Chapter 6: Using Entropy


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1 Chapter 6: Using Entropy Photo courtesy of U.S. Military Academy photo archives. Combining the 1 st and the nd Laws of Thermodynamics
2 ENGINEERING CONTEXT Up to this point, our study of the second law has been concerned primarily with what it says about systems undergoing thermodynamic cycles. In this chapter means are introduced for analyzing systems from the second law perspective as they undergo processes that are not necessarily cycles. The property entropy plays a prominent part in these considerations. The objective of the present chapter is to introduce entropy and show its use for thermodynamic analysis. The word energy is so much a part of the language that you were undoubtedly familiar with the term before encountering it in early science courses. This familiarity probably facilitated the study of energy in these courses and in the current course in engineering thermodynamics. In the present chapter you will see that the analysis of systems from a second law perspective is conveniently accomplished in terms of the property entropy. Energy and entropy are both abstract concepts. However, unlike energy, the word entropy is seldom heard in everyday conversation, and you may never have dealt with it quantitatively before. Energy and entropy play important roles in the remaining chapters of this book.
3 Introducing Entropy Corrollary of Second Law is introduced: Clasius Inequality Expands last chapter treatment of Two heat reservoirs to arbitrary number of heat reservoirs from which system receives energy by heat transfer or rejects energy by heat transfer Provides basis for two concepts of second law for analyzing closed and open systems: 1) Entropy as a Property ) Entropy Balance
4 Clasius Inequality For any thermodynamic cycle For any number of reservoirs: δq T b 0 δq Represents heat transfer at local system boundary location, b, at temperature T. Heat transfer may be positive (IN) or negative (OUT) T must be absolute temperature (Kelvin or Rankine) T always positive. Can NOT be negative (Celsius or Farenheit) Indicates integration over all processes and all parts of boundary
5 Clasius Inequality δq T b 0 Inequality has same meaning as with KelvinPlanck: Equality applies when no INTERNAL IRREVERSIBILITES are present Inequality applies when INTERNAL IRREVERSIBILITES are present
6 Example of Multiple Heat Transfers & Ts Note: δ Q j δ Q 1 δ Q < 0 T j T 1 δ Q 3 Because heat transfer is OUT T 3 T δ Q δq T < 0 We will see later that this is and entropy transfer OUT
7 Reversible Heat Transfers δq j δq 1 T j T 1 δq 3 T 3 T δq Note: T i for reservoir AND at boundary; T must approach zero
8 Developing Clasius Inequality Proof provided in text
9 Alternate Statements of Clasius Inequality δq T b 0 δq + σ generated = T b 0 δq + σ cycle = T b 0 δq = σ T b cycle
10 Alternate Statements of Clasius Inequality δq = σ T b cycle σ σ σ cycle cycle cycle = > < δq + σ generated = T b 0 no irreversibilities present within the system 0 irreversibilities present within the system 0 impossible 0 Unlike mass and energy, which are conserved in every process, entropy in the presence of irreversibilities, is always produced.
11 Defining Entropy Change Quantity is a Property if and only if change in value between two states is independent of process (path) Will now show that the quantity is a Property δq T Reversible Meaning: For a REVERSIBLE process, the integral of the heat transfer divided by the local absolute temperature does NOT depend on the process This integral is therefore a PROPERTY We will give this property the name ENTROPY
12 Defining Entropy Change Quantity is a Property if and only if change in value between two states is independent of process (path) Processes A, B and C are INTERNALLY REVERSIBLE: σ cycle = σ = generated 0 C Cycle AC: 1 δq δq + = σ T T 1 A C cycle 0 1 B A Cycle BC: 1 δq δq + = σ T T 1 B C cycle 0
13 Defining Entropy Change Quantity is a Property if and only if change in value between two states is independent of process (path) Subtract: From: 1 δq δq + = σ T T 1 A 1 δq δq + = σ T T 1 B C C cycle cycle 0 0 Yielding: δq δq = T T 1 A 1 B
14 Defining Entropy Change C Since A and B are arbitrary, it follows that the integral is same for ANY Internally Reversible process. B A Thus, the Integral is a Property 1 Property is ENTROPY, S Entropy Change between states 1 & can be calculated for ANY Internally Reversible process as: S δq S = T 1 1 Int Rev
15 Entropy Change 1 C B A Differential basis defining entropy change: ds Entropy, S, is an Extensive property Intensive form is: s = S m δq = T Int Rev Units for Entropy: Extensive (SI): kj/k Extensive (English): BTU/ o R Intensive (SI): kj/kg K Intensive (English): BTU/lb o R
16 Defining Entropy Change Entropy change between states 1 & can be calculated for ANY Internally Reversible process as: B C A S δq S = T 1 1 Int Rev 1 Once calculated, Entropy difference is known Important Note: Entropy change between states 1 & is SAME whether process between 1 & is REVERSIBLE or IRREVERSIBLE. Just can t calculate change for IRREVERSIBLE processes.
