Engineering Thermodynamics (MEE1003

Engineering Thermodynamics (MEE1003) Dr. M. Sreekanth Associate Professor, SMBS VIT University, Chennai Campus Email: manavalla.sreekanth@vit.ac.in MODULE-1 Basic Concepts in Thermodynamics Thermodynamics • The name The od a i s is derived by combining the Greek words Therme meaning heat and Dynamis meaning Power • The subject of Thermodynamics deals with the conversion of Heat to Work (Power Generation) and vice versa (Refrigeration) • Even the Relationship among properties of matter is dealt in Thermodynamics 3 Conservation of Energy • The principle of Co se atio of E e g is a fundamental law of nature • It states that E e g can neither be created nor dest o ed • Energy can only be changed from one form to another • A rock falling from a cliff has its potential energy converted to kinetic energy 4 1st and 2nd Laws of Thermodynamics • The 1st Law of Thermodynamics is simply as assertion of the Principle of Conservation of Energy • It deals with quantities of energy • The 2nd Law of Thermodynamics deals with quantity as well as quality of energy • It tells whether a process is possible or not, which the 1st Law cannot tell 5 Application of Thermodynamics • Thermodynamics is everywhere • The body heat generated due to metabolic activity is continuously rejected to the environment • A residential building having solar water heaters, refrigerator, humidifier etc is an example of application of Thermodynamics 6 System • A System is a fixed quantity or a region in space chosen for study • The mass or region outside the system is called the Surroundings • The real or imaginary surface that separates the System from the Surroundings is called the Boundary 7 Closed System (Control Mass) • A Closed System is one which does not permit mass to cross the system boundary. However, energy can cross • Energy can be in the form of Heat and Work • If even the energy is not permitted to cross the boundary, the system would be an Isolated System 8 Closed System with Moving Boundary • The s ste ou da does t ha e to e fi ed 9 Open System (Control Volume) • If a system permits both mass and energy across its boundary, it is called an Open System • A large number of engineering problems involve flow of mass in and out of a system • Any arbitrary region in space can be selected as a control volume • A proper choice of control volume can make the analysis simpler 10 Control Volumes 11 Property • A distinguishing characteristic of a system is called as a Property, E.g. P, T, V • Intensive Properties are those which are independent of mass, E.g. P, T, Density • Extensive Properties are those which depend on mass, E.g. Mass, Volume, Energy • Intensive Properties are denoted by capital letters while Extensive Properties are denoted by small letters 12 Method to Determine Extensive or Intensive Property 13 State • The state of a system is its condition • The state is completely described by a set of properties (e.g. T, P, v) • It is not necessary to specify all the properties to describe a state • The state postulate says that the state of a simple compressible system is completely specified by two independent intensive properties 14 Equilibrium • • • • Thermodynamics deals with equilibrium states Equilibrium implies balance There are several kinds of Equilibrium In an equilibrium state, there are no unbalanced potentials • Thermodynamics is concerned with Thermal, Mechanical and Chemical equilibrium 15 Thermal Equilibrium • A system is in Thermal Equilibrium if its temperature is equal throughout the system • After attaining equilibrium, there will be no thermal gradients responsible for heat transfer 16 Mechanical and Chemical Equilibrium • Mechanical equilibrium is said to have been attained if the pressure in the system does not change with time • The pressure may vary within the system • Chemical equilibrium is said to have been attained when the chemical composition does not change with time • For a system to be in Thermodynamic equilibrium, it must be in thermal, mechanical and chemical equilibrium at the same time 17 Processes and Cycles • Process is the change in state a system undergoes from one equilibrium state to the other • The series of states through which a system passes during a process is called the Path 18 Quasi-Static Process • It is a process which proceeds in such a manner that it is infinitesimally close to an equilibrium state at all times 19 Quasi-Stati P o ess… • It is an idealized process and is not a true representation of an actual process • However, most processes are close to quasistatic process • They are easy to analyze and • A work producing device produces maximum work when operating in a quasi-static process 20 Process Diagrams • Process diagrams plotted with thermodynamic properties as coordinates are useful in visualizing the process 21 Various Processes • Isothermal Process: Temperature remains constant, E.g. Phase change processes • Isobaric Process: Pressure remains constant, E.g. Heat addition in diesel engine • Isochoric Process: Volume remains constant, E.g. Heat addition in petrol engine • Isentropic Process: Entropy remains constant, E.g. Expansion in steam turbine 22 Cycle • A system is said to have undergone a cycle if it returns to its initial state at the end of the process 23 Steady State Process • A process during which a fluid flows through a control volume steadily 24 Zeroth Law of Thermodynamics • If two bodies are separately in thermal equilibrium with a third body, then they are in thermal equilibrium with each other • Two bodies are in thermal equilibrium if both have the same temperature reading even if they are not in contact 25 Pressure • Pressure is defined as the normal force exerted by a fluid per unit area • Absolute pressure is the actual pressure • Gauge (or Gage) pressure is the pressure over and above atmospheric pressure • Vacuum pressure is the pressure below atmospheric pressure • 1 atmosphere = 101325 Pa = 1.01325 bar 26 Absolute, Gauge and Vacuum Pressures Pgauge = Pabs – Patm Pvac = Patm - Pabs 27 Forms of Energy • Energy can exist in various forms • The sum of all forms of energy is known as Total Energy (E) • Thermodynamics deals with changes in total energy and not its absolute value • Macroscopic forms of energy are those a system possesses as a whole with respect to some outside reference frame (E.g KE, PE) • Microscopic forms of energy are those related to the molecular structure of a system and the degree of molecular activity. • Microscopic energy is independent of outside reference frames • The sum of all the microscopic forms of energy is called the 28 Internal Energy Fo s of E e g … • Energy can exist as thermal, mechanical, kinetic, potential, electric, magnetic, chemical, and nuclear forms • The energy possessed as a virtue of motion is called as Kinetic Energy • That energy that a system possesses as a result of its elevation is called Potential Energy 29 Heat Energy • Heat is the form of energy that is transferred between two systems (or a system and its surroundings) by virtue of temperature difference • Heat is energy in transition • It is recognized only as it crosses the system boundary • A process during which there is no heat transfer is called an Adiabatic System 30 Work • Work is an energy interaction between a system and its surroundings • Only heat and work cross system boundaries of a closed system • If the energy crossing the boundary of a closed system is not heat, then it must be work • Work is the energy transfer associated with a force acting through a distance, e.g. rising piston, rotating shaft etc. 31 Similarities between Heat and Work • Both are recognized at the boundaries of a system • Both are associated with processes and not states • Both are path functions 32 Path and Point Functions 33 Sign Convention (+) (-) (-) (+) 34 Total Energy • The magnetic, electrical, surface tension effects are significant only in few specialized cases only and are usually neglected • Then the total energy of a system is given as E = U + KE + PE • Most closed systems remain stationary during a process and have no change in KE or PE • Such systems are called as Stationary Systems 35 Mechanical Energy • Some engineering systems are designed to transport a fluid at a certain rate, velocity, pressure • The fluid may be used in a turbine to produce power • They may also consume energy (pump, fan) • Not all systems deal with heat • Such systems can be analyzed by considering only the mechanical form of energy 36 Me ha i al E e g … • Mechanical Energy is that form of energy that can be converted to mechanical work completely and directly by an ideal mechanical device such as an ideal turbine • KE and PE are forms of Mechanical Energy but not Thermal Energy emech Emech P V2    gz  2  P V2  m    gz    2 37 Change in Mechanical Energy emech  Emech P2  P1  V22  V12   g z 2  z1  2   P2  P1 V22  V12  m    g z 2  z1  2    The mechanical energy of a fluid does not change if the pressure, density, velocity and elevation remain constant 38 Mechanical Forms of Work • Shaft Work: • Spring Work: • Elastic Work: 2n Wshaft  60 Wspring   1 k x22  x12 2  Welastic    n Adx 2 1 39 MODULE 2 First Law of Thermodynamics 1st Law of Thermodynamics • Based on experimental observations, the 1st law states that energy can neither be created nor destroyed during a process; it can only change forms • This law is also known as the conservation of energy principle • It provides a sound basis for studying the relationships among various forms