Experimental methods for engineers 8th edition by holman solution manual

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 McGrawit without permission -Hill for their individual course preparation A student using this manual is using

Chapter 2

2-3

1

Amplitude ratio = Fk0 ìïïïïíỵï éêëê1 - ( ww 1n )2

ùúúûú+ éêêë2(ca c )( ww 1n

)úùúûïïïïþüýïï ê

=

amplitude ratio 0.99 (Use Figure 2-5)

F t( ) = F0 sin wt x t1 ;( ) = x0 sin(wt1 - )

time lag = t xmax - t Fmax

F t( ) = F0 = max (when sin wt1 = 1)

\ wt1 = sin- 11 = ; t Fmax

2

1

=

w1 2

ỉ1 ưỉ ưỉ t Fmax = çççè40 øè

øè÷÷÷÷çççπ2 ÷÷÷÷ççç21π ừ÷÷÷÷=

x t( ) = x0 = max (whensin(wt1 - ) = 1 \ (wt1 - ) = sin- 11 =

1 ỉçççèπ2 + ÷÷÷÷ừ

t xmax = w1

2(cc c )(ww 1n ) = tan- 1 2

= tan- 1

2 1 -

= 33.7° (Use Figure 2-6)

1 é ỉçççè180 ừ÷÷÷÷ùúúû= 0.054 sec

t xmax = 40 2êêë + 33.7

\ time lag = 0.54 - 0.000625 time lag = 0.0478sec

2-4

=

k

w1 = 0.306 which

gives

w n

w n = (100)(2 ) = 628 rad/sec

w1 = (0.306)(628) w1 = 192.1

rad/sec = 30.6 Hz

ïïỵíïï éêëê1 - ( ww 1n ) úúúûù + éêëê2(cc c

)( ww 1n ) úúûúùïïïïïïþý k ï

ï ê ê

4

For x F00 = 1.00 + 0.01 = 1.01we have ỉççççè

ww 1n ừ÷÷÷÷÷ k w1 ® imaginary

4

For x F00 = 1.00 - 0.01 = 0.99 we haveỉççççè

ww 1n ừ÷÷÷÷÷ -

2

0.04

ỉèçççç ww 1n ư÷÷÷÷÷ + 1 - çççèỉ1.011 øư÷÷÷÷2 = 0 and

ø

2

ỉèçççç ww 1n ư÷÷÷÷÷ + 1 - ççỉèç0.991÷÷÷÷ừ2 = 0

2-5

T - T¥ = e- ( RC1 )t

T - T

At t = 3sec,T = 200°F

T - T¥

= 0.435 At t

= T0 - T¥

T - T¥ =

0.1304 T0 - T¥

1 - 0.632 =

0.328 RC » 3.4

sec

2-6

P = E AB2

R

E AB = Eæççç

R ö÷÷÷÷÷ çè R +

R i ø

2

5sec, T = 270°F

P = RR i

0 ¥

1

1

R

R

R R+

ö

ç ÷ ø

è

2-7

®inch Readability

®inch Least count

2-8

t = RC = time constant t =

(106ohms)(10- 5f) = 10 sec

t = 10 sec

2-9

% error = éê(R + R i ) - R ùú´ 100 êêë

(R + R i) úúû

ç

% error = 20%

2-10

E2

P = AB ; E AB = E +RR i

E = 100 v

R = 20,000 ohms

R i = 5000 ohms

P = E2 æç R ÷÷ö2 = 104

é 2 ´ 104 ù = 0.32 Watts ê

Maximum power occurs when = 0 ® R = R i dR

P =

max E2 æçççè2PR øö÷÷÷÷2 = 2.010´ 4104 = 5000

Watts

R

When R = 1000 ohms and R i = 5000 ohms:

104 é

ê 103 ù 2

= 10 volts2 = 0.278 Watts

P = 3 ê6 ´ 103 úúû 36 ohm

10 êë

\ R = 5000 ohms

5

R Rèçç + R i ÷÷ø

2(10)4

êêë2.5´ 104

úúû dP

mx + kx = 0

® w n = k

x + k x = 0 where w n2 = k

m m m

From the static deflection: k = mg where = deflection = 0.5 cm

k g g 980 seccm 2

= ® w n = =

44.3 rad/sec

2-12

= 0.25 inch; g = 386 in/sec2

w n = g = 986 secin 2 = 39.4 rad/sec

0.25 in

2-14

w n = 39.4 rad/sec = 6.27 Hz

w x0 for c = 0

w w n F0 c c

k

2-11

60 9.57 0.011

2-15

dV

At = 0, V = 10 liters, = - 6

d

c = 0.6 hr- 1

2-16

(1 lbf/in )(4.448 N/lbf)(144 in /ft )(3.28 ft /2 2 2 2 2 m )2

= 6890 N/m2

1 lbf/in2 = (6890)(9.806) = 67570 kgf/m2 = 6.757 kp/cm2

2-17

(mi/gal)(5280 ft/mi)çççèỉ 2311 gal/in3÷

÷÷÷ừ(1728 in /ft3 3)

´ (35.313 ft /m33)ỉ

çççè1000 1 m /l3

ø÷÷÷÷ư´ (3.2808´ 10- 3 km/ft)

= 4.576 km/l

2-18

ỉ lbm ft ư

2-19

(kJ/kg·°C)çççè1.055 kJ

ø÷÷÷÷(0.454 kg/lbm) èççç9 °C/ F° ø÷÷÷÷= 0.2391 Btu/lbm· F°

(kJ/kg·°C)çççỉ4.1821 kcalkJ ø÷÷÷÷ưỉççèçç10001

kg

g ÷÷÷÷ưø= 2.391´ 10- 4 kcal/g- C° è

(lbf-s/ft2) 32.17çççè lbf s2 ø÷÷÷÷=

= 47.92 kg/ms·

(0.454 kg/lbm)(3.2808 ft/m)

7

2-20

3)æ çççè4541 lbm/gö æ

ø (g/m )(0.02832 m /ft3 3

è÷÷÷÷´ ççç32.171 slug/lbmö

ø÷÷÷÷

= 1.939 ´ 10- 6 slug/ft3

2-21

sec/hö÷

÷÷÷´

J/Btu° )ççæ 1 (Btu/h-ft- F)(1055

(107 erg/J)æ

èççç95° °F/ Cö

ø÷÷÷÷

çè3600 ø

æ çççè12 ´ 12.54 ft/cmøö÷÷÷÷

= 5.275 ´ 106 erg/s·ft· C° ´

= 1.731´ 105 erg/s·cm· C°

2-22

2

/s2 )æ çççè2.541´ 12 cmft ö÷÷÷÷ø = 1.076 ft /2 s

(cm

2-23

(W/m3) 3.413ççèçæçW·hBtu ø÷÷÷÷öèæççç3.28081 mft ö÷÷÷ø÷3 =

0.09664 Btu/h·ft3

2-24

(dyn s/cm· 2)(10- 5 N/dyn)(0.2248 lbf/N) ´ (2.54 ´ 12

cm/ft)2ççæ32.17 lbm ft2 ö÷

÷÷

= 0.0672 lbm/s ft· ´ 3600 s/h

lhm

= 241.8

h ft·

çè lbf s ø÷

2-25

W 3.413 Btu/W h·

´ c m

3

( 2.541 )2 cmin 22 ´ 1441

inft 22

W Btu/hr-ft2

cm2

2-26

R = 1545lbm molft-lbf·· R ´ 0.30480.454mftkg´

lbm ´ 9 R

2-27

cms 3ỉçççè ừ÷÷÷÷3 cmin33 ´ 2311

gal in2

´

cm3

gal/min s

2-28

R = K

J

2-29

= 105, T0 = 30 C, T¥ = 100 C

Rise time 90 = 2.303 = 23.03s

0.01 = e - t t

= 4.605

t(99 ) = 46.05 sec

2-30

A = 20 C = 0.01Hz = 0.0628 rad/s

= tan- 1 T

= tan-

1[(0.0628)(

10)]

= - 32.14 deg

0.561rad

t =

=

= 8.93sec

0.0628

2-31

c

n = 10,000 Hz

= 0.3, 0.4

c

c

c

For= 0.3, resonance

at = 0.9,

= 9000

n c

For= 0.4, resonance at = 0.8, = 8000 Hz

c c

2-32

n

c

= 0.2 and 0.4 for

c

At 2000 Hz

= 0.3

9

k

= tan- 1 éêêë(2)(0.3)(0.2)1 - 0.22 ùúúû=

7.13deg At 4000 Hz

k

= tan- 1

éêêë(2)(0.3)(0.4)

1 - 0.42 ùúúû=

15.9deg

Hz Hz

k

From Fig 2-6,

c

For = 1.0 c c

10

0.3

2-34

= - 50 = - tan- 1

= (2)(1.1918) = 2.3835

0.643

2-35

= 3 Hz = 18.85 rad/s = 0.5 sec

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

1/2 = 0.1055

21 / 2

1 [1 + 2.384]

0.3

n

>

<

21 / 2

1 [1 + 1.1918]

2-36

T0 = 35 C T¥ = 110 C (8 secT ) = 75 C

= e - t/

t

= 810.7621 = 10.497 sec 90 rise

time = 2.303 = 24.174 sec

2-38 static sens = 1.0

V/kgf

output = (10)(1.0) = 10.0 V

2-39 rise time =

0.003 ms e - t/ = e- Rc1

t = 0.1

1 = 7.86 ´ 105

RC

RC = 1.303 ´ 10- 6

R in ohm, C in farads

2-42

= 0.1sec

T t( ) = 17 C

T0 = 100 C T¥ = 15

C

= e - t/0.1

t

= 3.75

0.1

t = 0.375 sec

2-43

= 0.9

= 0.4843 to 4.84 rad/s

( ) = - tan- 1(0.4843) = - 25.84 = 0.451 rad

0.451

Dt = = 0.093 sec

4.84

11

2-44

=

500

Hz

3

0.98 = ( ) 2 ( ) ( ) 2üïïï1/2

ìïïïíïîïï êéë1 - 13 2 úûùú +

ëêêé(2) cc c 13 úûùú ýïïþï ê

c

= 0.619 c c

2-45

t = 1´ 10- 6 sec = 90% rise time

1

- (1 10´ - 6)

t = 2 hr = 3600sec = ( )