17 Recognize that Entropy is still an abstract concept for you Like enthalpy we defined earlier, to gain appreciation for Entropy, need to understand: HOW to use it and WHAT it is used for
18 Retrieving Entropy Data Chapter 3: Means for retrieving property data from Tables, graphs, equations (and software) Emphasis on properties: P, v, T, h Which are required for: 1 st Law (energy conservation) and mass conservation For application of nd Law, entropy values usually needed Thus, we need to understand how to retrieve entropy data
19 Entropy Reference Value? Similarity to Energy with regard to absolute values of Entropy In most cases, absolute values not important Only difference in Entropy between states is important B C A S =? S y y δq = Sx + T x Int Rev 1 S 1 =? S x is reference value S ref, S ref,1
20 Finding Entropy Data For Water and Refrigerants: Using Tables A to A18 and Tables AE to A18E
21 Tabulations of Water Properties Liquid Table A5, A5E Superheated Vapor Table A4, A4E T Sat Sat Liquid vapor f g LiquidVapor Mixture Table A, A3 (AE, A3E) Saturation Line v s
22 Two Saturated Tables for Each One for Saturated Temperature One for Saturated Pressure Sometimes we know T, sometimes P
23 Saturated Water Table A, A3, AE, A3E T f g s
24 T Computing Entropy Value Under Dome f s It is not enough to know T, P in order to establish state under dome g S = S + S liquid vapor S Sliq Svap s = = + m m m mliqsf mvapsg s = + m m mvap x = m mliq = 1 x m s = (1 xs ) + xs (x= Quality) f g Need T or P, and one other property ( ) s = s + x s s = s + xis f g f f fg
25 Superheated Vapor Table A4, A4E 500 F T f 93 F g s
26 Compressed or Subcooled Liquid Table A5, A5E T f g Note: v subcooled tables are sparse because it is accurate to use incompressible liquid model
27 Text Examples
28 Problem solving using property diagrams is important Two commonly used property diagrams are: Temperature Entropy (Ts diagram) Enthalpy Entropy (Mollier diagram)
29 Graphical Entropy Data TemperatureEntropy Diagram EnthalpyEntropy (Mollier) Diagram
30 Note general features See Figs A7 & A7E for water TemperatureEntropy Diagram
31 Figure A7 Temperature entropy diagram for water (SI units)
32 Figure A7E Temperature entropy diagram for water (English units).
33 EnthalpyEntropy (Mollier) Diagram Note general features See Figs A8 & A8E for water
34 Figure A8 Enthalpy entropy diagram for water (SI units)
35 Figure A8E Enthalpy entropy diagram for water (English units)
36 Examples as per text here
37 Using T ds Equations Although entropy changes can be determined from: S δq S = T 1 1 Int Rev This requires knowledge of how heat transfer and Q vary during a process. In practice, entropy changes calculated from changes in other properties. Let s see how.
