Fundamentals of Thermodynamics Chapter 8 Exergy
Exergy Availability, available energy Anergy Unavailable energy Irreversible energy, reversible work, and irreversibility Exergy analysis : Pure Thermodynamics EGM(Entropy Generation Minimization) Thermodynamics + Heat transfer + Fluid mechanics Thermal Engineering Lab. 2
8.1 Exergy, reversible work, and irreversibility Efficiency 1 2 W h = & Q& Wa ( h t ) s = W s ; 1 st law ; 2 nd law 3 exergy ; 2 nd law + environment (T, P ) * Possible work we can extract from a given physica l setup when it is allowed to interact with the ambi ent and the process end state is at T, P. Thermal Engineering Lab. 3
Exergy(Available energy) of heat 일정온도 T 에서 Q 만큼의열을받아얻을수있는최대 work Thermal Engineering Lab. 4
Wrev = Q - Q T = Q(1 - ) T Q W = Q -T DS º E rev Q T Q = = DS T Irreversibility : I º Wrev -W Thermal Engineering Lab. 5
- T 가변하는경우 W = Q - Q rev Q = TDS Q = òtds Thermal Engineering Lab. 6
dm dt de dt ds c. v. c. v. c. v. dt = m& - å i å m& e = Q& + m& h - m& h -W& å å å i tot, i e tot, e c. v. ac Q& = + & - & + å å mi s å i mese Sgen, ac T & Thermal Engineering Lab. 7
n de = Q & + Q & å - W & + å m & i h tot, i - å m & e h tot, e dt = 1 i e ds Q& Q& dt T T n = + å + åm& isi - åm& ese + S& gen = 1 i e Q & o 소거 Thermal Engineering Lab. 8
de ds Q& = T -T - T m s + T m s - T S + Q - W + m h - m h dt dt T n n o oå oå & i i oå & & & e e o gen å & å & i tot, i å & e tot, e = 1 = 1 ( - ) æ d E T n os To = 1- Q & å - W & + m i ( h tot, i - T o s ) - m e ( h tot, e - T o s ) - T o S gen dt ç = 1 T å & & å & è ( - ) W d E T S æ T ö Q m h è ø ö ø & n o o = - + 1- & å + i tot, dt ç = 1 T å & ( ) ( ) i -To s -å m& e htot, e -To s -To S& gen W& Þ T S& = rev o gen W& = W& rev -To S& gen { I& = W& -W& rev I& Thermal Engineering Lab. 9
General equation expressing d( E T S) æ T ö W& - = - + - Q + m& h -T s - m& h -T s -T S dt è ø n 1 & c. v., i ( tot, i i ) e ( tot, e e ) & åç gen = 1 T å å i e W& d( E -T S) æ T ö Q m& h T s m& h T s è ø n rev = - + - c v + i tot i - i - e tot e - e dt ç = 1 T i e d( E -T S) æ T ç 1 & å.., å (, ) å (, ) n & rev = - + 1- Q& å c. v dt ç T., + å m& i ( htot, i -T si ) -å m& e ( htot, e -T se ) = i e W I& = W& - W& = T S& rev gen ds T = T - Q - m T s + m T s n & å å & i i å & e e dt = T i e è ö ø W & æ n de ö ç QW& = - + Q& å c. v., + å m& ihtot, i -å m& ehtot, e è dt = i e ø Thermal Engineering Lab. 1
SSSF process W& æ T 1 - ö - è ø = Q& åç c v m&.., + i T å å i e ( htot i T si ) m& e( htot e T se ) T S&, - -, - gen & & & Assumption : Single inlet & single exit à m = m = m, i o T w = - q + h -T s - h -T s -T s å (1 ) ( tot, i i ) ( tot, e e) gen T T w q h T s h T s å rev = (1 - ) + ( tot, i - i ) - ( tot, e - e) T Thermal Engineering Lab. 11
SSSF process, Single inlet & Single exit T w = - q + h -T s - h -T s -T s å (1 ) ( tot, i i ) ( tot, e e) gen T T w q h T s h T s n rev = å(1 - ) + ( tot, i - i ) - ( tot, e - e) = 1 T T w q h T s h T s n rev = å(1 - ) + ( tot, i - i ) - ( tot, e - e) = T n rev i = w - w = T sgen = T ( se - si ) -å q = T ( Qw = h - h + q tot, i tot, e n å = ) T Thermal Engineering Lab. 12
Reversible work à Maximum or minimum work I & = W& rev -W& ac ³ Thermal Engineering Lab. 13
