Problem Solutions. Vehicle Energy and Fuel Consumption. Vehicle Energy Losses and Performance Analysis. Problem 2.1

Transcription

1 Problem Solutions Vehicle Energy and Fuel Consumption Vehicle Energy Losses and Performance Analysis Problem 2.1 For a vehicle with m v = 1500 kg, A f c d = 0.7m 2, c r = 0.012, a vehicle speed v = 120km/h and an acceleration a = 0.027g, calculate the traction torque required at the wheels and the corresponding rotational speed level (tires 195/65/15T). Calculate the road slope that is equivalent to that acceleration. F t = m c r g +1/2 ρ a A f c d v 2 +m v a = = ( ) = 1041 N 3.6 The information about the tires is explained by 195 }{{} width of the tire in cm / 65 }{{} ratio of sidewall height to tire width in % / 15 }{{} wheel diameter in inch T }{{} max. 190 km/h. Thus

2 2 Index r w = = = m 2 2 T t = r w F t = = 330 Nm ω w = v = = rad/s = 101 rpm r w 3.6 α = a = rad g α % = 100 tan(0.027) = = 2.7 % For the requested velocity, a traction torque of 330 Nm at a rotational speed of 101 rpm is required. This is equivalent to the acceleration caused by a slope of 2.7%. Problem 2.2 Find the road slope α that is equivalent to a step of height h for a car with (a rigid) wheel radius r w on a flat terrain. Calculate the result for h/2r w = {0.01, 0.02, 0.05, 0.1, 0.2}. When in contact with the step, the car wheel will rotate around the contact point. Thus the reaction force R will be directed from the contact point to the wheel center (neglect slip here) and the reaction force at the contact point on the terrain becomes null. R is balanced by the weight N and the traction force F t that has to be calculated. One can write two equations F t = R x N = R y where R x and R y are the two components of R, with R x = R y tan(α). Butcos(α) = (r w h)/r w = 1 h/r w (seefigure).thussin(α) = 1 (1 h/r w ) 2 and finally F t = N tan(α) = N sin(α) cos(α) = N 1 ( 1 h ) 2 r w r w r w h. One could obtain the same result with a balance of torques acting around the contact point: T t = F t (r w h) = N a, where a 2 +(r w h) 2 = r 2 w. The equivalent slope is calculated by equating m v g sin(α) r w = T t,

3 Index 3 r h F N R r r a α r-h h from whence ( ) a α = arcsin N r w m v g In a first approximation, N = m v g/2 (for center of gravity centered along the wheelbase) and thus ( ) a ( ) α = arcsin = arcsin z (1 z), where z = h ( z 1 ) 2 r w 2 r w 2 The exact result for rigid wheels would be slightly different, ( ) b α = arcsin a 2 (a+b) r w that tends to the former value if a b, where b is the wheelbase. For flexible wheel, the result is lower and decreases with the decrease of the wheel stiffness. z [m] α [rad] (44%) (31%) (exact solution for b/r w = 6.67: 0.27 rad) (22%) (14%) (10%) Problem 2.3 Find an equation to evaluate the speed profile of an ICE vehicle under maximum engine torque. Assume a maximum torque curve of the type T e = a ω 2 e +b ω e+c. Include the engine inertia and a traction efficiency η t. where v = ( γ η t T e m v g c r 1 ) r w 2 ρ a A f c d v 2 1 m eq

4 4 Index ( ) 2 γ m eq = m v +Θ e r w T e = a ωe 2 +b ω e +c where ω e = (γ/r w ) v (if the clutch is closed). Thus the law of motion is of the type v = A v 2 +B v +C, where ( A = 1 ( ) 3 γ 2 ρ 1 a A f c d + η t a) r w ( ) 2 γ 1 B = η t b r w m eq ( C = g m v c r + γ ) η t c r w 1 m eq To solve the differential equation an integral of the type t t 0 = t t 0 dτ = v(t) v 0 dν A ν 2 +B ν +C must be generally solved between v 0 = v(t 0 ) and v(t). As it is to find in any integral table, the solution to this integral depends on the sign of D = B 2 4 A C: If D > 0, where and If D < 0, where v(t) = S (1+K(t))+B (K(t) 1) 2 A (1 K(t)) m eq K(t) = 2 A v 0 +B S 2 A v 0 +B +S es (t t0) = K 0 e S (t t0) v(t) = ( S tan S = D arctan ( 2 A v 0+B S 2 A S = D ) ) + S (t t 0) 2 B If B = 0 and C < 0 (coasting), then D < 0 and one finds back (2.18) for the coasting velocity. Note that γ/r w generally varies along the acceleration, so a solution must be obtained step by step.

5 Index 5 Problem 2.4 Evaluate the km/h time precisely using the result of Problem 2.3. Use the following data: engine launch speed = 2500 rpm, engine upshift speed ω e,max = 6500rpm, γ/r w = {46.48,29.13,20.39,15.04,11.39}, a = Nms 2, b = Nms, c = 80Nm, and the following vehicle data: m v = 1240 kg, A f c d = 0.65 m 2, c r = 0.009, η t = 0.9. Further assume that a slipping clutch transmits all the torque. Since A, B, and C depend on γ, which changes along the speed trajectory, the calculation of v(t) must be separated in segments according to the gear and the clutch status. The target speed will be reached in the gear whose γ/r w is immediately lower than ω e,max v = π/60 100/3.6 = 24.5 Thus the target speed is reached in the third gear. Four segments (including takeoff) must be considered. 1 st segment: takeoff, v 0 = 0, t 0 = 0 The result of Problem 2.6 is used, with ω e = 2500 rpm, and T e = a ωe 2 +b ω e +c = 142 Nm A = 1 2 ρ 1 a A f c d = 1 m v = B = 0 ( C = g m v c r + γ ) 1 η t T e = r w m v = = S = 4 A C = = The synchronization time between the engine and the vehicle (clutch closed) is when v = 2500 π r w = 5.63 m/s = km/h, 30 γ thus when K = 2 A v S 2 A v +S = = 1.09 The synchronize time is thus t = ln(1.09) = 1.14 s

6 6 Index 2 nd segment, 1 st gear, v 0 = 5.36 m/s, t 0 = 1.14 s Here γ/r w = 46.48, thus ( ) 2 γ m eq = m v +Θ e = = 1517 kg r w ( A = 1 ( ) 3 γ 2 ρ 1 a A f c d + η t a) = r w m eq = ( 12 ) = ( ) 2 γ 1 B = η t b = = 0.45 m eq 1517 C = r w ( g m v c r + γ r w η t c ) 1 m eq = = = D = B 2 4 A C = = S = D = K 0 = 2 A v 0 +B S 2 A v 0 +B +S = = This phase ends when ω e = ω e,max, thus when and K = v = π = m/s 2 A v+b S 2 A v+b +S = = thus at time ( K t = t 0 +ln K 0 ) 1 S = 1.14+ln ( ) rd segment, 2 nd gear, v 0 = m/s, t 0 = 3.6 s Here γ/r w = = 3.6 s