38 Using T ds Equations Note that T ds equations are also important with respect to: 1) Deriving other important properties ) Constructing property tables (Chapter 11)
39 Simple Compressible Substance Energy balance in differential form: IN = STORED + OUT Neglecting KE and PE δq = du + δw ( ) ( ) Int Rev Int Rev δ Q = ( ) Int Re v TdS δ W = ( ) Int Re v PdV First TdS Equation: TdS = du + PdV
40 Now consider second TdS equation Recall that, by definition: H = U + PV Form differential: dh = du + d( PV ) = du + PdV + VdP Rearrange: Substitute: du + PdV = dh VdP TdS = du + PdV Second TdS Equation: TdS = dh VdP
41 TdS = du + PdV Per Unit Mass: Tds = du + Pdv Tds = du + Pdv Per Mole: TdS = dh VdP Tds = dh vdp Per Unit Mass: Tds = dh vdp Per Mole:
42 Although internally reversible processes used to derive these equations, they apply generally do not require reversible processes. Since all terms are properties, applies to irreversible and reversible processes (pathindependent) Change in Entropy is independent of details of process
43 Text Example
44 Entropy Determination For Incompressible Substances Incompressible substance model is generally used with liquids and solids only, and assumes constant specific volume. ds ct ( ) dt du P du = + dv = = T T T T Specific heat depends on Temperature only. No difference between c P and c v c= c = c P v T s( T) s1( T1) = c( T) T T Specific heat may be constant: ( ) ( ) 1 dt s T s1 T1 = cln T T 1
45 Entropy Change of an Ideal Gas: TdS equations used to compute entropy change between states of Ideal Gas Tds = du + Pdv Tds = dh vdp For an Ideal Gas: du c ( T ) dt ds ds dh = c ( T ) dt Pv = = v p RT du = + T dh = T p P T v T dv dp c T = c T + R ( ) ( ) v Equation of State
46 Substitute Ideal Gas Relations into TdS equations: du = c ( T ) dt v dh = c ( T ) dt p Pv = RT ds ds du P = + dv T T dh v = dp T T dt ds c T R T = ( ) + s = stv v (, ) dt ds = c ( ) p T R T dv v dp P s = (, ) s T P
47 Can integrate these equations: dt dv ds = cv ( T) + R T v dt dp ds = c ( ) p T R T P T dt v (, ) 1( 1, 1) = v ( ) + ln T v T 1 s T v s T v c T R 1 T dt P ( ) ( ) ( ), 1 1, 1 = P ln T P T 1 s T P s T P c T R 1
48 Where Using Ideal Gas Tables: o Let s define a new variable: ( ) ( ) T Where is an arbitrary reference temperature Thus, in the equation: ( ) T o p s T dt = T c T T T dt P ( ) ( ) ( ), 1 1, 1 = P ln T P T 1 s T P s T P c T R T T T dt dt dt c T = c T c T = s T s T T T T s ( ) ( ) ( ) o( ) o( ) P P P T T T T 1
49 T dt P ( ) ( ) ( ), 1 1, 1 = P ln T P T 1 s T P s T P c T R Becomes: ( ) ( ) o( ) o P ( ) s T, P s1 T1, P1 = s T s T1 Rln P1 Where: 1 ( T) c s T s T dt T 0 0 p ( ) ( 1) = (kj/kg K or BTU/lb o R) T T This integral can be tabulated (Ideal Gas Tables) Tables A, AE: 1
50 Table A: Ideal Gas Properties of Air
51 Table AE: Ideal Gas Properties of Air
52 Using Ideal Gas Tables: Similar approach for per mole basis s( T, P) s( T1, P1) s ( T) s ( T1) Rln P = P1 This integral can be tabulated (Ideal Gas Tables) Tables A3, A3E: What if we have T and v information, rather than T and P? T dt v (, ) 1( 1, 1) = v ( ) + ln T v T 1 s T v s T v c T R 1
53 Ideal Gases with Constant Specific Heats T dt P (, ) 1( 1, 1) = P ( ) ln T P T 1 s T P s T P c T R T P st (, P) st ( 1, P1) = cp ln Rln T P T dt v (, ) 1( 1, 1) = v ( ) + ln T v T 1 s T v s T v c T R T v st (, v) stv ( 1, 1) = cv ln + Rln T v Use Specific heat data in Tables A0 and A1
54 Table A0: Ideal Gas Specific Heats of Some Common Gases (kj/kg K)
55 Entropy Change for Closed Systems Internally Reversible Processes Entropy transfer ACCOMPANIES heat transfer For Internally Reversible Processes, Entropy change can be calculated from: This links heat transfer and entropy transfer ds δq = T Int Rev Note that for Closed, Internally Reversible system: Entropy INCREASES when heat transfer IN Entropy DECREASES when heat transfer OUT Entropy DOES NOT CHANGE when no heat transfer occurs
