Chapter 6: Using Entropy


 Magnus Burke
 1 years ago
 Views:
Transcription
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
88
89
90
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
Second Law of Thermodynamics Alternative Statements
Second Law of Thermodynamics Alternative Statements There is no simple statement that captures all aspects of the second law. Several alternative formulations of the second law are found in the technical
More informationChapter 6 Energy Equation for a Control Volume
Chapter 6 Energy Equation for a Control Volume Conservation of Mass and the Control Volume Closed systems: The mass of the system remain constant during a process. Control volumes: Mass can cross the boundaries,
More informationLesson 5 Review of fundamental principles Thermodynamics : Part II
Lesson 5 Review of fundamental principles Thermodynamics : Part II Version ME, IIT Kharagpur .The specific objectives are to:. State principles of evaluating thermodynamic properties of pure substances
More informationEntropy. Objectives. MAE 320  Chapter 7. Definition of Entropy. Definition of Entropy. Definition of Entropy. Definition of Entropy + Δ
MAE 320  Chapter 7 Entropy Objectives Defe a new property called entropy to quantify the secondlaw effects. Establish the crease of entropy prciple. Calculate the entropy changes that take place durg
More informationApplied Thermodynamics for Marine Systems Prof. P. K. Das Department of Mechanical Engineering Indian Institute of Technology, Kharagpur
Applied Thermodynamics for Marine Systems Prof. P. K. Das Department of Mechanical Engineering Indian Institute of Technology, Kharagpur Lecture  7 Ideal Gas Laws, Different Processes Let us continue
More informationFUNDAMENTALS OF ENGINEERING THERMODYNAMICS
FUNDAMENTALS OF ENGINEERING THERMODYNAMICS System: Quantity of matter (constant mass) or region in space (constant volume) chosen for study. Closed system: Can exchange energy but not mass; mass is constant
More informationSOLUTION MANUAL SI UNIT PROBLEMS CHAPTER 9 SONNTAG BORGNAKKE VAN WYLEN. FUNDAMENTALS of. Thermodynamics. Sixth Edition
SOLUTION MANUAL SI UNIT PROBLEMS CHAPTER 9 SONNTAG BORGNAKKE VAN WYLEN FUNDAMENTALS of Thermodynamics Sixth Edition CONTENT SUBSECTION PROB NO. Correspondence table ConceptStudy Guide Problems 20 Steady
More informationTHE CLAUSIUS INEQUALITY
Part IV Entropy In Part III, we introduced the second law of thermodynamics and applied it to cycles and cyclic devices. In this part, we apply the second law to processes. he first law of thermodynamics
More informationChapter 3: Evaluating Properties
Chapter 3: Evaluating Properties Lava flowing into the Pacific Ocean in Hawaii. Photo courtesy of Mike Benson. Property Relations in Engineering hermodynamics ENGINEERING CONEX o apply the energy balance
More informationThe First Law of Thermodynamics: Closed Systems. Heat Transfer
The First Law of Thermodynamics: Closed Systems The first law of thermodynamics can be simply stated as follows: during an interaction between a system and its surroundings, the amount of energy gained
More information(b) 1. Look up c p for air in Table A.6. c p = 1004 J/kg K 2. Use equation (1) and given and looked up values to find s 2 s 1.
Problem 1 Given: Air cooled where: T 1 = 858K, P 1 = P = 4.5 MPa gage, T = 15 o C = 88K Find: (a) Show process on a Ts diagram (b) Calculate change in specific entropy if air is an ideal gas (c) Evaluate
More informationEsystem = 0 = Ein Eout
AGENDA: I. Introduction to Thermodynamics II. First Law Efficiency III. Second Law Efficiency IV. Property Diagrams and Power Cycles V. Additional Material, Terms, and Variables VI. Practice Problems I.
More informationEsystem = 0 = Ein Eout
AGENDA: I. Introduction to Thermodynamics II. First Law Efficiency III. Second Law Efficiency IV. Property Diagrams and Power Cycles V. Additional Material, Terms, and Variables VI. Practice Problems I.
