Today lecture. 1. Entropy change in an isolated system 2. Exergy

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1 Today lecture 1. Entropy change in an isolated system. Exergy - What is exergy? - Reversible Work & Irreversibility - Second-Law Efficiency - Exergy change of a system For a fixed mass For a flow stream - Exergy transfer by mass - Exergy transfer by work - Exergy transfer by heat transfer - Exergy destruction - Exergy balance 1

2 Total entropy change in an isolated system The principal of entropy: The total entropy change of an isolated system during a process always increases or, in the limiting case of a reversible process, remains constant. Consider a general system exchanging mass as well as energy with its surroundings. Sum of entropy change = Entropy generation S = ΔS = ΔS + ΔS 0 gen total sys surr where = holds for the totally reversible process > holds for the irreversible process < impossible process

3 Example A heat source at 800 K loses 000 kj of heat to a sink at (a) 500 K. Determine which heat transfer process is more irreversible. (a) For heat transfer process to sink at 500 K Qsource 000 kj Δ Ssource = = =.5 kj/k T 800 K source Δ Qsink 000 kj S = sink 4kJ/K T = 500 K = Sgen =Δ Stotal =Δ Ssource +Δ Ssink = = 1.5 kj/k sink Entropy generation Irreversible process! 3

4 From Quiz 4 A 50 kg copper block initially at 80 O C is dropped into an insulated tank that contains 10 L of water at 5 O C. Determine the final equilibrium temperature and the total entropy change for this process. Note 1 L = 10-3 m 3, at 5 O C c p,water = 4.18 kj/kg O C, ρ water = 997 kg/m 3, and c p,copper = kj/kg O C. Ein Eout =ΔEsys 0 = Δ Ucu +ΔUw ( mc T ) ( mc T ) 0 p cu p = Δ + Δ w ( ) ( ) 0= m c T 80 + m c T 5 cu p, cu w p, w O T = 7 C 4

5 (b) Total entropy change Δ Stotal = Δ Scu +ΔSw ( ) Δ S = m s s cu cu 1 cu ( ) Δ S = m s s w w 1 w Δ = = 0.04 kj/k S total T = mcu cp, cu = T1 cu T = mw cp, wln = kj/k T1 ln 3.14 kj/k w Misunderstanding ( ) Q = m c T T cu cu p, cu 1 ( ) Q = m c T T w w p, w 1 Δ S = cu Δ S = w Q T Q T cu w 5

6 ds dq = T dq Δ s = s s1 = T 1 For ideal gas (pure substance) dt s s1 = cv ( T) + Rln T 1 dt s s1 = cp ( T) Rln T 1 v v 1 P P 1 T v s s1 = Cvav, ln + Rln T v 1 1 T s s1 = Cpav, ln Rln T P P 1 1 6

7 Exergy: A Measure of Work Potential The objectives are Examine the performance of engineering devices in term of the second law of thermodynamics. Define exergy, which is the maximum useful work that could be obtained from the system at a given state in a specified environment. Define reversible work, which is the maximum useful work that can be obtained as a system undergoes a process between two specified states. Define the exergy destruction, which is the wasted work potential during a process as a result of irreversibilities. Define the second-law efficiency. Develop the exergy balance relation. Apply exergy balance to closed systems and control volumes. 7

8 Exergy: A Measure of Work Potential Exergy and the Dead State -The useful work potential of a system is the amount of energy we extract as useful work. -The useful work potential of a system at the specified state is called exergy. - A system that is in equilibrium with its surroundings has zero exergy and is said to be at the dead state. price Investment No profit 8

9 Exergy Exergy is the work potential of a system in a specified environment and represents the maximum amount of useful work that can be obtained as the system is brought to equilibrium with the environment. Unlike energy, the value of exergy depends on the state of environment as well as the state of the system. 100 O C 1000 kj Total energy Income 90 O C 80 O C 70 O C 60 O C 800 kj 600 kj 400 kj 00 kj Exergy Environment (dead state) Available money Investment 9

