STEM Problem Detail

Problem #39

Electrical Engineering Hard

Problem Statement

Prompt

Give me the answer to question 3.46 step by step.

Solution Rendering...

**Solution:**
To evaluate hysteresis and eddy current losses, Swinburne's test is performed on a shunt D.C. generator. The machine is run as a D.C. shunt motor at no-load at rated speed. Field current is adjusted to a rated value. The power absorbed by the motor during test is equal to no-load losses which include hysteresis, eddy current and windage losses.

$$R_f = 110\ \Omega, \quad R_a = 0.25\ \Omega$$

$$I_f = \frac{250}{110} = 2.27\ \text{A}$$

$$P_{\text{out}} = 10\ \text{kW}$$

$$I_a = 50 - I_f = 47.73\ \text{A}$$

$$E_a = V_t - I_a \times R_a = 250 - 47.73 \times 0.25 = 238.068\ \text{V}$$

Since,

$$P_{\text{rotational}} = E_{a_1} \times I_{a_1} - P_{\text{out}} = 238.068 \times 47.73 - 10000 = 1362.37\ \text{W} \quad \text{(treated as constant)}$$

Now, when $N = 1200\ \text{rpm}$, $T_{\text{shaft}} = \text{constant}$ (as $T_{\text{load}} = \text{constant}$).

Since load torque is constant,

$$P_{\text{out}} = T_{\text{load}} \times \omega_m \implies P_{\text{out}} \propto \omega_m$$

$$\frac{P_{\text{out}(2)}}{P_{\text{out}(1)}} = \frac{N_2}{N_1} \implies P_{\text{out}(2)} = 10000 \times \frac{1200}{1500} = 8000\ \text{W}$$

Now,

$$\text{Power at armature} = P_{\text{out}} + P_{\text{rotational}} = 8000 + 1362.37 = 9362.37\ \text{W}$$

Since $E_a = K\phi\omega \implies E_a \propto N$,

$$\frac{E_{a_2}}{E_{a_1}} = \frac{N_2}{N_1} \implies \frac{E_{a_2}}{238.062} = \frac{1200}{1500} = 0.8$$

$$E_{a_2} = 0.8 \times 238.062 = 190.454\ \text{V}$$

$$P_{\text{arm}} = E_{a_2} \times I_{a_2} = 9362.37 \implies I_{a_2} = \frac{9362.37}{190.454} = 49.16\ \text{A}$$

$$E_{a_2} = V_t - I_{a_2}(R_a + R_{\text{ext}})$$

$$190.454 = 250 - 49.16\,(0.25 + R_{\text{ext}})$$

Solving: $R_{\text{ext}} = 0.96135\ \Omega$

Power lost in armature circuit:

$$= (R_a + R_{\text{ext}}) \times I_{a_2}^2 = 1.211 \times (49.16)^2 = 2926.63\ \text{W} = 2.9266\ \text{kW}$$

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**Solution (c):**

Load torque $T_L \propto n^2$. Electromagnetic torque $T_e = k_a \phi I_a$, with $\phi = \text{constant}$.

Under steady state, $T_e = T_L$, so

$$I_a \propto n^2 \tag{i}$$

$$E = V_t - I_a r_a = k_a \phi n \tag{ii}$$

From (i) and (ii), the armature current would be approximately one fourth of the normal value — option (c).

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**Solution:**

(a) Since the generated voltage (230 V) is less than the terminal voltage (240 V), the machine works as a motor.

(b)
$$V_t = E_a + I_a r_a \implies 240 = 230 + I_a r_a \implies 10 = 40 \times r_a \implies r_a = 0.25\ \Omega$$

$$T_e = \frac{E_a \cdot I_a}{\omega_m} = \frac{E_a \cdot I_a}{\dfrac{2\pi n}{60}} = \frac{230 \times 40 \times 60}{2\pi \times 1200} = 73.21\ \text{N·m}$$

(d) If the load is thrown off, the generated voltage equals the terminal voltage, i.e., 240 V:

$$E_{a0} = K_a' \phi N_0 \implies \frac{230}{240} = \frac{1200}{N_0} \implies N_0 = 1252\ \text{rpm}$$

**Various Losses in the D.C. Shunt Machine:**

(a) **No-load Rotational Loss:** Comprises (i) iron loss at working flux and speed, and (ii) mechanical loss (friction and windage) at operating speed.

