Loss modeling and analysis of the nuna solar car drive system

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International Conference on

Ecologic Vehicles & Renewable Energies

March 29 – April 1

Loss Modeling and Analysis of the Nuna Solar Car Drive

System

George C.M. Arkesteijn

CSE ENGI EAD Drives, Siemens Nederland NV Prinses Beatrixlaan 800, 2595 BN Den Haag, the Netherlands

E-mail: george.arkesteijn@siemens.com Erik C.W. de Jong

KEMA Consulting, Utrechtseweg 310, P.O. Box 9035, 6800 ET Arnhem, the Netherlands

E-mail: Erik.deJong@kema.nl

Henk Polinder

Electrical Power Processing, Delft University of Technolgy Mekelweg 4, 2628 CD Delft, The Netherlands

Abstract: The paper is devoted to the analysis of the losses in the drive system of the Nuna. The Nuna is a solar car that won the World Solar Challenge (WSC) a few times. The objective of this paper is to model and quantify the losses in the drive system of the Nuna, to validate this model by means of measurements, and to use this model to investigate if a decrease of the losses is possible. The losses in the drive are modelled to determine the optimal solar car parameters. In this model the losses of various components of the drive are described with equations that are a function of determinable and measurable quantities. In this way the model is suitable for calculating the losses that occur during the driving of a trajectory with a certain load to the drive.

Keywords: Solar car, permanent magnet motor, losses, efficiency, drive optimization.

1. Introduction

The Nuna is a solar car and like all solar cars, it is powered only by the energy that is radiated by the sun. Fig. 1 depicts the car. The Nuna won the World Solar Challenge (WSC) in Australia [10] three times (2001, 2003 and 2005). The WSC is known as the major race for solar cars. In this race, solar cars have to drive from Darwin to Adelaide (3020 km) as fast as possible. During this race, the Nuna drove at an average speed of about 100 km/h at an average power level of 2 kW.

Since the amount of energy that is gained from the sun is limited to around 2 kW, the available

energy must be wisely spent in order to be fast. In this way the design of the shape of the car is a trade off between a (small) aerodynamic shape and a large solar array surface. Besides, the car is made light to reduce the rolling resistance and to reduce the required power to accelerate and to climb up a hill.

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of the total solar car. The focus in this paper is on the losses in the drive of the car.

The drive train that is used in Nuna is designed at the Fachhochschule of Biel. Companies such as CSIRO in Australia and NGM in the USA also work on drive systems for solar cars. The rated power, torque, rotational speed and working points of the NGM motors and CSIRO motors are comparable to the Nuna drive. NGM is known to have motors with an efficiency of around 95%, CSIRO claims that the efficiency of the motor they developed is above 97.5% [4]. The objective of this paper is to model and quantify the losses in the drive system of the Nuna, to validate this model by means of measurements, and to use this model to investigate if a decrease of the losses is possible. The losses in the drive are modeled to determine the optimal solar car parameters. In this model the losses of various components of the drive are described with equations that are a function of determinable and measurable quantities. For instance the weight of the stator iron and the air gap length can be determined and torque and rotational speed can be measured. In this way the model is suitable for calculating the losses that occur during the driving of a trajectory with a certain load to the drive.

The paper starts with a description of the drive. Next, the loss components are estimated with the data from datasheets or determined by means of measurements. Subsequently, the model is compared to overall efficiency measurements. Finally, some conclusions are drawn about the efficiency and the performance of the drive.

Figure 1: Photograph of the Nuna.

2. Drive characteristics

When the drive operates as motor electrical power is converted into mechanical power and heat or losses. Various components produce various losses. The objective of the model derivation is to describe the major loss components of the drive with equations so the

losses can be analyzed for a wide range of working points.

A The Biel Drive

The Biel drive consists of a three-phase inverter connected to a permanent magnet brushless DC (BLDC) motor. The motor is a direct-drive motor in the rear wheel of the solar car. The electrical circuit of the drive is shown in fig. 2. This drive is capable of operating in three modes: forward motor mode, forward generator mode and backward motor mode. These different modes have consequences for the operation of the inverter and therefore on the amount of losses. On long tracks the drive operates mainly in forward motor mode and this mode will be described in the model.

The idealized voltage and current waveforms in a BLDC motor are shown in fig. 3.

Figure 2: Electrical circuit of the drive (power part only).

