Chapter Five Applications of Thermodynamic Cycle

Transcription

1 Chapter Five Applications of hermodynamic Cycle Updated on 4// Power cycles: Power generation by converting heat to wor. Uses: Power generation, Propulsion. Cycles: Carnot cycle, Otto cycle, Diesel cycle, Brayton cycle, Stirling cycle, Ranine cycle. Real applications: Power plant, Automobile engine, Jet engine, Waste heat recovery Refrigeration cycles: ow temperature generation by transfer heat from low temperature reservoir to high temperature reservoir. Uses: ow temperature generation, Cryogenics Cycles: Carnot cycle, Brayton cycle, Stirling cycle, Ranine cycle, inde cycle, Real applications: Air conditioning, Refrigeration, Cryogenics. Gas cycles: ransport of fluid. Real applications: Compressor, blower, vacuum pump Advanced hermodynamics, ME Dept NCU 頁

2 (5.). Carnot Cycle (5..). Carnot Power Cycle Reflections on the Motive Power of Fire and on Machines Fitted to Develop that Power by Sadi Carnot, 84. Fig. 5.. Carnot cycle v reversible isothermal compression, =, q = Rln v v3 3 reversible adiabatic compression, cv ln + Rln = v v4 3 4 reversible isothermal expansion, =, q = R ln v3 v 4 reversible adiabatic expansion, cv ln + Rln = v4 (5...). hermal efficiency Carnot cycle is an ideal cycle. It has the highest efficiency that no other cycle can beat it. owever, it has never been realized in real world. Not because the ideal isentropic process can never be reached, but because there are some intrinsic features that prohibit the realization of this cycle. v v v = v 3 4 Advanced hermodynamics, ME Dept NCU 頁

3 v v4 = v v3 Q = Q Q η = = Q he thermal efficiency is not related to the stroe. It is determined by the temperature of reservoirs only. Carnot cycle efficiency / Fig. 5..: Efficiency of Carnot cycle efficiency Example:Find the thermal efficiency of a Carnot cycle with = K and = 3 K. Assignment 5.:A Carnot cycle starts with P = bar, =3 K, V =. m 3. he high temperature is K. Calculate the properties of each state and the wor output if the woring fluid is air. P (bar) (K) V () Advanced hermodynamics, ME Dept NCU 頁 3

4 (5...) Wor output he wor output is determined by the stroe of the isothermal process. v4 W = Qη = mr ln ( ) v 3 owever, the stroe of the isothermal process can be expressed as v4 v4 v v4 v c = =, 4 v ln = ln = ln v3 v v3 v v R v Where = is the compression ratio of the engine. v3 v4 v4 ln = ln + ln = ln + ln v v 3 W = mr ln + ln ( ) = mr ln + ln ( ) It is noted that power output is affected by temperature ratio as well as compression ratio. owever, since compression ratio is in the logarithm form, its effect is small compared with temperature. Specific wor: the wor output per unit mass of woring fluid. W w = = R ln + ln ( ) m Non dimensional wor: W w = = ln + ln ( ) mr Mean effective pressure (MEP): MEP is an index to show how much wor can be done in a specific volume of displacement. he higher the value of MEP, the more wor that the engine can deliver. W W W P mep = = = = w P ( V = ) V3 V ( ) m R Advanced hermodynamics, ME Dept NCU 頁 4

5 mep w ( >> ) P Example:Find the wor output and the mean effective pressure of a Carnot cycle with = K, = 3 K, P = Pa,V =. m 3, and compression ratio of. (5...3). Maximum wor W w = = ( X ) ln ln X mr X = For a given value of compression ratio ( ), the specific wor is determined by the temperature ratio( X ) only. dw = ln ln X ( X ) = dx X ln = ln X + X his is the temperature ratio that will produce the highest power output provided that and are given. Example:Find the value of that will produce imum output wor of a Carnot cycle with = 3 K, and compression ratio of. (5...4). Maximum pressure Carnot cycle is a pressure limiting cycle. he imum pressure increases sharply with power for a given set of reservoirs. v v v v v 4 = 3 = = = = v3 v v3 v v3 P P P P P P Advanced hermodynamics, ME Dept NCU 頁 5

6 v v 4 3 W mr η = e, P 3 P e W mr η = P 3 is the imum pressure in the cycle. If P is the limiting factor in the design of Carnot cycle, the wor output is as follows. 4 = 3 v P v P Where P = P is the reference pressure. P W = mr ( ) ln[ ( ) ] P P = mr X X P ( )[ln ln ] P w = X X P ( )[ln ln ] dw p = + X = dx p X ln ( ln ) P ln ln X P + X =,the value of X can be obtained by solving this equation. P v = = P v3 = P P Example:For a Carnot cycle with P = bar, P = bar, and = 3 K, find the thermal efficiency and the specific wor output. v Ans:x=.7, = 65.5 K,η=.54, v =,w=557 J/mole 3 V = liter,n = mole,w = nw =.6 J =.6 N-m q π = W, q = 35.3 Nm in torque Mep = W/V =.6 Pa Advanced hermodynamics, ME Dept NCU 頁 6

7 Example:For a Carnot cycle with P = 3 bar, P = bar, V =.m 3, and = 3 K, find the efficiency, the amount of heat addition, the wor output, the mean effective pressure, and the properties of each state. PV m = =.6 g R p + ln x = ln =.697 x p x=.7, = 83 K,η= = 3 v p v p Q p =9.56, = p = (K) P (bar) V (m 3 ) η Q (J) W (J) IMEP (Pa) v 4 = mr ln = J v3 W = Qη = J W W mep = = = Pa V V 3 V Assignment 5.:For a Carnot cycle with P = bar, P = bar, and = 3 K, find the thermal efficiency and the specific wor output. With constrain of imum pressure, the thermal efficiency of Carnot cycle is no longer the highest. he mean effective pressure is low compared with other cycle. As a result, the engine displacement should be very large for Carnot cycle in order to obtain the same output power. (5...5). imited heat transfer rate Advanced hermodynamics, ME Dept NCU 頁 7

8 Isothermal process can only be implemented in two ways. he first way is that the system undergoes an infinitely slow process such that the system and its thermal reservoir may be at the same temperature. he second way is that the heat transfer coefficient between the system and its thermal reservoir is infinitely large such that heat transfer may proceed at regular speed. If the engine rotates at a regular speed, and the heat transfer coefficient between the system and its thermal reservoir is NO infinitely large, there should exist temperature deviation between reservoir and system, and the cycle is no longer isentropic. In order to enhance heat transfer rate, temperature of reservoir should be higher than that of system in heating period, and during the cooling period, temperature of reservoir should be lower that that of system. he resulting system is no more a reversible system, and the thermal efficiency would be getting lower. q = ha( R ) ω q = ha( R) ω Q ( R) η = = Q ( ) R he thermal efficiency of the cycle can be shown above in which R is the high thermal reservoir temperature, R is the low thermal reservoir temperature, is the high temperature of the cycle, and is the low temperature of the cycle. It is noted that there is a temperature difference between thermal reservoir and system such that heat transfer may occur. owever, the efficiency of Carnot cycle depends on the temperature ratio only. η = As a result, the cycle temperature and the thermal reservoir temperature can be related as the following. ( R) = ( ) R R = R Advanced hermodynamics, ME Dept NCU 頁 8