56 Entropy Change for Closed Systems Internally Reversible Processes Q & Entropy IN Q & Entropy OUT Q=0 S=0 Process where Q = 0 is: ADIABATIC (Insulated) For process that is ADIABATIC AND REVERSIBLE, The ENTROPY DOES NOT change. A constant Entropy process is called ISENTROPIC Thus, an ADIABATIC AND REVERSIBLE process is an ISENTROPIC process
57 Entropy Change Internally Reversible Processes For internally reversible processes, the area under the process curve represents the process Heat Transfer. ds δq = T Int Rev δ Q = ( ) Int Re v TdS Q Int Rev = TdS 1
58 Entropy Change Internally Reversible Processes 1) Heat transfer is entire area under TS diagram ) Temperature must be absolute scale (Kelvin or Rankine), >0 δq ds T 3) Not valid for IRREVERSIBLE processes. Then area is NOT equal to heat transfer ds δq T δ Q TdS
59 Entropy Change Internally Reversible Processes Internally reversible processes are idealizations, but are found in all Carnot cycles. Carnot cycles on the TS diagram. The diagram on the left represents a power cycle, and on the right, a refrigeration/heat pump cycle.
60 Internally reversible processes are idealizations, but are found in all Carnot cycles. Power Cycle Refrigeration Cycle Areas (+ or ) is magnitude of work: cycle η = 1 W = Q IN Q OUT W Q IN = T T C H Note how separation of T H and T C related to more work
61 Example in text
62 Entropy Balance: Closed Systems Rev Irrev, b Cycle consists of process, Irrev, where Internal Irreversibilities are present Followed by process, Rev, which is Internally Reversible 1 Note dotted line 1 δq δq + = σ T T 1 b IntRev cycle Subscript, b, designates evaluation of integral at boundary. Not necessary for second case because temperature is uniform throughout system
63 Entropy Balance: Closed Systems Rev Irrev Cycle consists of process, Irrev, where Internal Irreversibilities are present Followed by process, Rev, which is Internally Reversible 1 1 δq δq + = σ cycle = σ T T 1 b IntRev Where: S 1 δq S = T 1 Int Rev δq + ( S1 S) = σ T 1 b S δq S = + σ T ( ) 1 1 b Entropy Change = Entropy + Entropy Transfer Production
64 Entropy Balance: Closed Systems S δq S = + σ T ( ) 1 1 b Entropy Change = Entropy + Entropy Transfer Production Total entropy change associated with heat transfer (+ or ) AND entropy generated by IRREVERSIBILITIES (ALWAYS + or 0) Can INCREASE or DECREASE entropy by heat transfer IRREVERSIBILITY will ALWAYS INCREASE entropy ZERO entropy generated for a REVERSIBLE process
65 Entropy Balance: Closed Systems S 1 δq S = + σ T 1 b Entropy Change Entropy Transfer Entropy Production Since σ measures the effect of irreversibilities present within the system during a process, its value depends on the nature of the process, and thus is NOT a property S 1 δq S T 1 b
66 Some Examples Reversible, Adiabatic Process: S δq S = + σ T ( ) 1 1 b Entropy Change = Entropy + Entropy Transfer Production 0 Adiabatic 0 Reversible A Reversible, Adiabatic Process: Is an Isentropic Process
67 Some Examples Is an Isentropic Process always a Reversible, Adiabatic Process? S δq S = + σ T ( ) 1 1 b Entropy Change = Entropy + Entropy Transfer Production Suppose Entropy generation occurs by Irreversibility, Is there a way to decrease entropy to produce an Isentropic Process?
68 S Some Examples δq S = + σ T ( ) 1 1 b Entropy Change = Entropy + Entropy Transfer Production All Reversible and Adiabatic Processes are Isentropic All Isentropic Processes are NOT Reversible and Adiabatic
69 S Some Examples δq S = + σ T ( ) 1 1 b Entropy Change = Entropy + Entropy Transfer Production What happens to Entropy for an Adiabatic, Irreversible Process?