More informationEntropy and The Second Law of Thermodynamics
The Second Law of Thermodynamics (SL) Entropy and The Second Law of Thermodynamics Explain and manipulate the second law State and illustrate by example the second law of thermodynamics Write both the
More informationClausius Inequality (contd )
Module 7 Entropy Suppose we have an engine that receives from several heat reservoirs and rejects heat to several reservoirs, we still have the equation valid. Clausius Inequality Clausius Inequality (contd
More informationMass and Energy Analysis of Control Volumes
MAE 320Chapter 5 Mass and Energy Analysis of Control Volumes Objectives Develop the conservation of mass principle. Apply the conservation of mass principle to various systems including steady and unsteadyflow
More informationME 201 Thermodynamics
ME 0 Thermodynamics Second Law Practice Problems. Ideally, which fluid can do more work: air at 600 psia and 600 F or steam at 600 psia and 600 F The maximum work a substance can do is given by its availablity.
More informationAn introduction to thermodynamics applied to Organic Rankine Cycles
An introduction to thermodynamics applied to Organic Rankine Cycles By : Sylvain Quoilin PhD Student at the University of Liège November 2008 1 Definition of a few thermodynamic variables 1.1 Main thermodynamics
More information4.1 Introduction. 4.2 Work. Thermodynamics
Thermodynamics 41 Chapter 4 Thermodynamics 4.1 Introduction This chapter focusses on the turbine cycle:thermodynamics and heat engines. The objective is to provide enough understanding of the turbine
More informationExpansion and Compression of a Gas
Physics 6B  Winter 2011 Homework 4 Solutions Expansion and Compression of a Gas In an adiabatic process, there is no heat transferred to or from the system i.e. dq = 0. The first law of thermodynamics
More informationApplied Thermodynamics for Marine Systems Prof. P. K. Das Department of Mechanical Engineering Indian Institute of Technology, Kharagpur
Applied Thermodynamics for Marine Systems Prof. P. K. Das Department of Mechanical Engineering Indian Institute of Technology, Kharagpur Lecture  10 Steam Power Cycle, Steam Nozzle Good afternoon everybody.
More informationAPPLIED THERMODYNAMICS TUTORIAL 1 REVISION OF ISENTROPIC EFFICIENCY ADVANCED STEAM CYCLES
APPLIED THERMODYNAMICS TUTORIAL 1 REVISION OF ISENTROPIC EFFICIENCY ADVANCED STEAM CYCLES INTRODUCTION This tutorial is designed for students wishing to extend their knowledge of thermodynamics to a more
More informationChapter 11. Refrigeration Cycles
Chapter 11 Refrigeration Cycles The vapor compression refrigeration cycle is a common method for transferring heat from a low temperature space to a high temperature space. The figures below show the objectives
More informationUNIT 5 REFRIGERATION SYSTEMS
UNIT REFRIGERATION SYSTEMS Refrigeration Systems Structure. Introduction Objectives. Vapour Compression Systems. Carnot Vapour Compression Systems. Limitations of Carnot Vapour Compression Systems with
More informationThermodynamics  Example Problems Problems and Solutions
Thermodynamics  Example Problems Problems and Solutions 1 Examining a Power Plant Consider a power plant. At point 1 the working gas has a temperature of T = 25 C. The pressure is 1bar and the mass flow
More informationExergy: the quality of energy N. Woudstra
Exergy: the quality of energy N. Woudstra Introduction Characteristic for our society is a massive consumption of goods and energy. Continuation of this way of life in the long term is only possible if
More informationESO 201A/202. End Sem Exam 120 Marks 3 h 19 Nov Roll No. Name Section
ESO 201A/202 End Sem Exam 120 Marks 3 h 19 Nov 2014 Roll No. Name Section Answer Questions 1 and 2 on the Question Paper itself. Answer Question 3, 4 and 5 on the Answer Booklet provided. At the end of
More informationFundamentals of Thermodynamics Applied to Thermal Power Plants
Fundamentals of Thermodynamics Applied to Thermal Power Plants José R. SimõesMoreira Abstract In this chapter it is reviewed the fundamental principles of Thermodynamics aiming at its application to power
More information= T T V V T = V. By using the relation given in the problem, we can write this as: ( P + T ( P/ T)V ) = T
hermodynamics: Examples for chapter 3. 1. Show that C / = 0 for a an ideal gas, b a van der Waals gas and c a gas following P = nr. Assume that the following result nb holds: U = P P Hint: In b and c,
More informationProcess chosen to calculate entropy change for the system
Chapter 6 Exaple 6.33 3.  An insulated tank (V 1.6628 L) is divided into two equal parts by a thin partition. On the left
More informationLesson 10 Vapour Compression Refrigeration Systems
Lesson 0 Vapour Compression Refrigeration Systems Version ME, II Kharagpur he specific objectives of the lesson: his lesson discusses the most commonly used refrigeration system, ie Vapour compression
More informationRefrigeration and Air Conditioning Prof. M. Ramgopal Department of Mechanical Engineering Indian Institute of Technology, Kharagpur
Refrigeration and Air Conditioning Prof. M. Ramgopal Department of Mechanical Engineering Indian Institute of Technology, Kharagpur Lecture No. # 18 Worked Out Examples  I Welcome back in this lecture.
