Thermodynamics Full Lectures PDF

1 5ENT1129 Thermodynamics for Aerospace  2 Lecture 1 - Introduction to Thermodynamics Dr Burhan Saeed Contact Me 3 Dr. Burhan Saeed University of Hertfordshire – School of Physics, Engineering...

1 5ENT1129 Thermodynamics for Aerospace  2 Lecture 1 - Introduction to Thermodynamics Dr Burhan Saeed Contact Me 3 Dr. Burhan Saeed University of Hertfordshire – School of Physics, Engineering and Computer Science Office Location – 2nd Floor SPECTRA Email: [email protected] Semester A – Schedule 4 Lectures are up to 2 hours every week (Tuesday 1100 – 1300 SPECTRA SP1006) Tutorials are up to one hour every week ( Friday 1200 – 1300 Lindop Building A154) Reading Material – Reference Books 5 Lecture notes are mostly based on these two books: Cengel Y.A., Boles M.A., Thermodynamics – An Engineering Approach, 8th edition (SI Units), McGraw Hill (2007) ISBN [978-007-1257718] Serway R.A., Jewett J. W., Physics for Scientist and Engineers, Volume 1 , 8th edition, Brooks Cole (2009) ISBN [978-1439048382]. However you can select other books on basic physics, thermodynamics and heat transfer: Course Content 6 The definition of energy (to include units of measure). Properties of a system: system boundaries, open vs. closed systems Temperature and the Zeroth Law of Thermodynamics, Pressure Energy transfer: heat and work. First Law (Conservation of Energy) Ideal Gas Law Boundary Work Specific heats & internal energy Enthalpy and its relationship with boundary work Assessment Overview 7 Computer-Based test on your learning the first half Timed Individual of this module. 12th of November 2024 20% Assessment 1 Heat Pump experiments to be covered in the second half of this module. You will do the experiments in groups, but need to collect data and calculate & Lab coursework Individual analyse all individually to write up your individual 20% report. You will be answering a combination of multiple- choice and open questions about the range of topics Final Timed covered in this module. Individual 60% assessment Feedback will be in the form of a score and notes shown in the online assessment results screen following the release of marks. Introduction to Thermodynamics 8 Thermodynamics derive from the Greek words ”therme” (that means heat) and ”dynamis” (that means power). Thermodynamic is encountered commonly in many aspect of life and in many engineering systems. So it is important to understand its principles. Introduction to Thermodynamics 9 Thermodynamics describe the process of converting heat into power, interpreted broadly to include all aspects of energy and its transformation, and relationship among properties of matter. There are typically two approaches in thermodynamics: Classical (macroscopic approach) Statistical (microscopic approach) Do not require study of the Study the average behaviour of behaviour of individual particles large group of individual particles Introduction to Thermodynamics – System 10 System is a region in the space that encloses quantity of matter. Surroundings Boundary System A system can be: open, closed or isolated. Boundary has zero thickness and it may move. Introduction to Thermodynamics – System 11 Systems with moving boundary Open Systems 12 An open system can exchange energy and mass with the surroundings. Boiler Human body Rocket Nozzle Closed Systems 13 A closed system can exchange energy but not mass with the surroundings. Pressure Cooker Home Heat System Open vs. Closed System 14 Open system Closed system Matter does Matter not go in or out Isolated System 15 An isolated system can not exchange energy and mass with the surroundings. Universe Dewar flask or Vacuum flask Classification of Properties 16 Any Characteristic of a system is called Property - Pressure, Temperature, Volume, Mass Intensive (Mass independent) – temperature, pressure, density etc. Extensive (depend on the size or extent of the system) – total mass, total volume etc. Specific (extensive properties per unit mass {commonly}, volume or another property ) - specific volume, specific total energy, specific heat capacity etc. Density and Specific Gravity 17 State and Equilibrium 18 State: all the properties have fixed values throughout the system. Equilibrium: State of balance, no unbalanced potential or driving force that can change the system Processes and Cycles 19 Path: Series of states through which the system passes. Processes and Cycles 20 Isothermal: Temperature remains constant Isobaric: Pressure remain constant Isochoric: specific volume remains constant Processes and Cycles 21 Isovolumetric Process Isobaric Process or Isochoric Process P P P Isothermal Process 1 2 2 1 2 1 V1 V2 V V1  V2 V V1 V2 V We will see later, how the area enclosed represents the work. Processes and Cycles 22 P Cycle Initial state (1) and Final state (2) coincide. P1  P2 V V1  V2 V A Simple Cycle 23 P PL = 100000 Pa PH = 200000 Pa 2 3 PH VL = 1m 3 VH = 2m 3 PL 1 4 VL VH V Which point on the diagram is represented by these pistons? T=614K T=1228K T=307K T=614K A Simple Cycle 24 P PL = 100000 Pa PH = 200000 Pa 2 3 PH VL = 1m 3 VH = 2m 3 PL 1 4 VL VH V Which point on the diagram is represented by these pistons? T=614K T=1228K T=307K T=614K 4 3 1 2 Steady and Uniform Process 25 Steady: No change of properties with time Unsteady or transient: Opposite of steady Uniform: No change with location within the specified region Reverse and Irreversible Process 26 A reversible process is a process where the effects of following a thermodynamic path can be undone by exactly reversing the path. It is a process that is always at equilibrium even when undergoing a change. Reverse and Irreversible Process 27 Unit of Measure for Temperature 28 Dimension Temperature (Ө) International System of Units (S.I.) Système international d'unités William Thomson 1st Baron Kelvin (1824 – 1907) Unit of measure degree Kelvin (K) Temperature Scales – Celsius and Fahrenheit 29 Anders Celsius Scale (1742): 0 oC is the freezing point of water 100 oC is the boiling point of the water (Originally Celsius in 1742 proposed the scale in reverse order: 100 oC was the temperature for the freezing point of the water. This is due to the fact that Celsius wanted a positive value to describe a typical Swedish temperature! The scale as we know now (reversed respect to original Celsius’s choice) was established by another Swedish scientist Linnaeus in 1744, after Celsius’s death.) Fahrenheit Scale (1724): 32 oF is the freezing point of water 212 oF is the boiling point of the water  C  = 95 ( F − 32)    F  = 95  C + 32   Conversion Formulae Temperature Scale – Kelvin 30 Kelvin Scale (1848): 0 K is the absolute zero = -273.15 oC 273.15 K = 0 oC is the freezing point of water 373.15 K = 100 oC is the boiling point of the water In this scale we have no negative temperature. The lowest temperature is 0 (absolute zero). At this temperature all thermal motion ceases according to the classical description of thermodynamics. At very low temperatures in the vicinity of absolute zero, matter exhibits properties likes superconductivity and superfluidity. https://www.youtube.com/watch?v=9FudzqfpLLs The current world record was set in 1999 at 100 picokelvins (pK), or 0.000 000 000 Kelvin Scale is just a shifting of Celsius 1 of a Kelvin.     