Thermofluids II (ME2401) Module Notes PDF

Module Overview i Thermofluids II (ME2401) SINGAPORE POLYTECHNIC School of Mechanical & Aeronautical Engineering Thermofluids II (ME2401) Thermodynamics Second Law of Thermodynamics Gas Power Cycles Steam Power Cycles Air Compressor Fluid Mechanics Con...

Module Overview i Thermofluids II (ME2401) SINGAPORE POLYTECHNIC School of Mechanical & Aeronautical Engineering Thermofluids II (ME2401) Thermodynamics Second Law of Thermodynamics Gas Power Cycles Steam Power Cycles Air Compressor Fluid Mechanics Conservation of Momentum Conservation of Energy DARE/ DME/ DMRO Version 1.1 (Apr 2022) 2022/23 Sem1 Rev 1.11 WKH/TFII/Overview  Module Overview ii Thermofluids II (ME2401) MODULE OVERVIEW 1. Contents Title Pages Module Overview i - iv Topic 1 : Review of Basic Thermodynamics 1-1 to 1-11 Topic 2 : Second Law of Thermodynamics 2-1 to 2-16 Topic 3 : Gas Power Cycles 3-1 to 3-16 Topic 4 : Steam Power Cycles 4-1 to 4-16 Topic 5 : Air Compressor 5-1 to 5-15 Topic 6 : Fluid Mechanics 6-1 to 6-16 2. Introduction This module is a second year module and is a continuation of Thermofluids I for first year module. It introduces to students the thermodynamics of power cycles, air compress cycle and cover some fundamental of fluid mechanics. 3. Aim To provide students with basic knowledge of gas and steam powered cycles, air compressor cycle, and some fundamental in fluid mechanics. Rev 1.11 WKH/TFII/Overview  Module Overview iii Thermofluids II (ME2401) 4. Teaching Plan Week Lecture Tutorial Tutorial/Laboratory 1 Topic 1:Review of Basic Topic 1 Thermofluids 2 Topic 1 3 Topic 2:The Second Topic 2 Heat Engine Efficiency 4 Law of Topic 2 Thermodynamics and Entropy 5 Topic 2 6 Topic 3: Gas Power Topic 3 Steam Plant Cycles 7 MST 8 ~ 10 3 weeks Semester Break 11 Topic 3 12 Topic 4: Steam Power Air Compressor 13 Cycles Topic 4 14 Topic 5: Air Topic 4 15 Compressors Topic 5 Bernoulli’s Theorem 16 Topic 6: Fluid Topic 5 17 Mechanics Topic 6 18 Revision Topic 6 19 Exam 20 Rev 1.11 WKH/TFII/Overview  Module Overview iv Thermofluids II (ME2401) 5. Assessment Component Weightage Format Term Test (1 hour )[Wk 7] Answer ALL 3 questions Mid Term Test 15 % Total Marks : 100 Pre-Tutorial Submissions of Tutorial Assignment for Chapter 1, 2, 15 % Assignment 3, 4, 5 & 6 Laboratory Reports Laboratory 15 % Semester Examination (2 hours): Semestral Exam 55 % 4 Long questions, answer all 4 @ 25 marks each. Total Marks : 100 Total 100 % 6. Recommended Reads 1) T.D. Eastop and A. McConkey, Applied Thermodynamics for Engineering Technologists, (5th Ed) Longman 2) Rayner Joel, Basic Engineering Thermodynamics, (5th Ed) Longman. Rev 1.11 WKH/TFII/Overview Module Overview v Thermofluids II (ME2401) Blank Page Rev 1.11 WKH/TFII/Overview  Topic 1: Review of Basic Thermofluids 1-1 Thermofluids II (ME2401) 1 Review of Basic Thermofluids This is a review of the basic concepts taught in the module, Thermofluids I. You are advised to have a copy of that notes with you. 1.1 Thermodynamic System A system may be defined as a quantity of matter enclosed by a boundary for the purpose of investigation. Boundary Surroundings System Fig 1-1: System, boundary and surroundings The matter in the system is the working fluid e.g. air in compressors, water in pumps or steam in boilers and turbines. The boundary may be real e.g. the walls of a vessel (fixed volume) or the walls of an engine cylinder (variable volume). The boundary may also be imaginary e.g an imaginary boundary enveloping the whole steam plant. Energy in the form of heat (𝑄𝑄) or/and work (𝑊𝑊) may cross the boundary. The space across the boundary and surrounding the system is known as the surroundings. 1.1.1 Types of Thermodynamic systems Closed system Boundary 𝑾𝑾 Surroundings 𝑸𝑸 Fig 1-2: Example of a closed system, a piston Rev 1.11 CYF/MES/Review TF1 Topic 1: Review of Basic Thermofluids 1-2 Thermofluids II (ME2401) Open system steam in Boundary 𝑾𝑾 Surroundings Turbine steam out Fig 1-3: Example of an open system, a steam turbine 1.1.2 Sign Conventions Work input is -ve Heat rejected is -ve System Heat supplied is +ve Work output is +ve Fig 1-4: Sign conventions of heat and work of a system Rev 1.11 CYF/MES/Review TF1 Topic 1: Review of Basic Thermofluids 1-3 Thermofluids II (ME2401) 1.1.3 Thermodynamic properties The table below shows the commonly used are thermodynamic properties: Property of the fluid Units Remarks Pressure (𝒑𝒑) 𝑏𝑏𝑏𝑏𝑏𝑏 1 𝑏𝑏𝑏𝑏𝑏𝑏 = 105 𝑁𝑁/𝑚𝑚2 = 100 𝑘𝑘𝑘𝑘/𝑚𝑚2 𝑁𝑁/𝑚𝑚2 Intensive property Volume (𝑽𝑽) 𝑚𝑚3 1𝑙𝑙 = 10−3 𝑚𝑚3 Extensive property Temperature (𝑻𝑻) 𝐾𝐾 (kelvin) 𝐾𝐾 = 273 + ℃ Intensive property Internal Energy (𝑼𝑼) 𝐽𝐽 (joules) Extensive property Enthalpy (𝑯𝑯) 𝐽𝐽 (joules) 𝐻𝐻 = 𝑈𝑈 + 𝑃𝑃𝑃𝑃 Extensive property Note: Intensive properties: physical properties that does not depend on the size or mass of the system Extensive properties: physical properties that are proportional to the mass of the system Specific quantities Units (all intensive properties) Specific volume (𝒗𝒗) 𝑚𝑚3 /𝑘𝑘𝑘𝑘 Specific internal energy (𝒖𝒖) 𝐽𝐽/𝑘𝑘𝑘𝑘 Specific enthalpy (𝒉𝒉) 𝐽𝐽/𝑘𝑘𝑘𝑘 1.1.4 First Law of Thermodynamics The First Law of Thermodynamics is another statement of the Principle of the Conservation of Energy with particular emphasis on heat and work. The First Law for a closed system or non-flow process Non Flow Energy Equation (NFEE): 𝑄𝑄 − 𝑊𝑊 = 𝑈𝑈2 − 𝑈𝑈1 The First Law for a thermodynamic cycle When a fluid undergoes a series of processes and finally returns to its initial state, it assumes the values of its initial properties. Therefore, 𝑈𝑈2 = 𝑈𝑈1 and hence 𝑄𝑄 = 𝑊𝑊 The First Law for an open system or a steady flow process Steady Flow Energy Equation (SFEE): 1 𝑄𝑄 − 𝑊𝑊 = (𝐻𝐻2 − 𝐻𝐻1 ) + 2 𝑚𝑚 (𝐶𝐶22 − 𝐶𝐶12 ) + 𝑚𝑚𝑚𝑚(𝑍𝑍2 − 𝑍𝑍1 ) 1 Or 𝑞𝑞 − 𝑤𝑤 = (ℎ2 − ℎ1 ) + 2 (𝐶𝐶22 − 𝐶𝐶12 + 𝑔𝑔(𝑍𝑍2 − 𝑍𝑍1 ) per kg mass Rev 1.11 CYF/MES/Review TF1 Topic 1: Review of Basic Thermofluids 1-4 Thermofluids II (ME2401) 1.1.5 The Perfect Gas Equations for the perfect gas: Non-flow processes for the perfect gas 𝑝𝑝𝑝𝑝 𝑝𝑝1 𝑉𝑉1 𝑝𝑝2 𝑉𝑉2 a. = 𝐶𝐶 or = 𝑇𝑇 𝑇𝑇1 𝑇𝑇2 𝑝𝑝1 𝑉𝑉1 𝑝𝑝2 𝑉𝑉2 b. 