System Dynamics
System Dynamics
3rd Edition
ISBN: 9780073398068
Author: III William J. Palm
Publisher: MCG
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 6, Problem 6.41P
To determine

(a)

The step response of ia(t) and ω(t) for the armature voltage va=10V.

Expert Solution
Check Mark

Answer to Problem 6.41P

The responses are as follows:

ia(t)=0.125e201t2cos(40399t2)+2487.6875240399e201t2sin(40399t2)+25202

ω(t)=5000101e201t2cos(40399t2)100500010140399e201t2sin(40399t2)+5000101.

Explanation of Solution

Given:

The given parameters for an armature-controlled motor are as:

Kb=KT=0.2Nm/A, c=5×104Nms/rad, Ra=0.8Ω, La=4×103H, I=5×104kgm2 and va=10V.

Concept Used:

For an armature-controlled motor, the output responses armature current ia(t) and motor speed ω(t) is related to input response armature voltage va as follows:

Ia(s)Va(s)=Is+cLaIs2+(RaI+cLa)s+cRa+KbKT

Ω(s)Va(s)=KTLaIs2+(RaI+cLa)s+cRa+KbKT.

Calculation:

Since,

Ia(s)Va(s)=Is+cLaIs2+(RaI+cLa)s+cRa+KbKT

Keeping the values of all the parameters, we get

Ia(s)Va(s)=Is+cLaIs2+(RaI+cLa)s+cRa+KbKTIa(s)Va(s)=5×104s+5×1044×103×5×104s2+(0.8×5×104+5×104×4×103)s+5×104×0.8+0.2×0.2Ia(s)Va(s)=5×104(s+1)2×106s2+(4×104+2×106)s+4×104+0.04Ia(s)Va(s)=(s+1)0.004s2+0.804s+80.8

Now, since input armature voltage is a unit step response of strength 10V.

Therefore,

Ia(s)=(s+1)(0.004s2+0.804s+80.8)Va(s)Ia(s)=(s+1)(0.004s2+0.804s+80.8)10s

On using the partial fraction expansion in order to simplify above expression, we have

Ia(s)=(s+1)s(0.004s2+0.804s+80.8)10s=As+Bs+C(0.004s2+0.804s+80.8)10(s+1)s(0.004s2+0.804s+80.8)=A(0.004s2+0.804s+80.8)+s(Bs+C)s(0.004s2+0.804s+80.8)

On comparing the numerator on both sides, we get

10s+10=(0.004A+B)s2+(0.804A+C)s+80.8A A=25202,B=0.0005,C=9.9005

Thus,

Ia(s)=As+Bs+C(0.004s2+0.804s+80.8)Ia(s)=252021s+9.90050.0005s(0.004s2+0.804s+80.8)Ia(s)=252021s0.125s+2012(s+2012)2+403994+2487.68751(s+2012)2+403994

On taking the inverse Laplace of above obtained expression, the response ia(t) is,

ia(t)=0.125e201t2cos(40399t2)+2487.6875240399e201t2sin(40399t2)+25202

Similarly, for the response ω(t),

Since,

Ω(s)Va(s)=KTLaIs2+(RaI+cLa)s+cRa+KbKT

Keeping the parameters in this transfer function,

Ω(s)Va(s)=KTLaIs2+(RaI+cLa)s+cRa+KbKTΩ(s)Va(s)=0.24×103×5×104s2+(0.8×5×104+5×104×4×103)s+5×104×0.8+0.2×0.2Ω(s)Va(s)=0.22×106s2+(4×104+2×106)s+4×104+0.04

Ω(s)Va(s)=1105s2+(2×103+105)s+2×103+0.2Ω(s)Va(s)=105s2+201s+20200

Now, since input armature voltage is a unit step response of strength 10V.

