fundamentals of electric circuits 3rd edition solutions manual chapter 8
Fundamentals of Corporate Finance 8th edition: Solutions Manual
... 15, 183 35,612 62, 182 1,002 1,514 2,671 U U U $ 16, 185 37,126 64 ,85 3 $ 112,977 5, 187 U $1 18, 164 327,156 $ 440,133 31,007 36,194 U U 3 58, 163 $476,327 $ 78, 159 46, 382 $ 124,541 60,000 – 18, 850 1, 786 ... 3 58, 163 $ 476,327 75.19% 100% 1.09 48 1. 082 2 1.0116 1.0000 $ 78, 159 46, 382 $ 124,541 60,000 17.76% 10.54% 28. 30% 13.63% $ 59,309 48, 1 68 $ 107,477 75,000 12.45% 10.11% 22.56% 15.75% 0.7 588 1.0 385 ... = $ 58, 6 78 / $9 ,87 5 = 5.94 times Cash coverage ratio Cash coverage ratio = (EBIT + Depreciation) / Interest = ($ 58, 6 78 + 21,950) / $9 ,87 5 = 8. 16 times Profitability ratios: Profit margin Profit...
Ngày tải lên: 01/07/2014, 14:22
fundamentals of electric circuits
... 3 08 Step Response of a Series RLC Circuit 314 8. 6 Step Response of a Parallel RLC Circuit 319 8. 7 General Second-Order Circuits 322 8. 8 Second-Order Op Amp Circuits 327 8. 9 PSpice Analysis of ... Networks 18. 1 18. 2 18. 3 18. 4 18. 5 † 18. 6 18. 7 18. 8 Circuit Applicatons 727 Average Power and RMS Values 730 Exponential Fourier Series 734 Fourier Analysis with PSpice 740 16.7.1 16.7.2 789 707 Introduction ... DC CIRCUITS Chapter Basic Concepts Chapter Basic Laws Chapter Methods of Analysis Chapter Circuit Theorems Chapter Operational Amplifier Chapter Capacitors and Inductors Chapter First-Order Circuits...
Ngày tải lên: 08/05/2014, 15:37
fundamentals of heat and mass transfer solutions manual phần 1 docx
... 0.1m × 25m ) = 7 .85 m Hence, ( ) q = 7 .85 m2 10 W/m2 ⋅ K (150 − 25) K + 0 .8 × 5.67 × 10 8 W/m2 ⋅ K 4234 − 2 984 K q = 7 .85 m2 (1, 250 + 1, 095) w/m = ( 981 3 + 85 92 ) W = 18, 405 W < (b) ... condition of ε = 0.05, Ts = Tsur + 10 = 35°C = 3 08 K and Tsur = 25°C = 2 98 K, find that h r = 0.05 × 5.67 × 10 8 W m ⋅ K (3 08 + 2 98 ) 3 082 + 2 982 K = 0.32 W m ⋅ K ) ( h r,a = × 0.05 × 5.67 ×10 8 W ... ×10 8 W m ⋅ K ((3 08 + 2 98 ) ) K3 = 0.32 W m ⋅ K < < The free convection coefficient with Ts = 35°C and T∞ = Tsur = 25°C, find that h = 0. 98 T1/ = 0. 98 (Ts − T∞ ) 1/ = 0. 98 (3 08 − 2 98 ) 1/ < = 2.1W...
Ngày tải lên: 08/08/2014, 17:20
fundamentals of heat and mass transfer solutions manual phần 2 docx
... − T∞ ) W q′=0 .8 ( 0.2m ) 5.67 × 10 8 486 4 − 2 984 K K4 m ⋅ W +20 (π × 0.2m ) ( 486 -2 98) K m2 ⋅ K ( ) q′= (1365+2362 ) W/m=3727 W/m < With the insulation, the thermal circuit is of the form Continued ... 10−3 kW W = 625 kWh With a unit electric power cost of $0. 08/ kWh, the annual cost of the heat loss is C = ($0. 08/ kWh)625 kWh = $50.00 Hence, an insulation thickness of δ = 25 mm < will satisfy the ... 298K = 20 ⋅K ln ( 0.3m/0.2m ) m 2π ( 0.0 58 W/m ⋅ K ) W π (0.3m ) Ts,o − 2 984 K +0 .8 × 5.67 ×10 -8 ⋅ K4 m ( ( ) ) ( ) ( ) ( ) By trial and error, we obtain Ts,o ≈ 305K in which case q′= ( 486 -305)...