𝑝𝑝𝑝𝑝 = 𝑚𝑚𝑚𝑚𝑚𝑚 or 𝑚𝑚 = = 𝐶𝐶𝑝𝑝 = 1.005 𝑘𝑘𝑘𝑘/𝑘𝑘𝑘𝑘𝑘𝑘 𝑅𝑅𝑇𝑇1 𝑅𝑅𝑇𝑇2 𝐶𝐶𝑣𝑣 = 0.718 𝑘𝑘𝑘𝑘/𝑘𝑘𝑘𝑘𝑘𝑘 c. 𝐶𝐶𝑝𝑝 − 𝐶𝐶𝑣𝑣 = 𝑅𝑅 𝑅𝑅 = 287 𝐽𝐽/𝑘𝑘𝑘𝑘𝑘𝑘 𝐶𝐶𝑝𝑝 d. = 𝛾𝛾 𝐶𝐶𝑣𝑣 1.1.6 Non-flow processes for the perfect gas p-V diagrams 𝑝𝑝 𝑝𝑝 2 1 1 Expansion 2 Heating Cooling 2 Compression 1 1 2 𝑣𝑣 𝑣𝑣 Constant pressure process Constant volume process 𝑝𝑝 1 2 Expansion 2 Compression 1 𝑣𝑣 Isothermal, adiabatic or polytropic process Fig 1-5: 𝑝𝑝 − 𝑉𝑉 diagrams for various thermodynamic processes Rev 1.11 CYF/MES/Review TF1 Topic 1: Review of Basic Thermofluids 1-5 Thermofluids II (ME2401) Fig 1-6: 𝑝𝑝 − 𝑉𝑉 diagrams for isothermal, polytropic and adiabatic processes Rev 1.11 CYF/MES/Review TF1  Topic 1: Review of Basic Thermofluids 1-6 Thermofluids II (ME2401) 1.1.7 Table of formulae Perfect Gas Processes Gas Constant P Constant V Constant T Polytropic Adiabatic p1V1n = p 2V2n p1V1γ = p 2V2γ p1V1 p V p1V1 p 2V2 T=C = 2 2 = P=C V=C T1 T2 T1 T2 Equation of p1V1 = p 2V2 n −1 γ −1 V1 V2 p1 p 2 T2 p T2 p state = T1 T2 = T1 T2 =( 2) n =( 2) γ T1 p1 T1 p1 V V = ( 1 ) n −1 = ( 1 ) γ −1 V2 V2 ∆U mCv(T2 – T1) mCv(T2 – T1) 0 mCv(T2 – T1) mCv(T2 – T1) V p1V1 − p 2V2 p1V1 − p 2V2 p(V2 − V1 ) p1V1 ln ( 2 ) V1 n −1 γ −1 W mR(T2 − T1 ) 0 V mR(T1 − T2 ) mR(T1 − T2 ) mRT ln ( 2 ) V1 n −1 γ −1 Q mCp(T2 – T1) mCv(T2 – T1) W ∆U + W 0 T2 T2 V2 T2 V ∆S mCpln( ) mCvln( ) mRln( ) mCvln( )+ mRln( 2 ) 0 T1 T1 V1 T1 V1 For air, take Cv = 0.718 kJ/kgK, Cp = 1.005 kJ/kgK, R = 0.287 kJ/kgK, γ = 1.4 1.1.8 Steam Steam Closed System Processes ( per kg mass) Steam Constant P Constant V Constant T Polytropic Adiabatic p1v1 − p 2 v 2 w p(v 2 − v1 ) 0 q-∆u u1 – u2 n −1 q h2 – h1 u2 – u1 T(s2 – s1) w + ∆u 0 Rev 1.11 CYF/MES/Review TF1  Topic 1: Review of Basic Thermofluids 1-7 Thermofluids II (ME2401) 1.1.9 Properties of steam p ( bar or kPa ) critical point saturated liquid line saturated vapour line dry saturated steam superheated steam Sub - cooled water wet steam constant pressure line dryness fraction (x) constant temperature line saturated water x=0 x x=1 v ( m3/kg Fig 1-7: 𝑝𝑝 − 𝑣𝑣 diagrams for Steam Saturated water Wet Saturated steam Dry saturated steam (x = 0) (0 < x < 1) (x = 1) 𝑣𝑣 = 𝑣𝑣𝑓𝑓 ≈ 0.001𝑚𝑚3/𝑘𝑘𝑘𝑘 < 𝑣𝑣 = 𝑥𝑥𝑣𝑣𝑔𝑔 < 𝑣𝑣 = 𝑣𝑣𝑔𝑔 𝑢𝑢 = 𝑢𝑢𝑓𝑓 < 𝑢𝑢 = 𝑢𝑢𝑓𝑓 + 𝑥𝑥(𝑢𝑢𝑔𝑔 − 𝑢𝑢𝑓𝑓 ) < 𝑢𝑢 = 𝑢𝑢𝑔𝑔 ℎ = ℎ𝑓𝑓 < ℎ = ℎ𝑓𝑓 + 𝑥𝑥ℎ𝑓𝑓𝑓𝑓 < ℎ = ℎ𝑔𝑔 𝑠𝑠 = 𝑠𝑠𝑓𝑓 < 𝑠𝑠 = 𝑠𝑠𝑓𝑓 + 𝑥𝑥𝑠𝑠𝑓𝑓𝑓𝑓 < 𝑠𝑠 = 𝑠𝑠𝑔𝑔 where x is the dryness fraction of the wet steam Let 𝑦𝑦 be ℎ, 𝑢𝑢 𝑜𝑜𝑜𝑜 𝑠𝑠 𝑦𝑦𝑓𝑓𝑓𝑓 = 𝑦𝑦𝑔𝑔 − 𝑦𝑦𝑓𝑓 𝑦𝑦 = 𝑥𝑥 𝑦𝑦𝑔𝑔 + (1 − 𝑥𝑥)(𝑦𝑦𝑓𝑓 ) 𝑣𝑣 = 𝑥𝑥𝑣𝑣𝑔𝑔 Rev 1.11 CYF/MES/Review TF1  Topic 1: Review of Basic Thermofluids 1-8 Thermofluids II (ME2401) Equation for interpolation For properties which are not given exactly in the steam tables, the following equation may be used to interpolate between the tabulated values. 𝑥𝑥−𝑥𝑥1 𝑦𝑦−𝑦𝑦1 = where 𝑦𝑦 is the unknown value 𝑥𝑥2 −𝑥𝑥1 𝑦𝑦2 −𝑦𝑦1 Boundary of saturated steam 𝑥𝑥 will determine how much liquid and gas in mixture T Fully gas 𝑓𝑓 𝑔𝑔 Water Steam (Superheat) Fully liquid Wet Steam (Saturated) s Fig 1-8: 𝑇𝑇 − 𝑠𝑠 diagrams for steam Rev 1.11 CYF/MES/Review TF1  Topic 1: Review of Basic Thermofluids 1-9 Thermofluids II (ME2401) 1.1.10 Fluid Mechanics A fluid is defined as a substance which cannot resist a shear stress without moving. 1.1.11 Properties of fluids Density (ρ) = mass/volume, kg/m3 Specific volume (v) = 1/ρ, m /kg 3 Relative density (RD) or Specific Gravity (SG) RD (or SG) = density of substance/density of water Viscosity of a fluid is a measure of its resistance to flow. Ideal fluids have no viscosity and are said to be inviscid. Real fluids have viscosity and are said to be viscous. Pressure (static pressure implied) (p) = force/area, N/m2 or bar Atmospheric pressure (patm) = ρgh. This is measured by a Barometer. Gauge pressure (pg) is the pressure in a enclosed vessel and is measured by a pressure gauge such as the Bourdon gauge or a manometer. Absolute pressure (pabs) = patm + pg Static pressure (p) is the pressure acting at a point in a fluid Pressure Head (h) = p/ρg, m. This is the height of the column of fluid producing the pressure. volume,V Volume Flow Rate (Volumetric Flow Rate) (Q) = , m3/s or l/min. time, t Average velocity (U) = Q/A, m/s : A is the X-sectional area of the flow in m2. Mass flow rate ( m ) = ρQ, kg/s 1.1.12 The Continuity Equation The mass flow rate of a fluid entering a controlled volume is equal to the mass flow rate leaving it, i.e. m 1 = m 2 or ρ1Q1 = ρ2Q2 For incompressible fluids where ρ = constant, Then Q1 = Q2 or U1A1 = U2A2 Rev 1.11 CYF/MES/Review TF1 Topic 1: Review of Basic Thermofluids 1-10 Thermofluids II (ME2401)  Challenge! Using the continuity equation determine the value of g, the acceleration due to gravity. Refer to the figure on the right. Data z1-z2 = 40 cm; d1 = 2cm, d2 = 0.5cm. A 250ml cup collected the water in 4.5 secs. What are your assumptions? z1-z2 Prove that the error is 1.8%.  Rev 1.11 CYF/MES/Review TF1  Topic 1: Review of Basic Thermofluids 1-11 Thermofluids II (ME2401) Tutorial 1 on Review of Basic Thermofluids 1. A rigid vessel contains 0.1 𝑚𝑚3 of air at 5 𝑏𝑏𝑏𝑏𝑏𝑏 and 200℃. If the vessel is heated to 250℃, calculate the final pressure and the heat supplied. (5.53 bar, 13.22 kJ) 2. 