7 Index 7 ( ) 2 γ m eq = m v +Θ e = = 1349 kg r w ( A = 1 ) = B = = C = D = = S = 0.29 = 1.48 K 0 = = 1.15 This phase ends when ω e = ω e,max, thus when and thus at time v = π = m/s = 84.1 km/h K = = t = 3.6+ ln( ) = 7.1 s th segment, 3 rd gear, v 0 = m/s, t 0 = 7.1 s Here γ/r w = ( ) 2 γ m eq = m v +Θ e = = 1293 kg r w ( A = 1 ) = B = = C = ( )/1293= 1.05 D = = S = 0.15 K 0 = = 1.58 We want to calculate the time to reach v = 100 km/h = 27.8 m/s that will take place in this segment. Thus K = = 2.305

8 8 Index thus at time t = 7.1+ ln( ) = 9.6 s 0.15 Problem 2.5 Consider again Problem 2.4 for an engine with negligible inertia Θ e. Compare the result with (2.16). 1 st segment v 0 = 0, t 0 = 0 The same result as in Problem nd segment, 1 st gear v 0 = 5.36 m/s, t 0 = 1.14 s Here γ/r w = 46.48, thus A = ( 12 ) = B = = C = D = = S = 0.8 = K 0 = = This phase ends when ω e = ω e,max, thus when and thus at time v = π = m/s K = = 2.92 t = ln( ) = 3.14 s rd segment, 2 nd gear, v 0 = m/s, t 0 = 3.14 s Here γ/r w = 29.13

9 A = Index 9 ( 12 ) = B = = C = D = = S = = K 0 = = This phase ends when ω e = ω e,max, thus when and thus at time v = π = m/s = 84.1 km/h K = = t = ln( ) = 6.4 s th segment, 3 rd gear, v 0 = m/s, t 0 = 6.4 s Here γ/r w = A = ( 12 ) = B = = C = D = = S = = 1.09 K 0 = = We want to calculate the time to reach v = 100 km/h = 27.8 m/s that will take place reasonably in this segment. Thus thus at time K = = t = 6.4+ ln( ) = 8.82 s 0.156

10 10 Index Comparison Equation (2.16) gives t = v2 m v = = 11.9 s, P e,max which means a substantial overestimation.in reality, the relation P = P max /2 should be corrected in this case as where With this formula, P = P e,max +P e,min 2 P e,min = ω e,launch T e,launch = 2500 π 142 = 37.2 kw 30 which is an underestimation of 9%. P = 58.8 kw and t = 8.1 s Problem 2.6 Find an equation to calculate the takeoff time (=time to synchronise the speed before and after the clutch) as a function of engine launch speed and torque. Assume that the clutch is slipping but transmitting the whole engine torque. The motion law of the vehicle is ( γ v = η t T e m v g c r 1 ) r w 2 ρ a A f c d v 2 Thus the law of motion is of the type v = A v 2 +C, where A = 1 2 ρ a A f c d 1 m eq C = 1 m eq ( g m v c r + γ r w η t T e The solution can be obtained from the result of Problem 2.3. Now A < 0 and B = 0, thus D > 0. Moreover, t 0 = 0 and v 0 = 0. Therefore, where v(t) = S 1+K(t) 2 A (1 K(t)) K(t) = e S t and S = 4 A C ) 1 m eq

11 Index 11 The synchronization time is obtained by imposing that v = ω e r w /γ, and then calculating K as 2 A v S K = 2 A v +S and subsequently t takeoff = ln( K) S Problem 2.7 Evaluate the coasting speed and the roll-out time without acting on the brakes for a vehicle with an initial speed v 0 = 50 km/h and m v = 1200 kg, A f c d = 0.65m 2, c r = Assume the clutch open (no engine friction). The coasting speed as a function of time (2.18 and Problem 2.3 ) is v(t) = β ( ) } α {arctan α tan β v 0 α β t where 1 α = 2 ρa A f c d 1 = m v = β = g c r = = The braking time at which v = 0 is calculated as t rollout = α β arctan 1 ( ) α β v 0 = arctan Mechanical Energy Demand in Driving Cycles Problem 2.8 ( ) = s 3.6 Evaluate the traction energy and the recuperation energy for the MVEG 95 for the vehicle examples of Fig. 2.8, left and right, assuming perfect recuperation.

12 12 Index The traction energy is given by (2.31). With the data in Fig. 2.8 left, one obtains Ē = ( ) = = 43 MJ /100 km x tot = = 4.78 MJ. The total energy is given by (2.35), Ē rec = ( ) = = 33 MJ /100 km x tot = 3.64 MJ Ē = Ē Ērec = 1.14 MJ For the data of Fig. 2.8 right, (24% of )Ē. Ē = ( ) = = 20 MJ /100 km x tot = 2.21 MJ Ē rec = ( ) = = 15 MJ /100 km x tot = 1.61 MJ Ē = = 0.60 MJ (27% of Ē) The potential of regenerative braking is more important for the smaller vehicle here. Problem 2.9 Calculate the mean force and fuel consumption data shown in Fig. 2.8 left. Case 1: No recuperation F trac,a = 1 2 ρ a A f c d 319 = = 134 N 2 F trac,r = m v g c r = = 151 N F trac,m = m v = = 150 N The mechanical energy per 100km that corresponds to 1N is Therefore, 1 N 10 5 m = 10 5 J = 10 5 /3600 = Wh. V a = Wh /N 100 km 10 4 Wh/l F trac,a = = l /100 km V r = = l /100 km V m = l /100 km V = = 1.2 l /100 km

13 Index 13 Case 2: Perfect recuperation F a = 1 2 ρ a A f c d 363 = = 152 N 2 F r = m v g c r 1 = = 177 N V a = = l /100 km V r = = l /100 km V = = 0.91 l /100 km Evaluation of the differences rusults in F m,r = ( ) = 44 N F m,b = = 106 N Problem 2.10 Calculate the data in Fig. 2.9, left and right. Full-sized vehicle The cycle energy assuming no recuperation is given by (2.31), Ē = = kj/100 km, thus Light-weight vehicle thus S(A f c d ) = = 0.31 S(c r ) = = 0.35 S(m v ) = ( ) = 0.70 Ē = = kj/100 km S(A f c d ) = = 0.38 S(c r ) = = 0.25 S(m v ) = ( ) = 0.63

14 14 Index Problem 2.11 Calculate which constant vehicle speed on a flat road is responsible for the same energy demand at the wheels along a MVEG 95 cycle, in the case of no recuperation and of perfect recuperation, respectively. Assume the light-weight vehicle data of Fig. 2.8, left and right: {A f c d,m v,c r } = {0.7 m 2,1500 kg,0.012}. No recuperation The cycle energy is calculated in Problem 2.8 and it is Ē = kj/100 km. The mean traction force is F = To have the same F, find v such that Ē 100 km = 200 N 1 2 ρ a A f c d v 2 +m v g c r = F = v = v = Perfect recuperation Here, Ē = kj/100 km, thus F = 150 N v = 19.5 m/s = 70 km/h The equivalent speed is found by equating 2 = 24.2 m/s = 87 km/h F trac m v g c r ρ a A f c d F(m v ) = Ē 100 = A f c d m v c r g (from 2.35) and F v (m v ) = 1 2 ρ a A f c d v 2 +m v g c r. The two second right-hand side terms are equal, thus the equivalent constant speed is 1 2 ρ a v 2 = = v = = 19.5 m/s. Problem 2.12 Calculate the maximum mass allowed for a recuperation system with η rec = 40%.UsethevehicleparametersofFig.2.12:{A f c d,m v,c r } = {0.7 m 2,1500 kg,0.012}.