of energy and energy interactions 41 Case Study 1: Falling Rock • The decrease in PE exactly equals the increase in KE, if the air resistance is neglected 42 Case Study 2: System undergoing Adiabatic Processes • A system is undergoing a series of adiabatic processes from state 1 to state 2 • The net work has been found to be same regardless of the nature of the closed system and the details of the process • That means the value of net work depends on the end state only and it must correspond to a change in property of the system • This property is the Total Energy 43 Case Study 3: Baked Potato 44 Case Study 4: Heating Water 45 Case Study 5: Electric Room Heater 46 Case Study 6: Compressed Air 47 Case Study 7: System Involving Heat and Work Interaction 48 Energy Balance • The net change in the total energy of the system during a process is equal to the difference between the total energy entering and the total energy leaving the system during the process • Ein – Eout = Esystem • Esystem = Efinal – Einitial • Energy is a property of a system • E = U + KE + PE • For stationary system, KE and PE are equal to zero and hence E = U 49 Mechanisms of Energy Transfer • Energy can be transferred to or from a system in three forms: Heat, Work and Mass Flow • For a closed system, energy transfer is by heat and work transfer only • For an open system, the mass flow is included along with heat and work transfer • Ein – Eout = (Qin-Qout)+(Win-Wout)+(Emass, in-Emass,out) = Esystem • The above equation is known as the Energy Balance Equation 50 Other Forms of Energy Equation • Compact Form: Ein – Eout = Esystem • Rate Form: E in  E out  dEsystem dt • Specific Form: ein – eout = esystem • Differential Form: ein  eout  desystem 51 Closed System in a Cycle • For a closed system undergoing a cycle, the initial and final states are same and hence E = E1-E2 = 0. •  Wnet= Qnet 52 Problem 1 A classroom that normally contains 40 people is to be air-conditioned with a window A/C units of 5 kW cooling capacity. A person at rest may be assumed to dissipate heat at a rate of about 360 kJ/h. There are 10 light bulbs in the room, each with a rating of 100 W. The rate of heat transfer to the classroom through the walls and the windows is estimated to be 15,000 kJ/h. If the room air is to be maintained at a constant temperature of 21 0C, determine the number of window A/C units required. 53 Problem 2 Consider a room that is initially at the outdoor temperature of 20 0C. The room contains a 100 W light bulb, a 110 W TV set, a 200 W refrigerator, and a 1000 W iron. Assuming no heat transfer through the walls, determine the rate of increase of the energy content of the room when all of these devices are on. 54 Problem 3 A fan is to accelerate quiescent air to a velocity of 10 m/s at a rate of 4 m3/s. Determine the minimum power that must be supplied to the fan. Take the density of air to be 1.18 kg/m3. 55 Moving Boundary Work (Pdv Work) • In closed systems like a gas in a piston-cylinder arrangement, a part of the boundary moves back and forth • This is the form of work in automobile engines 56 Moving Boundary Work in Quasi Equilibrium Process Wb  Fds  PAds  Pdv Wb   Pdv 2 1 57 PdV Work-Graphical Representation The area under the process curve on a P-V diagram is equal, in magnitude, to the work done during a quasi-equilibrium expansion or compression process of a closed system 58 PdV Work for Various Processes • Constant Volume Process: Wb   Pdv  0 as dv  0 2 1 • Constant Pressure Process: Wb   Pdv  P0  dv  P0 v2  v1  2 2 1 1 59 PdV Wo k fo Va ious P o esses… • Isothermal Process: Pv=mRT0=C or P=C/v C dv v v v Wb   Pdv   dv  C   C ln 2  P1V1 ln 2  P2V2 ln 2 v v v1 v1 v1 1 1 1 2 2 2 • Polytropic Process: PVn=C or P=CV-n  n 1  n 1  v v P v  Pv Pv  P v 1 Wb   Pdv   Cv n dv  C 2  2 2 11 11 2 2  n 1 1 n n 1 1 1 2 2 60 Joule s E pe i e t • Measurements: Weight and Temperature • Variables: Materials and Weights • Conclusion:  Q   W 61 E te sio of Joule s ‘esult to a P o ess For cycle 1 - A - 2 - B - 1  Q   Q   W   W 2 1 2 A 1 B 1 A 2 1 B (i) 2 For cycle 1 - C - 2 - B - 1  Q   Q  W   W 2 1 2 C 1 B 1 C 2 1 B (ii) 2 S ubtracting (ii) from (i)  Q   Q 2 2 A   WA   WC 2 C 2  Q  W   Q  W  1 1 1 2 1 2  Q  W  depends only on end states A 1 C 1  Q  W  is a property Q  W  dE 62 Energy Balance in Closed Systems • The Energy Balance for any system undergoing any kind of process is Ein-Eout=Esystem • For a closed system undergoing a cycle, the initial and final states are same, Esystem=0 • Therefore Ein=Eout • Since in a closed system (in a cycle), no mass flows in or out, only heat and work interactions are involved, Wnet out=Qnet in • Similarly, for a closed system undergoing a process, Qnet in – Wnet out = Esystem 63 Specific Heats • Specific Heat is the energy required to raise the temperature of a unit mass of a substance by one degree  du  Cv     dT  v  dh  Cp    ; h  u  pv  dT  P 64 Problem 4 An engine is tested by means of a water brake at 1000 rpm. The measured torque of the engine is 10,000 m-N and the water consumption of the brake is 0.5 m3/s, its inlet temperature being 20 0C. Calculate the water temperature at exit, assuming that the whole of the engine power is ultimately transformed into heat which is absorbed by the cooling water. Ans: 20.5 0C 65 Problem 5 A mass of 8 kg gas expands within a flexible container so that the P-v relationship is of the form Pv1.2=const. The initial pressure is 1000 kPa and the initial volume is 1 m3. The final pressure is 5 kPa. If specific internal energy of the gas decreases by 40 kJ/kg, find the heat transfer in magnitude and direction. Ans: +2615 kJ 66 Problem 6 The properties of a certain fluid are related as follows: u=196 + 0.718 T pv=0.287(T+273) Where u is the specific internal energy (kJ/kg), T is in 0C, p is pressure (kN/m2), and v is specific volume (m3/kg). Find the Cv and Cp for this fluid. Ans: 0.718, 1.005 kJ/kg-K 67 Problem 7 A gas of mass 1.5 kg undergoes a quasi-static expansion which follows a relationship p=a+bv, where a and b are constants. The initial and final pressures are 1000 kPa and 200 kPa respectively and the corresponding volumes are 0.2 m3 and 1.2 m3. The specific internal energy of the gas is given by the relation: u=1.5 pv – 85 kJ/kg Where p is in kPa and v is in m3/kg. Calculate the net heat transfer and the maximum internal energy of the gas attained during expansion. Ans: 660 kJ, 503.3 kJ 68 Conservation of Mass • Mass cannot be created nor destroyed • Mass and energy can be converted to each other according to E=mc2 • For most energy interactions, the change in mass is extremely small, except for nuclear reactions • For Closed Systems, the mass of the system remains enclosed and hence constant • For an Open System (Control Volume), there are mass in and out flows and so mass should be tracked carefully 69 Conservation of Mass Principle • The net mass transfer to or from a control volume during a time interval t is equal to the net change in the total mass within the control volume during t min – mout = mcv dmcv m in  m out  dt • The above equations are known as Mass balance Equations 70 Mass Balance in Steady-Flow Processes • For Steady Flow Processes, mcv is constant and hence mcv=0 Steady Flow:  m  m ( The total rate of mass entering the CV is equal to the total rate of mass leaving it) in out Steady Flow (Single Stream): 1  m  2  1V1 A1  2V2 A2 m V1 A1  V2 A2 Steady Incompressible Flow (Single Stream): 71 Examples of Mass balance 72 Flow Work or Flow Energy • Unlike Closed Systems, control volumes involve mass flow across their boundaries, and some work is required to push the mass into or out of the CV • This work is known as Flow Work or Flow Energy 73 Expression for Flow Work F  PA W flow  FL  PAL  PV w flow  Pv 74 Total Energy of Flowing Fluid (θ) • • • • e = u + ke + pe θ = pv + e = pv + (u + ke + pe) But h = u + pv Therefore θ = h + ke + pe V2   h  gz 2 • The flow work is automatically taken care by Enthalpy, h 75 Energy Transport by Mass   V2 Amount of Energy Transport Emass  m  m h   gz  2   2   V  Rate of Energy Transport Emass  m   m  h   gz  2   When KE and PE are negligible , E mass  mh and h E mass  m 76 Energy Analysis for Steady Flow Systems • During a steady flow process, the total energy of a control volume remains constant (Ecv=const)  Ecv=0 • Therefore, the amount of energy entering a control volume in all forms (heat, work and mass) must be equal to the amount of energy leaving it 77 Energy Analysis for Steady Flow S ste s… E in  E out Q in  Win   m  Q out  Wout   m  2 2     V V Q in  Win   m  h   gz  Q out  Wout   m  h   gz  2 2 in out     in out 2 2     V V Q  W   m  h   gz   gz    m  h  2 2 out    in  For single stream devices,   V22  V12    Q  W  m h2  h1   g z 2  z1  2   In the absence of KE and PE changes, Q  W  m h  h  or q  w  h2  h1 2 1 78 Some Examples of Steady Flow Processes • • • • Nozzle and Diffusor Throttling Device Turbine and Compressor Heat Exchanger 79 Nozzle and Diffuser • A Nozzle is a device which increases the velocity of a fluid at the expense of its pressure drop • Diffuser increases the pressure of a fluid at the expense of its velocity 80 Nozzle a d Diffuse … • Assumptions: No heat transfer, no change in PE, for nozzle V2>>V1 • Work done is zero as nozzle and diffuser do not involve work Q  W  h  KE  PE   1 2 0  0  h2  h1  V2  V12  g z 2  z1  2 1 2 0  h2  h1  V2  0  0 2 V2  2h1  h2    81 Throttling Device • When a fluid flows through a constricted passage, like a partially opened valve, an orifice, or a porous plug, there is an