(b) **$I^2R$ Losses:**
- (i) Armature circuit loss $I_a^2 r_a$: where $r_a$ includes brush contact resistance, armature winding, interpole winding, and compensating winding (if present).
- (ii) Field ohmic loss $V_f I_f$: includes field-rheostat loss.

(c) **Stray Load Losses:** Produced by (i) distortion of air-gap flux due to armature reaction, and (ii) currents in commutated coils.

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**Solution:**

$$V_t = 400\ \text{V},\quad I_L = 30\ \text{A},\quad R_a = 1\ \Omega$$

Before plugging:

$$I_f = \frac{400}{250} = 1.6\ \text{A}$$

$$I_a = I_L - I_f = 30 - 1.6 = 28.4\ \text{A} \quad \text{(rated current)}$$

$$E_b = V_t - I_a R_a = 400 - 28.4 \times 1 = 371.6\ \text{V}$$

In plugging, the armature of the motor is reversed. Let $R_b = R_{\text{ext}} + R_a$ be the total armature-circuit resistance during plugging.

At the instant of plugging:

$$I_a = 1.5 \times 28.4 = 42.6\ \text{A}$$

$$V_t - I_a R_b + E_b = 0 \implies R_b = \frac{V_t + E_b}{I_a} = \frac{400 + 371.6}{42.6} = 18.11\ \Omega$$

$$R_{\text{ext}} = 18.11 - R_a = 18.11 - 1 = 17.11\ \Omega$$

$$T_{\text{plug}} = \frac{E_b I_a}{\omega} = \frac{371.6 \times 42.6}{120} = 131.92\ \text{N·m}$$

$$K\phi = \frac{E_b}{\omega} = \frac{371.6}{120}$$

$$T_{\text{load}} = K\phi \cdot I_a = \frac{371.6}{120} \times 28.4 = 87.94\ \text{N·m}$$

$$T_{\text{braking}} = T_{\text{plugging}} + T_{\text{load}} = 131.92 + 87.94 = 219.86\ \text{N·m}$$

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**Solution:**

Given: $V = 240\ \text{V}$, torque constant $K = 5\ \text{N·m/A}$

$$T = KI \implies 250 = 5 \times I \implies I = 50\ \text{A}$$

In regenerative braking, the direction of $I_a$ reverses, imposing a torque opposite to rotation:

$$E_b = V - (-I_a)\,r_a = 240 + 50 \times 0.5 = 265\ \text{V}$$

$$E_b = K\omega_m \implies 265 = 5\,\omega_m \implies \omega_m = 53\ \text{rad/s} = \frac{53 \times 60}{2\pi}\ \text{rpm} = 506.11\ \text{rpm}$$

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**Solution:**

Initial back-EMF:

$$E_a = V_t - I_a r_a = 400 - 60 \times 0.2 = 388\ \text{V}$$

With a sudden 20% increase in field flux, speed remains at 1000 rpm (due to motor inertia), so the new back-EMF becomes:

$$E_a' = 388 \times 1.2 = 465.6\ \text{V}$$

New armature current:

$$I_a' = \frac{400 - 465.6}{2} = -32.8\ \text{A}$$

Since $I_a'$ is negative, the machine momentarily acts as a generator, feeding power back to the supply at the expense of its kinetic energy.

$$\frac{|I_a'|}{I_a} = \frac{32.8}{60}$$

Corresponding torque:

$$T' = \frac{32.8}{60} \times 1.2 \times T_{\text{initial}} \approx 0.656 \times T_{\text{initial}}$$