15 Index 15 The maximum weight is the one that leads to an energy demand equal to the energy demand without recuperation. Thus, from (2.41) Ē(η rec,m rec ) = 0.7 ( ) m v ( ) m v = = ( ) m v ( kj /100 km) where m v = m v +m rec. That number must equal Ē, that has been calculated in Problem 2.8 as kj/100 km. Thus m v = ( ) 103 = 1750 kg m rec = 1720 kg 1500 kg = 250 kg. IC-Engine-Based Propulsion Systems Gear-box Models Problem 3.1 Improve (3.9) in order to take into account the engine inertia and the transmission efficiency. Including the engine inertia Neglecting other forces, Thus m v,eq = m v +m e,eq ( γ m e,eq = Θ e F t = T e η t a = g sin(α) = F t m v,eq = ( γ r w r w γ r w ) 2 γ T e η t r w ( ) 2 m v +Θ e γ r w ) 2 T e η t Θ e a γ + m v = 0, r w Θ e a quadratic equation in γ. Evaluate γ 1 by imposing that a = a max. With Θ e = 0 and η t = 1, obtain back equation (3.9).

16 16 Index Problem 3.2 Dimension the first gear of an ICE-based powertrain not based on a given gradability as in (3.9) but in order to obtain a given acceleration at vehicle take-off. Do the calculations according to Problem 3.1, using the following data: m v = 1100kg, payload m p = 100kg, equivalent mass of the wheels m r,w = 1/30 of m v, c r = 0.009, transmission efficiency η t = 0.9, T e = 142Nm at ω takeoff, desired acceleration a = 4m/s 2, engine inertia Θ e = 0.128kgm 2, r w = 30cm. Compare with the approximate solution of (3.9). Substitute a to g sin(α) in the result of Problem 3.1. Then solve ( γ r w ) γ r w The two mathematical solutions are γ/r w = and Choose the latter solution as the physical one (the former would lead to a too short gear with lower fuel efficiency; it would be too distant from γ 5 and would cause a bad sizing; during vehicle launch it would be held for a too short time in detriment of drivability). For a common wheel with r w = 30 cm, γ 1 = Using (3.9), thusneglectingtheengineinertia,onewouldhaveγ/r w = m v a/(t e η t ) = 38.7 (instead of 47.9), or γ 1 = 11.6 (instead of 14.4). Problem 3.3 Dimension the fifth gear in a six-gear transmission for maximum power using (3.10) for a vehicle with the following characteristics: curb m v = 1100kg, performance mass = 100kg, c r = 0.009, A f c d = 0.65m 2, r w = 30cm, transmission efficiency = 0.9, P e,max = 80.4kW at 6032rpm. The maximum power at the wheels is = 72.4 kw. Equation (3.10) is written as ( ) 1 P e,max = v max v2 max , which is solved by v max = 55.5 m/s = km/h. The fifth gear ratio is γ 5 = π 6032 r w = For a common wheel radius of 30 cm, γ 5 = 3.4.

17 Index 17 Problem 3.4 Consider again the system of Problems Calculate the vehicle speed values, v j at which the engine is at its maximum speed, for all the gears j. The 2 nd to 4 th gear can be chosen according to a geometric law with from whence R = = 4.2, 1 κ = R1/4 = 1.43, γ 1 = γ 2 = γ 3 = γ 4 = 1 γ 2 γ 3 γ 4 γ 5 κ, γ 2 = γ 1 κ = 33.5, r w r w γ 3 = γ 2 κ = 23.4, r w r w γ 4 = γ 3 κ = 16.3, r w r w γ 5 = γ 4 κ = 11.4 (confirmed). r w r w An alternative to the geometric sizing law is the arithmetic law with ( 1 R = ) = , from whence γ 2 r w = γ 3 r w = γ 4 r w = γ 5 r w = 1 r w γ1 +R = 26.6, 1 r w γ2 +R = 18.4, 1 r w γ3 +R = 14.1, 1 r w = 11.4 (confirmed). γ4 +R The shifting speeds are thus the values in the following table. v k = π 6032, 30 γk r w

18 18 Index Geometric Arithmetic v m/s = 47.5 km/h v m/s= 68.0 km/h 23.7 m/s= 85.5 km/h v m/s= 97.2 km/h 34.3 m/s=123.6 km/h v m/s=139.7 km/h 44.8 m/s=161.3 km/h v m/s = km/h Problem 3.5 Calculate the approximate efficiency of a clutch during a vehicle takeoff maneuver. In a first approximation, the transmitted torque could be considered as equal to the engine torque while the clutch is slipping. The engine speed is approximately constant at the launch value ω e, while the speed downstream of the clutch is given by the differential equation Θ gb dω gb dt = T e T losses, with Θ gb = Θ v /γ 2, which can be approximated by neglecting the losses. Thus ω gb (t) = T e Θ gb t. The launch maneuver ends when ω gb (t f ) = ω e. The corresponding synchronization speed is v = ω e γ. r w The launch time is t f = ω e Θgb T e. The energy provided by the engine during this time is E e = tf 0 T e ω e dt = T e ω e t f = Θ gb ω 2 e. Theenergytransferredtothevehicleisthe kineticenergyatthe endoflaunch, which is E v = 1 2 Θ gb ω 2 e. Thus the energy lost is E e E v which coincides with (3.16) since ω w,0 = ω e /γ and therefore, the efficiency is E v /E e = 0.5.

19 Index 19 Fuel Consumption of IC Engine Powertrains Problem 3.6 Find the CO 2 emission factor (g/km) as a function of the fuel consumption rate (l/100 km) for gasoline and diesel fuels. Use these average fuels (gasoline, diesel) data: density ρ = {0.745, 0.832} kg/l, carbon dioxide to fuel mass fraction m = {3.17, 3.16}. Consider the stoichiometric fuel burning reaction ( CH a + 1+ a ) ( a O 2 CO 2 + H 2 O 4 2) whereafuelofmolarcompositionch a isassumed.themolco 2 /molemission factor is 1. The kg CO 2 /kg emission factor is thus m = M CO 2 = = 44 M fuel 12+a 12+a The factor a can be found by using m for gasoline, and for diesel, a = = 12 = 1.88, m 3.17 a = = The kg/l factor is m ρ, where ρ (kg/l) is the fuel density. For gasoline we get and for diesel, The overall factor is for gasoline, and m ρ = = 2.36 kg/l m ρ = = 2.63 kg/l. g CO 2 /km = g CO 2 /l C g CO 2 /km = 26.3 C = 23.6 C for diesel, where C is the fuel consumption ( l 100 km ).