9 R = R, / R = / R R η = = = = + R R R R R R R ha R R W = ηq = + ( R ) ω R R R W R R / R = + ( ) har / ω R R R / R he value of R R R η = + R R R R ( ) + R R should be confined to the region that efficiency is nonnegative. It is noted that for a given value of /, the output wor varies as / increases from. to its imum allowable value, and peas a certain value of R / R. he efficiency decreases all the way as / R increases, and the value of decreases too. If we raise the value of /, the allowable value of / extends to a greater range, and the wor output increases too. R R R R R R Advanced hermodynamics, ME Dept NCU 頁 9

10 Performance of Carnot cycle (R/R=3.) performance wor /R efficiency /R Fig Performance of Carnot cycle with limited heat transfer rate ha R R W = ηq = + ( R ) ω R R R dw ha R R R R = ( R ) + ( ) d ω R R R R R ( R) ha R = + ( ) + R RR R ω R ( R) R R ha = ω R ( R) ( ) R R R R R ha = ω ( R) = ha [ 4 4 R RR RR ] ω ( ) + + = R 4 4 ( ) = R R R R [ ] R R R R R R R R R ( ( ) ) ( ) = ± + = ± 8 R R R R R R R R R R = ± R R Advanced hermodynamics, ME Dept NCU 頁

11 Since, the only choice is = + R R R R R R + + R R ( R = = = = + ) R R R R R + R R R + ( + χ R ) = = = = χ R χ + + R χ R R R R η = + = + + = R R R R R R R R R R R W ha R = ηq = ω R R Wor of Carnot cycle wor R/R=7. R/R=5. R/R= /R Fig. 5..4: Wor of Carnot cycle with limited heat transfer rate Example: A Carnot engine operates between two thermal reservoirs at K and 3 K. Find the efficiency if heat transfer rate is considered. R = K, R = 3 K, = 75 K, = 35 K, η =.45 Assignment 5.3: A Carnot engine operates between two thermal reservoirs at 3K Advanced hermodynamics, ME Dept NCU 頁

12 and 3 K. Find the efficiency if heat transfer rate is considered. Efficiency of Carnot cycle efficiency emperature ratio Carnot cycle eat transfer Fig. 5..5: Efficiency of Carnot cycle with finite temperature heat transfer (5...6). Entropy generation For ideal Carnot cycle, the process are reversible, no entropy is generated. owever, if limited heat transfer is considered, entropy will be increased. Entropy generation of Carnot cycle: q q ha( R ) ha( R) ha s = + = + = + R R R ω R ω ω R R ha ha s = + = ω R R ω R R 4 4 R R R R Example: A Carnot engine operates between two thermal reservoirs at K and 3 K. Find the entropy generated if heat transfer rate is considered. (5..). Carnot Refrigerator he Carnot refrigerator is composed of four components, including evaporator, Advanced hermodynamics, ME Dept NCU 頁

13 condenser, compressor, and expander. It s woring principle is just the reverse of Carnot heat engine. Fig. 5..6: he woring principle of Carnot Refrigerator q = h h, constant pressure heat addition, evaporation of refrigerant 4 q = h h, constant pressure heat removing, condensation of refrigerant C 3 w = h h, isentropic compression with compressor E w = h h, isentropic expansion with expander q q = 3 4 wnet = wc we = qnet = q q Coefficient of performance : COP COP q q = = = = w q q, COP In most applications of refrigerator, a. he lower the value of, the more difficult to remove heat from thermal reservoir. As a result, the value of COP would be lower. Advanced hermodynamics, ME Dept NCU 頁 3

14 =, =.5 5K, COP = > 5K, COP >, < 5K, COP <, If oad s w< q w> q =, and = a, then q q q q s = = = = a, oad q q s = net s + s = = Carnot refrigeration cycle is an ideal cycle in which net entropy generation is zero. Example: A Carnot refrigerator uses R34a as the refrigerant. It is nown that =6.7, and =-4, find the amount of heat absorption and the COP. P =.564 bar, =-4,P =7. bar, =6.7 h =6.85,s =.98,s =.98,x =.95,h =.69 h 3 =86.78,s 3 =.34,s 4 =.34,x 4 =.339,h 4 =75.58 q =h -h 4 =36.,q =h -h 3 =75.7,w C =h -h =5.6,w E =h 3 -h 4 =. COP=3.49 Bac wor ratio = w E /w C =.3 In real applications, < oad, and > a, then q q s = <, s = a > oad q q s = net s + s = > a q oad q Advanced hermodynamics, ME Dept NCU 頁 4

15 (5..7). he reversible wor of refrigeration he minimum wor to cool down an object from room temperature to the desired low temperature in a reversible process is the reversible wor of refrigeration. he net entropy will not increase during this process. a Q W Q Q S = Ssys + Sev = Ssys + = Q S S S = a sys = ( ) a Q = h h W = Q Q = h h S S Fig. 5..7: he reversible refrigeration process ( ) a Q h h ( ) COP = = W h h a S S a Example: Find the reversible wor to cool down g of brass bloc from 5 to -4. Example: Find the reversible wor to mae g of ice at -4 from water at 5 Advanced hermodynamics, ME Dept NCU 頁 5

16 . Assignment 5.4: Find the reversible wor to liquefy mae g of hydrogen from 5. Advanced hermodynamics, ME Dept NCU 頁 6

17 (5. ). Otto Cycle (5..). Ideal Otto cycle v reversible adiabatic compression, cv ln + Rln =, v 3 constant volume heat addition, q = c ( ) v reversible isothermal expansion, 4 = 3 4 constant volume heat rejection, q = c ( ) v 4 = Fig.5..: Otto cycle 4 q cv( 4 ) η = = = q cv( 3 ) = =, =, η = 4 Wor of Otto cycle w= q η = c ( ) w 3 = ( ) c v v 3 Unlie the Carnot cycle in which the thermal efficiency is determined by the high end temperature, the thermal efficiency of Otto cycle does not depend on the operating temperature of the cycle. owever, the output wor does depend on the high end temperature of the cycle. he higher the value of 3, the more output Advanced hermodynamics, ME Dept NCU 頁 7

