Gelöste Aufgaben/JUMP/E-Motor and Drive-Train: Unterschied zwischen den Versionen

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System parameters for the motor are ''k<sub>T</sub>, k<sub>e</sub>'' and ''R<sub>M</sub>'' . For the stationary (<math>\frac{\displaystyle d \; I_M}{\displaystyle dt} = 0</math>) - condition, we derive
System parameters for the motor are ''k<sub>T</sub>, k<sub>e</sub>'' and ''R<sub>M</sub>'' . For the stationary (<math>\frac{\displaystyle d \; I_M}{\displaystyle dt} = 0</math>) - condition, we derive


::<math>M_M=\frac{{U_M}\, {k_T}-{k_T}\, {k_e} \tilde{\omega}_M }{{R_M}}</math>,
::<math>M_M=\frac{{U_M}\, {k_T}-{k_T}\, {k_e} \tilde{\omega}_M }{{R_M}}</math>.


so that we see:
For our application, it will be easier to replace ''k<sub>e</sub>'', ''k<sub>R</sub>'' and ''R'' by mechanical properties as


<ol>
<ol>
<li>under a "no-load"-condition ''M<sub>M</sub>=0'', we have  
<li>under a "no-load"-condition ''M<sub>M</sub>=0'', we have  


::<math>\tilde{\omega}_{NL} = \frac{U_M}{k_e}</math></li>
::<math>k_e= \frac{U_M}{\tilde{\omega}_{NL}}</math></li>
<li>under the "stalled motor"-condition <math>\tilde{\omega}_{St}=0</math>, we have
<li>under the "stalled motor"-condition <math>\tilde{\omega}_{St}=0</math>, we have


::<math>M_{St} = k_T \cdot I_{St}</math> and <math>R = \frac{\displaystyle U_M}{\displaystyle I_{St}}</math>.</li>
::<math>k_T = \frac{\displaystyle M_{St}}{\displaystyle I_{St}}</math> and <math>R = \frac{\displaystyle U_M}{\displaystyle I_{St}}</math>.</li>
</ol>
</ol>


Thus - for ''U<sub>M</sub> = 12 V'' - we can import measured performance parameters and then replace ''k<sub>T</sub>, k<sub>e</sub>'', ''R<sub>M</sub>'' by <math>\tilde{\omega}_{NL}</math>, ''I<sub>St</sub>'' and ''M<sub>St</sub>'', which make the system parameters to be selected more descriptive.
Thus for ''U<sub>M</sub> = 12 V'' we can import measured performance parameters and then replace ''k<sub>T</sub>, k<sub>e</sub>'' and ''R<sub>M</sub>'' by <math>\tilde{\omega}_{NL}</math>  
''I<sub>St</sub>'' and ''M<sub>St</sub>''.


 
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<table style="border-collapse: collapse; border: 1px solid gray;">
!style="border-right: solid 0px;"|name
<tr style="border: none; background-color:gray"><th>name</th><th>symbol</th></tr>
<tr style="border: none;"><td style="border-right: solid 1px gray; border-left: solid 1px #555;">name</td><td style="border-right: solid 1px gray; border-left: solid 1px #555;">symbol</td></tr>
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{| class="wikitable"
!name
!symbol
!symbol
!value
!value
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|stall-current
|stall-current
|''I<sub>St</sub>''
|''I<sub>St</sub>''
|0.25
|8.6
|A
|A
|-
|-
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|1
|1
|}
|}
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[[Gelöste Aufgaben/JUMP|Back to Start →]]



Aktuelle Version vom 24. März 2021, 15:54 Uhr

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Scope

Diagram of drive-train components.

The Drive-Train consists of a DC/DC-converter, a DC Motor and a gear-box.

  • DC/DC-converter: is supplied with the battery voltage UB, the output voltage is controlled by the driver via setpoint “p“.
  • motor: is a standard DC brushed motor, the manufacturer provides only few information on its characteristics - we’ll need to improvise.
  • gearbox: has a gear ratio of ratio of nG=100, its shaft rotates at speed ωW and delivers a torque MW to the front wheels.

The task is: provide a mathematical model for the drive train that accounts for load-alterations imposed by the driver. And we assume losses in the two converters - DC/DC and gearbox - to be negligible.

Structure

Block diagram

The drive train receives a "gas"-pedal position "p" from the driver and a battery-voltage UB.

It delivers a torque MW on the wheel and creates an electric current IM through the motor.

Drive-train components.

The sub-model consists of  DC/DC-converter, Motor and gear-box:

DC/DC Converter

Losses in the DC/DC converter shall be small - so for input port “1“ and output port “2“ we obtain

UBIB=UMIM .

Let the “gas”-pedal-indicator “p“ control

UM=pUB0p1.

with

0p1 and U1=UB

Motor

Brushed-DC motor

We use a common electric circuit representation for a series wound motor, the field coils are connected electrically in series with the armature coils, resistance R sums up all electrical losses in the motor.

Gearbox

Losses in the gearbox shall be small - so for input (ωM, MM) and output (ωW, MW) we obtain the fixed relation

ωMMM=ωWMW.

Fortunately, this relationship is already "hardwired" in the Principle of Virtual Work. And we have only one differential equation for the electrical components:

dIBdt=ULL,

the remaining equations are algebraic.

Model

Electrical Components

For the motor, we find with Kirchhoff's law that

UM=UR+UL+e

with UR, UL being the differential voltage over resistance R and inductance L respectively. “e” is the back electromagnetic force with

e=keω~M

and the electromotive force constant ke. Note the ω~M is the differential rotational velocity between rotor and stator, i.e.

