Question 11.6: Consider the 100-hp dc motor of Examples 11.3 and 11.4 to be......

a. Neglecting any rotational losses, the armature current can be found from Eq. 11.4 by setting T_{\text {mech }}=T_{\text {load }}

T_{\mathrm{mech}}=\frac{E_aI_a}{ω_m}=K_{\mathrm{f}}I_{\mathrm{f}}I_a \quad \quad \quad(11.4)

I_{\mathrm{a}}=\frac{T_{\text {load }}}{K_{\mathrm{f}} I_{\mathrm{f}}}

Substituting

T_{\text {load }}=\left(\frac{n}{n_{\mathrm{f}}}\right) T_{\mathrm{fl}}

where n is the motor speed in \mathrm{r} / \mathrm{min}, n_{\mathrm{f}}=2500  \mathrm{r} / \mathrm{min} and T_{\mathrm{fl}}=285 \mathrm{~N} \cdot \mathrm{m} gives

I_{\mathrm{a}}=\frac{n T_{\mathrm{fl}}}{n_{\mathrm{f}} K_{\mathrm{f}} I_{\mathrm{f}}}

Solving for V_{\mathrm{a}}=E_{\mathrm{a}}+I_{\mathrm{a}} R_{\mathrm{a}} then allows us to complete the following table:

b. The dynamic equation governing the speed of the motor is

J \frac{d \omega_{\mathrm{m}}}{d t}=T_{\text {mech }}-T_{\text {load }}

Substituting \omega_{\mathrm{m}}=(\pi / 30) n and \omega_{r}=(\pi / 30) n_{\mathrm{f}} we can write

T_{\text {load }}=\left(\frac{T_{\mathrm{fl}}}{\omega_{\mathrm{f}}}\right) \omega_{\mathrm{m}}

Under armature-voltage control,

\begin{aligned}T_{\text {mech }} & =K_{\mathrm{f}} I_{\mathrm{f}} I_{\mathrm{a}}=K_{\mathrm{f}} I_{\mathrm{f}}\left(\frac{V_{\mathrm{a}}-E_{\mathrm{a}}}{R_{\mathrm{a}}}\right)\\ \\& =K_{\mathrm{f}} I_{\mathrm{f}}\left(\frac{V_{\mathrm{a}}-K_{\mathrm{f}} I_{\mathrm{f}} \omega_{\mathrm{m}}}{R_{\mathrm{a}}}\right)\end{aligned}

and thus the governing differential equation is

J \frac{d \omega_{\mathrm{m}}}{d t}=K_{\mathrm{f}} I_{\mathrm{f}}\left(\frac{V_{\mathrm{a}}-K_{\mathrm{f}} I_{\mathrm{f}} \omega_{\mathrm{m}}}{R_{\mathrm{a}}}\right)-\left(\frac{T_{\mathrm{fl}}}{\omega_{\mathrm{f}}}\right) \omega_{\mathrm{m}}

or

\begin{array}{c}\frac{d \omega_{\mathrm{m}}}{d t}+\frac{1}{J}\left(\frac{T_{\mathrm{fl}}}{\omega_{\mathrm{f}}}+\frac{\left(K_{\mathrm{f}} I_{\mathrm{f}}\right)^{2}}{R_{\mathrm{a}}}\right) \omega_{\mathrm{m}}-\frac{K_{\mathrm{f}} I_{\mathrm{f}} V_{\mathrm{a}}}{J R_{\mathrm{a}}} \\ \\=\frac{d \omega_{\mathrm{m}}}{d t}+48.4  \omega_{\mathrm{m}}-24.7 V_{\mathrm{a}}=0\end{array}

From this differential equation, we see that with the motor initially at \omega_{\mathrm{m}}=\omega_{\mathrm{i}}= 209  rad/sec, if the armature voltage V_a is suddenly switched from V_i = 413  V  \text{to}  V_f = 513  V, the speed will rise exponentially to \omega_{\mathrm{m}}=\omega_{\mathrm{f}}=262  \mathrm{rad} / \mathrm{sec} as

\begin{aligned}\omega_{\mathrm{m}} & =\omega_{\mathrm{f}}+\left(\omega_{\mathrm{i}}-\omega_{\mathrm{f}}\right) e^{-t / \tau} \\& =262-53 e^{-t / \tau}  \mathrm{rad} / \mathrm{sec}\end{aligned}

where \tau=1 / 48.4=20.7  \mathrm{msec}. Expressed in terms of r/min

n=2500-50 e^{-t / \tau}  \mathrm{r} / \mathrm{min}

The armature current will decrease exponentially with the same 20.7 msec time constant from an initial value of \left(V_{\mathrm{f}}-V_{\mathrm{i}}\right) / R_{\mathrm{a}}=1190 \mathrm{~A} to its final value of 149 A.

Thus,

I_{\mathrm{a}}=149+1041 e^{-t / \tau} \mathrm{A}

Notice that it is unlikely that the supply to the dc motor can supply this large initial current (eight times the rated full-load armature current) and, in addition, the high current and corresponding high torque could potentially cause damage to the dc motor commutator, brushes, and armature winding. Hence, as a practical matter, a practical controller would undoubtedly limit the rate of change of the armature voltage to avoid such sudden steps in voltage, with the result that the speed change would not occur as rapidly as calculated here.

c. The dynamic equation governing the speed of the motor remains the same as that in part (b) as does the equation for the load torque. However, in this case, because the motor is being operated from a current controller, the electromagnetic torque will remain constant at T_{\text {mech }}=T_{\mathrm{f}}=285 \mathrm{~N} \cdot \mathrm{m} after the current is switched from its initial value of 119 A to its final value of 149 A.

Thus

J \frac{d \omega_{\mathrm{m}}}{d t}=T_{\text {mech }}-T_{\text {load }}=T_{\mathrm{f}}-\left(\frac{T_{\mathrm{f}}}{\omega_{\mathrm{f}}}\right) \omega_{\mathrm{m}}

or

\begin{array}{l}\frac{d \omega_{\mathrm{m}}}{d t}+\left(\frac{T_{\mathrm{f} 1}}{J \omega_{\mathrm{f}}}\right) \omega_{\mathrm{m}}-\frac{T_{\mathrm{f}}}{J} \\ \\\quad=\frac{d \omega_{\mathrm{m}}}{d t}+1.18 \omega_{\mathrm{m}}-310=0\end{array}

In this case, the speed will rise exponentially to \omega_{\mathrm{m}}=\omega_{\mathrm{f}}=262  \mathrm{rad} / \mathrm{sec} as

\begin{aligned}\omega_{\mathrm{m}} &=\omega_{\mathrm{f}}+\left(\omega_{\mathrm{i}}-\omega_{\mathrm{f}}\right) e^{-t / \tau} \\& =262-53 e^{-t / \tau}  \mathrm{rad} / \mathrm{sec}\end{aligned}

where now the time constant \tau=1 / 1.18=845  \mathrm{msec}.

Clearly, the change in motor speed under the current controller is much slower. However, at no point during this transient do either the motor current or the motor torque exceed their rated value. In addition, should faster response be desired, the armature current (and hence motor torque) could be set temporarily to a fixed value higher than the rated value (e.g., two or three times rated as compared to the factor of 8 found in part (b)), thus limiting the potential for damage to the motor.