Scale  C = K  − 273.15 K  =  C + 273.15 Conversion Formulae A Quick Question 31 The temperature of the lubricating oil in an automobile engine is measured as 150 oF. What is the temperature of this oil in oC and in K? (from Y.A. Cengel – M.A. Boles – Thermodynamics – An Engineering Approach – 6th ed. 2007) (Problem 1.36) A Quick Question 32 The temperature of the lubricating oil in an automobile engine is measured as 150 oF. What is the temperature of this oil in oC and in K? (from Y.A. Cengel – M.A. Boles – Thermodynamics – An Engineering Approach – 6th ed. 2007) (Problem 1.36)  C= 5  9 (F − 32 ) 150F = 65.56C K = C + 273.15 65.56C = 338.71K Temperature and Zeroth Law of Motion 33 The Zeroth Law of Thermodynamics states: Two bodies are in thermal equilibrium if both have the same temperature reading (even if they are not in contact). This principle, states a very basic concept, but it cannot be deducted by the other laws of Thermodynamics. It was formulated by R.H. Fowler in 1931 after the first and second law were formulated (more than half a century before). Logically it should be the first law to be enunciated so it is why it is called zeroth law. 20 oC 50 oC 50 oC Heat flux 75 oC 50 oC 50 oC No equilibrium Equilibrium (different temperatures) Equilibrium Pressure 34 Pressure is defined as the ratio between force, F, and area, A: Formula to remember Pressure Dimension:    P = M  L−1  T −2 P= F A N Unit of measure: 2 = Pa (Pascal) m Pressure 35 Pressure 36 Absolute pressure - The actual pressure at a given position is called the absolute pressure, and it is measured relative to absolute vacuum (i.e., absolute zero pressure). Gage pressure - Gage pressure is the pressure relative to the atmospheric pressure. In other words, how much above or below is the pressure with respect to the atmospheric pressure. Vacuum pressure - Pressures below atmospheric pressure are called vacuum pressures and are measured by vacuum gages that indicate the difference between the atmospheric pressure and the absolute pressure. Atmospheric pressure - The atmospheric pressure is the pressure that an area experiences due to the force exerted by the atmosphere. Pressure 37 On the sea surface is acting atmospheric pressure Patm Patm h1 Formula to remember P =   g  h1 + P0 P =   g  h + Patm h2 P =   g  h2 + P0 h3 P =   g  h3 + P0 Hydrostatic Pressure 38 https://socratic.org/questions/what-is-the-difference-between-hydrostatic-pressure-and-atmospheric-pressure from http://www.britishdams.org/about_dams/gravity.htm Hydrostatic Pressure - Example 39 Suppose that the pressure acting on the back of a swimmer’s hand is 1.2*105 Pa (a realistic value near the bottom of the diving end of a pool). The surface area of the back of the hand is 8.4*10-3 m2. (a) Determine the magnitude of the force that acts on it. (b) Discuss the direction of the force. (from Cutnell & Johnson 2012) Hydrostatic Pressure - Example 40 F If we recall the definition of pressure: P= A We can obtain the value of the force: F P= → F = P A A  5 N   −3 1  F = P  A = 1.2 10 2    8.4 10 2  = 1008 N  m   m  Water applies a force perpendicular to each surface within the water, including the walls and bottom of the swimming pool, and all parts of the swimmer’s body Change of Atmospheric Pressure with Altitude 41 Air is also a fluid and hence we can apply the same concept of hydrostatic pressure. At higher altitude the value of the pressure is reduced. Pressure is maximum at the Sea Level ( h= 0 ) and minimum at the outer space. Change of Atmospheric Pressure with Altitude 42 This explains why aircraft need to be pressurized during the flight. To maintain a comfortable situation for the passengers during the flight. Poutside = 20000 Pa Pinside = 101325 Pa  43 Thank you herts.ac.uk 5ENT1129 Thermodynamics for Aerospace  Lecture 2 Thermodynamics: Energy Conservation University of Hertfordshire– SPECS – Thermodynamics for Aerospace – Dr. Burhan Saeed 2024-2025 2  Table of contents Energy Conservation Different form of Energy. Definition of Heat. Definition of Work. Formal sign convention on Heat and Work. First law of Thermodynamics and its applications. Adiabatic process. 3 Conservation of Energy 4  Conservation of Energy What happens to the room temperature? a) Decreases b) Increases c) Remains constant 5  Conservation of Energy What happens to the room temperature? a) Decreases b) Increases c) Remains constant The electric energy consumed by the device is converted into heat – the room temperature will rise. 6  Conservation of Energy What happens to the room temperature? a) Decreases b) Increases c) Remains constant 7  Conservation of Energy What happens to the room temperature? a) Decreases b) Increases c) Remains constant The electric energy consumed by the device is converted into heat – the room temperature will rise. 8 Forms of energy 9  Forms of energy Macroscopic form: system possesses as a whole with respect to some external reference, related to motion and/or influence of external effects such as gravity, magnetism, electricity etc. – e.g. potential energy, kinetic energy Microscopic form: related to the molecular structure of the system and the degree of molecular activity. – The sum of all microscopic forms of energy is called internal energy (U or Eint) of the system 10 Forms of energy 11  Forms of energy 1 Kinetic Energy KE = 2 𝑚𝑣 2 (kJ) 1 Kinetic energy per unit mass ke = 2 𝑣 2 (kJ/kg) Potential energy PE = 𝑚𝑔𝑧 (kJ) Mass and energy flow rates associated with Potential energy per unit mass pe = 𝑔𝑧 (kJ/kg) the flow of steam in a pipe of inner diameter D with an average velocity of Vavg. Total energy of a system For an open system or a control volume 1 𝐸 = 𝑈 + 𝐾𝐸 + 𝑃𝐸 = 𝑈 + 𝑚𝑣 2 + 𝑚𝑔𝑧 (𝑘𝐽) Mass flow rate 𝑚ሶ is the amount of mass 2 flowing through a cross-section per unit time. 