Thermodynamica 1. college 8 boek hoofdstuk 5. Bendiks Jan Boersma Thijs Vlugt Theo Woudstra. March 4, 2010

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1 Thermodynamica Bendiks Jan Boersma Thijs Vlugt Theo Woudstra March 4, 00 college 8 boek hoofdstuk 5

2 College 7 Massa en energie behoud open systemen Dimensie analyse Warmte transport Geleiding Convectie Straling March 4, 00

3 Dimensie analyse E f(, R, t) 5 kg 3 E[ kgm / s ] c R t c m s m E R c t t = 0.06 s, R= 00 m t = 0.05 s, R= m t = s, R = 30 m E~x0^ ergs E ~ 000^ J (00 TJ) 5kTON TNT March 4, 00 3 G.I. Taylor, 947

4 st law for stationary open system from lecture 6 control volume V p W cv E cv z Q p V z general: Q E cv stationary: W cv E 0 and m cv 0 m m m cv m h V gz m h V gz Q W m h h V V g z z cv March 4, 00 4

5 Wet van Bernoulli Q W m h h V V g z z cv V V Q W cv m h h g z z hu pv Q W cv m u u pv pv V V gz z Q W cv U U m pv pv V V gz z Neem nu aan dat warmte en arbeid nul zijn en de interne energie constant pv pv V V gz z 0; v p V gz Const Wet van Bernoulli March 4, 00 5

6 Bernoulli, lift van een vleugel Hoge snelheid lage druk P V Const Lage snelheid hoge druk March 4, 00 6

7 Leeglopend vat, p 4.6 (editie 6) p V gz p V gz p p neem aan dat, V V V g( z z ) gz dy / y dt dy dt / y / y tc dmcv d Az m A gz dt dt dz A g z dt A dz z A A t 0, z z C z A z gt z A A gdt z gt C A vat leeg op tijd t als A A gt z 0 tleeg z/g A A March 4, 00 7

8 Problem 4.5 V V Q W mh h g( z z) T T vgl (3.43) ht ( ) ht ( ) CdT p m Q 0.0 kg/ s 0.033kW March 4, 00 8

9 The nd law of Thermodynamics March 4, 00 9

10 Motivation Observations the st law states that energy is conserved, but it says nothing about the direction of energy-conversion processes. Nature, however, very well prescribes a direction, in which processes occur (spontaneously). a nd law of Thermodynamics is needed whenever a process does occur spontaneously, then there is an imbalance between begin- and end-state there is a non-equilibrium. This nonequilibrium condition can be used to obtain work (Arbeidspotentieel). how much work can potentially be obtained from a process? (The nd law gives an answer) March 4, 00 0

11 Spontane processen en het Arbeidspotentieel Spontaneous processes have driving forces, so that process occurs spontaneously are thus in non-equilibrium the driving forces can be used to obtain work (Arbeidspotentieel) Spontaneous processes never reverse spontaneously and are hence called irreversible The reversed process is only possible with external influence High pressure air (driving force) can be used to drive a turbine and obtain net work W (Arbeidspotentieel) March 4, 00

12 A second law of Thermodynamics is needed in order to answer: what is the direction in which an energy conversion takes place spontaneously. (And what processes do not occur by themselves) what is the theoretical maximum value of work that can be obtained from a process what is the value of work that can be realized technically. And how can one approach the theoretical maximum March 4, 00

13 Relevance of these questions and the nd law any energy conversion system, like power plants and their efficiency biological (biochemical) conversions & muscular action astronomy Thermodynamics determines the processes that do or do not occur in nature. It determines the processes that can not be realized technologically. March 4, 00 3

14 (Historische) Formulering van de HW - Clausius Het is onmogelijk om, zonder meer, warmte van lagere naar hogere temperatuur te brengen Dit betekent dat om dit toch te bereiken, arbeid toegevoerd moet worden op het systeem. Warm reservoir Q H out=q C in no! (Clausius) Q H out cyc W=0 no! (Clausius) cyc W=Q H out-q C in Q C in Koud reservoir Q C in March 4, 00 4

15 (Historische) Formulering van de HW Kelvin-Planck Het is onmogelijk dat een systeem als een thermodynamisch kringproces functioneert en een netto hoeveelheid arbeid levert aan de omgeving terwijl het warmte ontvangt van één enkel warmte reservoir. Q H in Q H in cyc W=Q H in no! (Kelvin-Planck) Q C out cyc W=Q H in -QC out Q C out =0 no! (Kelvin-Planck) March 4, 00 5

16 (Historische) Formulering van de HW Kelvin-Planck Het is onmogelijk dat een systeem als een thermodynamisch kringproces functioneert en een netto hoeveelheid arbeid levert aan de omgeving terwijl het warmte ontvangt van één enkel warmte reservoir. Q H in Q C out cyc W=Q H in -QC out Q C out=0 no! (Kelvin-Planck) W Q c in altijd < 00%! March 4, 00 6

