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

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 Download Full Solution Manual for Experimental Methods for Engineers 8th Edition by Holman https://getbooksolutions.com/download/solution-manual-for-experimental-methods-for-engineers8th-edition-by-holman 2-3 x Amplitude ratio = F0 = ìï ï k í ïï ỵ é ê1 ê ëê 2ù ïü ( ) úúûú+ éêêë2( )( )ùúúûïýïï w1 wn a cc w1 wn ỵ = {[1 - (0.4)2 ]2 + [2(0.7)(0.4)]2}1 amplitude ratio  0.99 (Use Figure 2-5) F (t ) = F0 sin w1t; x(t ) = x0 sin(w1t -  ) time lag = t xmax - t Fmax   F (t ) = F0 = max (when sin w1t = 1) \ w1t = sin- 11 = ; t Fmax = w1 ỉ1 ưỉ π ưỉ ÷ ữ ữ ỗ ỗ t Fmax = ỗ = 0.00625sec ữ ữ ữ ỗ ỗ ỗ ữ ữ ữ ỗ 40 ứố ỗ ứố ỗ ứ ố x(t ) = x0 = max (when sin(w1t -  ) = \ (w1t -  ) = sin- 11 = t xmax = 1ổ ữ ỗ + ữ ỗ ữ ố2 ứ w1 ỗ = tan - ( )( ) = 1- ( ) cc c w1 wn w1 wn π 2(0.7)(0.4) tan- 1 - (0.4)2  = 33.7° (Use Figure 2-6) é ù ỉ ê ÷ ú = 0.054 sec t xmax = + 33.7 ỗ ữ ỗ ữ ỗ ố180 ứỳ 40 ở2 ỷ \ time lag = 0.54 - 0.000625 time lag = 0.0478 sec 2-4 ïìï ïí ïï ỵï é ê1 ê ëê For 2ù ( ) x0 F0 k w1 wn é ú + ê2 ú ê ú êë û 12 ùïü ï ( )( ) c cc w1 wn ỳùý ỳù ỳù ỷ ùỵ = x0 F0 k ổ w ử4 ổ w ửữ2 ỗỗ ữ + ữ = 1.00 + 0.01 = 1.01 we have ỗỗỗ ữ 0.04 ữ ữ ỗố wn ứữ ốỗỗ wn ứữ ổ ữ ỗỗ ữ ữ = and ốỗ 1.01 ứ w1 đ imaginary For x0 F0 k ổ w ử4 ổ w ửữ2 ỗỗ ÷ + ÷ = 1.00 - 0.01 = 0.99 we have ỗỗỗ ữ 0.04 ữ ữ ữ ốỗ wn ứ ốỗỗ wn ứữ which gives w1 = 0.306 wn ổ ữ ử2 ỗỗ = ữ çè 0.99 ÷ ø SM: Experimental Methods for Engineers Chapter wn = (100)(2 ) = 628 rad/sec w1 = (0.306)(628) w1 = 192.1 rad/sec = 30.6 Hz SM: Experimental Methods for Engineers Chapter 2-5 - ( )t T - T¥ = e RC T0 - T¥ At t = 3sec, T = 200°F T - T¥ = 0.435 At t = 5sec, T = 270°F T0 - T¥ T - T¥ = 0.1304 T0 - T¥ - 0.632 = 0.328 RC » 3.4 sec 2-6 P = E AB R ỉ R ư÷ ÷ E AB = E ỗỗỗ ữ ỗố R + Ri ứữ ổ R ử2 ỗỗ Ri ữ ữ P = R ỗỗ R ữ ữ ữ ỗ ữ Ri ç è Ri + ÷ ø SM: Experimental Methods for Engineers Chapter 2-7 inch 64 Least count ® inch 32 Readability ® 2-8 t = RC = time constant t = (106 ohms)(10- f ) = 10 sec t = 10 sec 2-9 é( R + R ) - R ù i ú´ 100 % error = êê ( R + Ri ) ú ú ëê û 5000 = ´ 100 25, 000 % error = 20% 2-10 P = E AB R ; E