70 Text Example
71 Other common forms of entropy balance: Q 1 For multiple heat transfers T 1 T Q T n Q n S Q j S1 = +σ j Tj (Note: Q could be + or ) S Q Q Q 1 n S1 = σ T1 T Tn Uniform Boundary Temperature, T b S Q S = +σ T 1 b
72 Other common forms of entropy balance: Q 1 Rate basis, multiple heat transfers T 1 T T n ds dt Q j = + σ j Tj Q Q n Q (Note: could be + or ) Differential form: ds δq = + δσ T b Note inexact differential notation for nonproperties: Q and σ
73 Evaluating Entropy Production and Transfer Absolute value of entropy production not as useful as relative values: Can use this information to focus attention on components where most irreversibilties are produced
74 Example 6.: Irreversible process of water Example 6.3: Eval Min theoretical compression work Example 6.4: Pinpointing Irrevs
75 Increase in Entropy Principle: Closed Systems Show that the summation of the entropy changes of surroundings AND system must always increase or remain the same System Energy Transfer is Zero Surroundings Mass Transfer is Zero The larger system is an ISOLATED SYSTEM E] = 0 isol
76 System Energy Transfer is Zero Surroundings Mass Transfer is Zero The larger system is an ISOLATED SYSTEM = E] + E] = 0 E] 0 isol system Conservation of energy constrains possible processes: Energy changes in system and surroundings must balance However, not all such processes are possible ( nd Law) surr
77 System Energy Transfer is Zero Surroundings Mass Transfer is Zero Adiabatic 0 δq S] = isol T + σ 1 b isol S] = σ > 0 isol isol ] ] 0 S + S = σ > system surr isol
78 System Energy Transfer is Zero Surroundings Mass Transfer is Zero ] ] 0 S + S = σ > system surr isol Only processes that can occur are when entropy of Isolated System INCREASES. Entropy is Extensive: Entropy need NOT increase for BOTH System AND Surroundings, but SUM must increase. Direction of process and feasibility constrained Spontaneous processes tend to reach equilibrium increasing S
79 Increase in Entropy Principle: Closed Systems The summation of the entropy changes of surroundings AND system must always increase (or remain the same for ideal) ] ] total S + S = σ system surr σ σ σ total total total = > < 0 ideal 0 actual 0 impossible
80 Examples in Text Example 6.5: Quenching metal bar Statistical Interpretation of Entropy
81 Statistical Interpretation of Entropy
82 Entropy Rate Balance: Control Volumes σ cv Inlets, i T j Q j Exits, e In + Gen = Stored + Out ds dt cv Q j = + ms i i ms e e + σ CV j T j i e Rate of entropy change Rates of entropy transfer Rate of entropy production
83 Entropy Rate Balance: Control Volumes When spatial variations occur, use Integral Form: S () cv t = ρsdv V j Q T j j q = A T b da ms = sρvda i i n i i A i d dt q ρsdv= da+ ( sρvda) ( sρvda) + σ n n gen V A T A i A i e e b
84 At SteadyState Entropy Rate Balance: Control Volumes Mass: Energy: 0 Entropy: i i = me e m Vi Ve = Qcv Wcv + mi hi + + gzi me he + + gze i e Q i i e e j Tj i e j 0 = + ms ms + σ CV
85 Single Inlet, Single Outlet SteadyState Control Volumes Mass: Energy: 0 m = m = m i e ( V ) i Ve Q cv W cv = + ( h h ) + + g z z m m ( ) i e i e Entropy: Q = + m ( s i s e) + j Tj j 0 σ CV 0 1 Qj σ = + ( si se) + m j T j m CV
86 Single Inlet, Single Outlet SteadyState Control Volumes Entropy: ( s s ) e Q i = + m j T j m 1 j σ CV Entropy passing from inlet to exit can: increase, decrease or remain constant. Second term RHS is positive or zero. Entropy/mass can only decrease if NET entropy flow OUT with heat transfer exceeds entropy Generation IN control volume When NO heat transfer (adiabatic): s ( ) CV e σ si = m
87 Single Inlet, Single Outlet SteadyState Control Volumes When NO heat transfer (Adiabatic): s ( ) CV e σ si = m Entropy increases when IRREVERSIBILTY present Entropy constant when REVERSIBLE: ISENTROPIC s e = s i
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91 Examples 6.6 Entropy production in a steam turbine 6.7 Evaluating a performance claim 6.8 Entropy production in heat pump components
92 Isentropic Processes Showing isentropic processes is rapid/easy on Ts or hs diagrams. Vertical lines: However, tabular data may still be used as well.
