Advanced Electric Drive Vehicles 1st edition Ali Emadi SOLUTION MANUAL

COMPLETE SOLUTION MANUAL

SOLUTIONS MANUAL FOR ADVANCED ELECTRIC DRIVE VEHICLES

by

Ali Emadi

SOLUTIONS MANUAL FOR ADVANCED ELECTRIC DRIVE VEHICLES

by

Ali Emadi

Boca Raton London New York

CRC Press is an imprint of the Taylor & Francis Group, an informa business

CRC Press Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742 © 2015 by Taylor & Francis Group, LLC CRC Press is an imprint of Taylor & Francis Group, an Informa business No claim to original U.S. Government works Printed on acid-free paper Version Date: 20150128 International Standard Book Number-13: 978-1-4665-9773-0 (Ancillary) This book contains information obtained from authentic and highly regarded sources. Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint.

Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers.

For permission to photocopy or use material electronically from this work, please access www.copyright.com (http://www.copyright.com/) or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. CCC is a not-for-profit organization that provides licenses and registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged.

Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe.

Visit the Taylor & Francis Web site at http://www.taylorandfrancis.com and the CRC Press Web site at http://www.crcpress.com

Problem 1: Aerodynamic Resistance:

1 2  ACd V  Vw  2 1 2 Fw  (1.2kg / m3 )(2.05m2 )(0.32)  27.8m / s  2 Fw  10.94 N Fw 

Grading Resistance:

Fg  m g sin   Fg  (1500kg )(9.81m / s 2 )sin  6  Fg  1538.14 N Rolling Resistance:

Fr  Fz f r cos   Fr  (1500kg )(9.81m / s 2 )(0.013) cos  6  Fr  190 N

Problem 2:

T i i   2  en t 0 p  mgf r    rd   V  ACd  (450 Nm)(0.9)(3.21)(0.88)  2  (1500kg )(9.81m / s 2 )(0.013)  0.66m   V 3 2 (1.2kg / m )(2.05m )(0.32) V  62.6m / s V  225.34km / h

Problem 3: a

(v2 2  v12 ) 2( x2  x1 )

a

((0m / s ) 2  (27.8m / s ) 2 ) 2(50m  0m)

a  7.7 m / s 2 Fzf 

mg  ha   lr   (l f  lr )  g 

Fzf 

(1500kg )(9.81m / s 2 )  (0.6m)(7.7 m / s 2 )  1.25 m    2.5m 9.81m / s 2  

Fzf  10129.81N Fzr 

mg  ha  lf   (l f  lr )  g 

Fzf 

(0.6m)(7.7 m / s 2 )  (1500kg )(9.81m / s 2 )  1.25 m    2.5m 9.81m / s 2  

Fzr  4585.2 N

3.1 BMEP is the brake mean effective pressure, calculated by using the brake power/work. IMEP is the indicated mean effective pressure, calculated by using the indicated power. The main difference between them is the friction loss: the friction loss is considered in BMEP. To achieve maximal BMEP the following conditions should be considered: 1) High fuel conversion efficiency; 2) High mechanical efficiency (low friction loss); 3) Low pumping loss.

3.2 1) Compression ratio; generally, the greater the compression ratio the more efficient is the engine.

2) Constant volume combustion process (or short combustion duration): more towards Otto cycle.

3) Less pumping loss (increase volumetric efficiency).

4) Reduce friction loss (Lubrication).

3.3 1) The fuel injection amount.

2) The amount of intake charge.

3) Larger displacement volume.

4) Turbocharger or super-charger.

3.4 By definition:

MEP(kPa ) 

6.28nRT ( N  m) Vd (dm3 )

Assuming the given engine is four-stroke, then nR  2 , thus:

(3S-1)

Vd (dm3 ) 

6.28nRT ( N  m) 6.28  2 160   2.12dm3 BMEP(kPa) 950

(3S-2)

The displaced volume is also given by:

Vd  4 

 4

B2 L  4 

 4

B3  2.12dm3

(3S-3)

Thus, B  88mm .

The maximum brake power is calculated as:

Pb max 

BMEP  Vd  N 950  2.12  2500 / 60   41.87kW 2 103 nR 103

(3S-4)

3.5 The volumetric efficiency and mean speed piston speed are calculated by:

2  Vair Vd N

(3S-5)

S p  2 LN

(3S-6)

v 

Combining: S V N 7 Vair  v d  vVd p  0.9  0.006   78.75L / s 2 4L 4 120 103

(3S-7)

3.6 The chemical equation with stoichiometric air/fuel ratios is: C8 H 18  12.5(O2  3.773 N 2 )  8CO2  9 H 2O  47.16 N 2

Therefor the air/fuel ratio for gasoline is:

(3S-8)

 mair  12.5  (32  3.773  28)  15.09      m 8 12 18 gas  s

(3S-9)

The stoichiometric equation for diesel is: C12 H 23  17.75(O2  3.773 N 2 )  12CO2  11.5 H 2 O  66.97 N 2

(3S-10)

Therefor the air/fuel ratio for diesel is:  mair  17.75  (32  3.773  28)  14.63    12  12  23  mdiesel  s

(3S-11)

3.7 Assuming the heat capacity ratio is 1.4, then 

1.4

pd 1.2  2   1  2 1.41   0.8    0.528    pu 1.5  1  1.4    1 

(3S-12)

Thus the air flow is choked and can be calculated by:

(3S-13)

where, Av is the throttle plate area and C d is the discharge coefficient which varies with Av .

Assume Cd Av  0.0002 , and the upstream temperature is 300K, then the mass flow rate is calculated as:

3.8

.

1) The most critical factor that contributes the formation of nitrogen oxides is the incylinder peak temperature. The higher peak temperature, the more nitrogen oxides will generate. Another factor influencing the formation of nitrogen oxides is the oxygen concentration.

2) The EGR or burned gas diluent in the cylinder will increase the heat capacity of the cylinder charge to reduce the peak temperature and reduce the oxygen concentration.

3) As the EGR ratio increases from zero, the nitrogen oxides decrease.

4.1: Considering equation (4.8) of the chapter 4 and voltage levels specified, d=0.2 for Vdc=250V, d=0.6 for Vdc=500V, and d=0.69 for Vdc=650V.

4.2: If Vbatt=200V, then the input current iin= 20kW/200V=100A. Then, equation (4.9) of the chapter 4 yields iout= 80A for Vdc=250V, iout= 40A for Vdc=500V, and iout= 31A for Vdc=650V. If the power is 40kW, then iin= 40kW/200V=200A, which yields iout= 160A for Vdc=250V, iout= 80A for Vdc=500V, and iout= 62A for Vdc=650V. Note: Output currents can also be determined using the relation iout=P/Vdc.

4.3: According to input current levels from S2, maximum ripple level is Δipp=10%×100A=10A. Then, considering (4.32), minimum inductances corresponding to each output voltage can be defined, and are listed in Table 4.18. To meet requirement in all cases, a minimum inductance of 1600uH has to be selected.

Table 4.18: Inductance values to meet current ripple requirement Output voltage

250V

500V

650V

P=20kW

L=625uH

L=1300uH

L=1600uH

4.4: Power dissipated in a resistor R is equal to P=V2/R. As a result, corresponding resistance is R=21.12Ω.

4.5: Allowed voltage ripple level is equal to 6.5V. Then, according to equation (4.35) of the chapter 4, the capacitance has to be higher than 473uF.

4.6: Loss generated by the switches can be evaluated considering equations (4.114), (4.115), (4.116), and (4.117). To determine conduction time of each switch, operating conditions are considered to define the duty cycle. Neglecting the resistivity of the inductor, d=1‐Vin/Vout=0.33 according to (4.8). As a result, the conduction time of the switch is equal to Tswitch_on=0.33/Fsw=0.033ms. Conduction time of the Diode is equal to Tdiode_on=(1‐0.33) /Fsw=0.067ms.

Also, considering equation (4.3) of the chapter 4, input current circulating through the switches is equal to i=P/Vin=100A. Hence, Pinductor=0.01Ω*100A2=100W, Pcond_switch=1.3V*100A*0.33ms*Fsw=4.29W and Pcond_diode=1.4V*100A*0.067ms*Fsw=9.38W.

Turn‐on, turn‐off, and recovery loss of the switch and diode are equal to: Psw= (Eon + Eoff + Erec)*Fsw=270W. Finally, total loss of the switches is equal to Ploss=42.9+93.8+270+100=383.67W. Efficiency of the converter for the considered operating point is equal to Eeff=(20kW‐383.67W)/20kW=98.08%.

4.7: According to the relation (4.56) of chapter 4, phase voltage of the motor generated by the inverter depends on the modulation index Vm of the PMW as well as the dc‐bus voltage. As a results, considering (4.56) and Vm=1, minimum value of the dc‐bus voltage is Vdcmin=566V.

4.8: As detailed in section 4.4.3.2, the junction temperature of each power module can be estimated using the following relation:

T j  Pdiss .( Rth _ jc  Rth _ ch )  Tc

(4.10)

Results obtained for the power module are given in Table 4.19. It appears that the estimated junction temperatures are lower than the maximum allowed ones. Hence, the power module can be selected for this application and operating conditions.

Table 4.19: Estimated junction temperatures

Power

Tj IGBT (°C)

Tj Diode (°C)

IGBT OK?

Diode OK?

Power Module OK?

Module PM#1

118

114

YES

YES

YES

5.1 When the coil with N turns in Figure 5.90 is energized with current, the direction of magnetic field intensity, H will be upwards as per right hand rule. Assuming a uniform flux density and no leakage flux, Ampere’s law in (5.2) can be simplified as:

H clc  H g g  Ni

(s5.1)

where Hc and Hg are the magnetic field intensities in the core and airgap. Equation (s5.1) shows that the total magnetomotive force  Ni  produces magnetic field intensity and, hence, magnetic field around the core and airgap to produce flux in the magnetic circuit. Magnetic flux density is the magnetic flux per unit cross section area of the flux path, Ac. The relationship between magnetic field intensity and magnetic flux density is dependent on the characteristics of the medium and it was defined in (5.4) and (5.5):

B  H

   r 0

  B Ac

where  is the magnetic flux, 0 is the permeability of free space, and r is the relative permeability. Since the relative permeability of air is one, (s5.1) can be reorganized as:

H clc  H g g  Ni Bc

lc 

Bg

g  Ni

r 0 0 lc  g    Ni r 0 Ac 0 Ac    l g    Ni  c  r 0 Ac  0 Ac        Rg  Rc 

(s5.2)

Equation (s5.2) shows that the magnetomotive force  Ni  is analogous to the voltage, which derives magnetic flux (analogous to current) by overcoming the total magnetic reluctance in the flux path (analogous to resistance). Therefore, the magnetic circuit equation resembles to Ohm’s law in (5.23) and the equivalent circuit can be given as in Figure 5.90(b).

Now the magnetic flux in the circuit can be calculated as:



Ni Rc  Rg

Rc  Rg 



lc

r 0 Ac



lc

r 0  w d 



110  103

2500  4  10

g g   0 Ac 0  w d   4  107 

7

  18  10  10  10   3

3

0.6  103

18  10  10  10  3

3

 0.195  106 At Wb

 2.653  106 At Wb

Ni 10  20   7.023  10 5 Wb 6 6 Rc  Rg  0.195  10    2.653  10 

The flux density now can be calculated as:

B

 Ac



7.023  105 Wb  0.39 2 T  3 3 m 18  10 10  10 

The core material in this case is assumed to have linear properties and this is why a constant value for the relative permeability, r is applied. Ferromagnetic materials generally have nonlinear properties as shown in Figure 5.10.

As explained in Section 5.2, magnetic circuit has an inductance. It represents the inertia of the system and also energy stored in the magnetic circuit. Inductance can be derived from the flux linkage in (5.26): L

 i



N  B Ac  N  B Ac  N   i i  H c lc  H g g  N

N 2 B Ac   H c lc  H g g

N 2 B Ac N2  B B 1  lc g  lc  g    r0 0 Ac   r  0  0  10 2 N2 N2     35.11  H 6 6 lc g  R R    0.195 10 2.653 10     c g   r  0 A c  0 Ac

5.2

With sinusoidal magnetic field, the magnetization of the core material follows the hysteresis curve. As it was explained in Section 5.3.3, ferromagnetic materials are made of domains whereby the magnetic dipoles are 100% aligned. When an external magnetic field is applied, these magnetic domains try to align themselves in parallel to the axis of the applied magnetic field. The area under the hysteresis curve is the energy spent to change the orientation of the domains. This is the hysteresis loss and it is dependent on the peak value of the magnetic flux density and the frequency.

When the magnetic flux density in the machine is distorted due to non-ideal conditions, higher order harmonics appear in its waveform. These harmonics have higher frequency than the fundamental waveform and, since they have lower peak flux density, their hysteresis loop is different. This is called minor loop. Depending on the shape of the minor loop, excessive losses can be generated in the machine.

When the core material is exposed to an alternating magnetic field, eddy currents are induced in the core, which create a magnetic field opposing the external field. Due to the finite conductivity of the core materials, these eddy currents cause losses in the core. As it was explained in Section 5.3.4, in the ideal case, the field in a synchronous machine rotates with the same speed as the rotor. For a sinusoidal field distribution, the flux density that the rotor circuit sees does not change in time and, therefore, no eddy currents would be induced. If the airgap flux density is distorted, the higher frequency harmonics will rotate faster than the rotor. They will induce eddy currents and, hence, losses in the magnetic circuit.

Magnetic materials are electrically conductive. When the core material is exposed to an alternating magnetic field, eddy currents induced in the core generate losses. In practice, the core materials are laminated to reduce the eddy current loss. As shown in Figure 5.11, this reduces the conductive path of eddy currents and increases the resistivity of the core. For a given induced emf due to the time changing magnetic field, the eddy currents will be lower in a laminated core, because of the higher equivalent core resistance. As explained in Section 5.5.1, although laminating the core helps reducing the eddy current loss, it reduces the stacking factor. Stacking factor depends on the lamination thickness and also on the manufacturing process like punching and stamping of the lamination steels. If the lamination thickness is too low, the manufacturing process might be more challenging. Especially in mass manufacturing of electrical machines with thin laminations, the pressure applied to the core material should be carefully evaluated not to cause additional stress on the core, which might increase the core losses.

5.3

Higher number of phases helps creating a more sinusoidal ampere-conductor distribution in the airgap. But, for integral slot windings, where the number of slots per phase per pole is an integer number, the total number of slots has to be higher when the number of phases increases. For the given stator outer diameter, this means that the slot area reduces. This makes the implementation of the winding more challenging. For the same input power, in electric machines with higher number of phases, the current in each phase reduces. Therefore smaller conductors can be used.