10 Exergy of kinetic energy Kinetic energy is a form of mechanical energy and can be converted directly into work. Kinetic energy itself is the work potential or exergy of kinetic energy independent of the temperature and pressure of the environment. Exergy of kinetic energy: x ke = ke= V (kj/kg) Velocity, which is less than 10 m/s, can not produce electricity 10

11 Example: Exergy or work potential of the blowing air The electric power needs of community are to be met by windmills with 10-mdiameter rotors. The windmills are to be located where the wind is blowing steadily at an average velocity of 10 m/s. Determine the number of windmills that need to be installed if the required power output is 600 kw. 1 atm and 5 O C N= P required output P available ( ) V 10 m/s Exergy = ke = = = 50 J/kg Find mass flow rate of Air; = 0.05kJ/kg ρ=,available power= mke πd m =ρ AV =ρ V 4 P RT wind From Table; gas constant of air at 1 atm and 5 O C, R = 0.87 kpa.m 3 /kg.k 11

12 Example: Exergy or work potential of the blowing air P 101 kpa ρ= = = RT Mass flow rate of Air; ( kpa.m /kg.k)( K) πd m =ρ AV =ρ Vwind 4 3 π 10 = ( 1.18 m /kg) m ( 10 m/s) = 96.8 kg/s 4 Available Power mke ( )( ) m /kg = = 96.8 kg/s 0.05 kj/kg = 46.3 kw Problem gives the required power output = 600 kw Thus minimum number of windmills Prequired output 600 kw N= = = windmills P 46.3 kw available 1

13 Wind Turbine Generation 150 kw Location of Promthep Peninsula, Alternative Energy Station, Phuket Island 13

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15 Wind speed and output 1 atm and 5 O C - Useful work at v 1 m/s - Useful work = Input work loss work 15

16 Exergy of potential energy Potential energy is a form of mechanical energy and can be converted directly into work. Potential energy itself is the work potential or exergy of potential energy independent of the temperature and pressure of the environment. Exergy of potential energy: x = pe= gz pe (kj/kg) 16

17 Example :Exergy of potential energy One method of meeting the extra electric power demand at peak periods is to pump some water from a lake to a water reservoir at higher elevation at times of low demand and to generate electricity at times of high demand by letting this water run down and rotate a turbine. For an energy storage capacity of kwh, determine the minimum amount of water that needs to be stored at an average elevation (relative to the ground level) of 75m. Exergy = PE = mgz ( 5 10 kwh) ( 9.81 m/s )( 75 m) (kj) 6 PE 3600 s 1000 m /s m = = gz 1 h 1 kw.s/kg 10 = kg 17

18 Useful Work The work done by work producing devices is not always entirely in a useable form. The work done by the gas expanding in the piston-cylinder device is the boundary work and can be written as δw = PdV = ( P P0) dv + P0dV = δw + P dv b, useful 0 The word done on the surroundings is The actual work done by the gas is W = W + P dv = W + P ( V V ) b, useful 0 b, useful 0 1 W = P dv = P ( V V ) surr Any useful work delivered by a piston-cylinder device is due to the pressure above the atmospheric level. W = W W u surr 18

19 Reversible Work & Irreversibility Reversible work W rev is defined as the maximum amount of useful work that can be produced (or the minimum work that needs to be supplied) as a system undergoes a process between the specified initial and final states. Total work, W Useful work, W u Surrounding work, W sur W = W W u surr 19

20 Irreversibility, I Reversible Work & Irreversibility The difference between the reversible work W rev and the useful work W u is due to the irreversibilities present during the process and is called the irreversibility I. Irreversibility, I = loss It is equivalent to the exergy destroyed and is expressed as I = X destroyed = T0Sgen where S gen is the entropy generated during the process. I = Wrev, out Wu, out = Wu, in Wrev, in Exergy destroyed represents the lost work potential and is also called the wasted work or lost work. 0

21 Second-Law Efficiency The second-law efficiency is a measure of the performance of a device relative to the performance under reversible conditions for the same end states and is given by η II η η th = = W W u th, rev rev for heat engines and other work-producing devices and COP η II = = COP rev W W rev u for refrigerators, heat pumps, and other work-consuming devices. In general, the second-law efficiency is expressed as Exergy recovered η II = = 1 Exergy supplied Exergy destroyed Exergy supplied 1