𝒎ሶ = 𝝆𝑽ሶ = 𝝆𝑨𝒄𝑽𝒂𝒗𝒈 (kg/s) Total energy of a system per unit mass 1 2 Energy flow rate 𝐸 = 𝑢 + 𝑘𝑒 + 𝑝𝑒 = 𝑢 + 𝑣 + 𝑔𝑧 (𝑘𝐽/𝑘𝑔) 2 𝑬ሶ = 𝒎𝒆 ሶ (kJ/s or KW) Total energy of a system per unit mass 𝐸 𝑒= (𝑘𝐽/𝑘𝑔) 𝑚 12 Forms of energy Sensible energy: The portion of the internal energy associated with the kinetic energies of the molecules. Latent energy: The internal energy associated with the phase of a system. Chemical energy: The internal energy associated with the atomic bonds in a molecule. Nuclear energy: The tremendous amount of energy associated with the strong bonds within the nucleus of the atom itself. Thermal = Sensible + Latent Internal = Sensible + Latent + Chemical + Nuclear 13  Forms of energy The total energy of system, can be contained or stored in a system, and thus can be viewed as the static forms of energy. The forms of energy not stored in a system can be viewed as the dynamic forms of energy or energy interactions. The dynamic forms of energy are recognised at the system boundary as they cross it, and they represent the energy gained or lost by a system during a process. The only two forms of energy associated with a closed system are heat transfer and work. An energy interaction is heat transfer if its driving force is a temperature difference. Otherwise it is work. 14  Forms of energy Mechanical energy: The form of energy that can be converted to mechanical work completely and directly by an ideal mechanical device such as a turbine. Mechanical energy of a flowing fluid per unit mass Note: Pressure itself is not a form of energy but a pressure force acting on a fluid through a distance produces work called Rate of mechanical energy of a flowing fluid 𝑃 flow work in the amount of per 𝜌 unit mass. It can also be viewed as flow energy of a flowing fluid. Mechanical energy change of a fluid during incompressible flow per unit mass Rate of mechanical energy change of a fluid during incompressible flow per unit mass 15  Forms of energy Mechanical energy example 1 – Ideal Turbine Mechanical energy is illustrated by an ideal hydraulic turbine coupled with an ideal generator. In the absence of irreversible losses, the maximum produced power is proportional to (a) the change in water surface elevation from the upstream to the downstream reservoir or (b) the drop in water pressure from just upstream to just downstream of the turbine. 16  Forms of energy Mechanical energy example 2 – Wind Energy A site evaluated for a wind farm is observed to have steady winds at a speed of 8.5 m/s. Determine the wind energy (a) per unit mass, (b) for a mass of 10kg, and (c) for a flow rate of 1154 kg/s for air. 17  Energy Transfer and Heat Heat is a form of energy that is transferred between two systems (or a system and its surroundings) by virtue of a temperature difference. Energy can cross the Temperature difference is the boundaries of a closed driving force for heat transfer. system in the form of The larger the temperature heat and work. difference, the higher is the rate of heat transfer. 18 Energy Transfer by Heat 19  A Historical Background of Heat Kinetic Theory: treats molecules as tiny balls that are in motion and thus possess kinetic energy. Heat: The energy associated with the random motion of atoms and molecules. Heat Transfer Mechanisms: 1. Conduction: The transfer of energy from the more energetic particles of a substance to the adjacent less energetic ones as a result of interaction between particles. 2. Convection: The transfer of energy between the solid surface and the adjacent fluid that is in motion, and it involves the combined effects of conduction and fluid motion. 3. Radiation: The transfer of energy due to the emission of electromagnetic waves (or photons) 20 Energy Transfer by Work 21  Energy Transfer by Work Wherever mechanical force is expended, an exact equivalent of heat is always obtained. — J.P. Joule, August, 1843 James Prescott Joule 22  Energy Transfer by Work Definition of Work and Dimensions:        W = F  s → W = M  L T −2  L  = M  L2  T −2 Unit of measure (Joule): m 1J = 1N  1m = 1kg  1 2  1m s Formula to remember   W = F s 23  Energy Transfer by Work  F Definition of work (scalar product) -   s Mechanic examples     F  s = F s cos    F  F F    s s s W 0 W =0 W 0 24  Energy Transfer by Work There are two requirements for a work interaction between a system and its surrounding to exist: 1. There must be a force acting on the boundary 2. The boundary must move 25  Energy Transfer by Work Shaft Work: A force F acting through a moment arm r generates a torque T The force acts through a distance s Shaft Work Shaft Power 26  Energy Transfer by Work Shaft Work Example: Determine the power transmitted through the shaft of a car when the torque applied is 200 N.m and the shaft rotates at a rate of 4000 revolution per minute (rpm). 27  Energy Transfer by Work S2 W =  F  dy Definition of work S1 F P= F = P A Use definition of pressure A Substitute expression for the force into definition of work: S2 W =  P  Ady From Serway, Jewett., Physics S1 for Scientist and Engineers dV V2 W =  P  dV V1 28  Energy Transfer by Work y P y(x ) P(V ) x1 x2 x V1 V2 V x2 A =  y (x )  dx V2 W =  P(V )  dV x1 V1 The area under the curve (in the PV diagram) represents the work 29  Energy Transfer by Work Example 1 Example 2 Cycle P P P1 P1 P2 P2 P1  P2 V1 V2 V V1 V2 V V1  V2 V W W W The work done by a gas as it is taken from an initial state (P1, V1) to a final state (P2, V2) depends on the path between these states. (Please, look at example 1 and 2) 30  Energy Transfer by Work Isobaric Process Isovolumetric Process Isothermal Process P P P V1 V2 V V1  V2 V V1 V2 V W = P  (V2 − V1 ) W =0 V2 W =  P  dV V1 nRT PV = nRT → P= V We will discuss the equation of state for an ideal V  V2 nRT gas in the next lecture W= dV = nRT ln  2  V1 V  V1  31  Heat vs. Work Both are recognised at the boundaries of a system as they cross the boundaries. That is, both heat and work are boundary phenomenon. Systems possess energy, but not heat or work. Both are associated with a process, not a state. Unlike properties, heat or work has no meaning at a state. Both are path functions (i.e. their magnitudes depend on the path followed during a process as well as the end states) Properties are point functions with exact differentials (d) Path functions have inexact differentials (𝜹) 32  Formal sign convention on Heat and Work Win  0 Wout  0 SYSTEM Qin  0 Qout  0 33  The First Law of Thermodynamics Our aim is to formulate the first law of thermodynamics. First law of thermodynamics Generalization of the law of conservation of the energy Internal energy Work Heat 34  The First Law of Thermodynamics Our aim is to formulate the first law of thermodynamics. U = Q − W U = Change in Internal Energy of the system Q = Heat added to the system W = Work done by the system 35 Adiabatic process 36  Some quick examples on the first law of thermodynamics Write down the first law of thermodynamics for: An isolated system (in an isolated system there is no exchange, hence….) U = Q − W During an adiabatic process (adiabatic means no heat exchange) Isobaric process Isovolumetric process Isothermal process 37  Some quick examples on the first law of thermodynamics Write down the first law of thermodynamics for: An isolated system (in an isolated system there is no exchange, hence….) U = Q − W U = 0 During an adiabatic process (adiabatic means no heat exchange) U = −W Isobaric process U = Q − P (V f − Vi ) Isovolumetric process U = Q Isothermal process  Vf  Joule’s Law stated that internal energy for an ideal gas is 0 = Q − nRT ln   only function of the temperature. If the temperature is  Vi  constant (isothermal process), the internal energy remains constant so that means the variation of energy ΔU = 0 38 First Law of Thermodynamics – Examples 39 First Law of Thermodynamics – Examples U = (Q ) − (W ) U = (0) − (− 10kJ ) = 10kJ 40 First Law of Thermodynamics – Examples What is the change in internal energy? 