thermo 7e sm chap04 1 (1)

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thermo 7e sm chap04 1 (1)

Solutions Manual forThermodynamics: An Engineering ApproachSeventh EditionYunus A. Cengel, Michael A. Boles McGraw-Hill, 2011Chapter 4ENERGY ANALYSIS

thermo 7e sm chap04 1 (1) OF CLOSED SYSTEMSPROPRIETARY AND CONFIDENTIALTins Manual is the proprietary property of The McGraw-Hill Companies. Inc. (“McGraw-Hill”) and protected

by copyright and other state and federal laws. By opening and using this Manual the user agrees to the following restrictions, and if the recipient d thermo 7e sm chap04 1 (1)

oes not agree to these restrictions, the Manual should be promptly returned unopened to McGraw-Hill: This Manual is being provided only to authorized

thermo 7e sm chap04 1 (1)

professors and instructors for use in preparing for the classes using the affiliated textbook. No other use or distribution of this Manual is permitte

Solutions Manual forThermodynamics: An Engineering ApproachSeventh EditionYunus A. Cengel, Michael A. Boles McGraw-Hill, 2011Chapter 4ENERGY ANALYSIS

thermo 7e sm chap04 1 (1) splayed or distributed in any form or by any means, electronic or otherwise, w ithout the prior w ritten permission of McGraw-Hill.PROPRIETARY X1ATEP*

• *..."• •...........'chers and educators i thermo 7e sm chap04 1 (1)

Pham Quang Huy (ebooUyou ortine@gnxaii com)Moving Boundary Work4-1C Yes.4-2C The area under (lie process curve, and thus (lie boundary wort done, is

thermo 7e sm chap04 1 (1)

greater in the constant pressure case.4-3 1 kPa - tn’ =1 k(N/nr )• tnJ = 1 kN • in = 1 kJ4-4 Helium is compressed 111 a piston-cylinder device. The in

Solutions Manual forThermodynamics: An Engineering ApproachSeventh EditionYunus A. Cengel, Michael A. Boles McGraw-Hill, 2011Chapter 4ENERGY ANALYSIS

thermo 7e sm chap04 1 (1) s The gas constant of helium is R = 2.0’69 kJ.-*kg K (Table A-l).Analysis The initial specific volume is ụ 7 m'p= 77 = 777- = 7 ni Jrtg(kp •/m 1 kgusr

a?Using the ideal gas equation.^0^Jl50kPaX7m?y = 5051K1 R 2.0769 kJ/kg KSince the pressure stays constant.T2 = ~rTi = ị^J-(5O5.IK) = 216.5 K<47 nrI--- thermo 7e sm chap04 1 (1)

>u (nr)7and the work integral expression givesFFỞ0W = Í 'ĩ™ = p

thermo 7e sm chap04 1 (1)

IETARY MATERIAL, c 2011 The McGraw-Hill Companies. Inc. Lanted dcstributian pcimitted only to teachers and educators foe course

Solutions Manual forThermodynamics: An Engineering ApproachSeventh EditionYunus A. Cengel, Michael A. Boles McGraw-Hill, 2011Chapter 4ENERGY ANALYSIS