20 20 Index Problem 3.7 Calculate the fuel consumption and the CO 2 emission rate for the MVEG 95 cycle for a vehicle having the following characteristics: m v = 1100kg, payload = 100kg, c d A f = 0.7, c r = 0.013, e gb = 0.98, P 0,gb = 3%, P aux = 250W, v launch = 3m/s, e = 0.4, P e,0 = 1.26kW, P e,max = 66kW, diesel fuel (H f = 43.1MJ/kg, ρ f = 832g/l), idle consumption V f,idle= 150 g/h. The declared CO 2 emission rate for this car is 99g/km. The mean traction force is calculated as F trac,r = = 120 N F trac,a = = 134 N 2 F trac,m = = 120 N F trac = = 374 N From this result the mean engine power can be calculated as in (3.23) (3.25), P trac = = 5.9 kw 0.6 P 1 = = 6.2 kw 0.98 P start = = 51 W 105 P e = = 6.5 kw Using the given Willans approach, the average fuel power can be calculated as η e = e P e P e +P 0,e = = 0.34 P f = 0.6 P e η e = 11.5 kw Remember: The average velocity on the MVEG-95 cycle is 9.5 m/s, the engine idle time is approximately 300 seconds and the cyle length is 11.4 kilometers. The equivalent fuel flow is therefore calculated as in (3.28) (3.32), V trac = P f = H l ρ f = l/s = l/100 km = l /100 km V idle = = = l /100 km V f = 3.5 l /100 km+0.1 l /100 km = 3.6 l /100 km = = 95 g/km

21 Index 21 The additional CO 2 not taken into account in this method amounts to 4 g/km (env. 5%). Problem 3.8 Evaluate in a first approximation the contribution of stop-and-start, regenerative braking (m rec = 20% m v, η rec = 0.5) and optimization of power flows in reducing the fuel consumption when the system of Problem 3.7 is hybridized. Assume an 80% charge-discharge efficiency for the reversible storage system. From the result of Problem 3.7, V idle / V tot = 4% Now consider regenerative braking. Redo calculations of Problem 3.7 with m v = = 1320 kg. F trac,r = = 144 N F trac,a = = 134 N 2 F trac,m = = 142 N F trac = = 420 N For ideal regenerative braking, F trac,r = = 168 N F trac,a = = 152 N 2 F trac,m = 0 N F trac = = 320 N For η rec = 0.5, the traction force is F trac = 320+(1 0.5) ( ) = 370 N Thus the potential gain due to regenerative braking seems rather limited,

22 22 Index P trac = = 5.8 kw 0.6 P 1 = = 6.1 kw 0.98 P start = = 51 kw 105 P e = = 6.4 kw η e = e P e P e +P 0,e = = 0.33 P f = 0.6 P e = 11.6 kw η e V trac = = l/s = = l/100 km = 3.4 l /100 km In summary we have 3.6 l /100 km w.r.t. 3.4 l /100 km. The benefit due to regenerative braking is equal to the benefit due to idle consumption suppression and they amount to 3%. To evaluate the potential benefit of engine operating point shifting, assume that the engine could be able to work always at its maximum efficiency point, thus at P e,max = 66 kw. The efficiency is η e = /( ) = Moreover, during a time t 1, the engine delivers its surplus power to the battery, to be reused later. An energy balance across the traction phase yields P e 0.6 = P e,max t 1 +(P e,max P e ) t 1 η acc, where η acc is the efficiency of the accumulation system (to be charged and then discharged). Assuming η acc = 0.8, from the latter one calculates thus t 1 = (66 6.4) 0.8 = 0.034, P f = = 5.7 kw V trac = = l/s = = l/100 km = 1.7 l /100 km Ideal gain due to optimization of power flows = ( )/3.6 = 47%.

23 Electric and Hybrid-Electric Propulsion Systems Electric Propulsion Systems Problem 4.1 Index 23 Design an electric powertrain for a small city car having the following characteristics: curb mass = 840kg, payload = 2 75kg, tires: 155/65/14T, c d A f = 1.85m 2, rolling resistance coefficient = 0.009, to meet the following performance criteria: (i) max speed = 65km/h, (ii) max grade = 16%, (iii) 100 km range. Assume perfect recuperation, overall efficiency of 0.6, and 85% SoC window. Choose motor size in a class with a maximum speed of 6000rpm and the number of battery modules having a capacity of 1.2kWh each. For v max = 65 km/h = 18.1 m/s, the required power is P max = m v g c r v max c d A f v 3 max = 8 kw. The max speed of the motor is ω m,max = v max γ/r w where γ is the reduction ratio and r w is the wheel radius. The wheel radius is obtained from the tire specifications (see Problem 2.1) as m = = 0.28 m. If one fixes ω m,max = 6000 rpm = 628 rad/s, then γ = 628/ = 9.7. The max torque is T m,max = r w /γ m v g sin(α) = 0.28/9.7 ( ) = 45 Nm. Thus the base speed is P m,max /T m,max = 8000/45 = 178 rad/s = 1700 rpm. The base to max speed ratio is 1:3.5, which is a reasonable design choice. Assuming an efficiency η = 0.6 and perfect recuperation, the mean traction force for an ECE drive cycle is (see (2.34)) F = m v g c r A f c d = 303 N. Thus the energy required is 303/ = 50.5 MJ = 14 kwh. Add an unused15%rangeandobtain16.1kwh. Using6V/200Ah (1.2kWh) modules, 14 modules would be needed for a stored energy of 16.8kWh. Problem 4.2 Find an equation for the AER D ev of a full electric vehicle as a function of its vehicle parameters, battery capacity and powertrain efficiency. Then evaluate the D ev for a bus with the following characteristics: η rec = 100%, η sys = 0.45 (including unused SoC), c r = 0.006, A f c d = , Q bat = 89Ah, U bat = 600V, m v = 14.6t, without payload and with a load of 60 passengers, respectively. Assume a MVEG-type speed profile.