93 Isentropic Processes State 1 is in superheated region P 1 and T 1 used to get s 1 State is in superheated region Where s = s 1. Use P, T, other Isentropic process: s 3 = s = s 1 Use P 3 or T 3 to get s sat. IF s 3 < s sat, then state 3 is in saturated region. Get quality, x 3, then can get other properties.
94 Saturated Water Table A, A3, AE, A3E T f g v
95 Superheated Vapor Table A4, A4E 500 F T f 93 F g v
96 Isentropic Processes: Ideal Gas o o P s( T, P) s1( T1, P1) = s ( T) s ( T1) Rln P1 s ( T ) s ( T ) Rln P = P1 s = s 1 If we have 3 of: T 1, T, P 1, P, then we can get the fourth Example: T 1 and two pressures, can get T : s T s T P 1exp R ( ) o( ) o 1 = P Use Tables A and A3 for AIR and AE and A3E (English) for other gases
97 Isentropic Processes: Ideal Gas P ( ) ( ) s T s T = R o o exp 1 P1 Define Relative Pressure, P r o s T exp P R = P s T exp R ( ) o 1 1 ( ) r ( ) P T ( T) o s = exp R P P = P P r 1 r1 (for s 1 = s, AIR only) Use Tables A for Relative Pressure as function of Temperature. Not truly a pressure. Don t confuse with Reduced Pressure (Compressibility Diagram)
98 Isentropic Processes: Ideal Gas v RT P 1 = v P RT 1 1 PG equation of state: v = RT P r ( ) Define Relative Volume, v r v v v v r 1 r1 ( T ) v RT P r1 1 = v1 P T RT1 RT vr ( T) = = f T Pr ( T) = (s 1 =s, AIR only) ( ) Use Tables A for Relative Volume as function of Temperature. Not truly a volume. Don t confuse with Pseudoreduced specific volume (Compressibility Diagram)
99
100 Isentropic Processes: Ideal Gas Assume constant specific heats and define ratio of specific heats, k: c p kr = k 1 R = k 1 T P st (, P) st ( 1, P1) = Cp ln Rln T P 1 1 T v st (, v) stv ( 1, 1) = Cv ln + Rln T v c v 1 1 k 1 k 1 T P v = = T1 P1 v ( k 1) k cp = 1 c v T P 0= Cp ln Rln T P 1 1 T v 0= Cv ln + Rln T v P P v = v k
101 Examples: 6.9 Isentropic process in Air 6.10 Air leaking from tank
102 Isentropic Efficiencies Comparing actual (adiabatic) and isentropic devices with  same inlet states  same exit pressure Turbines, compressors, nozzles and pumps Isentropic turbine efficiency: Actual versus Isentropic W Turbine 1 This helicopter gas turbine engine photo is courtesy of the U.S. Military Academy.
