Q: A four pole self excited DC shunt generator ruining at 104,66 rad/s, supplies resistive load 18 Kw 1.1 2 constant resistance, the flux per pole is 0.02 Wb .What will be the new generated power if the rotor speed increased to (20%) ?, and the flux per pole increased to (80%), armature resistance is 0.037 Q and shunt field resistance is 220 2. Assume that the armature current increased to (10%) when the rotor speed increased.
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- Of a shunt excited DC generator, Excitation winding internal resistance Rf = 2.5 Ω Armature winding internal resistance Ra = 0.5 Ω A rheostat has been placed in the excitation circuit to keep the machine's end voltage constant at Va = 30 V. Machine Voltage magnetization induced in the armature while operating at n = 1400 rpm at If = 7 A excitation current It is determined as Ea = 30 V from its characteristic. According to this, a) With peak voltage Va = 30 V, excitation current If = 7 A, power supplied to the load when the speed n = 1400 rpm What is the resistance of the rheostat? b) Spindle revolutions are reduced to n = 1000 rpm. In this case Va = 30 V and Ea = 37 V constant What value should the rheostat be adjusted to keep it? Calculate the power given to the load in this new case.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.A Four pole, 50kW, 250 Volt Long Shunt DC Machine Generator has a shunt field current of 1.5 A. It is Lap-winded with 872 conductors, and a rated speed of 1200 RPM. Where, Ra = 100 mohms, and Rse = 0.05 ohms. Solve for the following: Armature Current Generate EMF Total Flux per pole
- A 20KW, 440V, short shunt compound dc generator has full load efficiency of 87%. If the resistance of the armature and interpoles is 0.4ohm and that of the series and shunt fields 0.25ohm and 240ohms respectively, calculate the combined bearing friction, windage and core loss of the machine A 10 kW 250 volts self-excited generator, when delivering rated load, has an armature-circuit voltage drop of 5% of the terminal voltage and a shunt field current equal to 5% of the rated load current. The resistance of the armature is 0.02 ohm. Calculate the voltage generated and power output.The open-circuit characteristic of a shunt generator when driven at normal speed is as;Field current (A)0.51.01.52.02.53.03.5O.C. volts (V)54107152185210230240The resistance of armature circuit is 0.1 Ω. Due to armature reaction the effective field current is given by the relation Ish (eff.) = Ish − 0.003 Ia. Find the shunt field circuit resistance that will give a terminal voltage of 220 V with normal speed (a) on open circuit (b) at a load current of 100 A. Also find (c) number of series turns for level compounding at 220 V with 100 A armature current; take number of shunts turns per pole as 1200 and (d) No. of series turns for over-compounding giving a terminal voltage of 220 V at no-load and 230 V with 100 A armature current.The open-circuit characteristic of a shunt generator when driven at normal speed is as;Field current (A)0.51.01.52.02.53.03.5O.C. volts (V)54107152185210230240The resistance of armature circuit is 0.1 Ω. Due to armature reaction the effective field current is given by the relation Ish (eff.) = Ish − 0.003 Ia. Find the shunt field circuit resistance that will give a terminal voltage of 220 V with normal speed (a) on open circuit (b) at a load current of 100 A. Also find (c) number of series turns for level compounding at 220 V with 100 A armature current; take number of shunts turns per pole as 1200 and (d) No. of series turns for over-compounding giving a terminal voltage of 220 V at no-load and 230 V with 100 A armature current.please answer only part C and D
- 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 *The armature winding resistance of a long shunt connected DC generator is 0.4ohm, the resistances of the shunt and serial field windings are 125 ohms and 0.1 ohms, respectively. While the generator supplies 8KW of power to the load it feeds, which is the voltage it produces since the output voltage is 250V?21 A 25 kW, 240 V self excited dc shunt generator has a field circuit resistance of 62.5 Ω, and an armature resistance of 0.05 Ω. When delivering rated current at rated speed and voltage, calculate % Efficiency when rotational losses are 1200W, and iron losses are 600W. Note that Copper losses must be calculated too. No Units! Please answer in typing format please ASAP for
- With the aid of mathematical expressions where necessary, explain the following terms with respect to squirrel cage induction motorsi. Stray Lossesii. Air gap Poweriii. Rotational lossesiv. Friction and Windage lossesv. Stator Copper lossesvi. Rotor Copper lossesvii. Mechanical LossesQ23) A shunt generator supplies a 20KW load at 200V through cables of resistance, R = 100mΩ. If the field winding resistance ,Rf=50Ω and the armature resistance, Ra=40mΩ, i) Estimate terminal voltage across the load. ii) Calculate the emf generated in the armature.With the aid of mathematical expressions explain the following terms with respect to squirrel cage induction motorsi. Stray Lossesii. Air gap Poweriii. Rotational lossesiv. Friction and Windage lossesv. Stator Copper lossesvi. Rotor Copper lossesvii. Mechanical Losses