Figure 3: Waveforms per phase: line voltage, phase voltage, phase current, torque and power

The current that goes into a phase and comes out of another is called the motor current Im.

The user of the drive sets this current. The two phases that conduct Im are called the active

phases. The third phase that doesn’t current is called the inactive phase. Im is treated as a DC

current with a ripple.

Normally to control the current the DC voltage

UDC is applied to the phases either positive or

negative. However, in forward motor mode, if the induced voltage has a certain value, the two active phases are either connected to UDC or

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fig. 4. It reduces the switching losses. During unipolar switching, synchronous switching [8] is also used to bypass the diode (the diode in S4 in fig. 4). This reduces the conduction losses in the inverter.

Figure 4: Electrical circuit during unipolar switching. In this case only phase a is switched by S1 and S4. S5 is always on.

B Assumptions

Before the model is derived a list of assumptions is given:

- The inverter is assumed always to use synchronous switching and unipolar switching.

- The losses in the switches that are caused by switching the current and the voltage are assumed not to depend on the number of switches in parallel. In the model the mosfet driver power is assumed only to depend on the temperature, not on the number of mosfets in parallel per switch.

- For the calculation of the capacitor losses, the motor current ripple is assumed to be negligible.

- The resistance of the motor windings and the mosfets in the on state are measured with DC currents. This value is assumed not to depend on the frequency.

- The motor constants that are used to calculate the current required for the generated torque and the induced voltages in the windings are assumed not to depend on the stator current and the temperature. - The iron losses are divided into hysteresis

losses and eddy current losses. Hysteresis losses are assumed to be proportional to the frequency and eddy current losses are assumed to be proportional to the square of the frequency [1]. In reality, iron losses are more complex.

- The no load rotational loss (the power that is dissipated in the motor when the rotor rotates, Im is zero) is assumed to be only the

iron losses that are caused by the rotating

permanent magnets. Other losses are assumed to be zero.

- The phase current is assumed to be a 120° square wave current with a ripple for every working point and in phase with the induced voltage.

- The induced motor voltages are assumed to be trapezoidal.

- All flux is assumed to cross the air gap perpendicularly and is linked with the conductors of the windings for the calculation of the induced voltage in the windings.

- The magnetic field distribution is assumed not to change because of the magnetic field that is caused by the motor current.

- The flux density in the motor iron is assumed not to exceed the saturation flux density of the iron Bsat=1.48T.

C The motor constants

If there are no losses then during the 60° that two phases are active then the following equations apply: T m shaft I k T = , ₍₁₎ m E m k E = ω , ₍₂₎ E T k k = , (3) T k T k I E P _m T m E m m = ω =ω = (4)

where Tshaft is the shaft torque, kT is the motor

constant for the torque, kE is the motor constant for the induced voltage, Em is the induced motor

voltage, ωm is the rotational speed of the rotor

and P is the power. In the model the motor constant is used to determine the induced voltage in the windings and the motor current from the measured torque.

The motor constants can be determined using Faraday’s law

Blv

E= (5)

or the equation for the Lorentz force

Bil

F = . (6)

With both equations the motor constants become motor motor g T E k N pBl r k = =2 2 . ₍₇₎ where

N number of conductors per slot

P number of pole pairs

Bg air gap flux density

lmotor length of the motor (axial)

rmotor air gap radius of the motor

(4)

Fig. 5 shows one pole pair of the motor and the path of the magnetic flux from the permanent magnets. The flux from two magnets flows to the yoke through 3 teeth.

Figure 5: One pole pair of the motor and the path of the magnetic flux from one permanent magnet in the rotor.

Some assumptions are made to calculate the magnetic flux density in the air gap Bg. The first

assumption is that the magnetic field intensity is concentrated to the magnet and the air gap. The magnetic field intensity in the iron is neglected and the iron does not get saturated. The second assumption is that there is no leakage flux: All flux from the magnets crosses the air gap and links with the stator windings. The third assumption is that the recoil permeability of the magnet µr=1.

