Polytropic Process. A polytropic process is a quasiequilibrium process described by

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Wilfred Carter
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1 Polytropc Procss A polytropc procss s a quasqulbrum procss dscrbd by pv n = constant (Eq. 3.5 Th xponnt, n, may tak on any valu from to dpndng on th partcular procss. For any gas (or lqud, whn n = 0, th procss s a constant-prssur (sobarc procss. For any gas (or lqud, whn n = ±, th procss s a constant-volum (somtrc procss. For a gas modld as an dal gas, whn n = 1, th procss s a constant-tmpratur (sothrmal procss.

2 Chaptr 4 Control Volum Analyss Usng Enrgy

3 Mass Rat Balanc tm rat of chang of mass contand wthn th control volum at tm t tm rat of flow of mass n across nlt at tm t tm rat of flow of mass out across xt at tm t dm dt (Eq. 4.1 = m m

4 Enrgy Rat Balanc tm rat of chang of th nrgy contand wthn th control volum at tm t nt rat at whch nrgy s bng transfrrd n by hat transfr at tm t nt rat at whch nrgy s bng transfrrd out by work at tm t nt rat of nrgy transfr nto th control volum accompanyng mass flow de dt V = Q W m ( u gz m ( u V gz (Eq. 4.9

5 Evaluatng Work for a Control Volum Th xprsson for work s W = W m ( p v m ( p v (Eq. 4.1 whr W accounts for work assocatd wth rotatng shafts, dsplacmnt of th boundary, and lctrcal ffcts. m ( p v s th flow work at xt. m ( pv s th flow work at nlt.

6 de dt Control Volum Enrgy Rat Balanc (On-Dmnsonal Flow Form Usng Eq. 4.1 n Eq. 4.9 V = Q W m ( u pv gz m (Eq ( u p v V For convnnc substtut nthalpy, h = u pv gz de dt V = Q W m ( h gz m ( h V gz (Eq. 4.14

7 Control Volum Enrgy Rat Balanc (On-Dmnsonal Flow Form In practc thr may b svral locatons on th boundary through whch mass ntrs or xts. Multpl nlts and xts ar accountd for by ntroducng summatons: de dt = Q W V m ( h gz V m ( h gz (Eq Eq s th accountng balanc for th nrgy of th control volum.

8 Many mportant applcatons nvolv on-nlt, on-xt control volums at stady stat. Th mass rat balanc rducs to. Control Volum Enrgy Rat Balanc (Stady-Stat Form, On-Inlt, On-Ext m m m = = 1 = ( V (V ( z z g h h m W Q Eq. 4.0a or dvdng by mass flow rat ( V (V ( z z g h h m W m Q = Eq. 4.0b

9 Nozzls and Dffusrs Nozzl: a flow passag of varyng crosssctonal ara n whch th vlocty of a gas or lqud ncrass n th drcton of flow. Dffusr: a flow passag of varyng crosssctonal ara n whch th vlocty of a gas or lqud dcrass n th drcton of flow.

10 If th chang n potntal nrgy from nlt to xt s nglgbl, g(z 1 z drops out. If th hat transfr wth surroundngs s nglgbl, drops out. = ( V (V ( z z g h h m W Q Eq. 4.0a Nozzl and Dffusr Modlng W = 0. = V V ( h h (Eq. 4.1 Q

11 Turbns Turbn: a dvc n whch powr s dvlopd as a rsult of a gas or lqud passng through a st of blads attachd to a shaft fr to rotat.

12 Turbn Modlng (V V 0 = Q ( 1 W m h1 h g( z1 z If th chang n kntc nrgy of flowng mattr s nglgbl, ½(V 1 V drops out. Eq. 4.0a If th chang n potntal nrgy of flowng mattr s nglgbl, g(z 1 z drops out. If th hat transfr wth surroundngs s nglgbl, Q drops out. W = m ( h h 1

13 Comprssors and Pumps Comprssors and Pumps: dvcs n whch work s don on th substanc flowng through thm to chang th stat of th substanc, typcally to ncras th prssur and/or lvaton. Comprssor : substanc s gas Pump: substanc s lqud

14 Comprssor and Pump Modlng (V V 0 = Q ( 1 W m h1 h g( z1 z Eq. 4.0a If th chang n kntc nrgy of flowng mattr s nglgbl, ½(V 1 V drops out. If th chang n potntal nrgy of flowng mattr s nglgbl, g(z 1 z drops out. If th hat transfr wth surroundngs s nglgbl, Q drops out. W = m ( h h 1

15 Hat Exchangrs Drct contact: A mxng chambr n whch hot and cold strams ar mxd drctly. Tub-wthn-a-tub countrflow: A gas or lqud stram s sparatd from anothr gas or lqud by a wall through whch nrgy s conductd. Hat transfr occurs from th hot stram to th cold stram as th strams flow n oppost drctons.

16 0 = Q Hat Exchangr Modlng W V m ( h gz If th kntc nrgs of th flowng strams ar nglgbl, m (V / and m (V / drop out. If th potntal nrgs of th flowng strams ar nglgbl, m gz and m gz drop out. If th hat transfr wth surroundngs s nglgbl, Q drops out. 0 = m h m h V m ( h gz (Eq W = 0.