18 wor the system will have. W c mep = = V V v 3 ( ) 3 V ( ) Otto cycle thermal efficiency efficiency compression ratio Fig. 5..: Efficiency of Otto cycle for air Example:For an Otto cycle with P = bar, = 3 K, V =. m 3, q = J/g, =, find the properties at each state, the mean effective pressure, and the thermal efficiency. () Assume that the woring fluid is air, and air is an ideal gas with constant heat capacity. () he heat capacity of air is temperature dependent. Assignment 5.5: If the compression ratio has been raised to=5, find the properties at each state, and the thermal efficiency. (5..). he effect of compression ratio It can be shown that the higher the compression ratio, the higher the thermal efficiency. If the effect of heat capacity dependence on temperature is considered, the thermal efficiency still increases as the compression ratio is raised. owever, the simple relationship above does not hold any longer. Advanced hermodynamics, ME Dept NCU 頁 8

19 If dissociations of CO and O at high temperature are considered, the dependence of thermal efficiency on compression ratio is NO so clear at high compression ratios. Furthermore, if heat transfer between burned gas and cylinder wall is considered, it is not necessary that high compression ratio would result in high thermal efficiency. he thermal efficiency is related to the compression ratio only. No matter how high or how low the temperature is, the efficiency remains the same. owever, the output wor is proportional to the input heat transfer. he compression ratio of Otto engine has been raised from 3 to in the past years. he limiting factor in chemical related, not mechanical related. he anti noc property of fuel as well the concern on the exhaust emission of nitrogen oxides is the dominant factor to determine the compression ratio of current internal combustion engines. he compression ratio of internal combustion engine has been raised gradually since 9 th century due to the improvement of fuel anti-noc quality. owever, this tendency has been reversed at the late th century because the concern of air pollution has overwhelmed the search for high efficiency. Fig. 5..3: Evolution of compression ratio Advanced hermodynamics, ME Dept NCU 頁 9

20 (5..4). he effect of ratio of specific heat For a fixed value of compression ratio, the thermal efficiency gets higher if the value of increases. η = In the limiting case that as, then η. In the other limiting case that, then η. For monatomic gases such as helium and neon, the value of is.667, which is the highest value among all gases. For multi atomic gases such CO and O, the value of is much lower. Wor in an isentropic process PV = Ω = constant. Ω Wrev = PdV = dv =Ω V = ( PV PV ) V V PV For a compression process, =, W WC = ( ), C ( = ) V PV V PV WE For an expansion process, =, WE = ( ), = ( ) V PV cp cv + R R = = = + cv cv cv R cv = Since the gas constant R is a fixed value, the higher the value of c v is, the lower the value of will be. Specific heat capacity at constant volume c v is a measure of the capability of gases to absorb heat. If the molecules contain several energy modes, temperature rise caused by heat absorption would be lower because energy is distributed among different modes. he pressure rise associated with the temperature variation is thus diminished and the wor produced by volume expansion is reduced. Example: For an Otto cycle with P = bar, = 3 K, V =. m 3, q = J/g, =, find the properties at each state if (a). the gas is air, (b). the gas is helium, (c). Advanced hermodynamics, ME Dept NCU 頁

21 the gas is carbon dioxide, and (d). the gas is ethane. e air CO C M (g/mole) R (J/g-K) c v (J/g-K) P m P Q (J) 3 P 3 4 P 4 Q (J) W (J) η Assignment 5.6: For an Otto cycle with P = bar, = 3 K, V =. m 3, q = J/g, =, find the properties at each state if 5% of air is replaced with burned gases as the following reaction. C + ( O N ) CO + O + 7.5N 4 (5..5). he amount of heat addition q he amount of heat addition has no relation with the cycle efficiency. owever, the imum temperature as well as the imum pressure of the cycle is determined by the amount of heat addition. q = c ( ) v 3 q q 3 = + = + cv cv 3 q = + c v Advanced hermodynamics, ME Dept NCU 頁

22 q P = P = P ( + ) 3 3 c v P3 P = + q c v he amount of heat addition in real engine is determined by the fuel characteristic as well as the equivalence ratio. m m 3.76 m C n m + ( n+ )( O N) nco + O+ ( n+ ) N φ 4 φ 4 m m ( n + ) a A/ F = = 4 m φ. n +.8 m q f mq f f Qf = = m + m + A/ F Where mixture. f f a Q is the heating value of fuel (J/g) and A/ F is the air fuel ratio of he amount of energy for the mixture of air and fuel contained in a fixed volume depends on the heating value of fuel as well as the fuel type. If the fuel is of gaseous form lie propane or methane, the volume is filled with the gaseous mixture. owever, if the fuel is of liquid form lie gasoline of methanol, the volume is filled with air only because liquid fuel would occupy negligible fraction of volume only. Example: Find the amount of heat addition and the associated wor output for an Otto engine running with propane mixed with stoichiometric amount of air. he compression ratio is. he displacement volume is. he initial condition is 3 K and P. Assignment 5.7: Find the amount of heat addition and the associated wor output for an Otto engine running with the fuels of methane, methanol, gasoline, and hydrogen respectively mixed with stoichiometric amount of air. he compression ratio is. he displacement volume is. he initial condition is 3 K and P. It is noted that gasoline and methanol are in the form of fine liquid droplets in the air fuel mixture while methane and hydrogen are in the gaseous state in the mixture. Advanced hermodynamics, ME Dept NCU 頁

23 Fuel MJ/g A/F Methane Propane Methanol ydrogen gasoline (5..6). Otto cycle with temperature limit 3 q = + c v P q = + = P c v In the case of P = bar and = 3 K, if =3 K, the imum pressure would be around bars for normal compression ratio of Otto cycle. It is noted that 3 K is very high for normal combustion process, however, bars is not a severe running condition for a real engine. As a result, it is easier to exceed the imum temperature in an Otto engine. 3 is the imum temperature in the cycle. In order to have more power output, the higher value of 3 the better. owever, there is a physical limit. If is the limiting factor in the design of Otto cycle, the wor output is as follows. Otto cycle wor wor = K =3 K =5 K compression ratio Fig Wor of Otto cycle Advanced hermodynamics, ME Dept NCU 頁 3

24 w = ( ) c v = + he imum wor occurs at d w = ( ) + ( ) = d c v ( ) =, = ( ) ( ) w = + c v = + = Example:For an Otto cycle with P = bar, = 3 K, V =. m 3, =3 K, find the optimum compression ratio that would render imum output wor, the efficiency, the amount of heat addition, the wor output, the mean effective pressure, and the properties of each state. PV m = =.6 g R ( ) = η=.684 = (K) P (bar) V (m 3 ) η Q (J) W (J) IMEP (Pa) Q = mc ( ) v 3 = J W = Q η = 6.88 J Advanced hermodynamics, ME Dept NCU 頁 4