0.2 𝑘𝑘𝑘𝑘 of air at 7 𝑏𝑏𝑏𝑏𝑏𝑏 and 150℃ is expanded at constant pressure until its volume is doubled. Find the final temperature, work done and heat transferred. (846 K, 24.28 kJ, 85.02 kJ) 3. 1 𝑘𝑘𝑘𝑘 of steam at a pressure of 10 𝑏𝑏𝑏𝑏𝑏𝑏 and a temperature of 250℃ is cooled at constant volume to 4 𝑏𝑏𝑎𝑎𝑎𝑎. Determine the final condition of the steam and the heat transferred. (0.5037, -1124.3 kJ) 4. 1 𝑘𝑘𝑘𝑘 of air at 1.2 𝑏𝑏𝑏𝑏𝑏𝑏 and 30℃ in a cylinder is compressed according to the law, 𝑝𝑝𝑝𝑝 1.25 = 𝐶𝐶. to a pressure of 6 𝑏𝑏𝑏𝑏𝑏𝑏. Calculate the final temperature, work done and heat transferred. (418.06 K, -132.09 kJ, -49.48 kJ) 5. Steam at 9 𝑏𝑏𝑏𝑏𝑏𝑏 and 0.8 dryness fraction expands at constant pressure until the temperature is 230℃. Calculate the work done and the heat supplied per 𝑘𝑘𝑘𝑘 of the steam. (68.5 kJ/kg, 534 kJ/kg) Rev 1.11 CYF/MES/Review TF1 Topic 1: Review of Basic Thermofluids 1-12 Thermofluids II (ME2401) Blank Page Rev 1.11 CYF/MES/Review TF1  Topic 2. The Second law of Thermodynamics 2-1 Thermofluids II (ME2401) 2 The Second Law of Thermodynamics Keywords: Second Law, Hot reservoir, Cold reservoir, Reversible process, Irreversible process, Entropy, Change of entropy, Isentropic process. Objective: Students should be able to understand the concept of the Second Law and its implications. They should also be able to understand the concept of Entropy and calculate the change of entropy for a particular process. They should be able to: 1. State the Second Law of Thermodynamics. 2. State the implications of the Second Law. 3. Define ‘Reversible process’ and ‘Irreversible process.’ 4. Explain the concept of Entropy. 5. Determine the change of entropy for various processes. 6. Sketch the temperature-entropy diagrams for various processes. 2.1 Statements of the Law The First Law of Thermodynamics is concerned with heat and work as forms of energy, which may be converted from one to the other. It places no limits on the amount or direction of the conversion. The Second Law of Thermodynamics indicates that the direction of an energy transfer has limits. It enables us to decide whether particular processes or cycles are possible. There are two statements for the Second Law of Thermodynamics: Kelvin Planck Statement Clausius Statement Rev 1.11 CYF/MES/Second Law Topic 2. The Second law of Thermodynamics 2-2 Thermofluids II (ME2401) Kelvin Planck Statement “It is impossible to construct a device which will operate in a cycle, extract heat from a reservoir and do an equivalent amount of work on the surroundings.” Fig 2-1: Kelvin Planck example of an impossible device The Kelvin Planck statement essentially means that not all the heat energy may be converted into work or, alternatively, that a heat engine requires two reservoirs of different temperatures before it will operate. Fig 2-2: Kelvin Planck example of a realistic device Clausius Statement Fig 2-3: Clausius example of an impossible device Rev 1.11 CYF/MES/Second Law  Topic 2. The Second law of Thermodynamics 2-3 Thermofluids II (ME2401) “It is impossible to construct a device that, operating in a cycle, will produce no effect other than the transfer of heat from a cooler body to a hotter body.” This statement is easily verified by the experience of natural processes. Heat is never observed to flow from a cooler body to a hotter body. In a refrigerator, an input of energy is required (at the compressor) to extract heat from the cold chamber and reject it to higher temperature surroundings. Fig 2-4: Clausius example of a realistic device The two statements of the Second Law are equivalent and the violation of one implies a violation of the other. 2.2 Reversibility and Irreversibility 2.2.1 Reversible Process: A reversible process is one whereby, if the process is reversed, both the system and the surroundings comes back to the initial conditions. 2.2.2 Irreversible Process: An irreversible process is one whereby, if the process is reversed, either the system or the surroundings comes back to the initial conditions but not both. If the system comes back to the initial conditions then the system is said to be internally reversible but externally irreversible. If the surroundings comes back to the initial conditions then the system is said to be externally reversible but internally irreversible. Rev 1.11 CYF/MES/Second Law  Topic 2. The Second law of Thermodynamics 2-4 Thermofluids II (ME2401) 2.2.3 Criteria for reversibility: a) The process must be frictionless. The fluid must have no internal friction. In reality friction is always present when particles in a fluid move one relative to the other. b) The difference in pressure between the fluid and its surroundings during the process must be infinitely small. This means that the process must take place infinitely slowly. c) The difference in temperature between the fluid and its surroundings during the process must be infinitely small. This means that the heat supplied or rejected to or from the fluid must be transferred infinitely slowly. It is obvious from the above that no process is truly reversible. However, in many practical processes a very close approximation to internal reversibility can be obtained. 