17 Equivalence of Clausius and Kelvin-Planck I Warm reservoir Q H in Q C out Koud reservoir cyc W= Q H in-q C out II Q H out=q C in Q C in no! (Clausius) III Q H in W=Q H in no! (Kelvin-Planck) IV Q C Q H Q C W =Q H We know I exists. Because II violates Clausius we know that Kelvin-Planck (III) also violates Clausius because IV is equivalent to III (as the blue dashed system is equivalent to III) Formulation of Kelvin-Planck (III) is equivalent to formulation of Clausius (II) March 4, 00 7

18 For an irreversible process: the system and the surrounding never spontaneously return to the initial state Reversible processes The system and its surrounding can for a reversible process at all times return to their initial state If a reversible process is reversed, then the backward process path for the system and its surrounding is identical to the forward path In practice reversible processes can only be approached but never fully reached; the environment will always be altered to some extent A perfectly reversible process can not deviate from thermal equilibrium and would be infinitely slow March 4, 00 8

19 Irreversibility friction (in flow of fluids as well as sliding friction) any (non-utilized) heat transfer through a temperature difference (because the reverse is not possible, Clausius!) non-utilized expansions of a fluid to lower pressure electric current flow through a resistance mixing of substances Irreversibilities are those parts of a process that reduce the work that would theoretically be obtainable if the process was reversible Irreversibilities = lost work (the work is lost as heat) March 4, 00 9

20 Inwendige reversibele processen Voor alle toestanden van een inwendig reversibel proces dat een gesloten systeem ondergaat, blijven alle intensieve toestandsgrootheden uniform in elk van de aanwezige fasen. Temperatuur, druk, specifieke volume en andere intensieve grootheden veranderen niet met de plaats. Verschillen in intensieve grootheden zouden moeten leiden tot drijvende krachten terwijl reversibiliteit eist dat er geen spontane processen kunnen plaatsvinden. March 4, 00 0

21 Result of the nd law for cycle-processes Q H out Q H in cyc W=Q H out -QC in cyc W = 0 not feasible (Clausius) cyc W=Q H in -QC out Q C out = 0 not feasible (Kelvin-Planck) Q C in refrigeration/ heat pump cycle Q C out power cycle The thermal efficiency rev of reversible a power cycle, is rev cyc W Qout H H Q Q in C in cyc H C W Qin Qout cyc Q C 0 Kelvin-Planck thermal efficiency rev even of a reversible power cycle is lower than unity, rev < March 4, 00

22 Results of the nd law for cycle-processes even a reversible (i.e. idealized) power cycle has an efficiency of rev < heat can not fully be converted to work all reversible Carnot-like* power cycles operating between the same temperature of two thermal reservoirs have the same thermal efficiency rev rev does not depend on the working fluid rev does not depend on the process steps making up the power cycle an irreversible (i.e. real) power cycle has yet a lower efficiency real < rev < book page 87 * Carnot-like means that all external heat transfer is realized isothermally with the two thermal reservoirs March 4, 00

23 The Carnot power cycle process p W 3 4 T H T C A reversible cycle comprising isothermal and adiabatic process steps v - adiabatic compression, T increases from T C to T H -3 expansion of the gas at const. T=T H (isothermal) 3-4 adiabatic expansion, T decreases from T H to T C 4- compression of gas at const. T=T C (isothermal) March 4, 00 3

24 Carnot power cycle of an ideal gas p first law for closed systems du process steps Q W 34 March 4, 00 4 v4 0 4Q 4W mrt ln v 4 Q 4 U U mc T T 34 ˆv H C 0 W U mc T T ˆv C H W W pdv; Q TdS U QW W pdv 34 Q 0 v3 3Q 3W mrt ln v 0 W 3 3 Q 4 T H T C heat in v heat out

25 Entropy S Arbeidsdiagram P versus V Warmtediagram T versus S W Q pdv TdS Adiabatisch reversible proces is isentroop, dus S=constant Oppervlakken zijn precies even groot omdat Qcyc Wcyc March 4, 00 5 TdS PdV

26 Carnot vapor power cycle The most obvious isothermal expansion is given by an evaporation step. The isotherm is then also an isobar. Ditto for condensation as an isothermal compression T H T C The compression of a two-phase system has (technical) disadvantages and powercycles with full condensation will later be discussed v March 4, 00 6

27 Resume Hoofdstuk -5 zijn nu behandeld Maak opgaves! Aanbevolen opgaves: 5., 5.4, 5.6, 5.36, 5.4, 5.48 March 4, 00 7