25 W W = = mep V V 3 V = Pa Assignment 5.8: Compare the efficiency and the wor output of a Carnot engine and an Otto engine with the same condition of P = bar, = 3 K, and V =. m 3 if the imum temperature must not exceed 3 K, and the imum pressure should be under 3 bars. (5..7). Pressure is the limiting factor P 3 is the imum pressure in the cycle. If P is the limiting factor in the design of Otto cycle, the wor output is as follows. w 3 P P P = ( ) ( ) c v = P = + P P he imum wor occurs at d w P P = ( ) + + = d c v P P P P ( ) P = + P P ( ) ( ) P = P P = > < =.39 Example: he imum pressure of Otto engine is 3 bars, find the optimum compression ratio that would render imum output wor. Advanced hermodynamics, ME Dept NCU 頁 5

26 (5..8). Entropy generation of Otto cycle: q q s= + = cv + = cv =, 3 =, 4 = s = = cv ( ) = 4 4 s = c v he reason that entropy would be generated in the Otto cycle is that during the constant volume heat addition process, the system temperature must be lower than the thermal reservoir temperature such that heat transfer may occur. In addition, during the heat removing process, the system temperature must be higher than the thermal reservoir temperature for the same reason. Entropy would be generated if a finite temperature difference heat transfer occurs. Example: Find the entropy generated of an Otto engine running at = 3 K and = 3 K. (5..9). Exergy analysis of Otto cycle reversible adiabatic compression, w, rev = u + u + ( s s) w = u + u, a I = ( s s) = Advanced hermodynamics, ME Dept NCU 頁 6

27 3 constant volume heat addition 3, rev = ( 3 ) + ( ) w u u s s q w 3, a = = ( 3 ) + ( ) I u u s s q q u u c = 3 + = ( ) v 3 + s s = c + R I 3 v 3 3 v ln( ) ln( ) v = 3 3 c v [ln( ) ( )] reversible adiabatic expansion w3 4, rev = u4 + u3 + ( s4 s3) w = u + u 3 4, a 4 3 I = 3 4 ( s 4 s3) = 4 constant volume heat rejection w4, rev = u + u4 + ( s s4) w4, a = = ( 4) = v [ln( ) ( )] 4 I u u s s c Cycle Performance w = q ( ) rev w = u + u u + u a 4 3 q η = = q Advanced hermodynamics, ME Dept NCU 頁 7

28 wa wa q q ηnd = = = η = η = η w q w q ( ) rev rev Example: For an Otto engine running at P = bar, = = 98 K, =, P = bars, find the thermal efficiency and the second law efficiency of the cycle. = K,P = 5.9 bar, 3 = P 3 /P = 98 K, 4 = 86.4 K q =6. J/g q = J/g w= J/g w = q =44.99 J/g ( ) rev η = =.69 η I I nd w η w a = = = rev.6688 = =35.98 J/g 3 3 c v [ln( ) ( )] 3 = = J/g 4 4 c v [ln( ) ( )] 4 he major irreversibility occurs at the heat rejection process. (5..). Miller cycle he expansion stroe is longer than the compression stroe. he expansion ratio is higher than the compression ratio. Valve timing control: Intae valve closes at late timing, and exhaust valve opens at the end of expansion stroe. : Constant pressure process Advanced hermodynamics, ME Dept NCU 頁 8

29 3: Isentropic compression 3 4: Constant volume combustion 4 5: Isentropic expansion 5 : Constant volume heat rejection V3 Geometric compression ratio: g = V V3 Effective compression ratio: e = V Piston control: he compression process and the expansion process are taen place in different cylinder. Air is suced into the compression cylinder as the piston moves down, and then is compressed. At the same time, the piston in the expansion cylinder moves up to expel the burned gas out of the cylinder. he expansion piston is a few degrees ahead of the compression piston. When the expansion piston moves to the DC, all the burned gas has been pushed out. At the same time, the compression piston is close to DC, and the inlet air has been compressed to a high pressure. he valve connecting between these two cylinders then opens to let the compressed air flow from the compression side to the expansion side. he expansion piston moves downwards as the compression piston moves upwards such that the air Advanced hermodynamics, ME Dept NCU 頁 9

30 transfer process is undertaen at constant volume process. When the compression piston reaches DC, all the air has been transferred, and the connecting valve closes. Combustion occurs in the expansion side and the burned gas expands to the end of stroe. Since compression process and expansion process tae place in different cylinder, the stroe may be different for each process. Example: A piston control type Miller cycle is combined with Otto cycle. he compression ratio is. he imum temperature is 5K. (). Fill up the following table. (). Determine the mean effective pressure and the thermal efficiency. (3). Compare the efficiency of this cycle with a standard Otto cycle, and mae your own comments. Advanced hermodynamics, ME Dept NCU 頁 3

31 P (bar) (K) V () (5..). Real internal combustion engine Comparison of ideal engine and real engine ideal engine real engine Inlet process Constant pressure air inlet Pressure drop through valve Compression Polytropic compression eat transfer and leaage during compression Combustion Instantaneous combustion Finite rate of combustion Incomplete combustion Expansion Polytropic expansion eat transfer and leaage during expansion Outlet process Constant pressure air outlet Exhaust blow down Pressure drop through valve Incomplete gas exchange Advanced hermodynamics, ME Dept NCU 頁 3

32 Fig. 5..5: Ideal and real Otto engine Irreversibility in the cycle of an internal combustion engine Mixing: mixing of fuel and air, mixing of residual and fresh mixture hrottling: free expansion without constrain Flow restriction Friction eat loss eaage Chemical reaction Blow down Deviations from ideal cycle: Advanced hermodynamics, ME Dept NCU 頁 3

33 Fig. 5..6: Deviations from ideal cycle Advanced hermodynamics, ME Dept NCU 頁 33

34 (5.3). Diesel Cycle (5.3.). Ideal Diesel cycle v reversible adiabatic compression, cv ln + Rln =, v 3 constant pressure heat addition, q = c ( ) p 3, q + v3 c 3 p q = c = = = + v c p 4 v3 3 4 reversible isothermal expansion, ( ) = 3 v4 4 constant volume heat rejection, q = c ( ) v 4 = Fig Diesel cycle c η = ( ), v3 c = : cutoff ratio v c he thermal efficiency is related to the stroe and the heat addition as well. Diesel efficiency efficiency cut off ratio Fig Efficiency of Diesel cycle ( = 5) eff Advanced hermodynamics, ME Dept NCU 頁 34