Ngày tải lên: 08/08/2014, 17:20
fundamentals of heat and mass transfer solutions manual phần 3 pdf
... below k T1 T2 T3 T4 T5 T6 T7 T8 T9(°C) 75 75.0 75.7 76.3 76.3 76.6 76.6 75 76.3 76.9 77.0 77.3 77.3 77.5 80 80 .0 80 .0 80 .2 80 .2 80 .3 80 .3 85 86 .3 86 .3 86 .3 86 .3 86 .3 86 .4 90 92.5 93.2 93.2 93.2 ... 183 .9 185 186 .6 187 .2 182 .3 180 .8 180 .4 180 .3 180 .3 185 186 .6 167.0 163.3 162.5 162.3 162.3 162.2 185 105 .8 96.0 94.2 93 .8 93.7 93.6 93.6 ← initial estimate ← ε
Ngày tải lên: 08/08/2014, 17:20
fundamentals of heat and mass transfer solutions manual phần 4 docx
... T0 85 85 85 85 76.9 76.9 68. 8 68. 8 61.7 61.7 T1 85 85 85 76.9 76.9 68. 8 68. 8 61.7 61.7 55.6 T2 85 85 68. 8 68. 8 60.7 60.7 54.6 54.6 49.5 49.5 T3 85 52.5 52.5 44.4 44.4 40.4 40.4 37.3 37.3 34 .8 ... 186 .6 217.7 2 48. 8 T1 200 184 .1 175.6 1 68. 6 163.3 1 58. 8 155.2 152.1 145.1 T2 200 181 .8 166.3 154 .8 145.0 137.1 130.2 124.5 119.5 T3 200 181 .8 165.3 150.7 1 38. 8 1 28. 1 119.2 111.3 104.5 T4 200 181 .8 ... 610.0 699.7 788 .8 867.4 945.0 1014.0 1 081 .7 1141.5 1199 .8 1250.5 1299 .8 1341.2 1 381 .6 300 403.5 506.9 584 .1 661.1 724.7 787 .9 84 2.3 89 6.1 943.2 989 .4 1029.9 1069.4 1103.6 1136.9 1164 .8 10.4 20.7...
Ngày tải lên: 08/08/2014, 17:20
fundamentals of heat and mass transfer solutions manual phần 5 ppsx
... K , ν = 15 .89 × 10 −6 m / s, Pr = 0.707 (a) With Re D = VD o / ν = m / s × 0.1m / 15 .89 × 10 −6 m / s = 18, 88 0, application of the Churchill-Bernstein correlation yields 0.62 ( 18, 800 ) 1/ Nu ... emissivity of Nichrome wire Electrical current Temperature of air flow and surroundings Velocity of air flow (a) Surface and centerline temperatures of the wire, (b) Effect of flow velocity and electric ... 1.648E4 */ Aw 6.24 etaf 1.44 etaoc 0.9 78 /* Correlation results and air thermophysical properties at Tf NuLbarPr ReL Tf hLbar k nu uinf 2294 0.7035 1.63E6 325 53 .82 0.0 281 5 1 .84 1E-5 25 m 0. 981 4...
Ngày tải lên: 08/08/2014, 17:20
fundamentals of heat and mass transfer solutions manual phần 6 docx
... K 0.3 h w = w 0.023 Re0 .8 Prw = 0.023 (75, 700 )0 .8 (3.56 )0.3 = 2060 W / m2 ⋅ K Dw Di 0. 084 m With Re D,a = VD o / ν a = m / s × ( 0.1m ) /15 .89 × 10 −6 m / s = 18, 88 0, the Churchill-Bernstein ... -2 4.321 × 10 -2 5.041 × 10 -2 5.215 × 10 Eq (7) 3 48. 6 463.7 524.6 539.0 915.0 89 8 89 1.0 88 9.5 Hence, we find & m a = 5.22 ×10 −2 kg/s Tm,o = 89 0 K < COMMENTS: To achieve the same cooling rate ... 330 K): cp = 10 08 J/kg⋅K, µ = 1 98. 8 × 10-7 N⋅s/m2, k = 0.0 285 W/m⋅K, Pr = 0.703 ANALYSIS: (a) For m = 0.01 kg/s, ReD = 4m π Dµ = 0.04 kg/s/π(0.05 m)1 98. 8 × 10-7 N⋅s/m2 = 12 ,81 0 Hence, the...