22 Example (1) : Second law efficiency How much of the 100 kj of thermal energy at 800 K can be converted to useful work? Assume the environment to be at 5 O C Amount of heat can be converted to work is simply the amount that a reversible heat engine can be convert to work W = W =η Q max, out rev, out th, rev in Thus Maximum thermal efficiency from Carnot cycle TL η th, rev = 1 = 1 T H T T 98 = 1 = Wmax, out = Wrev, out = o H ( )( )

23 Example () : Second law efficiency A heat engine that receive heat from a furnace at 100 O C and rejects waste heat to a river at 0 O C has a thermal efficiency of 40%. Determine the second law efficiency of this power input. Thermal efficiency of a reversible heat engine is TL To η th, rev = 1 = 1 TH TH = 1 = ηth 0.4 η II = = = η th, rev 49.9% 3

24 Example (3) :Second law efficiency A heat engine receives heat from a source at 1500 K at rate of 700 kj/s, and it rejects the waste heat to a medium at 30 K. The measured power output of the heat engine is 30 kw, and the environment temperature is 5 O C. Determine (a) the reversible power, (b) the rate of irreversibility, and (c) the second-law efficiency of this heat engine. (a) The reversible power Carnot cycle gives the maximum efficiency T η =η = T W th,max th,rev 1 L rev, out th,rev H ( Q ) in =η 30 = 1 = = ( 0.787)( 700 kj/s) = kw (b) Rate of irreversibility I = W W rev, out u, out = = 30.7 kw 4

25 Example (3) :Second law efficiency (c) The second law efficiency η II W = W u,out rev,out 30 kw = 58.1% kw = 5

26 Exergy of change of a system At dead state, pressure is P 0 δ E δ E = de in out system 0 δq δw = du Consider HE work Consider Piston Work The work is the boundary work and can be written as δw = P dv = ( P P ) dv + P dv 0 0 = δw + P dv b, useful 0 Find HE work (which is of a reversible process) from heat transfer to Heat engine δwhe = ηthδq T0 = (1 ) δq= δq T T 0 δq T 6

27 Definition of entropy; HE work Exergy of change of a system ds δq T net = = δ Q T δw = δq+ TdS or δq= δw TdS HE 0 HE 0 From energy balance; δq δ W = du ( δ ) ( ) HE 0 δ b, useful 0 W T ds W + P dv = du δw δw = du P dv T ds HE b, useful 0 0 δw = δw + δw total useful b, useful HE = du P0dV + T0dS 7

28 Exergy change of a system δw = du P dv + T ds total useful 0 0 Integrating from the given state (no subscript) to the dead state (0 subscript) W = ( U U) P( V V) + T ( S S) total useful = ( U U ) + P( V V ) T ( S S ) This is the total useful work due to a system undergoing a reversible process from a given state to the dead state 8

29 Exergy of a fixed mass: Nonflow (or closed system) exergy Including the kinetic energy and potential energy, the exergy of a closed system is V X = ( U U0) + P0( V V0) T0( S S0) + m + mgz on a unit mass basis, the closed system (or nonflow) exergy is V φ = ( u u0) + P0( v v0) T0( s s0) + + gz = ( e e ) + P( v v ) T ( s s ) Here, u 0, v 0, and s 0 are the properties of the system evaluated at the dead state. If a system has energy same as environment or dead state, exergy is zero Note that the exergy of the internal energy of a system is zero at the dead state is zero since u = u 0, v = v 0, and s = s 0 at that state. 9

30 Exergy change between two states 1 V1 X1 = ( U1 U0) + P0( V1 V0) T0( S1 S0) + m + mgz1 1 V X = ( U U ) + P( V V ) T ( S S ) + m + mgz The exergy change of a closed system during a process is simply the difference between the final and initial exergies of the system, Δ X = X X1 = m( φ φ1) = ( E E ) + P( V V ) T ( S S ) V V = ( U U ) + P( V V ) T ( S S ) + m + mg( z z ) On a unit mass basis the exergy change of a closed system is Δ φ = ( φ φ ) 1 = ( e e1) + P0( v v1) T0( s s1) V V1 = ( u u1) + P0( v v1) T0( s s1) + + g( z z1) 30