When the number phases increase, the inverter topology also changes and the number of semiconductors increases. However, since each phase, carries lower current, the size of the semiconductors will be also lower. Higher number of phases also reduces the DC-link current ripple; therefore the DC-link capacitor requirements might be lower. However, when designing a motor drive system with higher number of phases, a comprehensive analysis has to be conducted to analyze the advantages and disadvantages in terms of the performance, manufacturing, volume, cost and control of the electric machine and the power converter.

As explained in Section 5.4, the rotational speed of the current linkage waveform is dependent on the frequency of the current waveform and the number of poles. For the given frequency, the machine with higher number of poles will have lower synchronous speed. In this case, to achieve higher speeds, the frequency of the applied current has to be increased. Depending on the design of the machine this might increase the losses and, for high speed operation, a higher switching frequency might be required for the converter. This would also increase the losses in the converter. On the other hand, as given in (5.33), the electromagnetic torque is directly proportional to the number of poles. Therefore, the selection of number of poles is highly related to the torque, speed, and efficiency requirements of the electric machine.

By increasing the number of slots per phase per pole, the number of coils is increased. This can be observed in the 5-phase windings in Figure 5.92. The 4-pole version has 4 slots per phase per pole; whereas the 8-pole version has 2 slots per phase per pole. The higher number of coils can be observed from the density of the end turns in the 4-pole version. When the number of slots per phase per pole is increased, a more sinusoidal current linkage waveform can be maintained as shown in Figure 5.15 (d). However, a very high value of number of slots per phase per pole leads to higher number of coils and higher manufacturing cost.

5.4

The permeance coefficient at point A is the slope of the load line:

PC A  

1

0.3313  0.4 0 656976.6572

When the temperature increases to T2, the operating point switches to the intersection of the load line and the demagnetization curve at T2. Therefore magnetic field intensity at point B is HB and the magnetic flux density is BB, then from the slope of the demagnetization curve (in figure 5.93 the slope of the line at point B was given 100μ0, it should be 100):

1

BB  100 0 H B  495000 and from the load line



1 BB  0.4 0 H B

Solving these two equations results in  H B  493027.888 A m  BB  0.278 T

When the permeance coefficient of the magnetic circuit changes to PC2, the magnet start operating at point C, where the field intensity is HC and the magnetic flux density BC. Similar to the previous case, from the slope of the demagnetization curve:

1 BC  BB  1.05 0 H C  H B and from the load line



1 BC 4 0 H C

Solving these two equations results in  H C  141562.463 A m  BB  0.712 T

The remanent magnetic flux density of the new magnetization line can be calculated using the slope of the recoil line: 1 BC  Br  1.05  Br  0.898 T 0 H C It can be observed that the remanent magnetic flux density has been reduced from 1.20 to 0.898.

As explained in Section 5.5.2, this means that the magnet has lost some of its magnetization.

This is an irreversible loss and the magnet should be remagnetized to get back to its original operating point.

5.5

The base speed of an electric machine is the point where the induced voltage equals to the phase voltage applied from the inverter. If a higher speed is commanded at this point without field weakening, the induced voltage becomes higher than the terminal voltage and the inverter cannot supply current to the motor (Please see Section 5.6.1.6 to get more information about field weakening). In this case the electromagnetic torque drops down and the shaft speed reduces due to the load torque. If a higher voltage is applied to the inverter, this means that the inverter can supply higher phase voltage to the motor and a higher base speed can be achieved. It should be noted that, this would be possible only if the inverter and the motor are designed to handle the higher voltage. Therefore, if the base speed of an inverter-driven motor is 2000rpm at 200 Vdc, its base speed will be 4000rpm at 400Vdc for ideal conditions. If the peak power of the motor is 50kW at 200Vdc, this means that it delivers the peak torque at the peak current at 2000rpm.

When the DC-link voltage of the inverter is increased to 400Vdc, the inverter can still supply the same current to the motor when the speed and, hence, the induced voltage increases. Therefore, the peak power of the motor will increase to 100kW, considering ideal conditions.

If the motor, MD1 is designed for the current I1 and voltage V1, then it is designed to deliver a certain torque at a defined rated speed. Consider a design MD2 with voltage V2 = 2 x V1 and the same current I1, the base speed of MD2 will be twice the base speed of MD1. Also the efficiency in high speed regions is greatly improved and the peak torque at maximum speed is doubled when compared to the design MD1. Consider a design MD3 with voltage V3 = 2 x V1, and number of turns doubled as compared to MD1. In this case the performance of MD3 is the same as the performance of MD1. However if the turns are doubled the current reduces by half.

Copper loss between MD1 and MD3 is the same. Hence the efficiency also remains the same between MD3 and MD1. If the turns are doubled, it is very essential to change the wire size and/or the number of strands to maintain the same slot fill factor. If the turns are doubled, the reduced current in the motor can have significant advantages to reduce the cable size (between batter and inverter, inverter and motor) cost and weight. In case of permanent magnet machines, when the voltage rating changes, demagnetization characteristics of the permanent magnets might not be the same anymore. Therefore, the size of the magnets should be reevaluated.

If design MD1 has a stack length L1, if we just change the design to MD2 by doubling the stack length L2 = 2 x L1, then the peak torque of MD2 will be twice the peak torque of MD1; however

the base speed of MD2 will be half of the base speed of MD1. The peak torque of MD2 will be half the peak torque of MD1 at maximum motor speed.

The magnet remanence is the magnetic flux density of the magnet under open circuit conditions.

If the magnet remanence is higher, for the given magnetic circuit, the magnet generates more flux, and at a given speed, this results in higher induced voltage. Number of turns in the stator windings has the similar effect. Higher number of turns increases the back EMF. As it was shown in Figure 5.24, iron losses dominate in high-speed region. In the high-speed region, the back EMF or induced voltage is higher than the phase voltage applied by the inverter for a certain DC-link voltage. If the back EMF is lower, this means that the phase advancement of the current waveform can be reduced and higher torque can be achieved at high-speed region.

Higher torque means higher flux, and at a given speed, higher rate of change of flux density.

Then the iron losses will increase, leading to lower efficiency.

If a motor design is capable of providing continuous torque of 50Nm and continuous power of 20kW at 1500C, and if the motor temperature decreases to 750C, then the coil resistance will reduce and the magnetic flux density of the permanent magnets will be higher. This means that the motor can deliver higher continuous torque and power.

In permanent magnet machine, cogging torque appears due the difference in the airgap permeance when the permanent magnet is aligned with the slot opening and the stator teeth.

Cogging torque can be reduced by skewing the continuously skewing the stator core or step skewing the rotor magnets, optimizing the size of the magnets or stator tooth geometry, optimizing the magnet pole arc length, and by special winding design techniques. When the slot opening is wider, the difference in airgap permeance and, hence, the cogging torque will be higher. Skewing reduces the peak torque, average torque, but it also reduces the torque ripple.

Problems: Problem 1: Part a) 1

𝜔𝜔(𝑠𝑠) = 𝐽𝐽𝐽𝐽+𝑏𝑏 𝑇𝑇𝑒𝑒𝑒𝑒 (𝑠𝑠) 𝑇𝑇𝑒𝑒𝑒𝑒 (𝑠𝑠) = 𝐾𝐾𝑡𝑡

1

𝐿𝐿𝑎𝑎 𝑠𝑠+𝑅𝑅𝑎𝑎

𝑒𝑒(𝑠𝑠) = 𝐾𝐾𝑒𝑒 𝜔𝜔(𝑠𝑠)

(P1-1)

(𝑉𝑉(𝑠𝑠) − 𝑒𝑒(𝑠𝑠))

(P1-2) (P1-3)

Substituting P1-3 into P1-2 we get: 𝑇𝑇𝑒𝑒𝑒𝑒 (𝑠𝑠) = 𝐾𝐾𝑡𝑡

1

𝐿𝐿𝑎𝑎 𝑠𝑠+𝑅𝑅𝑎𝑎

�𝑉𝑉(𝑠𝑠) − 𝐾𝐾𝑒𝑒 𝜔𝜔(𝑠𝑠)�

(P1-4)

�𝑉𝑉(𝑠𝑠) − 𝐾𝐾𝑒𝑒 𝜔𝜔(𝑠𝑠)�

(P1-5)

Substituting P1-4 into P1-1 we get: 1

1

𝜔𝜔(𝑠𝑠) = 𝐽𝐽𝐽𝐽+𝑏𝑏 𝐾𝐾𝑡𝑡

𝐿𝐿𝑎𝑎 𝑠𝑠+𝑅𝑅𝑎𝑎

Solving P1-5 for ω(s) we get: 𝜔𝜔(𝑠𝑠) = or:

𝜔𝜔(𝑠𝑠)

𝐾𝐾𝑡𝑡 𝑉𝑉(𝑠𝑠)

𝐽𝐽𝐽𝐽𝑎𝑎 𝑠𝑠2 +(𝐽𝐽𝑅𝑅𝑎𝑎 +𝑏𝑏𝐿𝐿𝑎𝑎 )𝑠𝑠+ 𝐾𝐾𝑒𝑒 𝐾𝐾𝑡𝑡 + 𝑏𝑏𝑏𝑏𝑎𝑎 𝐾𝐾

= 𝐽𝐽𝐽𝐽 𝑠𝑠2 +(𝐽𝐽𝑅𝑅 +𝑏𝑏𝐿𝐿 𝑡𝑡 )𝑠𝑠+ 𝐾𝐾 𝐾𝐾 + 𝑏𝑏𝑏𝑏 𝑉𝑉(𝑠𝑠) 𝑎𝑎

𝑎𝑎

𝑎𝑎

𝑒𝑒 𝑡𝑡

𝑎𝑎

(P1-6)

(P1-7)

Part b) 1

𝜔𝜔(𝑠𝑠) = 𝐽𝐽𝐽𝐽+𝑏𝑏 (𝑇𝑇𝑒𝑒𝑒𝑒 (𝑠𝑠) − 𝑇𝑇𝐿𝐿 (𝑠𝑠)) 1

𝑇𝑇𝑒𝑒𝑒𝑒 (𝑠𝑠) = 𝐾𝐾𝑡𝑡 𝐿𝐿 𝑠𝑠+𝑅𝑅 (−𝑒𝑒(𝑠𝑠)) 𝑎𝑎

𝑒𝑒(𝑠𝑠) = 𝐾𝐾𝑒𝑒 𝜔𝜔(𝑠𝑠)

𝑎𝑎

(P1-8) (P1-9) (P1-10)

Substituting P1-10 into P1-9 we get: 𝑇𝑇𝑒𝑒𝑒𝑒 (𝑠𝑠) = −𝐾𝐾𝑡𝑡

1

𝐿𝐿𝑎𝑎 𝑠𝑠+𝑅𝑅𝑎𝑎

𝐾𝐾𝑒𝑒 𝜔𝜔(𝑠𝑠)

(P1-11)

Substituting P1-11 into P1-8 we get: 1

𝜔𝜔(𝑠𝑠) = 𝐽𝐽𝐽𝐽+𝑏𝑏 �−𝐾𝐾𝑡𝑡 𝐾𝐾𝑒𝑒

1

𝐿𝐿𝑎𝑎 𝑠𝑠+𝑅𝑅𝑎𝑎

𝜔𝜔(𝑠𝑠) − 𝑇𝑇𝐿𝐿 (𝑠𝑠)�

(P1-12)

Solving P1-12 for ω(s) we get: −(𝐿𝐿 𝑠𝑠+ 𝑅𝑅 )𝑇𝑇 (𝑠𝑠)

𝜔𝜔(𝑠𝑠) = 𝐽𝐽𝐽𝐽 𝑠𝑠2 +(𝐽𝐽𝑅𝑅 𝑎𝑎+𝑏𝑏𝐿𝐿 𝑎𝑎)𝑠𝑠+𝐿𝐿𝐾𝐾 𝐾𝐾 + 𝑏𝑏𝑏𝑏 𝑎𝑎

or:

𝜔𝜔(𝑠𝑠)

𝑇𝑇𝐿𝐿 (𝑠𝑠)

=

𝑎𝑎

𝑎𝑎

𝑒𝑒 𝑡𝑡

𝑎𝑎

−(𝐿𝐿𝑎𝑎 𝑠𝑠+ 𝑅𝑅𝑎𝑎 )

𝐽𝐽𝐽𝐽𝑎𝑎 𝑠𝑠2 +(𝐽𝐽𝑅𝑅𝑎𝑎 +𝑏𝑏𝐿𝐿𝑎𝑎 )𝑠𝑠+ 𝐾𝐾𝑒𝑒 𝐾𝐾𝑡𝑡 + 𝑏𝑏𝑏𝑏𝑎𝑎

(P1-13)

(P1-14)

Problem 2: Part a)

𝐼𝐼 (𝑠𝑠) =

1

𝐿𝐿𝑎𝑎 𝑠𝑠+𝑅𝑅𝑎𝑎

𝑉𝑉(𝑠𝑠)

(P2-1) 𝐾𝐾

𝑉𝑉 (𝑠𝑠) = 𝐾𝐾𝑉𝑉 �𝐾𝐾𝑝𝑝 + 𝑖𝑖 � (𝐼𝐼 ∗ (𝑠𝑠) − 𝐼𝐼(𝑠𝑠)) 𝑠𝑠

(P2-2)

Substituting P2-2 into P2-1 we get:

𝐼𝐼 (𝑠𝑠) =

1

𝐿𝐿𝑎𝑎 𝑠𝑠+𝑅𝑅𝑎𝑎

𝐾𝐾𝑉𝑉 �𝐾𝐾𝑝𝑝 +

𝐾𝐾𝑖𝑖 𝑠𝑠

Solving P2-3 for I(s) we get: ∗

𝐾𝐾𝑉𝑉 �𝐾𝐾𝑝𝑝 𝑠𝑠+ 𝐾𝐾 �𝐼𝐼 (𝑠𝑠)

� (𝐼𝐼 ∗ (𝑠𝑠) − 𝐼𝐼(𝑠𝑠))

𝐼𝐼(𝑠𝑠) = 𝐿𝐿 𝑠𝑠2 +�𝑅𝑅 + 𝐾𝐾 𝐾𝐾𝑖𝑖 �𝑠𝑠+ 𝐾𝐾 𝐾𝐾 or:

𝑎𝑎

𝑎𝑎

𝑉𝑉 𝑝𝑝

𝑉𝑉 𝑖𝑖

𝐾𝐾𝑉𝑉 �𝐾𝐾𝑝𝑝 𝑠𝑠+ 𝐾𝐾𝑖𝑖 � 𝐼𝐼(𝑠𝑠) = ∗ 2 𝐿𝐿𝑎𝑎 𝑠𝑠 +�𝑅𝑅𝑎𝑎 + 𝐾𝐾𝑉𝑉 𝐾𝐾𝑝𝑝 �𝑠𝑠+ 𝐾𝐾𝑉𝑉 𝐾𝐾𝑖𝑖 𝐼𝐼 (𝑠𝑠)

(P2-3)

(P2-4)

(P2-5)

