A dc compound motor having a rating of 10 kW,1150 r/min, 230 V, 50 A, has the following losses atfull-load:bearing friction loss40 Wbrush friction loss==50 Wwindage loss=200 W(1)total mechanical losses=290 W(2)iron losses=420 W(3)copper loss in the shunt field=120 Wcopper losses at full-load:(4)a. in the armatureb. in the series fieldc. in the commutating windingtotal copper loss in the500 W25 W70 Warmature circuit at full-load=595 W Calculate the efficiency of the motor in Ex-ample 6-2 when it delivers an output of 1 hp.

Answered: Image /qna-images/answer/1acaf43b-9df7-4f26-95bc-ce79ee4495cd.jpg

Answered: A dc compound motor having a rating of…

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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A dc shunt generator delivers 50 kw at 250 v, its armature has a 4-pole wave winding with a resistance of 0.01 ohm and its field resistance is 125 ohm. if the number of turns in the armature is 201 and the flux per pole is 0.02 wb, find the voltage generated and the speed of operation. *
5. Calculate the flux per pole required on full-load for a 50 kW, 400 V, 8-pole, 600 r.p.m. d.c. shunt generator with 256 conductors arranged in a lap-connected winding. The armature winding resistances is 0.1 Ω, the shunt field resistance is 200 Ω and there is a brush contact voltage drop of 1 V at each brush on full-load. Subject Electrical Machine 1
Question 43
T-a. The direction of rotation of a single-phase split-phase motor can be change by........ a) reversing the connection of the start winding b) reversing the connection of the run winding c) reversing the connection of the start and run winding at the same time d) "a" or "b", not "c" T-b. Single-phase split-phase motors using an electrolytic capacitor in the start winding circuit should not be cycled more than about__________ time per hour. a) 8 b) 12 c) 24 d) none of the above T-c. When replacing the capacitor in a capacitor-start induction-run motor, one must be sure to replace the bad capacitor with__________ . a) a smaller one b) one of the same size c) a larger one d) none of the above
Please solve this problem step by step and illustrate your answer properly. Write your answer in a piece of paper please. Box the final answer.
A 120 V shunt generator running at 850 rpm has its armature and shunt field resistance of 0.15 ohm and 50 ohm respectively. It supplies 200 lamps each rated at 60 W, 100 V. The friction and windage and core loss of the machine is 400 W. its armature copper * loss on full load and shunt field loss is 2156.7 W, 240 W 2232.6 W, 240 W O 2232.6 W, 200 W 2156.7 W, 200 W O
Calculate the flux per pole required on full-load for a 50 kW, 400 V, 8-pole, 600 r.p.m. d.c. shunt generator with 256 conductors arranged in a lap-connected winding. The armature winding resistances is 0.1 Q, the shunt field resistance is 200 Q and there is a brush contact voltage drop of 1 V at each brush on full- [0.162 Wb] 5. load. Calculate the flux in a 4-pole dynamo with 722 armature conductors generating 500 V when running at 1000 r.p.m. when the armature is (a) lap connected (b) wave connected. [(a) 41.56 mWb (b) 20.78 mWb] (City & Guilds, London) 7. A 4-pole machine running at 1500 r.p.m. has an armature with 90 slots and 6 conductors per slot. The flux per pole is 10 mWb. Determine the terminal e.m.f. as d.c. Generator if the coils are lap-connected. If the current per conductor is 100 A, determine the electrical power. [810 V, 324 kW] (London Univ.) 8. An 8-pole lap-wound d.c. generator has 120 slots having 4 conductors per slot. If each conductor can carry 250 A and if…
Here it is the equivalent circuit of and induction motor and it is per phase circuit my question is name for me the variable for example R1 is the winding resistance and so an the all variable and please show me a pic of the real machine with the variable for example for R1 happens because of the winding i need to see the winding in the real machine, please help me to understand well
In a DC generator, the iron losses at 1000 rpm are 10 kW at a given field current. At a speed of 750 rpm and at the same field current, the total iron losses become 6 kW. Assume the hysteresis loss is directly proportional to the speed and the eddy current loss is proportional to the square of the speed. Determine the iron losses at 500 rpm.

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