31 Example: work potential of a fixed mass A 00-m 3 rigid tank contains compressed air at 1 MPa and 300 K. Determine how much work can be obtained from this air if the environment conditions are 100 kpa and 300 K. Find useful work or exergy Take control volume in rigid tank - No mass crosses the system boundary during the process - This problem is the work potential of a fixed mass Assumptions - Air is an ideal gas - KE and PE are negligible V X m m u u P v v T s s gz = φ = ( 1 0) + 0( 1 0) 0( 1 0) m Pv 3 ( 1000 kpa)( 00 m ) 1 1 = = = RT 1 3 ( 0.87 kpa.m /kg.k)( 300 K) 33 kg 31

32 RT1 RT 0 P 0 P0( v1 v0) = P0 = RT0 1 P1 P0 P1 T1 P 1 P 1 T0( s1 s0) = T0 cp ln Rln = RT0 T0 P0 P0 (T 1 = T 0 ) (T 1 = T 0 ) { ( ) ( )} X = mφ = m P v v T s s P = m RT 1 + RT ln P P1 P0 P P = mrt0 + P0 P1 1 0 ln kpa 100 kpa = ( 33 kg)( 0.87 kj/kg.k)( 300 K) ln kpa 1000 kpa 80,55 kj = = useful work can be obtained from the compressed air stored in the tank in the specified environment 3

33 Exergy of flow The energy needed to force mass to flow into or out of a control volume is the flow work per unit mass given by wflow = Pv w shaft exergy The exergy of flow work is the excess of flow work done against atmospheric air at P 0 to displace it by volume v. The useful work potential due to flow work is wflow, energy = Pv Pv 0 Thus, the exergy of flow energy is x = Pv P v = ( P P ) v flow energy

34 Flow Exergy Since flow energy is the sum of nonflow energy and the flow energy, the exergy of flow is the sum of the exergies of nonflow exergy and flow exergy. x = x + x flowing fluid nonflowing fluid flow exergy V = ( u u0) + P0( v v0) T0( s s0) + + gz+ ( P P0) v V = ( u+ Pv) ( u0 + Pv 0 0) T0( s s0) + + gz V = ( h h0) T0( s s0) + + gz 34

35 The flow (or stream) exergy is given by Flow Exergy V ψ = ( h h0) T0( s s0) + + gz The exergy of flow can be negative if the pressure is lower than atmospheric pressure. The exergy change of a fluid stream as it undergoes a process from state 1 to state is state 1 state V1 ψ 1 = ( h1 h0) T0( s1 s0) + + gz1 V ψ = ( h h0) T0( s s0) + + gz V V1 Δ ψ = ψ ψ1 = ( h h1) T0( s s1) + + g( z z1) 35

36 Example: Exergy change of a system A piston-cylinder device initially contains L of air at 100 kpa and 5 O C. Air is now compressed to a final state of 600 kpa and 150 O C. The useful work input is 1. kj. Assuming the surroundings are at 100 kpa and 5 O C, determine (a) the exergy of the air at the initial and the final states, Exergy of the air at the initial and the final states V X = ( U U0) + P0( V V0) T0( S S0) + m + mgz Assume: (1) Air is an ideal gas with constant specific heats. () KE and PE are negligible Since air initially is at the dead state, the initial state X 1 = 0 X = m cvavg, ( T T0) T0( s s0) + P0( V V0) 36

37 Example: Exergy change of a system From Table A-1, Gas constant R = 0.87 kpa.m3/kg.k. Specific heat of air at average temperature of (98+43)/ = 360 K are (from Table A-) c p,avg = kj/kg.k c v,avg = 0.7 kj/kg.k Constant mass and ideal gas m Pv 3 ( 100 kpa)( 0.00 m ) 1 1 = = = RT 1 3 ( 0.87 kpa.m /kg.k)( 98 K) kg ideal gas; Pv Pv 1 1 T = T 1 v Pv ( )( ) ( ) T 100 kpa L 43 K P 98 K 600 kpa 1 1 = = = T L T P s s0 = cpavg, ln Rln T P K 600 K = ( kj/kg.k) ln ( 0.87 kj/kg.k) ln 98 K 100 K = kj/kg.k 37