17 Throttlng Dvcs Throttlng Dvc: a dvc that achvs a sgnfcant rducton n prssur by ntroducng a rstrcton nto a ln through whch a gas or lqud flows. Mans to ntroduc th rstrcton nclud a partally opnd valv or a porous plug.

18 (V V 0 = Q ( 1 W m h1 h g( z1 z W = 0. If th chang n kntc nrgy of flowng mattr upstram and downstram of th rstrcton s nglgbl, ½(V 1 V drops out. If th chang n potntal nrgy of flowng mattr s nglgbl, g(z 1 z drops out. If th hat transfr wth surroundngs s nglgbl, drops out. Q Throttlng Dvc Modlng h = h 1 (Eq. 4. Eq. 4.0a

19 Systm Intgraton Engnrs cratvly combn componnts to achv som ovrall objctv, subjct to constrants such as mnmum total cost. Ths ngnrng actvty s calld systm ntgraton. Th smpl vapor powr plant of Fg 4.16 provds an llustraton.

20 Th Mass Balanc (Transnt Analyss Transnt: stat changs wth tm. Intgrat mass rat balanc (Eq. 4. from tm 0 to a fnal tm t. Ths bcoms t 0 dm dt dt = t 0 m m t m (0 ( dt t 0 m = m m dt (Eq. 4.3 whr m s amount of mass ntrng th control volum through nlt, from tm 0 to t. m s amount of mass xtng th control volum through xt, from tm 0 to t.

21 Th Enrgy Balanc (Transnt Analyss Intgrat nrgy rat balanc (Eq. 4.15, gnorng th ffcts of kntc and potntal nrgy, from tm 0 to a fnal tm t.! de $ t # &dt = t Q dt t t! $! $ W 0" dt % 0 dt # m 0 h &dt t # m 0 h &dt 0 " % " % Whn th spcfc nthalps at nlts and xts ar constant wth tm, ths bcoms E (t E (0 = Q W m h m h (Eq. 4.5

22 Consdr a typcal gardn hos Assum th prssur n th hos (stat 1 s 30 psg at a tmpratur of 70 o F wth a vlocty of 5 ft/sc. Th chld rcvs th watr at 65 o F What s th xt vlocty?

Consdr th fgur blow of a prfct gas stuaton. On klogram of ntrogn flls th cylndr of a pston- cylndr assmbly. Thr s no frcton btwn th pston and th cylndr walls, and th surroundngs ar at 1 atm.

23 Consdr th fgur blow of a prfct gas stuaton. On klogram of ntrogn flls th cylndr of a pston- cylndr assmbly. Thr s no frcton btwn th pston and th cylndr walls, and th surroundngs ar at 1 atm. Th ntal volum and prssur n th cylndr ar 1 m 3 and 1 atm, rspctvly. Hat transfr to th ntrogn occurs untl th volum s doubld. (1 atm = bar; 1 bar = 10 5 N/m a Dtrmn th work for th procss, n kj. b Dtrmn th hat transfr for th procss, n kj, assumng th spcfc hat (0.74 kj/(kgk s constant. Rcall: R = 8.314kJ/(kmol K; M N = 8.01kg/kmol

24 Molcular wght of N gas s M N = 8.01kg/kmol R = R(unvrsal constant/m (Molcular wght = 8.314/8.01 =.968 kj/(kg K PV = mrt (Know P 1, V 1, m 1, R Solv for T 1 = 341.4K PV = mrt (Know P, V, m, R Solv for T (But P =P 1, V = V 1, thrfor T = *T 1 =68.8K Q = m(u -u 1 W du/dt = c v Or (u -u 1 = c v (T -T 1 =.74(341.4K= 53.3 kj/kg Q = 1kg(53.3kJ/kg x10 5 N/m (m 3-1m 3 (1kJ/(10 3 N/m = kj

25 Two klograms of watr at 5 C ar placd n a pston cylndr dvc undr 100 kpa absolut prssur as shown n th dagram (Stat(1. Hat s addd to th watr at constant prssur untl th pston rachs th stops at a total volum of 0.4 m 3 (Stat (. Mor hat s thn addd at constant volum untl th tmpratur of th watr rachs 300 C (Stat (3. Draw a P-v and a T-v dagram of th stats and procsss of th problm and nclud all th rlvant nformaton on th dagram. In ths cas thr ar thr stats and two procsss (stat 1 to stat and stat to stat 3. Th dagrams do not hav to b drawn to scal. Dtrmn th qualty (x of th flud and th mass of th vapor at stat (. (Schmatc s not ncssarly to scal. Calculat th spcfc volum, spcfc ntrnal nrgy, and spcfc nthalpy of th stat.!!!

27 1 3 Constant Tmp.

ANALYSIS: The mass rate balance for the one-inlet, one-exit control volume at steady state is

ANALYSIS: The mass rate balance for the one-inlet, one-exit control volume at steady state is Problm 4.47 Fgur P4.47 provds stady stat opratng data for a pump drawng watr from a rsrvor and dlvrng t at a prssur of 3 bar to a storag tank prchd 5 m abov th rsrvor. Th powr nput to th pump s 0.5 kw.