Experimental Methods for Engineers · J. P. Holman Professor of Mechanical Engineering Southern...

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

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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

x 0 = 1

Amplitude ratio = F0 ìïïïïïî éêêë 1 - ( ww1n )2 úûùúú+ êéêë2(cac )( ww1 n )ûùúúþýïüïïïïï

k ïí ê

=

amplitude ratio 0.99 (Use Figure. 2-5)

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

1

time lag = txmax - tFmax

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

\ wt1 = sin- 11 = ; tFmax

2

1

= w1

2

æ1 öæ tFmax = çççè40

øè÷÷÷÷ççç

π2

öæøè

÷÷÷÷ççç2

1π

öø

÷÷÷÷= 0.00625 sec

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

1 æ txmax =

w çççèπ

2 +

÷÷÷÷

ö ø

1

2(cc )(w1 )

= tan- 1 c wn2 = tan- 1 2

1

= 33.7° (Use Figure 2-6)

1 é æ öù

txmax = 40 êêë 2 + 33.7çççè180 ø

÷÷÷÷û úú= 0.054 sec

\ time lag = 0.54 - 0.000625 time lag = 0.0478

sec

2-4

1 x 0

=

F0

ìï 2 2 üï

1 2 2

SM: Experimental Methods for Engineers Chapter 2

2

k w 1 = 0.306. which

gives

wn

wn = (100)(2 ) = 628 rad/sec w1

= (0.306)(628) w1 = 192.1

rad/sec = 30.6 Hz

ïïí éêê 1 - ( ww1n ) ùúúûú + ëêéêê2(ccc )( ww1 n ) ùúúûúïýïïïþïï k ïïïïî ê ë

4

For xF00 = 1.00 + 0.01 = 1.01we have æçèççç

ww1n öø÷÷÷÷÷ k w1 ® imaginary.

4

For xF00 = 1.00 - 0.01 = 0.99 we haveæçè ççç

ww1n ö ø÷÷÷÷÷ -

2

çç

0.04æè çç ww1n öø÷÷÷÷÷ + 1

çççèæ1.011 ø ö÷÷ ÷÷2 = 0 and

2

0.04ç

æèççç ww1n ö ø÷÷÷÷÷ + 1

çççèæ0.99 1 ÷÷÷÷öø 2 = 0


3

2-5

At t = 3sec,T = 200°F

T - T ¥ = 0.435. At t = 5sec, T = 270°F T0 -

T¥

T - T ¥ = 0.1304

T0 - T¥

1 - 0.632 = 0.328 RC

» 3.4 sec

2-6

EAB 2

P =

R

( ) 1

0

RC t T T e

T T

- ¥

¥

- =

-


4

æ R

ö ÷÷ çç

EAB = E çèç R + Ri ø÷÷÷

2-7

Readability ® inch

Least count ® inch

2-8

t = RC = time constant t =

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

t = 10 sec

2-9

= é(R + Ri) - R ùú´ 100 % error

ê

ê (R + Ri)

úúû êë

= ´ 100

2

1

1 i

i i

R R

R R R R

P

+

æ ö ÷ ç ÷ ç ÷ ç = ÷ ç ÷ ç ÷ ÷ ç ÷ ø è


5

R çççèR + Ri ÷÷ø 2(10)4 êêë2.5´

104 úúû dP

Maximum power occurs when = 0 ® R = Ri dR

E2 æçççè2PR øö÷÷÷÷2 = 2.010´ 4 104 = 5000 Watts

Pmax =

R

When R = 1000 ohms and Ri = 5000 ohms:

=

P 101043 éêêêë6 ´10103

3 ùú úúû

2 = 1036

voltsohm 2 = 0.278 Watts

2-11

\ R = 5000 ohms

mx + kx = 0

x + k x = 0 where wn2 =

k ® wn =

k m m

m

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

k = g ® wn = g = 980 seccm 2

m 0.5 cm wn =

44.3 rad/sec

% error = 20 %

2-10

E2

P = AB ; EAB = E +R Ri

R R

E = 100 v

R = 20,000 ohms

Ri = 5000 ohms

E2 æ

P = ç R ÷÷ö2 = 104

é 2 104 ù

ê ´ ú= 0.32 Watts


6

2-12

= 0.25 inch; g = 386 in/sec2

wn = g = 986 secin.2 = 39.4 rad/sec

0.25 in.

