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

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Version vom 10. März 2021, 14:28 Uhr

Scope

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 U_B, 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 n_G=100, its shaft rotates at speed ω_W and delivers a torque M_W 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

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

It delivers a torque M_W on the wheel and creates an electric current I_M 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

U_{1} \cdot I_{1} = U_{2} \cdot I_{2}

.

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

U_{2} = p \cdot U_{1}; 0 \leq p \leq 1 .

with

0 \leq p \leq 1 and U_{1} = U_{B}

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 (ω_M, M_M) and output (ω_W, M_W) we obtain the fixed relation

ω_{M} \cdot M_{M} = ω_{W} \cdot M_{W}

.

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

\frac{d I_{B}}{d t} = \frac{U_{L}}{L}

,

the remaining equations are algebraic.

Model

Electrical Components

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

U_{M} = U_{R} + U_{L} + e

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

e = k_{e} \cdot ω_{M}

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

ω_{M} = {\dot{ψ}}_{W} (t) + \dot{ϕ} (t) .

Employing

U_{R} = R \cdot I_{M}, U_{L} = L \cdot \frac{d I_{M}}{d t}

and using

M_{M} = k_{t} \cdot I_{M}

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

From the above, we find

L \cdot \frac{d}{d t} I_{M} (t) = U_{B} (t) \cdot p (t) - R I_{M} (t) - e

and additionally the algebraic equations

\begin{array}{ll} I_{B} & = I_{M} \cdot p (t), \\ U_{R} & = R \cdot I_{M} (t), \\ U_{L} & = U_{B} \cdot p (t) - R \cdot I_{M} (t) - e, \\ U_{M} & = U_{B} \cdot p (t) \end{array}

.

tmp

Mechanical Components

Text

Variables

Parameter

>

Back to Start →

References

...

@@ Zeile 22: / Zeile 22: @@
 Losses in the DC/DC converter shall be small - so for input port “1“ and output port “2“ we obtain
-<math>U_1 \cdot I_1 = U_2 \cdot I_2</math> .
+::<math>U_1 \cdot I_1 = U_2 \cdot I_2</math> .
 Let the “gas”-pedal-indicator “''p''“ control
-<math>U_2 = p \cdot U_1 \text{; } 0 \le p \le 1.</math>
+::<math>U_2 = p \cdot U_1 \text{; } 0 \le p \le 1.</math>
 with
-<math>0 \le p \le 1 \text{ and } U_1 = U_B</math>
+::<math>0 \le p \le 1 \text{ and } U_1 = U_B</math>
 ===Motor===
@@ Zeile 36: / Zeile 36: @@
 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 (''ω<sub>M</sub>'', ''M<sub>M</sub>'') and output (''ω<sub>W</sub>'', ''M<sub>W</sub>'') we obtain the fixed relation
-<math>\omega_M \cdot M_M = \omega_W \cdot M_W</math>.
+::<math>\omega_M \cdot M_M = \omega_W \cdot M_W</math>.
 And we have only one differential equation for the electrical components:
-<math>\frac{\displaystyle d I_B}{\displaystyle d t} = \frac{\displaystyle U_L}{\displaystyle L}</math>,
+::<math>\frac{\displaystyle d I_B}{\displaystyle d t} = \frac{\displaystyle U_L}{\displaystyle L}</math>,
 the remaining equations are algebraic.
 ==Model==
-<!------------------------------------------------------------------------->
-For the motor, we find with Kirchhoff's law that
+{{MyCodeBlock
+|title=Electrical Components
+|text=For the motor, we find with Kirchhoff's law that
-<math>U_M = U_R + U_L + e</math>
+::<math>U_M = U_R + U_L + e</math>
 with ''U<sub>R</sub>, U<sub>L</sub>'' being the differential voltage over resistance ''R'' and inductance ''L'' respectively. “''e''” is the back electromagnetic force with
-<math>e = k_e\cdot \omega_M</math>
+::<math>e = k_e\cdot \omega_M</math>
 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}_W(t) + \dot{\phi}(t).</math>
 Employing
-<math>U_R = R \cdot I_M, U_L = L\cdot \frac{\displaystyle d I_M}{\displaystyle dt}</math>
+::<math>U_R = R \cdot I_M, U_L = L\cdot \frac{\displaystyle d I_M}{\displaystyle dt}</math>
 and using
-<math>M_M = k_t \cdot I_M</math>
+::<math>M_M = k_t \cdot I_M</math>
 with the armature constant ''k<sub>t</sub>'', we have the complete set of equations.
@@ Zeile 74: / Zeile 74: @@
 From the above, we find
-<math>L \cdot \frac{\displaystyle d}{\displaystyle d t} {I_M}(t)= U_B (t) \cdot p(t) -R \, {I_M}(t) - e</math>
+::<math>L \cdot \frac{\displaystyle d}{\displaystyle d t} {I_M}(t)= U_B (t) \cdot p(t) -R \, {I_M}(t) - e</math>
 and additionally the algebraic equations
-<math>\begin{array}{ll}
+::<math>\begin{array}{ll}
 {I_B}&={I_M} \cdot \operatorname{p}(t),\\
 {U_R}&=R \cdot {I_M}(t),\\
@@ Zeile 84: / Zeile 84: @@
 {U_M}&={U_B} \cdot \operatorname{p}(t)
 \end{array}</math>.
+|code=
+<syntaxhighlight lang="lisp" line start=1>
+/*******************************************************/
+/* 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);
-==tmp==
+/*******************************************************/
+/* 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)];
-{{MyCodeBlock
+solve(kirchhoff[7], [diff(I[M](t),t)]);
-|title=Electrical Components
-|text=Text
-|code=
-<syntaxhighlight lang="lisp" line start=1>
-+1=2
 </syntaxhighlight>
 }}

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

Version vom 10. März 2021, 14:28 Uhr

Inhaltsverzeichnis

Scope

Structure

DC/DC Converter

Motor

Gearbox

Model

Electrical Components

tmp

Mechanical Components

Variables

Parameter

Navigationsmenü

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

Version vom 10. März 2021, 14:28 Uhr

Scope

Structure

DC/DC Converter

Motor

Gearbox

Model

Electrical Components

tmp

Mechanical Components

Variables

Parameter

Navigationsmenü

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