In tan" tan M- M+ M OCT Evaluate (130 points); Use 3 decimal places. 1. The annual load of a substation is given in the following table. During each month, the power is assumed constant at an average value. Find the (a) average load and (b) the annual load factor. Annual System Load January 8 February 8 March 4 April 2 May 6 2, T- June 12 sing July 14 pha: August 14 idea September 10 imp October 8 5Y5 November 6 mac December 8 00 deliv
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- Table Q2a Time (hours) 0-6 6-12 12-14 14-18 18-24 Load (MW) 45 90 135 150 75 Referring to the Table Q2a; i) Draw the load curve and the load duration curve [Use the graph paper provided] ii) Indicate the base load and the peak load on the graph. Determine the values for both loads. iii) Choose the suitable generating units to supply the load and list their operating schedules. Later, compute the total installed capacity. iv) Calculate the load factor and the plant capacity factor. Discuss the impact on the operating schedule if the plant capacity factor increases to 0.9.Power factor correction is usually done by adding _________ to the load circuit.Select the correct response:a. switchb. capacitorc. inductive loadsIn the system shown in Figure 1, the transformers are connected star-star with both star points grounded and the generators are connected in star with thier star points grounded. The system base is 15 MVA. The transformers all have reactances of 0.04 p.u. on this 15 MVA base. The reactances of all other elements are given in Table 1 (in 2) and the voltage levels are given in Table 2. p.u. G1 p.u. T1 jö Per-Unit Convert all values to p.u. on a 15 MVA base. Xa= p.u. Xc₂= XL = V BABE G1 2 X 9 T3 Figure 1: A section of the distribution system T1 L Table 1: Sequence reactances (2) 3 G1 L G2 0.3 0.59 0.01 4 L 9/10 10 Fault Voltage What is the voltage at bus 3 (in Volts) after the fault has occurred? Vp= V T2 5 T2 34 10/4 Table 2: Voltage bases (kV) G2 4 T3 10/9 | G2 Fault Current A three-phase fault with a fault reactance of 0.01 p.u. occurs at bus 3. Calculate the fault current flowing at the fault point in KA. Ip=-j KA So
- Unit commitment if the spinnind reserve not there in the power system what will happen?The peak load on a 50 MW power station is 39 MW. It supplies power through for transformers whose connected loads are 17, 12,9 and 10 MW. The maximum demands on these transformers are 15, 10, 8 and 9 MW respectively. The annual load factor is 50% and the plant is operating for 65% of the period in a year. 102. The average load on the station is (A) 50000 kW (C) 39000 kW (B) 19500 kW (D) 48000 kW. 103. The energy supplied per year is (A) 438 x 106 kWh (C) 170.82 x 106 kWh (B) 341.64 x 106 kWh (D) 420.48 x 106 kWh.Connect three single-phase transformers in Y-Y, with subtractive polarity, so that the line-to-line voltages on the primary are 30 degrees delayed to their respective line-to-line voltages on the secondary. It should show each step that is needed to make this connection (example: the phasor diagram) and should show the final connection of the three transformers including the polarity markings. Assume phase A on the primary is terminal "H1".
- A generator operating at 50 Hz delivers 1 p.u. power to an infinite bus through a transmission circuit in which resistance is ignored. A fault takes place reducing the maximum power transferable to 0.5 p.u. Where as before the fault, this power was 2.5 p.u. and after the clearance of the fault, it is 2 p.u. The value of 8 is maxWhat is the main direct cause of reactive power in AC system?A. Resistance of transmission linesB. Inductance and capacitance in the loadsC. Ideal transformer connected in the systemD. Power produced by generatorIn single phase power factor meter there is error due to frequency and waveforms but not in three phase power factor meter, I know the reason why?? Please elaborate properly
- A generating station is to supply four regions of load whose peak loads are10 MW, 5 MW, 8 MW and 7 MW. The diversity factor at the station is 1-5 and the average annual load factor is 60%. Then the installed capacity is (25 to26)MW. O (24 to25)MW. O (23 to24)MW. OA 400 V, 3-phase alternator is subjected to an unsymmetrical fault. The sequence currents are Ia1= -j10 A, Ia2 = j6 A and Iao = j4 A. The sequence impedances are Z1 = j10 ohm, Z2 = j5 ohm and z0 = j7.5 ohm. Find the fault current and the type of fault. Also find the terminal voltages (phase and line) of the generator.A parallel connection of a RL branch with a C branch is connected across a 100V AC mains. At first R = 10 ohms, L = 20mH and frequency of 1000rad / s, the current measured is 2.2361A at 89.44 leading power factor. A) Determine the initial capacitance. B) However, a fault occurs on the capacitor branch making its capacitance 20% lower and a resistance of 5 ohms is detected. If this faulty circuit is rerun but at a frequency of 500 rad / s, determine the new current that will flow through the circuit.