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Chapter 2 Solutions
Power System Analysis & Design
- . A 46KV 0.8 lagging power factor load is connected to the end of a short transmission line wherethe voltage is 230. If the line resistance and reactance are 0.06 and 0.08 ohm, respectively,calculate the voltage at the sending end.arrow_forwardQ5. A single phase overhead transmission line has a supply voltage of 34 kV and a sending end current of magnitude 40 A at 0-8 p.f. lagging is feeding a certain load. The total resistance and inductance of the line are 5 and 21 mH respectively. Draw the practical circuit used and the devices necessary to perform this experiment. Then calculate the sending end power, receiving end current, voltage, power factor, receiving end power and line efficiency Vs 34000 Is 40 Cosos 0.8 R 5 Ps Ir Vr Cosor Pr n 21e-3arrow_forwardA transmission line of impedance (0.05 + j0.02) pu interconnects the buses of a switchyard and a bulk supply point. The receiving end apparent power is (1.0 + j0.6) pu and sending end voltage, 1/0°pu. Estimate the following: i) Receiving end voltage after two iterations ii) Perform two further iterations to test the convergence of the value of receiving end voltage deduced in (i) iii) load currentarrow_forward
- S1) The serial impedance per unit length of a three-phase 140 km power transmission line is 0.09 + j0.88 ohm/ km and its admittance is j4.1x10-6 S / km. Power factor under 210 kV interphase voltage from the end of this energy transmission line A power of 150 MVA, which is 0.85 back, is drawn. Using this transmission line data and the T equivalent circuit model, the line Calculate the head voltage (V1), current (I1) and load angle.arrow_forwardIf a 3- Ø, Y-connected system has a line-to-line voltage of 1103Sin(377t) V, then the line-to-neutral (RMS) voltage would be:arrow_forwardAn alternating single-phase circuit describes the instantaneous values of the applied voltage and the corresponding current as: v = 360 sin (201,69 t + π/6) and i = 36 sin (201,69 t - π/9) Calculate: the frequency (1) the phase angle, in degrees, (1) the power factor of the circuit (1) the value of the voltage and current 10 ms after zero, (2) the impedance, resistance and reactance of the circuit, (3) the r.m.s.-values of the voltage and current, (2) the true-, apparent- and reactive power in the circuit (3) The time taken to reach -190 V for the second time. (3) Sketch the phasor diagram of this circuit (2)arrow_forward
- S.2) The serial impedance of the 300 km power transmission line is 23 + j75 ohm / phase and the shunt acceptance is j500 microS / phase. A power of 50 MW with a power factor of 0.88 under 220 kV interphase voltage from the end of the power transmission line being shot. Using these data of the energy transmission line, a) Characteristic impedance of the energy transmission line, b) Natural apparent power of the energy transmission line according to the 220 kV interphase operating voltage, c) The values of the line parameters A, B, C and D of the transmission line using hyperbolic functions, d) Calculate the maximum power that the power transmission line can transmit.arrow_forwardA 3-phase, 50 Hz overhead transmission line, 100 km long, 110 kV between the lines at the receiving end has the following constants : Resistance per km per phase = 0·153 Ω Inductance per km per phase = 1·21 mH Capacitance per km per phase = 0·00958 μF The line supplies a load of 20,000 kW at 0·9 power factor lagging. Calculate using nominal π representation, the sending end voltage, current, power factor, regulation and the efficiency of the linearrow_forwardA 3-phase, 50 Hz, 440 V motor develops 100 H.P, which works for 3000 hours per annum, the power factor being 0·75 lagging and efficiency 93%. A bank of capacitors is connected in delta across the supply terminals and power factor raised to 0·95 lagging. The tariff is Rs 100 per kVA plus 20 paise per kWh. If the power factor is improved to 0·95 lagging by capacitors costing Rs 60 per kVAR and having a power loss of 100 W per kVA. Compute the annual saving effected. Allow 12% per annum for interest and depreciation on capacitorsarrow_forward
- Design a power factor correction circuit with the value of parallel capacitance needed tocorrect a load of 179 kVAR at 0.85 lagging pf to unity pf. Assume that the load issupplied by a 110-V (rms), 60-Hz line.arrow_forwardA 3-phase, 50 Hz, 440 V motor develops 100 H.P. (74·6 kW), which works for 3000 hours per annum, the power factor being 0·75 lagging and efficiency 93%. A bank of capacitors is connected in delta across the supply terminals and power factor raised to 0·95 lagging. The tariff is Rs 100 per kVA plus 20 paise per kWh. If the power factor is improved to 0·95 lagging by means of loss-free capacitors costing Rs 60 per kVAR, Compute the annual saving effected. Allow 12% per annum for interest and depreciation on capacitors.arrow_forwardThe load on a 120-V, 60 Hz supply is 5KW (resistive), 8 KVAR (inductive), and 2 KVAR (capacitive). Find the following: a) total kilovolt-amperes, b)power factor of the combined loads, c) current drawn from the supply, d) capacitance needed to establish unity power factor, e) capacitive reactance needed to establish unity power factor, f) the current drawn from the supply at unity p.f.arrow_forward
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