2.3 Entropy (S) (kJ/kgK) Entropy is a property of a system which arises as a result of the Second Law, in much the same way as the internal energy arises from the First law. It is the measure of the randomness or disorder in a system. The change in entropy can be found from the knowledge of a quantity of heat transferred only during a reversible non-flow process. In Thermodynamics, entropy is defined as “A property whereby the area under the process line as drawn on a Temperature-Entropy (T-S) diagram represents the heat transferred during that process.” The Temperature-Entropy (T-S) diagram of a process from 1 to 2 is shown in Fig T 2 T2 Q 1 T 1 S S ∆S S2 1 Fig 2-5: T-S diagram of a process 1-2 Rev 1.11 CYF/MES/Second Law Topic 2. The Second law of Thermodynamics 2-5 Thermofluids II (ME2401) It is not possible to establish the absolute magnitude of entropy. However, as with internal energy, only changes of entropy are of interest, and it is by this difference that entropy is defined. The SI unit for specific entropy is J/kg K. In steam table, the unit kJ/kg K is commonly used. For all the process in Thermodynamics we will now examine the changes in entropy that occurs. Next, these same processes will be examined for vapours through worked examples using steam as the working fluid. Fig 2-6 Rev 1.11 CYF/MES/Second Law  Topic 2. The Second law of Thermodynamics 2-6 Thermofluids II (ME2401) 2.4 Change in Entropy for the Perfect Gas For any reversible process, the change in entropy is given by: 𝑑𝑑𝑑𝑑 𝑆𝑆2 − 𝑆𝑆1 = 𝑇𝑇 For gases going through non-flow processes. A. Constant Volume Process For a constant volume process: Work done, 𝑊𝑊 = 0. Therefore from the 1st Law: 𝑄𝑄 = (𝑈𝑈2 − 𝑈𝑈1 ) = 𝑚𝑚𝐶𝐶𝑣𝑣 (𝑇𝑇2 − 𝑇𝑇1 ) Change of entropy, 𝑑𝑑𝑑𝑑 for constant volume 𝑆𝑆2 − 𝑆𝑆1 = 𝑘𝑘𝑘𝑘/𝐾𝐾 𝑇𝑇 p1 p 𝑑𝑑𝑑𝑑 = 2 = 𝑚𝑚𝐶𝐶𝑣𝑣 T1 T2 𝑇𝑇 𝑇𝑇2 T2 p = 𝑚𝑚𝐶𝐶𝑣𝑣 ∙ 𝐿𝐿𝐿𝐿𝐿𝐿𝑒𝑒 = 2 𝑇𝑇1 T1 p1 𝑃𝑃2 = 𝑚𝑚𝐶𝐶𝑣𝑣 ∙ 𝐿𝐿𝐿𝐿𝐿𝐿𝑒𝑒 𝑃𝑃1 V =V p V =V p 1 2 1 2 1 T 2 ∆S ∆S T 2 p p 1 T 1 T 2 2 1 1 2 T T 1 2 S S S S S S 1 2 2 1 (a) Pressure increased (b) Pressure reduced Fig 2-7: T-S diagram for constant volume process Rev 1.11 CYF/MES/Second Law  Topic 2. The Second law of Thermodynamics 2-7 Thermofluids II (ME2401) B. Constant Pressure Process For a constant pressure process: 𝑄𝑄 = (𝐻𝐻2 − 𝐻𝐻1 ) = 𝑚𝑚𝐶𝐶𝑝𝑝 (𝑇𝑇2 − 𝑇𝑇1 ) Change of entropy, 𝑑𝑑𝑑𝑑 𝑆𝑆2 − 𝑆𝑆1 = 𝑘𝑘𝑘𝑘/𝐾𝐾 𝑇𝑇 for constant pressure 𝑑𝑑𝑑𝑑 = 𝑚𝑚𝐶𝐶𝑝𝑝 V1 V 𝑇𝑇 = 2 𝑇𝑇2 T1 T2 = 𝑚𝑚𝐶𝐶𝑝𝑝 ∙ 𝐿𝐿𝐿𝐿𝐿𝐿𝑒𝑒 𝑇𝑇1 V1 T 𝑉𝑉2 = 1 = 𝑚𝑚𝐶𝐶𝑝𝑝 ∙ 𝐿𝐿𝐿𝐿𝐿𝐿𝑒𝑒 𝑉𝑉1 V2 T2 V1 V2 V2 V1 T T 2 p1=p2 1 p1=p2 T2 T1 1 2 T1 T2 S S S1 ∆S S2 ∆S S2 S1 (a) Expansion (b) Contraction Fig 2-8: T-S diagram for constant pressure process Rev 1.11 CYF/MES/Second Law  Topic 2. The Second law of Thermodynamics 2-8 Thermofluids II (ME2401) C. Constant Temperature (Isothermal) Process For a constant temperature process, (𝑈𝑈2 − 𝑈𝑈1 ) = 0 Therefore, 𝑄𝑄 = 𝑊𝑊 Change of entropy, 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑆𝑆2 − 𝑆𝑆1 = = 𝑘𝑘𝑘𝑘/𝐾𝐾 𝑇𝑇 𝑇𝑇 𝑃𝑃𝑃𝑃𝑃𝑃 = 𝑇𝑇 for constant 𝑑𝑑𝑑𝑑 = 𝑚𝑚𝑚𝑚 temperature 𝑉𝑉 𝑉𝑉2 = 𝑚𝑚𝑚𝑚 ∙ 𝐿𝐿𝐿𝐿𝐿𝐿𝑒𝑒 p1 V 𝑉𝑉1 = 2 𝑃𝑃1 p2 V1 = 𝑚𝑚𝑚𝑚 ∙ 𝐿𝐿𝐿𝐿𝐿𝐿𝑒𝑒 𝑃𝑃2 V1 V2 V2 p2 T T V1 p1 p2 p1 1 2 2 1 T1 T1 S S S1 ∆S S2 ∆S S2 S1 (a) Expansion (b) Compression Fig 2-9: T-S diagram for constant temperature process Rev 1.11 CYF/MES/Second Law Topic 2. The Second law of Thermodynamics 2-9 Thermofluids II (ME2401) D. Polytropic Process (𝑷𝑷𝑽𝑽𝒏𝒏 = 𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪) From the First Law: 𝑄𝑄 = (𝑈𝑈2 − 𝑈𝑈1 ) + 𝑊𝑊 Therefore, 𝑄𝑄 = (𝑈𝑈2 − 𝑈𝑈1 ) + 𝑊𝑊 Change in entropy, 𝑑𝑑𝑑𝑑 𝑆𝑆2 − 𝑆𝑆1 = 𝑘𝑘𝑘𝑘/𝐾𝐾 𝑇𝑇 𝑑𝑑𝑑𝑑 𝑃𝑃𝑃𝑃𝑃𝑃 = 𝑚𝑚𝐶𝐶𝑣𝑣 + 𝑇𝑇 𝑇𝑇 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 = 𝑚𝑚𝐶𝐶𝑣𝑣 + 𝑚𝑚𝑚𝑚 𝑇𝑇 𝑉𝑉 𝑇𝑇2 𝑉𝑉2 = 𝑚𝑚𝐶𝐶𝑣𝑣 ∙ 𝐿𝐿𝐿𝐿𝐿𝐿𝑒𝑒 + 𝑚𝑚𝑚𝑚 ∙ 𝐿𝐿𝐿𝐿𝐿𝐿𝑒𝑒 𝑇𝑇1 𝑉𝑉1 p2 V1 p1 V2 T T 1 V2 2 V1 T1 p2 T2 p1 n n pV = C pV = C T2 T1 1 2 S S ∆S ∆S S1 S2 S1 S2 (a) Expansion (b) Compression Fig 2-10: T-S diagram for polytropic process Rev 1.11 CYF/MES/Second Law  Topic 2. The Second law of Thermodynamics 2-10 Thermofluids II (ME2401) E. Reversible Adiabatic (Isentropic) Process ( 𝑷𝑷𝑽𝑽𝜸𝜸 = 𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪 ) For a reversible adiabatic process, 𝑄𝑄 = 0. 𝑑𝑑𝑑𝑑 Change of entropy, 𝑆𝑆2 − 𝑆𝑆1 = ∫ 𝑇𝑇 𝑘𝑘𝑘𝑘/𝐾𝐾 =0 A process which occurs such that the change of entropy is zero is called an isentropic process or a process of constant entropy. A reversible adiabatic process is therefore an isentropic process. However, an irreversible adiabatic process is not isentropic. V2 V1 p1 p2 T T 1 2 V1 T1 V2 T2 pVγ = C pVγ = C p2 p1 T2 T1 2 1 S S S1 = S2 S1 = S2 ∆S = 0 ∆S = 0 (a) Expansion (b) Compression Fig 2-11: T-S diagram for Isentropic process 2.5 Change in Entropy for Vapour For steam, values of entropy are tabulated in the steam tables. To find changes in entropy, find/calculate the initial and final values of the entropy and subtract them. The following are simple worked examples showing the use of the steam tables for each of the processes. Rev 1.11 CYF/MES/Second Law Topic 2. The Second law of Thermodynamics 2-11 Thermofluids II (ME2401) A. Constant Volume Process Steam at 10 𝑏𝑏𝑏𝑏𝑏𝑏 absolute and 0.9 dry is expanded reversibly to 5 𝑏𝑏𝑏𝑏𝑏𝑏 absolute at constant volume. Calculate the change in entropy and sketch the process on a T-s diagram. Rev 1.11 CYF/MES/Second Law Topic 2. The Second law of Thermodynamics 2-12 Thermofluids II (ME2401) B. Constant Pressure Process Steam at 10 𝑏𝑏𝑎𝑎𝑟𝑟 absolute and 0.9 dry is heated at constant pressure to a temperature of 350℃. Calculate the change in entropy and sketch the process on a T-s diagram. Rev 1.11 CYF/MES/Second Law Topic 2. The Second law of Thermodynamics 2-13 Thermofluids II (ME2401) C. Constant Temperature Process. Steam at 10 𝑏𝑏𝑎𝑎𝑟𝑟 absolute and 250℃ is expanded at constant temperature to a pressure of 5 𝑏𝑏𝑎𝑎𝑟𝑟 absolute. Calculate the change in entropy and sketch the process on a T-s diagram. Rev 1.11 CYF/MES/Second Law Topic 2. The Second law of Thermodynamics 2-14 Thermofluids II (ME2401) D. Polytropic Process ( PV n = Constant ) Steam at 10 𝑏𝑏𝑎𝑎𝑟𝑟 absolute and 0.9 dry is expanded to 5 𝑏𝑏𝑏𝑏𝑏𝑏 absolute according to the law 𝑃𝑃𝑉𝑉 1.1 = 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝑎𝑎𝑛𝑛𝑛𝑛. Calculate the change in entropy and sketch the process on a T- s diagram. Rev 1.11 CYF/MES/Second Law Topic 2. The Second law of Thermodynamics 2-15 Thermofluids II (ME2401) E. Adiabatic Process Steam at 10 𝑏𝑏𝑎𝑎𝑟𝑟 absolute and 0.9 dry is expanded to 5 𝑏𝑏𝑏𝑏𝑏𝑏 during a reversible adiabatic process. Determine the change in entropy and the final dryness fraction. Sketch the process on a T-s diagram. Rev 1.11 CYF/MES/Second Law Topic 2. The Second law of Thermodynamics 2-16 Thermofluids II (ME2401) Tutorial 2 on Change of Entropy 1) State the Second Law of Thermodynamics. 