38 Example: Exergy change of a system Thus exergy of the air at the final states X = m cvavg, ( T T0) T0( s s0) + P0( V V0) ( ) ( ) ( ) 3 ( 100 kpa )( m ) X = kg 0.7 kj/kg.k (43 98 K) 98K ( kj/kg.k) + = kj 38

39 Example Refrigerant-134a is to be compressed from 0.14 MPa and -10 O C to 0.8 MPa and 50 O C steadily by a compressor. Taking the environment conditions to be 0 O C 95 kpa, determine the exergy change of the refrigerant during this process and the minimum work input that needs to be supplied to the compressor per unit mass of the refrigerant. Take control volume on compressor - Mass crosses the system boundary during the process - This problem is the exergy change of a fluid stream Assumptions - Steady flow condition - KE and PE are negligible Exergy change V V1 Δ ψ = ψ ψ1 = ( h h1) T0( s s1) + + g( z z1) 39

40 Δ ψ = ψ ψ = ( h h) T ( s s ) Inlet P 1 T 1 = 0.14 MPa = O 10 C h s1 1 = 43.4 kj/kg = kj/kg.k Outlet P T = 0.8 MPa = O 50 C h s = = kj/kg kj/kg.k Δ ψ = ψ ψ = ( h h) T ( s s ) = ( kj/kg) ( 93 K )( kj/kg.k) = 37.9 kj/kg Minimum work required for compressor w in,min = ψ ψ1 = 37.9 kj/kg 40

41 Exergy transfer by heat transfer Exergy can be transferred by heat, work, and mass flow, and exergy transfer accompanied by heat, work, and mass transfer are given by the following. From the second law, temperature can be converted into work. The maximum useful work is produced from it by passing this heat transfer through a reversible heat engine. Xheat = 1 T T Exergy transfer by heat: 0 Q Note that exergy transfer by heat is zero for adiabatic systems. 41

42 Exergy is the useful work potential, and the exergy transfer by work can simply be expressed as Exergy transfer by work: Exergy transfer by work X work W = W W surr (for boundary work) (for other forms of work) where W = P( V V ) surr 0 1 P 0 V 1 V = atmospheric pressure = the initial volumes of the system. = the final volumes of the system. The exergy transfer for shaft work and electrical work = the work W itself. Note that exergy transfer by work is zero for systems that have no work. 4

43 Exergy transfer by mass Mass flow is a mechanism to transport exergy, entropy, and energy into or out of a system. As mass in the amount m enters or leaves a system the exergy transfer is given by Exergy transfer by mass: Xmass = mψ Note that exergy transfer by mass is zero for systems that involve no flow. 43

44 The Decrease of Exergy Principle and Exergy Destruction The exergy of an isolated system during a process always decreases or, in the limiting case of a reversible process, remains constant. This is known as the decrease of exergy principle Δ X = ( X X ) 0 isolated 1 isolated Exergy Destruction Irreversibilities such as friction, mixing, heat transfer through finite temperature difference, always generate entropy, and anything that generates entropy always destroys exergy. The exergy destroyed is proportional to the entropy generated X = T S destroyed 0 gen X destroyed > = < 0 Irreversible proces 0 Reversible process 0 Impossible process 44

45 The Decrease of Exergy Principle and Exergy Destruction The exergy of an isolated system during a process always decreases or, in the limiting case of a reversible process, remains constant. This is known as the decrease of exergy principle Δ X = ( X X ) 0 isolated 1 isolated 45

46 Δ X = ( X X ) 0 isolated 1 isolated For isolated system, decrease of exergy is equal exergy destroyed. 46

47 Exergy Balances Exergy balance for any system undergoing any process can be expressed as Total Total Total Change in the exergy exergy exergy = total exergy entering leaving destroyed of the system General: X X X =ΔX in out destroyed system Net exergy transfer Exergy Change by heat, work, and mass destruction in exergy General, unit-mass basis: ( x x ) x =Δx in out destroyed system 47