The following equations are used to derive the equation for the air gap flux density Bg [1,2,9]:

∑

∫

Hdl= i (8) g g H B =µ₀ (9) m r r m B H B = +µ₀µ (10) C g geff l k l = ⋅ (11) g t s t s C l w w w w k γ − + + = ₍₁₂₎

( )

⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ₋ ₊ = 4 ₂ arctan ₂ log 1 ₂ 2 g s g s g s l w l w l w π γ (13) m g m geff r m g w w l l B l B + = 2 (14) where

Hg air gap magnetic field intensity

Bm flux density in the magnet

Br remanent flux density of magnet

Hm magnetic field intensity in the magnet

lg radial air gap length

kC Carter factor

lgeff effective air gap length

ws radial width of the slot

wt radial width of a tooth inside the air gap

lm circumferential magnet length

wm radial width of the magnet

wg circumferential air gap width.

wgap is half of the width of the iron between two

magnets. kE and kT become 1.365 Vs/rad or

Nm/A.

Bg and the motor constants are measured by

measuring the induced voltage at a certain speed. The voltages are shown in fig. 6 for

ωm=49.53 rad/s.

The average line voltage during the 60-degree period when two phases are active is 54.26 V. The corresponding motor constant kE is 1.095

Vs/rad. The deviation of 15% with the calculated value and the shape of the waveforms can be caused by the stator iron being saturated in every position. kE=1.08Vs/rad is used in the

model.

Figure 6: Line voltages of the motor at 49.53 rad/s.

3. Loss model of the drive

A No load rotating losses

The iron losses that are caused by the moving permanent magnets are described in two equations: Ph for the hysteresis losses and Pe for

the eddy current losses. m

h

h T

P = ω (15)

Where the measured Th=71.1·10-3Nm is the

torque that is caused by hysteresis losses.

2

m eddy

e k

P = ω ₍₁₆₎

Where the measured keddy=1.797·10-3Nms/rad is

the torque per rad/s that is caused by the eddy current losses.

The motor current Im can be determined:

T m eddy h shaft m k k T T I = + + ω (17)

B Conduction losses in the motor

The conduction losses in the motor Pmo,cond are

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ph m cond mo I R P 2 2 , = ⋅ . (18)

Where Rph is the phase resistance:

winding winding ph A l R =ρ (19) Where ρ resistivity (17.5·10-9_Ωm) lwinding length of the winding

Awinding surface of the winding

Equation (19) gives Rph=34mΩ. The measured

Rph=52mΩ, probably because each conductor in

each slot is connected to the next conductor after the conductors were put in their slots, sometimes with an additional connector. The temperature correction for Rph is 0.4%/K.

C Duty ratio and switching frequency

The duty ratio D is calculated with

DC ph m E m DC m U R I k U U D= =ω + 2 . ₍₂₀₎

Where Um is the induced voltage plus the

voltage across the resistance of two phases. The switching frequency fsw in the model is a

function of D and is based on measurements

⎪⎩ ⎪ ⎨ ⎧ + − − − = e e sw sw sw f f f D f f 6 6 . 0 1 ) 6 )( 1 ( ₀ 0 6 . 0 6 . 0 > ≤ D D (21) D Ripple power

The ripple on the motor current induces some losses in the motor. The modeled ripple loss Pr

is based on measurements: 5 . 1 2 ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⋅ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⋅ = swm sw ppm pp rm r f f I I P P (22) where

Prm measured ripple loss 12W

Ipp peak-to-peak value on Im 8A

Ippm measured peak-to-peak value on Im 8A

fsw switching frequency Hz

fswm measured switching frequency 8kHz

E Conduction losses in the switches

The conduction losses in the inverter Psw,cond is

given by p s on DS m cond sw sw sw R I P = 2⋅ ₍ ₎⋅ , . (23) Where

RDS(on) “on” resistance of a mosfet 30mΩ

sws number of switches in series 2

swp number of switches in parallel 4

The temperature correction is 1%/K for RDS(on).

F Mosfet switching losses

The switching losses Psw(W) are given by

(

)

⎟ ⎠ ⎞ ⎜ ⎝ ⎛ _⋅ _⋅ ₊ ₊ ₊ ⋅ = sw DC m ir uf ur if sw f U I t t t t P 2 1 _{. (24)} Where fsw switching frequency Hz

tir current rise time during turn on 40ns

tuf voltage fall time during turn on 70ns

tur voltage rise time during turn off 200ns

tif current fall time during turn off 40ns

G Diode switching losses

The reverse recovery charge in the diode

Qrr=1.4µC is taken from datasheets. The

dissipated power in the diode Pdiode is calculated

with sw DC rr diode Q U f P 2 1 = (25) H Capacitor losses

The capacitor losses are calculated with cap ES cap cap n R I P = 2 ⋅ _. (26) where