Part b) Modifying figure 6.1-7 to include the voltage amplifier gain K V and assuming the controller time constant is set equal to the electrical time constant, we get: i* +

-

Ki s

v*

v KV

1 Ra

i

i

𝐼𝐼 (𝑠𝑠) =

1

𝑅𝑅𝑎𝑎

𝑉𝑉(𝑠𝑠)

𝑉𝑉 (𝑠𝑠) = 𝐾𝐾𝑉𝑉

𝐾𝐾𝑖𝑖 𝑠𝑠

(P2-6)

(𝐼𝐼 ∗ (𝑠𝑠) − 𝐼𝐼(𝑠𝑠))

(P2-7)

Substituting P2-7 into P2-6 we get:

𝐼𝐼 (𝑠𝑠) =

1

𝑅𝑅𝑎𝑎

𝐾𝐾𝑉𝑉

𝐾𝐾𝑖𝑖 𝑠𝑠

(𝐼𝐼 ∗ (𝑠𝑠) − 𝐼𝐼(𝑠𝑠))

(P2-8)

Solving P2-8 for I(s) we get: 𝐼𝐼(𝑠𝑠) 𝐾𝐾𝑉𝑉 𝐾𝐾𝑖𝑖 = ∗ 𝑅𝑅𝑎𝑎 𝑠𝑠+ 𝐾𝐾𝑉𝑉 𝐾𝐾𝑖𝑖 𝐼𝐼 (𝑠𝑠)

(P2-9)

Replacing s  2πf j: 𝐼𝐼(2πf j) 𝐾𝐾 𝐾𝐾 = 𝑅𝑅 2πf𝑉𝑉j+𝑖𝑖𝐾𝐾 𝐾𝐾 𝐼𝐼∗ (2πf j) 𝑎𝑎 𝑉𝑉 𝑖𝑖

(P2-10)

Equating the real and imaginary parts of the denominator, we get: 𝑅𝑅𝑎𝑎 2πf = 𝐾𝐾𝑉𝑉 𝐾𝐾𝑖𝑖

(P2-11)

or:

𝐾𝐾𝑖𝑖 =

𝑅𝑅𝑎𝑎 2πf 𝐾𝐾𝑉𝑉

(P2-12)

And given that the electrical and control time constants are equal, we get: 𝐿𝐿 2πf

𝐾𝐾𝑝𝑝 = 𝑎𝑎𝐾𝐾

𝑉𝑉

(P2-13)

Problem 3: Part a) 𝑉𝑉𝑎𝑎

= 2.65 𝑟𝑟𝑟𝑟𝑟𝑟�

𝑃𝑃𝑟𝑟

= 610.4 𝑁𝑁𝑁𝑁

𝐾𝐾𝑒𝑒 =

𝜔𝜔𝑛𝑛𝑛𝑛

𝑇𝑇𝑟𝑟 =

𝜔𝜔𝑟𝑟

Part b)

𝐼𝐼𝑟𝑟 =

Part c)

𝑇𝑇𝑟𝑟

𝐾𝐾𝑡𝑡

𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣

𝑠𝑠𝑠𝑠𝑠𝑠

= 230.1 𝐴𝐴

From 6.1.3.3-2 and -3, we get:

𝐾𝐾𝑝𝑝 = 2𝜋𝜋𝑓𝑓𝑏𝑏𝑏𝑏 𝐿𝐿𝑎𝑎 = 0.377Ω 𝐾𝐾𝑖𝑖 = 2𝜋𝜋𝑓𝑓𝑏𝑏𝑏𝑏 𝑅𝑅𝑎𝑎 = 377

Ω

𝑠𝑠𝑠𝑠𝑠𝑠

Problem 4: Part a) From 6.2.2.2-10, 𝑖𝑖𝑠𝑠𝑞𝑞𝑞𝑞 (𝑡𝑡) =

2

𝑖𝑖𝑠𝑠𝑞𝑞𝑞𝑞 (𝑡𝑡) =

2

3

1

1

2 3

With no zero sequence term, 𝑖𝑖𝑏𝑏𝑏𝑏 (𝑡𝑡) + 𝑖𝑖𝑐𝑐𝑐𝑐 (𝑡𝑡) = −𝑖𝑖𝑎𝑎𝑎𝑎 (𝑡𝑡) 3

1

�𝑖𝑖𝑎𝑎𝑎𝑎 (𝑡𝑡) − 2 𝑖𝑖𝑏𝑏𝑏𝑏 (𝑡𝑡) − 2 𝑖𝑖𝑐𝑐𝑐𝑐 (𝑡𝑡)� = �𝑖𝑖𝑎𝑎𝑎𝑎 (𝑡𝑡) − 2 (𝑖𝑖𝑏𝑏𝑏𝑏 (𝑡𝑡) + 𝑖𝑖𝑐𝑐𝑐𝑐 (𝑡𝑡))� 1

�𝑖𝑖𝑎𝑎𝑎𝑎 (𝑡𝑡) − 2 �−𝑖𝑖𝑎𝑎𝑎𝑎 (𝑡𝑡)�� = 𝑖𝑖𝑎𝑎𝑎𝑎 (𝑡𝑡)

𝑖𝑖𝑠𝑠𝑑𝑑𝑑𝑑 (𝑡𝑡) =

2 3

√3

√3

�− 2 𝑖𝑖𝑏𝑏𝑏𝑏 (𝑡𝑡) + 2 𝑖𝑖𝑐𝑐𝑐𝑐 (𝑡𝑡)� =

1

√3

(𝑖𝑖𝑐𝑐𝑐𝑐 (𝑡𝑡) − 𝑖𝑖𝑏𝑏𝑏𝑏 (𝑡𝑡))

Part b) From figure 6.2-4, 𝑖𝑖𝑒𝑒𝑞𝑞𝑞𝑞 (𝑡𝑡) = 𝑖𝑖𝑠𝑠𝑞𝑞𝑞𝑞 (𝑡𝑡) cos(𝜃𝜃) − 𝑖𝑖𝑠𝑠𝑑𝑑𝑑𝑑 (𝑡𝑡) sin(𝜃𝜃) = 𝑖𝑖𝑎𝑎𝑎𝑎 (𝑡𝑡) cos(𝜃𝜃) − 𝑖𝑖𝑒𝑒𝑑𝑑𝑑𝑑 (𝑡𝑡) = 𝑖𝑖𝑠𝑠𝑞𝑞𝑞𝑞 (𝑡𝑡) sin(𝜃𝜃) + 𝑖𝑖𝑠𝑠𝑑𝑑𝑑𝑑 (𝑡𝑡) cos(𝜃𝜃) = 𝑖𝑖𝑎𝑎𝑎𝑎 (𝑡𝑡) sin(𝜃𝜃) +

Part c) (from MathCad) Im := 1⋅ A

amplitude (1 amp)

f := 50⋅ Hz

frequency (50 Hz)

ω := 2⋅ π ⋅ f

angular frequency (100π rad/sec)

1

√3 1

√3

(𝑖𝑖𝑐𝑐𝑐𝑐 (𝑡𝑡) − 𝑖𝑖𝑏𝑏𝑏𝑏 (𝑡𝑡)) cos(𝜃𝜃)

ias ( t) := Im⋅ cos ( ω⋅ t)

phase A current

2⋅ π  ibs ( t) := Im⋅ cos  ω⋅ t −  3  

phase B current

2⋅ π  ics ( t) := Im⋅ cos  ω⋅ t +  3  

phase C current

iqs ( t) := ias ( t)

q-axis stationary phase current

ids ( t) :=

1 3

(

⋅ ics ( t) − ibs ( t)

)

d-axis stationary phase current

iqse( t) := iqs ( t) ⋅ cos ( ω⋅ t) −

1

idse( t) := iqs ( t) ⋅ sin ( ω⋅ t) +

1

t := 0⋅ sec ,

1 100⋅ f

..

4⋅ π ω

(𝑖𝑖𝑐𝑐𝑐𝑐 (𝑡𝑡) − 𝑖𝑖𝑏𝑏𝑏𝑏 (𝑡𝑡)) sin(𝜃𝜃)

3

3

⋅ ics ( t) − ibs ( t) ⋅ sin ( ω⋅ t)

(

)

q-axis synchronous phase current

(

)

d-axis synchronous phase current

⋅ ics ( t) − ibs ( t) ⋅ cos ( ω⋅ t)

time (over two cycles)

Problem 5: Part a) The major difference between the machines is in their magnetic saliency. The surface PM machine is non-salient and therefore has equivalent values of q-axis and d-axis inductance. Without a difference in q- and d-axis inductances, the reluctance portion of torque equation is zero and therefore only BLI (Lorentz force) torque is produced. Conversely, in an interior PM machine, magnetic saliency is intentionally introduced in order to exploit the reluctance torque of the machine. Additionally, the resulting field can be reduced by commanding non-zero d-axis current allowing for field weakening operation.

Part b) Torque control in a surface PM machine is achieved by commanding q-axis current alone.

Typically, d-axis current is commanded to zero (because there is no reluctance portion of torque). The equation for torque of a surface PM machine is then: 𝑇𝑇𝑒𝑒 =

3 𝑃𝑃 𝑒𝑒 �Λ 𝑚𝑚𝑚𝑚 𝑖𝑖𝑞𝑞𝑞𝑞 � 22

For a interior PM machine, both the BLI and reluctance torque are used. The d-axis current is typically found as a function of q-axis current. The d-axis current (for maximum torque) and torque equations are: 𝑖𝑖𝑒𝑒𝑑𝑑𝑑𝑑 (𝑖𝑖𝑒𝑒𝑞𝑞𝑞𝑞 ) =

Part c)

𝑇𝑇𝑒𝑒 =

−Λ𝑚𝑚𝑚𝑚

2�𝐿𝐿𝑚𝑚𝑚𝑚 +𝐿𝐿𝑚𝑚𝑚𝑚

� �

Λ2𝑚𝑚𝑚𝑚

4�𝐿𝐿𝑚𝑚𝑚𝑚 +𝐿𝐿𝑚𝑚𝑚𝑚 �

2

+ 𝑖𝑖𝑒𝑒𝑞𝑞𝑞𝑞

2

3 𝑃𝑃 𝑒𝑒 𝑒𝑒 𝑒𝑒 + �𝐿𝐿𝑚𝑚𝑚𝑚 + 𝐿𝐿𝑚𝑚𝑚𝑚 �𝑖𝑖𝑞𝑞𝑞𝑞 𝑖𝑖𝑑𝑑𝑑𝑑 � �Λ 𝑚𝑚𝑚𝑚 𝑖𝑖𝑞𝑞𝑞𝑞 22

The figure below shows the speed vs. torque curves for typical surface and interior PM machines. Also shown is the bus current required from the inverter in order to achieve the speedtorque values. The figure highlights key characteristics of the two different machine types. In an ideal surface PM machine, the bus current (and therefore input electrical power) increases linearly with torque. In an interior PM machine, the field is weakened to allow operation above base speed. This allows the speed to be increased inversely proportional to torque maintaining constant power (and thus constant bus current).

surface PM speed

ω

interior PM speed

Idc Inverter current limit

surface PM bus current

interior PM bus current T An important take away from this is that the interior PM machine is a good candidate to use for constant power applications. Many EV traction applications have constant power loads as high torque is required at low speeds (for accelerating) and low torque is required at high speeds (highway cruising).

The surface PM machine is a good candidate for loads that require high speed and high torque simultaneously (such as servo applications).

Problem 6: The inductance of the field is significantly larger than the inductance of the armature. The armature current can be adjusted much faster and therefore produce higher transient response. The field current is adjusted under light loading conditions (to reduce flux levels and therefore hysteresis losses in the machine) and operation above base speed (field weakening).

Problem 7: From 6.4.1.2-1 (assumes cancelation between controller and plant time constants): 𝜔𝜔(𝑠𝑠)

𝜔𝜔∗ (𝑠𝑠)

=

𝐾𝐾𝑖𝑖𝑖𝑖

(8-1)

𝑏𝑏𝑠𝑠+𝐾𝐾𝑖𝑖𝑖𝑖

In a process analogous to solving for current loop controller gains: 𝑁𝑁𝑁𝑁

𝐾𝐾𝑖𝑖𝑖𝑖 = 𝑏𝑏2πf = 1.5 × 10−5 𝑟𝑟𝑟𝑟𝑟𝑟�

𝑠𝑠𝑠𝑠𝑠𝑠

2π50Hz = 4.712 × 10−3 𝑁𝑁𝑁𝑁

(P8-2)

And given that the electrical and control time constants are equal, we get: 𝑁𝑁𝑁𝑁

𝐾𝐾𝑝𝑝𝑝𝑝 = 𝑏𝑏2πf = 1.5 × 10−5 𝑟𝑟𝑟𝑟𝑟𝑟�

𝑠𝑠𝑠𝑠𝑠𝑠

𝑁𝑁𝑁𝑁

2π50Hz = 0.016 𝑟𝑟𝑟𝑟𝑟𝑟�

𝑠𝑠𝑠𝑠𝑠𝑠

(P8-3)

Note that the proportional gain takes a speed error as an input and outputs a torque command. So the 𝑁𝑁𝑁𝑁

units of 𝑟𝑟𝑟𝑟𝑟𝑟�

𝑠𝑠𝑠𝑠𝑠𝑠

are consistent with what we’d expect. The units of the integral gain are simply the time

integral of this.

Problem 8: IFO estimates the rotor flux angle based on the slip relationship of the machine. DFO directly measures (or estimates based on terminal quantities) the position of the rotor flux.

Problem 9: Benefits:

1. They are relatively inexpensive to manufacture due to the simple stamped laminations for both the stator and rotor (no castings or magnets) and they can use bobbin windings on the stator.

2. They are inherently rugged due to simple (yet structurally sound) rotor construction.

Drawbacks: 1. Complex torque regulation (no closed form mathematical solution, highly nonlinear).