appreciable drop in pressure, and the flow is said to be throttled 82 Th ottli g De i e… • Assumptions: Adiabatic, no change in PE, no change in KE • No work is involved during the throttling process Q  W  h  KE  PE 0  0  h2  h1  0  0 h2  h1 • Therefore, throttling is Isenthalpic process 83 Turbine and Compressor • Turbines and engines give positive power output whereas compressors and pumps require power input 84 Tu i e a d Co p esso … • Assumptions: Adiabatic, no change in KE and PE Q  W  h  KE  PE 0  W  h2  h1  0  0  W  h2  h1 W  h1  h2 85 Heat Exchanger • A Heat Exchanger is a device in which heat is transferred from one fluid to another 86 Heat E ha ge … • Assumptions: No change in KE and PE • No heat transfer to surroundings and no work involved E in  E out Q in  Win   m  Q out  Wout   m  2 2     V V      gz  Qout  Wout   m  h  Qin  Win   m  h   gz  2 2 in out     0  0  mc h1  ms h2   0  0  mc h3  ms h4   KE  PE in out mc h1  ms h2   mc h3  ms h4  87 Problem 8 Air flows steadily at the rate of 0.5 kg/s through an air compressor entering at 7 m/s, 100 kPa pressure, and 0.95 m3/kg volume, and leaving at 5 m/s, 700 kPa, and 0.19 m3/kg. The internal energy of the air leaving is 90 kJ/kg greater than that of air entering. Cooling water in the compressor jackets absorbs heat from the air at the rate of 58 kW. (a) Compute the rate of shaft work input to the air compressor in kW. (b) Find the ratio of the inlet pipe diameter to outlet pipe diameter. Ans: 122 kW, 1.89 88 Problem 9 A hot water stream at 80 0C enters a mixing chamber with a mass flow rate of 0.5 kg/s where it is mixed with a stream of cold water at 20 0C. If it is desired that the mixture leave the chamber at 42 0C, determine the mass flow rate of the cold water stream. Assume all the streams are at a pressure of 250 kPa. Ans: 0.865 kg/s 89 Problem 10 Air enters an adiabatic nozzle steadily at 300 kPa, 200 0C and 30 m/s and leaves at 100 kPa and 180 m/s. The inlet area of the nozzle is 80 cm2. Determine (a) the mass flow rate through the nozzle, (b) the exit temperature of the air, and (c) the exit area of the nozzle. Ans: 0.5303 kg/s, 184.6 0C, 38.7 cm2 90 Problem 11 The steam supply to an engine comprises two streams which mix before entering the engine. One stream is supplied at the rate of 0.01 kg/s with an enthalpy of 2952 kJ/kg and a velocity of 20 m/s. The other stream is supplied at the rate of 0.1 kg/s with an enthalpy of 2569 kJ/kg and a velocity of 120 m/s. At the exit from the engine the fluid leaves as two streams, one of water at the rate of 0.001 kg/s with an enthalpy of 420 kJ/kg and the other of steam; the fluid velocities at the exit are negligible. The engine develops a shaft power of 25 kW. The heat transfer is negligible. Evaluate the enthalpy of the second exit stream. Ans: 2402 kJ/kg 91 Problem 12 Five hundred kilograms per hour of steam drives a turbine. The steam enters the turbine at 44 atm and 450 0C at a linear velocity of 60 m/s and leaves at a point 5 m below the turbine inlet at atmospheric pressure and a velocity of 360 m/s. The turbine delivers shaft work at a rate of 70 kW, and the heat loss from the turbine is estimated to be 104 kcal/h. Calculate the specific enthalpy change associated with the process. Ans: -650 kJ/kg 92 MODULE 3 Second Law of Thermodynamics Limitations of the 1st Law of TD • Any process must satisfy the 1st Law of TD • But satisfying the 1st Law does not ensure that the process will actually take place Examples: • Hot Coffee • Electric Heater • Paddle Wheel 94 Hot Coffee • A cup of hot coffee does not get hotter in a cooler room 95 Electric Heater • An electric resistor, on heating, will not generate electricity 96 Paddle Wheel • Transferring heat to paddle wheel will not cause it to rotate 97 Limitations of the 1st la of TD… • Processes proceed in a certain direction only • The 1st Law places no restriction on the direction of the process • However, satisfying the 1st Law does not ensure that a process can actually occur • This drawback of the 1st Law is taken care by the 2nd Law of TD • The 1st Law is concerned with the quantity of energy while the 2nd Law is concerned with its quality too 98 Thermal Energy Reservoirs • A hypothetical body with a large thermal energy capacity (mCp) that can supply or absorb finite amounts of heat without undergoing any change in temperature is called as a Thermal Energy Reservoir • Examples are oceans, lakes, atmosphere, two phase system, industrial furnace etc • A reservoir that supplies energy in the form as heat is called as Source and one that absorbs heat is called as Sink 99 Heat Engines • Work can be converted to other forms of energy but the reverse is not a straight forward process • Work can be converted to heat directly and completely but converting heat to work needs a special device called Heat Engine 100 Characteristics of a Heat Engine • They receive heat from a high temperature source • They convert part of this heat to work • They reject the remaining waste heat to a low temperature sink • They operate on a cycle 101 Heat E gi es… • Heat engines involve a fluid to and from which heat is transferred while undergoing a cycle • That fluid is called a Working Fluid • The term Heat Engine is used broadly to include work producing devices that do not operate on Thermodynamic Cycles (IC engines) • The working fluid is purged and replaced by a fresh air-fuel mixture 102 Steam Power Plant • A Steam Power Plant is a device that fits the definition of a Heat Engine 103 Thermal Efficiency • For the steam power plant, Wnet,out=Wout-Win Wnet,out=Qin-Qout • The fraction of the heat input that is converted to net work output is a measure of the performance of the heat engine and is called as Thermal Efficiency 104 The al Effi ie … Net Work Output Thermal Efficiency ,  th  Total Heat Input Wnet,out  th  Q in Wnet,out  Qin  Qout Qin  Qout Qout  th   1 Qin Q in 105 The al Effi ie … • For a Heat Engine, Wnet,out  QH  QL th  Wnet,out QH QH  QL QL th   1 QH QH • QH and QL are positive quantities 106 2nd Law of Thermodynamics: Kelvin-Planck Statement • It is impossible for any device that operates on a cycle to receive heat from a single reservoir and produce a net amount of work A heat engine that violates the Kelvin-Planck statement of the 2nd Law 107 Refrigerators • The transfer of heat from a low temperature medium to a high temperature medium is facilitated by a device called Refrigerator • They operate in cycles • The working fluid used in the refrigeration cycle is called a Refrigerant 108 Basic Components of a Refrigeration System 109 Coefficient of Performance (COP) • The performance of a refrigerator is evaluated by the COP • The objective of a refrigerator is to remove heat from a low temperature space and expel it to a high temperature space COPR  Desired Effect 1 QL QL    Work Input Wnet,in QH  QL QH 1 QL • COP can be greater than 1 110 Heat Pumps • Heat Pump is another device that pumps heat from low temperature to a high temperature medium • The objective of a heat pump is to maintain a heated space at a high temperature • This is done by absorbing heat from a low temperature source 111 COP of a Heat Pump Desired Effect QH COPHP   Work Input Wnet,in  1 QH  QH -QL 1  QL QH COPHP  COPR  1 112 2nd Law of Thermodynamics: Clausius Statement • It is impossible to construct a device that operates in a cycle and produces no effect other than the transfer of heat from a lower temperature body to a higher temperature body A refrigerator that violates the Clausius Statement of the 2nd law 113 Equivalence of Kelvin-Planck and Clausius Statements Violation of Kelvin-Planck Statement Violation of Clausius Statement 114 Perpetual Motion Machine (PMM) • Any device that violates either the 1st or the 2nd Law of Thermodynamics is called as a Perpetual Motion Machine • Perpetual Motion Machines do not work • PMM1 violates the 1st Law while PMM2 violates the 2nd Law of Thermodynamics 115 PMM1 and PMM2 PMM1 PMM2 116 Reversible and Irreversible Processes • A Reversible Process is one that can be reversed without leaving any trace on the surroundings • Both the system and surroundings are returned to their original states at the end of the reverse process • This is possible only if the net heat and work exchange between the system and the surroundings is zero for the combined process • Processes that are not reversible are called as Irreversible processes 117 ‘e e si le a d I e e si le P o esses… • Reversible processes do not occur in nature • They are merely idealizations of actual processes • Engineers are interested in Reversible processes because work producing devices produce most work and work consuming devices consume least work when operating on reversible processes 118 Irreversibilities • The factors that cause a process to be irreversible are called Irreversibilities • They include Friction, Unrestrained Expansion, Mixing of two fluids, Heat Transfer across a finite temperature difference, Electric Resistance, Inelastic Deformation of Solids and Chemical Reactions 119 Internally and Externally Reversible Processes • A process is called Internally Reversible if no irreversibilities occur within the boundaries of the system during the process • A process is called Externally Reversible if no irreversibilities occur outside the system boundaries during the process • A process is called Totally Reversible or simply Reversible if it involves no irreversibilities within the system or its surroundings 120 Internally and Externally Reversible P o esses… Reversible