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 2-3 x0 = Amplitude ratio = Fk0 ï ìïïïï íỵï éêëê1 - ựỳỳỷỳ+ ( ộờờở2 cac )ỳựỳỷùùùùỵỹýùù ( ww1n )( ) ww1n 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 = w1 ỉ1 ưỉ ưỉ tFmax = ỗỗỗố40 ứố ữ ữ ữ ứố ữữữỗỗỗ ữữữỗỗỗ2 ứ ữữữ= 0.00625sec x t( ) = x0 = max (whensin(wt1 - ) = \ (wt1 - ) = sin- 11 = ổỗỗỗố2 + ÷÷÷÷ừ txmax = w1 ( = tan- )(ww ) = tan- ccc 1n 2 1= 33.7 (Use Figure 2-6) ộ ổỗỗỗố180 ửứữữữữựỳỳỷ= 0.054 sec txmax = 40 2êêë + 33.7 \ time lag = 0.54 - 0.000625 time lag = 0.0478sec 2-4 x0 = SM: Experimental Methods for Engineers ïïỵíïï éêëê1 - )( ww1n ( ww1n ) ỳỳỳỷự + ộờởờ2( ) ỳỳỷỳựùùùùùùỵý k ï ïê Chapter 12 F0 ccc ìï ê 0.04 For xF00 = 1.00 + 0.01 = 1.01we have ổỗỗỗỗố ổốỗỗỗỗ ww1n ửữữữữữ + ỗỗỗốổ1.011 ứửữữữữ2 = and ứ ww1n ửứữữữữữ k w1 đ imaginary For xF00 = 1.00 - 0.01 = 0.99 we haveổỗỗỗỗố ww1n ửứữữữữữ - 0.04 ứ ổốỗỗỗỗ ww1n ửữữữữữ + ỗỗổốỗ0.991ữữữữửứ2 = 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 2 ïü SM: Experimental Methods for Engineers Chapter 2-5 T - T¥ ¥ = e- ( RC1 )t T-T T - TƠ ổ R ửữ çç ÷ ÷ ÷ R ç çç ÷ Ri è R + ữ ỗ i ữ ứ = 0.435 At t = T0 - T¥ T - T¥ = 0.1304 T0 - T¥ - 0.632 = 0.328 RC » 3.4 sec 2-6 P = EAB2 R EAB = Eổỗỗỗ R ửữữữữữ ỗố R + Ri ứ 5sec, T = 270°F At t = 3sec,T = 200°F P= RRi SM: Experimental Methods for Engineers Chapter 2-7 Readability ®inch Least ®inch 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 % error = 20% 2-10 E2 P= AB ; EAB = E +RRi R R E = 100 v R = 20,000 ohms Ri = 5000 ohms P = E2 ổỗ R ỗ ữữử2 = 104ộ Maximum power occurs when P max ´ 104 ú ù = 0.32 Watts ê = ® R = Ri dR \ R = 5000 ohms = E2 ổỗỗỗố2PR ứửữữữữ2 = 2.010 4104 = 5000 Watts R When R = 1000 ohms and Ri = 5000 ohms: é 104 ê 103 P= ê6 ù ú = 10 volts2 = 0.278 Watts ´ 103 úúû 36 ohm 10 êë SM: Experimental Methods for Engineers Chapter 2-11 + Ri ÷÷ø R Rốỗỗ 2(10)4 ờờở2.5 104 ỳỳỷ dP mx + kx = x + k x = where wn2 = k m m m ® wn = From the static deflection: k = mg where k g = g ® wn = k = deflection = 0.5 cm 980 seccm2 = 0.5 cm wn = m 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 c=0 x0 for w wn 20 3.19 0.108 40 6.38 0.025 F0 k cc SM: Experimental Methods for Engineers 60 9.57 Chapter 0.011 2-15 dV V = - cV -c =e d V0 dV At = 0, V = 10 liters, =-6 d c = 0.6 hr- 2-16 (1 lbf/in )(4.448 N/lbf)(144 in /ft )(3.28 ft / = 6890 N/m2 2 2 ÷ 2311 gal/in3 ÷÷÷ừ(1728 in /ft3 m )2 kgf = 9.806 N lbf/in2 = (6890)(9.806) = 67570 kgf/m2 = 6.757 kp/cm2 2-17 (mi/gal)(5280 ft/mi)ỗỗỗốổ ổ (35.313 ft /m33) ỗỗỗố1000 = 4.576 km/l m /l3 ø÷÷÷ ) ÷ư ´ (3.2808´ 10- km/ft) 2-18 ữ lbf s2 ứ ữữữ= (lbf-s/ft2) 32.17ỗỗỗố 32.17 lbm/sft · (0.454 kg/lbm)(3.2808 ft/m) = 47.92 kg/ms· æ lbm ft 2-19 ổ1 Btu ổ5 (kJ/kgãC)ỗỗỗố1.055 ữ ứ ữữữ(0.454 kg/lbm) kJ ữ ốỗỗỗ9 C/ F ứ ữữữ= 0.2391 Btu/lbmã F ữ ửổỗ (kJ/kgãC)ỗỗỗổ4.1821 kcalkJ ứ ữữữ ỗốỗỗ1000 kg ữ g ữữ ữ ứ= 2.391 10- kcal/g- C° è SM: Experimental Methods for Engineers Chapter 2-20 ổ ửổ ) ỗỗỗố454 lbm/g ø (g/m )(0.02832 m /ft3 ÷ ÷ ốữữ ữ ỗỗỗ32.17 slug/lbm ứ ữữữ = 1.939 10- slug/ft3 2-21 (Btu/h-ft- F)(1055 J/Btu° ỉ ữ (107 erg/J) ốỗỗỗ F/ C ứ ữữữ ỗố3600 = 5.275 ´ 106 erg/s·ft· C° ´ ÷ sec/hư ÷÷÷´ )ỗỗổ ứ ổ ửữ ỗỗỗố12 2.54 ft/cmứ ÷÷÷ = 1.731´ 105 erg/s·cm· C° 2-22 æ ft ửữ /s2 ) ỗỗỗố2.54 12 cm ữữữứ = 1.076 ft /2 s (cm 2-23 (W/m3) 3.413ỗỗốỗổỗWãhBtu ứữữữữửốổỗỗỗ3.28081 0.09664 Btu/h·ft3 mft ư÷÷÷ø÷3 = 2-24 (dyn s/cm· )(10- N/dyn)(0.2248 lbf/N) ´ (2.54 ´ 12 ỉ lbm ft ư÷ cm/ft)2ỗỗ 32.17 ữữ = 0.0672 lbm/s ftã 3600 s/h lhm = 241.8 h ftã ỗố lbf s ứữ ´ 2-25 W 3.413 Btu/W h· c m SM: Experimental Methods for Engineers ( ) 2.541 cmin22 inft22 W Chapter [(0.0628)( 10)] Btu/hr-ft2 ´ tan- = ´ 1441 = - 32.14 deg = = 0.561rad cm2 2-26 t = = R = 1545lbm molft-lbf·· R ´ 0.30480.454mftkg´ 4.4485 ΚlbfN = = 8.93sec 0.0628 lbm ´ R 2-31 2-27 cms 3ổỗỗỗố c ửứữữữữ3 cmin33 2311 n = 10,000 Hz = 0.3, 0.4 gal in2 ´ c cm3 ´ gal/min s = c c For= 0.3, resonance 2-28 R= K at Hz J kg mol · K 8305 2-29 = 0.9, = 9000 cc n c cc n 2-32 = 105, T0 = 30 C, T¥ Rise time c= 0.3 = 100 C = 0.2 and 0.4 for 90 = 2.303 = 23.03s 0.01 = e- t t = 4.605 n c c At 2000 Hz For= 0.4, resonance at = 0.8, = 8000 Hz t(99 ) = 46.05 sec 2-30 A = 20 C = 0.01Hz = 0.0628 rad/s = tan- T SM: Experimental Methods for Engineers xF00 = = 1.034 k = tan- ê éê ë(2)(0.3)(0.2)1 - 0.22 ùúúû= Chapter = tan- ê éê ë(2)(0.3)(0.4) - 0.42 15.9deg ùúúû= 7.13deg At 4000 Hz xF00 = = 1.145 k 2-33 xF00 = 0.4 21 == 1050 Hz Hz k From Fig 2-6, c > = 1.0 cc For 0.3 n 10 < = 33 Hz n 0.3 2-34 = - 50 = - tan- = 1.1918 = (2)(1.1918) = 2.3835 At 0.643 amp [1 + 1.1918]21 /2 response= At 21 /2 amp response [1 + 2.384] = 0.387 2-35 = Hz = 18.85 rad/s = 0.5 sec ( ) = - tan - 1[(18.85)(0.5)] = - 8.39 1/2 = 0.1055 SM: Experimental Methods for Engineers Chapter 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 = 7.86 ´ 105 RC RC = 1.303 ´ 10- R in ohm, C in farads 2-42 = 0.1sec T0 = 100 C T t( ) = 17 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 10 SM: Experimental