2) A rigid vessel of 0.2 𝑚𝑚3 contains air at 5 𝑏𝑏𝑏𝑏𝑏𝑏 and 300℃. If the vessel is cooled to 3 𝑏𝑏𝑏𝑏𝑏𝑏, determine the heat transferred and change of entropy. (-100.1kJ, -0.223kJ/K) 3) 0.2 kg of air at 7 𝑏𝑏𝑏𝑏𝑏𝑏 and 150℃ expands at constant pressure until the volume is doubled. Calculate the work done, heat transferred and change of entropy. (24.28kJ, 85.02kJ, 0.1393kJ/K) 4) 0. 1 𝑚𝑚3 of air at a pressure of 2 𝑏𝑏𝑏𝑏𝑏𝑏 and a temperature of 120℃ is compressed isothermally to a pressure of 4 𝑏𝑏𝑏𝑏𝑏𝑏. Find the change of entropy, heat transferred and work done. (-0.0353 kJ/K, -13.83 kJ, -13.83 kJ) 5) 1 kg of air at a pressure of 10 𝑏𝑏𝑏𝑏𝑏𝑏 and a temperature of 250℃ is expanded to 1 𝑏𝑏𝑏𝑏𝑏𝑏 according to the law, 𝑝𝑝𝑣𝑣 1.25 = 𝐶𝐶. Determine the work done, heat transferred and change of entropy. (221.56kJ, 82.99kJ, 0.198kJ/K) 6) Determine the following: a) The specific entropy of water at 100℃. b) Determine the specific entropy of steam at: i. 20 𝑏𝑏𝑎𝑎𝑟𝑟 and 0.8 dry ii. 15 𝑏𝑏𝑎𝑎𝑟𝑟 and 300℃ iii. 15 𝑏𝑏𝑎𝑎𝑟𝑟 and 320℃ iv. 12 𝑏𝑏𝑎𝑎𝑟𝑟 and 320℃. (1.307 J/kgK, 5.561 kJ/kgK, 6.919 kJ/kgK, 6.992kJ/kgK, 7.114kJ/kgK) 7) Steam at 10 𝑏𝑏𝑏𝑏𝑏𝑏 and 300℃ is cooled at constant volume to 6 𝑏𝑏𝑏𝑏𝑏𝑏. Determine the heat transferred and change of entropy per kg of the steam. (-572.6kJ/kg, -1.244kJ/kgK) 8) A quantity of steam with a dryness fraction of 0.7 at 5 𝑏𝑏𝑏𝑏𝑏𝑏 undergoes a constant pressure process until its temperature becomes 250℃. Find the work done, heat transferred and change of entropy per kg of the steam. (106.1kJ/kg, 845.7kJ/kg, 1.938kJ/kgK) Rev 1.11 CYF/MES/Second Law Topic 2. The Second law of Thermodynamics 2-17 Thermofluids II (ME2401) Blank Page Rev 1.11 CYF/MES/Second Law  Topic 3. Gas Power Cycles 3-1 Thermofluids II (ME2401) 3 Gas Power Cycles Keywords: Carnot cycle, Air standard cycle, Otto cycle, compression ratio, stroke, swept volume, clearance volume, Thermal efficiency, Diesel cycle, cut-off ratio, Dual combustion cycle, heat addition pressure ratio. Objectives: Students should be able to understand the concept of the Air standard Cycles and their applications. They should be able to: 1. State the assumptions for the Air standard cycles. 2. List in order the processes that make up the Carnot, Otto, Diesel and Dual combustion cycles. 3. Sketch the P-v and T-s diagrams for these cycles. 4. State the applications of these cycles. 5. Solve problems involving these cycles. 6. State the limitations of the Carnot cycle. 3.1 The Carnot Cycle The most efficient heat engine conceivable is one in which all the heat supplied is supplied at one fixed temperature, and all the heat rejected is rejected at a lower fixed temperature. This cycle, known as the Carnot Cycle, consists of two reversible isothermal processes joined by two reversible adiabatic processes. The boundary can be fixed or movable depending on the required conditions. An example of a fixed boundary is the enclosure of a room in which an air-conditioner is to be installed. An example of a movable boundary is an engine cylinder with a movable piston. Rev 1.11 CYF/MES/Gas Cycles Topic 3. Gas Power Cycles 3-2 Thermofluids II (ME2401) 𝑇𝑇 𝑄𝑄𝑠𝑠 4 1 𝑇𝑇1 𝑇𝑇2 2 3 𝑄𝑄𝑟𝑟 𝑆𝑆 Fig 3-1: T-s diagram of a Carnot cycle Process 1 2: Isentropic expansion 2 3: Isothermal heat rejection 3 4: Isentropic compression 4 1: Isothermal heat supply Thermal efficiency of the Carnot cycle Thermal efficiency of any cycle, 𝑁𝑁𝑁𝑁𝑁𝑁 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝜂𝜂 = 𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑄𝑄𝑟𝑟 = 1− 𝑄𝑄𝑠𝑠 𝑇𝑇2 (𝑆𝑆2 − 𝑆𝑆3 ) =1− 𝑇𝑇1 (𝑆𝑆1 − 𝑆𝑆4 ) Since for Carnot 𝑆𝑆1 − 𝑆𝑆4 = 𝑆𝑆2 − 𝑆𝑆3 , the Carnot efficiency will be: 𝑇𝑇2 𝜂𝜂 = 1 − 𝑇𝑇1 Rev 1.11 CYF/MES/Gas Cycles  Topic 3. Gas Power Cycles 3-3 Thermofluids II (ME2401) 3.2 Limitations of the Carnot cycle In practice, it is difficult for a system to receive and reject heat at constant temperature. Hence, it would not be possible to operate a heat engine on the Carnot cycle using a gas as working substance. It is more practical to heat a gas at approximately constant pressure or at constant volume. Furthermore, the Carnot cycle has a low work ratio. For these reasons, the Carnot cycle is not employed in a power plant. 