AB = E R R + Ri E = 100 v R = 20,000 ohms Ri = 5000 ohms ù ư2 104 é ÷ ê ´ 10 ú = 0.32 Watts ÷ = ÷ ê 4ú ÷ ø 2(10)4 ê ë2.5 ´ 10 ú û dP Maximum power occurs when = ® R = Ri \ R = 5000 ohms dR E2 ổ R ỗ P = ỗ ỗ ỗ R ố R + Ri E2 ổ P 104 ữ ỗ = = 5000 Watts ữ ỗ ữ ố 2R ứ R ỗ 2.0 ´ 104 When R = 1000 ohms and Ri = 5000 ohms: Pmax = P = ù 104 é ê 10 ú = 10 volts = 0.278 Watts ê 3ú 36 ohm 10 ê ë6 ´ 10 ú û 2-11 mx + kx = k k k x = where wn2 = ® wn = m m m From the static deflection: k  = mg where  = deflection = 0.5 cm x+ k g = ® wn = m  g =  980 cm sec2 0.5 cm wn = 44.3 rad/sec SM: Experimental Methods for Engineers Chapter 2-12  = 0.25 inch; g = 386 in/sec2 986 g =  wn = in sec2 0.25 in = 39.4 rad/sec 2-14 wn = 39.4 rad/sec = 6.27 Hz x0 w w wn 20 3.19 0.108 40 6.38 0.025 60 9.57 0.011 2-15 dV = - cV d for F0 k c = cc V = e- c V0 At  = 0, V = 10 liters, dV = - d c = 0.6 hr- 2-16 (1 lbf/in ) (4.448 N/lbf )(144 in /ft ) (3.282 ft /m2 ) = 6890 N/m2 kgf = 9.806 N lbf/in = (6890)(9.806) = 67570 kgf/m2 = 6.757 kp/cm2 2-17 ỉ1 3 (mi/gal)(5280 ft/mi) ỗ gal/in3 ữ ữ(1728 in /ft ) ỗ ữ ỗ ố 231 ứ ổ1 (35.313 ft 3/m3 ) ỗ m3/l ữ (3.2808 10- km/ft) = 4.576 km/l ữ ỗ ữ ỗ ố1000 ứ 2-18 ổ lbm ft ữ (lbf-s/ft ) ỗỗ 32.17 ữ = 32.17 lbm/sãft (0.454 kg/lbm)(3.2808 ft/m) ỗố ÷ lbf s2 ø = 47.92 kg/m·s 2-19 æ Btu ổ5 ữ (kJ/kgãC) ỗ (0.454 kg/lbm) ç °C/ °F÷ = 0.2391 Btu/lbm· °F ÷ ÷ ç ç ÷ ÷ ç1.055 kJ ø ç9 è è ø ổ kcal ữ ửổ ỗ kg ữ ữ (kJ/kgãC) ỗ = 2.391 10- kcal/g-C ữỗ ç ÷ ç ÷ ç è 4.182 kJ øè1000 g ữ ứ 2-20 ổ1 ổ ỗ (g/m3 )(0.02832 m3 /ft ) ỗ lbm/g ữ slug/lbm ữ ữ ữ ỗ ỗ ữ ố ữ ỗ 454 ỗ 32.17 è ø ø = 1.939 ´ 10- slug / ft SM: Experimental Methods for Engineers Chapter 2-21 ổ ổ9 (Btu/h-ft-F)(1055 J/Btu) ỗ sec/h ữ (107 erg/J) ỗ F/C ữ ữ ữ ç ç ÷ ÷ ç ç è3600 ø è5 ø æ ö = 5.275 ´ 106 erg/s·ft ·°C ´ ç ft/cm ÷ ÷ ç ÷ ç è12 ´ 2.54 ø = 1.731 ´ 105 erg/s·cm·°C 2-22 æ ft ữ (cm2 /s) ỗ = 1.076 ft 2/ s ữ ỗ ỗ ố 2.54 12 cm ữ ø 2-23 ỉ ưỉ m ư3 Btu ÷ ữ ỗ ữ (W /m3 ) ỗ 3.413 ỗ ữ = 0.09664 Btu/h ãft ỗ ữ ỗ ữỗ ỗ ố3.2808 ft ÷ ø W·h ø è 2-24 ỉ lbm ft ö ÷ (dyn ·s/cm2 )(10- N/dyn)(0.2248 lbf/N) ´ (2.54 12 cm/ft)2 ỗ 32.17 ữ ỗ ỗ ữ lbf s ø è = 0.0672 lbm/s ·ft ´ 3600 s/h = 241.8 2-25 W cm W cm lhm h ·ft 