Solution manual for experimental methods for engineers 8th edition by holman

Instructor’s Solutions Manual to accompany Experimental Methods for Engineers Eighth Edition J.. Holman Professor of Mechanical Engineering Southern Methodist University Boston Bur

Instructor’s Solutions Manual

to accompany

Experimental Methods

for Engineers

Eighth Edition

J P Holman

Professor of Mechanical Engineering Southern Methodist University

Boston Burr Ridge, IL Dubuque, IA Madison, WI New York San Francisco St Louis

Bangkok Bogotá Caracas Lisbon London Madrid Mexico City Milan New Delhi Seoul Singapore Sydney Taipei Toronto

PROPRIETARY MATERIAL © 2012 The McGraw-Hill Companies, Inc All rights reserved No part of this Manual may be displayed,

reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation A student using this manual is using

it without permission.

Chapter 2

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2-3

( ) ( )( )

0

0

2

2 2 2 1 2

1 Amplitude ratio

1

{[1 (0.4) ] [2(0.7)(0.4)] }

F

k

x

=

amplitude ratio  0.99 (Use Figure 2-5)

max max

max

1

1

time lag

1

F

F t F w t x t x w t

w



max

1

1 π 1

0.00625sec

40 2 2π

π ( ) max (when sin( ) 1 ( ) sin 1

2

-æ ö-æ ö-æ ö÷ ÷ ÷

ç ç ç

= çç ÷÷÷çç ÷÷÷çç ÷÷÷=

è øè øè ø

F

t

( )( )

( )

max

1

1

1

1 π

2

2

2(0.7)(0.4)

1 (0.4) 1

33.7 (Use Figure 2-6)

ç

= ççè + ÷÷÷ø

-= °

n

x

w c

w w

t





max

1

33.7 0.054 sec

40 2 180

time lag 0.54 0.000625

time lag 0.0478 sec

é æ öù

ç

= êë + ççè ÷÷÷øúû=

-=

x

2-4

( ) ( )( )

0

0

0

0

1 2 2

1

0

1

1

1.01 imaginary

F k

F

k

F

k

x

w

x

=

®

1

1

0.99

w

w

=

1

(100)(2 ) 628 rad/sec

(0.306)(628)

192.1 rad/sec 30.6 Hz

n

w

w

w



=

2-5

( 1 )

0

RC t

T T

e

T T

-¥

¥

-=

-0

0

At 3sec, 200 F

0.435 At 5sec, 270 F

0.1304

1 0.632 0.328

3.4 sec

RC

¥

¥

¥

¥

-=

»

2-6

2

2

1

i i

AB

AB

i R

R

E

P

R

R

R R

P

+

=

= çççè + ÷÷÷ø

ç

÷÷

2-7

1 Readability inch

64 1 Least count inch

32

®

®

2-8

time constant

(10 ohms)(10 f ) 10 sec

10 sec

t RC

t

t

-= =

=

2-9

( )

( )

5000

100

25, 000

% error 20%

i i

é + - ù

+

=

2-10

2

2

;

100 v

20,000 ohms

5000 ohms

10 2 10

0.32 Watts 2(10) 2.5 10

+

=

=

=

= çç ÷÷ = ê ú=

ç +

è ø êë ´ úû

AB

AB

i

i

i

E

R

R

P

2

2

10

5000 Watts

0.278 Watts

36 ohm

i

i

dP

dR

E P

P

R R

P

æ ö÷

ç

2-11

2

2

cm sec

0

0 where

From the static deflection: where deflection 0.5 cm

980 0.5 cm 44.3 rad/sec

n

n

mx kx

k mg

w

m

w

=

2-12

2

2 in.

sec

0.25 inch; 386 in/sec

986

39.4 rad/sec 0.25 in

n

g

g

w





2-14

0

0

39.4 rad/sec 6.27 Hz

for 0

20 3.19 0.108

40 6.38 0.025

60 9.57 0.011

n

F

k

w

x

w

=

2-15

0

1

At 0, 10 liters, 6

0.6 hr

c

dV V

d c









-=

2-16

(1 lbf/in ) (4.448 N/lbf )(144 in /ft ) (3.28 ft /m ) 6890 N/m

1 kgf 9.806 N

1 lbf/in (6890)(9.806) 67570 kgf/m 6.757 kp/cm

=

=

2-17

1 (mi/gal)(5280 ft/mi) gal/in (1728 in /ft )