0.1 = e RC

RC = 4.34 ´ 10- 7

R in ohm, C in farads

2-46

t = 2 hr

1

cyc/hr 2 9.27 10- 5rad/sec

1

n

n

= 0.2679

0.2679

= = 3685sec = 1.024 hr

7.27 ´

Amp response = = 0.966

2-47 m = 1.3 kg k = 100

N/m

k æ100 ö1/2

n = m = çççè1.3 ø÷÷÷÷= 8.77 rad/s

c c = 2 mk = 2[(1.3)(100)]1/2 = 72.11 c

= 1.0

c c

From Figure 2-8, n t = 3.6 for 90

3.6

t = = 0.41sec

8.77

2-48

c

= 0.1

c c From Figure 2-9, n t = 3.1

3.1

t = = 0.353 sec

8.77

2-49

x t( ) c

= 0.9 = 1.5

x0 c c

From Figure 2-9 n t = 6.2,

2-50

t = 1sec, n t = 8.77 c 5.7

c = 5.7 = = c c

72.11

x

From Figure 2-9, » 1.8

x0

2-51

Rise time = 10- 12 s

6.2

t = = 0.71sec

8.77

13

0.79

2-53

T0 = 20°C T¥ = 125°C e - t/ =

0.1

= 34.54 sec

= exp

T t20( ) 125125 ỉçççè- t

ừ÷÷÷÷=

T t( ) =

20.15°C

2-54

t = 0.05 sec

e- ( )

1 = 0.02088 lbf sec/ft× m

s×

2-55

English units

= lbm/ft ,3 u = ft/sec x =

ft, = lbm/s-ft

SI units

= kg/m ,3 u = m/s

x = m, = kg/m-s

2-56

SI system

g = m/s ,2 = 1 / °C, = kg/m3

T = °C, x = m, = kg/m-s

At = 1.0 c c and x0 n t 3.6, n = 3.7 ´ 10 rad/s = 0.9,

12

f = 5.7 ´ 1011 Hz = 570 GHz

2-52

m = 1000 lbm = 2203 kg

1000 lbf

÷÷÷÷ư1/2 = 7.29 rad/s

1.5

12

0

lbf/ft

8000 117,000 N/m

117,000

2203

3.6

0.495 s

7.29

n

n c

k

k

m

c

x

t

c

x

t

ỉ

ç

=

ø

è

=

=

=

(5 9

English system

g = ft/s ,2 = 1/ F,° = lbm/ft3

T = °F, x = ft, = lbm/ft-s

2-57

W-cm 0.01 m/cm

´

in - F2 ° (2.54 cm/in.) (0.01 m/cm2)2 °C/ F° )

W-cm in - F2

´ W/m-

C°

2-58

T0 = 45, T¥ = 100 rise time = 0.2 s T(0.1s) = ?

0.2 = 2.303 , = 0.0868 s

(T - 100)/(45 - 100) = exp( 0.1/0.0868)- = 0.316 T =

82.6ºC

2-59

m = 6000/4 = 1500 lbm = 3303 kg

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

n = (263250/3303)1/2 = 8.93 rad/s

For critically damped system:

0.9 = 1 - (1 + n t)exp( - n t)

Solution is n t = 3.8901

t = (3.8901)(8.93) = 0.435s

2-60

= 400 Hz, n = 1200 Hz / n = 400/1200 = 1/3

0.98 = 1/{[1 - (1/3) ]2 2 + [(2)(c c/ c)(1/3)] }2 1/2 c c/ c = 0.619

2-61

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

2-62

t = 1.5 h, = 1/24 cyc/h = 2 /(24)(3600) = 7.27 ´ 10- 5 rad/sec

(1.5)(3600) = 5400 s = ( )/

( ) = (5400)(0.0000727) = 0.3926 = - tan- 1( )

= 0.414

15

= 0.414/7.27 ´ 10- 5 = 5695 s = 1.58 h

2-63

T0 = 45 T¥ = 100 T(6s) = 70

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

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

= 8 s, T0 = 40, T¥ = 100

Rise time = 2.303 = 18.424 s

T(99%) = 99 C

(99 - 100)/(40 - 100) = exp(- t/8) t =

32.75 s

2-66

= 5 Hz, = 0.6 s

= (2π)(5) = 31.4 rad/s

( ) = - tan- 1( ) = - 8.7°

1/[1 + ( ) ]2 1/2 = 0.053

2-67

A = 15, = 0.01 Hz = 0.0628 rad/s

( ) = - tan- 1 = - tan- 1(0.0628)(8) = - 26.7°

Attenuation = 1/[1 + ( ) ]2 1/2 = 0.894

SM: Experimental Methods for Engineers Chapter 2

13