Experimental Methods for Engineers - Instructor Solutions Manual

184 Pages·2011·4.689 MB·English

by  Jack P. Holman

Instructor’s Solutions Manual

to accompany

Experimental Methods
for Engineers

Eighth Edition
J. P. Holman
Professor of Mechanical Engineering
Southern Methodist University

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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 McGraw-Hill for their individual course preparation. A student using this manual is

using it without permission.

TABLE OF CONTENTS

Chapter 2 ............................................................................................................................................... 1
Chapter 3 ............................................................................................................................................. 12
Chapter 4 ............................................................................................................................................. 62
Chapter 5 ............................................................................................................................................. 75
Chapter 6 ............................................................................................................................................. 82
Chapter 7 ............................................................................................................................................. 96
Chapter 8 ........................................................................................................................................... 114
Chapter 9 ........................................................................................................................................... 131
Chapter 10 ......................................................................................................................................... 141
Chapter 11 ......................................................................................................................................... 152
Chapter 12 ......................................................................................................................................... 167
Chapter 13 ......................................................................................................................................... 171
Chapter 14 ......................................................................................................................................... 176

Chapter 2
2-3
x 1
Amplituderatio = 0 =
Fk0 ìïïïíéê1-(w1 )2ùú + éê2( a )( w1 )ùúüïïïý
ïïïîêêë wn úúû êë cc wn úûïïïþ
1
=
{[1-(0.4)2]2 +[2(0.7)(0.4)]2}12
amplitude ratio = 0.99 (Use Figure. 2-5)
F(t) = F sinwt; x(t) = x sin(wt −φ)
0 1 0 1
timelag = t −t
x F
max max
F(t) = F = max(when sinwt = 1) ∴wt = sin−11= π;t = 1 π
0 1 1 F
2 max w 2
1
⎛ ⎞⎛ ⎞⎛ ⎞
tFmax = ⎜⎜⎜⎝410⎟⎟⎟⎟⎠⎜⎜⎜⎝π2⎟⎟⎟⎟⎠⎜⎜⎜⎝21π⎟⎟⎟⎟⎠ = 0.00625sec

x(t) = x = max(whensin(wt −φ) = 1 ∴ (wt −φ) = sin−11= π
0 1 1
2
⎛ ⎞
txmax = w1 ⎜⎜⎜⎝π2 +φ⎟⎟⎟⎟⎠
1
( )( )
2 c w1
φ= tan−1 cc wn = tan−12(0.7)(0.4)
1−( w1 )2 1−(0.4)2
w
n
φ= 33.7° (UseFigure 2-6)
txmax = 410⎢⎣⎡⎢π2 +33.7⎜⎜⎝⎛⎜1π80⎟⎟⎟⎠⎞⎟⎥⎦⎤⎥ = 0.054sec
∴ timelag = 0.54−0.000625
timelag = 0.0478sec

2-4
1 x
= 0
ìïïïïíéê1-(w1 )2ùú2 + éê2( c )(w1 )2ùúüïïïïý12 Fk0
ïê w ú ê c w úï
ïêë n úû êë c n úûï
ïïî ïïþ
For xF00 = 1.00 + 0.01= 1.01wehaveæççççèww1nö÷÷÷÷÷ø4 -0.04æççççèww1nö÷÷÷÷÷ø2 +1-æçççè1.101ö÷÷÷÷ø2 = 0and
k
w  imaginary.
1
For xF00 = 1.00-0.01 = 0.99wehaveæççççèww1nö÷÷÷÷÷ø4 -0.04æççççèww1nö÷÷÷÷÷ø2 +1-æçççè0.199ö÷÷÷÷ø2 = 0
k
w
whichgives 1 = 0.306.
w
n
w = (100)(2π) = 628rad/sec
n
w = (0.306)(628)
1
w = 192.1rad/sec = 30.6Hz
1
1
Chapter 2 SM: Experimental Methods for Engineers
2-5
T -T¥ = e-(R1C)t
T -T
0 ¥

Att = 3sec,T = 200°F
T −T∞ = 0.435.Att = 5sec,T = 270°F
T0 −T∞
T −T∞ = 0.1304
T0 −T∞
1−0.632 = 0.328
RC ≈ 3.4sec

2-6
E2
P = AB
R
EAB = EççççèæR +R R ö÷÷÷÷÷ø
i
æ R ö2
P = R1R çççççççè RRR+i1÷÷÷÷÷÷÷÷ø
i i

2
SM: Experimental Methods for Engineers Chapter 2
2-7
1
Readability → inch
64

1
Leastcount → inch
32

2-8
t = RC = timeconstant
t = (106ohms)(10-5f) = 10sec
t = 10sec

2-9
é(R + R)- Rù
%error = ê i ú´100
ê ú
êë (R + Ri) úû
5000
= ´100
25,000
%error = 20%

2-10
E2 R
P = AB;E = E
R AB R + R
i
E = 100v
R = 20,000ohms
R = 5000ohms
i
P = ER2⎜⎝⎜⎜⎜⎛R +R Ri⎟⎟⎠⎟⎟⎞2 = 21(1004)4 ⎢⎣⎢⎢⎡22.5××110044⎥⎦⎥⎥⎤ = 0.32Watts
dP
Maximumpower occurs when = 0 → R = R ∴R = 5000ohms
i
dR
Pmax = ER2⎛⎜⎜⎜⎝2PR⎞⎟⎟⎟⎟⎠2 = 2.01×04104 = 5000Watts
When R = 1000ohmsand R = 5000ohms:
i
⎡ ⎤2
104 ⎢ 103 ⎥ 10 volts2
P = ⎢ ⎥ = = 0.278Watts
103 ⎢⎣6×103⎥⎦ 36 ohm