Process Totally Reversible Internally Reversible 121 Reversible Cycles • Heat engines are cyclic devices that consume heat and deliver work • Work is done by the working fluid during one part of the cycle and work is done on the fluid during the other part • The net work is the difference between the two • The efficiency of a heat engine depends on the net work • The net work can be maximized by using processes which require least amount of work and deliver maximum work, i.e. Reversible Processes 122 ‘e e si le C les… • Therefore, the most efficient cycles are Reversible Cycles, which have reversible processes • Reversible cycles cannot be achieved in practice as the irreversibilities cannot be eliminated • But Reversible Cycles provide the upper limit of performance of real cycles 123 The Carnot Cycle • The best known reversible cycle is the Carnot Cycle, proposed by Sadi Carnot • The theoretical heat engine that operates on the Carnot Cycle is called the Carnot Heat Engine • It consists of four reversible processes-two isothermal and two adiabatic 124 The Ca ot C le… Reversible Isothermal Expansion Reversible Adiabatic Expansion Reversible Isothermal Compression Reversible Adiabatic Compression 125 The Ca ot C le… Carnot Cycle on P-V Coordinates 126 The Reversed Carnot Cycle The Reversed Carnot Cycle or The Carnot Refrigeration Cycle 127 Carnot Principles • The efficiency of an irreversible heat engine is always less than the efficiency of a reversible one operating between the same two reservoirs • The efficiencies of all reversible heat engines operating between the same two reservoirs are the same 128 Proof of the Carnot Principles 129 Carnot Efficiency • The Thermal Efficiency of any heat engine, reversible or irreversible is given as: QL th  1  QH • The Thermal Efficiency of a Carnot Engine (Carnot Efficiency) can be written as: th,rev QL TL  1  1 QH TH • This is the highest efficiency a heat engine operating between the two thermal energy reservoirs at temperatures TL and TH can have 130 Actual and Carnot Efficiencies th < th,rev Irreversible Heat Engine = th,rev Reversible Heat Engine > th,rev Impossible Heat Engine 131 Improving the Efficiency of Heat Engines • The Thermal Efficiency of a Carnot Heat Engine is: th,rev QL TL  1  1 TH QH • The efficiency increases as TH is increased or TL is decreased • This is also true for actual engines • The Thermal Efficiency of actual heat engines can be maximized by supplying heat to the engine at the highest possible temperature and rejecting heat from the engine at the lowest possible temperature 132 Carnot Refrigerator and Carnot Heat Pump • A refrigerator or a heat pump that operates on the reversed Carnot Cycle is called a Carnot Refrigerator or a Carnot Heat Pump COPR  1 QH COPHP  QL 1 1  QL 1  1 TH  QH TL 1 1 1  TL TH • These are the highest coefficients of performance that a refrigerator or a heat pump operating between the temperature limits of TL and TH can have 133 Actual and Carnot COPs COPR < COPR,rev Irreversible Refrigerator = COPR,rev Reversible Refrigerator > COPR,rev Impossible Refrigerator 134 Problem 1 A steam power plant with a power output of 150 MW consumes coal at a rate of 60 tons/hr. If the heating value of the coal is 30,000 kJ/kg, determine the overall efficiency of this plant. Ans: 30% 135 Problem 2 A coal burning steam power plant produces a net power of 300 MW with an overall thermal efficiency of 32%. The actual gravimetric air-fuel ratio in the furnace is calculated to be 12 kg air/kg fuel. The heating value of the coal is 28,000 kJ/kg. Determine (a) the amount of coal consumed during a 24-hour period and (b) the rate of air flowing through the furnace. Ans: (a) 2.89 X 106 kg, (b) 402 kg/s 136 Problem 3 A household refrigerator with a COP of 1.2 removes heat from the refrigerated space at a rate of 60 kJ/min. Determine (a) the electric power consumed by the refrigerator and (b) the rate of heat transfer to the kitchen air. Ans: (a) 0.83 kW, (b) 110 kJ/min 137 Problem 4 A household refrigerator that has a power input of 450 W and a COP of 2.5 is to cool five large watermelons, 10 kg each, to 8 0C. If the watermelons are initially at 20 0C, determine how long it will take for the refrigerator to cool them. The watermelons can be treated as water whose specific heat is 4.2 kJ/kg-0C. Ans: 2240 s 138 Problem 5 A completely reversible heat engine operates with a source at 800 K and a sink at 280 K. At what rate must heat be supplied to this engine, in kJ/h, for it to produce 4 kW of power. Ans: 22,150 kJ/h 139 Problem 6 A Carnot refrigerator operates in a room in which the temperature is 25 0C. The refrigerator consumes 500 W of power when operating and has a COP of 4.5. Determine (a) the rate of heat removal from the refrigerated space and (b) the temperature of the refrigerated space. Ans: (a) 135 kJ/min, (b) -29.2 0C 140 Problem 7 Consider two Carnot heat engines operating in series. The first engine receives heat from the reservoir at 1800 K and rejects the waste heat to another reservoir at temperature T. The second engine receives this energy rejected by the first one, converts some of it to work, and rejects the rest to a reservoir at 300 K. If the thermal efficiencies of both engines are the same, determine the temperature T. Ans: 735 K 141 Problem 8 A refrigeration cycle having a coefficient of performance of 3 maintains a computer laboratory at 18 C on a day when the outside temperature is 30 C. The thermal load at steady state consists of energy entering through the walls and windows at a rate of 30,000 kJ/h and from the occupants, computers, and lighting at a rate of 6000 kJ/h. Determine the power required by this cycle and compare with the minimum theoretical power required for any refrigeration cycle operating under these conditions, each in kW. Ans: 3.33 kW, 0.41 kW 142 Clausius Inequality • The Clausius inequality is given as: Q T 0 • The cyclic integral of Q/T is always less than or equal to zero • It is valid for all cycles, reversible or irreversible • The equality in the Clausius inequality holds for totally or just internally reversible cycles and the inequality for the irreversible ones 143 Entropy (S) • Entropy is a property designated by the symbol S  Q  dS   kJ / K   T int rev • Entropy change of a system during a process can be determined by integrating the above equation  Q  S  S 2  S1     T int rev 1 2 144 E t op … • Entropy is a property and hence has fixed values at fixed states • The entropy change between two specified states is the same irrespective of the path, irreversible or reversible, is followed 145 Internally Reversible Isothermal Heat Transfer Process • Isothermal heat transfer processes are internally reversible since there is no finite temperature gradient 2  Q  1  Q    S        T int rev 1  T0 int rev T0 1 2 Q    Q   int rev 2 1 T0 • The above equation is useful in determining the entropy change of thermal reservoirs • Heat transfer to the system increases the entropy while heat transfer from it decreases the entropy 146 Increase of Entropy Principle • Consider a cycle having two processes: 1-2 is any arbitrary process and 2-1 is a reversible process • From the Clausius inequality, Q T 0 2 1 Q  Q  0    T 2  T int, rev 1 147 I ease of E t op P i iple…  2 Q T  S1  S 2  0 S 2  S1   1 2 dS  Q 1 Q T Q • S2-S1 is the Entropy Change and  T is called the Entropy Transfer • The Entropy change is always greater than Entropy Transfer T 2 1 148 I ease of E t op P i iple… • The Entropy Change can be written as S System  S 2  S1   2 1 Q T  S gen • Sgen is always a positive quantity or is equal to zero • Sgen depends on the process and thus is not a property of the system • In the absence of entropy transfer, the entropy change of a system is equal to the entropy generation 149 I ease of E t op P i iple… • For an isolated system or an adiabatic closed system, the heat transfer is zero Sisolated  0 • The entropy of an isolated system during a process always increases or, in the limiting case of a reversible process, remains constant • It never decreases • This is known as the Increase of Entropy Principle 150 I ease of E t op P i iple… • As no actual process is truly reversible, it is concluded that some entropy is generated during a process • The universe can be assumed as an isolated system and hence the entropy of the universe is continuously increasing S gen > 0 Irreversible Process = 0 Reversible Process < 0 Impossible Process 151 Some Remarks about Entropy • A process must proceed in the direction that complies with the increase in entropy principle, i.e. Sgen  0 • Entropy is a non-conserved property. It is conserved during the idealized reversible processes only and increases during all actual processes • The performance of engineering systems is degraded by the presence of irreversibilities, and entropy generation is a measure of their magnitudes 152 Isentropic Process • Entropy of a system can be changed by heat transfer and irreversibilities • So, entropy does not change during a process that is reversible and adiabatic • Such a process where entropy remains constant is known as Isentropic Process • Many engineering devices like pumps, nozzles, turbines are adiabatic and they perform well when the irreversibilities such as friction are minimized • Therefore isentropic processes enable us to compare with actual processes 153 TdS Relations • For a closed system, the energy balance is Qint rev  Wint rev,out  dU dS  Qint rev  Qint rev  TdS Wint rev,out  PdV T TdS  dU  PdV Gibbs Equation  h  u  Pv  