2. Acoustic noise due to high radial reluctance forces tending to distort the stator housing.

1. a) Lead‐Acid theoretical Wh/kg 96400 Faraday Const 26.777778 Ah/mol Pb O S H

0.2072 0.016 0.032 0.001

2

2V

107.11111 Wh/mol

2 10 2 4

0.4144 0.16 0.064 0.004 0.6424 kg/mol

166.74 Wh/kg

b) Ni‐MH theoretical Wh/kg 96400 Faraday Const 26.77778 Ah/mol Ni O M H

0.0587 0.016 0.4325 0.001

MH6 M=LaNi5 La

1 1 2 0.167 2

1.35 V

36.15 Wh/mol

0.0587 0.032 0.0721 0.002 0.1648 kg/mol

219.38 Wh/kg

is the hydride 0.4325 0.139

Note: the amount of maximum hydrogen in the metal hydride needs to be considered

c) LiCo02 theoretical Wh/kg 96400 Faraday 26.777778 Ah/mol

1

3.7 V

99.077778 Wh/mol

Li Co O C

1 1 2 6

0.007 0.059 0.032 0.072 0.17 kg/mol

582.81 Wh/kg

2.58 V

138.173 Wh/mol

0.007 0.059 0.016 0.012

Note LiC6 structure used here

d) NaNiCl theoretical Wh/kg 96400 Faraday Const 26.77778 Ah/mol Ni Na Cl

0.0587 0.023 0.0354

2 1 2 2

0.0587 0.046 0.0708 0.1755 kg/mol

787.31 Wh/kg

2. a) Assumptions: Vo, Ro, R1, C1 constant, zero‐state hysteresis model, time horizon interval of Δt is used. This is part of dynamic peak‐power availability reporting.

b) The constant term Vo is replaced with nonlinear function of the form Vo(SOC0 +IpeakΔt/CAP) root finding methods need to be employed to solve for peak current Ipeak since it cannot be factored out of the nonlinear function. A partial solution is to assume a first order taylor series expansion around OCV. Assuming variations in RC parameters poses additional problems, analytical expressions will no longer result in simple exponential terms even if linear assumptions of variations are assumed. For best results assuming first‐order Taylor series expansion is to do so around resistance and time‐constant values.

3.

Let zi be the ith SOC data point, using the combined model data takes the form

 k0   Vo   1 z 1 z log(z )  log(1 z )   k  1 1 1 1  1    1           k2   V   1 z 1 z log(z )  log(1  z )   n n n n  on     k3        H v  k4  

min(Hk  v)  Hk  v   min k H Hk  2v Hk T

T

T

k

T

least squares pseudoinverse



k  HT H

 Hv 1

T

Matlab Optimization solver approach k = quadprog(H'*H,‐H'*v); Extension can include adding extra power terms, e.g. the custom Fit k0‐k1/z+k2z+k3log(z)‐k4log(1‐z)+k5z2+k6z3+k7z4+k8z5

Extra monotoniciity constrain nts can be ad dded easily uusing Matlab b “quadprog ”. Use of interp polation can n be exploite ed to ensure monotonici ty over the iinter‐samplee points is achie eved.

Sample Matlab code: c SOCra ange=linsp pace(min(SO OCdata),ma ax(SOCdata) ),150)'; Hq=[ [sign(SOCr range), -1. ./SOCrange e, -SOCrang ge, log(SO OCrange), l log(1SOCra ange) SOCr range.^2 SO OCrange.^3 3 SOCrange. .^4 SOCran nge.^5]; Hd=[ [0*SOCrang ge 1./SOCra ange.^2 -S SOCrange./S SOCrange 1 1./SOCrange e -1./(1SOCra ange) 2*SO OCrange 3*S SOCrange.^ ^2 4*SOCran nge.^3 5*S SOCrange.^4 4]; k2 = quadprog g(Hq'*Hq,-H Hq'*interp p1(SOCdata, ,Vdata,SOC Crange),Hd,ze eros(size( (Hd,1),1))

e derivative constraint oof the form o of A,b of con nstraint form m The laast term is related to the Ax<=b. Also possible to add equality e constraints of foorm Aeqx=beq. (e.g. to set some e k’s to zero) Sample:

For extra comparrison, both liinear and cu ubic interpolaation approaaches are alsso shown.

4.

Matlab Simulink with SimPowerSystems can be employed.

γ=1

γ=0.1

γ=0.01 Ro= 1 mΩ R1= 1 mΩ, C1= 10000 F R2= 0.5 mΩ, C2=100000 F Vhmax= 50 mV RC time constants can be changed to influence relaxation effect, a case with zero hysteresis can also be considered.

5.

Simulink with SimPowerSystems or SimScape can be employed. To focus on the RC asymmetry, symmetric behavior is assumed for the ohmic resistance.

It is possible to create custom variable resistor and capacitor blocks in SimScape whose input is dependent on charge or discharge, e.g. see the paper by T. Huria, M. Ceraolo, J. Gazzarri, and R. Jackey, "High Fidelity Electrical Model with Thermal Dependence for Characterization and Simulation of High Power Lithium Battery Cells," in IEEE Int. Electric Vehicle Conf. (IEVC), Pisa, Italy, 2012, pp. 1‐8.

The rest of the solution will show models using standard circuit elements.

Candidate #1:

Vrc Scope Sample Output:

This model produces the desirable asymmetrical time‐constants between charging and discharging. However, an undesirable discontinuous voltage is observed across the

diode/RC. Such behavior is already modeled via the ohmic resistor and should not be part of the asymmetric RC block.

Candidate #2:

Vrc Scope Sample Output:

This model shows the desirable continuous voltage response with some asymmetrical dynamic behavior. A subtle flaw is that the asymmetrical dynamics are not occurring during the desired times, i.e. the time durations that current flows through the two different resistors does not match battery cell charge/discharge times.

Candidate #3:

Vrc Scope Sample Output:

Modeling approaches using diodes for asymmetric behavior in the RC block contained fundamental flaws, therefore an alternative candidate employing ideal switches can be used. It shows desirable behavior of a continuous signal with asymmetrical dynamic behavior matching charge/discharge current times. It is noted here this model can represented by 4 parameters, i.e. Ro, Ca, Racharge, Radischarge.

6.

The following reference provides many example topologies: Cao, Jian, Nigel Schofield, and Ali Emadi. "Battery balancing methods: A comprehensive review." Vehicle Power and Propulsion Conference, 2008. VPPC'08. IEEE. IEEE, 2008.

Sample Topology #1 PWM controlled shunting. Detect the voltage difference of the two neighboring cells The current flows through the higher voltage cell lower than lower voltage cell 2(n‐1) switches and n‐1 inductors

Sample Topology #2 Single switched Capacitor Advanced control can be employed in a cell‐to‐cell balancing approach to improve speed of balancing.

Sample Topology #3 Multi‐winding transformer Energy converter method.

Secondary taps for each cell Difficult for modular design Design of multi‐winding transformer is complex

7.

Specific

Voltage

Capacity

vs Li+/Li

Tap Density (kg/L)

LiCoO2

135‐160

3.6

2.6‐3

LiMn2O4

100‐130

3.7

1.8‐2.4

LiFePO4

140‐160

3.3

0.8‐1.4

Additional information can be found in

Comments Popular for consumer products, expensive, highest risk of thermal runway, high volumetric capacity Low cost, poor cycle and calendar life Low cost, improved abuse tolerance, low calendar life, OCV‐SOC curve flatness

Tao, Haisheng, Zhizhong Feng, Hao Liu, Xianwen Kan, and P. Chen. "Reality and future of rechargeable lithium batteries." Open Mater Sci J 5 (2011): 204‐214.

8.

PNGV Model

Continuous Time:

1 I VCo  C0 1 I V1   V1  R1C1 C1 V  Vo  IR0 VCo V1 Discrete Time using Zero Order Hold:

VCo  VCo  k 1

k

1 I t C0 k 1

V1  e R1C1V1  R1  1 e R1C1  I k 1 k k1    t

 t

Vk  Vo  I k R0 VCo V1 k

k

RC Saft Model:

 1  V   C  R  R  c  b  b b   1  Vc      Cc  Rb  Rc   Rc VL    Rb  Rc

   1 Rc    Cb  Rb  Rc    Vb   Cb  Rb  Rc    V     IL 1 Rb  c       Cc  Rb  Rc   Cc  Rb  Rc      

  Vb   RR  Rb     R0  b c  I L Rb  Rc   Vc   Rb  Rc   

The above can be placed into the form:

x = Ax + Bu y = Cx + Du The discrete time model can be obtained using the equation:

 (tk , tk1, x k1, u)  e



e At  L1  sI  A 

t  A(tk tk1 )

1



tk

x k   e A(tk  )Bu( )d tk1

Can use "ilaplace" matlab function with symbolic toolbox

9.

The following equation is used. The input “u” is a linear function during the interval.

Integration by parts is employed to evaluate it.

 (tk , tk1, x k1, u)  e

t  A(tk tk1 )

tk

x k   e A(tk  )Bu( )d tk1

e.g. for a 1RC model with no hysteresis the following form is obtained

Vka  e

 t 

 t    t     a a  t  1  e 1  e a a a  I k  Ra   I k1 Vk1  Ra  1  e t    t  a  a   

oc a Vk  V  Ro I k  Vk

One‐state Hysteresis ‐ Continuous‐time model



dh   i sgn(i)h  h max dt



When current is positive

i0





max

 h0  h

dh   i h  h max dt  h(t)  h



max



 t  exp    i(t)dt   0 

When current is negative

i0



dh   i h  h max dt  h(t)  h

max





 h0  h

max



 t  exp    i(t)dt   0 

Combining the negative and positive cases leads to

h(t)  sgn i(t) h

max



 h0  sgn  i(t) h

max



t   exp  sgn i(t)   i(t)dt    0

Not a full proper solution, valid as long as i(t) doesn’t intersect zero, but sufficient to derive a discrete model Use trapezoidal rule integration. Equivalent to first‐order like hold for current input. t

 i(t)dt 

i(0)  i(t) t 2

0 Correction is needed for zero crossing to avoid exponential become greater than 1.

 ik1  , sgn  ik1ik   1 1   ik  ik1  1, otherwise   ik  , sgn  ik1ik   1  2   ik  ik1  1, otherwise 





hko  sgn ik1  h max  hk1  sgn  ik1  h max exp   12  ik1 1t 





hk  sgn  ik  h max  hko  sgn ik  h max exp   12  ik  2 t 

10.

Note: hysteresis is ignored or lumped into Voc. % Matlab Script to Derive Regression Equations assuming ZOH and 2RC % Network clear; clc; syms R1 R2 Ro dT tau1 tau2 Voc %I_0, I_1, and I_2 are at timestep "k", "k-1", and "k-2", likewise for V_1, V_2 syms I_0 I_1 I_2 V_1 V_2 a1=(1-exp(-dT/tau1)) a2=1-a1 b1=(1-exp(-dT/tau2)) b2=1-b1 bside=[Voc - Ro*I_1-V_1; Voc - Ro*I_2-V_2 + R1*(a1/a2)*I_2+R2*b1/b2*I_2] Aside=[R1 R2; R1/a2 R2/b2] Ip_1=inv(Aside)*bside V_0=Voc-Ro*I_0-R1*(a1*I_1+a2*Ip_1(1))-R2*(b1*I_1+b2*Ip_1(2)) collect(simplify(V_0),[V_1 V_2 I_0 I_1 I_2 Voc]) pretty(ans)

11.

Matlab Simulink with SimPowerSystems is employed.

The following cell 1RC ECM subsystem is used for all cases

This subsystem contains the following:

RLS Simulink Model:

Parameters m‐file: P_0=eye(4); x_hat_0=[0.99 -0.001 lambda=0.99; Ts=1; Ro=0.01; R1=0.0075; C1=4000;

RLS Subsystems:

0.001 2.5];

RLS Sample Results:

KF Simulink Model:

Parameters m‐file: Ro=0.01; R1=0.0075; C1=4000; P_0=eye(4)*0.01; x_0=[0 2.5 0 0]; Ts=1; theta=exp(-Ts/(R1*C1)); Q=diag([0.01^2 0.1^2 0 0]); R=0.002^2;

KF Subsystems:

KF Sample Output:

Q(2,2)=0.1^2

Q(2,2)=0.01^2

EKF Simulink Model: NOTE: This model estimates Vo instead of SOC

Parameters m‐file: Ro=0.01; R1=0.0075; C1=4000; P_0=eye(5)*0.01; x_0=[0 2.5 0 0 0.93]; Ts=1; Q=diag([0.01^2 0.1^2 0 0 0]); R=0.002^2;

EKF Subsystems:

EKF Sample Output:

Q(2,2)=0.1^2

Q(2,2)=0.01^2

12.

There is plenty of literature and technologies to reference, a select few are mentioned below with some comments.

Flywheel: Stores energy mechanically in a rotating disk. Has been used in race car applications.

Precession effects can be problematic as the vehicles moves since the flywheel angular momentum generates a precession rotations as its rotation axis is moved. Energy loss through disk friction is a type of self‐discharge that can be high.

Compressed Air Storage A thermodynamic and mechanical energy storage technology. Gas explosion is a safety concern for automotive applications. The compressed air tank heats up then cools to ambient temperature, in effect a self‐discharge loss. Decompression of air may need to be reheated to be usable. This cold air can be exploited for automotive AC applications.

Superconducting Magnetic Energy Storage Stores energy in a magnetic field via a flowing current. Superconducting materials are needed to avoid current drop. In its operation mode it is extremely efficient and fast responding. A parasitic loss is the need to maintain super‐conducting material cool.

Costly technology and still in experimental stage. Possible applications can be for synergetic use in cases where low‐temperature liquid storage of gaseous (room temperature) fuel is used.

Flow Battery An electrochemical energy storage device that typically employs two storage tanks for two liquid electrolyte solutions (catholyte and anolyte) , one for each electrode. Both of which are pumped across the electrodes. A membrane is present to separate the electrolyte solutions but enable ion transport. In principal the energy and power can be scaled independently. An experimental technology being purseing by for example GE. An interesting advantage for automobiles is recharging can be rapid and in a familiar fashion via transferring liquids in the storage tanks. Chemistries include: Zinc‐bromine, Polysulfide‐bromide, Vandium Redox batteries.

Other experimental technologies include: Lithium‐air and Lithium Sulfur batteries.

1 Problem 1: E stored 

1 CV 2 [Ws] 2

1 CV 2 [Wh]  2 3600

Estored 

0.5  63 1252  136.72 [Wh] 3600

Problem 2: Emax_ specific 

Estored 136.72   2.26 [Wh/kg] Mass 60.5

Problem 3: Per maximum voltage and current rating of the ultra-capacitor, the maximum continuous power and the specific power of the module based on the maximum continuous current rating can be found as:

Pmax  Vmax  I max_ cont  125  240  30000 [W] and Pmax  Vmax  I max_ cont / mass  125  240  495.86 [W/kg]

Per IEC 62391-2, maximum power is given by:

2

Pmax 

0.12  V 2 0.12  1252 0.12  V 2   104166 .

6 [W] and P   1721.75 d ESRDC 18  10 3 ESRDC  mass

[W/kg]

According to the impedance match approach, the maximum power and specific power of the ultra-capacitor module can be calculated as: Pmax 

Pd 

V2 125 2   217013.88 [W] and 4  ESRDC 4  18  10 3

V2  3587 [W/kg] 4  ESRDC  mass

Problem 4:

I max_1sec 

1 CV 0.5  63  125 2   1845 [A] C  ESRDC  1 63  18  10 3  1

E disch arg e 

1 2 2 C Vinitial  V final  177.187 [kW-seconds] = 49.21 [Wh] 2

Problem 5:





Problem 6: 10 seconds after the discharge starts;

3 V  Vinitial 

I 100 dt  IRDC  125  10  100  0.018  107.33 [V] C 63

50 seconds after the discharge starts; V  Vinitial 

I 100 dt  IRDC  125  50  100  0.018  43.84 [V] C 63

Problem 7:

t

C  (V1  V2 ) 63  (125  45)   50 [seconds] 100 I disch arg e

Problem 8: I

C  (V1  V2 ) 63  (125  75)   10.5 [A] t 300

Problem 9: Ploss  RDC  I 2  0.018  10 2  1.8 [W]

Problem 10: t









1 1 2 2 C Vinitial  V final  63 125 2  5 2  491.4 [s] 2P 2000

4

Problem 11: Pd _ battery 

Pd _ hybrid 

Pbattery _ max



M battery

100kW  340.22 [W/kg] 293.93kg

Pd _ UC  M UC  Pd _ battery  M battery M UC  M battery



1721.75  60.5  340.22  293.93  576.04 60.5  293.93

[W/kg]

P   576.04   1100  69.31 [%], increase in power density Pd %   d _ hybrid  1100   P  340 .