Therefore,

Ω(s)Va(s)=105s2+201s+20200Ω(s)=105(s2+201s+20200)Va(s)Ω(s)=106(s2+201s+20200)1s

On using the partial fraction expansion in order to simplify the above expression, we have

Ω(s)=106(s2+201s+20200)1s=As+Bs+C(s2+201s+20200)106s(s2+201s+20200)=A(s2+201s+20200)+(Bs+C)ss(s2+201s+20200)

Compare the denominator of both sides

Thus,

106=A(s2+201s+20200)+(Bs+C)s A=5000101,B=5000101andC=1005000101

Ω(s)=106(s2+201s+20200)1s=5000s1005000101(s2+201s+20200)+5000101s

Ω(s)=5000101s+2012(s+2012)2+4039945025001011(s+2012)2+403994+5000101s

On taking the inverse Laplace of the above obtained expression, the response ω(t) is,

ω(t)=5000101e201t2cos(40399t2)100500010140399e201t2sin(40399t2)+5000101.

Conclusion:

The responses ia(t) and ω(t) for the armature-controlled motor is,

ia(t)=0.125e201t2cos(40399t2)+2487.6875240399e201t2sin(40399t2)+25202

ω(t)=5000101e201t2cos(40399t2)100500010140399e201t2sin(40399t2)+5000101.

To determine

(b)

The step response of ia(t) and ω(t) for the load torque TL=0.2Nm.

Expert Solution
Check Mark

Answer to Problem 6.41P

The responses are follows:

ia(t)=100101e201t2cos(40399t2)2010010140399e201t2sin(40399t2)+100101

ω(t)=3.9604e201t2cos(40399t2)+3.960440399e201t2sin(40399t2)+400101.

Explanation of Solution

Given:

The given parameters for an armature-controlled motor are as:

Kb=KT=0.2Nm/A, c=5×104Nms/rad, Ra=0.8Ω, La=4×103H, I=5×104kgm2 and va=10V.

Concept Used:

For an armature-controlled motor, the output responses armature current ia(t) and motor speed ω(t) is related to input load torque TL as follows:

Ia(s)TL(s)=KbLaIs2+(RaI+cLa)s+cRa+KbKT

Ω(s)TL(s)=Las+RaLaIs2+(RaI+cLa)s+cRa+KbKT.

Calculation:

Since,

Ia(s)TL(s)=KbLaIs2+(RaI+cLa)s+cRa+KbKT

Keeping the values of all the parameters, we get

Ia(s)TL(s)=KbLaIs2+(RaI+cLa)s+cRa+KbKTIa(s)TL(s)=0.24×103×5×104s2+(0.8×5×104+5×104×4×103)s+5×104×0.8+0.2×0.2

Ia(s)TL(s)=0.22×106s2+(4×104+2×106)s+4×104+0.04Ia(s)TL(s)=1105s2+(2×103+105)s+2×103+0.2Ia(s)TL(s)=105s2+201s+20200

Now, since input load torque is a unit step response of strength 0.2Nm.

Therefore,

Ia(s)TL(s)=105s2+201s+20200Ia(s)=105s2+201s+20200TL(s)Ia(s)=2×104s(s2+201s+20200)

On using the partial fraction expansion in order to simplify above expression, we have

Ia(s)=2×104s(s2+201s+20200)Ia(s)=2×104s(s2+201s+20200)=As+Bs+C(s2+201s+20200)2×104s(s2+201s+20200)=A(s2+201s+20200)+(Bs+C)ss(s2+201s+20200)

On comparing the numerator on both sides, we get

2×104=A(s2+201s+20200)+(Bs+C)s A=100101,B=100101,C=20100101

Thus,

Ia(s)=106(s2+201s+20200)1s=100s20100101(s2+201s+20200)+100101sIa(s)=100101s+2012(s+2012)2+403994100501011(s+2012)2+403994+100101s

On taking the inverse Laplace of the above obtained expression, the response ia(t) is as shown

ia(t)=100101e201t2cos(40399t2)2010010140399e201t2sin(40399t2)+100101

Similarly, for the response ω(t),

Since,

Ω(s)TL(s)=Las+RaLaIs2+(RaI+cLa)s+cRa+KbKT

Keeping the parameters in this transfer function,

Ω(s)TL(s)=Las+RaLaIs2+(RaI+cLa)s+cRa+KbKTΩ(s)TL(s)=4×103s+0.84×103×5×104s2+(0.8×5×104+5×104×4×103)s+5×104×0.8+0.2×0.2Ω(s)TL(s)=4×103s+0.82×106s2+(4×104+2×106)s+4×104+0.04Ω(s)TL(s)=(s+200)0.0005s2+0.1005s+10.1

Now, since input load torque is a unit step response of strength 0.2Nm.