3.413 Btu/W·h ´ ( 2.54 ) in cm ´ ft 144 in ´  = Btu/hr-ft 2-26 R = 1545 N 0.3048 m ´ 4.448 lbf ft-lbf J ft ´ = 8305 kg  Κ lbm ·mol ·R kg mol · K 0.454 lbm ´ R 2-27 cm3 ỉ  in gal ÷ ỗ ữ ỗ ữ ỗ ố ứ cm3 231 in s cm3 ´  = gal/min s 2-28 R = K SM: Experimental Methods for Engineers Chapter 2-29  = 105, T0 = 30C, T¥ = 100C 90 = 2.303 = 23.03s Rise time 0.01 = e- t  t  = 4.605 t (99) = 46.05 sec 2-30 A = 20 C   =  tan  = 0.01 Hz = 0.0628 rad/s - T - =  tan [(0.0628)(10)] = - 32.14 deg = - 0.561 rad t =    = - 0.561 = 8.93 sec 0.0628 2-31  n = 10, 000 Hz c = 0.3, 0.4 cc For c  = 0.3, resonance at = 0.9,  = 9000 Hz cc n For c  = 0.4, resonance at = 0.8,  = 8000 Hz cc n 2-32  c = 0.2 and 0.4 for = 0.3 n cc At 2000 Hz x0 F0 k = {(1 - 0.22 ) + [(2)(0.3)(0.2)]2}1/2 = 1.034 é(2)(0.3)(0.2) ù ú = 7.13deg  = tan- ê ê ë - 0.2 ú û At 4000 Hz x0 F0 k = {[1 - 0.42 ]2 + [(2)(3)(0.4)]2}1/2 = 1.145 é(2)(0.3)(0.4) ù ú = 15.9 deg  = tan- ê ê ë - 0.4 ú û SM: Experimental Methods for Engineers 2-33 x0 F0 k Chapter  = 10 Hz = 0.4  = 50 Hz From Fig 2-6, For  > 0.3 n c = 1.0 cc n < 10 = 33 Hz 0.3 2-34   = - 50 = - tan- 1   = 1.1918  = (2)(1.1918) = 2.3835      At   amp response At   amp response [1 + 1.19182 ]1/2 [1 + 2.3842 ]1/2 2-35  = Hz = 18.85 rad/s = 0.643 = 0.387  = 0.5 sec ( ) = - tan - 1[(18.85)(0.5)] = - 8.39      1/2 2-36 T0 = 35 C = 0.1055 T¥ = 110 C T (8 sec) = 75 C 75 - 110 = e- t/ 35 - 110 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 t Rc = 0.1 = 7.86 ´ 105 RC RC = 1.303 ´ 10- R in ohm, C in farads SM: Experimental Methods for Engineers 2-42  = 0.1sec T0 = 100 C Chapter T¥ = 15 C T (t ) = 17 C 17 - 15 = e- t / 0.1 100 - 15 t = 3.75 0.1 t = 0.375 sec 2-43 [1 + ( )2 ]1/2  = 0.4843 ( ) = - tan Dt = = 0.9  to 4.84 rad/s - (0.4843) = - 25.84 = 0.451 rad 0.451 = 0.093 sec 4.84 2-44  = 500 Hz 0.98 =  = n n = 1500 Hz ïìï í ïï ỵï é ê1 êë 2ù é ú + ê(2) ú êë û ( ) ( )( c cc ü1/2 ự2 ùù ỳý ỳ ỷ ùùù ỵ ) c = 0.619 cc 2-45 t = ´ 10- sec = 90% rise time - 0.1 = e RC (1´ 10- 6) RC = 4.34 ´ 10- R in ohm, C in farads 2-46 t = hr 2 cyc/hr = = 9.27 ´ 10- rad/sec 24 (24)(3600) ( ) t = hr = 3600sec =  =  ( ) = 0.2618 rad = - tan( )  = 0.2679  = 0.2679 