experimental methods for engineers - solution manual testbank … · 2018-09-20 · instructor’s...

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,

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using this manual is using

Chapter 2 2-3

x0 = 1

1

Amplitude ratio = Fk0 ìïïïïïíîï éêëê1 - ( ww1n )2

ùúúûú+ éêêë2(cac )( ww1n

)úùúûïïïïþüýïï ê

=

amplitude ratio 0.99 (Use Figure. 2-5)

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

time lag = txmax - tFmax

F t( ) = F0 = max (when sin wt1 = 1) \ wt1 = sin- 11 = ; tFmax

2

1

=

w1 2

æ1 öæ öæ tFmax = çççè40 øè

øè÷÷÷÷ççç

π2

÷÷÷÷ççç2

1π

öø

÷÷÷÷=

0.00625sec

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

1 æçççèπ2 + ÷÷÷÷öø

txmax = w1

2(ccc )(ww1n ) = tan- 1 2 = tan- 1

2 1 -

= 33.7° (Use Figure 2-6)

1 é æçççè180 öø÷÷÷÷ùúúû= 0.054 sec

txmax = 40 2êêë + 33.7

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

2-4

1 x0

=

SM: Experimental Methods for Engineers Chapter 2

2

F0

ìï 2 2 ïü

k

w1 = 0.306. which

gives

wn

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

w1 = (0.306)(628) w1 = 192.1

rad/sec = 30.6 Hz

ïïîíïï éêëê1 - ( ww1n ) úúúûù + éêëê2(ccc

)( ww1n ) úúûúùïïïïïïþý k ïï ê ê

4

For xF00 = 1.00 + 0.01 = 1.01we have æççççè

ww1n öø÷÷÷÷÷ k w1 ® imaginary.

4

For xF00 = 1.00 - 0.01 = 0.99 we haveæççççè

ww1n öø÷÷÷÷÷ -

2

0.04

æèçççç ww1n ö÷÷÷÷÷ + 1 -

çççèæ1.011 øö÷÷÷÷2 = 0 and

ø

2

0.04 ø æèçççç ww1n ö÷÷÷÷÷ + 1 -

ççæèç0.991÷÷÷÷öø2 = 0

12 2


3

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 = EAB2

R

EAB = Eæççç

R

ö÷÷÷÷÷ çè R +

Ri ø

2

5sec, T = 270°F

P = RRi

0 ¥

1

1

i i

R

R R R +

ö æ ÷ ç ÷ ç ÷ ç ÷ ç ÷ ç ÷ ÷ ç ÷ ø è


4

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 + Ri) - R ùú´ 100 êêë

(R + Ri) úúû

=´ 100

ç

% error = 20%

2-10

E2

P = AB ; EAB = E +RRi

R R

E = 100 v

R = 20,000 ohms

Ri = 5000 ohms

P = E2 æç R ÷÷ö2 = 104

é 2 ´ 104 ù

ú= 0.32 Watts ê

Maximum power occurs when = 0 ® R = Ri dR

P =

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

Watts

R

When R = 1000 ohms and Ri = 5000 ohms:

104 éê 103 ùú2 = 10 volts2 = 0.278 Watts

P = 3 ê6 ´ 103 úúû 36 ohm

10 êë

\ R = 5000 ohms


5

R Rèçç + Ri ÷÷ø

2(10)4

êêë2.5´ 104

úúû dP

mx + kx = 0

® wn = k

x + k x = 0 where wn2 = k

m m m

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

k g g 980 seccm2

= ® wn = =

m 0.5 cm wn =

44.3 rad/sec

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 x0 for c = 0

w wn F0 cc

k

20 3.19 0.108

40 6.38 0.025

2-11


6

60 9.57 0.011

2-15

dV V - c

= - cV = e

d V0

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 kgf = 9.806 N

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

2-17

(mi/gal)(5280 ft/mi)çççèæ 2311 gal/in3÷÷÷÷öø(1728 in /ft3 3)

´ (35.313 ft /m33)æ

çççè10001

m /l3 ø÷÷÷

÷ö´ (3.2808´ 10- 3 km/ft)