More information Know basic of refrigeration  Able to analyze the efficiency of refrigeration system 
Refrigeration cycle Objectives  Know basic of refrigeration  Able to analyze the efficiency of refrigeration system  contents Ideal VaporCompression Refrigeration Cycle Actual VaporCompression Refrigeration
More informationESO201A: Thermodynamics
ESO201A: Thermodynamics Instructor: Sameer Khandekar First Semester: 2015 2016 Lecture #1: Course File Introduction to Thermodynamics, Importance, Definitions Continuum, System: closed, open and isolated,
More informationENGINEERING COUNCIL CERTIFICATE LEVEL THERMODYNAMIC, FLUID AND PROCESS ENGINEERING C106 TUTORIAL 5 STEAM POWER CYCLES
ENGINEERING COUNCIL CERTIFICATE LEVEL THERMODYNAMIC, FLUID AND PROCESS ENGINEERING C106 TUTORIAL 5 STEAM POWER CYCLES When you have completed this tutorial you should be able to do the following. Explain
More informationUNITIII PROPERTIES OF PURE SUBSTANCE AND STEAM POWER CYCLE
UNITIII PROPERTIES OF PURE SUBSTANCE AND STEAM POWER CYCLE Pure Substance A Pure substance is defined as a homogeneous material, which retains its chemical composition even though there may be a change
More informationThermochemistry. r2 d:\files\courses\111020\99heat&thermorans.doc. Ron Robertson
Thermochemistry r2 d:\files\courses\111020\99heat&thermorans.doc Ron Robertson I. What is Energy? A. Energy is a property of matter that allows work to be done B. Potential and Kinetic Potential energy
More informationc. Applying the first law of thermodynamics from Equation 15.1, we find that c h c h.
Week 11 homework IMPORTANT NOTE ABOUT WEBASSIGN: In the WebAssign versions of these problems, various details have been changed, so that the answers will come out differently. The method to find the solution
More information1.5. Which thermodynamic property is introduced using the zeroth law of thermodynamics?