22   d _ battery  

Problem 12: Ed _ battery 

Ed _ hybrid 

E stored _ max M battery



24000W  81.65 [Wh/kg] 293.93kg

Ed _ UC  M UC  Ed _ battery  M battery M UC  M battery



2.26  60.5  81.65  293.93  68.1 60.5  293.93

[Wh/kg]

E   68.1  Ed %   d _ hybrid  1100    1100  -16.59 [%], decrease in energy density E   81.65   d _ battery  due to additional ultra-capacitor mass.

9-1. 𝑉𝑉𝑜𝑜 = 𝑉𝑉𝑖𝑖𝑖𝑖 (

12 = 𝑉𝑉𝑖𝑖𝑖𝑖 (

𝐷𝐷 N2 )( ) 1 − 𝐷𝐷 N1

0.2 2 )( ) 1 − 0.2 15

V in = 360V

N1 ) N2 15 Vstress = 360 + 12 × ( ) 2 Vstress = Vin + Vo × ( Vstress = 450𝑉𝑉

9-2.

(1)

N1 ) N3 N1 300 = 200 × (1 + ) N3 N1 1 = N3 2 N3 𝐷𝐷(1 + ) < 1 N1

Vstress = Vin × (1 +

𝐷𝐷(1 + 2) < 1 𝐷𝐷 <

(2)

𝐷𝐷(1 +

1 3

N3 )<1 N1

1 N3 (1 + ) < 1 N1 2 N3 <1 N1

𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉 = 200 × (1 + 𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉 > 400

9-3.

N1 ) N3

When Switch Sw1 is on, V p1 =V in , the voltage across P1 is transformed to the three other windings, resulting in, 𝑉𝑉𝑠𝑠1 = 𝑉𝑉𝑠𝑠2 = 𝑉𝑉𝑖𝑖𝑖𝑖 ( 𝑉𝑉𝑃𝑃2 = 𝑉𝑉𝑖𝑖𝑖𝑖

𝑁𝑁𝑠𝑠 ) 𝑁𝑁𝑃𝑃

𝑉𝑉𝑆𝑆𝑆𝑆2 = 2𝑉𝑉𝑖𝑖𝑖𝑖 = 800𝑉𝑉

D1 is conducting and D2 is reverse-biased, then the inductor voltage is 𝑁𝑁𝑠𝑠 𝑉𝑉𝐿𝐿 = 𝑉𝑉𝑖𝑖𝑖𝑖 � � − 𝑉𝑉𝑜𝑜 𝑁𝑁𝑃𝑃

𝑁𝑁 𝑉𝑉𝑖𝑖𝑖𝑖 �𝑁𝑁𝑠𝑠 � − 𝑉𝑉𝑜𝑜 𝑃𝑃 ∆𝑖𝑖𝐿𝐿|𝑜𝑜𝑜𝑜 = � � 𝐷𝐷𝐷𝐷 𝐿𝐿

When Switch Sw2 on, V p2 = -V in , the voltage across P2 is transformed to the three other winding, 𝑉𝑉𝑠𝑠1 = 𝑉𝑉𝑠𝑠2 = −𝑉𝑉𝑖𝑖𝑖𝑖 (

𝑁𝑁𝑠𝑠 ) 𝑁𝑁𝑃𝑃

𝑉𝑉𝑆𝑆𝑆𝑆1 = 2𝑉𝑉𝑠𝑠 = 800𝑉𝑉

When both switches off, two diodes are freewheeling, which results in, 𝑉𝑉𝐿𝐿 = −𝑉𝑉𝑜𝑜

∆𝑖𝑖𝐿𝐿|𝑜𝑜𝑜𝑜𝑜𝑜 = �

−𝑉𝑉𝑜𝑜 1 � � − 𝐷𝐷� 𝑇𝑇 𝐿𝐿 2

For steady state, the average current in the inductor is zero,

Thus, obtains the Vo/Vin,

∆𝑖𝑖𝐿𝐿|𝑜𝑜𝑜𝑜 + ∆𝑖𝑖𝐿𝐿|𝑜𝑜𝑜𝑜𝑜𝑜 = 0 𝑁𝑁𝑠𝑠 𝑉𝑉𝑜𝑜 = 2𝑉𝑉𝑖𝑖𝑖𝑖 � � 𝐷𝐷 𝑁𝑁𝑃𝑃 𝑁𝑁

3

Since, V o =12 and V in =400V, D=50%, get the 𝑁𝑁 𝑠𝑠 = 100. 𝑃𝑃

9-4.

Ds1

Ds2

T = 1 fsw

60o

Ds3 Ds4

Vtransformer

iL1

IL,max

IL,min

IL,max

iL2

IL,min

io

Io

9-5.

Pdiode conduction loss = V × I = 0.6 × 100 = 60W PMOSFET conduction loss = I2 × R ds < 60W R ds < 0.006Ω

9-6.

(1)

Ds1

T = 1 fsw

D

α

Ds2 Ds3 Ds4

Dα-

Vtransformer

t1

t2

t3

t4

During t2, when S1 and S4 both turn on, the primary voltage is V p =V in. The current I L1 ramps up linearly while I L2 ramps down linearly. The two inductor voltages are, 𝑉𝑉𝐿𝐿1 = 𝑉𝑉𝑖𝑖𝑖𝑖 �

𝑁𝑁𝑠𝑠 � − 𝑉𝑉𝑜𝑜 𝑁𝑁𝑃𝑃

𝑉𝑉𝐿𝐿2 = −𝑉𝑉𝑜𝑜

𝑁𝑁 𝑉𝑉𝑖𝑖𝑖𝑖 �𝑁𝑁𝑠𝑠 � − 𝑉𝑉𝑜𝑜 𝑃𝑃 ∆𝑖𝑖𝐿𝐿1|𝑜𝑜𝑜𝑜 = � � (𝐷𝐷𝐷𝐷 − 𝛼𝛼𝛼𝛼) 𝐿𝐿

During t1 and t3, the primary voltage is zero, both L1 and L2 are freewheeling. 𝑉𝑉𝐿𝐿1 = 𝑉𝑉𝐿𝐿2 = −𝑉𝑉𝑜𝑜

During t4, S2 and S3 both turn on, the primary voltage is V p =-V in. The current I L2 ramps up linearly while I L1 ramps down linearly. The two inductor voltages are, 𝑉𝑉𝐿𝐿2 = 𝑉𝑉𝑖𝑖𝑖𝑖 �

𝑁𝑁𝑠𝑠 � − 𝑉𝑉𝑜𝑜 𝑁𝑁𝑃𝑃

𝑉𝑉𝐿𝐿1 = −𝑉𝑉𝑜𝑜

Hence, L1 or L2 are energized only during the corresponding positive or negative primary voltage, and de-energized during the rest of the period. −𝑉𝑉𝑜𝑜 −𝑉𝑉𝑜𝑜 ∆𝑖𝑖𝐿𝐿1|𝑜𝑜𝑜𝑜𝑜𝑜 = � � (1 − 𝐷𝐷)𝑇𝑇 + � � 𝛼𝛼𝑇𝑇 𝐿𝐿 𝐿𝐿

For steady state, the average current in the inductor is zero, ∆𝑖𝑖𝐿𝐿1|𝑜𝑜𝑜𝑜 + ∆𝑖𝑖𝐿𝐿1|𝑜𝑜𝑜𝑜𝑜𝑜 = 0

𝑁𝑁 𝑉𝑉𝑖𝑖𝑖𝑖 �𝑁𝑁𝑠𝑠 � − 𝑉𝑉𝑜𝑜 −𝑉𝑉𝑜𝑜 −𝑉𝑉𝑜𝑜 𝑃𝑃 � � (𝐷𝐷𝐷𝐷 − 𝛼𝛼𝛼𝛼) + � � (1 − 𝐷𝐷)𝑇𝑇 + � � 𝛼𝛼𝑇𝑇 = 0 𝐿𝐿 𝐿𝐿 𝐿𝐿 Thus, obtains the Vo/Vin,

𝑁𝑁

𝑁𝑁𝑠𝑠 𝑉𝑉𝑜𝑜 = 𝑉𝑉𝑖𝑖𝑖𝑖 � � (𝐷𝐷 − 𝛼𝛼) 𝑁𝑁𝑃𝑃

1

Since, V o =12 and V in =400V, 𝑁𝑁 𝑠𝑠 = 10, and D is 0.5, get 𝛼𝛼 is equal to 0.2. 𝑃𝑃

(2). In order to achieve the highest efficiency, S5 and S6 need to be kept turned-on during the period when the voltage across the transformer is null. The control scheme is shown as following, T = 1 fsw VP 0.2

Ds1 72o

Ds2

Ds3 Ds4

Ds5 Ds6

9-7.

Sa

S1

S3 Np:Ns

VinDC

C S2

L1

S5

Vo

Vp S4

L2

Co S6 Sb

For APM, due to the low output voltage and high current requirements, conduction losses dominate on the secondary side. Paralleling MOSFETs can reduce the Rds, and therefore the conduction loss, to half. As the result, higher efficiency is achieved.

1:

54mpg, 2 million barrels a day

2:

System DC Voltage < 60VDC System AC Voltage < 30VAC

3:

Belt driven starter generator system

4:

Cost, packaging, wiring, safety

5:

Start:

Max torque, Constant torque regions

Assist:

Peak power curve, duration at peak power

Regen Power: Peak power, current Generator:

6:

Continuous power, Efficiency, Operations temp range

Brake and gear position is valid, Ambient and coolant (liquid or air) temperatures within range for all components, Battery state of charge above threshold, Engine and vehicle speed below threshold

7:

Current 12V power steering systems are one of the largest loads on the auxiliary system. Designing to meet edge to edge steering (~50+ Amps) has driven the DC-DC to be much larger than required for normal operation. By switching to a 48V, the current required can be reduced and a DC-DC converter will not be required to support it.

8:

Motor, Battery, Inverter, DC-DC, FEAD and Controls

9:

Accelerated aging for some batteries and less ability to use the battery for regenerative energy recapture, such as from braking. The 48V system operates at a mid-level state of charge, allowing for significant regenerative energy recovery, thereby significantly improving fuel economy over a 12V stop-start system.

10:

Vnominal = 48V, Vmin=38V, Pmax = 11kW. R

= (Vnominal – Vmin) x Vmin / Pmax = (48-38) x 38/ (11 x 1000) = 0.0345ohm, or 34.5 mohm at end of life.

11:

Specific Energy = (5kW x 2.5 min/60min/hr =)208Wh + 105Wh =313Wh. Installed energy = 313Wh/.5 = 616Wh. Specific Energy = 616/8 = 77Wh/kg, Specific Power = 11kW/8kg = 1.4kW/kg

12:

TEM

= 150/2.5 = 60Nm

SPEEDEM = 750*2.5=1875

11.1. To be done by the student.

11.2. To be done by the student.

11.3. To be done by the student.

11.4. To be done by the student.

Question 12.1 To answer Question 12.1, a simulation study needs to be performed and this should be performed by the student. Therefore, the guideline to perform this study and answer the questions is provided here.

120

Vehicle velocity (kmph)

100

80

60

40

20

0

0

200

400

800 600 time (s)

1000

Figure 12.21. NEDC driving cycle

1200

1400

Table 12.7. NEDC Driving Cycle

time (s) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36

Vehicle velocity (km/h) 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

Vehicle velocity (m/s) 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3.8 7.5 11.2 15 15 15 15 15 15 15 15 15 12.5 10 6.7 3.3 0 0 0 0 0 0 0 0 0 0

0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 1.06 2.08 3.11 4.17 4.17 4.17 4.17 4.17 4.17 4.17 4.17 4.17 3.47 2.78 1.86 0.92 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118

0 0 0 0 0 0 0 0 0 0 0 0 3 6 9 12 15 15 15 18.4 21.8 25.2 28.6 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32

0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.83 1.67 2.50 3.33 4.17 4.17 4.17 5.11 6.06 7.00 7.94 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.89

119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159

32 32 32 32 32 32 32 29.2 26.5 23.8 21 18.2 15.5 12.8 10 6.7 3.3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 6

8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.11 7.36 6.61 5.83 5.06 4.31 3.56 2.78 1.86 0.92 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.83 1.67

160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200

9 12 15 15 15 17.2 19.4 21.7 23.9 26.1 28.3 30.6 32.8 35 35 35 36.9 38.8 40.6 42.5 44.4 46.2 48.1 50 50 50 50 50 50 50 50 50 50 50 50 50 48.1 46.2 44.4 42.5 40.6

2.50 3.33 4.17 4.17 4.17 4.78 5.39 6.03 6.64 7.25 7.86 8.50 9.11 9.72 9.72 9.72 10.25 10.78 11.28 11.81 12.33 12.83 13.36 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.36 12.83 12.33 11.81 11.28

201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 240 241

38.8 36.9 35 35 35 35 35 35 35 35 35 35 35 35 35 35 33.5 32 28.9 25.7 22.6 19.4 16.3 13.1 10 6.7 3.3 0 0 0 0 0 0 0 0 0 0 0 0 0 0

10.78 10.25 9.72 9.72 9.72 9.72 9.72 9.72 9.72 9.72 9.72 9.72 9.72 9.72 9.72 9.72 9.31 8.89 8.03 7.14 6.28 5.39 4.53 3.64 2.78 1.86 0.92 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

242 243 244 245 246 247 248 249 250 251 252 253 254 255 256 257 258 259 260 261 262 263 264 265 266 267 268 269 270 271 272 273 274 275 276 277 278 279 280 281 282

0 0 0 0 0 3.8 7.5 11.2 15 15 15 15 15 15 15 15 15 12.5 10 6.7 3.3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0.00 0.00 0.00 0.00 0.00 1.06 2.08 3.11 4.17 4.17 4.17 4.17 4.17 4.17 4.17 4.17 4.17 3.47 2.78 1.86 0.92 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

