Hw6

ME781 Power tr ain D ynamic Homew or k #6

Ming Feng (Alex) Hsieh

Part 1

(a) The kinematical equation of the planetary geartrain can be expressed as: Si Ri Sr Rr RSi = ; RRi = ; RSr = ; RRr = S i + Ri Si + Ri S r + Rr S r + Rr

ω Cr = RSrωSr + RRrω Rr ω Ci = RSiωSi + RRiω Ri 1st Gear: Speed ω Ri =ω Cr ;ω Ci = 0; ωSi = ωSr

ωSr = ωSi = − ω Cr (1 +

RRi R ω Ri = − i ω Cr RSi Si

S r Ri Rr )= ω Rr S i ( S r + Rr ) S r + Rr

=>

ω Cr =

Rr Si ω Rr S i S r + S i Rr + S r Ri

ωSr = −

Ri Rr Si ω Rr 2 Si S r + S i Rr + S i S r Ri 2

The lever diagram is:

Rr

ω ,ω

ω Sr , ω Si Torque TSi = TSr = 0 To = TCr + TRi Ti = TRr TCi =

S r Si + S r Ri Ti Rr Si

To = (1 +

S r Si + S r Ri )Ti Rr Si

The lever diagram is:

ω CiCr, Ri Cr Ri Ci Si, Sr

ω Rr

Rr

Ti

Cr, Ri

To

Ci Si, Sr

TCi

2nd Gear Speed ωSi = ωSr = 0

ω Cr =

Rr ω Rr S r + Rr

The lever diagram is:

ω Rr

Rr

Cr

ωCr

Si, Sr Torque S + Rr To = r Ti Rr The lever diagram is:

Rr

Ti

Cr, Ri

To

Si, Sr 3rd Gear

TSr

Speed ωSi = ωSr = ω Rr

ω Cr = RSrω Rr + RRrω Rr = ω Rr The lever diagram is:

ω Rr

Rr

ωCr

Cr

ωSr

Si, Sr Torque To = TSr + TRr = Ti The lever diagram is:

Rr

TRr

Cr, Ri

To

Si, Sr

TSr

(b) The free-body diagram is shown below. T0

Tf 2 ω sr

Tin

I Si + I S r

I in

ω Cr

Tf 2 ω sr

Ti

ω Rr

I Cr + I R i

T0ut ω Cr

I Rr

Where I in is the inertia of the input shaft, I Rr is the inertia of the reaction ring gear, I Si + I S r is the combination of sun gear inertia of input geartrain and reaction geartrain, and I Cr + I R i is the equilibrium inertia of input ring gear, reaction carrier gear, and output shaft gear. The differential equation of the system according to Tin , Tin , and ωCr can be obtained as follow:

Tin − Ti = ( I in + I Rr )ω Rr  Tout − To = ( I Cr + I R i )ω Cr Ti = − A12T f 2 + A22To To = Tout − ( I Cr + I R i )ω Cr T f 2 = −( I Si + I S r )ω sr

ω Cr A22 ω Ti = A12 ( I Si + I S r ) A12 Cr + A22 (Tout − ( I Cr + I R i )ω Cr ) A22 ω ω Tin − [ A12 ( I Si + I S r ) A12 Cr + A22 (Tout − ( I Cr + I R i )ω Cr )] = ( I in + I Rr ) Cr A22 A22 ω sr = A12ω Rr = A12

2

Tin − A22Tout = [

I in + I Rr + A12 ( I Si + I S r ) − A22 ( I Cr + I R i )]ω Cr A22

Part 2

The load torque TL can be linearized as follow: TL = TL 0 + δT

ωv =ω v 0 +δω TL = TL 0 + 2Cω v 0 δω = TL 0 − 2Cω v 0 +2Cω v 0 ωv 2

The close loop transfer function can be solved as follow 1 2 [Ts − (TL 0 − 2Cω v 0 +2Cω v 0 ωv )] = ωv Iv s Ts (

1 ) = ωv I v s + 2Cω v 0

(Tt − Tt (

I t1 s 1 1 ωv )( )( ) = ωv d 21 Rd d 21 Rd I v s + 2Cω v 0

1 ) = ωv I t1 ( I v d 21 Rd + ) s + 2Cd 21 Rd ω v 0 d 21 Rd

With time constant τ : I v (d 21 Rd ) 2 + I t1 τ= 2C (d 21 Rd ) 2 ω v 0 It is obvious that the time constant will decrease with the increase of the tire angular velocity ω v 0 . Which physically means the vehicle speed will stablize faster at high speed than slower speed as the engine torque changes. Assuming all parameters are 1, the transfer function becomes: 1 Tt = ωv 2 s + 2ωv 0 Now let ω v 0 equals to 10, the time constant becomes τ = 0.1 . The following picture shows the frequency response of the system with time constant equals to 0.1 (convergence rate equals to 10).

Part 3

(a) To investigate the affection of clutch pleasure on shift quality, the pleasure of C2 clutch is modified from 1000KPa, as the blue line in the figures, to 500KPa, as the red line in the figures, to compare the differences.

The “flag” here is the index value indicating the shift status. 1 means the 1 st gear status, 1.2 means the torque phase of the 1st to 2nd shift, 1.7 means the inertia phase of the 1st to 2nd shift, and 2 means the 2nd gear status. All the following simulation results will be marked by the “flag” for the shifting phases. In the above picture it can be found that the lower clutch pressure caused a longer shifting time. It is obvious that with half of the original clutch pressure the shifting duration increased more than 5 times comparing to the original shifting duration. From this picture it can be concluded that to shorten the shifting time the clutch pleasure should be increased.