dh  du  Pdv  vdP du  dh  Pdv  vdP dU  dH  PdV  VdP TdS  dU  PdV Tds  du  Pdv TdS  dH  VdP Tds  dh  vdP TdS  dH  PdV  VdP  PdV TdS  dH  VdP 154 Important Relation • For an ideal gas undergoing as Isentropic Process T2  P2     T1  P1   1   v1      v2   1 155 Entropy Change of an Incompressible Substance Tds  du  pdv du p  dv ds  T T For incompressible substances, dv  0 du c p dT  ds   T T On integratin g over a process1 - 2, dT 1 ds  c p 1 T 2 2 T2 s2  s1  c p ln T1 156 Entropy Change of an Ideal Gas Tds  du  pdv du p  dv ds  T T p R pv  RT   T v du Also cv  dT dT R  ds  cv  dv T v For a process1 - 2, dT dv   ds c R v 1  T v 1 1 2 2 2 T2 v2 s2  s1  cv ln  R ln T1 v1 Tds  dh  vdp ds  dh v  dp T T v R pv  RT   T p dh Also c p  dT dT R  ds  c p  dp T p For a process1 - 2, dT dp 1 ds  c p 1 T  R 1 p 2 2 2 T2 p2 s2  s1  c p ln  R ln T1 p1 157 Entropy Balance  Change      Total   Total   Total   in the Total        Entropy    Entropy    Entropy      Entering   Leaving   Generated   Entropy of    the System         Sin  S out  S gen  S System S System  S final  Sinitial  S 2  S1 Entropy Transfer by Heat : S heat   2 Q 1 T  Q T Entropy Transfer by Work : S work  0 Entropy Transfer by M ass : S mass  ms 158 Entropy Balance for Closed System Q  T  S gen  S system  S2  S1 For an adiabatic system: S gen  S system  S 2  S1 159 Entropy Balance for Open System Q  T  mi si   me se  S gen  S 2  S1 For steady flow systems,S 2  S1 Q  T  mi si   me se  S gen  0 In rate form Q  T  m i si   m e se  Sgen  0 For steady flow, single stream Q  T m si  se   S gen  0 For steady flow, single stream, adiabatic m s  s   S  0 i e gen 160 Problem 9 Air is compressed by a 12 kW compressor from P1 to P2. The air temperature is maintained constant at 25 0C during the process as a result of heat transfer to the surrounding medium at 10 0C. Determine the rate of entropy change of the air. Ans: -0.0403 kW/K 161 Problem 10 Heat is transferred at a rate of 2 kW from a hot reservoir at 800 K to a cold reservoir at 300 K. Calculate the rate at which the entropy of the two reservoirs change. Ans: 0.00417 kW/K 162 Problem 11 A completely reversible heat pump produces heat at a rate of 100 kW to warm a house maintained at 21 0C. The exterior air, which is at 10 0C, serves as the source. Calculate the rate of entropy change of the two reservoirs and determine if this heat pump satisfies the second law according to the increase of entropy principle Ans: Yes 163 Problem 12 A 20 kg aluminium block initially at 200C is brought into contact with a 20 kg block of iron at 100C in an insulated enclosure. Determine the final equilibrium temperature and the total entropy change for this process. CpAl=0.93, CpFe=0.45 kJ/kg-K Ans: 167.4 C, 0.175 kJ/K 164 Problem 13 A piston-cylinder device contains 1.2 kg of nitrogen gas at 120 kPa and 27C. The gas is now compressed slowly in a polytropic process during which PV1.3=constant. The process ends when the volume is reduced by one half. Determine the entropy change of nitrogen during this process. Ans: -0.0617 kJ/K 165 MODULE 4 Concept of Exergy Availability (Exergy) • The useful work potential of a given amount of energy at some specified state is called Exergy or Availability or Available Energy • The Work Potential of the energy contained in a system at a specified state is simply the maximum useful work that can be obtained from the system • A system delivers the maximum possible work as it undergoes a reversible process from the specified initial state to the state of its environment, that is the Dead State 167 Some Terms • Dead State: The state when a system is in thermodynamic equilibrium with the surroundings. (T0, P0) • Surroundings: Everything outside the system boundary • Immediate Surroundings: Portion of the surroundings that are affected by the process • Environment: Region beyond the immediate surroundings whose properties are not affected by the process at any point 168 So e Te s… 169 Exergy Associated With KE and PE • Kinetic Energy is a form of mechanical energy and it can be converted to work entirely • Exergy of the KE of a system is its KE itself V2 xke  2 • Exergy of the PE of a system is its PE itself x pe  gz • Therefore, the KE and PE are entirely available for work • However, internal energy and enthalpy are not entirely available for work 170 Surroundings Work • Surroundings work is the work done by or against the surroundings during a process Wsurr  P0 V2  V1  Wu  W  Wsurr  W  P0 V2  V1  171 Su ou di gs Wo k… • When a system is expanding and doing work, part of the work done is used to overcome the atmospheric pressure • When a system is compressed, the atmospheric pressure helps the compression process • Wsurr is significant only is systems involving moving boundary work 172 Reversible Work (Wrev) • Wrev is the maximum amount of useful work that can be produced (or the minimum work that needs to be supplied) as a system undergoes a process between the specified initial and final states • This is the useful work output (or input) obtained (or expended) when the process between the initial and final states is executed in a reversible manner 173 Irreversibility • Irreversibility is the difference between the reversible work and useful work • Irreversibility is equivalent to the exergy destroyed W I W rev ,out u ,out or I  Wu ,in W rev,in • For a reversible process, the irreversibility is zero • Irreversibility is a positive quantity 174 2nd Law Efficiency • Thermal Efficiency and COP are defined based on the 1st Law of TD • They make no reference to the best possible performance • Therefore 2nd Law efficiency is defined in order to compare performances of devices based on their working potentials 175 2nd La Effi ie … • Engine B has a greater potential • So Engine B is performing poorly  rev, A  TL  1   TH   300   1   50% 600 A T   rev, B  1  L   1   70% T 1000  H B 300 176 2nd La Effi ie … • 2nd Law Efficiency is defined as the ratio of the actual thermal efficiency to the maximum possible thermal efficiency under the same conditions  II  th th,rev  II , A  0.3 0.3  0.6 and  II , B   0.43 0.5 0.7 177 General Definition of 2nd Law Efficiency Wu  II  Wrev Wrev  II  Wu For work producing devices  For work consuming devices  COP  II  COPrev For refrigerat ors and heat pumps 178 Exergy of a Closed System 179 E e g of a Closed S ste … Exergy : V2  gz   u  u0   P0 v  v0   T0 s  s0   2 Exergy Cha nge: V22  V12   2  1  u2  u1   P0 v2  v1   T0 s2  s1    g z 2  z1  2 For Statio nary Systems :   u  u0   P0 v  v0   T0 s  s0  180 Exergy of an Open System Flow Exerg y : V2   h  h0   T0 s  s0    gz 2 Flow Exerg y Change : V22  V12    2 - 1  h2  h1   T0 s2  s1    g z2  z1  2 181 Exergy Transfer By Heat :  T0  X heat   1  Q  T  By Work : X work  W  Wsurr W By M ass : X mass  m 182 Decrease of Exergy Principle • Exergy of an isolated system during a process always decreases • Only in a reversible process, exergy remains constant • Exergy never increases and in an actual process it is destroyed and is given by the Gouy-Stodola Theorem X destroyed  T0 S gen 183 Exergy Destruction X destroyed  T0 S gen X destroyed  0 Irreversible Process X destroyed  0 Reversible Process X destroyed  0 Impossible Process 184 Exergy Balance Exergy in  Exergy out  Exergy destroyed  Exergy Cha nge xin  xout  xdes  x For Closed System : xheat-xwork-xdes  Δx For Open System : xheat-xwork  xmass,in -xmass,out  xdes  Δxcv 185 Summary Energyin  Energyout  Energy Entropy in  Entropy out  Entropy gen  Entropy Exergy in  Exergy out  Exergy des  Exergy 186 Problem 12 A heat engine receives heat from a source at 1500 K at a rate of 700 kJ/s, and it rejects the waste heat to a medium at 320 K. The measured power output of the heat engine is 320 kW, and the environment temperature is 25 0C. Determine (a) the reversible power, (b) the rate of irreversibility, and (c) the second law efficiency of this heat engine. Ans: (a) 550.7 kW, (b) 230.7 kW, (c) 58.1% 187 Problem 13 A house that is losing heat at a rate of 80,000 kJ/h when the outside temperature drops to 15 0C is to be heated by electric resistance heaters. If the house is to be maintained at 22 0C at all times, determine the reversible work input for this process and the irreversibility. Ans: 0.53 kW, 21.69 kW 188 Problem 14 A freezer is maintained at -7 0C by removing heat from it at a rate of 80 kJ/min. The power input to the freezer is 0.5 kW, and the surrounding air is at 25 0C. Determine (a) the reversible power, (b) the irreversibility, and (c) the second-law efficiency of this freezer. Ans: 0.16 kW, 0.34 kW, 32% 189 Problem 15 A cylinder of an internal combustion engine contains 2450 cm3 of gaseous combustion products at a pressure of 7 bar and a temperature of 867 0C just before the exhaust valve opens. Determine the specific exergy of the gas, in kJ/kg. Ignore the effects of motion and gravity and model the combustion products as air as ideal gas. Take T0=300 K and P0=1.103 bar. Ans: 325.2 kJ/kg 190 Problem 16 Show that the power produced by a wind turbine is proportional to the cube of the wind velocity and to the square of the blade span diameter 191 MODULE 5 Properties of Pure Substance Pure Substance • A substance that has a fixed chemical composition throughout is called a Pure Substance • A pure substance does not have to be of a single chemical element or compound • A mixture of various chemical elements or compounds is a pure substance as long as the mixture is homogeneous, e.g. Air • A mixture of oil and water is not a pure substance as oil is not soluble in water 193 Pu e Su