231 1 (35.313 ft /m ) m /l (3.2808 10 km/ft) 4.576 km/l

1000

ç

2-18

2

2

lbm ft (lbf-s/ft ) 32.17 32.17 lbm/s ft (0.454 kg/lbm)(3.2808 ft/m)

lbf s

47.92 kg/m s

=

·

·

2-19

4

4.182 kJ 1000 g



2-20

(g/m )(0.02832 m /ft ) lbm/g slug/lbm

454 32.17 1.939 10- slug / ft

2-21

7

6

5

(Btu/h-ft- F)(1055 J/Btu) sec/h (10 erg/J) F/ C

1 5.275 10 erg/s·ft · C ft/cm

12 2.54 1.731 10 erg/s·cm· C

° çç ÷÷÷´ çç ° ° ÷÷÷

ç

= ´ ° ´ ççè ´ ÷÷÷ø

2-22

2

2-23

3

(W /m ) 3.413 0.09664 Btu/h ·ft

W ·h 3.2808 ft

2-24

2

lbm ft (dyn s/cm )(10 N/dyn)(0.2248 lbf/N) (2.54 12 cm/ft) 32.17

lbf s 0.0672 lbm/s ft 3600 s/h

lhm

241.8

h ft

=

·

·

·

2-25

2 2

1 in 1

2.54 cm 144in

2 2

3.413 Btu/W h

cm

Btu/hr-ft cm

W

W 

´

´

·

2-26

N m

ft lbf

kg 5 Κ lbm 9 R

R





´

´

2-27

3

3

cm

gal/min s







ç

´ ççè ÷÷÷ø ´

2-28

9

5

2-29

0

Rise time 90 2.303 23.03s

0.01

4.605

(99 ) 46.05 sec

t

e

t

t













¥

=

=

=

2-30

1

1

tan

tan [(0.0628)(10)]

32.14 deg

0.561 rad

0.561

8.93 sec 0.0628

T

t









  



 



=

=

=

=

2-31

10, 000 Hz 0.3, 0.4

For 0.3, resonance at 0.9, 9000 Hz

For 0.4, resonance at 0.8, 8000 Hz

n

c

c c c

c

c

c







2-32

0

0

0

1

2

0

2 2 2 1/2

1

2

0.2 and 0.4 for 0.3

At 2000 Hz

1

1.034 {(1 0.2 ) [(2)(0.3)(0.2)] }

(2)(0.3)(0.2)

At 4000 Hz

1

1.145 {[1 0.4 ] [(2)(3)(0.4)] }

(2)(0.3)(0.4)

tan

F

k

F

k

c c

x

x









é

=

ù

2-33

0

1 0

2

10 Hz 0.4

50 Hz From Fig 2-6,

10

33 Hz 0.3

F

k

n

x

c

c











=

=

=

< =

2-34

1

50 tan

1.1918

(2)(1.1918) 2.3835

 

 









   

-= - =

-=

2 1/2

2 1/2

1

[1 1.1918 ] 1

[1 2.384 ]









= +

= +

2-35

1/2

3 Hz 18.85 rad/s 0.5 sec

( ) tan 1[(18.85)(0.5)] 8.39

0.1055



 



    

-=

2-36

0

/

35 C 110 C (8 sec) 75 C

0.7621

810.7621 10.497 sec

90 rise time 2.303 24.174 sec

t

e

t











¥

-=

-=

2-38

static sens 1.0 V/kgf

output (10)(1.0) 10.0 V

=

2-39

1

/

5 6

rise time 0.003 ms

0.1 1

7.86 10

1.303 10

Rc t

t

RC

RC

-=

2-42

0 / 0.1

3.75

0.1

0.375 sec

t

T t

e

t

t



¥

=

-=

-=

=

2-43

2 1/2

1

1

0.9

0.451

0.093 sec 4.84

t





-=

+

=

2-44

1/2

2

1

500 Hz 1500 Hz

3 1

0.98

0.619

c

n

n

c c

c

c

c





=

ïê - ú + ê úï

ïêë úû êë úûï

=

2-45

6

1

(1 10 6)

