ME 466 PERFORMANCE OF ROAD VEHICLES 2016 Spring Homework …me.metu.edu.tr/courses/me466/ME466_SP16_HW6_Solution.pdf · ME 466 PERFORMANCE OF ROAD VEHICLES 2016 Spring Homework 6

ME 466

PERFORMANCE OF ROAD VEHICLES

2016 Spring Homework 6 Solution

Assigned by Dr. Kerem Bayar

PROBLEM 1

There is actually a typo in this problem. The engine speed corresponding to the max

vehicle speed is given wrong. It should be corrected as:

Max vehicle speed: 126 mph at 6284 rpm. The solution below for Part 1a is given

taking 5930 rpm into account for simplicity.

a) Vmax = 126 kph @ ne = 5930 rpm

K = Vmax.it5 / ne = 126 . 0,92 / 5930

= 0.01955

Maximum speed @ 1st gear Minimum speed @ 1st gear

V1max = K . ne

max / it1 = 0,01955 . 6500 / 3,55 V1min = K . ne

min / it1 = 0,01955 . 1000 / 3,55

= 35,79 mph = 5,5 mph

In 2nd gear @ this vehicle speed

ne = V1max.it2 / K = 35,79 . 2,24 / 0,01955

= 4101 rpm

V2max = K . ne

max / it2 = 0,01955 . 6500 / 2,24

= 56,72 mph

With the same approach, the gear shift diagram can be prepared as follows:

GEAR 1st 2nd 3rd 4th 5th

Vlow [mph] 5,5 35,8 56,7 83,6 108,6

Vhigh [mph] 35,8 56,7 83,6 108,6 126

nlow [rpm] 1000 4101 4411 5003 5111

nhigh [rpm] 6500 6500 6500 6500 5930

b) rw = 0,96 ((15 . 25,4) / 2000 + (195/1000) 0,55) = 0,286 m

Vmax5 = 126 . 1,609 = 202,7 kph

idref = 0,286 . 6100 / (2,65 * 0,92 * 202,7) = 3,53 < 3,56 (given) which means the vehicle

is overgeared.

An alternative to this method, with the corrected engine speed corresponding to the max

vehicle speed would be simply 6284 rpm > 6100 rpm indicating overgeared. For the ones

who have concluded 5930 rpm < 6100 rpm indicating undergeared, no points have been

broken.

c) Wf = 846,09 kg Wr = 496,91 kg af = 0.01187 ar = 9,934 10-3

Ra = 0,047 . 0,32 (0,8*1,425*1,67) V2

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Ra = 0,02863 V2

Rr = 146,95 + 0,1976 V

Rtotal = 0,02863V2 + 0,1976V + 146,949

V.Rt = P1.t

V/3,6 (0,02863V2 + 0,1976V + 146,949) = 147 * 745,7 * 0,88

V = 220 kph

id = 0,286 . 6100 / (2,65*0,92*220)

id = 3,25

PROBLEM 2 (15 pts)

Max speed would be observed at max power. Thus

(Vmax / 3,6) (150 + 0,2 Vmax + 0,031 Vmax2) = 90000.0,9

Solving the above equation yields

Vmax = 201,4 kph

rw = 0,96 (15*25,4/2000+195/1000*0,5) = 0,276 m

id = 0,276 . 5800 / (2,65*1*201,4)

id = 3

PROBLEM 3 (15 pts)

a) n = 12

S = nH/nL = 2000/1600 = 1,25

iT = Sn-1

iT = 1,25(12-1)

iT = 11,642

iTt = it1/itn

11,642 = it1 / 1 → it1 = 11,642

b) We can check by the following rule of thumb: Speed in 1st gear corresponding to min

engine speed must be below 2,5 kph for heavy commercial vehicles. @ ne = 500 rpm

ntire = 500 / it id = 500 / (11,462*3,31) = 12,975 rpm; that means one rotation lasts 1/12,975

minutes. From Table III-5 of Chapter 3 in the lecture notes 11/80 R 22,5 14/G → 320

revolution per kilometer. Then 1 revolution corresponds to 1/320 km; i.e. V = (1/320) km

/ (1/12,975) min = 2,433 kph < 2,5 kph → therefore 12 gears would be sufficient for this

vehicle.

PROBLEM 4 (15 pts)

Writing the relative speed equations for the two

meshes highlighted on the figure in the left:

1 1 lhs cage rhs cagein in

in cage lhs in cage rhs

N N

N N

Dividing the first equation by the second equation

yields:

1lhs cage

rhs cage

this yields cage = 0.5(lhs+rhs)

Knowing also cage = em / imain yields em = 0.5imain(lhs+rhs). The free body diagram

of the electric motor output shaft and the input gear is shown below:

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1

em

em

input

T T

I (1) where Iinput = Iem + Iinput with Iem electric motor

output shaft inertia and Iinput input gear inertia.