283 284 285 286 287 288 289 290 291 292 293 294 295 296 297 298 299 300 301 302 303 304 305 306 307 308 309 310 311 312 313 314 315 316 317 318 319 320 321 322 323

0 0 3 6 9 12 15 15 15 18.4 21.8 25.2 28.6 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 29.2 26.5 23.8

0.00 0.00 0.83 1.67 2.50 3.33 4.17 4.17 4.17 5.11 6.06 7.00 7.94 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.11 7.36 6.61

324 325 326 327 328 329 330 331 332 333 334 335 336 337 338 339 340 341 342 343 344 345 346 347 348 349 350 351 352 353 354 355 356 357 358 359 360 361 362 363 364

21 18.2 15.5 12.8 10 6.7 3.3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 6 9 12 15 15 15 17.2 19.4 21.7 23.9 26.1

5.83 5.06 4.31 3.56 2.78 1.86 0.92 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.83 1.67 2.50 3.33 4.17 4.17 4.17 4.78 5.39 6.03 6.64 7.25

365 366 367 368 369 370 371 372 373 374 375 376 377 378 379 380 381 382 383 384 385 386 387 388 389 390 391 392 393 394 395 396 397 398 399 400 401 402 403 404 405

28.3 30.6 32.8 35 35 35 36.9 38.8 40.6 42.5 44.4 46.2 48.1 50 50 50 50 50 50 50 50 50 50 50 50 50 48.1 46.2 44.4 42.5 40.6 38.8 36.9 35 35 35 35 35 35 35 35

7.86 8.50 9.11 9.72 9.72 9.72 10.25 10.78 11.28 11.81 12.33 12.83 13.36 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.36 12.83 12.33 11.81 11.28 10.78 10.25 9.72 9.72 9.72 9.72 9.72 9.72 9.72 9.72

406 407 408 409 410 411 412 413 414 415 416 417 418 419 420 421 422 423 424 425 426 427 428 429 430 431 432 433 434 435 436 437 438 439 440 441 442 443 444 445 446

35 35 35 35 35 35 33.5 32 28.9 25.7 22.6 19.4 16.3 13.1 10 6.7 3.3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3.8 7.5 11.2 15 15

9.72 9.72 9.72 9.72 9.72 9.72 9.31 8.89 8.03 7.14 6.28 5.39 4.53 3.64 2.78 1.86 0.92 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 1.06 2.08 3.11 4.17 4.17

447 448 449 450 451 452 453 454 455 456 457 458 459 460 461 462 463 464 465 466 467 468 469 470 471 472 473 474 475 476 477 478 479 480 481 482 483 484 485 486 487

15 15 15 15 15 15 15 12.5 10 6.7 3.3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 6 9 12 15 15 15 18.4

4.17 4.17 4.17 4.17 4.17 4.17 4.17 3.47 2.78 1.86 0.92 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.83 1.67 2.50 3.33 4.17 4.17 4.17 5.11

488 489 490 491 492 493 494 495 496 497 498 499 500 501 502 503 504 505 506 507 508 509 510 511 512 513 514 515 516 517 518 519 520 521 522 523 524 525 526 527 528

21.8 25.2 28.6 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 29.2 26.5 23.8 21 18.2 15.5 12.8 10 6.7 3.3 0 0 0

6.06 7.00 7.94 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.11 7.36 6.61 5.83 5.06 4.31 3.56 2.78 1.86 0.92 0.00 0.00 0.00

529 530 531 532 533 534 535 536 537 538 539 540 541 542 543 544 545 546 547 548 549 550 551 552 553 554 555 556 557 558 559 560 561 562 563 564 565 566 567 568 569

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 6 9 12 15 15 15 17.2 19.4 21.7 23.9 26.1 28.3 30.6 32.8 35 35 35 36.9 38.8 40.6 42.5

0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.83 1.67 2.50 3.33 4.17 4.17 4.17 4.78 5.39 6.03 6.64 7.25 7.86 8.50 9.11 9.72 9.72 9.72 10.25 10.78 11.28 11.81

570 571 572 573 574 575 576 577 578 579 580 581 582 583 584 585 586 587 588 589 590 591 592 593 594 595 596 597 598 599 600 601 602 603 604 605 606 607 608 609 610

44.4 46.2 48.1 50 50 50 50 50 50 50 50 50 50 50 50 50 48.1 46.2 44.4 42.5 40.6 38.8 36.9 35 35 35 35 35 35 35 35 35 35 35 35 35 35 33.5 32 28.9 25.7

12.33 12.83 13.36 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.36 12.83 12.33 11.81 11.28 10.78 10.25 9.72 9.72 9.72 9.72 9.72 9.72 9.72 9.72 9.72 9.72 9.72 9.72 9.72 9.72 9.31 8.89 8.03 7.14

611 612 613 614 615 616 617 618 619 620 621 622 623 624 625 626 627 628 629 630 631 632 633 634 635 636 637 638 639 640 641 642 643 644 645 646 647 648 649 650 651

22.6 19.4 16.3 13.1 10 6.7 3.3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3.8 7.5 11.2 15 15 15 15 15 15 15 15 15 12.5 10 6.7

6.28 5.39 4.53 3.64 2.78 1.86 0.92 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 1.06 2.08 3.11 4.17 4.17 4.17 4.17 4.17 4.17 4.17 4.17 4.17 3.47 2.78 1.86

652 653 654 655 656 657 658 659 660 661 662 663 664 665 666 667 668 669 670 671 672 673 674 675 676 677 678 679 680 681 682 683 684 685 686 687 688 689 690 691 692

3.3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 6 9 12 15 15 15 18.4 21.8 25.2 28.6 32 32 32 32 32 32 32

0.92 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.83 1.67 2.50 3.33 4.17 4.17 4.17 5.11 6.06 7.00 7.94 8.89 8.89 8.89 8.89 8.89 8.89 8.89

693 694 695 696 697 698 699 700 701 702 703 704 705 706 707 708 709 710 711 712 713 714 715 716 717 718 719 720 721 722 723 724 725 726 727 728 729 730 731 732 733

32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 29.2 26.5 23.8 21 18.2 15.5 12.8 10 6.7 3.3 0 0 0 0 0 0 0 0 0 0 0 0 0

8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.89 8.11 7.36 6.61 5.83 5.06 4.31 3.56 2.78 1.86 0.92 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

734 735 736 737 738 739 740 741 742 743 744 745 746 747 748 749 750 751 752 753 754 755 756 757 758 759 760 761 762 763 764 765 766 767 768 769 770 771 772 773 774

0 0 0 0 0 0 0 0 0 3 6 9 12 15 15 15 17.2 19.4 21.7 23.9 26.1 28.3 30.6 32.8 35 35 35 36.9 38.8 40.6 42.5 44.4 46.2 48.1 50 50 50 50 50 50 50

0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.83 1.67 2.50 3.33 4.17 4.17 4.17 4.78 5.39 6.03 6.64 7.25 7.86 8.50 9.11 9.72 9.72 9.72 10.25 10.78 11.28 11.81 12.33 12.83 13.36 13.89 13.89 13.89 13.89 13.89 13.89 13.89

775 776 777 778 779 780 781 782 783 784 785 786 787 788 789 790 791 792 793 794 795 796 797 798 799 800 801 802 803 804 805 806 807 808 809 810 811 812 813 814 815

50 50 50 50 50 50 48.1 46.2 44.4 42.5 40.6 38.8 36.9 35 35 35 35 35 35 35 35 35 35 35 35 35 35 33.5 32 28.9 25.7 22.6 19.4 16.3 13.1 10 6.7 3.3 0 0 0

13.89 13.89 13.89 13.89 13.89 13.89 13.36 12.83 12.33 11.81 11.28 10.78 10.25 9.72 9.72 9.72 9.72 9.72 9.72 9.72 9.72 9.72 9.72 9.72 9.72 9.72 9.72 9.31 8.89 8.03 7.14 6.28 5.39 4.53 3.64 2.78 1.86 0.92 0.00 0.00 0.00

816 817 818 819 820 821 822 823 824 825 826 827 828 829 830 831 832 833 834 835 836 837 838 839 840 841 842 843 844 845 846 847 848 849 850 851 852 853 854 855 856

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 6 9 12 15 15 15 17.2 19.4 21.7 23.9 26.1 28.3 30.6 32.8 35

0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.83 1.67 2.50 3.33 4.17 4.17 4.17 4.78 5.39 6.03 6.64 7.25 7.86 8.50 9.11 9.72

857 858 859 860 861 862 863 864 865 866 867 868 869 870 871 872 873 874 875 876 877 878 879 880 881 882 883 884 885 886 887 888 889 890 891 892 893 894 895 896 897

35 35 36.9 38.8 40.6 42.5 44.4 46.2 48.1 50 50 50 51.5 53.1 54.6 56.2 57.7 59.2 60.8 62.3 63.8 65.4 66.9 68.5 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70

9.72 9.72 10.25 10.78 11.28 11.81 12.33 12.83 13.36 13.89 13.89 13.89 14.31 14.75 15.17 15.61 16.03 16.44 16.89 17.31 17.72 18.17 18.58 19.03 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44

898 899 900 901 902 903 904 905 906 907 908 909 910 911 912 913 914 915 916 917 918 919 920 921 922 923 924 925 926 927 928 929 930 931 932 933 934 935 936 937 938

70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 67.5 65 62.5 60 57.5 55 52.5

19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 18.75 18.06 17.36 16.67 15.97 15.28 14.58

939 940 941 942 943 944 945 946 947 948 949 950 951 952 953 954 955 956 957 958 959 960 961 962 963 964 965 966 967 968 969 970 971 972 973 974 975 976 977 978 979

50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50

13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89

980 981 982 983 984 985 986 987 988 989 990 991 992 993 994 995 996 997 998 999 1000 1001 1002 1003 1004 1005 1006 1007 1008 1009 1010 1011 1012 1013 1014 1015 1016 1017 1018 1019 1020

50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 51.5 53.1 54.6 56.2 57.7 59.2 60.8 62.3 63.8 65.4 66.9 68.5

13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 13.89 14.31 14.75 15.17 15.61 16.03 16.44 16.89 17.31 17.72 18.17 18.58 19.03

1021 1022 1023 1024 1025 1026 1027 1028 1029 1030 1031 1032 1033 1034 1035 1036 1037 1038 1039 1040 1041 1042 1043 1044 1045 1046 1047 1048 1049 1050 1051 1052 1053 1054 1055 1056 1057 1058 1059 1060 1061

70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70

19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44

1062 1063 1064 1065 1066 1067 1068 1069 1070 1071 1072 1073 1074 1075 1076 1077 1078 1079 1080 1081 1082 1083 1084 1085 1086 1087 1088 1089 1090 1091 1092 1093 1094 1095 1096 1097 1098 1099 1100 1101 1102

70 70 70 70 70 70 70 70 70 70 70.9 71.7 72.6 73.4 74.3 75.1 76 76.9 77.7 78.6 79.4 80.3 81.1 82 82.9 83.7 84.6 85.4 86.3 87.1 88 88.9 89.7 90.6 91.4 92.3 93.1 94 94.9 95.7 96.6

19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.69 19.92 20.17 20.39 20.64 20.86 21.11 21.36 21.58 21.83 22.06 22.31 22.53 22.78 23.03 23.25 23.50 23.72 23.97 24.19 24.44 24.69 24.92 25.17 25.39 25.64 25.86 26.11 26.36 26.58 26.83

1103 1104 1105 1106 1107 1108 1109 1110 1111 1112 1113 1114 1115 1116 1117 1118 1119 1120 1121 1122 1123 1124 1125 1126 1127 1128 1129 1130 1131 1132 1133 1134 1135 1136 1137 1138 1139 1140 1141 1142 1143

97.4 98.3 99.1 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 101 102 103 104 105 106 107

27.06 27.31 27.53 27.78 27.78 27.78 27.78 27.78 27.78 27.78 27.78 27.78 27.78 27.78 27.78 27.78 27.78 27.78 27.78 27.78 27.78 27.78 27.78 27.78 27.78 27.78 27.78 27.78 27.78 27.78 27.78 27.78 27.78 27.78 28.06 28.33 28.61 28.89 29.17 29.44 29.72

1144 1145 1146 1147 1148 1149 1150 1151 1152 1153 1154 1155 1156 1157 1158 1159 1160 1161 1162 1163 1164 1165 1166 1167 1168 1169 1170 1171 1172 1173 1174 1175 1176 1177 1178 1179 1180 1181 1182 1183 1184

108 109 110 111 112 113 114 115 116 117 118 119 120 120 120 120 120 120 120 120 120 120 120 117.5 115 112.5 110 107.5 105 102.5 100 97.5 95 92.5 90 87.5 85 82.5 80 76.2 72.5

30.00 30.28 30.56 30.83 31.11 31.39 31.67 31.94 32.22 32.50 32.78 33.06 33.33 33.33 33.33 33.33 33.33 33.33 33.33 33.33 33.33 33.33 33.33 32.64 31.94 31.25 30.56 29.86 29.17 28.47 27.78 27.08 26.39 25.69 25.00 24.31 23.61 22.92 22.22 21.17 20.14

1185 1186 1187 1188 1189 1190 1191 1192 1193 1194 1195 1196 1197 1198 1199 1200 1201 1202 1203 1204 1205 1206 1207 1208 1209 1210 1211 1212 1213 1214 1215 1216 1217 1218 1219 1220

68.8 65 61.2 57.5 53.8 50 45 40 35 30 25 20 15 10 5 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

19.11 18.06 17.00 15.97 14.94 13.89 12.50 11.11 9.72 8.33 6.94 5.56 4.17 2.78 1.39 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

Figure 12.21 and Table 12.7 provide the NEDC driving cycle data. Use this data to perform the simulation study (this data can also be obtained from Autonomie software). To perform this study, model the point mass vehicle dynamic as shown in the equation 12.4 in MATLAB Simulink.

It has been assumed that both the IC engine and the electric propulsion system are achieving 100% efficiency. Therefore, during acceleration and constant speed driving, there is no energy serving. However, the regenerative braking can recover the braking energy and it can enhance the total energy consumption of the driving cycle. Since we consider a parallel hybrid electric vehicle with 30% electric propulsion system, only 30% of the peak traction power can be recovered during the braking mode of the vehicle (which is 60 kW since the peak traction power of the vehicle is 200 kW). Therefore, if the peak braking power requirement is less than 60 kW, then the energy efficiency can be maximized. In case, if the peak braking power requirement is greater than 60 kW, then part of the braking power should be dissipated in the disc brake and it would reduce the energy efficiency of the vehicle. In this condition, series hybrid electric vehicle is more energy efficient than the parallel hybrid electric vehicle. From this analysis, the student could determine which kind of vehicle configuration is suitable for this particular driving cycle.