RES equivalent series resistance 90mΩ

ncap number of capacitors parallel 6

The effective capacitor current depends on D and Im and is calculated with

(

)

(

I D

)

D

(

DI

) (

D

)

Icap = m 1− + m 1− 2 2 (27) I Driver Power

The driver circuitry consumes 8W at 25°C and 12W at 60°C. The driver circuitry is fed from a separate (12V) supply and left out of comparison of the model and overall efficiency measurements. In the model Pdriver=8W with a

temperature correction of 1,4%/K.

J Total losses

The power going into the motor Pmotor,in is

calculated in the model with ripple m m in motor U I P P , = + . (28)

The total power that goes into the power part of the inverter Pin is

.

,

,in sw swcond cap diode

motor

in P P P P P

P = + + + + ₍₂₉₎

The total modelled loss Pmodel is given by

mech in del

mo P P

(6)

4. Overall loss measurements

Two overall loss measurements were performed to compare the model with reality. The following equations apply.

DC DC in U I P = (31) shaft m out T P =ω (32) out in loss P P P = − (33) % 100 / ⋅ =P_in P_out η (34) where

Pin Input power from power supply W

Pout Shaft power shaft of the Biel motor W

Ploss The measured loss W

η The measured efficiency %

The first set of measurement was done with a varying DC bus voltage and a duty ratio close to one. The second set of measurements was done under more practical situations. In both

measurements Pdriver is left out of the

comparison and calculation because the driver is fed from a separate supply. The drive is quite efficient with these measurements.

Table 1 gives measurement results. Table 2 gives the results of the model and compares them to the measurements. There is a large difference between modelled and measured losses. One phenomenon that might explain this difference is the current waveform. An example is given in fig. 7. Relatively high current peaks occur during these measurements. These peaks increase the motor conduction losses. However, the higher conduction losses are not enough to explain the differences between the modelled and measured losses.

Table 3 and Table 4 show the measurement results of a more practical situation with constant DC bus voltage. The measurements show the highest efficiency at torques around 18 Nm and speeds around 125 rad/s, which is close to the main operating points during the WSC. The measurements again show a big difference between the modelled and the measured losses, even compared with the first measurement. Also the overall efficiency is lower than the efficiency of the first measurements.

The two measurements differ in two ways. One difference is the way in which the current is controlled. Another difference is the ripple on the motor current.

In the first set of measurements, the switches in the inverter to the active phases were always open, UDC limited the rotational speed of the

rotor. This may have introduced some not modelled losses.

Table 1: First measurement input and output power. The driver power is left out of consideration. UDC

(V) (A) IDC T(Nm) shaft ω(rad/s) m (W) Pin P(W) out P(W) mloss η (%) 70.4 5.67 5.9 62.8 399 369 30.3 92.41 70.4 5.77 6.0 62.8 406 376 30.2 92.56 121.0 9.19 10.0 104.8 1112 1045 66.5 94.02 131.8 9.18 10.0 115.2 1210 1146 63.8 94.72 142.7 9.30 10.0 125.7 1327 1257 70.2 94.71 72.2 11.28 12.2 62.8 815 764 50.6 93.79 118.3 13.27 14.7 102.6 1570 1504 66.6 95.76 130.4 13.66 14.9 113.0 1781 1687 93.9 94.73 143.6 13.23 14.8 121.5 1899 1793 106.4 94.40 119.6 17.44 19.4 102.4 2085 1991 94.0 95.49 Table 2 First measurement model results. The column “Diff” gives the difference between total modeled losses and measured losses in W. The last two columns separate the losses in the motor and the inverter. UDC (V) IDC (A) Tshaft (Nm) ωm (rad/s) Pmodel (W) Diff. (W) Invert. Losses Motor losses 70.4 5.67 5.9 62.8 16.2 14.1 0.64 15.60 70.4 5.77 6.0 62.8 16.1 14.1 0.62 15.47 121.0 9.19 10.0 104.8 40.5 26.0 1.94 38.60 131.8 9.18 10.0 115.2 45.4 18.4 1.97 43.39 142.7 9.30 10.0 125.7 50.8 19.4 2.01 48.76 72.2 11.28 12.2 62.8 28.7 21.9 2.31 26.38 118.3 13.27 14.7 102.6 51.8 14.8 3.66 48.09 130.4 13.66 14.9 113.0 57.8 36.1 3.91 53.85 143.6 13.23 14.8 121.5 62.6 43.8 4.17 58.44 119.6 17.44 19.4 102.4 69.4 24.6 6.16 63.21

Figure 7: UDC=115V, Tshaft=10Nm, ωm=100rad/s

CH2: Motor voltage, CH3 Motor current with peaks of 15% higher than the average motor current.