The free body diagram of the cage is shown below:

1 1 6main

cage

cage

T .i rF rF

I

(2) where

cage em main/ i (3)

The free body diagram of the upper inner gear, right gear, lower inner gear and left gear

respectively, are shown below:

Force balance F1 = F2 + F3 (4)

Moment balance 3 2 1in inF F r I (5)

Force balance F2 = F5 (6)

Moment balance 5 2 rhs r rhsF F r T I (7)

Force balance F6 = F7 + F5 (8)

Moment balance 7 5 2in inF F r I (9)

Force balance F3 = F7 (10)

Moment balance 3 7 lhs l lhsF F r T I (11)

From (1) we have 1 em input emT T I and substituting cage using Equation 3 yields

1 em input cage mainT T I i Substituting this equation into Eqution 2 yields:

1 6em input cage main main cage cageT I i i rF rF I . rearranging terms yields

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2

1 6main em cage input main cagei T I I i rF rF

we also know from the kinematic relationship that 0 5cage lhs rhs

. which yields

2 2

1 62 2

cage input main cage input main

main em lhs rhs

I I i I I ii T rF rF

(12)

Equations 4, 6, 8, 10 yield F1 = F6 and1in

= 2in

. Therefore Equations 4-11 boil down to

F1 = F2 + F3 (13)

3 2 1in inF F r I (14)

22

rhs r rhsF r T I (15)

32

lhs l lhsF r T I (16)

Equations 13 and 14 yield

1 1

32 2

in inI F

Fr

substituting this equation in Equations 14-16 yields:

1

20

2 2

in inIF

Fr

(17)

1lhs in in l lhsT I Fr I (18)

22

rhs r rhsT F r I (19)

From Equation 17 we have

1

22 2

in inI F

Fr

and substituting this in Equation 19 yields:

1rhs in in r rhsT I Fr I adding this equation with Equation 18 side by side yields:

12

lhs rhs r rhs l lhsFr T T I I (20)

On the other hand subtracting the same equation from Equation 18 yields

2rhs lhs in in l lhs r rhs

T T I I I (21)

Substituting Equation 20 into Equation 12 yields 2 2

2 2


main em lhs rhs l lhs r rhs

I I i I I ii T T T I I

and substituting

Equation 21 into this equation yields: 2 2

0 54 4


main em lhs l lhs rhs in in

I I i I I i. i T T I I

(22)

Finally, from the very first kinematic relationships above, one can write

2

l r

in

or

2

l r

in

and substitutuing this equation into Equation 22 yields

2 2

0 54 2 4 2

cage input main cage input mainin in

main em lhs l lhs rhs

I I i I I iI I. i T T I

This concludes the derivation, with the other equation being symmetric, i.e. 2 2

0 54 2 4 2

cage input main cage input mainin in

main em rhs r rhs lhs

I I i I I iI I. i T T I

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PROBLEM 5 (40 pts)

1a)

em

r

sICE

r

rsout

n

n

n

nn

1b)

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1c)

1d) The maximum vehicle speeds (kinematically) for different electric motor speeds are

seen on the following figure:

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So the range of vehicle speeds kinematically possible to achieve is -39.2 - 180.1 kph.

1e) Pure electric mode (wICE=0) is imposed as an arrow on the following figure:

So the range of speeds for pure electrical mode (wICE=0) is 0-39.1 kph

1f) The arrow imposed on the figure below shows 0 → idle speed while the vehicle is

stationary.

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As can be seen on the figure, the electric motor speed is increased till the engine idle

speed is reached.

1g) The arrows on the figure below show the way to change the direction of the vehicle

speed with changing wem with the engine at idle.

2a) The speed relation was found to be:

em

r

sICE

r

rsout

n

n

n

nn

(1)

Similarly

ICE

rs

rout T

nn

nT

(2)

Like CVT, an EVT system is capable of continuously modulating input/output speed

ratios but the difference of EVT from CVT as can be realized from the above relations

that it brings the benefit of controllable distribution of power from two different sources,

namely engine and motor, to one output. This relation can be realized if equations (1) and

(2) are multiplied to find the output power:

emICE

rs

sICEICEICEoutoutout T

nn

nPTPT

)(

b, c, d, e, f) The results below were taken from the workspace after running the m-file:

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wem wice Tmax_engine Tmax_em Tem_equivalent Tice_actual

0 1596.4616 279.6743 -124 446.4 279.6743

500 1735.3505 309.7299 -124 446.4 309.7299

1000 1874.2394 319.5095 -101.2225 364.4012 319.5095

1500 2013.1283 325.8534 -67.4817 242.9341 242.9341

2000 2152.0172 322.1867 -50.6113 182.2006 182.2006

2500 2290.9061 318.8964 -40.489 145.7605 145.7605

3000 2429.795 316.5075 -33.7408 121.4671 121.4671

3500 2568.6838 315.0802 -28.9207 104.1146 104.1146

4000 2707.5727 314.6358 -25.3056 91.1003 91.1003

4500 2846.4616 313.4582 -22.4939 80.978 80.978

*The angular velocities are in rpm and torque values are in Nm.