Ex. 8.1 A feedwater heater has 5 kg/s water at 5 MPa and 4 flowing through it, being heated from two sources, as shown in Fig. 8.6. One source adds 9 kw from a 1 reservoir, and the other source transfers heat from a 2 reservoir such that the water exit condition is 5 MPa, 18. Find th e reversible work and the irreversibility. Thermal Engineering Lab. 14
Ex. 8.2 Consider an air compressor that receives ambient air at 1 kpa and 25. It compresses the air to a pressure of 1 MPa, where it exits at a temperatu re of 54 K. Since the air and compressor housing are hotter than the amb ient surroundings, 5 kj per kilogram air flowing through the compressor are lost. Find the reversible work and the irreversibility in the process. Thermal Engineering Lab. 15
Control mass process æ T ö d W& å Q& E T S è ø rev = 1 - - ( - ) ç c. v. T dt æ T ö W = åç 1 - Q - E - E -T ( S - S ) è ø T I = W - W = T ( S - S ) - Q [ ] 1 2, rev ç 1 2 2 1 2 1 T å 1 2 1 2, rev 1 2 2 1 1 2 T Thermal Engineering Lab. 16
Ex. 8.3 An insulated rigid tank is divided into two parts, A and B, by a diaphrag m. Each part has a volume of 1 m 3. Initially, part A contains water at roo m temperature, 2, with a quality of 5 %, while part B is evacuated. T he diaphragm then ruptures and the water fills the total volume. Determin e the reversible work for this change of state and the irreversibility of the process. Thermal Engineering Lab. 17
The transient process Uniform ; E=me, V=mv, S=ms W& d E T S æ T ö Q è ø n rev = - ( - ) c. v. + 1- & å c. v., dt ç = T å i (, ) å & (, ) + m& h -T s - m h -T s i tot i i i tot e o e i Integration : 1 2 T W = (1 - ) Q + m ( h -T s ) - m ( h -T s ) å å å 1 2, rev 1 2 i tot, i i e tot, e e T -[ m e - m e -T ( m s - m s )] 2 2 1 1 2 2 1 1 1 I = W - W = T [( m s - m s ) + m s - m s - Q ] å å å 1 2 1 2, rev 1 2 2 2 1 1 i i e e 1 2 T Thermal Engineering Lab. 18
Ex. 8.4 A 1 m 3 rigid tank, Fig. 8.8, contains ammonia at 2 kpa and ambient te mperature 2. The tank is connected with a valve to a line flowing satu rated liquid ammonia at -1. The valve is opened, and the tank is charg ed quickly until the flow stops and the valve is closed. As the process hap pens very quickly, there is no heat transfer. Determine the final mass in th e tank and the irreversibility in the process. Thermal Engineering Lab. 19
8.2 Exergy and second-law efficiency W& d E T S æ T ö Q è ø n rev = - ( - ) c. v. + å 1- & c. v., dt ç = T i ( ) å & ( ),, + m& h -T s - m h -T s å i tot i i i tot e o e i Exergy of heat æ T ö F & q = å 1 - ç T è ø Q& Thermal Engineering Lab. 2
Flow exergy 2 æ V ö y = ç h + + gz -T s - h + gz -T s è 2 ø ( h T s) ( h T s ) = - - - tot tot, ( ) ( h T s ) ( h T s ) y - y = - - - i e tot, i i tot, e e Thermal Engineering Lab. 21
Nonflow exergy W& surr o d W& = W& - W& = - ( E - T S + PV ) dt W = -[ E - E -T ( S - S ) + P ( V -V )] avail rev rev surr o c. v. avail rev o c. v. F = E - E -T ( S - S ) + P ( V -V ) W W& avail rev avail rev = PV& = -F = -F& o éæ V ö ù f = ê + + + - - + + + - ëè 2 ø û 2 ç u gz P v T sú éë ( u gz ) P v T s ùû e e [ e P v T s ] [ e P v T s ] f - f = + - - + - 2 1 2 2 2 1 1 1 Thermal Engineering Lab. 22
Exergy balance equation W& d E T S PV æ T ö Q è ø n avail rev = - ( - + ) c. v. + å 1- & c. v., dt ç = T i ( ) å & ( ),, + m& h -T s - m h -T s å i tot i i i tot e o e i W& = F & + m& y - m& y - F& å å avail rev q i i e e C. V. i e F & = F& - W& + m& y - m& y å å avail C. V. q rev i i e e i e Thermal Engineering Lab. 23