41 First Law of Thermodynamics – Examples What is the change in internal energy? U = (Q ) − (W ) U = (15kJ − 3kJ ) − (− 6kJ ) = 18kJ 42 First Law of Thermodynamics 43 First Law of Thermodynamics 44  First Law of Thermodynamics Mechanisms of Heat Transfer A closed mass energy Work Transfer involves only transfer: Mass Flow heat transfer and work 45  First Law of Thermodynamics Example: A system is subjected to a cycle composed by 3 processes as summarised in the following table. Calculate the missing values assuming all the quantities are measured in kJ. 46 U First Law of Thermodynamics Example: A piston in a circular cylinder (radius 10cm) moves up 5cm and the heat provided is 200J. The piston has a mass of 60kg (see figure) and on the top is acting a spring with an elastic constant k = 50kN/m. The atmospheric pressure as 100 kPa, calculate the variation of internal energy for the gas contained in the cylinder. 47 Thermodynamics for Aerospace Properties of Fluids Thermodynamics for Aerosapce – Dr. Burhan Saeed 1 Recap ∆U = Q − W 2 Example 1 3 Example 1 4  Example 2 Q9- A piston in a circular cylinder (radius 10cm) moves up 5cm and the heat provided is 200J. The piston has a mass of 60kg (see figure) and on the top is acting a spring with an elastic constant k = 50kN/m. The atmospheric pressure as 100 kPa, calculate the variation of internal energy for the gas contained in the cylinder. 5  Example 2 Q9- A piston in a circular cylinder (radius 10cm) moves up 5cm and the heat provided is 200J. The piston has a mass of 60kg (see figure) and on the top is acting a spring with an elastic constant k = 50kN/m. The atmospheric pressure as 100 kPa, calculate the variation of internal energy for the gas contained in the cylinder. Ans: The work done by the gas is the work necessary to lift up the weight (against the atmospheric pressure) and to compress the elastic spring. Hence: 1 W = mgh + kh 2 + Patm ⋅ A ⋅ h 2 1 W = 60 ⋅ 9.81 ⋅ 0.05 + ⋅ 50000 ⋅ 0.052 + 100000 ⋅ π ⋅ 0.12 ⋅ 0.05 = 249 J 2 The application of the first law of thermodynamics leads to: ∆U = Q − W = 200 J − 249 J = −49 J 6  Lecture Content Pure substance (phase and phase changes) Saturation temperature and pressure Latent heat Phase diagram 7  Pure Substance Pure substance: A substance that has a fixed chemical composition throughout. e.g. Nitrogen gas Air is a mixture of several gases, but it is considered to be a pure substance Nitrogen and gaseous air are pure substances. 8  Pure Substance A pure substance may exist in more than one phase, if the chemical composition is the same in all phases. A mixture of liquid and gaseous water is a pure substance, but a mixture of liquid and gaseous air is not. https://www.youtube.com/watch?v=pWZlICXw3Ng [Explaining Pure Substance] 9  Phases of a Pure Substance Solid Fluid Liquid Gas Molecules oscillates about Molecules can translate Molecules are not arranged their equilibrium position. and rotate freely. in particular order, they In a solid, shape and In a liquid, volume is fixed collide each other. volume are fixed. but shape depends on In a gas, volume and shape container. depend on container. 10  Phase Change Process If a liquid is well below its boiling point it is called a compressed liquid If we have a solid at a low temperature and start to add heat to it, the temperature of the solid will increase until it begins to melt If we continue to add heat to it, there will be no change in temperature until all the solid has melted After this point, addition of further heat will increase the temperature until it is just about to start vapourising A liquid in this state is called a saturated liquid 11 Phase Change Process At 1 atm and 20°C, water exists in the liquid phase (compressed liquid). At 1 atm pressure and 100°C, water exists as a liquid that is ready to vapourise (saturated liquid). 12  Phase Change Process Once we have a saturated liquid, addition of heat will cause all of the liquid to eventually turn into gas. The temperature of substance does not increase in this time. Once all the liquid has become gas, then the gas is called a saturated vapour. The gas can condense easily if no further heat is added. Addition of further heat will cause a temperature rise, and the gas will move into a superheated state. In this state, condensation is not about to occur. 13  Phase Change Process As more heat is At 1 atm pressure, the As more heat is transferred, part of the temperature remains transferred, the saturated liquid vapourises constant at 100°C until the temperature of the (saturated liquid–vapour last drop of liquid is vapour starts to mixture). vapourised (saturated rise (superheated vapour). vapour). 14  Phase Change Process: P-V and T-V diagrams Changes in phase under different conditions are often represented on temperature-volume (T-V) or pressure- volume (P-V) diagrams (volume is specific volume) 15  Phase Change Process: T-V diagram If the entire process between state 1 and 5 described in the figure is reversed by cooling the water while maintaining the pressure at the same value, the water will go back to state 1, retracing the same path, and in so doing, the amount of heat released will exactly match the amount of heat added during the heating process. T-v diagram for the heating process of water at constant pressure. 16  Phase Change Process: T-V diagram This T-V diagram is based on the pressure being 1 atm, but what would happen to the diagram if the pressure increased? T-v diagram for the heating process of water at constant pressure. 17  Saturation temperature and saturation pressure The temperature at which water starts boiling depends on the pressure; therefore, if the pressure is fixed, so is the boiling temperature. Water boils at 100°C at 1 atm pressure. Saturation temperature Tsat: The temperature at which a pure substance changes phase at a given pressure. Saturation pressure Psat: The pressure at which a pure substance changes phase at a given temperature. The liquid–vapor saturation curve of a pure substance (numerical values are for water). 