The fuel -cost characteristic equations for two thermal plants in OMR/Hr are given bị C1 = 20+4.8 P1+0.012P1 ², C2 = 40+2.85P2+0.05P2 ², Where P1, P2 are in MW Neglect line losses and generator limits. a) find the total load on power system when generators are operating, under economi load dispatch condition, at an equal incremental cost of 15 OMR/ MWhr. b) for the same total load calculated above, determine the saving in OMR/Hr economic load dispatch is followed over equal distribution of load between generators
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- Consider a three-phase generator rated 300MVA,23kV, supplying a system load of 240 MA and 0.9 power factor lagging at 230 kV through a 330MVA,23/230Y-kV step-up transformer with a leakage reactance of 0.11 per unit. (a) Neglecting the exciting current and choosing base values at the load of 100 MVL and 230 kV. Find the phasor currents IA,IB, and IC supplied to the load in per unit. (b) By choosing the load terminal voltage IA as reference, specify the proper base for the generator circuit and determine the generator voltage V as well as the phasor currents IA,IB, and IC, from the generator. (Note: Take into account the phase shift of the transformer.) (C) Find the generator terminal voltage in kV and the real power supplied by the generator in MW. (d) By omitting the transformer phase shift altogether, check to see whether you get the same magnitude of generator terminal voltage and real power delivered by the generator.Consider two interconnected voltage sources connected by a line of impedance Z=jX, as shown in Figure 2.27. (a) Obtain expressions for P12 and Q12. (b) Determine the maximum power transfer and the condition for it to750kW alternators P₁ = 750kW, P₂=750kW are operated in parallel. Speed regulation of one machine is 100% to 103%. Speed regulation of second machine is 100% to 104%. Shared load, (P₁)' + (P₂)' arrow_forward Step 2 = (P₂) 1000kW....... (A) So the shared load equation for alternator-1 will be, 103-100-103- 103- (B)' = 103- (Pi)...... (₁) The shared ad equation for alternator-2 will be, 104 100 104 (P₂) 750 = 104-1 = 104 (P₂)'...... (²) Now from equation (1) and (2), 103 (P₁) 104- (P₂)' 103-104 (P)' - (P₂)' (P₁)-(P₂)' = -1 3(P₁)-4(P₂)-750kW....... On solving equation (A) and (B), (P₁) 464. 28kW (P₂) 535. 714kW How did he get the value inside the circle? Can you explain step by step and clear line please?
- In a given system of base power of 250 MW, and bus 3 is taken as reference. The per-unit reactances are X12 = 0.2 p.u., X13 = X23 = 0. 1 p.u. The power flow in the system is given as: PF12 = 50 MW, PF13 = 150 MW, PF23 = 50 MW. Based on readings from 2 meters (not including M12 meter), M13 = 145 MW (not calibrated), and M23 = 50 MW (well calibrated), deduce the flow on line 1-2. Select one: O a. PF12 = - 47.5 MW. O b. None of these O c. PF12 = 67.5 MW. O d. PF12 = 47.5 MW. O e. PF12 =- 57.5 MW. O f. PF12 = 57 MW.Two generating units rated for 500 MW and 400 MW have governor speed regulation of 6.0 and 6.4 percent, respectively, from no-load to full-load, respectively. The generators are operating in paral lel and share a load of 600 MW. Use a common base of 1000 MVA and assuming free governor action, justify: i) the new governor speed regulation. ii) the load shared by each unit.The figure below shows the one-line diagram of a four- bus power system. The voltages, the scheduled real power and reactive powers, and the reactances of transmission lines are marked at this one line diagram (The voltages and reactances are in PU referred to 100 MW base. The active power P2 in MW is the last three digits (from right) of your registration number (i.e for the student that has a registration number 202112396, P2 =396). [10] Starting from an estimated voltage at bus 2, bus 3, and bus 4 equals V2 (0) = 1.15<0°, V3 = 1.15 < 0°, V4 1.1< 0°. 1- Specify the type of each bus and known & unknown quantities at each bus. 2- Find the elements of the second row of the admittance matrix (i.e. [Y21 Y22 Y23 Y24]). 3- Using Gauss-Siedal fınd the voltage at bus 2 after the first iteration. 4- Using Newton-Raphson, calculate: |- The value of real power (P2), at bus 2 after the first iteration. Il- The second element in the first row of the Jacobian matrix after the first iteration. 2 P2…