24 24 Index Equation (2.30) is used for the energy at the wheels. To have energy demand in Wh/km, divide the outcome of (2.30) by the factor Now, if the battery capacity is expressed in Ah, D ev = Q bat U bat E rec,mv EG η sys For the numerical case without payload, Ē = = kj/100km = = 500 Wh/km, D ev = = 48 km For a payload of kg = 4500 kg, Ē = = kj/km = = 570 Wh/km, D ev = = 42 km Problem 4.3 The 2011 Nissan Leaf electric vehicle has been rated by the EPA as achieving 99 mpg equivalent or 34 kwh/100 miles. Justify this rating. Just consider that the energy content of one U.S. liquid gallon of gasoline is 33.41kWh. Then 33.4 kwh gal Hybrid-Electric Propulsion Systems Problem gal 99 miles 100 = 34 kwh 100miles. Classify the five different parallel hybrid architectures, (1) single-shaft with single clutch between engine and electric machine (E-c-M-T-V), (2) singleshaft with single clutch between engine electric machine and transmission (E-M-c-T-V) or (M-E-c-T-V), (3) two-clutches single-shaft (E-c-M-c-T-V), (4) double-shaft (E-c-T-M-V), (5) double-drive, with respect to the following features:

25 Index 25 - regenerative braking: optimized (without unnecessary losses) / not optimized - ZEV mode: optimized (without unnecessary losses) / not optimized - stop-and-start: optimized (independent from vehicle motion) / not optimized / not possible - battery recharge at vehicle stop: possible / not possible - gear synchronization: optimized (no additional inertia on the primary shaft) / not optimized - compensation of the torque holes during gear changes: possible / not possible - active dampening of engine idle speed oscillations: possible / not possible - Solution Architecture (1): - compensation of the torque holes during gear changes not possible - reg. braking ideal - ZEV mode ideal - stop/start compromised - battery recharge at vehicle stop impossible - gear synchronization compromised (but compensation through the electric machine itself possible) - active dampening impossible Architecture (2), e.g., an Honda IMA-type system (E-M-c-T-V), or a belt starter-alternator case (M-E-c-T-V): - reg. braking compromised - ZEV mode compromised - stop/start ideal - active dampening possible - battery recharge at vehicle stop possible - gear synchronization ideal Architecture (3): - reg. braking ideal - ZEV mode ideal - stop/start ideal - active dampening possible - battery recharge at vehicle stop possible - gear synchronization ideal Architecture (4): - reg. braking compromised - ZEV mode compromised - stop/start compromised

26 26 Index - active dampening impossible - battery recharge at vehicle stop impossible - gear synchronization ideal - compensation of the torque holes during gear changes possible Architecture (5): - reg. braking ideal - ZEV mode not ideal - stop/start impossible (would need an additional starter machine) - active dampening impossible (see stop/start) - battery recharge at vehicle stop impossible - gear synchronization ideal - compensation of the torque holes during gear changes possible Find a summary of these features in the table below. E-c-M-T-V E-M-c-T-V E-c-M-c-T-V E-c-T-M-V E-c-T-V-M RB ZEV S/S rech. at stop gear sync. comp. holes act. dmp. Problem 4.5 Determine the overall degrees of freedom u in modeling (i) a parallel hybrid, (ii) a series hybrid, (iii) a combined hybrid, with the quasistatic approach. For (ii) and (iii) use both the generator causality depicted in Figs and the alternative causality introduced in Sect Parallel hybrid There are 6 blocks {V,T,E,M,P,B} and 7 relationships: 1. f V (v,f t ) = 0 2. f T,1 (F t,t e,t m,γ) = 0 3. f T,2 (v,ω e,γ) = 0 4. f T,3 (v,ω m,γ) = 0 5. f E (u e,t e,ω e ) = 0 (u e is the engine control vector) 6. f M (T m,ω m,p m ) = 0 7. f P (P m,p b ) = 0

27 Index 27 between the 10 variables v,f t,t e,t m,ω e,u e,ω m,γ,p m,p b. Consider γ as fixed. Thus there are two independent variables. In the quasistatic approach, v is known, thus the remaining degree of freedom is, e.g., the torque split ratio at the torque coupler u (needed to solve the f T,1 equation). Series hybrid There are 7 blocks and relationships 1. f V (v,f t ) = 0 2. f T,1 (F t,t m ) = 0 3. f T,2 (v,ω m ) = 0 4. f M (T m,ω m,p m ) = 0 5. f P (P m,p g,p b ) = 0 6. f G (P g,ω g,t g ) = 0 7. f E (u e,t e,ω e ) = 0 between the 10 variables v,f t,t m,ω m,p m,p b,p g,t g = T e,ω g = ω e,u e. Thus there are three independent variables. In the quasistatic approach, v is known, thus the remaining degrees of freedom are the power split ratio u (needed to solve the f P equation) and the generator speed ω g (needed to solve the f G equation). In the alternative causality of the generator block, generator speed and torque T g are used to solve the f G equation. Combined hybrid There are 8 blocks and 11 relationships 1. f V (v,f t ) = 0 2. f T,1 (F t,t f ) = 0 3. f T,2 (v,ω f ) = 0 4. f PSD,1 (ω f,ω g,ω e ) = 0 5. f PSD,2 (ω f,ω g,ω m ) = 0 6. f PSD,3 (T f,t g,t e ) = 0 7. f PSD,4 (T f,t g,t m ) = 0 8. f M (T m,ω m,p m ) = 0 9. f P (P m,p g,p b ) = f G (P g,ω g,t g ) = f E (u e,t e,ω e ) = 0 betweenthe14variablesv,f t,ω f,t f,t m,ω m,p m,t g,ω g,p g,p b,t e,ω e,u e.thus there are three independent variables. The degrees of freedom are the same as for the series hybrid case. Problem 4.6 Perform the same analysis as in Problem 4.5 with the dynamic approach. Calculate the number n v of variables in the flowcharts of Figs Then calculate the number n e of the equations available using the simple models presented in this chapter. Finally evaluate the manipulated variables that are necessary to realize the degrees of freedom (DOF) determined in Problem 4.5.

28 28 Index Parallel hybrid There are 6 blocks and n e = 10 relationships: 1. f V (v,f t ) = 0 2. f T,1 (F t,t e,t m,γ) = 0 3. f T,2 (v,ω e,γ) = 0 4. f T,3 (v,ω m,γ) = 0 5. f M,1 (T m,i m ) = 0 6. f M,2 (ω m,u m,u m ) = 0 (u m is the motor control vector) 7. f P,1 (U m,u b ) = 0 8. f P,2 (I m,i b ) = 0 9. f E (u e,t e,ω e ) = f B (I b,u b ) = 0 between the n v = 10 variables represented in the figure. If γ is fixed, the control inputs u e, u m determine the vehicle speed and the torque split ratio. Series hybrid There are 6 blocks and n e = 12 relationships: 1. f V (v,f t ) = 0 2. f T,1 (F t,t m ) = 0 3. f T,3 (v,ω m ) = 0 4. f M,1 (T m,i m ) = 0 5. f M,2 (ω m,u m,u m ) = 0 6. f P,1 (U m,u b,u g ) = 0 7. f P,2 (I m,i b ) = 0 8. f P,3 (I m,i g ) = 0 9. f G,1 (T g,i g ) = f G,2 (ω g,u g,u g ) = 0 (u g is the generator control vector) 11. f E (u e,t e,ω e ) = f B (I b,u b ) = 0 between the n v = 12 variables represented in the figure. The control inputs u e, u m, and u g determine the vehicle speed, the power split ratio, and the generator speed. Combined hybrid There are 8 blocks and n e = 16 relationships: 1. f V (v,f t ) = 0 2. f T,1 (F t,t f ) = 0 3. f T,2 (v,ω f ) = 0 4. f PSD,1 (ω f,ω g,ω e ) = 0 5. f PSD,2 (ω f,ω g,ω m ) = 0 6. f PSD,3 (T f,t g,t e ) = 0 7. f PSD,4 (T f,t g,t m ) = 0 8. f M,1 (T m,i m ) = 0 9. f M,2 (ω m,u m,u m ) = f P,1 (U m,u b,u g ) = f P,2 (I m,i b ) = f P,3 (I m,i g ) = 0