48 General, rate form: Exergy Balances X X X = ΔX in out destroyed system Rate of net exergy transfer Rate of exergy Rate of change by heat, work, and mass destruction of exergy where X heat = 1 T T 0 Q X work = W useful X mass = m ψ Δ X = dx dt system system / For a reversible process, the exergy destruction term, X destroyed, is zero. 48

49 Exergy Balances Considering the system to be a general control volume and taking the positive direction of heat transfer to be to the system and the positive direction of work transfer to be from the system, the general exergy balance relations can be expressed more explicitly as 1 T Q W P ( V V ) + m m X X X i i eψe = Tk 0 [ ] ψ Rate form k 0 1 destroyed 1 T Q W P dv + m m X = dx 0 CV CV 1 k 0 iψi eψe destroyed Tk dt dt where the subscripts are i = inlet, e = exit, 1 = initial state, and = final state of the system. For closed systems, no mass crosses the boundaries and we omit the terms containing the sum over the inlets and exits. 49

50 Example Oxygen gas is compressed in a piston-cylinder device from an initial state of 0.8 m 3 /kg and 5 o C to a final state of 0.1 m 3 /kg and 87 o C. Determine the reversible work input and the increase in the exergy of the oxygen during this process. Assume the surroundings to be at 5 o C and 100 kpa. We assume that oxygen is an ideal gas with constant specific heats. From Table A-, R = kj/kg K. The specific heat is determined at the average temperature T av T1+ T (5 + 87) C 156 = = = C = ( ) K = 49 K Table A-(b) gives C v, ave = kj/kg K. 50

51 T v s s1 = cpavg, ln Rln T v Example (cont.) m 0.1 kj ( ) K kj kg = 0.69 ln ln 3 kg.k ( ) K kg.k m 0.8 kg kj = kg.k We calculate the reversible work input, which represents the minimum work input W rev,in in this case, from the exergy balance by setting the exergy destruction equal to zero. Xin Xout Xdestroyed =ΔXsystem Net exergy transfer Exergy Change by heat, work, and mass destruction in exergy Wrev,in = X X1 51

52 Therefore, the change in exergy and the reversible work are identical in this case. Substituting the closed system exergy relation, the reversible work input during this process is determined to be w = φ φ rev,in 1 = ( u u1) + P0( v v1) T0( s s1) = C ( T T) + P( v v ) T ( s s ) v,ave kj 3 m kj 3 = (87 5)K + 100kPa( ) kg K kj (5 + 73)K( ) kg K = Example (cont.) kj 14.1 kg kg m kpa The increase in exergy of the oxygen is x x = φ φ = w = 1 1 rev,in kj 14.1 kg 5

53 Example Steam enters an adiabatic turbine at 6 MPa, 600 C, and 80 m/s and leaves at 50 kpa, 100 C, and 140 m/s. The surroundings to the turbine are at 5 C. If the power output of the turbine is 5MW, determine (a)the power potential of the steam at its inlet conditions, in MW. (b) the reversible power, in MW. (c)the second law efficiency. We assume steady-flow and neglect changes in potential energy. 53

54 Example (cont.) The mass flow rate of the steam is determined from the steady-flow energy equation applied to the actual process, E in E out = ΔE systems 0 (steady) Rate of net energy transfer Rate of change by heat, work, and mass of energy V1 V m 1 ( h1 + ) m ( h + ) W out = 0 Conservation of mass for the steady flow gives m in m out =Δm system Rate of net mass transfer m 1 1 Rate of change of mass m = 0 m = m = m 54

55 Example (cont.) The work done by the turbine and the mass flow rate are V V W out = m ( h1 h ) + W out m = ( h h ) Δke 1 1 where V V 1 Δ ke = (140m/s) (80m/s) 1kJ/kg = 1000 m /s kj = 6.6 kg 55