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So the first three data points which correspond to electric motor speeds of 0, 500, and

1000 rpm determine the limit on torque, i.e. engine is the limiting part for these speeds.

For the rest of the range of interest, it can be seen that electric motor is the limiting part.

Note that the region bounded by the pink curve is the region where it is possible to operate

the engine and the electric motor while respecting their torque limits.

g, h) The results below were taken from the workspace after running the m-file:

Tice Tem Toutput

279.6743 -77.6873 201.987

309.7299 -86.0361 223.6938

319.5095 -88.7526 230.7569

242.9341 -67.4817 175.4524

182.2006 -50.6113 131.5893

145.7605 -40.489 105.2714

121.4671 -33.7408 87.7262

104.1146 -28.9207 75.1939

91.1003 -25.3056 65.7947

80.978 -22.4939 58.4841

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Pice Pem Poutput

46.7562 0 46.7562

56.2858 -4.5048 51.781

62.7101 -9.2942 53.4159

51.214 -10.6 40.614

41.0605 -10.6 30.4605

34.9684 -10.6 24.3684

30.907 -10.6 20.307

28.006 -10.6 17.406

25.8302 -10.6 15.2302

24.138 -10.6 13.538

Bonus question (bonus.m provided in the zip file)

With the new case, all the above results and figures are modified as follows after running

the m-file (bonus.m provided in zip file)

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wem wice Tmax_engine Tmax_em Tem_equivalent Tice_actual

-4500 1623.6309 285.5537 -22.4939 80.978 80.978

-4000 1762.5198 313.5661 -25.3056 91.1003 91.1003

-3500 1901.4087 320.9549 -28.9207 104.1146 104.1146

-3000 2040.2976 325.1361 -33.7408 121.4671 121.4671

-2500 2179.1865 321.4695 -40.489 145.7605 145.7605

-2000 2318.0754 318.4291 -50.6113 182.2006 182.2006

-1500 2456.9643 316.0402 -67.4817 242.9341 242.9341

-1000 2595.8531 314.9933 -101.2225 364.4012 314.9933

-500 2734.742 314.5488 -124 446.4 314.5488

0 2873.6309 313.1648 -124 446.4 313.1648

500 3012.5198 311.3042 -124 446.4 311.3042

1000 3151.4087 305.8042 -101.2225 364.4012 305.8042

1500 3290.2976 299.8045 -67.4817 242.9341 242.9341

2000 3429.1865 292.5823 -50.6113 182.2006 182.2006

2500 3568.0754 285.6596 -40.489 145.7605 145.7605

3000 3706.9643 279.0485 -33.7408 121.4671 121.4671

3500 3845.8531 270.7504 -28.9207 104.1146 104.1146

4000 3984.742 261.6948 -25.3056 91.1003 91.1003

4500 4123.6309 252.2436 -22.4939 80.978 80.978

*The angular velocities are in rpm and torque values are in Nm.

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Tice Tem Toutput

80.978 -22.4939 58.4841

91.1003 -25.3056 65.7947

104.1146 -28.9207 75.1939

121.4671 -33.7408 87.7262

145.7605 -40.489 105.2714

182.2006 -50.6113 131.5893

242.9341 -67.4817 175.4524

314.9933 -87.4981 227.4951

314.5488 -87.3747 227.1742

313.1648 -86.9902 226.1746

311.3042 -86.4734 224.8308

305.8042 -84.9456 220.8586

242.9341 -67.4817 175.4524

182.2006 -50.6113 131.5893

145.7605 -40.489 105.2714

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121.4671 -33.7408 87.7262

104.1146 -28.9207 75.1939

91.1003 -25.3056 65.7947

80.978 -22.4939 58.4841

Pice Pem Poutput

13.7684 10.6 24.3684

16.8144 10.6 27.4144

20.7308 10.6 31.3308

25.9526 10.6 36.5526

33.2631 10.6 43.8631

44.2289 10.6 54.8289

62.5052 10.6 73.1052

85.6269 9.1628 94.7896

90.081 4.5749 94.6559

94.2394 0 94.2394

98.2072 -4.5277 93.6795

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100.9199 -8.8955 92.0244

83.7052 -10.6 73.1052

65.4289 -10.6 54.8289

54.4631 -10.6 43.8631

47.1526 -10.6 36.5526

41.9308 -10.6 31.3308

38.0144 -10.6 27.4144

34.9684 -10.6 24.3684

So the electric motor starts acting like a motor instead of a generator as its speed is

reversed.

Documents

ME 466 PERFORMANCE OF ROAD VEHICLES 2016 Spring Homework …me.metu.edu.tr/courses/me466/ME466_SP16_HW6_Solution.pdf · ME 466 PERFORMANCE OF ROAD VEHICLES 2016 Spring Homework 6