29 13. f G,1 (T g,i g ) = f G,2 (ω g,u g,u g ) = f E (u e,t e,ω e ) = 0 (u e is the engine control vector) 16. f B (I b,u b ) = 0 Index 29 between the n v = 16 variables represented in the figure. The control inputs u e, u m, and u g determine the vehicle speed, the power split ratio, and the generator speed. Problem 4.7 Perform the same analysis as in Problems for an electric powertrain powered by a battery and a supercapacitor. Quasistatic approach There are 6 blocks {V,T,M,P,B,SC} and 5 relationships: 1. f V (v,f t ) = 0 2. f T,1 (F t,t m ) = 0 3. f T,2 (v,ω m ) = 0 4. f M (T m,ω m,p m ) = 0 5. f P (P m,p b,p sc ) = 0 between the 7variablesv,F t,t m,ω m,p m,p b,p sc. Thusthere aretwoindependent variables. In the quasistatic approach, v is known, thus the remaining degree of freedom is, e.g., the power split ratio at the DC link u (needed to solve the f P equation). Dynamic approach There are 6 blocks and n e = 10 relationships 1. f V (v,f t ) = 0 2. f T,1 (F t,t m ) = 0 3. f T,2 (v,ω m ) = 0 4. f M,1 (T m,i m ) = 0 5. f M,2 (ω m,u m,u m ) = 0 6. f P,1 (U m,u b,u sc ) = 0 7. f P,2 (I m,i b ) = 0 8. f P,3 (I m,i sc ) = 0 9. f B (I b,u b ) = f SC (I sc,u sc ) = 0 between the n v = 10 variables. If there is only one control input u m the power split ratio cannot be chosen. Thus a second controllable component is needed, typically a DC DC converter on either the supercapacitor or the battery side.

30 30 Index Problem 4.8 For a plug-in hybrid, the fuel consumption according to UN/ECE regulation [91] is C = D e C 1 +D av C 2 D e +D av, where C 1 is the fuel consumption in charge-depleting mode, C 2 is the consumption in charge-sustaining mode, D e is the electric range, and D av is 25 km, the assumed average distance between two battery recharges. Estimate the fuel consumption of the electric system of Problem 4.1 equipped with a range extender having a max power of 5kW and an efficiency of 0.4. D e = 100 km, C 1 = 0, D av = 25 km. To evaluate the fuel consumption in charge-sustaining mode, divide the cycle into two phases, with (i) APU on, and (ii) APU off. The mean force is the same. During phase (i), E bat = F r e b x on, where F r is the mean traction force to recharge the battery, e b = e is the battery efficiency, and x on is the distance coveredduring the phase (i). During phase (ii) E bat = F e (x tot x on ). By equalizing these two energy terms (charge sustaining), x on = x tot F F +F r e e. The APU mean power during phase (i) is P apu = (F r + e F ) xon = F +F r e t tot F +F r e e v e and the average fuel power is Numerically, F r = P apu,max v F e = P apu = P f = P apu e apu xon x tot = 135 Nm = 4.1 kw P f = = 10.4 kw V f = = l/s = = l /100 km = C 2

31 Index 31 Thus the combined fuel consumption is C = D e C 1 +D av C 2 = = 0.68 D e +D av 125 l /100 km Motor and Motor Controller Problem 4.9 Consider a separately-excited DC motor having the following characteristics: R a = 0.05Ω, battery voltage = 50V (neglect battery resistance), rated power = 4kW, nominal torque constant κ a = κ i = 0.25Wb, aimed at propelling a small city vehicle. Calculate the motor voltage and current limits, then the flux weakening region limit (maximum torque and base speed). Calculate the step-down chopper duty-cycle α for the following operating points: (i) ω m = 100rad/s and T m = 15Nm; (ii) ω m = 300rad/s and T m = 8Nm. The maximum voltage is U max = 50V. The maximum admissible current is calculatedbyforcingu a = U max and ω m T m = P max. Thefollowingquadratic equation is obtained, U max I max = R a I 2 max +P max, 0.05 Ω I 2 max 50 V I max W = 0, or whosesolution is I max = 88A. Thus the maximum torqueis = 22Nm. The flux weakening region limit occurs when U a = U max thus R a T m κ a +κ a ω m = U max, or 0.2 T m ω m = 50, extending from T m = 250Nm on the torque axis to ω m = 200rad/s on the speedaxis.thebasespeedisω b = 4000/22= 182rad/s.Forthefirstoperating point, I a = T m = 15 κ a 0.25 = 60 A U a = R a I a +κ a ω m = = 3+25 = 28 V. Both current and voltage limits are respected. The chopper duty-cycle is α = 28/50 = 56%. The second operating point belongs to the flux weakening region. In fact, if one were to calculate the current and voltage with the above equations, I a = 8/0.25 = 32A would be obtained, but U a = = 77V that is beyond the admissible voltage. Thus κ a must be reduced. To find κ a such that U a = 50V (α = 100%), the following equation is used

32 32 Index U max = R a T m κ a +κ a ω m, whichleadstoκ a = 0.16Wb.Anapproximatedvalueisobtainedbyneglecting the resistance as κ a = U max /ω m = 50/300= 0.17Wb. Problem 4.10 For the DC motor of Problem 4.9, evaluate the approximation of mirroring the efficiency from the first to the fourth quadrant, for the two operating points (i) ω m = 50rad/s and T m = 22Nm; (ii) ω m = 300rad/s, T m = 8Nm. Assume further that P l,c = 0. From (4.14), for T m > 0 1 = 1+ R a T m η m κ 2 a ω, m P l = R a Tm 2 κ 2. a For T m < 0 For the point (i) η m = 1+ R a T m κ 2 a ω m = η m (50,22) = = 0.74, η m (50, 22) = = 0.65, ( ) 2 22 P l = 0.05 = 387 W In the field weakening region, for the point (ii), 1 η m (300,8) = = 0.98, η m (300, 8) = = 0.98, ( ) 2 8 P l = 0.05 = 51 W For the given values, the approximation of mirroring the efficiency is better as the losses decrease.