CHAPTER INTRODUCTION AND BASIC PRINCIPLES. (Tutorial). Determine if the following properties of the system are intensive or extensive properties: Property Intensive Extensive Volume Density Conductivity
More informationThere is no general classification of thermodynamic cycle. The following types will be discussed as those are included in the course curriculum
THERMODYNAMIC CYCLES There is no general classification of thermodynamic cycle. The following types will be discussed as those are included in the course curriculum 1. Gas power cycles a) Carnot cycle
More informationTHE SECOND LAW OF THERMODYNAMICS
1 THE SECOND LAW OF THERMODYNAMICS The FIRST LAW is a statement of the fact that ENERGY (a useful concept) is conserved. It says nothing about the WAY, or even WHETHER one form of energy can be converted
More informationENTROPY AND THE SECOND LAW OF THERMODYNAMICS
Chapter 20: ENTROPY AND THE SECOND LAW OF THERMODYNAMICS 1. In a reversible process the system: A. is always close to equilibrium states B. is close to equilibrium states only at the beginning and end
More information6.5 Simple Vapor Compression Refrigeration System:
6.5 Simple Vapor ompression Refrigeration System: A simple vapor compression refrigeration system consists of the following equipments: i) ompressor ii) ondenser iii) Expansion valve iv) Evaporator. B
More informationSecond Law of Thermodynamics
Thermodynamics T8 Second Law of Thermodynamics Learning Goal: To understand the implications of the second law of thermodynamics. The second law of thermodynamics explains the direction in which the thermodynamic
More informationa) Use the following equation from the lecture notes: = ( 8.314 J K 1 mol 1) ( ) 10 L
hermodynamics: Examples for chapter 4. 1. One mole of nitrogen gas is allowed to expand from 0.5 to 10 L reversible and isothermal process at 300 K. Calculate the change in molar entropy using a the ideal
More informationThermodynamics Answers to Tutorial # 1
Thermodynamics Answers to Tutorial # 1 1. (I) Work done in free expansion is Zero as P ex = 0 (II) Irreversible expansion against constant external pressure w = P ex (V 2 V 1 ) V 2 = nrt P 2 V 1 = nrt
More informationk is change in kinetic energy and E
Energy Balances on Closed Systems A system is closed if mass does not cross the system boundary during the period of time covered by energy balance. Energy balance for a closed system written between two
More informationOUTCOME 4 STEAM AND GAS TURBINE POWER PLANT. TUTORIAL No. 8 STEAM CYCLES
UNIT 61: ENGINEERING THERMODYNAMICS Unit code: D/601/1410 QCF level: 5 Credit value: 15 OUTCOME 4 STEAM AND GAS TURBINE POWER PLANT TUTORIAL No. 8 STEAM CYCLES 4 Understand the operation of steam and gas
More informationTHERMODYNAMICS NOTES  BOOK 2 OF 2
THERMODYNAMICS & FLUIDS (Thermodynamics level 1\Thermo & Fluids Module Thermo Book 2ContentsDecember 07.doc) UFMEQU201 THERMODYNAMICS NOTES  BOOK 2 OF 2 Students must read through these notes and
More informationChapter 17. For the most part, we have limited our consideration so COMPRESSIBLE FLOW. Objectives
Chapter 17 COMPRESSIBLE FLOW For the most part, we have limited our consideration so far to flows for which density variations and thus compressibility effects are negligible. In this chapter we lift this
More informationPROPERTIES OF PURE SUBSTANCES
Thermodynamics: An Engineering Approach Seventh Edition in SI Units Yunus A. Cengel, Michael A. Boles McGrawHill, 2011 Chapter 3 PROPERTIES OF PURE SUBSTANCES Mehmet Kanoglu University of Gaziantep Copyright
More informationGibbs Free Energy and Chemical Potential. NC State University
Chemistry 433 Lecture 14 Gibbs Free Energy and Chemical Potential NC State University The internal energy expressed in terms of its natural variables We can use the combination of the first and second
More informationSupplementary Notes on Entropy and the Second Law of Thermodynamics
ME 4 hermodynamics I Supplementary Notes on Entropy and the Second aw of hermodynamics Reversible Process A reversible process is one which, having taken place, can be reversed without leaving a change
More informationEntropy and the Second Law of Thermodynamics. The Adiabatic Expansion of Gases
Lecture 7 Entropy and the Second Law of Thermodynamics 15/08/07 The Adiabatic Expansion of Gases In an adiabatic process no heat is transferred, Q=0 = C P / C V is assumed to be constant during this process
More informationThermodynamic Systems
Student Academic Learning Services Page 1 of 18 Thermodynamic Systems And Performance Measurement Contents Thermodynamic Systems... 2 Defining the System... 2 The First Law... 2 Thermodynamic Efficiency...