3.3 Air Standard Power Cycles These are ideal gas cycles where the working substance is assumed to be air. There are three important air standard cycles which form the basis of all practical reciprocating engines: 1. Otto cycle - used to approximate the Petrol engine cycle 2. Diesel cycle - used to approximate the low speed Diesel engine cycle 3. Dual Combustion cycle - used to approximate the high speed diesel engine cycle The air standard cycles represented on the p-v diagrams discussed here are based on some assumptions and therefore differ from the indicator diagrams taken from the actual engines. The assumptions are: 1. The cycle is a complete thermodynamic cycle 2. The working substance is air 3. The air is an ideal gas with a constant specific heat capacity 4. All processes in the cycle are reversible Rev 1.11 CYF/MES/Gas Cycles  Topic 3. Gas Power Cycles 3-4 Thermofluids II (ME2401) 3.4 The Otto Cycle 𝑝𝑝 𝑇𝑇 3 3 𝑸𝑸𝒔𝒔 𝑠𝑠 = 𝑐𝑐 𝑣𝑣 = 𝑐𝑐 𝑠𝑠 = 𝑐𝑐 2 𝑣𝑣 = 𝑐𝑐 4 4 2 𝑣𝑣 = 𝑐𝑐 𝑠𝑠 = 𝑐𝑐 𝑸𝑸𝒓𝒓 𝑠𝑠 = 𝑐𝑐 1 1 𝑣𝑣 = 𝑐𝑐 𝑣𝑣 𝑠𝑠 𝑣𝑣𝑐𝑐 𝑣𝑣𝑠𝑠 Fig 3-2: p-V and T-s diagram of an Otto cycle Process 12: Isentropic compression 𝑑𝑑2 Swept volume, 𝑣𝑣𝑠𝑠 = 𝜋𝜋 𝑙𝑙 4 where 𝑑𝑑 is the diameter and 𝑙𝑙 is the stroke of the engine cylinder. The volume compression ratio, 𝑣𝑣1 𝑟𝑟𝑣𝑣 = 𝑣𝑣2 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 + 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 𝑣𝑣𝑠𝑠 + 𝑣𝑣𝑐𝑐 = 𝑣𝑣𝑐𝑐 Also, γ 𝛾𝛾 𝑷𝑷𝟐𝟐 𝑣𝑣2 = 𝑃𝑃1 𝑣𝑣1 𝑻𝑻𝟐𝟐 𝑣𝑣1 𝛾𝛾−1 𝛾𝛾−1 = = 𝑟𝑟𝑣𝑣 𝑇𝑇1 𝑣𝑣2 Rev 1.11 CYF/MES/Gas Cycles Topic 3. Gas Power Cycles 3-5 Thermofluids II (ME2401) Process 23: Constant volume heat supply 𝒗𝒗𝟑𝟑 = 𝑣𝑣2 𝑃𝑃3 𝑃𝑃2 = 𝑻𝑻𝟑𝟑 𝑇𝑇2 Heat supplied, 𝑄𝑄𝑠𝑠 = m 𝐶𝐶𝑣𝑣 (𝑇𝑇3 − 𝑇𝑇2 ) Process 34: Isentropic expansion γ 𝛾𝛾 𝑷𝑷𝟒𝟒 𝑣𝑣4 = 𝑃𝑃3 𝑣𝑣3 𝑻𝑻𝟒𝟒 𝑣𝑣3 𝛾𝛾−1 𝑣𝑣2 𝛾𝛾−1 1 𝛾𝛾−1 = = = 𝑇𝑇3 𝑣𝑣4 𝑣𝑣1 𝑟𝑟𝑣𝑣 Process 41: Constant volume heat rejection 𝒗𝒗𝟒𝟒 = 𝑣𝑣1 Heat rejected, 𝑄𝑄𝑟𝑟 = m 𝐶𝐶𝑣𝑣 (𝑇𝑇4 − 𝑇𝑇1 ) Thermal efficiency of the Otto cycle The thermal efficiency of the cycle, 𝑊𝑊𝑛𝑛𝑛𝑛𝑛𝑛 𝜂𝜂 = 𝑄𝑄𝑠𝑠 𝑄𝑄𝑟𝑟 = 1− 𝑄𝑄𝑠𝑠 𝑚𝑚 𝐶𝐶𝑣𝑣 (𝑇𝑇4 − 𝑇𝑇1 ) =1− 𝑚𝑚 𝐶𝐶𝑣𝑣 (𝑇𝑇3 − 𝑇𝑇2 ) 𝛾𝛾−1 From process 12, 𝑇𝑇2 = 𝑇𝑇1 𝑟𝑟𝑣𝑣 𝑇𝑇 𝑣𝑣 𝛾𝛾−1 And process 34, 𝑇𝑇3 = 𝑣𝑣4 or 𝑇𝑇3 = 𝑇𝑇4 𝑟𝑟𝑣𝑣 𝛾𝛾−1 4 3 𝑇𝑇4 −𝑇𝑇1 1 Hence, 𝜂𝜂 = 1 − 𝛾𝛾−1 𝛾𝛾−1 =1− 𝛾𝛾−1 𝑇𝑇4 𝑟𝑟𝑣𝑣 −𝑇𝑇1 𝑟𝑟𝑣𝑣 𝑟𝑟𝑣𝑣 The thermal efficiency of the Otto cycle depends only on the compression ratio, 𝑟𝑟𝑣𝑣. Rev 1.11 CYF/MES/Gas Cycles  Topic 3. Gas Power Cycles 3-6 Thermofluids II (ME2401) 3.5 The Diesel Cycle In the Diesel cycle, the heat is added at constant pressure instead of constant volume. 𝑝𝑝 𝑇𝑇 𝑝𝑝 = 𝑐𝑐 3 3 2 𝑸𝑸𝒔𝒔 𝑠𝑠 = 𝑐𝑐 𝑠𝑠 = 𝑐𝑐 𝑝𝑝 = 𝑐𝑐 4 4 2 𝑸𝑸𝒓𝒓 𝑠𝑠 = 𝑐𝑐 𝑣𝑣 = 𝑐𝑐 𝑠𝑠 = 𝑐𝑐 𝑣𝑣 = 𝑐𝑐 1 1 𝑣𝑣 𝑠𝑠 𝑣𝑣𝑐𝑐 𝑣𝑣𝑠𝑠 Fig 3-3 p-V and T-s diagram of an diesel cycle Process 12: Isentropic compression γ 𝛾𝛾 𝑷𝑷𝟐𝟐 𝑣𝑣2 = 𝑃𝑃1 𝑣𝑣1 𝑻𝑻𝟐𝟐 𝑣𝑣1 𝛾𝛾−1 𝛾𝛾−1 = = 𝑟𝑟𝑣𝑣 𝑇𝑇1 𝑣𝑣2 Process 23: Constant pressure heat supply 𝑷𝑷𝟑𝟑 = 𝑃𝑃2 𝑻𝑻𝟑𝟑 𝑇𝑇2 = 𝑣𝑣3 𝑣𝑣2 The cut-off ratio, 𝛽𝛽 is given by, 𝑣𝑣3 𝛽𝛽 = 𝑣𝑣2 it is known as such as heat supply is cut off at state 3. Heat supplied, 𝑄𝑄𝑠𝑠 = m 𝐶𝐶𝑝𝑝 (𝑇𝑇3 − 𝑇𝑇2 ) Rev 1.11 CYF/MES/Gas Cycles Topic 3. Gas Power Cycles 3-7 Thermofluids II (ME2401) Process 3  4: Isentropic expansion γ 𝛾𝛾 𝑷𝑷𝟒𝟒 𝑣𝑣4 = 𝑃𝑃3 𝑣𝑣3 𝑻𝑻𝟒𝟒 𝑣𝑣3 𝛾𝛾−1 𝑣𝑣3 𝑣𝑣2 𝛾𝛾−1 𝑣𝑣3 𝑣𝑣2 𝛾𝛾−1 𝛽𝛽 𝛾𝛾−1 = = × = × = 𝑇𝑇3 𝑣𝑣4 𝑣𝑣2 𝑣𝑣4 𝑣𝑣2 𝑣𝑣1 𝑟𝑟𝑣𝑣 Process 4  1: Constant volume heat rejection 𝒗𝒗𝟒𝟒 = 𝑣𝑣1 Heat rejected, 𝑄𝑄𝑟𝑟 = m 𝐶𝐶𝑣𝑣 (𝑇𝑇4 − 𝑇𝑇1 ) Thermal efficiency of the Diesel cycle Thermal efficiency of the cycle, 𝑤𝑤𝑛𝑛𝑛𝑛𝑛𝑛 𝑄𝑄𝑟𝑟 𝜂𝜂 = = 1− 𝑄𝑄 𝑄𝑄𝑠𝑠 The thermal efficiency can also be expressed in terms of 𝑟𝑟, 𝛽𝛽 and 𝛾𝛾 as follows: 𝛽𝛽 𝛾𝛾 − 1 𝜂𝜂 = 1 − 𝛾𝛾−1 𝛾𝛾(𝛽𝛽 − 1)𝑟𝑟𝑣𝑣 The thermal efficiency of the diesel cycle depends on the cut-off ratio, 𝛽𝛽 and hence on the amount of heat added as well as the compression ratio, 𝑟𝑟𝑣𝑣. Rev 1.11 CYF/MES/Gas Cycles  Topic 3. Gas Power Cycles 3-8 Thermofluids II (ME2401) 3.6 Dual Combustion Cycle The cycle is so called because the heat is supplied in two parts. The first part is supplied at constant volume and the remainder at constant pressure. 𝑃𝑃 𝑇𝑇 𝑝𝑝 = 𝑐𝑐 𝑸𝑸𝒔𝒔(𝟑𝟑𝟑𝟑) 3 4 4 𝑣𝑣 = 𝑐𝑐 𝑝𝑝 = 𝑐𝑐 𝑠𝑠 = 𝑐𝑐 2 𝑠𝑠 = 𝑐𝑐 𝑸𝑸𝒔𝒔(𝟐𝟐𝟐𝟐) 3 5 5 𝑣𝑣 = 𝑐𝑐 𝑠𝑠 = 𝑐𝑐 2 𝑸𝑸𝒓𝒓 𝑣𝑣 = 𝑐𝑐 𝑠𝑠 = 𝑐𝑐 𝑣𝑣 = 𝑐𝑐 1 1 𝑣𝑣 𝑠𝑠 Vc Vs Fig 3-4: p-V and T-s diagram of an Otto cycle Process 12: Isentropic compression γ 𝛾𝛾 𝑷𝑷𝟐𝟐 𝑣𝑣2 = 𝑃𝑃1 𝑣𝑣1 𝑻𝑻𝟐𝟐 𝑣𝑣1 𝛾𝛾−1 𝛾𝛾−1 = = 𝑟𝑟𝑣𝑣 𝑇𝑇1 𝑣𝑣2 Process 23: Constant volume heat supply 𝒗𝒗𝟑𝟑 = 𝑣𝑣2 𝑃𝑃3 𝑃𝑃2 = 𝑻𝑻𝟑𝟑 𝑇𝑇2 The heat addition pressure ratio is given by: 𝑃𝑃3 𝑘𝑘 = 𝑃𝑃2 Heat supplied, 𝑄𝑄23 = m 𝐶𝐶𝑣𝑣 (𝑇𝑇3 − 𝑇𝑇2 ) Rev 1.11 CYF/MES/Gas Cycles Topic 3. Gas Power Cycles 3-9 Thermofluids II (ME2401) Process 34: Constant pressure heat supply 𝑷𝑷𝟒𝟒 = 𝑃𝑃3 The cutoff ratio is given by: 𝑣𝑣4 𝛽𝛽 = 𝑣𝑣3 𝑣𝑣4 𝑣𝑣3 𝑣𝑣4 = , ⟹ 𝑇𝑇4 = 𝑇𝑇 = 𝛽𝛽𝑇𝑇3 𝑻𝑻𝟒𝟒 𝑇𝑇3 𝑣𝑣3 3 Heat supplied, 𝑄𝑄34 = m 𝐶𝐶𝑝𝑝 (𝑇𝑇4 − 𝑇𝑇3 ) Process 4  5: Isentropic expansion 𝑷𝑷𝟓𝟓 𝑣𝑣4 𝛾𝛾 = 𝑃𝑃4 𝑣𝑣5 𝑻𝑻𝟓𝟓 𝑣𝑣4 𝛾𝛾−1 𝑣𝑣4 𝑣𝑣3 𝛾𝛾−1 𝑣𝑣4 𝑣𝑣2 𝛾𝛾−1 𝛽𝛽 𝛾𝛾−1 = = × = × = 𝑇𝑇4 𝑣𝑣5 𝑣𝑣3 𝑣𝑣5 𝑣𝑣3 𝑣𝑣1 𝑟𝑟𝑣𝑣 Process 51: Constant volume heat rejection 𝒗𝒗𝟓𝟓 = 𝑣𝑣1 Heat rejected, 𝑄𝑄𝑟𝑟 = m 𝐶𝐶𝑣𝑣 (𝑇𝑇5 − 𝑇𝑇1 ) Thermal efficiency of the Dual combustion cycle Total heat supplied, 𝑄𝑄𝑠𝑠 = 𝑄𝑄23 + 𝑄𝑄34 = m 𝐶𝐶𝑣𝑣 (𝑇𝑇3 − 𝑇𝑇2 ) + m 𝐶𝐶𝑝𝑝 (𝑇𝑇4 − 𝑇𝑇3 ) Heat rejected, 𝑄𝑄𝑟𝑟 = m 𝐶𝐶𝑣𝑣 (𝑇𝑇5 − 𝑇𝑇1 ) Net work, 𝑊𝑊𝑛𝑛𝑛𝑛𝑛𝑛 = 𝑄𝑄𝑠𝑠 − 𝑄𝑄𝑟𝑟 Hence Thermal efficiency is given by, 𝑊𝑊𝑛𝑛𝑛𝑛𝑛𝑛 𝜂𝜂 = 𝑄𝑄𝑠𝑠 Or, 𝑄𝑄𝑟𝑟 𝜂𝜂 = 1 − 𝑄𝑄𝑠𝑠 The efficiency can also be expressed in terms of the compression ratio, 𝑟𝑟𝑣𝑣 , the cutoff ratio, 𝛽𝛽, and the heat addition pressure ratio, 𝑘𝑘. Rev 1.11 CYF/MES/Gas Cycles Topic 3. Gas Power Cycles 3-10 Thermofluids II (ME2401) Thermal efficiency will be, 𝑘𝑘𝛽𝛽 𝛾𝛾 − 1 𝜂𝜂 = 1 − 𝛾𝛾−1 𝑟𝑟𝑣𝑣 [(𝑘𝑘 − 1) + 𝛾𝛾𝛾𝛾(𝛽𝛽 − 1)} Thus the efficiency of the dual combustion cycles depends on 𝑟𝑟𝑣𝑣 , 𝛽𝛽, and 𝑘𝑘. To calculate the thermal efficiency of the Otto, Diesel or Dual Combustion cycle, it is best to use the simple equation, 𝑤𝑤𝑛𝑛𝑛𝑛𝑛𝑛 𝜂𝜂 = 𝑄𝑄𝑠𝑠 or 𝑄𝑄𝑟𝑟 𝜂𝜂 = 1 − 𝑄𝑄𝑠𝑠 Rev 1.11 CYF/MES/Gas Cycles Topic 3. Gas Power Cycles 3-11 Thermofluids II (ME2401) Example 1 An engine operating on the Ideal Otto cycle has a cylinder bore of 68 𝑚𝑚𝑚𝑚 and a stroke of 80 𝑚𝑚𝑚𝑚. Its suction conditions are 103 𝑘𝑘𝑁𝑁/𝑚𝑚2 and 28℃ while the maximum cycle temperature is 1350℃. The clearance volume is 10% of the swept volume. Sketch the P- v and T-s diagrams and calculate the: a. compression ratio, b. net work output per 𝑘𝑘𝑘𝑘, c. ratio of the maximum to minimum pressure in the cycle, d. thermal efficiency. Rev 1.11 CYF/MES/Gas Cycles Topic 3. Gas Power Cycles 3-12 Thermofluids II (ME2401) Example 2 A Diesel cycle has a