Ngày tải lên: 08/08/2014, 17:20
fundamentals of heat and mass transfer solutions manual phần 7 pot
... 125°C = 3 98 K, h b = 6.517 (3 98 − 300 ) 0. 188 ) ( + 3.402 ×10 8 (3 98 + 300 ) 3 982 + 3002 = 21.33 W m ⋅ K q f = 21.33W m 2⋅ K × 1 .85 5 × 10−2 m ( 0.0025 m )(3 98 − 300 ) K +14.3 W m ⋅ K × 2 .82 7 × 10−5 ... or h r = 0.95 × 5.67 × 10 8 W m ⋅ K (3 18 + 288 ) 3 182 + 288 2 K = 6.01W m ⋅ K Neglecting radiation: For the horizontal cylinder, Eq 9.34 yields ( ) gβ Ts,i − T∞ Di 9 .8 m s (1 303 K )( 45 − 15 ... NuLbar Pr RaD RaL 244.7 80 1.3 219.2 3.665E9 1.357E11 */ /* Results - properties, Ts = 200 C; Tf = 383 K Pr alpha beta deltaT k nu Tf 219.2 7. 188 E -8 0.0007 180 0.1357 1. 582 E-5 383 /* Correlation description:...
Ngày tải lên: 08/08/2014, 17:20
fundamentals of heat and mass transfer solutions manual phần 8 ppsx
... Cmin/Cmax = (5,000 × 41 78) /(10,000 × 4 188 ) = 0.499 Combining Eqs (1) and (2), find Th,o = 80 °C − 0.7 × 3 48. 2 × 103 W / (10,000/3600 ) k g / s × 4 188 J / k g ⋅ K ( Th,o = ( 80 − 21.0 ) ° C = 59°C ... ⋅ K " 1 787 2250 2690 3112 ' h o W / m2 ⋅ K " 38. 4 38. 4 38. 4 38. 4 ' U W / m2 ⋅ K " ' 37.6 37 .8 37.9 37.9 Note that while hi varies nearly 50%, there is a negligible effect on the value of U COMMENTS: ... = 0. 088 = ηo,h mLf 11.33 Substituting into Eq (2), 0.01 UA = + + 0. 088 × 5000 × 0. 785 0.706 × 52.7 × 0. 785 ( −1 W W = 25.6 K K ) Hence, with Cmin = m cp = 0.03 kg / s × 10 08 J /...
Ngày tải lên: 08/08/2014, 17:20
fundamentals of heat and mass transfer solutions manual phần 9 ppt
... 10 8 W Gd = 1.51×10 8 W / ×10 −6 m = 7.57 ×10 −3 W / m < (b) Since λmax T = 289 8 µm⋅K, it follows that λmax(e) = 289 8 µm⋅K/400 K = 7.25 µm and λmax(r) = 289 8 µm⋅K/273 K = 10.62 µm < -4 -8 λ ... 0 .8( 1 − 0. 68) = 0.460 For the maximum allowable value of Ts = 180 0 K, it follows that ho = ho = 0.391 × 5.67 × 10 8 (2400) + 0 .8 × 5.67 × 10 8 (300)4 − × 0.46 × 5.67 × 10 8 ( 180 0)4 ( 180 0 − 300) ... −11 2.4504 ×10 K −1 −1 ⋅s ) (29 28 K ) 3 ln 29 28 + 2741 29 28 + 300 − ln 29 28 − 2741 29 28 − 300 2741 −1 300 +2 tan −1 − tan 29 28 29 28 t = 0.40604 ( 3.4117...
Ngày tải lên: 08/08/2014, 17:20
fundamentals of heat and mass transfer solutions manual phần 10 doc
... 13 .89 (Cont.) 0* 0.2071 0.1713 0.1997 0 .82 84 0. 585 8 0 .82 87 0 .80 03 0.1696 0.2051 0* 0* 0.001997* 0.002001 0* 0* The Fij shown with an asterisk were independently determined From knowledge of the ... appropriate expressions for each of the processes, find ( ) −62 .84 W / m + 18. 85W − 1.069 × 10−9 700 − Tc − 2 38 W = < Tc = 84 2.5 K From Eq (4), with Tc = 81 5 K, the electrical power required to maintain ... − T2 ) Accordingly, 0 .85 − 1 + + 0.01 R rad,hc = 8 4 (0.01 m − 0.002 m ) × 5.67 × 10 W / m ⋅ K ( 288 K )2 + ( 2 68 K )2 ( 288 + 2 68 ) K where, again, it is assumed...