2-14

wn = 39.4 rad/sec = 6.27 Hz

w x 0 for c = 0

w wn F0 cc

k

20 3.19 0.108

40 6.38 0.025

60 9.57 0.011

d V0

dV

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

c = 0.6 hr- 1

2-16

(1 lbf/in2)(4.448 N/lbf)(144 in2/ft2)(3.282 ft2/m2) = 6890 N/m2

1 kgf = 9.806 N

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

2-17

æ1 ö

(mi/gal)(5280 ft/mi)çççè231 gal/in3÷÷÷÷ø(1728 in3/ft3)

´ (35.313 ft3/m3)æ

çç 1

m3/l÷÷÷÷

ö´ (3.2808´ 10- 3 km/ft) = 4.576

km/l

çè1000 ø

2-18

2-15

dV

= - cV

V = e - c


7

(lbf-s/ft

2)æçççè32.17 lbmlbf sft2 øö÷÷÷÷= 32.17 lbm/sft · ´ (0.454

kg/lbm)(3.2808 ft/m)

= 47.92 kg/ms·

2-19

(kJ/kg·°C)çççèæ1.0551 BtukJ øö ÷÷÷÷(0.454 kg/lbm) èæççç9 5 °C/°Fö

ø÷÷÷÷= 0.2391 Btu/lbm·° F

æ 1 kcal (kJ/kg·°C)ççèç4.182 kJ

ø÷÷÷÷öæççççè10001 kgg ÷÷÷÷öø = 2.391´ 10- 4 kcal/g-°C

2-20

(g/m3)(0.02832 m3/ft3)æ

çççè4541

lbm/göø÷÷÷

÷´ æè ç

çç32.17 1

slug/lbmöø ÷÷÷÷

= 1.939 ´ 10- 6 slug/ft3

2-21

(Btu/h-ft-°F)(1055 J/Btu)çççèæ36001 sec/høö ÷÷÷÷´ (10

7

erg/J)èæ

ççç9

5 °F/°Cöø

÷÷÷ ÷

= 5.275 ´ 106 erg/s·ft·°C ´ çæççè12 ´

12.54 ft/cmøö ÷÷

÷÷

= 1.731´ 105 erg/s·cm·°C

2-22 2

(cm2/s)æ

çççè 2.541´ 12 cm

ft ö÷÷÷÷ø = 1.076 ft2/s

2-23

(W/m3)çèççæ3.413 W·hBtu ø÷÷÷÷öèæççç3.28081

mft ö÷÷÷ø÷3 = 0.09664 Btu/h·ft3 ç


8

2-24

(dyn·s/cm2)(10

- 5 N/dyn)(0.2248 lbf/N) ´ (2.54 ´ 12 cm/ft)

2æçç32.17

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

´ cm 3 ( 2.541 )2

cm in22 ´ 1441 inft2 2

W Btu/hr-ft 2

´ =

cm 2

2-26

R = 1545

lbmft-lbf·mol· R ´ 0.30480.454mftlbmkg´ ´4.44895

ΚR lbfN =

2-27

´ cms 3 æçççè

öø÷÷÷÷3 cmin33 ´ 2311

gal in2

cm3

´ = gal/min s

2-28

R = K

J

8305 kg mol ·

K

2-29

= 105, T0 =

30 C, T¥

= 100 C

Rise time 90

= 2.303 = 23.03

s

0

.

0

1

=

e

-

t

t


9

= 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

= - 0.561 rad

t = = = 8.93 sec

0.0628

2-31

c

n = 10,000 Hz = 0.3, 0.4

c

c c

For = 0.3, resonance at = 0.9, = 9000 Hz cc

n c

For

= 0.4,

resonance at = 0.8, = 8000 Hz

xF00 = =

1.034

êë 1 -

0.22 úû

At 4000 Hz

x F00 =

= 1.145 k

= t

a

n- 1

éê(

2

)

(

0

.

3

)

(

0

.

4

)

ùú= 1

5

.

9

d

e

g

ê

ë 1

- 0

.