= 4.576 km/l

2-18

æ lbm ft ö

2-19

æ 1 Btu ö æ5 ö

(kJ/kg·°C)çççè1.055 kJ

ø÷÷÷÷(0.454 kg/lbm) èççç9 °C/ F° ø

÷÷÷÷= 0.2391 Btu/lbm· F°

(kJ/kg·°C)çççæ4.1821 kcalkJ ø÷÷÷÷

öæççèçç1000

1

kgg ÷÷

÷÷öø= 2.391´ 10

- 4 kcal/g- C° è

(lbf-s/ft2) 32.17çççè lbf s2 ø÷

÷÷÷=

32.17 lbm/sft · ´

= 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÷÷÷÷´ ççç32.17

1 slug/lbm

öø

÷÷÷÷

= 1.939 ´ 10- 6 slug/ft3

2-21

sec/hö÷÷÷÷´ J/Btu° )ççæ 1(Btu/h-ft- F)(1055

(107

erg/J)æ

èççç9

5° °F/ Cöø

÷÷÷÷

çè3600 ø

æçççè12 ´

12.54 ft/cmø

ö÷÷÷÷= 5.275 ´ 10

6 erg/s·ft· C° ´

= 1.731´ 105 erg/s·cm· C°

2-22 2

/s2 )æ

çççè2.541´ 12 cm

ft 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)2çç

æ32.17

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

m3


8

( 2.541 )2 cmin22 ´ 1441

inft22

W Btu/hr-ft2

´ =

cm2

2-26

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

4.4485 ΚlbfN =

lbm ´ 9 R

2-27

cms 3æçççè 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 = 23.03s

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

Hz cc

n c

For= 0.4,

resonance at =

0.8, = 8000 Hz

cc

2-32

n

c

= 0.2 and 0.4 for

n c

c

At 2000 Hz

= 0.3


9

xF00 = = 1.034 k

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

7.13deg At 4000 Hz

xF00 = = 1.145

k

= tan- 1

éêêë(2)(0.3)(0.4)

1 - 0.42 ùúúû=

15.9deg

2-33 xF00 = 0.4 21 == 1050

Hz Hz

k

From Fig. 2-6,

c

For = 1.0 cc

10

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

21 2 / 1

[1 2.384] +

0.3 n

>

<

21 2 / 1

[1 1.1918] +


10

2-36

T0 = 35 C T¥ = 110 C (8 secT ) = 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

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

= 1500 Hz = 1

3

0.98 = ( ) 2 ( )( ) 2üïïï1/2

ìïïïíïîïï êéë1 - 13 2 úûùú +

ëêêé(2) ccc 13 úûùú ýïïþï ê

c

= 0.619 cc

2-45

t = 1´ 10- 6

sec = 90% rise time

1

- (1 10´ - 6)

24 (24)(3600)

t = 2 hr = 3600sec = ( )

( ) = 0.2618 rad = - tan( )

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- 5

rad/sec

= = = ´

1

n n


12

= 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

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 = = 0.41sec

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

From Figure 2-9 nt = 6.2,

2-50

t = 1sec, nt = 8.77 c 5.7

c = 5.7 = = cc

72.11

x

From Figure 2-9, » 1.8.

x0

2-51

Rise time = 10- 12 s

c x

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 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/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/m

3

T = °C, x = m, = kg/m-s

At = 1.0 cc and x0 nt 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

÷÷÷÷ö1/2 = 7.29

rad/s

1.5 12

0

lbf/ft 8000 N/m 117,000

117,000

2203

0.9 1.0 At 3.6

3.6 0.495 s

7.29

n

n c

k

k

m

c x t

c x

t

= = =

æ ç = = ç ç ø è

= = »

= =


14

( 5 9

English system

g = ft/s ,2 = 1/ F,° = lbm/ft

3

T = °F, x = ft, = lbm/ft-s

2-57

W-cm 0.01 m/cm

´

in - F2 ° (2.54 cm/in.) (0.01 m/cm

2)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 + nt)exp( - nt)

Solution is nt = 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/ )-

= 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

= (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

experimental methods for engineers - solution manual testbank … · 2018-09-20 · instructor’s...

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