Methods for Engineers Chapter 2-44 = 500 Hz n n = 1500 Hz =1 ( )( ) ()2 0.98 = ìïïïíïỵïï êéë1 - 13 ëêêé(2) ccc 2üïïï1/2 úûùú + 13 ỳỷựỳ ýùùỵù c = 0.619 cc 2-45 t = 1´ 10- sec = 90% rise time 0.1 = e RC RC = 4.34 ´ 10- R in ohm, C in farads 2-46 t = hr = = - 24 10- 5rad/sec 9.27 cyc/hr = ´ (1 10´ - 6) (24)(3600) t = hr = 3600sec = () ( ) = 0.2618 rad = - tan( ) 11 SM: Experimental Methods for Engineers = 0.2679 0.2679 = Chapter = 3685sec = 1.024 hr 7.27 ´ Amp response = = 0.966 2-47 m = 1.3 kg N/m k n= k = 100 ổ100 ử1/2 m = ỗỗỗố1.3 ứữữữữ= 8.77 rad/s 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 Rise time = 10- 12 s t= = 0.41sec c x 8.77 2-48 c = 0.1 6.2 t= = 0.71sec 8.77 cc From Figure 2-9, nt = 3.1 3.1 t= = 0.353 sec 8.77 2-49 x t( ) c = 0.9 x0 = 1.5 cc From Figure 2-9 nt = 6.2, 2-50 t = 1sec, nt = 8.77 c c = 5.7 = = cc 72.11 x From Figure 2-9, x0 5.7 » 1.8 2-51 12 SM: Experimental Methods for Engineers Chapter T t( ) = 20.15°C 0.79 2-53 T0 = 20°C T¥ = 125°C e- t/ = 2-54 t = 0.05 sec 0.1 = 34.54 sec = 1.0 cc and At 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 k = = 8000 lbf/ft = 117,000 N/m 1.5 12 k ổ117,000 = ỗỗ ỗố 2203 ø m x c At = 0.9 = 1.0 x0 cc n 1000 lbf = t = ÷÷÷÷ư1/2 = 7.29 rad/s nt » 3.6 3.6 = 0.495 s 7.29 = exp T t20( ) 125125 ổỗỗỗố- t e- ( ) ửứữữữữ= kg = 0.02088 lbf sec/ftì 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 = / °C, = kg/m3 T = °C, x = m, = kg/m-s 13 SM: Experimental Methods for Engineers Chapter 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 ´ (5 in - F2 ° (2.54 cm/in.) (0.01 m/cm2)2 9°C/ F° W-cm ) in - F2 ´ W/m- C° 2-58 T0 = 45, T¥ 0.2 = 2.303 = 100 rise time = 0.2 s T(0.1s) = ? , = 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 (263250/3303)1/2 = 8.93 rad/s For critically damped system: n= 0.9 = - (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)(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 = /(24)(3600) (1.5)(3600) = 5400 s = = 7.27 ´ 10- rad/sec ( )/ ( ) = (5400)(0.0000727) = 0.3926 = - tan- 1( ) = 0.414 14 SM: Experimental Methods for Engineers Chapter = 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/ )= 7.61s 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 = 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( 1/[1 + ( 2-67 A = 15, ()=- ) = - 8.7° ) ]2 1/2 = 0.053 = 0.01 Hz = 0.0628 rad/s tan- =- Attenuation = 1/[1 + ( tan- (0.0628)(8) = - 26.7° ) ]2 1/2 = 0.894 15 SM: Experimental Methods for Engineers Chapter 13