DC Motor Braking & Dynamic Response — Solved Problems

Step 1: Denominator coefficients

\[JL_a = 0.0167 \times 0.003 = 5.01 \times 10^{-5}\] \[JR_a + B_l L_a = (0.0167 \times 0.5) + (0.01 \times 0.003)\] \[= 0.00835 + 0.00003 = 0.00838\] \[B_l R_a + K_b^2 = (0.01 \times 0.5) + 0.8^2 = 0.005 + 0.640 = 0.645\]

Step 2: Transfer function

Divide all coefficients by \(JL_a = 5.01\times10^{-5}\):

\[\frac{0.00838}{5.01\times10^{-5}} = 167.3, \qquad \frac{0.645}{5.01\times10^{-5}} = 12{,}874\] \[\frac{K_b}{JL_a} = \frac{0.8}{5.01\times10^{-5}} = 15{,}968\] \[\boxed{\frac{\Omega(s)}{V_a(s)} = \frac{15{,}968}{s^2 + 167.3\,s + 12{,}874}}\]

Step 3: Pole analysis

\[\omega_n = \sqrt{12{,}874} = 113.5\,\text{rad/s}\] \[\zeta = \frac{167.3}{2 \times 113.5} = 0.737 \qquad \text{(underdamped, } \zeta < 1\text{)}\] \[\sigma = \zeta\,\omega_n = 0.737 \times 113.5 = 83.6\,\text{rad/s}\] \[\omega_d = \sqrt{\omega_n^2 - \sigma^2} = \sqrt{12{,}874 - 6{,}989} = \sqrt{5{,}885} = 76.7\,\text{rad/s}\]

Step 4: Time-domain response (unit-step \(V_a = 220\,\text{V}\))

\[\varphi = \arctan\!\left(\frac{\omega_d}{\sigma}\right) = \arctan\!\left(\frac{76.7}{83.6}\right) = 0.742\,\text{rad}\] \[\boxed{\omega(t) = 272.9\!\left[1 - 1.48\,e^{-83.6t}\sin(76.7t + 0.742)\right]\,\text{rad/s}}\]

Step 5: Time to reach 100 rad/s

Solving \(\omega(t^*) = 100\,\text{rad/s}\) numerically (first crossing):

\[\boxed{t^* \approx 10\,\text{ms}}\]

Step 1: Motor constant \(K_b\phi\)

\[E_r = V_r - I_{ar} R_a = 230 - 100\times0.1 = 220\,\text{V}\] \[\omega_r = \frac{500\times2\pi}{60} = 52.36\,\text{rad/s}\] \[K_b\phi = \frac{E_r}{\omega_r} = \frac{220}{52.36} = 4.202\,\text{V·s/rad}\]

Step 2: Braking armature current (reversed)

The overhauling load drives the motor; to hold it, the motor produces braking torque equal to \(T_L\):

\[I_a = \frac{T_L}{K_b\phi} = \frac{800}{4.202} = 190.4\,\text{A}\quad\text{(reversed)}\]

Step 3: Back-EMF during regeneration

With reversed current, KVL gives \(E_b = V_a + I_a R_a\):

\[E_b = 230 + 190.4 \times 0.1 = 230 + 19.04 = 249.0\,\text{V}\]

Step 4: Regenerative speed

\[\omega_m = \frac{E_b}{K_b\phi} = \frac{249.0}{4.202} = 59.3\,\text{rad/s}\] \[\boxed{N = \frac{59.3\times60}{2\pi} \approx 566\,\text{rpm}}\]

Step 1: Braking armature current

Constant flux \(\Rightarrow T \propto I_a\):

\[I_{a2} = 0.8 \times I_{ar} = 0.8 \times 200 = 160\,\text{A}\]

Step 2: Back-EMF at rated speed (800 rpm)

\[E_1 = V_r - I_{ar}\,R_a = 220 - 200\times0.06 = 208\,\text{V}\]

Step 3: Back-EMF at 600 rpm

Constant flux \(\Rightarrow E_b \propto N\):

\[E_2 = \frac{600}{800} \times E_1 = 0.75 \times 208 = 156\,\text{V}\]

Step 4: Source internal voltage

In regenerative braking, armature current (reversed) flows into the source. KVL through source:

\[E_2 = V_{source} + I_{a2}(R_a + R_s)\] \[V_{source} = E_2 - I_{a2}(R_a + R_s)\] \[= 156 - 160 \times (0.06 + 0.04) = 156 - 16\] \[\boxed{V_{source} = 140\,\text{V}}\]

Step 1: Angular speeds

\[\omega_{600} = \frac{2\pi\times600}{60} = 62.83\,\text{rad/s}\] \[\omega_{500} = \frac{2\pi\times500}{60} = 52.36\,\text{rad/s}\]

