3 Energy Non-Flow Energy Equation (NFEE) Internal Energy. MECH 225 Engineering Science 2

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1 MECH 5 Egieerig Sciece 3 Eergy 3.3. No-Flow Eergy Equatio (NFEE) You may have oticed that the term system kees croig u. It is ecessary, therefore, that before we start ay aalysis we defie the system that we are lookig at. o do this we costruct a imagiary boudary aroud what we are iterested i for examle, the cricket ball (struck by Nasser Hussei) or the water i the kettle). he dealig with a o-flow situatio, the the system will be of fixed mass - o matter crosses the boudary - so it is useful to defie a cotrol mass. If we are cosiderig a flow situatio, the a cotrol volume through which the fluid flows is more useful. I hoe these ideas will become clearer whe we cosider some examles. A tyical closed system is a gas eclosed i a cylider by meas of a isto. he gas iside the cylider is the cotrol mass. Iteral Eergy From the First Law, + E + E k + E + U But for a o-flowig gas, its velocity will be 0 to start with, ad whe it has settled dow after the rocess, its velocity will agai be 0, so the chage i kietic eergy, E k 0. Similarly, there is o sigificat chage i otetial eergy, so E 0. here will, however, be a chage i the iteral eergy, U. So, the o-flow eergy equatio (NFEE) becomes, simly, + U or + U U where U is the iteral eergy i state, after the rocess ad U is the iteral eergy i state, before the rocess. Examle Durig a comlete cycle, a system is subjected to the followig heat trasfers: 800 kj from the surroudigs ad 500 kj to the surroudigs. At two oits i the cycle, work is trasferred to the surroudigs of 96 kj ad 0 kj. At a third oit there is a further work trasfer. Determie its magitude ad sese. Examle I a air comressor, the comressio takes lace at costat iteral eergy ad 00 kj of eergy are rejected to the coolig water er kg of air. Determie the secific work trasfer durig the comressio stroke.

2 Solutio Here, the system is defied for us o details are give as to its ature. For a comlete cycle, we kow that: Σ + Σ 0 Σ kJ Σ where is the value of the ukow work trasfer i kj kj he egative sig tells us that it is a work trasfer to the surroudigs. Solutio he system is the air i the comressor. e do ot kow its mass, so let us suose that it is m kg. he iteral eergy remais costat, so U, the chage i iteral eergy is 0. he o-flow eergy equatio becomes + 0 he eergy trasferred by heatig is from the air i the comressor to the coolig water, ad is therefore egative. It is 00 kj er kg, so for m kg: - 00m kj Alyig the NFEE, -00m m kj You are asked to fid the secific work trasfer. his meas the work trasfer for every kilogram, ad this is give the symbol, w. herefore, w 00 kj m hat does the ositive sig for the work trasfer tell you? Is this what you would exect? Ca you exlai why it is ositive?

3 3.4 Alicatios of the NFEE he o-flow eergy equatio ca be alied to ay o-flow thermodyamic rocess. e ca classify rocesses deedig o the coditios uder which they take lace, ad the tye of fluid. yical rocesses that are erformed o gases are at costat volume, costat ressure, costat temerature (isothermal) ad with o eergy trasfer by heatig (adiabatic). Calculatig eergy trasfers (a) Eergy trasfer by heatig, e saw i.6 that we ca calculate the eergy trasferred by heatig usig the equatio: mc( - ) (b) Eergy trasfer by workig, How ca we calculate the eergy trasferred by workig? Force x distace moved ad Force ressure x area herefore, ressure x area x distace moved. Cosider a gas eclosed by a isto: he force o the isto A area, A he distace moved dx herefore the work doe by the gas is d Adx d dx he total work doe, whe the isto moves so that the volume icreases from to is: isto moves d his is the area uder the - diagram betwee states ad. he egative sig is there, because as the gas exads it does work o the surroudigs. e eed to itegrate d because as the gas exads the ressure may ot remai costat, i geeral, so we eed to add u all the small amouts of work doe at each of the iterveig ressures whe the isto moves by a small distace, dx. For a erfect gas, we kow the relatioshis betwee ressure, volume ad temerature. If we kow the coditios of the rocess, the we ca calculate the work doe ad the heat trasferred. But these quatities, ad, deed o the rocess, ot oly o the begiig ad ed states.