35 Cutoff ratio is an index of how much heat is added, or in practice much fuel is burned, into the engine during the isothermal expansion process. c he minimum value of c is one, which indicates that no fuel is burned. he imum value of is, which indicates that fuel is burned all the way as piston moves down. c c > ( ) c c, η = < ( c ) It is noted that the efficiency of Diesel cycle is lower than that of Otto cycle provided that the compression ratio is the same. Why is that the fuel economy of Diesel car is better than that of gasoline car? If c, η Otto cycle as heat addition is very little.,, the thermal efficiency of Diesel cycle is approaching that of c c w= qη = cp( 3 ) c p ( c ) = ( c ) ( c ) w c = ( c ) c p d w = = c d c c p he imum wor occurs at c = Note: If there is no constraint on the operation of Diesel cycle, the way to obtain the highest efficiency is to add heat as little as possible, and the resulting power is minimum. On the other side, the way to obtain the highest power is to add heat as much as possible, and the resulting efficiency is minimum. Advanced hermodynamics, ME Dept NCU 頁 35

36 w c p c P c mep = = ( c ) ( c ) v = v v ( ) ( ) Example:For a Diesel cycle with P = bar, = 3 K, V =. m 3, q = J/g, =, find the properties at each state, the thermal efficiency, the wor output, and the mean effective pressure. (). Assume that the woring fluid is air, and air is an ideal gas with constant heat capacity. (). he heat capacity of air is temperature dependent. Assignment 5.9: If the compression ratio has been raised to=5, find the properties at each state, the thermal efficiency. (5.3.). emperature limiting cycle If there is any constraint on the operation of Diesel cycle, the most possible one is temperature limitation. Its imum temperature is subjected to a upper bound such that the engine would not be overheating. 3 is the imum temperature in the cycle. If is the limiting factor in the design of Diesel cycle, the wor output is as follows. c = = c =, c For example, if = 3 K and = 3 K, the value of ~5.8, and the corresponding values of are 36.~5.8. c varies in the range w = ( ) = + ( ) c p Advanced hermodynamics, ME Dept NCU 頁 36

37 d w d c =( ) + ( ) = ( ) + p ( ) + = ( )( ) + = + + c = =, = + ( )( ) + ( + ) + ( + ) + w = ( ) = + c p + + c c η = = = W ( + ) ( + ) + + = mc p + = PV + Example:For a Diesel cycle with P = bar, = 3 K, V =. m 3, =3 K, find the optimum compression ratio that would render imum output wor, the efficiency, the amount of heat addition, the wor output, the mean effective pressure, and the properties of each state. PV m = =.6 g R ( )( ) + = + c = =.6 = 8.73 Advanced hermodynamics, ME Dept NCU 頁 37

38 c η = ( ) =.67 c 3 4 (K) P (bar) V (m 3 ) η Q (J) W (J) IMEP (Pa) Q = mc ( ) p 3 = 5.87 J W = Qη = 45.6 J W W mep = = = 5.94 Pa V V 3 V Assignment 5.: In a turbocharged Diesel engine without intercooler, the inlet air is boosted to Pa. Assume that the compressor has an efficiency of 9%,and the imum allowable temperature is 3K, find the imum power output. (5.3.3). eat release rate analysis cpv p cp p V PV q = cp( 3 ) = ( c ) = ( c ) = ( c ) mr mr m he amount of heat addition is related to the compression ratio as well as the cut off ratio. Q = mq = PV ( c ) du Q + mh i i= + mh e e+ W dt m i = m e = dv W = P dt m= m + m b u Advanced hermodynamics, ME Dept NCU 頁 38

39 U = mu = mbub + muuu du b b u u = m du b + u dm du dm b + mu + uu dt dt dt dt dt dmb dmu = dt dt du b b u = mc d b b + ( ub uu) dm + mc d u u dt dt dt dt Assume the mixture of burned and unburned gas is uniformly distributed inside the cylinder. b = u = du b b = ( mbcb + mucu) d + ( ub uu) dm = mc d v + ( ub uu) dm dt dt dt dt dt b Q = du + W = du = mc d v + ( ub uu) dm + P dv dt dt dt dt dt b ( ub uu) dm = Q mc d dv v P dt dt dt PV = mr d P dv V dp = + dt mr dt mr dt dmb P dv V dp dv cp dv cv dp ( ub uu) = Q mcv + P = Q P V dt mr dt mr dt dt R dt R dt cp = R c v = R dmb dv dp ( ub uu) = Q P V dt dt dt his is the apparent heat release rate of internal combustion engine. It is determined by the heat transfer rate and the variations of pressure. For ideal Diesel cycle, we have Q = P= P = P dmb dv Q RR =( ub uu ) = P dt dt Advanced hermodynamics, ME Dept NCU 頁 39

40 V D V = V + ( + cos θ ) V + V = D V dv VD dθ VD = sinθ = ωsinθ dt dt PV D Q RR = ωsinθ Q = Q RRdt = P( V3 V) (5.3.4). Entropy generation of Diesel cycle q ( ) q cp 3 cv( 4 ) s = + = + =, =, = 4 c ( + ) s = + + = + + cp + + = + + Example: Find the entropy generated of a Diesel engine running at = 3 K and = 3 K. (5.3.5). Exergy analysis of Diesel cycle reversible adiabatic compression, w, rev = u + u + ( s s) Advanced hermodynamics, ME Dept NCU 頁 4

41 w = u + u, a I = ( s s ) = 3 constant pressure heat addition 3, rev = ( 3 ) + ( ) w u u s s q w = 3, a u + 3 u + q q q I = ( s s ) = ( s s ) q h h c = 3 + = ( ) p 3 + s s = c I 3 3 p ln( ) = 3 3 c p [ln( ) ( )] reversible adiabatic expansion w3 4, rev = u4 + u3 + ( s4 s3) w = u + u 3 4, a 4 3 I = 3 4 ( s 4 s3) = 4 constant volume heat rejection w4, rev = u + u4 + ( s s4) w4, a = I u u s s c = ( 4) = v [ln( ) ( )] 4 3 Cycle Performance w = rev q ( ) Advanced hermodynamics, ME Dept NCU 頁 4

42 w = u u + q a 4 wa u u4 + q ηnd = = = η = η w rev q ( ) Example: For a Diesel engine running at P = bar, = 98 K, =, =5K, find the thermal efficiency and the second law efficiency of the cycle. = 98 K, P = Pa. (5.3.6). Intae and exhaust process constant pressure intae process, = =, P = P = P, 3 reversible adiabatic compression 3 4 constant pressure heat addition 4 5 reversible isothermal expansion 6 P 6 = 5 P5 6 7 constant pressure push out process, 6 = 7, P6 = P7 = P 5 6 blow down process, he wor required to suc air into cylinder: Wi = PV ( V) he wor required to push air out of cylinder: Wo = PV ( 7 V6) W i + W = o No net wor is required for the gas exchange process theoretically. Practically, the cylinder pressure is lower than the atmospheric pressure during the intae stroe, and is higher than the atmospheric pressure during the exhaust stroe. As a result, the net wor is negative, indicating that wor is required to complete the gas exchange process. his amount of wor is called the pumping wor. W+ W= PV ( V) + PV ( V) = ( PP) V< i o e 7 6 e i e he exergy loss during the blow down process: Advanced hermodynamics, ME Dept NCU 頁 4