Step 2: Compute torque at magnetization curve points (at 600 rpm)

\[K\phi_{50} = \frac{E_{50}}{\omega_{600}} = \frac{437}{62.83} = 6.955,\quad T_{50} = 6.955\times50 = 347.8\,\text{N·m}\] \[K\phi_{60} = \frac{E_{60}}{\omega_{600}} = \frac{485}{62.83} = 7.719,\quad T_{60} = 7.719\times60 = 463.1\,\text{N·m}\]

Step 3: Interpolate for \(T = 400\,\text{N·m}\)

\[\text{fraction} = \frac{400 - 347.8}{463.1 - 347.8} = \frac{52.2}{115.3} = 0.453\] \[I_a = 50 + 10\times0.453 = 54.5\,\text{A}\] \[K\phi = 6.955 + 0.453\times(7.719 - 6.955) = 6.955 + 0.346 = 7.301\,\text{V·s/rad}\]

Step 4: Back-EMF at 500 rpm

Scale \(K\phi\) (which is fixed by the current) to the new speed:

\[E_{500} = K\phi \times \omega_{500} = 7.301 \times 52.36 = 382.4\,\text{V}\]

Step 5: External braking resistance

In dynamic braking: \(E_{500} = I_a(R_m + R_{ext})\)

\[R_{ext} = \frac{E_{500}}{I_a} - R_m = \frac{382.4}{54.5} - 1 = 7.017 - 1\] \[\boxed{R_{ext} \approx 6.0\,\Omega}\]

Step 1: Braking current

Series motor, linear mag. circuit: \(T_e \propto I_a^2\).

\[\frac{T_2}{T_1} = \frac{I_{a2}^2}{I_{a1}^2} = 2 \quad\Rightarrow\quad I_{a2} = I_{a1}\sqrt{2} = 100\sqrt{2}\] \[\boxed{I_{a2} = 141.4\,\text{A}}\]

Step 2: Back-EMF at rated conditions (1000 rpm)

\[E_1 = V - I_{a1}\,R = 220 - 100\times0.1 = 210\,\text{V}\]

Step 3: Back-EMF at 800 rpm under braking

Series motor (linear): \(E_b = M_{af}I_a\omega_m \Rightarrow E \propto I_a \cdot N\):

\[\frac{E_2}{E_1} = \frac{I_{a2}}{I_{a1}}\times\frac{N_2}{N_1} = \frac{141.4}{100}\times\frac{800}{1000} = 1.131\] \[E_2 = 210 \times 1.131 = 237.6\,\text{V}\]

Step 4: Braking resistance

In dynamic braking: \(E_2 = I_{a2}(R_B + R)\)

\[R_B = \frac{E_2}{I_{a2}} - R = \frac{237.6}{141.4} - 0.1 = 1.680 - 0.1\] \[\boxed{R_B \approx 1.58\,\Omega}\]

Step 1: Back-EMF at rated and initial speeds

\[E_{970} = V_a - I_{ar}\,R_a = 220 - 100\times0.05 = 215\,\text{V}\]

Scale to 1000 rpm (constant flux \(\Rightarrow E_b \propto N\)):

\[E_{1000} = \frac{1000}{970}\times215 = 221.6\,\text{V}\]

Step 2: External resistance for \(I_{plug} = 2\times100 = 200\,\text{A}\)

In plugging, supply and back-EMF are additive:

\[R_{total} = \frac{V_a + E_{1000}}{I_{plug}} = \frac{220 + 221.6}{200} = \frac{441.6}{200} = 2.208\,\Omega\] \[\boxed{R_B = R_{total} - R_a = 2.208 - 0.05 = 2.16\,\Omega}\]

Step 3: Motor constant \(K_b\phi\)

\[\omega_{1000} = \frac{1000\times2\pi}{60} = 104.72\,\text{rad/s}\] \[K_b\phi = \frac{E_{1000}}{\omega_{1000}} = \frac{221.6}{104.72} = 2.116\,\text{V·s/rad}\]

Step 4: Braking torque at 1000 rpm

\[T_{1000} = K_b\phi \times I_{plug} = 2.116 \times 200\] \[\boxed{T_{1000} = 423.2\,\text{N·m}}\]

Step 5: Torque at zero speed (\(E_b = 0\))

At \(\omega = 0\), only \(V_a\) drives the current:

\[I_{a0} = \frac{V_a}{R_{total}} = \frac{220}{2.208} = 99.6\,\text{A}\] \[\boxed{T_0 = K_b\phi \times I_{a0} = 2.116 \times 99.6 = 210.7\,\text{N·m}}\]