More informationLesson 12 MultiStage Vapour Compression Refrigeration Systems. Version 1 ME, IIT Kharagpur 1
Lesson MultiStage Vapour Compression Refrigeration Systems Version ME, IIT Kharagpur The objectives of this lesson are to: Discuss limitations of single stage vapour compression refrigeration systems
More informationPhys 2101 Gabriela González
Phys 2101 Gabriela González If volume is constant ( isochoric process), W=0, and Tp 1 is constant. pv=nrt If temperature is constant ( isothermal process), ΔE int =0, and pv is constant. If pressure is
More information= Q H Q C Q H Q C Q H Q C. ω = Q C W =
I.D The Second Law The historical development of thermodynamics follows the industrial olution in the 19 th century, and the advent of heat engines. It is interesting to see how such practical considerations
More informationThe Second Law of Thermodynamics
The Second aw of Thermodynamics The second law of thermodynamics asserts that processes occur in a certain direction and that the energy has quality as well as quantity. The first law places no restriction
More informationIntroduction to Thermodynamics
Introduction to Thermodynamics Training Objectives The participant will be introduced to: 1.1 basic concepts and definitions. 1.2 the properties of a pure substance. 1.3 work and heat. 1.4 the fist law
More informationTHE UNIVERSITY OF TRINIDAD & TOBAGO
THE UNIVERSITY OF TRINIDAD & TOBAGO FINAL ASSESSMENT/EXAMINATIONS JANUARY/APRIL 2013 Course Code and Title: Programme: THRM3001 Engineering Thermodynamics II Bachelor of Applied Sciences Date and Time:
More informationCompressor Efficiency Definitions
Compressor Efficiency Definitions K. Ueno, PhD, and R. E. Bye, VAIREX Corporation K. S. Hunter, PhD, University of Colorado May 12th, 2003 Many standard efficiency definitions exist that qualify the mass
More informationChapter 15: Thermodynamics
Chapter 15: Thermodynamics The First Law of Thermodynamics Thermodynamic Processes (isobaric, isochoric, isothermal, adiabatic) Reversible and Irreversible Processes Heat Engines Refrigerators and Heat
More informationBasic Thermodynamics Prof. S. K. Som Department of Mechanical Engineering Indian Institute of Technology, Kharagpur
Basic Thermodynamics Prof. S. K. Som Department of Mechanical Engineering Indian Institute of Technology, Kharagpur Lecture  10 Second Law and Available Energy  I Good morning, I welcome you to this
More informationThe First Law of Thermodynamics
The First Law of Thermodynamics (FL) The First Law of Thermodynamics Explain and manipulate the first law Write the integral and differential forms of the first law Describe the physical meaning of each
More informationIn a cyclic transformation, where the final state of a system is the same as the initial one, U = 0
Chapter 4 Entropy and second law of thermodynamics 4.1 Carnot cycle In a cyclic transformation, where the final state of a system is the same as the initial one, U = 0 since the internal energy U is a
More informationThe Second Law of Thermodynamics
Objectives MAE 320  Chapter 6 The Second Law of Thermodynamics The content and the pictures are from the text book: Çengel, Y. A. and Boles, M. A., Thermodynamics: An Engineering Approach, McGrawHill,
More informationChapter 19 Thermodynamics
19.1 Introduction Chapter 19 Thermodynamics We can express the fundamental laws of the universe which correspond to the two fundamental laws of the mechanical theory of heat in the following simple form.
More informationEngineering Software P.O. Box 2134, Kensington, MD 20891 Phone: (301) 9199670 Web Site:
Engineering Software P.O. Box 2134, Kensington, MD 20891 Phone: (301) 9199670 EMail: info@engineering4e.com Web Site: http://www.engineering4e.com Brayton Cycle (Gas Turbine) for Propulsion Application
More informationwhere V is the velocity of the system relative to the environment.
Exergy Exergy is the theoretical limit for the wor potential that can be obtaed from a source or a system at a given state when teractg with a reference (environment) at a constant condition. A system
More informationEnergy Conservation: Heat Transfer Design Considerations Using Thermodynamic Principles
Energy Conservation: Heat Transfer Design Considerations Using Thermodynamic Principles M. Minnucci, J. Ni, A. Nikolova, L. Theodore Department of Chemical Engineering Manhattan College Abstract Environmental
More informationEngineering Problem Solving as Model Building
Engineering Problem Solving as Model Building Part 1. How professors think about problem solving. Part 2. Mech2 and BrainFull Crisis Part 1 How experts think about problem solving When we solve a problem
More informationC H A P T E R T W O. Fundamentals of Steam Power
35 C H A P T E R T W O Fundamentals of Steam Power 2.1 Introduction Much of the electricity used in the United States is produced in steam power plants. Despite efforts to develop alternative energy converters,
More informationTHERMODYNAMIC PROPERTIES AND CALCULATION. Academic Resource Center
THERMODYNAMIC PROPERTIES AND CALCULATION Academic Resource Center THERMODYNAMIC PROPERTIES A quantity which is either an attribute of an entire system or is a function of position which is continuous and
More informationLaws of Thermodynamics
Laws of Thermodynamics Thermodynamics Thermodynamics is the study of the effects of work, heat, and energy on a system Thermodynamics is only concerned with macroscopic (largescale) changes and observations
More informationChapter 2 Classical Thermodynamics: The Second Law
Chapter 2 Classical hermodynamics: he Second Law 2.1 Heat engines and refrigerators 2.2 he second law of thermodynamics 2.3 Carnot cycles and Carnot engines 2.4* he thermodynamic temperature scale 2.5
More informationThe first law: transformation of energy into heat and work. Chemical reactions can be used to provide heat and for doing work.