Question 12.2 To answer Question 12.2, a simulation study needs to be performed and this should be performed by the student. Therefore, the guideline to answer the question is provided here. a. Follow the procedure explained in case study 2 in section 12.4.1 to obtain the peak traction power, acceleration energy requirement, constant speed propulsive energy requirement, and energy dissipated during braking of the vehicle (since it is a conventional vehicle, braking energy cannot be recovered). By integrating 15% of the peak power with the vehicle idling time, energy dissipated during idling can be obtained. Adding the acceleration, constant speed, and idling energy requirements, the total energy requirement of the vehicle can be obtained. b. Perform the same task as above with the road inclination information. This will provide different power and energy requirements to achieve this driving cycle. c. The peak maximum braking power of the vehicle will be the electric propulsion system requirement to capture the entire regenerative braking energy of the vehicle.

13.1 The AC chargers are commonly on-board the vehicle, where an AC power is supplied to the vehicle, and charger supplies DC to the battery.

But the DC charging systems are mounted at fixed locations, like the garage or dedicated charging stations. Built with dedicated wiring, these chargers can handle much more power and charge the batteries more quickly.

13.2 The AC charging systems: Charge Method Electrical Ratings - SAE EV AC Charging Power Levels

Charge Method AC Level 1 AC Level 2

Nominal Supply Voltage (V) 120 V AC, 1-phase 120 V AC, 1-phase 208 to 240 V AC, 1-phase 208 to 240 V AC, 1-phase 208 to 240 V AC, 1-phase

Maximum Current (A) 12 A 16 A 16 A 32 A ≤ 80 A

Branch Circuit Breaker Rating (A) 15 A 20 A 20 A 40 A Per NEC 635

Output Power Level (W) 1080 W 1440 W 3300 W 6600 W ≤ 14.4 kW

The DC charging systems: Charge Method Electrical Ratings - SAE EV DC Charging Power Levels

Charge Method DC Level 1 DC Level 2 DC Level 3

Supplied DC Voltage Range (V) 200-450 V DC 200-450 V DC 200-600 V DC

Maximum Current (A)

Power Level (W)

≤ 80A DC ≤ 200A DC ≤ 400A DC

≤ 36 kW ≤ 90 kW ≤ 240 kW

13.3 a. b. c. d. e. f.

Conventional Boost PFC converter Interleaved Boost PFC converter Bridgeless Boost PFC converter Dual Boost PFC converter Semi-bridgeless Boost PFC converter Bridgeless Interleaved Boost PFC converter

13.4 g. h. i. j.

Zero Voltage Switching Full Bridge Phase Shifted converter Zero Voltage Switching Full Bridge Trailing Edge PWM converter Zero Voltage Switching Full Bridge with Capacitive Output Filter converter Interleaved Zero Voltage Switching Full Bridge with Capacitive Output Filter converter, Operating in BCM k. Interleaved Zero Voltage Switching Full Bridge with Voltage Doubler, Operating in BCM l. Full Bridge LLC Resonant DC-DC Converter

13.5

𝑃𝑃𝑃𝑃 = 𝐾𝐾𝑑𝑑 𝐾𝐾𝜃𝜃

1 1+𝑇𝑇𝑇𝑇𝑇𝑇 2

𝐾𝐾𝑑𝑑 = �

𝐾𝐾𝜃𝜃 = 𝑐𝑐𝑐𝑐𝑐𝑐𝜃𝜃1

Therefore: 𝑃𝑃𝑃𝑃 = �

1

1+𝑇𝑇𝑇𝑇𝑇𝑇2

× 𝑐𝑐𝑐𝑐𝑐𝑐𝜃𝜃1

13.6 In order to find minimum required inductance, first the minimum duty cycle under low line must be calculated: Dmin _LL = 1 −

√2Vin_min √2 × 85 =1− = 0.699 VPFC_bus 400

Assuming the inductor current ripple is 25% of input current at low line:

∆IL−LL = 0.25 ×

Po 1650 = 0.25 × = 5.108 [A] Vin_min × ηPFC 85 × 0.95

The minimum required boost inductor value to maintain 25% input ripple current is given by: LB_min =

√2Vin_min Dmin _LL √2 × 85 × 0.699 = = 235 [µH] fs ∆IL−LL 70 × 103 × 5.108

The low frequency and high frequency ripple current in the output capacitors are given by: Io =

Po

VPFC_bus

=

1650 = 4.125 [A] 400

16VPFC_bus 16 × 400 IC_HF = Io � − ηPFC 2 = 4.125 × � − 0.952 = 5.719 [𝐴𝐴] 6π√2 × 85 6π√2Vin_min IC_LF =

Io 4.125 = = 2.062 [𝐴𝐴] 2 2

In order to calculate required boost capacitor, the minimum hold-up time should be calculated: THold_Up =

1 1 1 1 = = 2.5 [ms] 4 2fLine 4 2 × 50

And finally the boost output capacitance value is: CPFC =

2 × 1650 × 2.5 × 10−3 = ≈ 530 [µF] 4002 − (400 − 20)2 VPFC_bus 2 − (VPFC_bus − ∆VPFC_bus )2 2Po THold_Up

13.7 It is similar to problem 6. To be done by student.

13.8

Although the proposed converter can operate in either discontinuous conduction mode (DCM), boundary conduction mode (BCM), or continuous conduction mode (CCM), only the DCM and BCM modes are desirable.

Assuming the effective duty cycle is 0.96, the transformer turn ratio is given by: n=

Vo_max 450 = = 1.172 Deff Vin 0.96 × 400

Referring to [33] and [34], the converter DC gain in BCM is given by: MBCM =

Vo_nom =D×n Vin

Replacing V o-nom , V in and n, one could find Duty Cycle D, and M BCM : D = 0.64

MBCM = 0.75

The load resistance is given by: Ro =

2 Vo−nom 3002 = = 30 Ω Po 3300

The converter DC gain in DCM is given by: MDCM =

Vo_nom = Vin

2n

1 + �1 +

4k D2

Where the normalized time constant of converter is: k=

4 × n2 × LR Ro × T

Therefore:

k = 0.36, and the resonant inductor can be calculated from above equation:

LR ≈ 10 µH

The output capacitor RMS current is given by:

1 Tp ICo(rms) = � � (irec (t) − Io )2 dt = 6.8 𝐴𝐴 Tp 0

The output capacitor value is given by: Co =

ICo(rms) ≈ 20 µF 4 × π × fs × Vripple

13.9 To be done by the student.

13.10 To be done by the student.

14.1 PHEV-X stands for X miles of all-electric range (AER) in the corresponding PHEVs. PHEV-20 is capable of 20 miles of all-electric range and utilizes the smallest energy storage devices compared to the other two. PHEV-40 has higher all-electric range by utilizing larger battery pack. Both PHEV-20 and PHEV-40 are designed to meet the typical daily urban driving range by using electricity as the major energy source. As shown in Figure. 14.1, PHEV20 satisfies 50% of the typical U.S. daily driving distance and PHEV-40 satisfies 73% of the typical U.S. daily driving distance. In contrast, PHEV-100 has the highest all-electric range and has the largest energy storage devices on board. It greatly increases the all-electric driving range and meets the majority of the daily driving need by using just electricity. However, more energy storage devices are required from a much larger battery pack, increasing the volume and weight, and hence the manufacturing cost. More sophisticated battery management system (BMS) is required to monitor the cell health and cell balances. Higher requirements on cooling systems are also demanded to control the battery cell temperatures and pack temperature. PHEV-20 and PHEV-40 demand less on the added electric components and require less vehicular level changes compared to PHEV-100. On the other hand, PHEV-100 achieves the highest degrees of both petroleum fuels reduction and emissions reduction.

14.2 As can be shown in Figure 14.16, when simulating the same vehicle by the same drive cycle under the same conditions using (a) AER control strategy and (b) blended control strategy, the SOC depleting rate of the blended control strategy is much smaller than the AER control strategy. The SOC of vehicle under AER control strategy drops from 0.7 to 0.3 around 1200 seconds while the SOC of the same vehicle under the blended control strategy remains above 0.5 after 5500 seconds. In AER mode, the engine does not start until it hits the minimum SOC threshold and then runs in the hybrid mode thereafter. In comparison, blended mode takes advantage of both the engine and the electric machines together to operate the system in a more efficient way, i.e. neither the engine nor the electric motor has to operate at the peak power alone and less demands on the battery power is required, hence less losses on the overall system. Therefore, the blended mode provides longer range compared to the AER mode.

The fuel economy and the amount of emissions depend on the travel distance. Within short ranges of daily commute, AER mode is adequate to run the vehicle purely on electricity, and thus has maximum fuel economy and generates no emissions. However, with longer travel distance, the SOC hits the minimal threshold and engine starts to end the AER mode. Thus, fuel economy comes down and more emissions will be generated compared to the blended mode.

The table below compares the fuel economy and CO 2 emissions for the simulated vehicle under AER mode and blended mode for 4 UDDS cycles:

Fuel economy CO 2 emissions Electric consumption

AER mode 94.8 miles/gallon, 2.48L/100km 125.91 g/mile 143.75 W h/ mile

Blended mode 98.15 miles/gallon, 2.4L/100km 121.62 g/mile 143.76 W h/ mile

14.3 By adjusting the minimal engine start power in the vehicle propulsion control panel, the engine start threshold value can be set differently. The lower the threshold value, the more frequent engine assist will be provided to help the electric motor to meet the maximum road load, while the higher engine start threshold value enables the vehicle to run more on electricity in the charge depleting mode. In the study below, 4 different minimal engine start threshold values are used as initial variable parameters: 10 kW, 25 kW, 50 kW, and 75 kW. The same vehicle are simulated by using 4 UDDS drive cycle under the same all other conditions. The following figure compares the SOC status, indicating that the higher the engine start threshold value, the more rapid charge depleting is used so that the vehicle runs more on electric mode. The engine operation power under 10 kW and 75 kW start threshold are also compared in the following figure.

0.7

SOC

0.6 0.5 75 kW 50 kW 25 kW 10 kW

0.4 0.3 0

1000

2000

3000

4000

5000

6000

Time (s)

60 10 kW 75 kW

Engine power (kW)

50 40 30 20 10 0 -10

0

1000

2000

3000

4000

Time (s)

5000

6000

14.4 Open question for students.

1. To be done by the student.

2. To be done by the student.

3. To be done by the student.

4. To be done by the student.

1.

EVs as “peakers” A “peaker” is a small, but nimble generating unit that can supply the grid with relatively fast response. Historically, natural gas turbines or small hydroelectric plants were the devices of choice for this task. They are active for only a few hours every day and therefore provide only limited energy. Thus, a substantial fleet of EVs can carry out this task as a highly distributed resource without significantly depleting their batteries. Unfortunately, as long as peak power is not considered a "service", the utility operator will compensate the car owner solely for the energy sold, albeit at a higher peak-demand rate [31]. This may not constitute a strong enough incentive to the car owner who has to consider other factors, such as the additional battery and power electronic wear and tear for his vehicle. Nevertheless, future adjustments in energy market models are under study to address this among other issues.

EVs as Spinning and Non-spinning Reserve One of the most lucrative ancillary service is the spinning and non-spinning reserve.

The former consists of generators that are on line, but normally run at very low capacity. In the case of a disruption, such as a failure in baseload generation or transmission, these generators are commanded to provide the missing power. They must be able to ramp up in less than 10 minutes and provide power for as long as 1 hour or more. Non-spinning reserves are not on line and are required to ramp up to full power within 30 minutes.

Because this is a service, the utility company will pay for the availability of the power as well as its amount. In fact, this service is paid even when no power is ever delivered. An EV owner can provide this service naturally and be reimbursed starting at the time he plugs his vehicle to the grid even if the battery is never discharged. Also, it must be noted that Plugin Hybrid Electric Vehicles (PHEV) have smaller battery capacity than all-electric vehicles, but contain an internal combustion engine that can be started on a V2G command to generate electricity and function as a spinning reserve as well.

EVs as Voltage/Frequency Regulation Agents An ancillary service that is even better tailored for EVs is regulation. It consists of delivering or absorbing limited amounts of energy on demand and in real time. Normally the request is automated in order to match exactly the instantaneous power generation with the instantaneous load. Failure to do so results in dangerous shifts in line frequency and voltage.

The dispatched amount of energy has short duration - in the order of few minutes - but it is requested relatively frequently. Therefore this is a continuous service. It is important to underline that the amount of energy involved is relatively small and changes direction quite rapidly and regularly, implying minimal EV battery discharge for any reasonably short time interval. The near instantaneous response time and the distributed nature of the EV fleet explains why regulation is probably the most competitive application for V2G from the point of view of the utility operators.

EVs as Reactive Power Providers Most electronic topologies used for the inverter/rectifier function in the interface of the EV to the grid are fully capable of shaping the line current to have low distortion and varying

amounts of phase shift with respect to the AC line voltage. This implies that reactive power can be injected into the grid on demand and in real time [32]. Furthermore, since reactive power translates in no net DC currents, this service can be provided without any added stress to the EV battery.

2. As noted earlier, only Level 1 and 2 are feasible within the confines of a residential setting. This can be accomplished through integrated chargers when available or an external EVSE. In the latter case, the most obvious circuit configuration is a single-phase bidirectional rectifier/inverter powered by a 240VAC/60A circuit that is readily available from the distribution transformer. The DC-link voltage is then processed by a bi-directional DC/DC converter that performs the isolation function. This simple topology, shown in Fig. 1, can be called the canonical topology, with minor changes, for most grid-tied systems, irrespective of power rating.

Fig. 1. Canonical single-phase EVSE configuration.

In North America, the 240V from the residential distribution transformer is in the form of a split 120V supply, suggesting small modifications to the canonical topology. Fig. 2 shows two possibilities.

Fig. 2. Split phase sourced EVSE configurations.

The two topologies in Fig. 2 are similar, but the one on the right has better voltage utilization, and is better equipped to counter unbalanced loads on the split supply. For the DC/DC converter, many bi-directional isolated circuit topologies have been proposed.

Typical circuits are shown in Fig. 3.

Fig. 3. Typical isolated bidirectional buck-boost DC/DC converter topologies.

When the two controlled bridges are independently driven in phase-shift modulation (PSM), these are generally referred to as Dual Active Bridge (DAB) topologies. In their simplest operation mode, when power needs to be transferred from the left-side circuit to the right-side circuit, for instance, the right-side IGBT switches are left undriven, leaving their antiparallel diodes in the form of a regular diode bridge. Under these circumstances, the topology becomes identical to a regular PSM converter, which is simple to operate, but not very flexible in terms of voltage gain. On the other hand, when both bridges are modulated, power transfer can be accomplished in both directions and with great variability ranges on the input and output voltages. In addition, zero voltage switching (ZVS) can be assured for all switches for reduced switching loss and generated electrical noise (EMI). Other topologies based on the DAB have been proposed with purported additional benefits, such as better switch utilization, extended ZVS operating range, and more flexible voltage amplification.