2-11
mx+ kx = 0
k k k
x+ x = 0wherew2 = → w =
n n
m m m
From the static deflection: kΔ = mg where Δ = deflection= 0.5cm
980 cm
k = g → w = g = sec2
m Δ n Δ 0.5cm
w = 44.3 rad/sec
n

3
Chapter 2 SM: Experimental Methods for Engineers
2-12
Δ = 0.25 inch; g = 386 in/sec2
986 in.
w = g = sec2 = 39.4 rad/sec
n Δ 0.25in.

2-14
w = 39.4 rad/sec = 6.27 Hz
n
w x c
w 0 for = 0
wn F0 cc
k
20 3.19 0.108
40 6.38 0.025
60 9.57 0.011

2-15
dV = −cV V = e−cτ
dτ V
0
dV
Atτ= 0, V = 10 liters, = −6
dτ
c = 0.6hr−1

2-16
(1 lbf/in2)(4.448N/lbf)(144in2/ft2)(3.282 ft2/m2) = 6890N/m2
1 kgf = 9.806 N
1 lbf/in2 = (6890)(9.806) = 67570kgf/m2 = 6.757kp/cm2

2-17
⎛ ⎞
(mi/gal)(5280 ft/mi)⎜⎜⎜⎝2131 gal/in3⎟⎟⎟⎟⎠(1728 in3/ft3)

⎛ ⎞
×(35.313 ft3/m3)⎜⎜⎜⎝10100 m3/l⎟⎟⎟⎟⎠×(3.2808×10−3km/ft) = 4.576 km/l

2-18
(lbf-s/ft2)ççççèæ32.17llbbmf sf2t ö÷÷÷÷ø = 32.17 lbm/s·ft ´(0.454 kg/lbm)(3.2808 ft/m)
= 47.92kg/m·s

2-19
⎛ ⎞ ⎛ ⎞
(kJ/kg·°C)⎜⎜⎝⎜1.0155 BktJu⎟⎟⎟⎠⎟(0.454 kg/lbm) × ⎜⎜⎝⎜95°C/°F⎠⎟⎟⎟⎟ = 0.2391 Btu/lbm·°F

⎛ ⎞⎛ ⎞
(kJ/kg·°C)⎜⎜⎜⎝4.1182 kkcJal⎟⎟⎟⎟⎠⎜⎜⎜⎜⎝10100 kgg⎟⎟⎟⎟⎠ = 2.391×10−4 kcal/g-°C

2-20
⎛ ⎞ ⎛ ⎞
(g/m3)(0.02832m3/ft3)⎜⎜⎜⎝4154 lbm/g⎟⎟⎟⎟⎠×⎜⎜⎜⎝321.17 slug/lbm⎟⎟⎟⎟⎠
= 1.939×10−6 slug/ft3

4
SM: Experimental Methods for Engineers Chapter 2
2-21
⎛ ⎞ ⎛ ⎞
(Btu/h-ft-°F)(1055 J/Btu)⎜⎜⎜⎝36100 sec/h⎟⎟⎟⎟⎠×(107erg/J)⎝⎜⎜⎜95°F/°C⎟⎟⎟⎟⎠
⎛ ⎞
= 5.275×106 erg/s·ft·°C×⎜⎜⎜⎝12×12.54 ft/cm⎟⎟⎟⎟⎠
= 1.731×105 erg/s·cm·°C

2-22
⎛ ⎞2
(cm2/s)⎜⎜⎜⎝2.541×12 cfmt ⎟⎟⎟⎟⎠ = 1.076 ft2/s

2-23
⎛ ⎞⎛ ⎞3
(W/m3)⎜⎜⎜⎜⎝3.413WBt·uh⎟⎟⎟⎟⎠⎜⎜⎜⎝3.21808 mft ⎟⎟⎟⎟⎠ = 0.09664 Btu/h·ft3

2-24
⎛ ⎞
(dyn(cid:8836)s/cm2)(10−5N/dyn)(0.2248 lbf/N)×(2.54×12 cm/ft)2⎜⎜⎜⎝32.17llbbmf sf2t ⎟⎟⎟⎟⎠
= 0.0672 lbm/s(cid:8836)ft×3600 s/h
lhm
= 241.8
h(cid:8836)ft

2-25
W 3.413 Btu/W(cid:8836)h
×
cm3 ( 1 )2 in2 × 1 ft2
2.54 cm2 144 in2
W
×3170 = Btu/hr-ft2
cm2

2-26
ft-lbf 0.3048 m ´4.448 N J
R = 1545 ´ ft lbf = 8305
lbm(cid:8836)mol(cid:8836)°R 0.454 kg ´ 5 °Κ kg mol (cid:8836)(cid:8720)°K
lbm 9 °R

2-27
cms3 ×⎛⎜⎜⎜⎝2.154⎞⎠⎟⎟⎟⎟3cimn33 × 2131gina2l

cm3
×0.01585 = gal/min
s

2-28
9
°R = °K
5

5