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**Solution:**

**Case 1:**

$$I_{sh} = \frac{V}{R_{sh}} = \frac{220}{176} = 1.25\ \text{A}$$

$$I_M = I_L - I_{sh} = 20 - 1.25 = 18.75\ \text{A}$$

$$E_1 = V - I_M R_a = 220 - 18.75 \times 0.4 = 212.5\ \text{V}$$

Since torque is constant: $\phi_1 I_{a1} = \phi_2 I_{a2}$

$$\phi_1 \times 18.75 = 0.89\,\phi_1 \times I_{a2} \implies I_{a2} = 21.06\ \text{A}$$

**Case 2:**

$$E_2 = V - I_{a2}(R_a + R_s) = 220 - 21.06 \times (0.4 + 0.1) = 209.47\ \text{V}$$

Since $E \propto \phi N$:

$$\frac{E_1}{E_2} = \frac{\phi_1 N_1}{\phi_2 N_2} \implies N_2 = \frac{E_2 \,\phi_1\, N_1}{E_1\, \phi_2} = \frac{209.47 \times 2500\,\phi_1}{212.5 \times 0.89\,\phi_1} = 2768.88\ \text{rpm}$$

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**Solution:**

$$E_{a_1} = E_{a_2} = V - R_a I_a = 500 - (0.36)(50) = 482\ \text{V}$$

$$N_{m_1} = 700\ \text{rpm},\quad N_{m_2} = 750\ \text{rpm}$$

Since $E_a = K\phi N_m$:

$$\frac{\phi_1}{\phi_2} = \frac{N_{m_2}}{N_{m_1}} = \frac{750}{700} = \frac{15}{14}$$

When the two motors are connected in series and mechanically coupled:

$$E_{a_1} + E_{a_2} = 500 - 50(0.72) = 464\ \text{V}$$

Since $N_{m_1} = N_{m_2} = N_m$:

$$\frac{E_{a_2}}{E_{a_1}} = \frac{\phi_2}{\phi_1} = \frac{14}{15}$$

$$E_{a_1} + \frac{14}{15}E_{a_1} = 464 \implies E_{a_1} = 240\ \text{V}$$

Comparing with separate operation for machine 1:

$$\frac{E'_{a_1}}{E''_{a_1}} = \frac{N'_{m_1}}{N''_{m_1}} \implies \frac{240}{482} = \frac{N}{700} \implies N = 348.55\ \text{rpm}$$

Verification with machine 2: if $E'_{a_2} = 224\ \text{V}$,

$$\frac{224}{482} = \frac{N}{750} \implies N = 348.55\ \text{rpm} \checkmark$$

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**Solution:**

We know:
$$E_b = k\phi\omega_m, \quad E_b = V - I_a R_a, \quad T = k\phi I_a$$

Substituting $I_a = \dfrac{T}{k\phi}$:

$$\omega_m = \frac{V}{k\phi} - \frac{T R_a}{k^2\phi^2}$$

Differentiating $\omega_m$ with respect to $\phi$:

**Case (i):** $\dfrac{d\omega_m}{d\phi} > 0$

$$-\frac{V}{k\phi^2} + \frac{2TR_a}{k^2\phi^3} > 0 \implies 2I_a R_a > V \implies 2(V - E_b) > V \implies E_b < \frac{V}{2}$$

$\therefore$ When $E_b < \dfrac{V_t}{2}$, speed **increases** with rising flux.

**Case (ii):** $\dfrac{d\omega_m}{d\phi} < 0$

$$-\frac{V}{k\phi^2} + \frac{2TR_a}{k^2\phi^3} < 0 \implies E_b > \frac{V}{2}$$

$\therefore$ When $E_b > \dfrac{V_t}{2}$, speed **decreases** with rising flux.