43 PV = mr m m P = P 5 5 P 5 5 dp dm = m P P P m = = 5 P5 m5 P P dp dp V5 d = cpdm = cp5 m5 = cp5m 5 = dp P5 P5 P P5 he enthalpy contained in the exhaust flowing out of the cylinder during the blow down process is as the following. V5( P5 P6) = he imum wor that can be recovered bac is as the following. dw = c ( ) p 6 dm PV 6 6 PV 5 5 W = V5( P5 P6) c p 6( m5 m6) V5( P5 P6) c p 6( ) = + R R PV 5 5 P6 P 6 = V5( P5 P6) + PV 6 6 PV 5 5 = ( ) + ( ) 5 P5 P5 If P6 P P 6 = 5 P 5 =, the final temperature in cylinder would be PV 5 5 P P PV 5 5 P W = ( ) + ( ) = PV 5 + ( ) P5 P5 P5 owever, more wor is needed to push out the remaining gas. W = PV o 5 Advanced hermodynamics, ME Dept NCU 頁 43

44 As a result, the net wor during the exhaust process would be less than the value shown above. PV 5 5 P W = ( ) P5 his is the imum wor that can be recovered bac. recovered by means of turbocharger. Usually, this part of wor is igh pressure blow down he piston pushes the gas out of cylinder such that the pressure inside cylinder remains the same. P= P = P 5 6 mcvd + udm = hdm PdV mcvd = Rdm PdV mr P = = const. V dm d dv + = m V dm d dv = V + V m dm d mcvd Rdm + P V + V = m 5 mcvd Rdm + Rdm + mrd = mcpd = = 5 = const. dw = c dm ( ) p 5 6 Advanced hermodynamics, ME Dept NCU 頁 44

45 P W = cp( 5 6) m5 = PV 5 5 P 5 owever, piston has to do extra wor to push gas out. he required wor is W = PV p 5 5 P PV 5 5 P W = PV 5 5 PV 5 5 = P5 P 5 It can be seen that the wor of high pressure blow down equals to that of the natural blow down. PV 5 5 P P W = mcp( ) mcp = = P 5 P Advanced hermodynamics, ME Dept NCU 頁 45

46 (5.3.7). Comparison of Otto cycle and Diesel cycle Comparison of Otto cycle and Diesel cycle:( the same) Otto cycle Diesel cycle ( ) ( + )( ) c + w c v + ( + ) + η + + Example: ( =3K) =3K =4K Otto Diesel Otto Diesel c.6.94 w c v η Advanced hermodynamics, ME Dept NCU 頁 46

47 If Otto cycle and Diesel cycle have the same value of imum temperature and the same amount of heat addition, then the required compression ratio would be different. q q = + = + c Otto,3 Otto cv q q = + = + c Diesel,3 Diesel cp = Diesel,3 Otto,3 v p q + = + Diesel Otto cp q c v q Diesel = Otto + cv c p One the other side, if Otto cycle and Diesel cycle have the same value of imum pressure and the same amount of heat addition, then the required compression ratio would be also different. q + 3 c v q q POtto,3 = P = P Otto = P Otto + = P Otto + c v c v Otto P = P = P P Diesel,3 = P Otto,3 Diesel,3 P Diesel + q = P Otto c v Otto q = + Diesel Otto Otto c v = + q Diesel Otto c v Otto Diesel Advanced hermodynamics, ME Dept NCU 頁 47

48 Example: For an Otto cycle running with the following condition, =., q = 8 J/g, = 3 K, P = Pa Find the compression ratio of a Diesel cycle if (). Both the imum temperature (). Both the imum pressure are the same. = 36.3 K,P = 8.74 bar Otto (=.) Diesel(=8.48) Diesel(=53.79) Same P Same (K) P(Bar) (K) P(bar) (K) P(Bar) η η Otto Assignment 5.: If the imum allowable temperature is 3K for both the Otto cycle and the Diesel cycle, find the imum wor output for each cycle. Advanced hermodynamics, ME Dept NCU 頁 48

49 (5.4). Brayton Cycle (5.4.). he ideal Brayton cycle P reversible adiabatic compression, cv ln Rln =, = φ P q + 3 constant pressure heat addition, q ( ) c 3 p q = cp 3, = = + c 4 P3 3 4 reversible isothermal expansion, = ( ) 3 P4 4 constant pressure heat rejection, q = c ( ) p 4 p Fig Brayton cycle q cp( ) η = = = q cp( 3 ) = φ =, = 3 4 η = = ( ) / ( P/ P) φ et X = φ, η = X he thermal efficiency is related to the pressure ratio only. No matter how high or how low the temperature is, the efficiency remains the same. owever, the output wor is proportional to the input heat transfer. Advanced hermodynamics, ME Dept NCU 頁 49

50 w= q = c = c X X X η p( 3 ) p ( 3 ) w c X p 3 3 = X + Example:For a Brayton cycle with P = bar, = 3 K, P = bar, q = J/g, find the properties at each state, the thermal efficiency, and the wor output. (5.4.). emperature limiting Brayton cycle 3 is the imum temperature in the cycle. If is the limiting factor in the design of Brayton cycle, the wor output is as follows. w c X p = + d w = = φ dφ c p φ + X ( ) + + = φ = φ φ ( ) =, X = η = = = X p w = + = c Advanced hermodynamics, ME Dept NCU 頁 5

51 Example: For a Brayton cycle with P = bar, = 3 K, P = bar, = 5 K, find the properties at each state, the thermal efficiency, and the wor output. (5.4.3). Entropy generation of Brayton cycle q q s= + = cp + = cp =, 3 ( ) / = φ, 4 = ( ) / φ s = φ = cp ( ) / ( ) / ( ) / ( ) / φ φ φ φ ( ) =, 4 4 s = cp Example: Find the entropy generation in the previous example. (5.4.4). Brayton Cycle with regenerator Advanced hermodynamics, ME Dept NCU 頁 5

52 Fig Brayton cycle with regenerator Fig he s diagram of Brayton cycle with regenerator = φ qr = cp 5 = cp 4 6 q = c ( ) p ( ) ( ) = ( ) φ ( ) ( ) 3 qr = cp 5 = cp 4 6 q = c ( ) p In the limiting case, the regenerator wors ideally such that 4 = 5, and = 6. q cp( ) X η = = = = = = q ( ) cp X X Advanced hermodynamics, ME Dept NCU 頁 5

53 It is noted that the condition that efficiency with regenerator is greater than efficiency without regenerator is that X > 3 X > X X X 3 < 3 Fig he Efficiency of Brayton cycle with regenerator Example: he imum temperature of Brayton engine is 5 K, and the pressure ratio is 5.. Find the improvement in cycle efficiency if a regenerator is adopted. Advanced hermodynamics, ME Dept NCU 頁 53

