Transformer & Electrical Engineering Questions PDF

Here is the conversion of the document/image into a structured markdown format: ### Question 1 The current, $i(t) = 237\cos{(\omega t + 75^\circ)}A$. Find $I_{max}$ and $I_{rms}$ in phasor and rectangular representation. $I_{max} = 237A$ $I_{rms} = \frac{I_{max}}{\sqrt{2}} = \frac{237}{\sqrt{2}}...

Here is the conversion of the document/image into a structured markdown format: ### Question 1 The current, $i(t) = 237\cos{(\omega t + 75^\circ)}A$. Find $I_{max}$ and $I_{rms}$ in phasor and rectangular representation. $I_{max} = 237A$ $I_{rms} = \frac{I_{max}}{\sqrt{2}} = \frac{237}{\sqrt{2}} = 167.6A$ From the equation, phase angle = $75^\circ$. Phasor representation in polar form: $I = 167.6 \angle 75^\circ A$. Phasor representation in rectangular form: real part $\implies 167.6\cos{75^\circ} = 43.4$ imaginary part $\implies 167.6\sin{75^\circ} = 161.9$ $\therefore I = 43.4 + j161.9 A$ ### Question 2 The voltage, $v(t) = 321.7\cos{(\omega t + 25^\circ)}V$. Calculate $V_{rms}$ in phasor and rectangular presentation. $V_{max} = 321.7$ $V_{rms} = \frac{V_{max}}{\sqrt{2}} = \frac{321.7}{\sqrt{2}} = 227.5V$ From the equation, phase angle $= 25^\circ$ Phasor representation in polar form: $V = 227.5\angle 25^\circ$ Phasor representation in rectangular form: real part $\implies 227.5\cos{25^\circ} = 206.2$ imaginary part $\implies 227.5\sin{25^\circ} = 96.1$ $\therefore V = 206 + j96.1 V$ ### Question 3 Which components in an electrical circuit contribute to the imaginary part of the impedance? Explain briefly with a suitable sketch. In an electrical circuit, the imaginary part of the impedance is contributed by the reactive components which are inductors and capacitors. The impedance of these components involves a phase shift between voltage and current, which leads to the imaginary part of the total impedance. Inductor: ~~~ capacitor: ~~~ ### Question 4 A 4500kW, 350V variable-speed motor is powered by a 1500kW generator using a Ward-Leonard control system. The combined resistance of the motor and generator armature circuit is $17m\Omega$. The motor operates at its nominal speed of 570 r/min when the output voltage, $E_o = 1000V$. a) Calculate the motor torque and speed when $E_s = 750V$, $E_o = 420V$. b) Calculate the motor torque and speed when $E_s = 370V$, $E_o = 400V$. Armature current, $I = \frac{E_s - E_o}{R} = \frac{750-420}{17 \times 10^{-3}} = \frac{330}{0.017} = 19411.8 A$ Power to the motor armature, at $E_o = 420V$, $P = E_o I = (420)(19411.8) = 8152956 W$ Motor speed at $E_o = 420V$, $n = \frac{420}{350} (570) = 684 r/min$ Motor torque at $E_o = 400V$, $T = \frac{9.55 P}{n} = \frac{(9.55)(8153 \times 10^3)}{684} = 113832.29 N.m = 113.8 kN.m$ Armature, $I = \frac{E_s - E_o}{R} = \frac{370-400}{17 \times 10^{-3}} = -2941.2A$ (negative means the current flows in reverse direction) Power to the motor armature, $P = E_o I = (420)(2941.2) = 1235304W = 1235.3kW$ at $E_o = 400V$ Motor speed at $E_o = 420V$, $n = (\frac{420}{350}) (570) = 684 r/min$ Motor torque at $E_o = 420V$, $T = \frac{9.55 P}{n} = \frac{(9.55)(1235.3 \times 10^3)}{684} = 17247.24 N.M = 17.2 kNm$ ### Question 5 As a maintenance engineer, you are asked to test a large transformer A. Figure 1 shows the circuits diagram of the ideal transformer with an imperfect core and its parameters are listed in table 2. a) Determine