In the second set of measurements, connecting

UDC to the active phases or short-circuiting the

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voltage in the third phase, a current could flow in the third inactive phase when the two active phases are short-circuited.

Table 3 Second measurement results of input and output power without the power that is required for the driver UDC (V) IDC (A) Tshaft (Nm) ωm (rad/s) Pin (W) Pout (W) Ploss (W) η (%) 162.6 2.05 2.7 104.3 334 282 52.4 84.3 162.5 5.24 14.9 51.6 851 771 79.9 90.61 162.4 10.37 14.9 104.3 1684 1552 132.0 92.16 162.4 10.27 14.7 104.7 1668 1539 128.7 92.29 162.4 10.57 15.0 105.9 1716 1588 128.4 92.52 162.3 12.42 15.1 124.9 2016 1889 127.3 93.69 162.4 12.31 14.9 125.8 1999 1879 119.8 94.01 162.4 6.46 18.1 51.7 1049 934 115.1 89.03 162.3 12.65 17.7 106.0 2054 1876 177.9 91.34 190.0 10.77 17.9 105.1 2046 1886 160.0 92.18 162.2 14.81 18.0 125.5 2402 2258 143.6 94.02 162.0 20.78 29.4 104.4 3366 3070 296.3 91.20 162.0 21.01 30.0 103.7 3403 3110 293.1 91.39 Table 4 Comparisons of second measurements and model

UDC

(V) (A) IDC T(Nm) shaft (rad/s)ωm P(W) model Diff. (W) Invert.lossesMot. losses 162.6 2.05 2.7 104.3 41.8 10.6 1.0 40.9 162.5 5.24 14.9 51.6 58.7 21.2 8.0 50.7 162.4 10.37 14.9 104.3 72.2 59.8 7.3 65.0 162.4 10.27 14.7 104.7 71.5 57.2 7.1 64.4 162.4 10.57 15.0 105.9 72.8 55.6 7.2 65.6 162.3 12.42 15.1 124.9 76.2 51.1 6.1 70.1 162.4 12.31 14.9 125.8 75.8 44.0 5.9 69.9 162.4 6.46 18.1 51.7 73.0 42.1 10.6 62.4 162.3 12.65 17.7 106.0 84.7 93.2 9.3 75.5 190.0 10.77 17.9 105.1 92.6 67.4 11.5 81.1 162.2 14.81 18.0 125.5 89.1 54.5 8.0 81.1 162.0 20.78 29.4 104.4 158.2 138.1 21.1 137.1 162.0 21.01 30.0 103.7 162.9 130.2 21.9 141.0

One active phase becomes the inactive phase if it is disconnected from the plus or minus rail of the inverter. If it was connected to the minus rail then the induced voltage in that phase was negative and changes to positive in the following 60°. If the minus rail is used to short-circuit the active phases and the star point of the motor gets the same potential as the minus rail then the induced voltage in the inactive phase pushes the voltage between its two diodes in the inverter below the potential of the minus rail. In this way a current can flow in the inactive phase. This current either increases or decreases

the current in the other phases. This situation is illustrated in fig. 8 when phase b is the inactive phase.

Fig. 9 shows the voltages between the switches and the phase currents during unipolar switching.

Figure 8: Electrical circuit when the active phases are short-circuited. Phase b is the inactive phase, a current flows into this phase.

Figure 9: UDC =162V, Im =15A, and ωm =52rad/s.

Measured voltages between the switches of the inverter (above) and current waveforms in the motor.

The influence of the current in the inactive phase on the efficiency was not quantified, but it decreases the efficiency and reduces the torque.

This phenomenon can be prevented if the active phases are short-circuited via the correct rail. If the former active phase is disconnected from the minus rail then, during the first 30°, the plus rail should be used to short-circuit the active phases. During the second 30° the minus rail should be used to short-circuit the active phases and vice versa if the former active phase is disconnected from the plus rail.

