List FOUR advantages of a DC series motor.

Here are the solutions to the questions: QUESTION 1: DC MACHINES 1.1 FOUR advantages of a DC series motor are: • High starting torque, making them suitable for heavy-duty applications like cranes and traction. • Ability to operate on both AC and DC supplies, often referred to as universal motors. • Simple construction due to the series connection of field and armature windings. • Automatic speed adjustment with load changes, as speed decreases significantly with increasing load. 1.2 Given: V_T = 240 V, R_sh = 120 , R_se = 0.15 , R_a = 0.20 . Assume a long shunt DC compound motor. 1.2.1 The back EMF at no load No-load: I_L = 2 A. Step 1: Calculate the shunt field current. I_sh = (V_T)/(R_sh) = 240 V120 = 2 A Step 2: Calculate the armature current. I_a = I_L - I_sh = 2 A - 2 A = 0 A Step 3: Calculate the back EMF at no load. E_b = V_T - I_a (R_a + R_se) E_b = 240 V - 0 A × (0.20 + 0.15 ) = 240 V E_b = 240 V 1.2.2 The torque developed at full load Full-load: I_L = 40 A, N = 1000 r/min. Step 1: Calculate the shunt field current (assumed constant). I_sh = (V_T)/(R_sh) = 240 V120 = 2 A Step 2: Calculate the armature current. I_a = I_L - I_sh = 40 A - 2 A = 38 A Step 3: Calculate the back EMF at full load. E_b = V_T - I_a (R_a + R_se) E_b = 240 V - 38 A × (0.20 + 0.15 ) = 240 V - 38 A × 0.35 E_b = 240 V - 13.3 V = 226.7 V Step 4: Calculate the power developed in the armature. P_d = E_b I_a = 226.7 V × 38 A = 8614.6 W Step 5: Calculate the angular speed. = (2 N)/(60) = 2 × 1000 r/min60 ≈ 104.72 rad/s Step 6: Calculate the torque developed. T_d = (P_d)/() = 8614.6 W104.72 rad/s ≈ 82.26 Nm T_d = 82.26 Nm 1.2.3 The shaft power at full load Assuming rotational losses are negligible or not provided, the shaft power is approximately equal to the mechanical power developed in the armature. P_shaft = P_d = E_b I_a = 8614.6 W P_shaft = 8.61 kW 1.3 A DC generator with a differential compound winding has its series field winding connected such that its magnetic flux opposes the magnetic flux of the shunt field winding. This connection results in a decreasing terminal voltage with increasing load. [scale=0.8] % Armature [thick] (0,0) circle (1.5cm); at (0,0) Armature; [thick] (0,1.5) -- (0,2); [thick] (0,-1.5) -- (0,-2); % Brushes [fill=black] (-0.5,1.5) rectangle (0.5,1.7); [fill=black] (-0.5,-1.5) rectangle (0.5,-1.7); % Shunt Field [thick] (0,2) -- (2,2) -- (2,0.5); [thick] (0,-2) -- (2,-2) -- (2,-0.5); [thick, decorate, decoration=coil,aspect=0.3,segment length=3mm,amplitude=2mm] (2,0.5) -- (2,-0.5) node[right, midway] R_sh; % Series Field [thick] (0,2) -- (-2,2) -- (-2,0.5); [thick] (0,-2) -- (-2,-2) -- (-2,-0.5); [thick, decorate, decoration=coil,aspect=0.3,segment length=3mm,amplitude=2mm] (-2,0.5) -- (-2,-0.5) node[left, midway] R_se; % Load [thick] (2,2) -- (3,2) node[right] V_T; [thick] (2,-2) -- (3,-2); [thick] (3,2) -- (3,-2); [thick] (3.5,0) node[right] Load; % Arrows for current direction (indicating differential connection) [->] (2,1.5) -- (2,0.5); % Shunt current down [->] (-2,1.5) -- (-2,0.5); % Series current down (opposing shunt) [->] (0,1.7) -- (0,2); % Armature current out [->] (0,-1.7) -- (0,-2); % Armature current out [->] (2.5,2) -- (3,2); % Load current [->] (2.5,-2) -- (3,-2); % Load current Note: The diagram shows a long shunt connection, where the series field is in series with the armature, and this combination is in parallel with the shunt field and the load. For differential compounding, the winding direction of the series field is such that its magnetic field opposes the shunt field. 1.4 If the armature voltage of a DC shunt motor is increased while the load torque remains constant and the field flux is constant, the speed of the motor will increase. This is because with constant load torque and constant field flux, the armature current remains constant. An increase in armature voltage leads to an increase in the back EMF (E_b = V_T - I_a R_a), and since E_b is proportional to speed (for constant flux), the speed must increase. QUESTION 2: AC CIRCUIT THEORY Given: Delta-connected load, Z_ph = (6 - j8) per phase, V_L = 400 V, f = 50 Hz. 2.1 The line current Step 1: For a delta connection, the phase voltage is equal to the line voltage. V_ph = V_L = 400 V Step 2: Calculate the magnitude of the phase impedance. |Z_ph| = sqrt(6^2 + (-8)^2) = sqrt(36 + 64) = sqrt(100) = 10 Step 3: Calculate the phase current. I_ph = V_ph|Z_ph| = 400 V10 = 40 A Step 4: Calculate the line current for a delta connection. I_L = sqrt(3) I_ph = sqrt(3) × 40 A ≈ 69.28 A I_L = 69.28 A 2.2 The power factor Step 1: Calculate the power factor. = R_ph|Z_ph| = (6 )/(10 ) = 0.6 Since the impedance has a negative imaginary part (capacitive reactance), the power factor is leading. = 0.6 leading 2.3 The active power consumed Step 1: Calculate the active power. P = 3 I_ph^2 R_ph = 3 × (40 A)^2 × 6 = 3 × 1600 × 6 = 28800 W P = 28.8 kW 2.4 The reactive power consumed Step 1: Calculate the reactive power. Q = 3 I_ph^2 X_ph = 3 × (40 A)^2 × (-8 ) = 3 × 1600 × (-8) = -38400 VAR The negative sign indicates that the load is supplying reactive power (or consuming negative reactive power), which is characteristic of a capacitive load. Q = -38.4 kVAR 2.5 The apparent power (kVA) Step 1: Calculate the apparent power. S = sqrt(3) V_L I_L = sqrt(3) × 400 V × 69.28 A ≈ 48000 VA Alternatively, using P and Q: S = sqrt(P^2 + Q^2) = sqrt((28800)^2 + (-38400)^2) = sqrt(829440000 + 1474560000) = sqrt(2304000000) = 48000 VA S = 48.0 kVA Last free one today — make it count tomorrow, or type /upgrade for unlimited.