7

1 10 sec 90% rise time

0.1

4.34 10

in ohm, in farads

RC

t

e

RC

´

=

2-46

5

2 1/2

2 hr

( )

2 hr 3600 sec

0.2679

0.2679

3685sec 1.024 hr 7.27

1

[1 ( ) ]

t

t



























-=

-=

´

+

2-47

1/2

1/2

1.3 kg 100 N/m

100

8.77 rad / s 1.3

1.0

From Figure 2-8, 3.6 for 90

3.6

0.41 sec

8.77

ç

= = ççè ÷÷÷ø =

=

=

n

c

c

n

k

m

c

c

t

t



2-48

0.1

From Figure 2-9, 3.1

3.1

0.353 sec

8.77

=

=

c

n

c

c

t

t



2-49

0

( )

0.9 1.5

6.2 From Figure 2-9 6.2, 0.71 sec

8.77

c

n



2-50

0

5.7

72.11 From Figure 2-9, 1.8

»

n

c

c

c

c

x x



2-51

12

12 0

11

Rise time 10 s

At 1.0 and 0.9, 3.6, 3.7 10 rad/s

5.7 10 Hz 570 GHz

c

t

f

-=

2-52

1.5

12

1/2

1000 lbm 2203 kg

1000 lbf

8000 lbf/ft 117, 000 N/m

117, 000

7.29 rad/s 2203

n

m

k

k

m

t





ç

2-53

(0.05)

34.54

0

/

0.1

34.54 sec

( ) 125

exp

( ) 20.15 C

t

e

e

T t







¥

=

=

÷÷

2-54

2

kg

1 0.02088 lbf sec/ft

×

2-55

3

English units

lbm/ft , ft/sec

ft, lbm/s-ft

u

x





3

SI units

kg/m , m/s

m, kg/m-s

u

x





2-56

SI system

m/s , 1 / C, kg/m

C, m, kg/m-s

English system

ft/s , 1/ F, lbm/ft

F, ft, lbm/ft-s

g

g









= ° = =

= ° = =

2-57

9 2

in - F (2.54 cm/in.) (0.01 m/cm) C/ F

- cm

W/m- C

in - F

W

W

 



´

2-58

0 45, 100 rise time 0.2 s (0.1s) ?

0.2 2.303 , 0.0868 s

( 100) /(45 100) exp( 0.1/ 0.0868) 0.316

82.6ºC

T

T

 

¥

=

2-59

1/ 2

6000/4 1500 lbm 3303 kg

1500/(1/12) 18, 000 lb/ft 263, 250 N/m

(263250/3303) 8.93 rad/s

n

m

k



For critically damped system:

Solution is 3.8901

(3.8901)(8.93) 0.435 s

-=

n

t

t



2-60

400 Hz, 1200 Hz / 400/1200 1/3

0.98 1/{[1 (1/3) ] [(2)( / )(1/3)] }

/ 0.619

c c

c c

c c

=

2-61

Insert function in Equations (2-38) and (2-39), manipulate algebra and the indicated result will be given

2-62

5

1

5

1.5 h, 1/24 cyc/h 2 /(24)(3600) 7.27 10 rad/sec

(1.5)(3600) 5400 s ( )/

0.414

  





-=

2-63

0 45 100 (6s) 70

(70 45)/(100 45) exp( 6/ )

7.61 s

For 90% rise time exp( /7.61) 0.1

Rise time 17.52 s







¥

-=

=

2-64

Insert function in Equations (2-38) and (2-39), manipulate algebra to give the indicated result

2-65

0

8 s, 40, 100

Rise time 2.303 18.424 s

(99%) 99 C

(99 100)/(40 100) exp( /8)

32.75 s

T

t t







¥

=

-=

2-66

1

2 1/ 2

5 Hz, 0.6 s

(2π)(5) 31.4 rad/s

( ) tan ( ) 8.7

1/[1 ( ) ] 0.053





= - = - °

2-67

15, 0.01 Hz 0.0628 rad/s