The first law: transformation of energy into heat and work Chemical reactions can be used to provide heat and for doing work. Compare fuel value of different compounds. What drives these reactions to proceed
More informationChapter 18 Temperature, Heat, and the First Law of Thermodynamics. Problems: 8, 11, 13, 17, 21, 27, 29, 37, 39, 41, 47, 51, 57
Chapter 18 Temperature, Heat, and the First Law of Thermodynamics Problems: 8, 11, 13, 17, 21, 27, 29, 37, 39, 41, 47, 51, 57 Thermodynamics study and application of thermal energy temperature quantity
More informationFundamentals of THERMALFLUID SCIENCES
Fundamentals of THERMALFLUID SCIENCES THIRD EDITION YUNUS A. CENGEL ROBERT H. TURNER Department of Mechanical JOHN M. CIMBALA Me Graw Hill Higher Education Boston Burr Ridge, IL Dubuque, IA Madison, Wl
More informationThermodynamics Basics: Enthalpy, Entropy, Mollier Diagram and Steam Tables
Thermodynamics Basics: Enthalpy, Entropy, Mollier Diagram and Steam Tables Course No: M08005 Credit: 8 PDH S. Bobby Rauf, P.E., CEM, MBA Continuing Education and Development, Inc. 9 Greyridge Farm Court
More informationEngineering Software P.O. Box 2134, Kensington, MD 20891 Phone: (301) 9199670 Web Site:
Engineering Software P.O. Box 2134, Kensington, MD 20891 Phone: (301) 9199670 EMail: info@engineering4e.com Web Site: http://www.engineering4e.com Carnot Cycle Analysis Carnot Cycle Analysis by Engineering
More informationREFRIGERATION (& HEAT PUMPS)
REFRIGERATION (& HEAT PUMPS) Refrigeration is the 'artificial' extraction of heat from a substance in order to lower its temperature to below that of its surroundings Primarily, heat is extracted from
More informationChemical Thermodynamics
Chemical Thermodynamics David A. Katz Department of Chemistry Pima Community College Tucson, AZ 85709, USA First Law of Thermodynamics The First Law of Thermodynamics was expressed in the study of thermochemistry.
More informationPhysics 5D  Nov 18, 2013
Physics 5D  Nov 18, 2013 30 Midterm Scores B } Number of Scores 25 20 15 10 5 F D C } A A A + 0 059.9 6064.9 6569.9 7074.9 7579.9 8084.9 Percent Range (%) The two problems with the fewest correct
More informationChapter 6 The first law and reversibility
Chapter 6 The first law and reversibility 6.1 The first law for processes in closed systems We have discussed the properties of equilibrium states and the relationship between the thermodynamic parameters
More informationis the stagnation (or total) pressure, constant along a streamline.