33 Index 33 Problem 4.11 Using the PMSM model calculate a static control law, i.e., a selection ofreferencevaluesi d,i q asafunction oftorqueandspeed,suchthatthe stator current intensity is minimized. Do the calculation in the case (i) I s I max, U s U max = mu m (maximum torque region) and (ii) when the voltage constraint is active (flux weakening region). Neglect the stator resistance R s and consider a machine with p = 1. The stator current and voltage intensities are defined as I 2 s = I2 q +I2 d, U2 s = U2 q +U2 d. Evaluate the base speed. If R s is neglected, the static counterparts of (4.26)-(4.28) are U q = ω m (ϕ m +L s I d ), U d = ω m L s I q, T m = 2 3 T m = ϕ m I q. To obtain the desired torque, set I q = T m /ϕ m. To minimize I s without constraints, I d = 0. Under these conditions, I s = I q = T m/ϕ m and ( ) ) 2 U 2 s = ω2 m ϕ 2 m + ( Ls T m ϕ m Such a situation is valid in (i), i.e., if I s I m, i.e., if T m ϕ m I max, and if ( ) ) 2 U 2 s = ω2 ϕ 2 m + ( Ls T m ϕ m. U 2 max. The base speed is obtained from the intersection of the latter limits, i.e., for ω b = U max ϕ 2 m +(L s I max ) 2. If a negative I d is allowed, points above the base speed are obtained. In case (ii), U s = U max andt m = ϕ m I q andi d is calculated(a second-orderequation is obtained) as (Umax ) 2 ( ) T 2 I d = m ϕ m. L s ω m ϕ m L s

34 34 Index Problem 4.12 Evaluate the torque limit curve for a PMSM both (i) in the maximum torque region and (ii) in the flux weakening region, see Problem 4.11, assuming R s = 0.Evaluatethe transitioncurvebetweenthese tworegions.assume that ϕ m > L s I max (why is that important?). The torque limit in the maximum torque region is simply T max = ϕ m I max. In the flux weakening region, the torque limit is generally lower than ϕ m I max and is calculated using a graphical construction. The maximum current ( I ) curve is a circle in the I d I q plane, with center at the origin and radius I max. The maximum voltage ( U ) curve is a circle with center [ ϕ m /L s,0] and radius U max /(ω m L s ). Under the assumption that ϕ m > L s I max, the center of U-curve is found outside the I-curve. Thus the largest value of I q that fulfills both constraints is where the two curves I and U intersect. In this case, the coordinates of the intersection are ( ) 2 Us ω m ϕ 2 L 2 s I2 max I d =, 2 ϕ m L s I q = 1 I 2 max 4 ϕ 2 m L 2 s ( (Umax ω m ) 2 ϕ 2 m (L s I max ) 2 ) 2, from whence T max (ω m) = ϕ m I q (ω m ). The maximum speed at which the maximum torque is null is ω max = U max ϕ m L s I max. The transition curve is the locus of the torque points that can be still achieved with I d = 0. It is given by the intersection of U-curve with the y-axis, resulting in ( ) 2 L 2 s I2 q = Umax ϕ 2 m ω, m that is the transition curve sought with T m = ϕ mi q. In particular, for I q = I max, obtain the base speed as Problem 4.13 U 2 max ωb 2 = ϕ 2 m +L 2 s Imax 2 Using the same assumptions as in Problem 4.11, evaluate the maximim power curve as a function of speed.

35 Index 35 The general expression for the maximum power is P max = ω m T max. Let us calculate the maximum of P max. dp max dω m = 0 T max +ω m dt max dω m = 0. But T max = ϕ m I q and I 2 q = f(x) as given by Problem 4.11, where X = U max /ω m. Consequently, the maximum condition is given by 2 I q di q = df dω m dx X ω ϕ m I q ϕ m df dx X 2 I q = 0 2 f = df dx X. After having calculated the derivative df/dx, obtain a 2nd-order equation in the variable X, whose solution is X 2 = ϕ 2 m L2 s I2 max, from whence ω P = U max. ϕ 2 m L 2 s I2 max For this speed, f = 4 L 2 s I2 max (ϕ2 L 2 s I2 max ) = 4 L2 s I2 max (U max ω m ) 2, I 2 q = f/(4 ϕ 2 L 2 s) I q = U max I max ϕ m ω m, and finally T m = Umax Imax ω m or P max = U max I max. Problem 4.14 Equation 4.42 is only valid when L d = L q = L s. In the general case in which L d L q, the correct equation is T m = 3 2 p I q (ϕ m p (L q L d ) I d ). Consider again Problem 4.11 and derive a static control law I d, I q that minimizes the current intensity (MTPA), assuming that the constraints over current and voltage are not active, for a motor where p = 4, R s = 0.07Ω, L q = H, L d = H, ϕ m = 0.185Wb and T = 50Nm. Then compare the result with that obtained for L = L q L d = 0.

36 36 Index To evaluate I d, a procedure similar to that of Problem 4.11 is used. Now minimize I 2 d +I2 q, subject to the condition ϕ m I q L I d I q = T m. This is a parameter optimization problem in the two parameters I d, I q. Build the Hamiltonian H = I 2 d +I 2 q +µ (ϕ m I q L I d I q T m). Pontryagin s Minimum Principle reads from whence dh di d = 2 I d µ L I q = 0, dh di q = 2 I q +µ ϕ m µ L I d = 0, L I 2 d ϕ m I d L I 2 q = 0, that is, a quadratic equation is found. Now combine this equation with the torque equation to have I d and I q as a function of T m, L (T m )2 = ( L I 2 d ϕ m I d ) (ϕ m L I d ) 2 = = L 3 I 4 d 3 ϕ m L 2 I 3 d +3 ϕ2 m L I2 d ϕ3 m I d = = I d ( L I d ϕ m ) 3 For the numerical case, L = H, T m = 50/(3/2 4) = 8.33Nm, thus 0 = ( ) 3 I 4 d ( ) 2 I 3 d I 2 d I d I d = 17 A, 0 = I 2 q I q = 34 A. Verify that T m = ( ) = 50 Nm. With L = 0, one would have obtained I d = 0 and I q = 50/4/(3/2)/0.185= 45 A. Problem 4.15 Calculate the torque characteristic curve T m (ω m,u s ) of a PMSM having the following characteristics: R s = 0.2Ω, L = 0.003H, 3 2 ϕ m = 0.89Wb, p = 1, for a voltage intensity (see definition in Problem 4.11) U s = 30V. Derive an affine approximation of the DC-motor type, T m,lin (ω m,u s ). Evaluate the torque error ε(ω m ) U 2 s (T m) U 2 s (T m,lin) and calculate its maximum value.