18  Liquid-Vapour saturation curve You can see as pressure increases, and the that boiling point of the substance increases specific volume decreases. When we get to 22.09 MPa pressure, atm, the isobar has the characteristic that it becomes flat at only one single point - a horizontal inflection point. For water, this is the highest pressure for which a liquid and a vapour can coexist (and only at the single point in which the isobar is flat). For any higher pressure, water will always be in a single phase - no matter T-v diagram of constant-pressure phase-change processes of a pure what the temperature or specific volume substance at various pressures (numerical values are for water). are. The inflexion point is known as the critical point. 19  Liquid-Vapour saturation curve Note that for any state inside the solid red line dome-shape, vapour and liquid are both present. Anywhere outside the dome, only a single phase is present. Note also that the isobar that became flat at a single point coincides with the highest point of the two-phase dome. i.e. at the critical point. At the critical point, the temperature is known as the critical temperature and the pressure is known as the critical pressure. 20  Liquid-Vapour saturation curve Isobars below the isobar which shows the critical point are known as subcritical isobars. To the left of the dome region, the substance will exist as a liquid only. This is given various names: - Compressed liquid region - Single phase liquid region - Subcooled liquid region To the right of the dome region, the substance will exist as a vapour only, and is usually called the superheated vapour region Isobars above the critical point are known as supercritical isobars. The material in this region has properties somewhat intermediate between what most people would call a gas and what most would call a liquid. It is usually referred to as a supercritical fluid. 21  P-V Diagram A similar P-V diagram can be produced. In this case the lines are isothermal rather than isobaric. 22  The Phase diagram Critical Point – the temperature and pressure above which there is no distinction between the liquid and vapour phases. Triple Point – the temperature and pressure at which all three phases can exist in equilibrium. Sublimation – change of phase from solid to vapour. Vapourisation – change of phase from liquid to vapour. Condensation – change of phase from vapour to liquid. Fusion or Melting – change of phase from solid to liquid. Note! Vapour: British English Vapor: American English 23 The Phase diagram 24  Demo Videos Supercooled water demo https://www.youtube.com/watch?v=ph8xusY3GTM Superheated water demo https://www.youtube.com/watch?v=1_OXM4mr_i0 Superheated steam demo https://www.youtube.com/watch?v=R9uvIhgVz04 Supercritical fluid demo https://www.youtube.com/watch?v=QHcqyFm0i9M 25  Latent Heat The amount of energy required to cause a phase change is known as “latent heat” [https://www.youtube.com/watch?v=q6aDz6kY__s] The fact that the temperature does not change during melting is due to the fact that the heat energy added is required to actually melt the substance. This is known as the latent heat of fusion (melting) Similarly during boiling, the heat is required to vapourise the substance. This is known as the latent heat of vapourisation (boiling) The latent heat required depends on the pressure that heating or cooling is carried out at. 26  Latent Heat Inside the dome region, the “saturated liquid- vapour region”, liquid and vapour exist as a mixture in equilibrium. Let vf be the specific volume of liquid and vg be the specific volume of gas The total volume will be: mv = mfvf +mgvg The ratio of mass of saturated vapour to the total v = (m-mx)vf +(mx)vg m mass is called the quality of the mixture, designated by the symbol x. v = mvf -mxvf + mxvg m mg v = vf -xvf + xvg x= m v = vf +x(vg-vf) 27  Latent Heat (a simple example) Example: Calculate the total heat required to convert 7 kg of water at 100 °C into steam at 100 °C at standard atmospheric pressure. Latent heat of vaporization of water is 2256.5 kJ/kg. 28  Latent Heat (a simple example with solution) Example: Calculate the total heat required to convert 7 kg of water at 100 °C into steam at 100 °C at standard atmospheric pressure. Latent heat of vaporization of water is 2256.5 kJ/kg. Total amount of heat required is = 7 kg * 2256.5 kJ/kg = 15795.5 kJ 29  Thermodynamic Property Tables Extensive tabulations of thermodynamic property data have been made for many substances. When dealing with water, these are commonly referred to as “steam tables”. Booklets of steam tables are available to purchase, or can be found in appendices to textbooks etc. 30  Thermodynamic Property Tables Saturated water - Temperature Table – Gives the saturation properties of water as a function of saturation temperature Saturated water - Pressure Table – gives the saturation properties of water as a function of saturation pressure (Note that in the saturation region temperature and pressure are dependant!) Superheated Water – Gives the properties of superheated steam (Note that outside of the saturation region temperature and pressure are independent, and both are needed to specify other properties!) Ideal Gas Properties of Air 31  Thermodynamic Property Tables Enthalpy – A Combination Property In steam turbine analysis a combination of properties 𝑢𝑢 + 𝑃𝑃𝑃𝑃 are often used, therefore, for simplicity a new property is defined ‘enthalpy’. Specific enthalpy: ℎ = 𝑢𝑢 + 𝑃𝑃𝑃𝑃 (kJ/kg) Total Enthalpy: 𝐻𝐻 = 𝑈𝑈 + 𝑃𝑃𝑃𝑃 (kJ) 32  Thermodynamic Property Tables Saturated Liquid and Saturated Vapour States There are two forms of tables , one give properties under temperature and the other under pressure – both give same information. Subscript g: saturated vapour Subscript f: saturated liquid Subscript fg: the difference between the saturated vapour and saturated liquid values 33  Thermodynamic Property Tables – Linear Interpolation The steam tables provide saturation properties for certain intervals of Pressure/Temperatures. To obtain values in between these intervals the following linear interpolation method is deployed. 𝑦𝑦 − 𝑦𝑦2 𝑦𝑦3 − 𝑦𝑦2 = 𝑥𝑥 − 𝑥𝑥2 𝑥𝑥3 − 𝑥𝑥2 𝑦𝑦3 − 𝑦𝑦2 𝑦𝑦 = 𝑦𝑦2 + (𝑥𝑥 − 𝑥𝑥2 ) 𝑥𝑥3 − 𝑥𝑥2 34  Thermodynamic Property Tables – Linear Interpolation If the saturation temperature is 2120C, the corresponding specific volume can be found using the linear interpolation as following. 