3. A public parking/charging installation would deliver only Level 2 power, given the relatively long plugin times. Because there are several parking locations in close proximity, the power configuration used for residential use may not optimal. Rather, a single transformer can be installed at the grid, delivering isolated power to all vehicles in the facility. This way, cheaper and more efficient non-isolated DC/DC converters can be used without violating safety rules. Fig. 4 illustrates this configuration for each charging station.

For the whole installation, the architectures shown in Fig. 5 are possible.

Fig. 4. Configuration with isolation at the grid.

Fig. 5. Centralized architecture (top); Distributed architecture (bottom).

In the centralized architecture, a single, large, poly-phase, supply frequency step-down transformer connects to the grid, providing isolation for the whole facility. A large bidirectional rectifier that produces a single high-voltage DC bus follows this. Each parking station uses inexpensive high-efficiency, non-isolated DC/DC converters to process this bus voltage into the appropriate charging current for the individual EVs. Because isolation is either desirable or required, especially on PV panels depending on local electrical codes, additional storage or generating resources, such as wind turbines and fuel cells, can also benefit from simpler interface to the DC bus. Moreover, the single transformer connection guarantees that no DC current is injected into the grid, doing away with complicated active techniques to achieve the same purpose.

The advantages just noted for centralized configuration are somewhat offset by the following drawbacks: 1) The need for a bulky and usually inefficient line-frequency transformer; 2) An expensive high-power poly-phase inverter/rectifier; 3) Single-fault vulnerability in the transformer and central inverter rectifier; 4) Lack of voltage amplification in each non-isolated DC/DC converter (otherwise afforded by the turns ratio of the high frequency transformer in isolated topologies).

The first issue can be partially countered by brute force methods such as the development of advanced sub-milliohm connectors and minimizing cable lengths by placing the grid step-down transformer in physical proximity of the vehicle. It is obvious that any intervening power conditioning electronic circuit should be added only when absolutely necessary. This immediately suggests that the architecture of the charging station should be distributed rather than central. As can be seen from Fig. 5, a distributed architecture could potentially reduce the number of processors from grid to battery from two to one. To be fair, this single stage may not be feasible when managing large input-output voltage ranges, especially if buck-boost operation is required. Nevertheless, if an additional DC/DC stage should prove necessary, it will be easily integrated locally with the inverter for improved

efficiency. Furthermore, a central processor, besides constituting a single point of failure as already noted, would have to be rated for the full service station power, which could be of the order of a megawatt. On the contrary, a distributed architecture benefits from repeated circuitry (economies of scale), redundancy for higher reliability, and the possibility of power conditioning in physical proximity to the vehicles, reducing Ohmic loss.

4. For the DC/DC converter, all common basic topologies, that is, boost, buck-boost, buck, Cuk, SEPIC, and ZETA can be used, as long as they are rendered bidirectional. In this case, these topologies function differently, depending on the direction of power flow (see Fig. 6).

Fig. 6. Basic bidirectional non-isolated topologies.

Different design requirements might suggest different topologies, but some of these are objectively more difficult to justify. For instance, using the buck-boost/buck-boost (bottom left in Fig. 6 produces a voltage inversion from positive to negative that may be undesirable.

It also places higher electrical stress on the switches; it requires a more sophisticated design for the inductor and draws pulsed current from the battery. Similarly, the ZETA/SEPIC topology has a higher part count, including a capacitive, rather than inductive energytransferring element. On the other hand, as long as the DC bus is guaranteed to exceed the battery voltage - a requirement that is assured by the use of the controlled bridge discussed earlier - the Buck/Boost topology (top left in the figure) is quite attractive.

5. The buck-boost DC/DC topology is readily modified in order to divide the task of handling a very large power flow among paralleled modules. This is shown in Fig. 7; the

amount of converted power can be split among n identical sections and the battery ripple current greatly reduced by the well-known technique of phase-shift interleaving.

Fig. 7. Interleaved modular approach for the DC/DC converter.

TO BE DONE BY THE STUDENT: Consider fast charging a Nissan Leaf® battery pack (48 modules, Li-ion battery cells, 24.0 kWh), within using the DC/DC interleaved buck-boost converter circuit, with n = 3, switching frequency, f s = 2.0 kHz, and P = 125.0 kW. It would be interesting to calculate charging efficiency for this particular case. Make suitable assumptions, as deemed suitable.

6. When relatively large energy production from intermittent sources is to be tied to the grid, a statistically predictable EV presence could serve the purpose of minimizing on-site dedicated storage. This would be the case for municipal carports powered by wind and/or solar generation and where the vehicles must be able to interact intelligently with both locally generated and grid distributed power at the same time. The possible scenario described in Fig. 5 (top) may not be ideal when the renewable resource is meant to generate the dominant share of EV charging energy. Rather, by realizing the advantages of the distributed configuration, as in Fig. 5 (bottom), one stage of conversion can be eliminated so long as a conversion topology with wide input-output voltage range capability can be found.

Fig. 8. Possible configurations for solar carport.

Fig. 8 above shows some possible configurations for one of the several charging stations in a solar carport. The architecture depicted on the left has the disadvantage of inserting a

DC/DC converter into the main intended power flow, from PV to battery. Moreover, the power drawn from a single-phase connection is pulsed at twice the line frequency. This pulsating power takes the form of an undesirably high ripple current into the battery. The configuration shown in the middle of Fig. 8 removes the ripple issue, but adds an additional conversion stage between the grid and the battery. The configuration on the right requires a converter that is capable of bidirectional flow between the EV and the grid, as well as steering of PV power to either the EV or the grid in a controlled fashion. Furthermore, this should ideally be achieved by a single conversion stage for all power flow paths and with wide voltage range capability.

7. Renewable energy resources are not easily harnessed. This simple fact makes energy conservation through high conversion efficiency a priority. As mentioned earlier, the number of conversion stages should be minimized, at least for the most utilized power flow paths. In the case of renewable charging, the power will flow most frequently from the PV/Wind to the EV battery and from the AC grid to the EV battery. Therefore, given a choice, the flow paths should be optimized in line with the following priorities: in decreasing order of importance, renewable source to battery, grid to battery, renewable resource to grid, and battery to grid. A schematic, showing possible power flow paths is depicted in Fig. 9, specifically for a PV/AC grid interface. A proposed structure with PV, AC grid-connected home, and an EV battery system is shown in Fig. 10.

Fig. 9. Possible power flow paths with specified power level.

Fig. 10. EV, AC grid connected home, and PV interconnection.

The configuration of Fig. 10 has unique advantages. First, the PV power can be directly injected to the battery pack and not to the grid then from the grid to the load, which reduces the overall losses. Secondly, this configuration provides great flexibility for different power flows. Depending on the amount of available power from PV panels and required power by the battery pack, different modes of operation may occur. If available power from PV is more than the required power for charging the battery pack the remaining can be injected to the grid (mode 1). If PV power is not enough, grid will be involved to supply the remaining power (mode 2). If there is no power available from the PV panels battery pack can be charged solely by the grid (mode 3). If the battery pack is not connected to the system, PV power can be injected to the grid to reduce the electricity bill of the house (mode 4). If needed, the battery pack can be discharged to the grid during specified times performing a configuration supporting vehicle-to-grid (V2G) (mode 5). Even in the case of blackouts, the EV battery pack can perform as an energy source, and supply the power to the house for some time, depending on the battery pack capacity (mode 6).

TO BE DONE BY THE STUDENT: Consider the test-case of Fig. 9 for the second part of the question. The power flow scenarios depicted in Fig. 9 show a typical home PV/Grid/EV interface.

Solution to Exercises of Chapter 17 Electronic Control Units and Energy Management by Ilse Cervantes E.1 Implement heuristic EMS given in Section 1.2.3.1 in a PHEV using Table 1.1.

R.1 A nice implementation of these heuristic rules can be found in [2] for UDDS and EPA driving cycles and, several values of vehicle parameters.

E.2 What is the effect of the battery size in the fuel consumption for the optimization problem in Section 1.4.1?

R.2 The size of the battery has a significant effect in the fuel economy since the solution depends on the feasible region. As stated in Section 1.4.1 the feasible region can be reduced as the SOC level decreases. Larger battery capacities lead to an increment of the energy available and therefore, the optimization is possible. Large values of C lead to large values of α (lower fuel consumption) and viceversa. As C decreases the feasible region becomes smaller and the fuel economy decreases leading to smaller fuel economies. This fact can be clearly seen from the dynamic restriction of the battery ˙ −CVbat ηe f f ,bat SOC(t) = α, and it is illustrated in Figure 1, where c1 = C−1 has been used.

Pload 90 c1=3.457e-5(1/(A.s)); c1=6.914e-5(1/(A.s)) c1=1.3828e-4(1/(A.s))

85

Fuel Economy (%)

80 75 70 65 60 55 50 45 40

0

200

400

600

800

1000

1200

PFCmin(W)

Figure 1: Fuel Economy for different battery capacities as a function of PFC,min . Notice that c1 = C−1 (City II Driving cycle).

E.3 Introduce the terminal constraint (17.15) to the problem of Section 17.4.2. i) Compute the solution. ii) What is the expression of the adjoint equation?

R.3 The minimization problem can be formulated as follows: 1

min J3 = min α

α

∫ T 2 2 2 a K Pload (σ)[1 − α(σ)]2

η2e f f ,FC

t0

dσ = min u

∫ T 2 2 2 a K Pload (σ)[1 − u]2

η2e f f ,FC

t0

dσ (1)

with terminal condition SOC(T ) ≥ SOC(t0 )

(2)

which can be rewritten as SOC(T ) = SOC∗ ≥ SOC(t0 ) The optimization problem is also subjected to restrictions (1.53). Let u(t) = α(t) and x(t) = SOC(t); therefore, the functions in Theorem 2 are given by:

f (t, x, u) = B(t)u M(t, x(t), u(t)) =

(3)

2 2 a2 K Pload (σ)[1 − u]2 η2e f f ,FC

(4)

G(x(T )) = 0 ϕ(T, x(T )) = x(T ) − SOC∗ 

load (t) PFCmin − (1−u)P ηe f f ,FC

(5) (6) 

  load (t)   −PFCmax + (1−u)P ηe f f ,FC     SOC − x min R(t, x, u) =     x − SOCmax     −u u−1

(7)

load (t) where B(t) , CV−P and bat ηe f f ,bat

∇x f (t, x, u) = 0 ∇u f (t, x, u) = B(t) ∇x M(t, x, u) = 0

(8) (9) (10) 2

2 (t)(1 − u) 2a2 K Pload ∇u M(t, x, u) = − η2e f f ,FC

(11)

∇x ϕ(t, x) = 1 ( ) ∇′x R(t, x, u) = 0 0 −1 1 0 0 ∇′u R(t, x, u) =

(

ρ(t, u) −ρ(t, u) 0 0 −1 1 2

(12) (13) )

(14)

(t) with ρ(t, u) = Pηload . According to Theorem 2, necessary conditions for the optimum e f f ,FC are:

λ̇(t) = −ν3 + ν4 λ(T ) = 1 0 = B(t) + Ξ(t, u) + ν1 ρ(t, u) − ν2 ρ(t, u) − ν5 + ν6 with Ξ(t, u) = −

(15) (16) (17)

2

2 (1−u) −2a2 K Pload and 2 ηe f f ,FC

{

νj =

> 0, R j (t, x, u) = 0 = 0, R j (t, x, u) < 0

Observe that Equation (15) can be rewritten as :   < 0, = 0, if λ̇ =  > 0,

if SOC = SOCmin SOCmin < SOC < SOCmax if SOC = SOCmax

Let θ(t, u) = λB(t) + Ξ(t, u) + ν1 ρ − ν2 ρ then Equation (17.62) can be rewritten as: θ(t, u) > 0 θ(t, u) = 0 θ(t, u) < 0

u=0 0<u<1 u=1

if if if

(18) (19) (20)

As in Section 17.4.2. the optimal solution resides at the boundary of the feasible region; therefore, the following cases must be analyzed since conditions in Theorem 2 are only necessary: Case (i). SOCmax < SOC(t) < SOCmin and PFCmin = PFC (t), Case (ii). SOCmax < SOC(t) < SOCmin and PFCmax = PFC (t), Case (iii). SOC(t) = SOCmin and PFCmin = PFC (t), Case (iv). SOCmax = SOC(t) and PFCmin = PFC (t), Case (v). SOCmin = SOC(t) and PFCmax = PFC (t), and finally Case (vi). SOCmax = SOC(t) and PFCmax = PFC (t).

Case (i) For the first case, we have that the restriction (19) becomes, 2

−

2 (1 − u) 2a2 K Pload Pload (t) + ν1 =0 2 ηe f f ,FC ηe f f ,FC

which leads to u = 1− 3

ν1 ηe f f ,FC 2

2a2 K Pload

(21)

Case (ii) For PFC = PFC,max we have that restriction (19) becomes, ν2 ηe f f ,FC u = 1+ 2 2a2 K Pload

(22)

Notice that since the restrictions on the x are not active, the solution is the same as the free terminal point problem; that is, the solution of the optimization problem (for SOCmin < SOC(t) < SOCmax ) is in the curve of the minimum FC power PFC,min . However this is not the case where such restrictions are active Case (iii) For SOC = SOCmin and PFCmin = PFC (t) we have that the restriction (19) becomes, 2 2 2a2 K Pload (1 − u) Pload (t) λB(t) − + ν1 =0 2 ηe f f ,FC ηe f f ,FC with λ̇ = −ν3 with λ(T ) = 1 which leads to u = 1−

ν1 ηe f f ,FC ηe f f ,bat CVbat − λη2e f f ,FC 2

2Pload ηe f f ,bat Ca2 K Vbat

(23)

Case (iv) For SOC = SOCmax and PFCmin = PFC (t) then λ̇ = ν4 , with λ(T ) = 1 and restriction (19) is given also by (23).

Case (v) For SOC = SOCmin and PFCmax = PFC (t), then λ̇ = −ν3 , with λ(T ) = 1 and restriction (19) becomes, ν1 ηe f f ,FC ηe f f ,bat CVbat + λη2e f f ,FC u = 1+ (24) 2 2Pload ηe f f ,bat Ca2 K Vbat Case (vi) For SOC = SOCmax and PFCmax = PFC (t), then λ̇ = ν4 , with λ(T ) = 1 and restriction (19) becomes (24) also.

By comparing the control functions of Cases (iii) and (v) as well as the control function of Cases (iv) and (vi), it is possible to observe that the solution of the optimization problem is at the PFCmin = PFC (t) restriction for both at SOC(t) = SOCmax and SOC(t) = SOCmin . However, in contrast to the terminal free point problem, the adjoint equation (the dynamic of λ) requires of backward integration in order to satisfy the terminal condition λ(T ) = 1.

E.4 Consider the problem of Section 1.4.2. i) How regeneration can be taken into account?

R.4 The terminal condition for a given driving cycle, will determine the presence or not of the energy regeneration process.

4