103 Isentropic Efficiencies (Adiabatic, Reversible) Turbines 1 st Law: W cv = h h m 1 nd Law: σ cv = s s m 1 0 This helicopter gas turbine engine photo is courtesy of the U.S. Military Academy. Typical: η T W Turbine 1
104 Isentropic Turbine Efficiency p Mollier chart (Ts) s (h 1 h ) Actual expansion s s Isentropic expansion (h 1 h s ) Note: W cv m ( h1 h) η T = = 1 W cv ( h1 h) s m s Accessible states: Q = 0; s > 0
105 Isentropic Turbine Efficiency (Ts) (Mollier chart better) p (h 1 h ) s Actual expansion (h 1 h s ) s s Isentropic expansion Note: W cv m ( h1 h) η T = = 1 W cv ( h1 h) s m s Accessible states: Q = 0; s > 0
106 Compressors and Pumps Reciprocating compressor Rotating compressors Common Form of 1 st Law: W V V = h1 h + g z1 z + m ( ) 1 ( ) [ W < 0 ]
107 Isentropic Compressor Efficiency (Gases) h s s T s p p 1 Actual compression (h h 1 ) Liquid (h s h 1 ) Isentropic compression s 1 T 1 Note: W cv m s ( hs h1) ηc = = 1 W cv ( h h1) m Accessible states: Q = 0; s > 0
108 Isentropic Efficiencies: Compressors (Gases) and Pumps (Liquids) Typical Isentropic Efficiencies of Compressors and Pumps: η c Pumps, assuming incompressible model: (will consider more detail later) η P = v P P ( ) h 1 h 1 s
109 Nozzles and Diffusers Common Form of 1 st Law: 0 V V ( ) 1 = h1 h +
110 Isentropic Nozzle Efficiency V 1 Nozzle V η = nozzle V V s / / Common for: ηnozzle 0.95
111 Examples: 6.11 Eval turbine work using Isentropic efficiency 6.1 Eval Isentropic turbine efficiency 6.13 Eval Isentropic nozzle efficiency 6.14 Eval Isentropic compressor efficiency
112 Special Cases: Evaluate Q & W for NO Internal Irreversibility Internally Reversible, SteadyState Flow Oneinlet, OneOutlet: cv Heat Transfer: Q = m Int Rev 0 S.S Q ms s T cv 0 = + ( ) + σ CV Q cv ( s ) = T s m Tds T=T(s)
113 Special Cases (Internally Reversible): Heat Transfer and Work Work Transfer for Reversible or Irreversible: W m cv ( V ) 1 V Q cv = + ( h h ) + + g z z m ( ) 1 1 Internally Reversible, SteadyState Flow Oneinlet, OneOutlet: W m cv ( ) ( V ) 1 V ( ) = Tds + h h + + g z z 1 1 Int Rev 1
114 Special Cases: Heat Transfer and Work Tds = dh vdp Thus, W m cv 1 Tds = h h vdp ( ) 1 1 ( V1 V ) Int Rev 1 ( ) = vdp + + g z z 1 KE and PE changes often negligible W m cv Int rev = 1 vdp
115 Note: magnitude of work per mass for gas or liquid is directly related to specific volume of fluid, v W m cv Int rev = 1 vdp Thus, for same pressure rise, magnitude of work per mass for liquid in pump (low v) is much smaller than for gas (larger v) in compressor Derived for Internally Reversible case, but qualitatively true for real, irreversible processes
116 Special Cases: Heat Transfer and Work Internally Reversible, SteadyState Flow: Q cv = m int rev 1 Tds W cv = m int rev 1 vdp (In many cases KE = PE = 0) Internally Reversible, SteadyState, Incompressible fluid: W cv = v P m int rev ( P) 1 Steady State, Reversible, no significant CV work terms (eg nozzles & diffusers): V V1 1 vdp + + g ( z z1) = This equation, used commonly in fluid mechanics, is known as the Bernoulli Equation 0
117 n PV = General: Work in Polytropic Processes constant Internally Reversible cv 1 n W = vdp = ( cons tan t ) m int 1 1 rev dp P 1 n W cv n = ( Pv Pv 1 1) (n 1) m n 1 int rev cv W P = ( Pv 1 1) m P int 1 rev ln (n=1)
118 Work in Polytropic Processes: Ideal Gas For Ideal Gases: n PV = W cv n = ( Pv Pv 1 1) (n 1) m n 1 int rev constant T T Equivalent to: Where: 1 1 k 1 k P = P W cv nr = ( T T 1) (ideal gas, n 1) m n 1 int rev ( n 1) n W nrt P = 1 (ideal gas, n 1) cv 1 m n 1 P1 int rev cv W P = RT ln (ideal gas, n=1) m P int 1 rev
119 Examples: 6.15 Polytropic compression of air
120 END
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