35  Thermodynamic Property Tables - Example Determine the volume change when 10 kg of saturated water is completely vapourised at a pressure of (a) 1 kPa, (b) 260 kPa, and (c) 10 000 kPa. 36  Thermodynamic Property Tables - Example Determine the volume change when 10 kg of saturated water is completely vaporized at a pressure of (a) 1 kPa, (b) 260 kPa, and (c) 10 000 kPa. Download Appendix1_SIunits.pdf (Canvas) provides the necessary values. The quantity being sought is ΔV = mvfg where vfg = vg – vf. Note that P is given in MPa. (a) 1 kPa = 0.001 MPa. Thus, vfg = 129.2 – 0.001 = 129.2 m3 /kg.... ΔV = 1292 m3 (b) At 0.26 MPa the values need to be interpolated from the tables. The tabulated values are used at 0.2 MPa and 0.3 MPa: The value for vf is, to four decimal places, 0.0011 m3 /kg at 0.2 MPa and at 0.3 MPa; hence, no need to interpolate for vf. We then have v fg = 0.718 − 0.0011 = 0.717 m3 / kg. ∴ ΔV = 7.17 m3 37  Thermodynamic Property Tables - Example Determine the volume change when 10 kg of saturated water is completely vaporized at a pressure of (a) 1 kPa, (b) 260 kPa, and (c) 10 000 kPa. Table Appendix B provides the necessary values. The quantity being sought is ΔV = mvfg where vfg = vg – vf. Note that P is given in MPa. (c) At 10 MPa, vfg = 0.01803 − 0.00145 = 0.01658 m3 /kg so that ΔV = 0.1658 m3 Notice the large value of vfg at low pressure compared with the small value of vfg as the critical point is approached. [http://scienceuniverse101.blogspot.com/2014/10/properties-of-pure-substancesthe-p-v-t.html#.X7P74Gj7QuW] 38 Thermodynamics for Aerospace– Lecture 4 Gas Laws and Ideal gas Dr Burhan Saeed 1  Table of contents Ideal gas, gas laws and equation of state. Specific heat and adiabatic process. 2  What is an ideal gas? Gas is formed by particles (molecules) that can be considered as small hard spheres. Motion is governed by Newton’s laws. All the particles moves frictionless. During the motion particle collide and collisions are perfectly elastic (no energy is lost). The average distance among molecules is larger than the size of the molecules. 3  What are the physical quantities affecting gas behavior? Pressure Temperature Volume Number of particles 4  Gas Laws Boyle’s Law Charles’ Law Gay-Lussac’s Law Avogadro’s Law (1662) (1780) (1809) (1811) Pressure Volume Pressure Volume Volume Temperature Temperature Particles Equation of state for an ideal gas - Emil Claperyon (1834) Pressure (P) Volume (V) Particles (n) Temperature (T) 5  Boyle’s Law This law was derived by Robert Boyle (1662) Pressure (P) Volume (V) P V = K Particles (n) Temperature (T) Boyle’s law states that at constant temperature for a fixed mass, the absolute pressure and the volume of a gas are inversely proportional. The law can also be stated in a slightly different manner, that the product of absolute pressure and volume is always constant. P V = K → P1 V1 = P2 V2 6 Boyle’s Law 7 Boyle’s Law P1 V1 = P2 V2 http://www.grc.nasa.gov/WWW/k-12/airplane/aboyle.html 8  Charles’ Law This law was derived by Jacques Charles (1780) Pressure (P) V1 V2 Volume (V) = Particles (n) T1 T2 Temperature (T) Charles’ law (or law of volumes) is a law that describes how gases tend to expand when heated. Charles law states that at constant pressure for a fixed mass, the volume of an ideal gas increases or decreases by the same factor as its temperature (i.e. the gas expands as the temperature increases). 9 Charles’ Law 10 Charles’ Law V1 V2 = T1 T2 http://www.grc.nasa.gov/WWW/k-12/airplane/glussac.html 11  Gay-Lussac’s Law This law was derived by Joseph Louis Gay-Lussac (1809) Pressure (P) P1 P2 Volume (V) = Particles (n) T1 T2 Temperature (T) Gay-Lussac’s law states that for a fixed mass and a fixed volume, the pressure of an ideal gas is proportional to the gas absolute temperature. P1 P2 = → P1  T2 = P2  T1 T1 T2 12 Gay-Lussac’s Law P P1 P2 = T1 T2 13  What is a mole? A mole of any substance it is the amount of substance that contains an Avogadro’s number of particles (i.e. atoms or molecules). N A = 6.022 1023 Example: 1 mole of water (H2O) contains N A = 6.022 1023 molecules of water (H2O) 1 mole of water is just 18 grams. Imagine how many molecules of water (H2O) you drink every day! H mass O number _ of _ moles = Molar _ mass m H n= M 14 Values of atomic and molecular masses 15  An example on how to calculate the number of moles What is the mass of 0.1934 kmol of CO2? 16  An example on how to calculate the number of moles What is the mass of 0.1934 kmol of CO2? We know from the previous table that 1 mol of CO2 has a mass of 44 grams. This means that 1kmol of CO2 has a mass of 44000 g or 44 kg. Hence the mass of 0.1934 kmol of CO2 is: 0.1934  44kg = 8.5096kg 17  In class problem 2.5 kg of air with an initial pressure (100 kPa), volume of 2m3 and temperature of 300K is compressed isothermally (temperature is kept constant) to a pressure 200 kPa. Calculate the number of moles of air and the final volume. 18  In class problem (solution) 2.5 kg of air with an initial pressure (100 kPa), volume of 2m3 and temperature of 300K is compressed isothermally (temperature is kept constant) to a pressure 200 kPa. Calculate the number of moles of air and the final volume. From the table we know that 1kmol (1000 moles) of air has a mass of 29kg. Hence 2.5kg represent 86.2069 moles: 2.5kg 1000  = 86.2069mol 29kg The quantity of the gas and temperature are kept constant. Hence, we can use the Boyle’s law: P1 V1 = P2 V2 P1 V1 100kPa  2m3 V2 = = = 1m3 P2 200kPa 19  Avogadro’s Law This law was derived by Amedeo Avogadro (1811) Pressure (P) V Volume (V) =K Particles (n) n Temperature (T) Avogadro’s law states that, under the same condition of temperature and pressure, equal volumes of all gases contain the same number of molecules V V1 V2 =K → = n n1 n2 20 Avogadro’s Law Avogadro’s law states that, under the same condition of temperature and pressure, equal volumes of all gases contain the same number of molecules (in this example number of molecules N = 1024). 21 Avogadro’s Law 22  Gas laws and equation of state The Equation of State (EoS) for an ideal gas includes all these laws together. Boyle, Charles, Gay-Lussac and Avogadro’s law can be retrieved from the EoS. Boyle’s Law Gay-Lussac’s Law (1662) (1809) Equation of state for an ideal gas Charles’ Law Avogadro’s Law (1780) (1811) 23  Ideal Gas – Equation of State (EoS) The equation of state for an ideal gas is: Formulae to remember PV = nRT Universal Gas Constant J R = 8.314 mol  K PV = nRT Numbers of moles Absolute temperature Pressure (measured in mol) (measured in K) (measured in Pascal) Volume (measured in m3) 24  An application of EoS – In class problem A spherical balloon with a diameter of 6 m is filled with helium at 20 oC and 200 kPa. Determine the mole number and the mass of the helium in the balloon. (molar mass of helium is 0.004 kg/mol) R = 3m (from Y.A. Cengel – M.A. Boles – Thermodynamics – An Engineering Approach – 6th ed. 2007) (Problem 3.71) 25  An application of EoS – In class problem (solution) A spherical