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

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and the electromotive force constant ''k<sub>e</sub>''. Note the ''ω<sub>M</sub>'' is the differential rotational velocity between rotor and stator, i.e.  
and the electromotive force constant ''k<sub>e</sub>''. Note the ''ω<sub>M</sub>'' is the differential rotational velocity between rotor and stator, i.e.  


::<math>\omega_M = \dot{\psi}_W(t) + \dot{\phi}(t).</math>
::<math>\omega_M = \dot{\psi}_M(t) + \dot{\phi}(t).</math>


Employing
Employing

Version vom 16. März 2021, 14:30 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

.

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

with

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

.

And we have only one differential equation for the electrical components:

,

the remaining equations are algebraic.

Model

Electrical Components

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

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

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

Employing

and using

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

From the above, we find

and additionally the algebraic equations

.

/*******************************************************/
/* 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

.

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

and on the motor-side

.

With the gear transmission ratio

and

we find

with

.

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:

where

and with this algebraic equation

.

Variables

name symbol unit
motor current IM A

Parameter

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

,

so that we see:

  1. under a "no-load"-condition MM=0, we have
  1. under the "stalled motor"-condition ωSt=0, we have

Thus - for UM = 8 V - we can replace kT / ke, RM by ωNL and MSt, which make the system parameters to be selected more descriptive.

name symbol value unit
no-load rotational speed ωNL 1033 rad/s
stall-torque MSt 0.042 N m
inductance L 0.05 H
gear tramission ratio nG 7 1

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References

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