Time [msec]

This picture shows the comparison of speeds in the transmission system. “Wrr” is the reaction ring gear speed, “Wsi” is the input sun gear speed, and “Wt” is the turbine speed. In the above picture the two “Wrr” and the two “Wsi” are almost match together. And the “Wt” matches with “Wsi” before inertia phase and matches with “Wrr” after inertia phase. The “inertia phase”, or the speed adjustment phase”, is the main topic for this picture. According to the indication of the “flag”, it can be found that during the inertia phase the turbine speed “Wt” starts to decrease from the original speed, the same with the input sun gear speed “Wsi”, to match with the new speed “Wrr”. Since such synchronization is caused by the clutch C2, it is clear that a higher clutch pleasure will cause a faster matching speed as we can see in the above picture, which is one of the reasons that the smaller C2 pressure caused a longer shifting duration. It can be found that the duration of the inertia phase with smaller Pc2l pressure is about five times longer than the original case.

Time [msec]

This picture compares torques of some parts in the transmission system. “Ts” is the output shaft torque, “RTc2” is the reaction torque of the C2 clutch, and “RTsp1” is the reaction torque of C1 clutch. The comparison of torques can understand the phenomena during torque phase. The torque phase is in the range where “Flag*200 is equal to 240. During the shifting process from 1st gear to 2nd gear, the C1 clutch is released and the C2 is engaged. The torque phase is from C1 release start to C1 is completely released. As we can see in the figure, the smaller clutch pressure leaded to a longer torque phase duration which causes a longer power interruption to the output shaft “Ts”. However, with the slower shifting movement, we can also find that during the inertia phase, where “Flag*200” is 240, the torque variation is smoother than the original one, it is especially obvious when the inertia phase finishes. With such phenomena, it can be concluded that the smaller clutch pressure will lead to a slower shifting. However, the shifting process will be smoother.

Time [msec]

This picture can compare the smoothness more clearly. The “Wv-d” is the wheel acceleration which directly related to the vehicle acceleration. As we can see, the higher clutch pressure has shorter shifting duration. However, it also causes a violent jerk. On the other hand, the less clutch pressure, which has longer shifting duration, has a smoother shifting process. Thus it can be understood that the clutch pressure will affect the shifting smoothness and shifting duration, and the two factors should be compensated.

(b) The affection of friction coefficient to the shifting process is investigated in this case. With the same clutch pleasure Pc2l=1000KPa, two cases, the original friction coefficient which marked in blue and a reduced friction coefficient case which marked in red, are simulated and compared. Since the friction coefficient will directly related to clutch torque, it will be found that the simulation results of the reduced friction coefficient case are very close to the case with reduced clutch pressure.

Comparing the “Flag”, which indicates the shifting status, it can be found that the reduced friction coefficient case will have longer torque phase and longer inertia phase, and of course longer shifting duration. It is reasonable since the smaller friction coefficient causes smaller clutch torque which is the same result with a less clutch pressure case.

The tendency of this picture is almost the one in case (a). Since the friction coefficient is reduced, the clutch torque is reduced directly. By which the clutch need more time to synchronize the turbine speed and reaction ring gear speed. As a result a longer inertia phase duration is caused as can be compared by the green lines and the yellow lines.

The same with the inertia phase, since the clutch torque is reduced, the clutch C2 need more time to

engage the input career gear to be able to completely transmit the turbine torque. With such reason the C1 clutch will disengage slower. As we can see in the picture that the torque phase duration with the reduced friction coefficient is longer than the original case. However, it can also be observed that the smaller friction coefficient can also reduce the torque variation during transient state as shown in the final stage of the inertial phase.

The above picture shows a clear view of the final output of the transmission system in the shifting transient state. The reduced friction coefficient case leads to longer shifting duration, however, the jerk is also smaller. In conclusion, if the shifting speed is required, clutch pressure should be increased or clutch with higher friction coefficient should be chosen. On the other hand, if smoothness is required, clutch pressure should be reduced and use a clutch with smaller friction coefficient. But it should be noted that the smoothness can also be reached by control but shifting speed will be restricted by the clutch characteristics.

(c) In this case the affection of the engine torque to the shifting quality is discussed. To change the engine torque, the throttle angle us reduced from 90 degree ramp to 40 degree ramp.

The above picture shows the shifting condition of the two cases. It is reasonable that since the reduced throttle case has smaller engine torque it need more time to arrive at the shifting speed point (0.71 second slower than the original case). To observe the shifting process more precisely, all the reduced throttle cases in the following are shifted forward 0.71 sec to match the shifting start point with the original case.

After shifting the reduced throttle case forward 0.71 second, we can find that the smaller engine torque can lead to a faster shifting. Because clutch torques used to synchronize the gears are the same, smaller engine torque will be synchronized faster than the large torque.

The same condition is obvious in the inertia phase. Since the engine torque is reduced, the torque transmitted through the geartrain is also reduced. As a result, the clutch C2 needs less time to synchronize the speed. By which a short shifting duration is required. It can also be noticed that the reaction torque of the C2 clutch, RTc2 marked with red lines, can reach the target torque faster with smaller engine torque.

The acceleration curve shows and interesting phenomena. It can be found that the differences between high and low pick points of the two cases are almost the same with 7.5. However, in the original case the acceleration drops 3 rad/sec^2 when torque phase starts, but only 2 rad/sec^2 in the case with smaller engine torque. Which means the shifting process loses more power with larger engine torque case than the smaller engine torque case. Thus it can be concluded that the shifting process with smaller engine torque can increase shifting speed and decrease power loses. Of course it will be better to reduce engine torque to shift gear. However it should be noted that the torque reduction will also leads to the un-smoothness of vehicle acceleration.