54 w q η cp( 3 4) c p 3 ( ) φ 3 φ φ = = = 3 3 ( ) w c p c p 3 = φ + = Χ + 3 φ 3 3 Χ 3 dw dχ = c p 33 + = 3 Χ Χ= = φ φ ( ) = he optimum value of φ for Brayton cycle with regenerator is identical to that without regenerator. η = φ = = It is noted that the imum wor occurs at the point that the efficiency with regenerator is the same with without regenerator. (5.4.5). Regenerative gas turbine with inter cooling and reheat Q Q Fig Regenerative gas turbine with inter cooling and reheat Advanced hermodynamics, ME Dept NCU 頁 54

55 = 3 = 6 = 8 = = 4 = X 7 = 9 = = 5 9 X W = c ( ) p 6 7, W = c ( ) p 8 9 WC = c ( ) p, WC = c ( ) p 4 3 Wnet = c p ( ) c p ( X ) X Q = cp( 5 4) + cp( 8 7) = cp( ) X c p ( ) c p ( X) Wnet η = = X = Q c p ( ) X Example: he imum temperature of Brayton engine is 5 K, and the pressure ratio is. Find the improvement in cycle efficiency if intercooler and re-heater are adopted. X If the compression process is divided into three stages with intercoolers installed in between, and the expansion process is also divided into three stages with re-heater installed in between, then the efficiency would be as the following. η = X 3 Advanced hermodynamics, ME Dept NCU 頁 55

56 he limiting case is that both the compression process and the expansion process are isothermal. In that case, the combustor is no longer needed because the regenerator may raise the inlet air to the temperature of W W C Q C Fig imiting case of regenerative gas turbine with inter cooling and reheat P = R ln P = R P3 ln P Wnet = W WC 4 Q Q = W Wnet η = = Q he efficiency of the limiting case is that of Carnot cycle. P V Fig he cycle of limiting case S Advanced hermodynamics, ME Dept NCU 頁 56

57 Assignment 5.: he imum temperature of Brayton engine is 5 K, and the pressure ratio is 6. Find the improvement in cycle efficiency if three sets of intercoolers and re-heaters are adopted. (5.4.6). Real gas turbine engine For a gas turbine equipped with a regenerator and burning jet fuel, the irreversibility of the cycle is obtained step by step analysis as the following. Compressor efficiency, turbine efficiency, intercooler effectiveness are considered in a real gas turbine engine. Example: he imum temperature of Brayton engine is 5 K, and the pressure ratio is. It is nown that compressor efficiency is 8%, turbine efficiency is 85%, intercooler effectiveness is 7%. Find the cycle efficiency. Example: A regenerative gas turbine with inter cooler and reheating is shown in the figure. he inlet air is at 3K and Pa. Air is compressed to the pressure of 4 Pa in the first stage and then is cooled down by the intercooler. Air is compressed again in the second stage to 6 Pa and then flows into the regenerator and the combustor consecutively. he imum temperature is 4K. hese two compressors are driven by the high pressure turbine. he low pressure turbine drives a generator. In the reheating chamber, air is heated to the imum temperature again. he compressor efficiency is 85%. he turbine efficiency is 9%. Both the regenerator and the intercooler have an effectiveness of.7. Assume that air is an ideal gas with constant heat capacity. (). Find the wor output of this system. (). Find the thermal efficiency. (3). Find the thermal efficiency if all the components are perfect with efficiency.. Advanced hermodynamics, ME Dept NCU 頁 57

58 Q Q Real case Ideal case (K) P(Pa) (K) P(Pa) Real case w= c ( ) p 8 9 =45 J/g q = c ( ) p 6 5 =57.9 q = c ( ) p 8 7 =37.4 q = q+ q=944.3 w η = =.479 q Ideal case w= c ( ) p 8 9 =599. J/g Advanced hermodynamics, ME Dept NCU 頁 58

59 q = c ( ) p 6 5 =599. q = c ( ) p 8 7 =6.6 q = q+ q=89.8 w η = =.673 q (5.4.7). Reversible wor of gas turbine engine without a regenerator,adiabatic compression w, rev = h + h + ( s s) w = h + h, a I = ( s s) = 3,constant pressure heat addition w h h s s q 3, rev = [ 3 ] + ( ) q h h = 3, 3 w = ( s s ) q = c ln( ) c ( ) 3 3, rev 3 p p 3 If = 3 w = c 3 3, rev p [ln( ) ( )] 3 w 3, a = I = 3 3 c p [ln( ) ( )] 3 3 4,isentropic expansion w3 4, rev = h4 + h3 + ( s4 s3) w = h + h 3 4, a 4 3 I = 3 4 ( s 4 s3) = Advanced hermodynamics, ME Dept NCU 頁 59

60 4,constant pressure heat rejection w4, rev = h + h4 + [ s s4] w4, a = = [ 4] = p [ln( ) ] + 4 I h h s s c Cycle performance cycle, rev = + + ( ) ( 3 ) + ( ) w h h s s h h s s q h + h + ( s s ) h + h + ( s s ) w q ( ) c[ (Pr) ][ ] cycle, rev = = p w = h + h h + h cycle, a 4 3 w w q q η η η a a nd = = = = = [ ( ) ] wrev q w rev q ( ) φ ( ) ( ) ( ) Icycle = q h + h h + h = h h h h = c [ ( ) ] c [ ] = c [ ( ) + ] p p φ p φ φ φ Example:Calculate the reversible wor of an ideal Brayton cycle with = K, and Pr=. Assume that air is an ideal gas with constant heat capacity P = Pa, = 98 K,P = Pa, = K P 3 = Pa, 3 = K,P 4 = Pa, 4 = 6.5 K w c = J/g,w =58. J/g,w a =3.55 J/g q = 67.5 J/g,q = J/g Advanced hermodynamics, ME Dept NCU 頁 6

61 w = =47.68 J rev q ( ) I = W W =69.3 J/g I I rev a = = = 64.4 J/g 3 3 c p [ln( ) ( )] 3 = + =4.93 J/g 4 4 c p [ln( ) ] 4 he major irreversibility occurs at the heat rejection process. η nd =.644, η =.48 Example:Calculate the reversible wor of an ideal Brayton cycle with = K, and Pr=. he heat capacity of air is not a constant. P = Pa, = 98 K,h = 98.8 J/g,s =.6958 J/g-K P = Pa, = 57.5 K,h = 576. Pa,s =.3553 J/g-K P 3 = Pa, 3 = K,h 3 = J/g,s 3 = J/g-K P 4 = Pa, 4 = 665. K,h 4 = J/g,s 4 =.5787 J/g-K w c = J/g,w =6.98 J/g,w a =34.5 J/g q = 7.68 J/g,q = J/g w = =57.43 J rev q ( ) I = W W =3.38 J/g rev a w s s q 3, rev = ( 3 ) = =7.7 J/g I = h + h + [ s s ] =3.5 J/g he major irreversibility occurs at the heat rejection process. η nd =.644, η =.468 Advanced hermodynamics, ME Dept NCU 頁 6