37 Index 37 If L d = L q, the MTPA (maximum torque per ampere) control is (see Problem 4.11) I d = 0, I q = T m/ϕ m. Consequently, the voltage is U q = ω m ϕ m +R s I q = ω m ϕ m + R s T m ϕ m, U d = ω m L s I q = ω m Ls T m ϕ m. The torque characteristiccurve T m = T m (ω m,u s ) for a givenu s = is given by U 2 d +U2 q U 2 s = ω2 m ϕ2 m + R2 s (T m )2 ϕ 2 m (ϕ m U s ) 2 ω 2 m ϕ4 m = (T m )2 +2 ω m R s T m +ω2 m L2 s (T m )2 ϕ 2 m ( ) (ω m L s ) 2 +Rs 2 +2 R s ω m ϕ 2 m T m. For ω m = 0, the breakaway torque is The zero torque speed is T br = U s ϕ m R s. ω 0 = U s ϕ m. The affine approximation of the characteristic curve is T lin(ω m,u s ) = T br + T m ω m ω m ωm=0 wheret m = T m(ω,u s ) is derivedfrom the equationabove.fromacomparison with the DC-motor characteristic curve, one can make the equivalence κ a = ϕ m, and T lin (ω m,u s ) = ϕ m R s U s ϕ2 m R s ω m. At ω m = ω 0 the approximated torque is T 0 = ϕ m R s U s ϕ m R s U s = 0. Thus, at ω0 the error is zero with respect to the nonlinear characteristic curve. To generally evaluate this error, calculate U s (T m ) from the nonlinear curve and U s (T m,lin ) from the affine curve. One obtains U 2 s(t m ) U 2 s(t m,lin ) = ε(ω m ). This term can be calculated using the results above, such that

38 38 Index ε(ω m ) = ( Rs 2 (T +ω2 m L2 s ) m) 2 ϕ 2 +2 ω m R s T m m ( ) + 1+ ω2 m L2 s Rs 2 (U s ω m ϕ m ) 2 2 ω m ϕ m (U s ω m ϕ m ) = R2 s (T m )2 + ω2 m L2 s (T m )2 +ωm 2 ϕ2 m +2 ω m R s T m U2 s ϕ 2 m ϕ 2 m ( ) 2 ωm L s + (U s ω m ϕ m ) 2 R s ( ) 2 ωm L s = (U s ω m ϕ m ) 2, R s which is zero for ω m = 0 and ω m = ω 0. Define E(ω m ) = ω 2 m (U s ω m ϕ m ) 2. The maximum value for E(ω m ) is obtained by differentiating w.r.t. ω m : de dω m = 2 ω m (U s ω m ϕ m ) 2 +2 ω 2 m (U s ω m ϕ m ) ( ϕ m ) = 0 U s ω m ϕ m = ω m ϕ m ω max,e = U s 2 ϕ m = ω 0 2. The maximum error is ε(ω max,e ) = ( U 2 s L s 4 ϕ m R s In relative terms ( ) 2 ε(ω max,e ) Us L s Us 2 =. 4 ϕ m R s With the numerical data ε(ω max,e ) = 14.4V 2 or in relative terms 14.4/30 2 = 1.6%. Problem 4.16 A simple thermal model of an electric machine reads ) 2. C t,m d dt ϑ m(t) = P l (t) ϑ m(t) ϑ a R th, where C t,m is an equivalent thermal capacity, R th is an equivalent thermal resistance, ϑ m (t) is the relevant motor temperature, and ϑ a is the external temperature. Derive the current limitation to I max from thermal considerations using the models ofsect (DC motor).how would the result change if other losses of the type β ω m (iron losses, mechanical losses) were taken into account?

39 Index ε [V 2 ] ω max [rad/s] v11 15 ε [V 2 ] ω max [rad/s] Fig Visualization of the results of Problem 4.15: (a) Absolute error over ω max,e, (b) exact ( ) and linear ( ) torque curve. Behind the limitation I a = I max leading to T m = T max (for ω m < ω b ) there is a temperature limitation ϑ m = ϑ max. Consider the DC motor model. Here the only loss is due to ohmic losses P l = R a I 2 a. The motor (windings) temperature varies according to this power dissipated into heat and according to heat exchange to the ambient, so C t,m dϑ m dt = P l ϑ m ϑ a R th To guarantee that ϑ m < ϑ max, it should be or I a < R a I 2 a < α (ϑ max ϑ a ), = R a I 2 a ϑ m ϑ a R th. (10.24) ϑmax ϑ a R th R a = const. = I max. If other losses of the type β ω m are considered (the factor β could in turn be dependent on ω m ), the condition on temperature reads

40 40 Index from whence I a < R a I 2 a +β ω m < ϑ m ϑ a R th, ϑmax ϑ a β ω m R th R a = I max (ω m ). Thus the maximum torque for ω m < ω b would decrease with ω m. Problem 4.17 Derive a (simplified) rule to express peak torque limits of an electric machine as a function of application time. Use the result of Problem Relax the condition on stationary temperature. The solution to the ODE (11.1) is then ϑ m (t) = ϑ a +(ϑ stat ϑ a ) (1 e t τ ), where ϑ stat = ϑ a + R th R a Ia 2, and τ = C t,mr th. Impose that ϑ(t) = ϑ max and obtain I max as a function of time, ( ) ϑ max ϑ a = R th R a Imax(t) 2 1 e t τ I max (t) = (ϑ max ϑ a ) ( ) R th R a 1 e t τ For t, one finds T max (t) = κ a I max (t) T max = κ a which is the result of Problem ϑmax ϑ amb R th R a, Problem 4.18 One PMSM has the following design parameter: external diameter d 1 = 0.145m, weight m 1 = 14kg, length l 1 = 0.06m. At 5500rpm it delivers a maximum torque T 1 = 12Nm. Predict the power P 2 and the weight m 2 for a similarly designed machine with a diameter d 2 = 0.2m and a length l 2 = 0.2m. Compare the cases in which the design is made (i) on the basis of constant tangential stress and peripheral speed, or (ii) of constant speed.

41 Index 41 Case (i). The power of machine 1 is The mean rotor speed is The mean pressure is c m = ω 1 d 1 2 P 1 = 5500 π 30 p me = T 1 2 V 1 = = 5500 π = 6.9 kw = 41.8 m/s (π ) = 6 kpa. Indeed, P 1 = π d 1 l 1 p me c m = π = 6.9 kw. For machine 2, P 2 = π = 31.5 kw, ( ) 2 ( ) 2 d2 m 2 = m 1 l = 14 = 89 kg, l d 1 assuming constant density. Case (ii). What changes now is that For machine 2, P = ω p me π d2 2 l. P 2 = π π = 43 kw, while m 2 is unchanged. The specific power is the same for machine one and two, = = 0.49 kw/kg, Problem 4.43 Evaluate ( the specific power of a motor and inverter assembly, knowing that P m) motor = 1.2kW/kg and ( ) P m inverter = 11kW/kg. The specific power of the motor system is simply ( ) P 1 1 = m 1 1 = + ( m) P motor ( m) P inverter = 1.08 kw/kg.