Therefore,

Ω(s)TL(s)=(s+200)0.0005s2+0.1005s+10.1Ω(s)=(s+200)0.0005s2+0.1005s+10.1TL(s)Ω(s)=0.2(s+200)s(0.0005s2+0.1005s+10.1)

On using the partial fraction expansion in order to simplify above expression, we have

Ω(s)=0.2(s+200)s(0.0005s2+0.1005s+10.1)=As+Bs+C(0.0005s2+0.1005s+10.1)0.2(s+200)s(0.0005s2+0.1005s+10.1)=A(0.0005s2+0.1005s+10.1)+s(Bs+C)s(0.0005s2+0.1005s+10.1)

Compare the denominator of both sides

Thus,

0.2s+40=(0.004A+B)s2+(0.804A+C)s+80.8A A=400101,B=0.0019,C=0.1980

Ω(s)=As+Bs+C(0.0005s2+0.1005s+10.1)Ω(s)=0.0019802s0.198020.0005s2+0.1005s+10.1+400101sΩ(s)=3.9604s+2012(s+2012)2+403994+1.98021(s+2012)2+403994+400101s

On taking the inverse Laplace of the above obtained expression, the response ω(t) is as shown

ω(t)=3.9604e201t2cos(40399t2)+3.960440399e201t2sin(40399t2)+400101.

Conclusion:

The responses ia(t) and ω(t) for the armature-controlled motor is as follows:

ia(t)=100101e201t2cos(40399t2)2010010140399e201t2sin(40399t2)+100101

ω(t)=3.9604e201t2cos(40399t2)+3.960440399e201t2sin(40399t2)+400101.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
##2# Superheated steam powers a steam turbine for the production of electrical energy. The steam expands in the turbine and at an intermediate expansion pressure (0.1 Mpa) a fraction is extracted for a regeneration process in a surface regenerator. The turbine has an isentropic efficiency of 90% Design the simplified power plant schematic Analyze it on the basis of the attached figure Determine the power generated and the thermal efficiency of the plant ### Dados in the attached images
### To make a conclusion for a report of an experiment on rockets, in which the openrocket software was used for the construction and modeling of two rockets: one one-stage and one two-stage. First rocket (single-stage) reached a maximum vertical speed of 100 m/s and a maximum height of 500 m The second rocket (two-stage) reached a maximum vertical speed of 50 m/s and a maximum height of 250 m To make a simplified conclusion, taking into account the efficiency of the software in the study of rockets
Determine the coefficients of polynomial for the polynomial function of Cam profile based on the boundary conditions shown in the figure. S a 3 4 5 C₁ (+) Ꮎ В s = q + q { + c f * + q € * + q ( +c+c+c 6 Ꮎ +C5 +C β В В 0 cam angle 0 B 7 (

Chapter 6 Solutions

System Dynamics

Knowledge Booster
Background pattern image
Mechanical Engineering
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Elements Of Electromagnetics
Mechanical Engineering
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Oxford University Press
Text book image
Mechanics of Materials (10th Edition)
Mechanical Engineering
ISBN:9780134319650
Author:Russell C. Hibbeler
Publisher:PEARSON
Text book image
Thermodynamics: An Engineering Approach
Mechanical Engineering
ISBN:9781259822674
Author:Yunus A. Cengel Dr., Michael A. Boles
Publisher:McGraw-Hill Education
Text book image
Control Systems Engineering
Mechanical Engineering
ISBN:9781118170519
Author:Norman S. Nise
Publisher:WILEY
Text book image
Mechanics of Materials (MindTap Course List)
Mechanical Engineering
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Cengage Learning
Text book image
Engineering Mechanics: Statics
Mechanical Engineering
ISBN:9781118807330
Author:James L. Meriam, L. G. Kraige, J. N. Bolton
Publisher:WILEY
Ch 2 - 2.2.2 Forced Undamped Oscillation; Author: Benjamin Drew;https://www.youtube.com/watch?v=6Tb7Rx-bCWE;License: Standard youtube license