Exergy efficiency and isentropic efficiency h = s w w a s isentropic h exergy = i w a y -y e, h exergy E = - & E& 1 loss input Thermal Engineering Lab. 24
Ex. 8.5 An insulated steam turbine (Fig. 8.1), receives 3 kg of steam per secon d at 3 MPa, 35. At the point in the turbine where the pressure is.5 M Pa, steam is bled off for processing equipment at the rate of 5 kg/s. The te mperature of this steam is 2. The balance of the steam leaves the turb ine at 15 kpa, 9 % quality. Determine the exergy per kilogram of the ste am entering and at both points at which steam leaves the turbine, the isent ropic efficiency and the second-law efficiency for this process. Thermal Engineering Lab. 25
Heat exchanger h exer = m& 1( y 2 -y1) m& ( y -y ) 3 3 4 Generally F& F& - I& hexer = = F& F& wanted source c. v. source source Thermal Engineering Lab. 26
Ex. 8.6 In a bolier, heat is transferred from the products of combustion to the stea m. The temperature of the products of combustion decreases from 11 to 55, while the pressure remains constant at.1 MPa. The average co nstant-pressure specific heat of the products of combustion is 1.9 kj/kg K. The water enters at.8 MPa, 15, and leaves at.8 MPa, 25. De termine the second-law efficiency for this process and the irreversibility p er kilogram of water evaporated. Thermal Engineering Lab. 27
8.3 Exergy balance equation F = mf = m( e - e ) + P m( v - v ) -T ( s - s ) 2 v e = u + + gz 2 1 df æ T ö = å - dt ç T è ø & 1 Qc. v. æ dv ö -ç Wc. v. - P è dt ø å + m& y - i i å m& y e e heat work flow - & exergy destruction ( E& D = T S& gen ) E D Thermal Engineering Lab. 28
Ex. 8.7 Let us look at the flows and fluxes of exergy for the feedwater heater in E xample 8.1. The feedwater heater has a single flow, two heat transfers, an d no work involved. When we do the balance of terms in Eq. 8.38 and ev aluate the flow exergies from Eq. 8.22, we need the reference properties (take saturated liquid instead of 1 kpa at 25 ); Thermal Engineering Lab. 29
Ex. 8.8 Assume a 5 W heating element in a stove with an element surface temp erature of 1 K. On top of the element is a ceramic top with a top surfa ce temperature of 5 K, both shown in Fig. 8.15. Let us disregard any he at transfer downward, and follow the flux of exergy, and find the exergy destruction in the process. Thermal Engineering Lab. 3
8.4 Engineering applications W = η HE HEI Q H W HE æ ç - è T = ηheii F H = ηheii ç1 TH ö Q ø H W HE æ ç - è T = ηheii F H = ηheii ç1 TH ö Q ø H F T ö ç1 è ø æ H η HPII = = QH / W ç - HP TH W HP Thermal Engineering Lab. 31
공기압축기의 Exergy analysis h 1 h 2 W & in Q & Thermal Engineering Lab. 32
내연기관의 Exergy analysis Expansion wor k Heat transfer combustion irreversibility availability (exergy) BDC TDC BDC Thermal Engineering Lab. 33
Adiabatic SSSF process Thermal Engineering Lab. 34
SSSF process with heat and work transfer Thermal Engineering Lab. 35
Heat transfer process Thermal Engineering Lab. 36
Isothermal compression process Thermal Engineering Lab. 37
Steam power plant Thermal Engineering Lab. 38
Combustion process Thermal Engineering Lab. 39
Gas turbine process Thermal Engineering Lab. 4