7.27 ´   Amp response = = 3685sec = 1.024 hr [1 + ( ) ]1/2 = 0.966 SM: Experimental Methods for Engineers 2-47 m = 1.3 kg n = Chapter k = 100 N/m 1/2 ổ100 k ữ = ỗ ữ ỗ ữ = 8.77 rad / s ỗ ố 1.3 ứ m cc = 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 = 0.41 sec 8.77 t = 2-48 c = 0.1 cc From Figure 2-9,  n t = 3.1 3.1 = 0.353 sec 8.77 t = 2-49 x(t ) = 0.9 x0 c = 1.5 cc From Figure 2-9  n t = 6.2, t = 6.2 = 0.71 sec 8.77 2-50 t = sec,  n t = 8.77 c 5.7 c = 5.7 = = 0.79 cc 72.11 From Figure 2-9, x » 1.8 x0 2-51 Rise time = 10- 12 s At c = 1.0 cc and x = 0.9,  n t x0 3.6,  n = 3.7 ´ 1012 rad/s f = 5.7 ´ 1011 Hz = 570 GHz 2-52 m = 1000 lbm = 2203 kg k = 1000 lbf 1.5 12 = 8000 lbf/ft = 117, 000 N/m æ117, 000 ữ ử1/2 k = ỗ = 7.29 rad/s ữ ỗ ç è 2203 ÷ ø m x c At = 0.9 = 1.0  n t » 3.6 x0 cc n = t = 3.6 = 0.495 s 7.29 10 SM: Experimental Methods for Engineers 2-53 T0 = 20°C T¥ = 125°C Chapter t = 0.05 sec e- t/ = 0.1  = 34.54 sec æ t T (t ) - 125 = exp ỗ ỗ ỗ ố 20 - 125 T (t ) = 20.15°C ö - ( 0.05 ) ÷ = e 34.54 ÷ ÷ ø 2-54 kg = 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/m3, u = m/s x = m,  = kg/m-s 2-56 SI system g = m/s ,  = / °C,  = kg/m3 T = °C, x = m,  = kg/m-s English system g = ft/s ,  = 1/°F,  = lbm/ft T = °F, x = ft,  = lbm/ft-s 2-57 W -cm in -°F W -cm in - F ´ 0.01 m/cm (2.54 cm/in.)2 (0.01 m/cm)2 95 °C/°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 = 82.6ºC 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/ = 8.93 rad/s 11 SM: Experimental Methods for Engineers Chapter For critically damped system: 0.9 = - (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- rad/sec (1.5)(3600) = 5400 s =  ( )/  ( ) = (5400)(0.0000727) = 0.3926 = - tan- 1( )  = 0.414  = 0.414/7.27 ´ 10- = 5695 s = 1.58 h 2-63 T0 = 45 T¥ = 100 T (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  = 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  = Hz,  = 0.6 s  = (2π)(5) = 31.4 rad/s  ( ) = - tan- 1( ) = - 8.7° 1/[1 + ( )2 ]1/ = 0.053 2-67 A = 15,  = 0.01 Hz = 0.0628 rad/s  ( ) = - tan-  = - tan- 1(0.0628)(8) = - 26.7° Attenuation = 1/[1 + ( )2 ]1/2 = 0.894 12 SM: Experimental Methods for Engineers Chapter 13