compression ratio of 18: 1. The heat supplied to the air is 1750 𝑘𝑘𝐽𝐽/𝑘𝑘𝑘𝑘. At the start of the cycle, the pressure and temperature are 1.0 𝑏𝑏𝑎𝑎𝑟𝑟 and 28℃. Sketch the pv and T-s diagrams for this cycle and calculate the: a. temperature for the remaining points in the cycle, b. cut-off ratio, c. net work done in 𝑘𝑘𝑘𝑘/𝑘𝑘𝑘𝑘, d. thermal efficiency. Rev 1.11 CYF/MES/Gas Cycles Topic 3. Gas Power Cycles 3-13 Thermofluids II (ME2401) Example 3 An engine working on the ideal Dual combustion cycle has a compression ratio of 15: 1. The suction conditions are 1.01 𝑏𝑏𝑎𝑎𝑟𝑟 and 32℃ (point 1). The maximum pressure of the cycle is 65 𝑏𝑏𝑎𝑎𝑟𝑟. The heat added at constant volume is 60% of the heat added at constant pressure. Sketch the P-v and T-s diagrams for this cycle and calculate the: a. volume cut-off ratio, b. heat supplied and heat rejected in 𝑘𝑘𝑘𝑘/𝑘𝑘𝑘𝑘, c. net work done in 𝑘𝑘𝑘𝑘/𝑘𝑘𝑘𝑘, d. thermal efficiency of the cycle. Rev 1.11 CYF/MES/Gas Cycles Topic 3. Gas Power Cycles 3-14 Thermofluids II (ME2401) Tutorial 3 on Gas-Powered cycles 1. An engine operating on the ideal air-standard Otto cycle has a bore of 10 𝑐𝑐𝑐𝑐 and a stroke of 12 𝑐𝑐𝑐𝑐. At the beginning of the compression process, the air is at 29℃ and 101 𝑘𝑘𝑘𝑘/𝑚𝑚2. The maximum temperature in the cycle is 1560℃ and the clearance volume is 12% of the swept volume. Sketch and label the p-v and T-s diagrams for this cycle. Determine the: a. compression ratio, b. work done by the engine in 𝑘𝑘𝑘𝑘/𝑘𝑘𝑔𝑔, c. cycle thermal efficiency. (9.33, 464.5, 59.1) 2. A diesel engine cycle has a compression ratio of 16: 1. The starting temperature is 30℃ and the pressure is 1 𝑏𝑏𝑎𝑎𝑟𝑟. The heat added at constant pressure is 2800 𝑘𝑘𝐽𝐽/𝑘𝑘𝑘𝑘. Sketch the P-v and T-s diagrams for this cycle and calculate the: a. volume cut-off ratio, b. nett work output in 𝑘𝑘𝐽𝐽/𝑘𝑘𝑘𝑘, c. cycle thermal efficiency. (4.03, 1485.3 kJ/kg, 0.53) 3. An air-standard Otto cycle has a bore of 8 𝑐𝑐𝑐𝑐 and a stroke of 10 𝑐𝑐𝑐𝑐. The intake conditions are 1 𝑏𝑏𝑏𝑏𝑏𝑏 and 30℃. The maximum pressure in the cycle is 40 𝑏𝑏𝑏𝑏𝑏𝑏. The clearance volume is 11 % of the swept volume. Sketch the p-V and T-s diagrams for this cycle and determine the: a. compression ratio, b. heat supplied in kJ/kg, c. nett cycle work in kJ/kg, d. thermal efficiency of the cycle. (10.09, 313.26, 188.97, 0.603) Rev 1.11 CYF/MES/Gas Cycles Topic 3. Gas Power Cycles 3-15 Thermofluids II (ME2401) 4. **An air-standard diesel cycle has a compression ratio of 20 and a cut-off ratio of 3. The inlet pressure and temperature are 100 𝑘𝑘𝑁𝑁/𝑚𝑚2 and 27℃. Sketch the p-V and T-s diagrams and determine the: a. heat added in kJ/kg, b. nett work in kJ/kg, c. thermal efficiency of the cycle. (1994.6, 1209.2, 0.606) 5. **A petrol engine working on the standard Otto Cycle operates with an inlet temperature of 27℃ and a pressure of 101 𝑘𝑘𝑘𝑘/𝑚𝑚2. The compression ratio is 8: 1 and the maximum pressure is 4800 𝑘𝑘𝑘𝑘/𝑚𝑚2. Sketch the p-V and the T-s diagrams for this engine cycle and determine the: a. remaining pressures (in 𝑘𝑘𝑘𝑘/𝑚𝑚2 ) and temperatures (in 𝐾𝐾) for the cycle. b. heat supplied, heat rejected and the net work output for the cycle (in 𝑘𝑘𝑘𝑘/𝑘𝑘𝑘𝑘). c. thermal efficiency of the Cycle. (1900kN/m2, 689.2K, 1800K, 261.2kN/m2, 775.7K; 784.7kJ/kg, 341.6kJ/kg, 443.2kJ/kg; 56.5 %) 6. An engine working on the air standard Dual Combustion Cycle has a single cylinder whose bore is 8 cm and stroke is 14 cm. The suction conditions are 1.01 bar and 32 oC while the maximum cycle pressure and temperature are 65 bar and 1500 oC. The clearance volume is 7.69 % of the swept volume. Sketch the p-V and T-s diagrams and determine the: a. compression ratio, b. heat supplied and heat rejected in 𝑘𝑘𝑘𝑘/𝑘𝑘𝑘𝑘, c. nett work done in 𝑘𝑘𝑘𝑘/𝑘𝑘𝑘𝑘, d. thermal efficiency of the cycle. (14, 750.2 kJ/kg, 267.6 kJ/kg, 482.6 kJ/kg, 64.33 %) Rev 1.11 CYF/MES/Gas Cycles Topic 3. Gas Power Cycles 3-16 Thermofluids II (ME2401) 7. An engine operating on the Dual Combustion Cycle has a compression ratio of 16:1. At the start of the compression stroke, p1 = 0.98 bar, T1 = 28o C. The maximum cycle pressure is 65 bar and the maximum cycle temperature is 1350 oC. Draw the p-V and T-s diagrams for this cycle and calculate a. work done per cycle in kJ/kg, b. heat input per cycle in kJ/kg, c. thermal efficiency. (407.3 kJ/kg, 617 kJ/kg, 66 %) 8. An engine operating on the ideal dual combustion cycle has a bore of 10 cm and a stroke of 20 cm. The suction conditions are 1.02 bar and 30o C while the maximum cycle pressure and temperature are 60 bar and 1400o C. The clearance volume is 8 % of the swept volume. Draw the p-V and T-s diagrams and determine the: d. compression ratio, e. heat supplied and heat rejected in kJ/kg, f. net work done in kJ/kg, g. thermal efficiency of the cycle. (13.5, 686.3 kJ/kg, 249.1 kJ/kg, 437.2 kJ/kg, 64 %) 9. **A diesel engine working on the ideal dual combustion cycle has a compression ratio of 15.5: 1. The inlet conditions of the engine are 1.01 𝑏𝑏𝑏𝑏𝑏𝑏 and 29℃. The maximum cycle pressure reached is 60 𝑏𝑏𝑏𝑏𝑏𝑏. The heat supplied at constant volume is equal to 70% of the heat supplied at constant pressure. Sketch the p-V and T-s diagrams of this cycle and determine the: a. maximum cycle temperature, b. cut off ratio, c. heat supplied in 𝑘𝑘𝑘𝑘/𝑘𝑘𝑘𝑘, d. heat rejected in 𝑘𝑘𝑘𝑘/𝑘𝑘𝑘𝑘, e. nett work done in 𝑘𝑘𝑘𝑘/𝑘𝑘𝑘𝑘, f. thermal efficiency. (1416K, 1.224, 442, 151.4, 290.6, 0.657) Rev 1.11 CYF/MES/Gas Cycles Topic 3. Gas Power Cycles 3-17 Thermofluids II (ME2401) Blank Page Rev 1.11 CYF/MES/Gas Cycles  Topic 4. Steam Power Cycles 4-1 Thermofluids II (ME2401) 4 STEAM POWER CYCLES Keywords: Boiler, Turbine, Condenser, Feed pump, Work ratio, Turbine work, Feed pump work, High pressure turbine work, Low pressure turbine work, Heat supplied, Thermal efficiency, Specific Steam Consumption, Isentropic work, Actual work, Isentropic efficiency, Carnot cycle, Rankine cycle, Super heat, Reheat. Objective: Students should be able to understand the concepts of the Steam power cycles and solve problems involving these cycles. They should be able to: 1. Sketch the component parts of a steam plant based on the Carnot, Rankine, Rankine with superheat and the reheat cycles. 