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from happening and measure the difference in losses. A new inverter should be made to do this. The current should be controlled in the same way that is done now but the short circuit rail should depend on the position of the rotor. Adjustments to the model showed that a higher motor current in the model and the same motor voltage matched the model to the measurements quite well. This means that more motor current is needed for the motor to produce the same amount of torque for the model at the same motor voltage. The adjustments were not in the model during the analyses because there was too little known about the not modeled losses to be certain that the adjustments were valid.

5. Drive loss analysis and optimization

One objective of this model is to see if the motor is optimal and how it can be used optimal. For this several analysis have been done with the (incomplete) model of the motor and a model of the car. The model for the required power of the car Pcar is [4]

incl rr rr ar car P P P P P = + ₁+ ₂+ . (35) Where Par Aerodynamic resistance W

Prr1 Mass dependent rolling resistance W

Prr2 Wheel windage resistance W

Pincl Required power to climb a hill W

These powers can be written as

2 2 1

2

1 rr car rr car wheel rr car

rr P M C v n C v P + = + (36) 3 2 1 car air car d ar C A v P = ρ (37) incl car car incl M gv P = α (38) where

Mcar mass of the car

Crr1 mass rolling resistance coefficient

vcar speed of the car

nwheel number of wheels

Crr2 wheel windage coefficient

Cd (air) drag coefficient

Acar frontal area of the car

ρair density of air

g gravitational constant

αincl inclination

Above equations only hold for vcar is larger than

0m/s and αincl is smaller than 6%.

The shaft torque is calculated with car

car wheel

shaft r P v

T = / (39)

Where rwheel=0.25 m is the radius of the wheel.

Analysis showed that this radius is quite optimal

for the situation in 2003. The rotational speed is calculated with

wheel car

m =v /r

ω (40)

Fig. 10 depicts the modelled losses at various speeds. The ripple losses decrease at higher duty ratio and vcar in Australia is mostly larger

than 27.5 m/s. The largest losses and possible improvements are in the motor.

Figure 10: The different modelled loss components distributed for various speeds at UDC=170V

A Yoke Thickness

Calculations with the model show that a thicker yoke reduces the overall losses. If the outer radius of the yoke ryoke,out=89.95 mm stays

constant and the inner radius of the yoke

ryoke,in=85.45 mm decreases to ryoke,in2=80.95 mm

then the mass of the yoke increases from

myoke=1.72 kg to myoke2=3.39 kg and the total

benefit is a little more than 2 W.

B Driving up a hill cleverly

An analysis is performed to see if a hill can be climbed with lower losses if it is not done with a constant speed. One boundary next to being more efficient is that it must take the same time. The idea is to give some extra kinetic energy to the car before the hill. This helps the car to maintain the average speed on the hill with a lower torque than when the hill was taken with a constant speed.

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Pcond in the drive according to the model would

be roughly given by

(

wheel T

)

(

ph sw

)

cond T k R R

P = / 2 2 + . (41)

If Twheel is 55 Nm then Pcond is 311 W.

If Twheel is 15 Nm then Pcond is 23.1 W.

The idea is to increase the torque applied by the motor with 20 Nm to 35 Nm 100 m prior to the start of the hill and maintain this torque until the average speed of 27.5 m/s is reached again after the hill is taken.

For Twheel is 35 Nm Pcond is 127.1 W

The acceleration of the car acar depends on the

additional force on the car Fadd according to

car add

car F m

a = / (42)

Fadd is 80 N 100 m before to the hill and acar

becomes 0.229 m/s2_{. This acceleration is}

assumed constant. Just before the hill vcar

becomes 28.3 m/s and the increase of Pcar is 120

W.

The decelerating force on the hill Fadd=–80 N.

This means that the car decelerates with acar=–

0.229 m/s2_{. The speed halfway the hill is} vcar=27.5 m/s and on top of the hill it is

vcar=26.7 m/s. The decrease in power required

by the car to drive 26.7 m/s is 114 W. The car requires roughly another 100 m or 3.64 seconds to reach 27.5 m/s again.

Due to the changing speeds, the average car power Pcar and the required time change very

little. The total time that the torque was kept constant is 4 times 3.64 s is 14.55 s.

