A 100hp, 250 V, 1200 rpm shunt DC motor with an armature resistance of 0.03 Ω and a field resistance of 41.67 Ω. The motor has compensating windings, so armature reaction can be ignored. Mechanical and core losses may be assumed to be negligible for the purposes of this problem. The motor is assumed to be driving a load with a line current of 126 A and an initial speed of 1103 rpm. To simplify the problem, assume that the amount of armature current drawn by the motor remains constant. what is the motor's speed if the field resistance is raised to 50 Ω? can you help me to find N2?

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...
icon
Related questions
Question

A 100hp, 250 V, 1200 rpm shunt DC motor with an armature resistance of 0.03 Ω and a field resistance of 41.67 Ω. The motor has compensating windings, so armature reaction can be ignored. Mechanical and core losses may be assumed to be negligible for the purposes of this problem. The motor is assumed to be driving a load with a line current of 126 A and an initial speed of 1103 rpmTo simplify the problem, assume that the amount of armature current drawn by the motor remains constant.

what is the motor's speed if the field resistance is raised to 50 Ω?

can you help me to find N2?

 

Pour= 100hp
V = 250V
N = 1200rpm
RA = 0.032
Rei = 41.672
ILI = 126A
N = 1103 rpm
R2 = 50 2
I= 120A
I2= 125A
Ip = V + Re1 = 250 + 41.67 = 6A
In = ILI -IF= 126 - 6 = 120A (constant)
En = Vr - I(RA) = 250 - 120(0.03) = 246.4V
I2 = Vr + Re2 = 250 - 50 = 5A|
I2=I, + I2 = 120 + 5 = 125A
E12= Vr - I(R) = 250 - 120(0.03) = 246.4V
EAI = kON = 246.4 = ko = 1103.
-(1)
E12= KON2 = 246.4 = ko = (N2).(2)
%3D
Transcribed Image Text:Pour= 100hp V = 250V N = 1200rpm RA = 0.032 Rei = 41.672 ILI = 126A N = 1103 rpm R2 = 50 2 I= 120A I2= 125A Ip = V + Re1 = 250 + 41.67 = 6A In = ILI -IF= 126 - 6 = 120A (constant) En = Vr - I(RA) = 250 - 120(0.03) = 246.4V I2 = Vr + Re2 = 250 - 50 = 5A| I2=I, + I2 = 120 + 5 = 125A E12= Vr - I(R) = 250 - 120(0.03) = 246.4V EAI = kON = 246.4 = ko = 1103. -(1) E12= KON2 = 246.4 = ko = (N2).(2) %3D
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 5 steps

Blurred answer
Knowledge Booster
Three phase Induction Motor
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, electrical-engineering and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
Electrical Engineering
ISBN:
9780133923605
Author:
Robert L. Boylestad
Publisher:
PEARSON
Delmar's Standard Textbook Of Electricity
Delmar's Standard Textbook Of Electricity
Electrical Engineering
ISBN:
9781337900348
Author:
Stephen L. Herman
Publisher:
Cengage Learning
Programmable Logic Controllers
Programmable Logic Controllers
Electrical Engineering
ISBN:
9780073373843
Author:
Frank D. Petruzella
Publisher:
McGraw-Hill Education
Fundamentals of Electric Circuits
Fundamentals of Electric Circuits
Electrical Engineering
ISBN:
9780078028229
Author:
Charles K Alexander, Matthew Sadiku
Publisher:
McGraw-Hill Education
Electric Circuits. (11th Edition)
Electric Circuits. (11th Edition)
Electrical Engineering
ISBN:
9780134746968
Author:
James W. Nilsson, Susan Riedel
Publisher:
PEARSON
Engineering Electromagnetics
Engineering Electromagnetics
Electrical Engineering
ISBN:
9780078028151
Author:
Hayt, William H. (william Hart), Jr, BUCK, John A.
Publisher:
Mcgraw-hill Education,