- Suppose two generators supplying a load. Generator-1 has a no-load frequency of 61.5 Hz and a slopesp1 of 1 MW/Hz. Generator-2 has a no-load frequency of 62 Hz and a slope sp2 of 1MW/Hz. The twogenerators are supplying a real load totaling 2.5 MW at 0.8 PF lagging. Draw the resulting systempower-frequency or house-diagrams. Determine (a) at what frequency is the system operating, andhow much power is supplied by each of the generators? (b) if an additional 0.75-MW load wereattached to this power system. What would the new system frequency be, and how much powerwould the each generator supply now? (c) with the system in the configuration described in part (b),what will the system frequency and generators power be if the governor set points on generator-1 areincreased by 0.5 Hz?(8) Q2. Correct the wrong sentence if you find it in only (all) of the following sentences: ual and same direct 1) T1 Porto at e Tated current med on low vo "aly mmf:. the. ideal trout -side. copper loss is not a vy ne vau and 10au. 12) Shunt field winding have a large number of turns compared with a series field winding. 13) The voltage regulation of generator A is 10% and that of generator B is 5%. The A is better than generator B. generator poTOTIS Us. 10 Un 14) The external characteristic of a dc generator is affected by armature reaction. 15) The commutating poles or interpoles is connected in parallel with the armature. 16) The de motor should be started by impressing its rated voltage across the armature terminals. at 17) The cumulative compound motor has the advantage of over a series Motor. 18) The torque in a differential compound motor increase and its speed is decreased with an increase in the armature current. 19) The flux in a machine does not vary with time. 20) After…A generator with constant 1.0 p.u. terminal voltage supplies power through a step-up transformer of 0.12 p.u.reactance and a double-circuit line to an infinite bus bas as shown in the following figure. The infinite bus voltage is maintained at 1.0 p.u.Neglecting the resistances and susceptances of the system, the steady statestability power limit of the system is 6.25 p.u. If one of the double-circuit istripped, the resulting steady state stability power limit in p.u. will be oi
- b) Two synchronous generators, G2 and G2, are connected parallelly supplying a load. Generator G1 has a no-load frequency of 50.5 Hz and a slope of 300 MW/Hz. Generator G2 has a no-load frequency of 50.2 Hz and a slope of 500 MW/Hz. The load consumes 250 MW real power. (1) At what frequency does this system operate, and how much power is supplied by each of the two generators? (ii) An additional 100 MW load is added to this power system. What is the new system frequency, and how much power do G1 and G2 supply? (iii) The governor set point of G2 is changed to control system frequency back to 50 Hz. Determine the G2 governor set point.In a short compound generator, the terminal voltage is 230 induced emf tion of this power. field, diverter and arnmature resistances are 92 2, 0.015 2, 0.03 2 and 0. 03 2 respectively. V when generator delivers b) total power generted and 160 A. Determine a) c) distribu- Given that the shunt field, seriesTwo similar 2,500 kW, 3 phase generators operate in parallel. The speed load characteristic of the first generator is such that is no-load frequency of 60.5 Hz drops uniformly to 58.5 Hz at full load, while for generator 2, its no-load frequency of 60.5 Hz drops uniformly to 59 Hz at full load. Determine the kW load of each generator if the combined load is 4,000 kW. a. 1,714.3, 2,285.7 b. 2,015.5, 1,984.5 c. 1,765.5, 2,234.5 d. 2,452, 1,548