62 For fixed values of and,the pressure ratio φ could be varied to minimize the minimum value of I cycle. d c p + =, X dx X [ X] = φ + = X X = = φ, φ = ( ) Icycle = c p [ ( ) + φ ] = φ η nd = = [ ] owever, at the pressure ratio obtained by the approach mentioned above, the outlet temperature of compressor equals to that of the imum temperature, and no heat addition is required. he output wor is thus zero. his is a trivial situation, because no wor can be obtained. It seems that the higher the pressure, the higher the total efficiency as well as the second law efficiency. owever, the higher the pressure ratio, the lower the heat addition and the resulting wor output is getting lower too. here is an optimized value of pressure that would imize the output wor. wa = ηq = [ ( ) ] cp[ (Pr) ] = c p ( )( X), X Pr X = (Pr) dwa = c p ( + ) = dx X X =, Pr = ( ) ( ) Advanced hermodynamics, ME Dept NCU 頁 6

63 w = c( )( ) = c( ) o a p p η = = Pr [ ( ) ] [ ] w c[ (Pr) ][ ] c[ ][ ] cycle, rev = p = p η = nd [ ] Example:Calculate the best pressure ratio for an ideal Brayton cycle with = K. ( ) Pr = ( ) =.447 P = Pa, = 98 K,P = 44.7 Pa, = 598. K P 3 = 44.7 Pa, 3 = K,P 4 = Pa, 4 = 598. K w c =-3.35 J/g,w =64.7 J/g,w a =33.36 J/g q = 64.7 J/g,q = J/g w = = J rev q ( ) I = W W =5.8 J/g η nd rev a =.6674, η =.57 (4.6.8). Reversible wor of real gas turbine engine For a gas turbine equipped with a regenerator and burning jet fuel, the irreversibility of the cycle is obtained step by step analysis as the following. Advanced hermodynamics, ME Dept NCU 頁 63

64 ,adiabatic compression w, rev = h + h + ( s s) For a compressor with efficiency of h h w, a = h + h = η s c η c, the actual wor is = + η c P P I = ( s s) = c p ln + φ ( )lnφ ηc 3,constant pressure combustion Air + fuel products R = P mh ( ) + mh( ) = mh( ) f f f a a p p Since is nown, the mass of fuel is determined by the energy balance of reaction. S = ms ( ) ms ( ) + ms( ) R p p f f a a S R is the increase of entropy associated with the combustion reaction. w 3, rev = h3 + h + ( s3 s) = sr,since h = h3 w 3, a = I = 3 sr 3 4,adiabatic expansion w3 4, rev = h4 + h3 + ( s4 s3) w = h + h 3 4, a 4 3 Advanced hermodynamics, ME Dept NCU 頁 64

65 I = 3 4 ( s 4 s3) 4,constant pressure gas exchange here is no process 4 in a real gas turbine engine, since the exhaust gas does not return bac to the inlet. he cycle is complete with a constant pressure gas exchange process. 4 5,constant pressure cooling he products are cooled in the atmosphere to without doing any wor. w = h + h + ( s s ) 4 5, rev w4 5, a = I = h + h + ( s s ) Cycle performance w = h + h + ( s s ) h + h + ( s s ) h + h + ( s s ) cycle, rev = h h + ( s s ) 5 5 Where h and s are the properties of air and fuel at state, and h 4 and s 4 are the properties of products at state 4. w = h + h h + h = h h,since h = h3 cycle, a η nd wa h h4 = = w h h + ( s s ) rev 5 5 I = h h + ( s s ) cycle Example:Calculate the reversible wor of a gas turbine engine with = K, and Pr=. he fuel is methane and methane is compressed separately to the same pressure of air. It is assumed that all the compression and expansion processes are Advanced hermodynamics, ME Dept NCU 頁 65

66 of the efficiency of C + ( O N ) CO + O + ( ) O + N φ φ φ 4 φ =, C O N CO + O + 6.4O N 4. P = Pa, = 98 K,h =-64.7 J/g,s =6.98 J/g-K P = Pa, = 6.5 K,h =66.65 J/g,s =7.66 J/g-K P 3 = Pa, 3 = 98 K,h 3 =66.65 J/g,s 3 =7.879 J/g-K P 4 = Pa, 4 = 757. K,h 4 = J/g,s 4 =8.8 J/g-K P 5 = Pa, 5 = 98 K,h 5 =-75.J/g,s 5 =7. J/g-K w c = J/g,w =5.955 J/g,w a =9.5 J/g w = h h + ( s s ) = J rev rev 5 5 I = W W =5.75 J/g a I = ( s s ) =5.3 J/g I = s =4.74 J/g 3 R I = ( s s ) =38.44 J/g I = h + h + ( s s ) =97.6 J/g he major irreversibility is accomplished by the reaction. he second one is the exhaust energy released to the atmosphere. It is noted that the loss of exergy in reaction can hardly be improved because that is the most convenient way to heat up gas to the required temperature. owever, one of the reason that the loss of exergy in reaction is so huge is that dilute air is too much. If the dilute air can be reduced, the loss of exergy can be improved. η nd =.759 he lower heating value of methane is 544 J/g, and the air fuel ratio is 7.4, so the averaged heating value of the mixture is about 683. J/g. Advanced hermodynamics, ME Dept NCU 頁 66

67 η =.83 Example 5.5.5: A regenerative gas turbine with inter cooling and reheat runs with the following condition: = 6 = 8 = K, = 3 = =98 K, 5 = 9, 4 =,Pr= P /P =P 4 /P 3,P 7 /P 6 =P 9 /P 8, = 4 = 44 K, 7 = 9 =863.6 K q =.45 (-863.6)= J/g w =.45 (-863.6)= J/g w C =.45 (44-98)=33.44 J/g w a = J/g w rev = J/g η = (Pr) =.655 w a η nd = =.876 qrev Single stage without regeneration: wrev = c p [ (Pr) ]( ) wa = c p [ (Pr) ] + c p [ ( ) ] Pr Advanced hermodynamics, ME Dept NCU 頁 67

68 q = c[ (Pr) ] p η =, η nd = η (Pr) Single stage with regeneration: w rev = c p [ ( ) ]( ) Pr wa = c p [ (Pr) ] + c p [ ( ) ] Pr q = c [ ( ) ] Pr p η = (Pr), η nd = η Double stage with regeneration: w rev = c p [ ( ) ]( ) Pr wa = c p [(Pr) ] + c p [ ( ) ] Pr q = c [ ( ) ] Pr p η = (Pr), η nd = η Advanced hermodynamics, ME Dept NCU 頁 68