balloon with a diameter of 6 m is filled with helium at 20 oC and 200 kPa. Determine the mole number and the mass of the helium in the balloon. (molar mass of helium is 0.004 kg/mol) PV = nRT R = 3m 200 10 Pa   (3m ) 4 3 3 PV 3 n= = = 9276mol RT 8.314 J  293.15K mol  K kg mass = 9276mol  0.004 = 37.10kg mol (from Y.A. Cengel – M.A. Boles – Thermodynamics – An Engineering Approach – 6th ed. 2007) (Problem 3.71) 26 Specific Heat 27  Specific Heat Q = m Cv (T2-T1) Q = Heat; m = Mass T2-T1 = Temperature change Q = m Cp (T2-T1) Q = Heat; m = Mass T2-T1 = Temperature change 28  Internal energy – Mono-atomic gas Internal energy is all the energy of a system associated with its microscopic components (atoms and molecules). Mono-atomic gas (Helium, Neon, Argon) z 1 uz = nRT 2 1 3 u y = nRT uint = nRT 2 2 y 1 x ux = nRT 2 Imagine that the atom is just a small point (not as represented in figure). The concept of rotation is meaningless for a point. This is why for the total energy we have only contributions of the 3 translations. 29  Internal energy – Di-atomic gas Di-atomic gas (Hydrogen) 1 uz = nRT 2 z 1 urot _ z = nRT 2 1 uy = nRT 5 2 uint = nRT y 2 u rot _ y = 0 1 x ux = nRT 2 1 urot _ x = nRT 2 This is why for the total energy we have only contributions of 3 translations and 2 rotations. 30  Some useful values and formula for cv Mono-atomic gas (Helium, Neon, Argon) Di-atomic gas (Hydrogen) z z 5 3 uint = nRT uint = nRT 2 2 y y x x uint = ncv T 3 5 cv = R cv = R 2 2 31  Some useful values and formula for cv and cp What about the molar specific heat at constant pressure cp? Mono-atomic gas (Helium, Neon, Argon) Di-atomic gas (Hydrogen) Mayer’s formula z z c p = cv + R y y x x Julius Robert von Mayer 3 5 cv = R cv = R 2 2 5 7 cp = R cp = R 2 2 32  Some useful values and formula for cv and cp 3 5 Mono-atomic gas (Helium, Neon, Argon) cv = R cp = R  = 1.67 2 2 5 7 Di-atomic gas (Hydrogen) cv = R cp = R  = 1.4 2 2 cp = cv 33  The adiabatic process of an ideal gas When an ideal gas undergoes a volume change for which no heat was allowed to be added or removed from the system PVk = Constant; k = Cp/Cv 34 Different process on the PV (Clapeyron) plane 35  Thermodynamics for Aerospace Lecture 5 Energy Analysis Steady Flow, Control Volume Dr. Burhan Saeed 1  Table of contents Volume and mass flow rate. Conservation of mass. Flow work or flow energy Energy analysis of a steady flow, control volume and first law of thermodynamics And Energy analysis of a steady flow with Bernoulli’s equation, Efficiency 2  Recap Specific Heat Specific Heat changes with Temperature!! Q = m Cv (T2-T1) or Q = m CP (T2-T1) Q = Heat; m = Mass T2-T1 = Temperature change cp uint = ncv T = cv Pioro, Igor & Zvorykin, C.. (2016). Appendix A3. Thermophysical properties of fluids at subcritical and critical/supercritical conditions1. 10.1016/B978-0-08-100149-3.15003-7. 3 Problem 1– In Class 4 Problem 1– In Class 5 Problem 1– In Class 6 Problem 2– In Class 7 Problem 2– In Class 8 Problem 2– In Class 9 Problem 2– In Class 10 Fluid flow: Volume flow rate (Q) t = 0s t = 0.2 s t = 0.5 s In 1 second this is the t = 1.0 s volume of fluid that passes through the line 11  Fluid flow: Volume flow rate (Q) v In 1 second this is the A t = 1.0 s volume of fluid that passes through the line L = v t Formulae to remember Volume A  L A  v  t Q= = = = Av Volume Time t t Q= Time The volume flow rate, is the ratio between fluid volume and time Q = Av 12  A simple example We assume that the flow in a river is uniform with a velocity v = 0.5 m/s. The geometric dimensions of the channel are given in the figure. Formulae to Calculate the volume flow rate. remember Calculate the volume of water collected in 5 seconds. Volume Q= Time Q = Av b = 10m v = 0.5 m s h = 5m 13  A simple example (solution) We assume that the flow in a river is uniform with a velocity v = 0.5 m/s. The geometric dimensions of the channel are given in the figure. Formulae to Calculate the volume flow rate. remember Calculate the volume of water collected in 5 seconds. Volume Q= Time Q = Av b = 10m v = 0.5 m s Volume flow rate. m m3 Q = A  v = 10m  5m  0.5 = 25 s s h = 5m Volume of water collected in 5 seconds. Volume m3 Q= → Volume = Q  Time = 25  5s = 125m 3 Time s Volume = 125m 3 = 125000liters 14  Fluid Flow: Mass flow rate ( m ) Mass = Density Volume v In 1 second this is the mass A t = 1.0 s of fluid that passes through the line L Formulae to Mass  Volume remember m= = =  Q =  v  A Mass Time Time m= Time The mass flow rate, m , is the ratio between fluid mass and time m =  Q 15  Mass and volume flow rate We can use average velocity Vavg over area Ac for getting mass flow rate Volume Flow rate in terms of average velocity over are Ac Then, 16 Conservation of mass 17  Conservation of mass The time rate change of mass within the control volume plus the net mass flow rate through the control surface is equal to zero 18  Conservation of mass The total rate of mass entering a control volume is equal to the total rate of mass leaving it 19 Conservation of mass 20 Example 21 Example (solution) 22 Example (solution) 23  Flow Work or Energy We know that an open system is one where mass can transfer across the boundary For a closed system we were able to neglect the kinetic and potential energy changes. For an open system the KE and the PE may change In an open system where there is flow, there is also some additional energy or work required to cause the flow. This is called flow energy. 24  Flow Work or Energy Consider flow into an open system A1 L1 L2 Flow enters at a pressure P1 and leaves at a pressure P2 The inlet volume V1=A1L1 and the outlet volume V2=A2L2 The resistance force that prevents entry of mass at the inlet is F=P1A1 The work done in moving mass along the inlet is W=P1A1L1=P1V1 This is the energy added to the system as mass flows in Similarly the energy leaving the system as mass flows out is P2V2 25  Flow Work or Energy A1 L1 L2 Change in system energy 26  Applying First Law to Open Systems.. d 1 2 Q− W = [ms ( cs + gz s + u s )] dt 2 Adding in the terms for flow energy, we now have:...... 1 2. 1 2 Q − W + m1 ( c1 + gz1 + u1 + Pv 1 1 ) − m 2 ( c2 + gz2 + u2 + P2v2 ) = 2 2 d 1 2 [ms ( c + gz + u )] dt 2 27  Applying First Law to Open Systems.... 1 2... 1 2 d 1 2 Q − W + m1 ( c1 + gz1 + u1 + P1v1 ) − m 2 ( c2 + gz2 + u2 + P2v2 ) = [ms ( c + gz + u )] 2 2 dt 2 Here it is useful to use enthalpy H=PV+U, or h=Pv+u, [Enthalpy is a property of a thermodynamic system, defined as the sum of the system's internal energy and the product of its pressure and volume. It is a convenient state function standardly used in many measurements in chemical, biological, and physical systems at a constant pressure.].... 1 2... 