sta e… • A mixture of two or more phases of a pure substance is still a pure substance as long as the chemical composition of all phases is the same, e.g. Mixture of ice and water • A mixture of liquid air and gaseous air is not pure substance as the compositions are different 194 Phases of a Pure Substance • A Phase is identified as having a distinct molecular arrangement that is homogeneous throughout and separated from the others by easily identifiable boundary surfaces, e.g. iced water • Substances are present in three phases-Solid, Liquid and Gas (vapour) 195 Phases of a Pu e Su sta e… • The molecules in a solid are arranged in 3-D pattern • Due to the small distances between the molecules in a solid, the attractive forces of molecules on each other are large and keep the molecules in fixed positions • The molecular spacing in the liquid phase is similar to that in solids except that the molecules are no longer at fixed positions relative to each other and can rotate and translate freely 196 Phases of a Pu e Su sta e… • In the gas phase, the molecules are far apart from each other and molecular order is nonexistent • Gas molecules move about random, continually colliding with each other and the walls of the container • Molecules in the gas phase are at a considerably higher energy level than they are in the liquid and solid phases 197 Phase Change Processes • Phase change processes take place in practical situations • Water exists as liquid and vapour in a boiler and a condenser • The refrigerant turns from liquid to vapour in a refrigerator • Here, water is used to explain the process but the principles are equally applicable to other pure substances 198 Phase Change Processes in a Pure Substance State 1: Compressed Liquid State 2: Saturated Liquid State 3: Saturated liquid + Vapour State 4: Saturated Vapour State 5: Superheated Vapour 199 Definitions • Compressed or Sub-Cooled liquid means that it is not about to vapourize • Saturated Liquid is a liquid that is about to vapourize • Saturated Vapour is a vapour that is about to condense • Saturated Liquid-Vapour Mixture is a condition where liquid and vapour coexist in equilibrium • Superheated Vapour is a vapour that is not about to condense 200 Defi itio s… 201 Defi itio s… • Saturation Temperature (Tsat) is the temperature (at a given pressure) at which a pure substance changes phase • Saturation Pressure (Psat) is the pressure (at a given temperature) at which a pure substance changes phase • Latent Heat of Vaporization is the amount of energy absorbed during evaporation 202 Liquid-Vapour Saturation Curve Tsat=f(Psat) 203 T-V Diagram 204 Critical Point • Critical Point is defined as the point at which the saturated liquid and saturated vapour states are identical • For water, Pcr=22.06 Mpa • Tcr=373.95 0C 205 T-V Diag a … 206 Liquid-Vapour Mixture mvapour Mass of Vapour mvapour Quality  Dryness Fraction , x    Total Mass mtotal mliquid  mvapour 207 Properties of Liquid-Vapour Mixture v  1  x v f  xv g u  1  x u f  xu g h  1  x h f  xhg s  1  x s f  xs g F (for fluid) =Liquid G (for gas) =Vapour y  1  x  y f  xy g 208 Characteristics of Superheated vapour • Compared to saturated vapour, superheated vapour is characterized by Lower Pressures P  Psat at a given T  Higher Temperatures T  Tsat at a given P  Higher Specific Volumes v  v g at a given P or T  Higher internal energies u  u g at a given P or T  Higher Enthalpies h  h g at a given P or T  209 Characteristics of Compressed Liquid • A compressed liquid is characterized by Higher Pressures P  Psat at a given T  Lower Temperatures T  Tsat at a given P  Lower Specific Volumes v  v f at a given P or T  Lower internal energies u  u f at a given P or T  Lower Enthalpies h  h f at a given P or T  • However, the properties of compressed liquid are not much different from the corresponding saturation liquid values 210 Problem Table 1 T, 0C P, kPa 50 V, m3/kg Phase Description 4.16 200 250 400 110 600 Saturated Vapour 211 Problem Table 2 T, 0C P, kPa h, kJ/kg 200 140 Phase Description 0.7 1800 950 80 x 0.0 500 800 3162.2 212 Problem 1 A rigid tank with a volume of 2.5 m3 contains 15 kg of saturated liquid-vapour of water at 75 0C. Now the water is slowly heated. Determine the temperature at which the liquid in the tank is completely vaporized. Also, show the process on a T-v diagram with respect to saturation lines. Ans: 187 0C 213 Problem 2 Superheated water vapour at 1.4 MPa and 250 0C is allowed to cool at constant volume until the temperature drops to 120 0C. At the final state, determine (a) the pressure, (b) the quality, and (c) the enthalpy. Also, show the process on a T-v diagram with respect to saturation lines. Ans: (a) 198.7 kPa, (b) 0.1825, (c) 905.7 kJ/kg 214 Problem 3 A sample of steam from a boiler drum at 3 MPa is put through a throttling calorimeter in which the pressure and temperature are found to be 0.1 MPa and 120 0C respectively. Find the quality of the sample taken from the boiler. Ans: 0.951 215 Problem 4 Steam enters a turbine operating at steady state with a mass flow rate of 4600 kg/h. The turbine develops a power output of 1000 kW. At the inlet, the pressure is 60 bar, the temperature is 400 C, and the velocity is 10 m/s. At the exit, the pressure is 0.1 bar, the quality is 0.9 (90%), and the velocity is 50 m/s. Calculate the rate of heat transfer between the turbine and surroundings, in kW. Ans: 61.3 kW 216 Ideal Gas • The equation Pv=RT is known as the Ideal Gas Equation or the Ideal Gas Equation of State • A gas which obeys this equation is called an Ideal Gas • P is absolute pressure, T is absolute temperature and v is specific volume R Ru kJ kg - K M Ru  Universal Gas Constant M  M olar M ass or M olecular Weight 217 Universal Gas Constant kJ  8.31447 kmol - K  J 8.31447  mol - K Ru   3 kPa m 8.31447  kmol - K  3 bar m 0.0831447  kmol - K 218 Molar Mass and Mole • Molar Mass is the mass of one mole of a substance in grams, or the mass of one kmol in kilograms • Number of Moles (n) is the ratio of the mass (m) to the molecular weight (M) m n M 219 Other Forms of the Ideal Gas Equation PV  mRT PV  nRuT PV  RuT Where P  Absolute Pressure, Pa For a gas undergoing a process1  2, the Ideal Gas Equation can be used to write P1V1 P2 V2  T1 T2 V  Volume, m 3 m  M ass,kg n  M oles R  Specific Gas Constant,kJ/kg  K Ru  Universal Gas Constant,kJ/kmol  K T  Absolute Temperature, K 220 Applicability of The Ideal Gas Equation • The Ideal Gas Equation closely approximates the P-V-T behaviour of Real gases at low densities • At low pressures and high temperatures, the density of a gas decreases, and the gas behaves as an ideal gas • Many familiar gases like air, nitrogen, oxygen, hydrogen, helium, argon, neon, krypton, carbon dioxide are ideal gases • Dense gases like water vapour, refrigerants must not be treated as ideal gases 221 Other Equations of State • The ideal gas equation is simple but range of applicability is limited • To cover a wider range, few other equations are suggested • Van der Waals equation (one of the earliest) • Beattie-Bridgeman equation (one of the best known and reasonably accurate) • Benedict-Webb-Rubin equation (one of the most recent and very accurate) 222 Van der Waals Equation a   P  2 v  b   RT v   27 R 2Tcr2 RTcr a and b  64 Pcr 8Pcr 223 Beattie-Bridgeman Equation RuT  c  A P  2 1  3 v  B   2 v  vT  v  b  a A  A0 1   and B  B0 1    v  v The constants A0, a,B0,b and c can be obtained from tables. The Equation is reasonably accurate for densities up to about 0.8 cr 224 Benedict-Webb-Rubin Equation RuT  C0  1 bRuT  a a c     / v 2 P   B0 RuT  A0  2  2   6  3 2 1  2 e 3 T v v v vT  v  v  The constants A0, a, B0, b, C0, c, α, and  can be obtained from tables. The Equation is reasonably accurate for densities up to about 2.5 cr 225 Compressibility Factor (Z) • The deviation from ideal gas behaviour at a given temperature and pressure can accurately be accounted by the Compressibility Factor Pv Z or Pv  ZRT RT 226 Co p essi ilit Fa to ) … • The value of Z can be found from charts using the values of Reduced Pressure (Pr) and Reduced Temperature (Tr) P PR  Pcr T and TR  Tcr • The value of Z is approximately the same at the same reduced pressure and temperature. • This is called the Principle of Corresponding States 227 Generalized Compressibility Chart 228 Some Observations • At very low pressures (PR<<1), gases behave as an ideal gas regardless of temperature • At high temperatures (TR>2), ideal gas behaviour can be assumed with good accuracy regardless of pressure (except when PR>>1) • The deviation of a gas from ideal gas is greatest in the vicinity of the critical point 229 So e O se atio s… 230 Problem 1 Determine the specific volume of superheated water vapour at 10 MPa and 400 0C, using (a) the ideal-gas equation, (b) Van der Waals equation, (c) the generalized compressibility chart and (d) the steam tables. Ans: (a) 0.03106 m3/kg, (b) and (d) 0.02644 m3/kg ,(c) 0.02609 m3/kg 231 Problem 2 Determine the specific volume of nitrogen gas at 10 MPa and 150 K based on (a) the ideal-gas equation and (b) the generalized compressibility chart. Compare these results with the experimental value of 0.002388 m3/kg, and determine the error involved in each case. For Nitrogen, Pcr=3.39 MPa, Tcr=126.2 K. Ans: (a) 0.004482 m3/kg, 86.4%, (b) 0.002404 m3/kg, 0.7% 232 MODULE 6 Thermodynamics Relations Why Thermodynamic Relations? • Properties like mass, volume, temperature, and pressure can be easily measured • Properties like density, specific volume can be calculated by