations and braking.
• Working cycles give us
• medium conditions
• of driving in urban and
• suburban circuits.

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• 8.1 CICLE SAE J-227
• Is composed by the following periods
• 1º acceleration, 2º cruise speed, 3º
  deceleration, 4º braking, 5º stop.
• The figure shows
• different cycle periods
• where speed is given
• in Km/h and time in
• seconds.

Time (s)
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• 8.2 CYCLE ECE - 15
• It is composed by the following periods, urban
  period which is repeated 4 times and a suburban
  period. Cycle length is 1180 seconds, 195 s per
  urban period and 400 s per suburban period.

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• Example 3. For a given vehicle with the following
  characteristics 650 Kg weight, 1350 mm height,
  and 1481 mm width, aerodynamic coefficient of
  0.38, rolling coefficient 0.018 and range in
  urban cycle 30 Km (transmission and electric
  motor efficiency are 85 and 87 respectively).
  Analyze the total weight of the batteries if the
  cycle used is the following and we assume that
  there are no regeneration braking system and the
  energy capacity of batteries is 30 wh/Kg.
• Acceleration from 0 to 47 Km/h 16 s
• Cruise speed 47 Km/h 16 s
• Braking from 47 to 0 Km/h 7 s
• Stop 0 Km/h 25 s
• cycle time 64 s
• cycle length 400 m

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• P 650 Kg hT 85 m 0.018
  hE 87 Cx 0.38 A 30 Km
• S 0.8 a b 0.8 1.35 1.481 1.6
  m2 (1.112)
• Because it is a urban cycle, we need to know how
  many times we can repeat the cycle for the given
  range, taking into account that every cycle has a
  length of 400 meters
• (1.113)
• Calculation of total resistance.
• RT Rr Ra Rp (1.114)
• (assume that Rp 0)
• RT P m 1/16 Cx S v2 (1.115)
• 650 0.018 0.0625 0.38 1.6
  (13.05)2
• 11.7 6.47 18.17 Kg

Range
30.000




of cycles
75
Length
driven
in one cycle
400
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• Calculation of the power at a speed of 47 Km/h.
• (1.116)
• Calculation of energies for the complete cycle.
  
• Eacceleration Pb taverage 3147.15 16/2
  25177.2 ws (1.117)
• Ecruise Pb t 3147.15 16 50354.4
  ws (1.118)
• Ecycle Eacceleration Ecruise Ebraking
  Estop (1.119)
• 25177.2 50354.4 0 0
  75531.6 ws
• ETotal 75 Ecycle 75 75531.6 5664870
  ws (1.120)
• ETotal 1573.575 wh (1.121)

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• Calculation of battery weight.
• Energy capability of the battery for a 5 hours
  discharge
• dE 30 wh/Kg (1.122)
• Discharge time of the battery in the total cycle
• t discharge t cycle cycles 64 75 4800
  s 80 min 1h 20 (1.123)
• For this battery, the discharge capability in 1h
  20 corresponds to 70 of the discharge
  capability in 5 hours.
• (1.124)
• To prevent a deep discharge of the battery, an
  over dimensioning is necessary. In this case we
  will take hdp 90.
• 74.93 / 0.9 83.25 Kg (1.125)
• The calculation of the battery capacity is done
  taking into account the motor working voltage.
  Assuming a voltage of V 24 Volts, the needed
  capacity will be
• CAh ETotal / V 1573.575 wh / 24 v 65.56
  Ah (1.126)

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• Example 4. The vehicle of the previous example is
  subjected now to a road cycle, where after to get
  the cruise velocity (60 Km/h), maintain this
  speed during 90 Km. Calculate the battery weight,
  assuming that there no exist a regenerative
  braking system and the energy capability of the
  battery is 30 wh/Kg.
• Acceleration 0 a 30 Km/h 6.4 s Pw 27.2 Kw
  (triangle)
• Acceleration 30 a 60 Km/h 6 s Pw 27.2 Kw
  (rectangular)
• Cruise speed60 Km/h tcr s
• Cycle time (12.4 tcr) s
• Cycle length 90.000 m

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• P 650 Kg hT 85 m 0.018 hE
  87 Cx 0.38 A 90 Km
• S 0.8 a b 0.8 1.35 1.481 1.6
  m2 (1.127)
• There are not repetitive cycles, but we need to
  calculate the time used at constant speed (cruise
  speed).
• A v tcr tcr A / v 90/60 1.5
  hours (1.128)
• Calculation of total resistance.
• RT Rr Ra Rp (1.129)
• (we assume Rp 0)
• RT P m 1/16 Cx S v2 (1.130)
• 650 0.018 0.0625 0.38 1.6
  (16.66)2

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• Calculation of motor power at a speed of 60
  Km/h.
• (1.131)
• Calculation of the energy of the total cycle.
• Eacceleration from 0 to 30 Km/h ½ Pw t1
  0.5 27200 6.4 87040 ws (1.132)
• Eacceleration from 30 to 60 Km/h Pw (t2 - t1)
  27200 (12.4 - 6.4)
• 163200 ws (1.133)
• Eacceleration from 0 to 60 Km/h 87040 163200
  250240 ws (1.134)
• Due to the necessity to consider the electric
  motor efficiency
• Eacceleration Eacceleration from 0 to 60 Km/h /
  hE 250240/0.87
• 287632.1 ws 79.89 wh (1.135)
• Ecruise Pb tcr 4910.2 1.5 7365.3
  wh (1.136)

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• Calculation of battery weight.
• Energy capability of the battery for a 5 hours of
  discharge
• dE 30 wh/Kg (1.138)
• Time needed to discharge battery for the given
  cycle
• tdischarge 1h 30 (1.139)
• For this battery, the discharge capability in 1h
  30 corresponds to a 78 of the discharge
  capability in 5 hours.
• (1.140)
• To prevent a deep discharge of the battery, we
  apply an over dimensioning factor of hdp 90.
• 318.17 / 0.9 353.52 Kg (1.141)
• The calculation of the battery capacity is done
  taking into account the working voltage of the
  motor. If we assume that our motor works at a
  voltage of V 24 Volts, the battery capacity
  will be.