the value of $Q_m$, $R_m$ and $X_m$. b) Determine the value of $I_f$, $I_m$ and $I_0$. A circuit diagram is shown, representing an ideal transformer with an imperfect core. There are labels for: - $E_g$ - $I_0$ - $I_f = 0$ - $I_2=0$ - $I_f$ - $I_m$ - $X_m$ - $R_m$ - $E_2$ | Transformer | Source & Frequency | Exciting current | Iron losses | Reactive power | Resistance | Magnetix Reactance | | --- | --- | --- | --- | --- | --- | --- | | A | 120V, 60Hz | 5A | 180W | $Q_m$ | $R_m$ | $X_m$ | Given data: Source voltage, $V_g = 120V$ Exciting current, $I_0 = 5A$ Core loss power, $P_{core} = 180W$ a) Determine $R_m$, $Q_m$ and $X_m$: (i) Core resistance ($R_m$): $R_m = \frac{V_g^2}{P_{core}} = \frac{(120)^2}{180} = 80.0\Omega$ (ii) Reactive Power ($Q_m$): Find S (Apparent); $S = V_g \cdot I_0 = (120)(5) = 600 VA$ $Q_m = \sqrt{S^2 - P_{core}^2} = \sqrt{(600)^2 - (180)^2} = 572.36 VAR$ (iii) Magnetizing Reactance ($X_m$): $X_m = \frac{V_g^2}{Q_m} = \frac{(120)^2}{572.36} = 25.16\Omega$ b) Determine $I_f$, $I_m$ and verify $I_0$: (i) Core losses current ($I_f$): $I_f = \frac{V_g}{R_m} = \frac{120}{80} = 1.5 A$ (ii) Magnetizing current ($I_m$): $I_m = \frac{V_g}{X_m} = \frac{120}{25.16} = 4.77 A$ (iii) Verify $I_0$: $I_0 = \sqrt{I_f^2 + I_m^2} = \sqrt{(1.5)^2 + (4.77)^2} = 5 A$ (Verified) ### Question 6 You are required to design an aerial transmission line. Suggest THREE (3) consideration factors and give one example for each with supporting arguments. 1. One of the considerations is the local environment which will affect the line design and materials used. In areas prone to high winds or extreme temperatures, conductors used. 2. One of the considerations of the transmission line should be designed to handle the required voltage and power capacity for the area it serves. It is because a higher voltage level should be used for long-distance transmission to minimise voltage drop and line losses. 3. The second consideration is the local environment which will affect the line design and materials used. In areas prone to high winds or extreme temperatures, conductors need to be designed for additional mechanical stress. For instance, in hot climates, the conductor material must be able to withstand thermal expansion without sagging excessively. 4. Third consideration is the choice of conductor material which affects the line's efficiency, durability and cost. Aluminium is commonly chosen for aerial transmission lines because it is lighter, less expensive and has good conductivity. It is often used in steel-reinforced aluminium conductor (ACSR), which are strong enough for long spans. ### Question 7 Given another transformer B has the same parameters as transformer A except for the iron losses, which is 200W. Compare the reactive power absorbed by the core of transformers A and B. $Q_m = \sqrt{S^2 - P_{core}^2}$ $S = V_g \cdot I_0$ | | Transformer A | Transformer B | | :-------------------- | :---------------------------------- | :---------------------------------- | | $P_{core}$ | $180W$ | $200 W$ | | S | $600 VA$ | $600 VA$ | | $Q_m$ | $\sqrt{600^2 - 180^2} = 572.36 VAR$ | $\sqrt{600^2 - 200^2} = 565.69 VAR$ | | **Conclusion** | | | The reactive power absorbed by the core of transformer B is slightly lower than that of transformer A because Transformer B has higher iron losses ($P_{core,B} = 200 W$) compared to Transformer A ($P_{core,A} = 180W$).