2. Sketch the T-s diagrams for these cycles. 3. Explain the limitations of the Carnot steam cycle. 4. Explain the limitations of the Rankine cycle. 5. State the advantages of the Superheat cycle. 6. State the advantages of the Reheat cycle. 7. Calculate the work ratio, net work, heat supplied, thermal efficiency, specific steam consumption, and power developed for the cycle. 8. Calculate the turbine isentropic efficiency. 4.1 The Carnot Vapour Cycle Carnot cycle is the most efficient cycle operating between two specified temperature levels. Thus it is natural to look at the Carnot cycle as a prospective ideal cycle for the steam plant. If we could, we would certainly adopt it as the ideal cycle. But as explained later, the Carnot cycle is an unsuitable model for the steam plant. Rev 1.11 CYF/MES/Steam Cycles  Topic 4. Steam Power Cycles 4-2 Thermofluids II (ME2401) Consider a steady flow Carnot cycle executed as shown in the following T-s diagram: 1 𝑇𝑇 4 1 Turbine 4 Boiler 2 3 2 Condenser 𝑠𝑠 3 Feed Pump Figure 4-1: Rankine cycle Process 12: Isentropic expansion in turbine 23: Isothermal heat rejection in condenser 34: Isentropic compression in feed pump 41: Isothermal heat supply in boiler 4.2 Thermal Efficiency of the Carnot Cycle 𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 − ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝜂𝜂 = ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 =1− ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑇𝑇2 (𝑠𝑠2 − 𝑠𝑠3 ) =1− 𝑇𝑇2 (𝑠𝑠1 − 𝑠𝑠4 ) 𝑇𝑇2 ∴ 𝜂𝜂 = 1 − 𝑇𝑇1 Rev 1.11 CYF/MES/Steam Cycles  Topic 4. Steam Power Cycles 4-3 Thermofluids II (ME2401) 4.3 Limitations of the Carnot Vapour Cycle The Carnot cycle is not used in the steam plant because: A large feed pump is required to compress the wet steam to boiler pressure. Heat addition and rejection at constant temperature is a very slow process. The work ratio of Carnot cycle is low. 𝑁𝑁𝑁𝑁𝑁𝑁 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 = 𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 4.4 The Rankine Cycle 1 𝑇𝑇 Turbine 1 2 4 Boiler 4 Condenser 3 2 3 𝑠𝑠 Feed pump Figure 4-2: Rankine cycle Process 12: Isentropic expansion in turbine (work output) 23: Heat rejection at constant pressure in condenser (change steam to water) 34: Isentropic compression in feed pump (pressurize the liquid) 41: Heat added at constant pressure in boiler (heat the water into steam) Rev 1.11 CYF/MES/Steam Cycles  Topic 4. Steam Power Cycles 4-4 Thermofluids II (ME2401) 4.5 Work and Heat Supplied Turbine (process 1-2) and Feed pump Work (process 3-4) Applying the S.F.F.E to the turbine (process 1-2) and ignoring change in the P.E. & K.E. Steam in 1 ℎ1 + 𝑞𝑞 = ℎ2 + 𝑤𝑤 work out No heat transfer, 𝑞𝑞 = 0 Turbine work output, ∴ 𝒘𝒘𝑻𝑻 = 𝒉𝒉𝟏𝟏 − 𝒉𝒉𝟐𝟐 Steam out 2 Similarly, Feed pump work input (process 3-4), 𝒘𝒘𝑭𝑭 = 𝒉𝒉𝟒𝟒 − 𝒉𝒉𝟑𝟑 or 𝒘𝒘𝑭𝑭 = 𝒗𝒗𝑭𝑭 (𝒑𝒑𝟒𝟒 − 𝒑𝒑𝟑𝟑 ) Figure 4-3: Turbine Note: 𝑣𝑣𝑓𝑓 ≈ 0.001 𝑚𝑚3 /𝑘𝑘𝑘𝑘 Heat supplied to Boiler (process 4-1) and heat rejected from condenser (process 2-3) Steam out 1 Applying the S.F.F.E to the boiler and ignoring change in the P.E. & K.E. water in 4 ℎ4 + 𝑞𝑞 = ℎ1 + 𝑤𝑤 No work transfer, 𝑤𝑤 = 0 Heat supplied to the Boiler, Heat in 𝒒𝒒𝒔𝒔 = 𝒒𝒒𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃 = 𝒉𝒉𝟏𝟏 − 𝒉𝒉𝟒𝟒 Similarly, condenser (process 2-3) heat Figure 4-4: Boiler rejection, 𝒒𝒒𝒓𝒓 = 𝒒𝒒𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄 = 𝒉𝒉𝟐𝟐 − 𝒉𝒉𝟑𝟑 Rev 1.11 CYF/MES/Steam Cycles Topic 4. Steam Power Cycles 4-5 Thermofluids II (ME2401) Work Ratio 𝑁𝑁𝑁𝑁𝑁𝑁 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 = 𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 − 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 = 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 (ℎ1 − ℎ2 ) − (ℎ4 − ℎ3 ) = (ℎ1 − ℎ2 ) Thermal Efficiency of the Rankine cycle 𝑁𝑁𝑁𝑁𝑁𝑁 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝜂𝜂 = 𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 (ℎ1 − ℎ2 ) − (ℎ4 − ℎ3 ) = (ℎ1 − ℎ4 ) In most cases, the Feed pump work is small when compared to the turbine work, therefore, if Feed pump work is neglected: ℎ4 ≈ ℎ3 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝜂𝜂 = 𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 ℎ1 − ℎ2 = ℎ1 − ℎ4 Specific Steam Consumption This is the mass of steam required to produce 1 kWh of work 3600 𝑆𝑆. 𝑆𝑆. 𝐶𝐶. = 𝑘𝑘𝑘𝑘/𝑘𝑘𝑘𝑘ℎ 𝑤𝑤 where 𝑤𝑤 = 𝑛𝑛𝑛𝑛𝑛𝑛 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤/𝑘𝑘𝑘𝑘 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 Rev 1.11 CYF/MES/Steam Cycles Topic 4. Steam Power Cycles 4-6 Thermofluids II (ME2401) Isentropic Efficiency of Turbine In practice, the actual expansion process in the turbine is not reversible and the actual turbine work, 𝑤𝑤 = ℎ1 − ℎ2′ where ℎ2′ can be obtained from the equation, 𝑇𝑇 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 1 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 = 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 ℎ1 − ℎ2′ 𝜂𝜂𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 = ℎ1 − ℎ2 2 2’ 𝑠𝑠 Figure 4-5: ℎ2?