simple equations • Properties like internal energy, enthalpy, and entropy cannot be measured directly or related to easily measurable properties by simple equations • Hence it is necessary to develop relations between commonly encountered properties and relate them to easily measurable properties 234 Gibbs and Helmholtz Functions • The Gibbs (g) and Helmholtz (a) functions are combination properties (like enthalpy) • Gibbs Function: g=h-Ts On differentiation, dg=dh-Tds-sdT (eqn 1) • Helmholtz Function: a=u-Ts On differentiation da=du-Tds-sdT (eqn 2) 235 Gibbs Relations • The four Gibbs relations are derived from the Tds relations and the Gibbs and Helmholtz functions • The Tds relations are Tds=du+Pdv  du=Tds-Pdv (eqn 3) Tds=dh-vdP  dh=Tds+vdP (eqn 4) • Combining eqns 1 to 4, we get da=-sdT-Pdv (eqn 5) dg=-sdT+vdP (eqn 6) 236 Gibbs Relations • All four Gibbs relations are given below: du=Tds-Pdv dh=Tds+vdP da=-sdT-Pdv dg=-sdT+vdP • The above equations are of the form dz=Mdx+Ndy with  M   y   N       x  x  y 237 Maxwell Relations The change in entropy can be easily calculated using the changes in pressure, volume and temperature  T   P        v  s  s  v  T   v       P  s  s  P  s   P       v T  T  v  s   v        P T  T  P 238 Clapeyron Equation • During phase change, the temperature remains constant • Also, the pressure depends only on temperature  P   s       T  v  v T  dP   s       dT  sat  v T 239 Clapeyron E uatio … • Between the two saturation states (saturated liquid and saturated vapour states), the equation can be integrated Clapeyron ' s equation a nd Trouton's rule combined :  dP   88  T  s s   v  v  P  101.325 exp  1   g f  dP    dT   sat  dT  sat s fg  v fg g f  R  T  TB : Boiling Po int at 1 atm B R : Universal Gas Constant  8.314 J/mol - K During a phase change process, the pressure remains constant dh  Tds  vdP  Tds  dh   Tds  h g g fg  Ts fg h fg  dP     --- The Clayperon Equation dT   sat Tv fg f f 240 Importance of Clapeyron Equation • The Clapeyron equation enables to determine the enthalpy of vaporization by measuring the slope of the saturation curve on a P-T diagram and the specific volume of the saturated liquid and vapour 241 Clausius-Clapeyron Equation at Low Pressures • At low pressures, vg>>vf and so vfg=vg • Also, at low pressures, vapour behaves ideally vg  RT P Substituting in the Clapeyron equation, we get Ph fg  dP     2  dT  sat RT h fg  dT   dP   2    P R  T  sat   sat For small temperature intervals, h fg can be treated as a constant at some average value On integratio n between the two saturation states (liquid and vapour) h fg  1 1  P     ln  2    P1  sat R  T1 T2  sat --- Clausius-Clapeyron Equation 242 General Relations for a Simple Compressible Systems • According to the state postulate, two independent intensive properties are sufficient to completely describe a simple compressible system • This can be used to determine the un-measurable properties from easily measurable ones • Using calculus, Tds relations and Maxwell relations, some useful relations can be obtained for property changes 243 Ge e al ‘elatio s…   P   u2  u1   Cv dT   T    P  dv T  v  T1 v1   T2 v2   v   h2  h1   C P dT   v  T    dP  T  P  T1 P1  T2 P2 2 Cv  P  s2  s1   dT     dV T  v T T1 v1  T2 CP  CP 0 T v   2v   T   2  T  P 0 2 CP  v   dT     dP T  P T T1 P1  T2 P P vT 2  v   P   v   P  C P  Cv  T        T      T  P  v T  T  P  T  v 2 1  v     ; v  T  p 1  v      v  P T 244 Internal Energy Changes Let u  u T , v   u   u  du    dT    dv  T  v  v T  u  But cv     T  v  u   du  cv dT    dv  v T Let s  s T , v   s   s  ds    dT    dv  v T  T  v TdS eqn : Tds  du  Pdv  du  Tds  Pdv   s    s  du  T   dT  T    P  dv  T  v   v T   s  cv  u   s  ; T         P  T  v T  v T  v T 245 I te al E e g Cha ges…  s   P  Using Maxwell's 3rd eqn :      v T   v  T   P   u  T    P    T  v  v T   P    du  cv dT  T    P  dv   T  v  Internal energy change in a simple compressible system for a state change from T1,v1  to T2 ,v2  is    P  u2  u1   cv dT   T    P  dv T  v  T1 v1   T2 v2 246 Enthalpy Changes Let h  hT , P   h   h   dT dh     dP   P T  T  P  h  dh  c p dT    dP  P T Let s  sT , P   s   s   dT ds     dP   P T  T  P Tds eqn : Tds  dh  vdP  dh  Tds  vdP   s   s   dh  T   dT  v  T    dP  T  P  P T   c p  h   s   s     ;    v T   T  P T  P T  P T 247 E thalp Cha ges…  v   s  Using 4th Maxwell's relation      T  P  P T  v   h    v T      T  P  P T   v    dh  c p dT  v  T    dP  T  P   Enthalpy change in a simple compressible system for a change of state from T1,P1  to T2 ,P2  is   v   h2  h1   c p dT   v  T    dP  T  P  T1 P1  Also, h2  h1  u2  u1  P2 v2  P1v1  T2 P2 248 Problem 1 Using the Maxwell relations, determine a  s  relation for  P  for a gas whose equation of state is P(v - b) = RT. Answer: -R/P T 249 Problem 2 Two grams of a saturated liquid are converted to a saturated vapour by being heated in a weighted piston-cylinder device arranged to maintain the pressure at 200 kPa. During the phase conversion, the volume of the system increases by 1000 cm3; 5 kJ of heat are required; and the temperature of the substance stays constant at 80C. Estimate the boiling temperature of this substance when its pressure is 180 kPa. Also, estimate the sfg at 80C. Ans: 352 K, 7.08 kJ/kg-K 250 Problem 3 Determine the change in the enthalpy of air, in kJ/kg, as it undergoes a change of state from 100 kPa and 20C to 600 kPa and 300C using the equation of state P(v-a)=RT where a=0.10 m3/kg, and compare the result to the value obtained by using the ideal gas equation of state. Ans: 335 kJ/kg, 285 kJ/kg 251 Problem 4 Determine the change in the entropy of helium, in kJ/kg-K, as it undergoes a change of state from 100 kPa and 20C to 600 kPa and 300C using the equation of state P(v-a)=RT where a=0.10 m3/kg, and compare the result to the value obtained by using the ideal gas equation of state. Ans: -0.239 kJ/kg-K, -0.239 kJ/kg-K 252 Problem 5 Find the approximate saturation pressure of water at 120 C. Ans: 1.73 bar 253 MODULE 7 Gas Power Cycles Carnot Cycle • Carnot Cycle is the most efficient cycle between a source at TH and a sink at TL • But the reversible isothermal heat transfer is very difficult to achieve as it needs very large heat exchangers and a very long time • So Carnot cycle is not practical • However, it can be used to compare the performance of an actual cycle 255 Gas Power Cycles • In Gas Power Cycles, the working fluid remains a gas throughout the cycle • The energy is provided by burning a fuel • The composition of the working fluid changes from air and fuel to combustion products during the cycle • As air consists mostly of Nitrogen which does not react, the working fluid closely resembles air 256 Air Standard Assumptions • The working fluid is air, which continuously circulates in a closed loop and always behaves as an ideal gas • All the processes in the cycle are internally reversible • The combustion is replaced by a heat-addition process from an external source • The exhaust is replaced by a heat-rejection process that restores the working fluid to its initial state • If the specific heats are assumed to be constant at 25 0C, the above assumptions are known as Cold Air Standard Assumptions 257 Otto Cycle 258 Otto Cycle Processes: 1-2: Isentropic Compression 2-3: Constant Volume Heat Addition 3-4: Isentropic Expansion 4-1: Constant Volume Heat Rejection 259 Analysis of an Otto Cycle • The Otto Cycle is executed in a closed system • The changes in KE and PE are neglected • No work is involved during the heat transfer processes q-w=u 260 Thermal Efficiency of Otto Cycle qin  u3  u2  Cv T3  T2  qout  u4  u1  Cv T4  T1  th,Otto Wnet qout   1 qin qin  T4  T1   1 T1  T4  T1   1  1 T3  T2  T3  T2   1  T2   v3  T1  v2  T        4 But T2  v1  T3  v4  V V v 1 where r  max  1  1 th,Otto  1   1 r Vmin V2 v2  1 r = Compression Ratio  1 261 The al Effi ie of Otto C le… The specific heat ratio and the efficiency decrease as the molecules of the gas get larger. In actual engines, the molecules are larger and hence give low efficiency 262 Diesel Engine and Differences from Petrol Engine 263 Diesel Cycle 264 Analysis of Diesel Cycle qin  w  u3  u2 qin  w  u3  u2  P2 v3  v2   u3  u2   h3  h2  C p T3  T2  qout  u4  u1  Cv T4  T1  th, Diesel  Wnet q  1  out qin qin Cutoff Ratio  rc  th, Diesel  T T1  4  1 T1  T4  T1   1  1  T3  T2   T3  T2   1  T2  v3 v2 1  rc  1   1   1   r   rc  1 265 Efficiency of Diesel Cycle For the same compression ratio, Otto Cycle is more efficient than the Diesel Cycle 266 Dual Cycle • Actual heat addition process in an IC engine is better approximated by a combination of constant volume and constant pressure processes •The cycle obtained in such a way is called the Dual Cycle •The analysis is similar to that of Otto and Diesel cycles 267 Mean Effective Pressure (MEP) • Mean Effective Pressure is a fictitious pressure, if acting on the piston during the entire power stroke would produce the same amount of net work as that produced during the actual cycle • The MEP is a parameter used to compare the performances of reciprocating engines of equal size • Engine with larger MEP delivers more work per cycle and hence better 268 Mea Effe ti e P essu e MEP … 269 Gas Turbine Open Cycle Gas Turbine Closed Cycle Gas Turbine 270 Brayton Cycle 1-2: Isentropic Compression 2-3: Constant Pressure Heat addition 3-4: Isentropic Expansion 4-1: Constant Pressure Heat Rejection 271 Efficiency of Brayton Cycle th, Brayton  1  1  1 rp  Pressure Ratio, rp  P2 P1 272 Back Work Ratio • Back Work ratio in a gas turbine power plant