70 Incompressible flow (page 60): Bernoulli s equation (steady, inviscid, incompressible): p 0 is the stagnation (or total) pressure, constant along a streamline. Pressure tapping in a wall parallel to
More informationTransient Mass Transfer
Lecture T1 Transient Mass Transfer Up to now, we have considered either processes applied to closed systems or processes involving steadystate flows. In this lecture we turn our attention to transient
More informationAn analysis of a thermal power plant working on a Rankine cycle: A theoretical investigation
An analysis of a thermal power plant working on a Rankine cycle: A theoretical investigation R K Kapooria Department of Mechanical Engineering, BRCM College of Engineering & Technology, Bahal (Haryana)
More informationProblem Set 1 3.20 MIT Professor Gerbrand Ceder Fall 2003
LEVEL 1 PROBLEMS Problem Set 1 3.0 MIT Professor Gerbrand Ceder Fall 003 Problem 1.1 The internal energy per kg for a certain gas is given by U = 0. 17 T + C where U is in kj/kg, T is in Kelvin, and C
More informationReversible & Irreversible Processes
Reversible & Irreversible Processes Example of a Reversible Process: Cylinder must be pulled or pushed slowly enough (quasistatically) that the system remains in thermal equilibrium (isothermal). Change
More informationAME 50531 Homework Solutions 1 Fall 2011
Homework 1 AME 50531 Homework Solutions 1 Fall 2011 1. CPIG air enters an isentropic nozzle at 1.30 atm and 24 Cwithavelocityof2.5m/s. The nozzle entrance diameter is 120 mm. The air exits the nozzle at
More informationQUESTIONS THERMODYNAMICS PRACTICE PROBLEMS FOR NONTECHNICAL MAJORS. Thermodynamic Properties
QUESTIONS THERMODYNAMICS PRACTICE PROBLEMS FOR NONTECHNICAL MAJORS Thermodynamic Properties 1. If an object has a weight of 10 lbf on the moon, what would the same object weigh on Jupiter? ft ft ft g
More informationProblem Set 4 Solutions
Chemistry 360 Dr Jean M Standard Problem Set 4 Solutions 1 Two moles of an ideal gas are compressed isothermally and reversibly at 98 K from 1 atm to 00 atm Calculate q, w, ΔU, and ΔH For an isothermal
More informationLesson. 11 Vapour Compression Refrigeration Systems: Performance Aspects And Cycle Modifications. Version 1 ME, IIT Kharagpur 1
Lesson Vapour Compression Refrigeration Systems: Performance Aspects And Cycle Modifications Version ME, IIT Kharagpur The objectives of this lecture are to discuss. Performance aspects of SSS cycle and
More informationThe First Law of Thermodynamics
Thermodynamics The First Law of Thermodynamics Thermodynamic Processes (isobaric, isochoric, isothermal, adiabatic) Reversible and Irreversible Processes Heat Engines Refrigerators and Heat Pumps The Carnot
More informationAPPLIED THERMODYNAMICS. TUTORIAL No.3 GAS TURBINE POWER CYCLES. Revise gas expansions in turbines. Study the Joule cycle with friction.
APPLIED HERMODYNAMICS UORIAL No. GAS URBINE POWER CYCLES In this tutorial you will do the following. Revise gas expansions in turbines. Revise the Joule cycle. Study the Joule cycle with friction. Extend
More informationThermodynamics I Spring 1432/1433H (2011/2012H) Saturday, Wednesday 8:00am  10:00am & Monday 8:00am  9:00am MEP 261 Class ZA
Thermodynamics I Spring 1432/1433H (2011/2012H) Saturday, Wednesday 8:00am  10:00am & Monday 8:00am  9:00am MEP 261 Class ZA Dr. Walid A. Aissa Associate Professor, Mech. Engg. Dept. Faculty of Engineering
More informationChapter 5 The Second Law of Thermodynamics
Chapter 5 he Second aw of hermodynamics he second law of thermodynamics states that processes occur in a certain direction, not in just any direction. Physical processes in nature can proceed toward equilibrium
More informationReading. Spontaneity. Monday, January 30 CHEM 102H T. Hughbanks
Thermo Notes #3 Entropy and 2nd Law of Thermodynamics Monday, January 30 CHEM 102H T. Hughbanks Reading You should reading Chapter 7. Some of this material is quite challenging, be sure to read this material
More informationWhen the fluid velocity is zero, called the hydrostatic condition, the pressure variation is due only to the weight of the fluid.
Fluid Statics When the fluid velocity is zero, called the hydrostatic condition, the pressure variation is due only to the weight of the fluid. Consider a small wedge of fluid at rest of size Δx, Δz, Δs
More information