Electric Motor Control
10th Edition
ISBN: 9781133702818
Author: Herman
Publisher: CENGAGE L
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Chapter 50, Problem 8SQ
To determine
Explain the necessity of resistor (R) in Figure 50-4(A).
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A 12-phase ∆-connected 3200 V (line-to-line), 12,000 kW, 50 Hz, twelve-pole squirrel-cage induction drive is designed for electric vehicle use. The drive has the following equivalent parameter values in Ω/phase referred to the stator: R1= 0.697 Ω R2= 0.328 Ω Rm1 = 226 Ω X1 = 0.549 Ω X2 = 0.456 Ω Xm1 = 31.573 Ω The total friction, windage, and core losses are assumed to be constant at Plosses=1156 W, independent of load.a)Draw the equivalent circuit for this drive and show the relevant parameters on the circuit.For a slip value of s = 0.032 and using the above equivalent circuit parameters, determine: b) The stator current and the power factorc) The synchronous speed and the rotor speed for this drived) The torque generatede) The input power for this operating conditionf) The efficiency when operated at rated voltage and frequencyg) Considering the design of this motor, state if it is possible to start this drive with a slip ring starter. Give reasons to justify your answer
Branch-circuit conductors supplying a single continuous-duty motor shall have an ampacity not less than _____ rating
A. 125% of the motor’s nameplate current
B. 125% of the motors full-load current as determined by 430.6(a)(1)
C. 125% of the motor’s full locked-rotor
D. 80% of the motors full-load current.
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Electric Motor Control
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- A 2,000-hp, unity power factor, three-phase,Y-connected, 2,300-V, 30-pole, 60-Hz synchronousmotor has a synchronous reactance of 1.95Ω perphase. Neglect all losses. Find the maximum powerand torque.arrow_forward9. Which motor you will use for all your domestic needs?10. Why we need to calibrate the measuring instruments?arrow_forwardA 6-phase ∆-connected 1600 V (line-to-line), 150 kW, 50 Hz, six-pole induction drive is used in an electric car. The drive has the following parameter values in Ω/phase referred to the stator: R1 =0.334 Ω R2 =0.168 Ω Rm1 = 125 Ω X1 = 0.273 Ω X2 = 0.287 Ω Xm1 = 16.78 Ω The total friction, windage, and core losses may be assumed to be constant at Prot =773 W, independent of load. For a slip value of s=0.015: a) Draw the equivalent circuitb) Compute the speed, output torque, power and stator currentc)Determine the power factor and whether leading or lagging d)Calculate the efficiency when the motor is operated at rated voltage and frequency.arrow_forward
- SOLVE THE PROBLEM, SHOW YOUR COMPLETE SOLUTION AND MAKE SURE THAT THE FINALS ANSWERS ARE IN 4 DECIMAL PLACES. A shunt generator delivers 450A at 230V and the armature resistance is 0.03 0. Calculate the shunt field resistance if the generated emf is measured at 243.6V.arrow_forwardM:47) A constant speed drive has the following duty cycle: - Load rising from 0 to 500 KW , 4 min. - Uniform load of 400 KW , 4min. - Regenerative power of 500 KW returned to the supply, 5 min. - Remains idle for 1 min. Estimate power rating of the motor. Assume losses to be proportional to (power)2.arrow_forwardTwo shunt generators and a battery are all connected to common bus-bars. The open circuit voltage of the battery is 247 V. The open circuit voltage, armature and field resistances of the generators are 250 v, 0.12 ohms and 100 ohms respectively. Both the generators supply the same current when the load on the bus-bars is 40 A. Calculate the internal resistance of the battery. ANSWER ASAP!arrow_forward
- Two shunt generators and a battery are all connected to common bus-bars. The open circuit voltage of the battery is 247 V. The open circuit voltage, armature and field resistances of the generators are 250 v, 0.12 ohms and 100 ohms respectively. Both the generators supply the same current when the load on the bus-bars is 40 A. Calculate the internal resistance of the batteryarrow_forwardTOPIC: Power Systems, Power Plants & IlluminationINSTRUCTIONS:- Answer in this format: Given, Illustration, Required Conversion, Solution, Final Answer.- Step-by-step solution, do not skip even simple calculations to avoid confusion.- If answered in written form, make sure it is readable.PROBLEM:In a group of 3 motors, one draws 15 A the other draws 40 A, and the third 52 A. What size of the conductor in terms of ampacity may be used for the feeder circuit? 107 A 120 A 134 A 160 Aarrow_forwardSHOW YOUR COMPLETE SOLUTION, ANSWER MUST BE IN 2 DECIMAL PLACES. SHOW WHAT IS GIVEN AND WHAT IS THE UNKNOWN. A 20KW short shunt generator with a terminal voltage of 250V has armature, series and shunt field winding have resistances of 0.05 ohms, 0.03 ohms and 100 ohms respectively. Calculate the total voltage generated in the armature when the brush drop is 1 V per brush and the feeder resistance is 0.5 ohm.arrow_forward
- According to the stator terms of a 3-phase Star connected 380 V, 5 Hp 50 H, 4-pole 1450 rpm induction motor, Rs= 0.3 ohm, Xs=0.5 ohm, R2= 0.15 ohm, X2=0.3 ohm. The idling power of the motor is 400 W. By drawing the equivalent circuit of the motor, find the power and efficiency taken from the motor when it is running at 1% slip and under full load.arrow_forward6-pole, 3-phase, 7.5 kW, 60 Hz, with 2% slip in a network with 220 V voltageparameters of a running ASM transferred to the stator; R1 = 0.294 Ω, X1 = 0.503Ω , R2′ =0.144 Ω, X2′ = 0.209 Ω, Xm = 13.25 Ω and stator core losses are neglected. Wind and friction losses are 400 W. Find the torque produced by the motor.arrow_forwardQ24) A shunt generator running at 1000rpm supplies 22KW at shunt field a terminal voltage of 220V.The resistance of the armature, shunt field are 0.05Ω and 110Ω respectively. The overall efficiency at the above load is 88%. i) Calculate the copper loss. ii) Calculate the Iron and friction losses.arrow_forward
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