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**Solution:**

Given: Full load output $= 20\ \text{kW}$ at $500\ \text{V}$, $\eta = 90\%$

$$\text{Input} = \frac{20000}{0.9} = 22222\ \text{W}$$

$$\text{Losses} = 22222 - 20000 = 2222\ \text{W}$$

Armature copper losses $= 0.4 \times 2222 = 888.8\ \text{W}$

Full load line current $= \dfrac{22222}{500} = 44.4\ \text{A}$

Field current $= \dfrac{500}{250} = 2\ \text{A}$

Full load armature current: $I_a = 44.4 - 2 = 42.4\ \text{A}$

$$R_a = \frac{888.8}{(42.4)^2} = 0.49\ \Omega$$

Minimum starting line current $= 1.2 \times 44.4 = 53.3\ \text{A}$; armature current $= 53.3 - 2 = 51.3\ \text{A}$

Maximum starting line current $= 2 \times 44.4 = 88.8\ \text{A}$; armature current $= 88.8 - 2 = 86.8\ \text{A}$

A 4-section starter is designed as follows:

$$K = \frac{I_{a2}}{I_{a1}} = \frac{51.3}{86.8} = 0.591$$

$$R_{\text{total}} = \frac{500}{86.8} = 5.76\ \Omega$$

$$R_1 = R_{\text{total}}(1 - K) = 5.76(1 - 0.591) = 2.356\ \Omega$$

$$R_2 = K R_1 = 0.591 \times 2.356 = 1.392\ \Omega$$

$$R_3 = K^2 R_1 = 0.8229\ \Omega$$

$$R_4 = K^3 R_1 = 0.486\ \Omega$$

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**Solution:**

Given: $V_t = 400\ \text{V}$, $r_a = 0.2\ \Omega$, $r_{fL} = 200\ \Omega$, $N_{NL} = N_{FL}$

At no load:

$$I_{a(NL)} = I_{L(NL)} - I_{sh} = 6 - \frac{400}{200} = 4\ \text{A}$$

$$E_{b(NL)} = V_t - I_{a(NL)}\,r_a = 400 - 4 \times 0.2 = 399.2\ \text{V}$$

At full load:

$$I_{a(FL)} = I_{L(FL)} - I_{sh} = 70 - \frac{400}{200} = 68\ \text{A}$$

$$E_{b(FL)} = V_t - I_{a(FL)}\,r_a = 400 - 68 \times 0.2 = 386.4\ \text{V}$$

Since $E_b \propto N\phi$:

$$\frac{E_{b(FL)}}{E_{b(NL)}} = \frac{N_{FL}\,\phi_{FL}}{N_{NL}\,\phi_{NL}}$$

With $N_{FL} = N_{NL}$:

$$\phi_{FL} = \frac{386.4}{399.2}\,\phi_{NL} = 0.9679\,\phi_{NL}$$

$$\% \text{ reduction in flux} = \frac{\phi_{FL} - \phi_{NL}}{\phi_{NL}} \times 100 = (0.9679 - 1) \times 100 = -3.20\%$$

So as the D.C. shunt motor is loaded from no-load to full-load at constant speed, the field weakens by **3.20%**.

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**Solution:**

(i) Additional resistance to obtain rated torque at starting:

At start, $N = 0$, $E_b = 0$. To obtain rated torque, $I_a = 40\ \text{A}$:

$$I_a = \frac{V - E_b}{R_a + R_{\text{ext}}} \implies 40 = \frac{240 - 0}{0.3 + R_{\text{ext}}} \implies R_{\text{ext}} = 6 - 0.3 = 5.7\ \Omega$$

Adding $5.7\ \Omega$ in series at start produces rated torque.

(ii) Back-EMF at 1500 rpm:

$$E_b = 240 - 40(0.3) = 228\ \text{V}$$

For a series motor ($\phi \propto I_a$): $E_b = K I_a N$

$$K = \frac{E_b}{I_a N} = \frac{228}{40 \times 1500} = 0.0038$$

At $N = 1000\ \text{rpm}$, $I_a = 40\ \text{A}$:

$$E_b = K I_a N = 0.0038 \times 40 \times 1000 = 152\ \text{V}$$

Required $R_{\text{ext}}$ for $E_b = 152\ \text{V}$:

$$E_b = V - I_a(R_a + R_{\text{ext}}) \implies 152 = 240 - 40(0.3 + R_{\text{ext}}) \implies R_{\text{ext}} = 1.9\ \Omega$$

Adding $1.9\ \Omega$ in series with the motor develops rated torque at 1000 rpm with $I_a = 40\ \text{A}$.

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