ω~M=ψ˙M(t)ϕ˙(t).

Employing

UR=RIM,UL=LdIMdt

and using

MM=kTIM

with the armature constant kT, we have the complete set of equations.

From the above, we find

LddtIM(t)=UB(t)p(t)RIM(t)e

and additionally the algebraic equations

IB=IMp(t),UR=RIM(t),UL=UBp(t)RIM(t)e,UM=UBp(t).

/*******************************************************/
/* MAXIMA script                                       */
/* version: wxMaxima 16.04.2                           */
/* author: Andreas Baumgart                            */
/* last updated: 2021-02-08                            */
/* ref: Modelling and Simulation (TUAS)                */
/* description: virtual work of drive train electrics  */
/*******************************************************/

/*******************************************************/
/* declarations                                        */
/*******************************************************/
declare("φ", alphabetic);
declare("ψ", alphabetic);
declare("ω", alphabetic);

/*******************************************************/
/* kinematics                                          */
/*******************************************************/
kirchhoff : [U[M] = p(t)*U[B],
	     U[B]*I[B] = U[M]*I[M],
	     U[M] = U[R] + U[L] + e,
             e = k[e]*omega[M],
	     ω[M] = diff(ψ[M](t),t)+diff(φ(t),t),
	     U[R] = R*I[M](t),
	     U[L] = L * diff(I[M](t),t),
	     M[M] = k[t]*I[M](t)];

solve(kirchhoff[7], [diff(I[M](t),t)]);





Mechanical Components

The Virtual Work of d'Alembert forces, motor torque MM and wheel torque MW is

δW=JMψ¨M(t)δψM+MM(ψM+ϕ(t))MWδψW=0.

Since the gearbox is built into the car, the wheel-side relative gear-box-angle is

ψ~W=ψW(t)+ϕ(t)

and on the motor-side

ψ~M=ψM(t)ϕ(t).

With the gear transmission ratio

nG=ψ~Mψ~W

and

q_E=(ϕ(t)ψW(t))

we find

δW=δq_E(M__Eq¨_E+(nGMM+nGMMMW))

with

M__E=((nG2+2nG+1)JM(nG2+nG)JM(nG2+nG)JMnG2JM).

We have thus "returned" all state variables to the Car-Body-submodel.


/*******************************************************/
/* MAXIMA script                                       */
/* version: wxMaxima 16.04.2                           */
/* author: Andreas Baumgart                            */
/* last updated: 2021-02-08                            */
/* ref: Modelling and Simulation (TUAS)                */
/* description: virtual work of drive train mechanics  */
/*******************************************************/

/*******************************************************/
/* declarations                                        */
/*******************************************************/
declare("δ", alphabetic);
declare("φ", alphabetic);
declare("ψ", alphabetic);

/*******************************************************/
/* kinematics                                          */
/*******************************************************/
kin : [ψ[M,r] = n[G]*ψ[W,r],
       ψ[W,r] = ψ[W](t) - φ(t),
       ψ[M,r] = ψ[M](t) + φ(t)];
Q : [ψ[M](t),ψ[M,r],ψ[W,r]];
sol: ratsimp(solve(kin,Q))[1];


/* Principle of Virtual Work */
PVW: δW = - J[M]*diff(ψ[M](t),t,2)*δψ[M] + M[M]*(δψ[M]+δφ) - M[W]*δψ[W];

/* minimal coordinates */
q : [ φ(t), ψ[W](t)];
δq: [δφ   ,δψ[W]   ];

varia: [ψ[M](t)=δψ[M],ψ[M,r]=δψ[M,r],ψ[W,r]=δψ[W,r],
        ψ[W](t)=δψ[W], φ(t) =δφ];
sol : expand(append(sol, subst(varia,sol)));

PVW : expand(subst(sol,PVW));
PVW : ev(PVW,nouns);

eom : makelist(coeff(rhs(PVW),δq[i]),i,1,2);
eom : ratsimp(eom);

MM : funmake('matrix,makelist(makelist(ratsimp(coeff(-eom[i],'diff(q[j],t,2))),j,1,2),i,1,2));

rest : expand(eom + makelist(sum(MM[i,j]*diff(q[j],t,2),j,1,2),i,1,2));

print(δq, "*(-",MM,"*",transpose(diff(q,t,2)),"+",transpose(rest),") = 0")$




Consequently, the drive train has only one state-variable: IM:

q_E(t)=(IM(t))

where

ddtIM(t)=UBp(t)RIM(t)eL

and with this algebraic equation

e=ke(ψ˙M(t)+ϕ˙(t))=ω~M=nGω~W.

Keep in mind that ω~M and ω~W are the rotational velocities of the gearbox-shafts relative to the car-body, thus

ω~W=dψ~dt=(dψdt=ωdϕdt).

Variables

name symbol unit
motor current IM A

Parameter

System parameters for the motor are kT, ke and RM . For the stationary (dIMdt=0) - condition, we derive

MM=UMkTkTkeω~MRM.

For our application, it will be easier to replace ke, kR and R by mechanical properties as

  1. under a "no-load"-condition MM=0, we have
    ke=UMω~NL
  2. under the "stalled motor"-condition ω~St=0, we have
    kT=MStISt and R=UMISt.

Thus for UM = 12 V we can import measured performance parameters and then replace kT, ke and RM by ω~NL ISt and MSt.

name symbol value unit
nominal voltage Uref 12 V
no-load rotational speed ω~NL 14800 rpm
stall-current ISt 8.6 A
stall-torque MSt 0.042 N m
inductance L 0.05 H
gear tramission ratio nG 25 1

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References

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