56 Example (cont.) From the steam tables: P 1 o T1 C s 1 P T P T 0 0 kj h1 = = 6MPa kg = 600 kj = kg K kj h = 68.4 = 50kPa kg = o 100 C kj s = kg K kj h0 h o = f@5 C = 100kPa kg = o 5 C kj s 0 s = o f@5 C kg K 56

57 Example (cont.) m = W out ( h h ) Δke 1 5MW 1000kJ/s = kj ( ) MW kg = kg 5.16 s 57

58 Example (cont.) The power potential of the steam at the inlet conditions is equivalent to its exergy at the inlet state. Recall that we neglect the potential energy of the flow. 0 V Ψ = = mψ 1 m ( h1 h0) T0( s1 s0) gz1 kj ( ) (98 K)( ) kg kg Ψ 1 = 5.16 s (80m/s) kj/kg m /s kg kj MW = s kg 1000kJ/s = 7.91MW kj kg K 58

59 Example (cont.) The power output of the turbine if there are no irreversibilities is the reversible power and is determined from the rate form of the exergy balance applied on the turbine and setting the exergy destruction term equal to zero. X in X out X destroyed = ΔX system Rate of net exergy transfer Rate of exergy Rate of change by heat, work, and mass destruction of exergy W = m ( ψ ψ ) rev, out 1 X in = X out m ψ = W + m ψ 1 rev,out = m h h T s s Δke Δpe [( ) ( ) ]

60 Example (cont.) W rev, out = 5.16 kg s kj ( ) (98 K)( ) kg kj 6.6 kg kj kg K kg kj MW = s kg 1000kJ/s = 5.81MW The second-law efficiency is determined from η = W 5MW II 86.1% W = 5.81MW = rev 60

61 Example Chickens with an average mass of. kg and average specific heat of 3.54 kj/kg. O C are to be cooled by chilled water that enters a continuous-flow-type immersion chiller at 0.5 O C and leaves at.5 O C. Chicken are dropped into the chiller at a uniform temperature of 15 O C at a rate of 500 chickens per hour and are cooled to an average temperature of 3 O C before they are taken out. The chiller gains heat from the surroundings at a rate of 00 kj/hr. Determine (a) the rate of heat removal from the chicken, in kw and (b) the rate of exergy destruction during this chilling process. Take T O = 5 O C. Chilled water input T w,in = 0.5 O C. chiller Chicken of. kg T chicken,in = 15 O C. T chicken,out = 3 O C. T w,out =.5 O C. Heat from surroundings = 00 kj/hr. 61

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67 Example An ordinary egg be approximated as a 5.5 cm diameter sphere. The egg is initially at a uniform temperature of 8 O C can is dropped into boiling water at 97 O C. Taking the properties of egg to be ρ = 100 kg/m 3 and c p = 3.3 kj/kg. O C, determine how much heat is transferred to the egg by the time average temperature of the egg rises to 70 O C and the amount of exergy destruction associated with this heat transfer process. Take T O = 5 O C. 67

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73 Example A 0.04-m3 tank initially contains air at ambient conditions of 100 kpa and O C. Now, a 15 liter tank containing liquid water at 85 O C is placed into the tank without causing any air to escape. After some heat transfer from the water to the air and the surroundings, both the air and water are measured to be at 44 O C. Determine (a) the amount of heat lost to the surroundings and (b) the exergy destruction during this process. 73

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80 Example Hot combustion gases enter the nozzle of a turbojet engine at 60 kpa, 747 O C, and 80 m/s and exit at 70 kpa and 500 O C. Assuming the nozzle to be adiabatic and the surroundings to be at 0 O C, determine (a) the exit velocity and (b) the decrease in the exergy of the gases. Take k = 1.3 and c p = 1.15 kj/kg. O C for the combustion gases. 80

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84 Example Liquid water at 15 O C is heated in a chamber by mixing it with saturated steam. Liquid water enters the chamber at the steam pressure at a rate of 4.6 kg/s. The mixture leaves the mixing chamber as a liquid at 45 O C. If the surroundings are at 15 O C, determine (a) the temperature of saturated steam entering the chamber, (b) the exergy destruction during this mixing process, and (c) the second-law efficiency of the mixing chamber. 84

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