The conduction losses in the drive in the constant speed situation are two intervals 23.1 W and two intervals 311 W. The average power is 178.6 W. In the constant torque situation the conduction losses are 126 W for four time intervals. The difference is more than 50 W. A small hill and a torque that is too low for constant speed up the hill do not change the speed of the car so much. This analysis shows that a model of the drive is useful in a strategy program in which the inclinations of the road are known. For this the strategy program is recommended to control the speed of the car.

6. Conclusions and recommendations

The efficiency of the drive in the practical working points of 15Nm and 18Nm and 125rad/s varies between 93% and 94%. The measurements show that the drive is the most efficient when the duty ratio or the rotational speed is quite high. The efficiency decreases with a lower duty ratio. The highest efficiencies

were measured in the first set of measurements when the duty ratio was very close to one. According to the calculations that are done on the capacitor, the switches and the diodes of the mosfets, the inverter is quite efficient, around 99%. The most losses are in the motor. The design of the inverter is relatively straightforward and the components are used are cheap compared to the motor. Little attention has been paid to the improvement of the inverter except for the way that the motor phases are short-circuited. The recommended way to short-circuit the phases should not decrease the efficiency of the inverter.

The influence of the current in the inactive phase on the losses was not determined but this current can only decrease the efficiency. It is recommended to determine influence of the unwanted current in the third inactive phase on the losses. One way to do this is by measuring the efficiency of the drive with an inverter that prevents the current in the inactive phase to flow.

Not all losses are modeled. The model differs somewhat with the first set of measurements but a lot with the second set of measurements. The model seems to match the measurements very well if a larger motor current and thus more input power is used. The model has to be improved to draw more accurate conclusions about the drive.

The measured phase resistance is 33% larger than the calculated phase resistance. The cause of this difference is not known but if it is because of the way that the windings in the motor are manufactured then a lower resistance and lower conduction losses should be easily possible if the windings are kept in one piece instead of 168 pieces.

The calculated maximum ratio between rotational speed and induced motor voltage is 1.365Vs/rad. The maximum measured ratio is 1.185Vs/rad. If the teeth of the stator iron were not saturated the voltage shapes would have a more trapezium shape. Therefore the teeth of the iron in the stator of the motor are always saturated.

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The influence of a higher series inductance in the motor should be investigated. Placing series inductances between the inverter and the motor can simulate a higher motor inductance.

The used model is based on quite simple measurements. With the correct equipment, a model like this for another motor of the same type should be quite easily determined.

References

[1] P.C. Sen, Principles of Electric Machines and Power Electronica, second ed., Wiley, 1997. [2] N. Mohan, T.M. Undeland, W.P. Robbis, Power

Electronics, Converters, application and design, third ed., Wiley, 2003.

[3] R. Richter, Elektrische Maschinen, Erster Band, 3rd ed. Basel, Switzerland: Birkhäuser, 1967. [4] D.J. Patterson, ‘An efficient Optimized

controller for a Brushless DC Machine, And Loss measurement Using a simple Caloric Technique’, in Conf. Rec. of the 26th_IEEE Power Electronics Specialists Conference, 18-22 June 1995, pp. 22-27, Vol. 1.

[5] D.J. Patterson, X. Yan, S. Camilleri, ‘A very high efficiency controller for an axial flux permanent magnet wheel drive in a solar powered vehicle’, in Proceedings of the Int. Conf. on Power Electronic Drives and Energy Systems for Industrial Growth, 1-3 Dec. 1998, pp. 886 – 891, Vol. 2.

[6] D.J. Patterson, ‘Electrical system design for a solar powered vehicle’, in Conf. Rec. of the 21st

annual IEEE power electronics specialist conference, 11-14 June 1990, pp. 618 – 622. [7] H.C. Lovatt, V.S. Ramsden, B.C. Mecrow,

‘Design of an in-wheel motor for a solar-powered electric vehicle’, in IEE Proceedings – Electric Power Applications, Vol. 145, pp. 402 – 408, 1998

[8] C.S. Mitter, ‘Device considerations for high current, low voltage synchronousbuck regulators (SBR)’, in Conf. Proc. of WESCON/97, 4-6 Nov. 1997, pp. 281-288

[9] T.J.E. Miller, Brushless permanent-magnet and reluctance motor drives, Oxford University Press New York, 1989.

[10] http://www.wsc.org.au/

[11] http://www.skf.nl