4

2

cc

2-32

n

c

= 0.2 and 0.4 for

n c

c

At 2000 Hz

= 0.3

k

= tan- 1 éê(2)(0.3)(0.2) ùú= 7.13deg


10

úû


11

2-33

For = 1.0 cc

n = 33 Hz

0.3

2-34

= - 50 = - tan- 1

= 1.1918

= (2)(1.1918) = 2.3835

At amp response =

0.643

At amp response =

0.387

2-35

= 3 Hz = 18.85 rad/s = 0.5 sec

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

1 / 2 = 0.1055

2-36

T0 = 35 C T¥ = 110 C T(8 sec) = 75 C

= e- t/

t

= 0.7621

= 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

x0 = 0.4 F0

k

From Fig. 2-6,

c

1 = 10 Hz

2 = 50 Hz

0.3

10

n

>

<

2 1 2 /

2 1 2 /

1

[1 1.1918 ]

1

[1 2.384 ]

+

+


12

2-39 rise time =

0.003 ms e- t/ = e-

Rc1 t = 0.1

1 = 7.86 ´ 10 5

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

2-44

= 500

Hz = 1500 Hz = 1

0.98 = 2 2üï1 / 2

ìïïïí ï éêê 1 - (13 )2 úúûù +

êëéê(2)(ccc )(13 )ûùúú ïïïïþ ý

ïïïî ë

c = 0.619 cc

2-45

3

1

n n


13

t = 1´ 10- 6

sec = 90% rise time

1

- (1´ 10- 6)

24 (24)(3600)

t = 2 hr = 3600sec = ( )

= 0.2679

0.2679

= = 3685sec = 1.024 hr

2-47 m = 1.3 kg k = 100

N/m

m k æçççè1001.3 øö÷÷÷÷

1 / 2 = 8.77

rad/s

1 .27 ´

Amp. response = = 0.966

0.1 = e RC

RC = 4.34 ´ 10- 1

R in ohm, C in farads

2-46

t = 2 hr

1

cyc/hr 2 9.27 10- 5

rad/sec

= = = ´

( ) = 0.2618 rad = - tan( )


14

n = =

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

= 1.0

cc

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

3.6

t = = 0.41 sec

8.77

2-48

c

= 0.1 cc

From Figure 2-9, nt = 3.1.

3.1

t = = 0.353 sec

8.77

2-49

x(t) c

= 0.9 = 1.5

x0 cc

6.2

From Figure 2-9 nt = 6.2, t = = 0.71 sec.

8.77

2-50

t = 1sec, nt = 8.77 c

5.7

c = 5.7 = = 0.79 cc

72.11 x

From Figure 2-9, » 1.8.

At = 1.0 and = 0.9, nt

3.6, n = 3.7 ´ 1012 rad/s cc x0

x0

2-51

Rise time = 10- 12 s c

x


15

f = 5.7 ´ 1011 Hz = 570 GHz

2-52

m = 1000 lbm = 2203 kg k = = 8000

lbf/ft = 117,000 N/m

n = ÷÷÷÷ö1 / 2 = 7.29 rad/s

At = 1.0 n t » 3.6

t =

1.5 12

0

1000 lbf

117,000

2203

0.9

3.6 0.495 s

7.29

c

k

m

x c

x c

æ ç = ç ç ø è

=

=


16

( 5 9

2-53

T0 = 20°C T¥ = 125°C e- t/ =

0.1

= 34.54 sec

T20(t)-- 125 125 = exp

æçççè- t öø÷÷÷÷=

T(t) = 20.15°C

2-54

t = 0.05 sec

e - ( )

kg 2

1 = 0.02088 lbf ×sec/ft m×s

2-55

English units

= lbm/ft3, u = ft/sec x =

ft, = lbm/s-ft

SI units

= kg/m3, u = m/s

x = m, = kg/m-s

2-56

SI system

g = m/s2, = 1 / °C, = kg/m

3

T = °C, x = m, = kg/m-s

English system

g = ft/s2, = 1/°F, = lbm/ft

3

T = °F, x = ft, = lbm/ft-s

2-57

W-cm 0.01 m/cm

´

in 2-°F (2.54 cm/in.)

2(0.01 m/cm)

2 °C/°F)

W- cm in2- F ´

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 = 82.6

ºC

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 + nt)exp( - nt)

Solution is nt = 3.8901

t = (3.8901)(8.93) = 0.435 s

2-60

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

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

= 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/ )

= 7.61s

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


18

= (2π)(5) = 31.4 rad/s - 1( )

= - 8.7°

( ) = - tan

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


13

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Experimental Methods for Engineers · J. P. Holman Professor of Mechanical Engineering Southern...

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