Ngày tải lên: 08/08/2014, 17:20
Fundamentals of engineering thermodynamics, 7th edition
... Composition 683 746 12 .8. 4 Humidification 750 12 .8. 5 Evaporative Cooling 752 688 11.9.6 Chemical Potential for Ideal 689 Chapter Summary and Study Guide 690 12 .8. 6 Adiabatic Mixing of Two Moist Air Solutions ... Mixtures 82 2 13.7.2 Standard Chemical Exergy of Other 82 5 13 .8 Applying Total Exergy 82 6 13 .8. 1 Calculating Total Exergy 82 6 Substances 13 .8. 2 Calculating Exergetic Efficiencies 82 9 Chapter Summary ... Inequality 264 Chapter Summary and Study Guide 266 Using Entropy 284 6.6 285 Introducing the T dS Equations 286 Entropy Change of an Incompressible Substance 288 Entropy Change of an Ideal Gas 289 6.5.1...
Ngày tải lên: 25/08/2013, 09:33
Principles of transistor circuits, ninth edition
... Iliffe Books Ltd 1959 Second edition 1961 Third edition 1965 Fourth edition 1969 Fifth edition 1975 Sixth edition 1 981 Seventh edition 1990 Eighth edition 1994 Ninth edition 2000 © S W Amos and ... generators Digital circuits Further applications of transistors and other semiconductor devices vi 22 53 65 78 90 112 134 1 58 180 205 227 255 281 293 3 28 Appendix A The manufacture of transistors ... Principles of T ransistor Circuits This Page Intentionally Left Blank Principles of Transistor Circuits Ninth Edition Introduction to the Design of Amplifiers, Receivers and Digital Circuits S...
Ngày tải lên: 28/12/2013, 22:19
Fundamentals of Ecological Modelling, Third Edition pot
... for ISBN: 0- 080 -44015-0 (Hardbound edition) ISBN: 0- 080 -440 28- 2 (Paperback edition) ~The paper used in this publication meets the requirements of ANSI NISO Z39. 48- 1992 (Permanence of Paper) Printed ... lay-out of the cabins, etc These details are, of course, irrelevant to the objectives of that model Other models of the ship will serve other aims: blue prints of the electrical wiring, lay-out of ... 8. 3 Characteristics and Structure of Ecotoxicological Models 8. 4 An Overview: The Application of Models in Ecotoxicology 8. 5 Estimation of Ecotoxicological...
Ngày tải lên: 23/03/2014, 01:20
unit operation of chemical engineering 6th ed, solutions manual
... Flow of Compressible Fluids 150 Transportation and Metering of Fluids 187 Pipe, Fittings, and Valves Pumps 187 133 134 140 145 147 1 48 149 157 162 171 182 183 185 194 mcca_fm.qxd 8/ 3/00 8: 42 ... Exchangers 83 9 mcca_fm.qxd xiv 8/ 3/00 8: 42 AM Page xiv CONTENTS Chromatography Symbols Problems References Membrane Separation Processes 85 7 Separation of Gases Separation of Liquids 26 84 5 85 7 85 2 85 4 ... Pastes / Dryers for Solutions and Slurries / Selection of Drying Equipment Symbols Problems References 25 80 7 Fixed-Bed Separations 81 2 Adsorption Adsorption Equipment 81 2 80 9 81 0 81 3 Equilibria;...
Ngày tải lên: 01/04/2014, 11:53
unit operation of chemical engineering 7th ed, solutions manual
... are a student using this Manual, you are using it without permission Page 28 - PROPRIETARY MATERIAL © The McGraw-Hill Companies, Inc All rights reserved No part of this Manual may be displayed, ... are a student using this Manual, you are using it without permission Page - PROPRIETARY MATERIAL © The McGraw-Hill Companies, Inc All rights reserved No part of this Manual may be displayed, ... 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...
Ngày tải lên: 01/04/2014, 11:53