1 2 d 1 Q − W + m1 ( c1 + gz1 + h1 )− m 2 ( c2 + gz 2 + h2 ) = [ms ( c 2 + gz + u )] 2 2 dt 2.... 1 2... 1 2 dEs Q − W + m1 ( c1 + gz1 + h1 )− m 2 ( c2 + gz 2 + h2 ) = 2 2 dt 28  Applying First Law to Open Systems...... 1 2. 1 2 dEs Q − W + m1 ( c1 + gz1 + h1 )− m 2 ( c2 + gz 2 + h2 ) = 2 2 dt 29 Energy analysis of a steady flow with Bernoulli’s equation Mechanical Efficiency 30 Steady flow 31  Bernoulli equation So far, we have used the Bernoulli equation as a “conservation of energy”. 1 1 P1 + V12 + gh1 = P2 + V22 + gh2 2   2   Input Output The motion of the fluid was driven just by difference in pressure, height and velocity. 32 Nozzles and diffusers 33  Energy analysis of steady flow- Bernoulli’s Equation We want to generalize the formulation with external mechanical inputs and outputs. Many times we will have just one input and one output. Input System Output Mechanical loss In reality we are not able to transform all the energy (provided by the input) in an useful energy used by the output. We have some mechanical loss (waste of energy). 34  Energy analysis of steady flow- Bernoulli’s Equation We want to generalize the formulation with external mechanical inputs and outputs. 1 1 P1 + V12 + gh1 = P2 + V22 + gh2 2   2   Input Output We can divide both members for the density:  P11 2 P2 1 2 + V1 + gh1 = + V2 + gh2    2     2 Input Output We can multiply both members for the mass flow rate: m  P1 1 2   P2 1 2  m + V1 + gh1  = m + V2 + gh2   2  2     Input Output 35  Energy analysis of steady flow- Bernoulli’s Equation Let’s see the unit of measure of the different terms in this equation.  P1 1 2   P2 1 2  m + V1 + gh1  = m + V2 + gh2    2    2     Input Output P1 For example the term: m  kg m is the mass flow rate and it is measured in : s N P is the pressure and it is measured in : Pa = 2 m kg  is the density and it is measured in : 3 m P Hence the term, m 1 , is measured in:  Joule (is the unit of measure of work & energy) kg N m3 N  m J  2 = = =W s m kg s s Watt (is the unit of measure of power) 36  Energy analysis of steady flow- Bernoulli’s Equation Substantially the equation is a balance of power (input vs. output)  P1 1 2   P2 1 2  m + V1 + gh1  = m + V2 + gh2    2    2     Input Output We can introduce a mechanical power as input (pump, fan, compressor) and a mechanical output (turbine)........off course we have also some power loss. Example of a centrifugal pump  P1 1 2  P 1 2  m + V1 + gh1  + w pump = m + V2 + gh2  + wturbine+ wloss 2  2  2         Input Output 37  Energy analysis of steady flow- Bernoulli’s Equation Turbines and compressors 38  Energy analysis of steady flow  P1 1 2  P 1  m + V1 + gh1  + w pump = m 2 + V2 + gh2  + wturbine+ wloss 2  2  2         Input Output Power of the flow (Output) Power of the flow  P2 1 2  m + V2 + gh2   2 (Input)  P1 1 2   m + V1 + gh1   2  Power of the turbine Power of the pump (Output) (Input) w pump wturbine Power loss w loss 39  Energy analysis of steady flow including heat- Bernoulli’s Equation  P1 1 2  P 1  m + V1 + gh1  + w pump = m 2 + V2 + gh2  + wturbine+ wloss 2  2  2         Input Output  P1 1 2  P 1  m + V1 + gh1  + w pump + Q in = m 2 + V2 + gh2  + wturbine+ wloss + Q out 2  2  2       Input Output 40 Efficiency 41 Efficiency 42  Application of energy analysis The pump of a water distribution system is powered by a 15-kW electric motor whose efficiency is 90 percent (see figure below). The water flow rate through the pump is 50 L/s. The diameters of the inlet and outlet pipes are the same, and the elevation difference across the pump is negligible. If the absolute pressures at the inlet and outlet of the pump are measured to be 100 kPa and 300 kPa, respectively, determine (a) the mechanical efficiency of the pump. from Cengel et al. “Fundamentals of Thermal-Fluid Science” 43  Application of energy analysis We can use the equation that express a balance of powers:  P1 1 2  P 1 2  m + V1 + gh1  + w pump = m + V2 + gh2  + wturbine+ wloss 2  2  2         Input Output As stated in the text: “...the elevation difference across the pump is negligible.”. That means h1 = h2. As stated in the text: “The diameters of the inlet and outlet pipes are the same...”. That means A1 = A2 and hence for the continuity equation v1 = v2. 44  Application of energy analysis We can simplify the previous formula:  P1 1 2  P 1 2  m + V1 + gh1  + w pump = m + V2 + gh2  + wturbine+ wloss 2  2   2   P1  P  m  + w pump = m  + wturbine+ wloss 2   There is no turbine in the system, hence: wturbine = 0 There is no mechanical loss for stage 2: wloss = 0 Hence the equation can be (further) simplified:  P1  P  m  + w pump = m 2    45  Application of energy analysis  P1  P  m  + w pump = m 2    From the text we know P1 = 100 kPa, P2 = 300 kPa and the volume flow rate: L m3 Q = 50 = 0.05 Mass flow rate m s s In this equation we know everything except: w pump 46  Application of energy analysis The pump provides a power of 10 kW (to the fluid). From the text we know: “The pump of a water distribution system is powered by a 15-kW electric motor whose efficiency is 90 percent....”. That means that the motor is not providing all the 15kW of power because its efficiency it is not 100%, but only 90%. Hence, the motor is providing a power: 90 wmotor = 15kW  = 13.5kW 100 The motor is providing a power of 13.5 kW to the pump However , the pump provides just 10 kW. We can calculate the efficiency of the pump: Power _ output 10kW  pump = 100 = = 74.1% Power _ input 13.5kW 47  Application of energy analysis 13.5kW Motor  motor = 100 = 90.0% 15kW 15kW 13.5kW 10kW Pump  pump = 100 = 74.1% 13.5kW 13.5kW 10kW 48 School of Physics, Engineering & Computer Science Processes and power cycles by Dr Chaoyue Ji 5ENT1129 Thermodynamics for Aerospace 19th November 2024 Thermodynamic Processes Define manner in which a thermodynamic system changes from an initial state to a final state Characterised according to what parameters are constant during process Pop-pop boat video 3 Enter the page title A closed system does not allow mass transfer A closed system can across the boundary have a moving boundary  Isobaric Process Occurs at constant pressure Classic example Gas heated in a cylinder fitted with a movable frictionless piston Pressure = pressure due to piston weight + external atmospheric pressure Real example Boiler Superheater Energy of steam is increased without increasing its associated pressure  Enter the page title Rankine cycle and P – V diagram Boiler: 1 - B, The working fluid (water) is heated until it reaches saturation (phase change / boiling point) in a constant-pressure process. Boiler: B - 2, Once saturation is reached, further heat transfer takes place at constant pressure, 1 until the working fluid is completely vaporised 2 (quality of 100% / dry steam) Turbine: 2 - 3, The vapour is expanded