4 A rocess ca be show o a - diagram. O the diagrams below, isotherms are marked as dashed lies. Isotherms are lies of costat temerature. For a erfect gas at costat temerature, costat. So these dashed lies rereset costat temerature rocesses. ut i the lies o the - diagrams, to rereset the relevat rocess i each of the followig ad see if you ca write dow, or calculate, ad for each. (a) Costat olume (b) Costat ressure cost force isto moves

5 (c) Isothermal (at costat temerature) (d) Adiabatic (o heat trasfer) erfectly isulated

6 You should have the followig: (a) Costat olume 0 U mc v ( - ) (b) Costat ressure cost force isto moves -( ) (shaded area uder the lie) For a erfect gas, -mr( ) U mc v ( - )+mr( ) m(c v + R) ( ) mc ( )

7 (c) Isothermal (at costat temerature) costat d mr d mr l U 0 - (d) Adiabatic (o heat trasfer) erfectly isulated c where γ. 4 for cool air cv 0 ( ) γ mr( ) γ mc U v ( )

8 Some questios for you to cosider: () hy is the work doe i a costat volume rocess equal to 0? () hy is the chage i iteral eergy i a isothermal rocess equal to 0? (3) Derive the exressio for the work doe i a olytroic rocess, that is a rocess which ca be modelled by the equatio costat, where ca take ay value, ad is ormally foud by exerimet. Examle (olytroic rocess) Air at.4 bar with a secific volume of. m 3 kg - is comressed to 0 bar accordig to v.3 costat. Fid: the ew secific volume the secific work doe the iitial ad fial temeratures the secific heat trasfer Reeat for a isothermal rocess. For air, take R 87 Jkg - K - ad c 005 Jkg - K -. Solutio:.4 bar v. m 3 kg - 0 bar v? v.3 costat v.3.3 v v.4 *..4 *.67 v v m 3 kg - w v v (see summary below) NOE: lower case letters are used to deote secific quatities, i.e. values for kg mass * 0 * * 0 * w * * w kj kg - he iitial ad fial temeratures we ca fid usig the equatio of state for a erfect gas (see.5 gas laws) 5 J

9 5 v.4 * 0 * K 3 C R 87 5 v 0 * 0 * 0.55 ad 080 K 807 C R 87 o fid the secific heat trasfer, aly the NFEE: q u w u c v ( ) (c R)( ) (005-87)*( ) 78*495 J u kj kg - q kj kg - For a isothermal rocess, we have K v.4 *. ad v costat, v v v m 3 kg - 0 v w R l 87 * l 87 * * (.659) v. w 447 kj kg - For a isothermal rocess u 0, so the NFEE gives q -w q -447 kj kg - he thermodyamic relatioshis for erfect gases are summarised below:

10 Summary of thermodyamic relatioshis for ideal gases For ay ideal gas: ressure (Nm - or a) mr volume (m 3 ) (Equatio of State) temerature (K) m mass (kg) R gas costat (Jkg - K - ) 87 Jkg - K - for air c γ cv γ the adiabatic idex.4 for air R c v γ c v secific heat at costat volume (Jkg - K - ) c c v R c secific heat at costat ressure (Jkg - K - ) For ay rocess: U mc v + U (First Law of hermodyamics) U is the chage i iteral eergy (J) is the chage i temerature is the work doe (J) is the eergy trasfer by heatig (J) For a adiabatic rocess: γ γ also mcv ( γ ) 0 γ ( γ ) / γ For a costat ressure rocess: ( - ) mr( ) mc ( )

11 For a costat temerature (isothermal) rocess: l l mr - For a costat volume rocess: 0 mc v ( ) For a olytroic rocess: is the olytroic idex / ) ( mc v ( ) Further readig: Baco ad Stehes, Mechaical echology Rogers, G ad Mayhew, Y, Egieerig hermodyamics ork ad Heat rasfer Ch 3 he Oe Uiversity, 36 Itroductio to thermofluid mechaics Block 4