R, R. ¿,(1) (1)7 R R. Fig. Equivalent Circuit of IM Consider the IM circuit in Figure. A V speed of 1750 rpm. Calculate its synchronous speed(n,), slip(s) and rotor frequency(f,). 150 volt, 140 Hz, seven pole, three phase IM has full load %3! 13 455 O a. n, = 2400 rpm, s = , fr = Hz %3D O b. ng 15 2420 грт, s %3 = Hz 48 f, 17 O .n, = 2400 rpm, s = fr= 12 Hz %3D 13 455 O d. n, = 2400 rpm, s = , fr = Hz 5 fr = Hz 13 450 O e. ng = 2450 rpm, s = 48 12
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- Q. A 3-phase, star connected, 10 kVA, 400 V, salient pole alternator with direct and quadrature axis reactances of 15 ohm and 8 ohm respectively, delivers full-load current at 0.8 power factor lagging.Neglect the armature resistance. Solve this question to chose the correct answer for the following: The direct axis and quadrature axis components of armature current are ........ and ......... respectively ?The synchronous reactance per phase of the 460 V, 60 Hz, Y-connected 6-pole synchronous generator is 1.2 Ω. full load armature current 30 A (at 0.8 reverse power-coefficient). Rotary losses 1.5 kW and core losses 1.0 kW (60 Hz at full load). The armature resistance winding losses are neglected. Excitation current, generator no load terminal voltage is set to 480 V while operating. According to this: a. What is the rotation speed of the generator? b. What will be the terminal voltages of the generator in the following cases? 1. At rated load and current and a reverse power-coefficient of 0.8; 2. At rated load and current and a power-coefficient of 1.0; 3. At rated load and current and a forward power-coefficient of 0.8;4. A 6 pole wave wound shunt generator delivers 200A at a terminal voltage of 240V. Ra and Rsh are 0.03Ω and 70Ω respectively. Neglecting brush drop, determine (i) armature current (ii) current per parallel path (iii) generated voltage and (iv) Power developed. armature current ? current per parallel path ? generated voltage ? Power developed ?
- A 3 phase, star connected synchronous generator supplies current of 10A having phase angle of 20º lagging at 400V. Find the load angle and the components of armature current Id and Iq if Xd = 10 ohm and Xq = 6.5 ohm Assume armature resistances to be negligibleA 3 phase 4MVA, 12 kV star connected synchronous generator has Zs= 0.8+J30 ohm. Using circle diagram only, find: 1. The output power developed at unity power factor with excitation equal to 1.25 of rated voltage. 2. The load angle and power factor if the generator supplies (1500 KW) at rated voltage and excitation lequal to (1.2) of rated voltage. 3. Repeat part (2) when .armature resistance is neglectedA shunt generator supplies a 31kW load at 230V through cables of resistance, R = 122mΩ. If the field winding resistance, Rf =50Ω and the armature resistance, Ra =51mΩ. Determine the following below: The terminal voltage is, Answer and unit for part 1 The e.m.f. generated in the armature is,
- Q. A 3-phase, star connected, 10 kVA, 400 V, salient pole alternator with direct and quadrature axis reactances of 15 ohm and 8 ohm respectively, delivers full-load current at 0.8 power factor lagging.Neglect the armature resistance. Solve this question to chose the correct answer for the following: The load angle?The open-circuit characteristic of a shunt generator when driven at normal speed is as follows :Field current : 0.5 1.0 1.5 2.0 2.5 3.0 3.5 AO.C. volts : 54 107 152 185 210 230 240 VThe resistance of armature circuit is 0.1 Ω. Due to armature reaction the effective field current is givenby the relation Ish (eff.) = Ish − 0.003 Ia. Find the shunt field circuit resistance that will give a terminal voltageof 220 V with normal speed (a) on open circuit (b) at a load current of 100 A. Also find (c) number of seriesturns for level compounding at 220 V with 100 A armature current ; take number of shunt turns per pole as1200 and (d) No. of series turns for over-compounding giving a terminal voltage of 220 V at no-load and230 V with 100 A armature current.The synchronous reactance per phase of the 460 V, 60 Hz, Y-connected 6-pole synchronous generator is 1.2 Ω. full load armature current 30 A (at 0.8 reverse power-coefficient). Rotary losses 1.5 kW and core losses 1.0 kW (60 Hz at full load). The armature resistance winding losses are neglected. Excitation current, generator no load terminal voltage is set to 480 V while operating. According to this: c. What is the efficiency of the generator when operating at rated current and a reverse power-coefficient of 0.8? D. How much shaft torque should be applied by the drive machine when the generator is running at full load? E. What happens to the induced torque?
- The following test data are obtained from a 1/4 hp, 1φ, 120V, 60Hz, 1730rpm induction motor.Stator winding (main) resistance = 2.9 ohmsBlocked rotor test: V =43 V, I=5A, P=140WNo-load test: V =120 V, I=3.5A, P=125Wa) Obtain the double revolving field equivalent circuit for the motor.b) Determine the rotational loss.If the armature resistance of a 3-phase, 318.75 kVA, 2300-V synchronous generator is 0.35 ohms per phase, the synchronous reactance is It is 1.2 ohms. Calculate the excitation voltage and voltage regulation for the following load cases: a) Full load, back power factor of 0.8 and rated voltage. b) Full load, forward power factor of 0.6 and rated voltage.Q/ dc shunt generator 4 pole with 500 condcutor , wave running at speed 1000 r.p.m the resistance of armature and field are 0.02 , 55 ohm and delivering load 43.12kwatt at terminal voltage 220 volt, the iron losses is 750 watt 1-Torque developed by the prime mover .........and The torque developed by the armature ........... when running at 1000 r.p.m * 2-The load current ,,,,,,,,,,,,when the speed drops to 800 r.p.m. if If is unchanged ? * 3-Output of the prime motor…….kwatt * 4-Mechanical , electrical and overall efficiency …….. * 5-Flux per pole is......... wb *