is the ratio of the compressor work to the turbine work • This ratio is very high • More than half of the turbine work output is used to drive the compressor • In steam power plants the back work ratio is very low 273 Isentropic Turbine and Compressor Efficiency ws h2 s  h1 C   wa h2 a  h1 wa h3  h4 a T   ws h3  h4 s 274 Problem 1 The temperature at the beginning of the compression process of an air standard Otto cycle with a compression ratio of 8 is 300 K, the pressure is 1 bar, and the cylinder volume is 560 cm3. The maximum temperature during the cycle is 2000 K. Determine (a) the temperature and pressure at the end of each process, (b) the thermal efficiency and (c) the mean effective pressure. Ans: (b) 56.5% (c) 7.05 bar 275 Problem 2 A gas turbine power plant operating on an ideal Brayton cycle has a pressure ratio of 8. The gas temperature is 300 K at the compressor inlet and 1300 K at the turbine inlet. Determine (a) the gas temperature at the exits of the compressor and the turbine, (b) the back work ratio and (c) the thermal efficiency. Compute the back work ratio and thermal efficiency if the compressor efficiency is 80% and turbine efficiency is 85%. Ans: (a) 540 K, 717 K (b) 0.42 (c) 44.8% (a) 0.592 (b) 26.6% 276 MODULE 8 Vapour and Refrigeration Cycles The Carnot Vapour Cycle The reversible involved are: • • • • processes Isothermal heat addition (1-2) Isentropic expansion (2-3) Isothermal condensation (3-4) Isentropic compression (4-1) 278 Impracticalities of Carnot vapour Cycle • Isothermal heat transfer possible but only in two-phase region and hence low cycle temperatures • Low steam quality during isentropic expansion • Isentropic compression difficult to control and handle two-phase fluid 279 Problems with Super-Critical Carnot Vapour Cycle • Compression to extremely high pressures • Isothermal heat transfer at variable pressure 280 Ideal Rankine Cycle The processes in an Ideal Rankine Cycle are: • Isentropic Compression in a Pump (1-2) • Constant Pressure Heat Addition in a Boiler (23) • Isentropic Expansion in a Turbine (3-4) • Constant Pressure Heat Rejection in a Condenser (4-1) 281 Ideal ‘a ki e C le… Schematic Diagram T-S Diagram 282 Energy Analysis of the Ideal Rankine Cycle • The four devices operate under steady flow conditions • KE and PE changes are negligible • Boiler and Condenser do not involve any work • Pump and Turbine operate Isentropically 283 Energy Analysis of the Ideal Rankine C le… SFEE : Q  W  h Pump q  0 : w pump  h2  h1   P2  P1  Boiler w  0  : qin  h3  h2 Turbine q  0 : wturbine  h3  h4 Condenser w  0  : qout  h4  h1 wnet wturbine  w pump qout   1 th  qin qin qin Heat Rate: Heat supplied in kJ to generate 1 kWh of electricity 284 Deviation of Actual Vapour Cycle from Ideal Cycle • Deviations are a result of irreversibilities • Fluid Friction causes pressure drop • Heat loss from the steam to the surroundings results in lower efficiency • Due to irreversibilities in pump and turbine, a pump requires more work and a turbine produces less work 285 Deviation of Actual Vapour Cycle from Ideal C le…  pump  wisentropic wactual h2 s  h1 ;  h2 a  h1 wactual h3  h4 a turbine   wisentropic h3  h4 s 286 Methods to Improve Efficiency of Rankine Cycle Basic Principles used are: • Increase the average temperature of heat addition • Decrease the average temperature of heat rejection Methods Adopted are: • Lowering the Condenser Pressure • Superheating the steam to high temperatures • Increasing the Boiler Pressure 287 Methods to Improve Efficiency of ‘a ki e C le… Lowering the Condenser Pressure (Lowers Tlow,avg) 288 Methods to Improve Efficiency of ‘a ki e C le… Superheating the steam to high temperatures (Increases Thigh,avg) 289 Methods to Improve Efficiency of ‘a ki e C le… Increasing the Boiler Pressure (Increases Thigh,avg) 290 Ideal Reheat Rankine Cycle qin  q primary  qreheat  h3  h2   h5  h4  wturbine  wturbineI  wturbineII  h3  h4   h5  h6  291 Ideal Regenerative Rankine Cycle 292 Problem 1 A simple Rankine cycle with water as the working fluid operates between the pressure limits of 3 MPa in the boiler and 30 kPa in the condenser. If the quality at the exit of the turbine cannot be less than 85 percent, what is the maximum thermal efficiency this cycle can have? Ans: 29.7% 293 Problem 2 A simple ideal Rankine cycle with water as the working fluid operates between the pressure limits of 15 MPa in the boiler and 100 kPa in the condenser. Saturated steam enters the turbine. Determine the work produced by the turbine, the heat transferred in the boiler, and thermal efficiency of the cycle. If the irreversibilities in the turbine cause the steam quality at the outlet of the turbine to be 70%, determine the isentropic efficiency of the turbine and thermal efficiency of the cycle. Ans: 699 kJ/kg, 2178 kJ/kg, 31.4% 87.7%, 27.4% 294 Problem 3 Consider a steam power plant that operates on the ideal reheat Rankine cycle. The plant maintains the boiler at 4000 kPa, the reheat section at 500 kPa, and the condenser at 10 kPa. The mixture quality at the exit of both turbines is 90 percent. Determine the temperature at the inlet of each turbine and the le s thermal efficiency. Ans: 292 0C, 283 0C, 33.5% 295 Problem 4 A steam power plant operates on the reheat Rankine cycle. Steam enters the high-pressure turbine at 12.5 MPa and 550 0C at a rate of 7.7 kg/s and leaves at 2 MPa. Steam is then reheated at constant pressure to 450 0C before it expands in the low-pressure turbine. The isentropic efficiencies of the turbine and the pump are 85 percent and 90 percent, respectively. Steam leaves the condenser as a saturated liquid at 10 kPa. If the moisture content of the steam at the exit of the turbine is not to exceed 5 percent, determine (a) the net power output, and (b) the thermal efficiency. Ans: (a) 10.2 MW, (b) 36.5% 296 Desirable Characteristics of a Working Fluid of a Vapour Power Cycle • High critical temperature and a safe maximum pressure • Low triple point temperature to avoid solidification • Not too low condenser pressure • High enthalpy of vaporization • Good heat transfer characteristics • Must be inert, inexpensive, readily available and non-toxic 297 Binary Vapour Cycles • Binary cycle is a power cycle that is a combination of two cycles, one at high temperature and other at low temperature • The condenser of the high temperature cycle (Topping cycle) serves as a boiler for the low temperature cycle (Bottoming Cycle) • The heat output of the high temperature cycle is used as the heat input to the low temperature cycle • Working fluids suitable for the high temperature cycle are mercury, sodium, potassium and sodium-potassium mixtures 298 Bi a Vapou C les… 299 Bi a Vapou C les… • The Binary Vapour Cycle approximates the Carnot cycle more closely than the Rankine cycle for the same temperature limits • Hence the thermal efficiencies are higher • Thermal efficiencies of 50% and higher are possible • They are not economically attractive due to high initial cost and competition from Combined Cycle power plants (Combined Gas-Vapour Power Cycles) 300 Combined Gas-Vapour Power Cycles • Also known as Combined Cycle • The combined cycle of greatest interest is the Gas Turbine Cycle (Brayton Cycle) topping a Steam Turbine Cycle (Rankine Cycle) • The exhaust of a gas turbine is above 500 0C and the heat in these gases can be used to generate steam and operate a steam turbine • The result is a Combined Gas-Steam cycle • Recent Combined Cycle power plants have efficiencies above 60% 301 Combined Gas-Steam Power Plant 302 Refrigeration Cycles For heat transfer from low to high temperatures, special devices called Refrigerators are needed COPR  Desired Effect QL  Work Input Wnet,in COPHP  Desired Effect Q  H Work Input Wnet,in COPHP  COPR  1 303 Cooling Capacity of a Refrigerator • Heat removal rate of a refrigerator is expressed in Tons of Refrigeration • The capacity of a refrigeration system that can freeze 1 ton (2000 lb) of water at 0C to ice at 0C in 24 hours is called 1 Ton of Refrigeration • Taking 1 ton as 907 kg and latent heat of water as 334 kJ, 1 ton of refrigeration=211 kJ/min = 3.5 kW 304 Reversed Carnot Cycle 305 Problems with the Reversed Carnot Cycle • The compressor must handle two phase fluid • The turbine must expand to low quality • The above problems can be solved if the compression and expansion processes are moved to superheat region • In that case, the isothermal heat absorption and heat rejection will be difficult as pressure does t remain constant 306 Ideal vapour Compression Refrigeration Cycle 307 Processes in an Ideal vapour Compression Refrigeration Cycle • 1-2: Isentropic compression in the compressor • 2-3: Constant-Pressure heat rejection in a condenser • 3-4: Throttling in an expansion device • 4-1: Constant-Pressure heat absorption in an evaporator 308 Ideal vapour Compression Refrigeration Cycle On applying the SFEE COPR  QL h h  1 4; Wnet,in h2  h1 COPHP  h h QH  2 3 Wnet,in h2  h1 309 Actual Vapour Compression Cycle 310 Problem 5 A refrigerator uses refrigerant-134a as the working fluid and operates on an ideal vapor-compression refrigeration cycle between 0.12 and 0.7 MPa. The mass flow rate of the refrigerant is 0.05 kg/s. Show the cycle on a T-s diagram with respect to saturation lines. Determine (a) the rate of heat removal from the refrigerated space and the power input to the compressor, (b) the rate of heat rejection to the environment, and (c) the coefficient of performance. Ans: (a) 7.41 kW, 1.83 kW, (b) 9.23 kW, (c) 4.06 311 Problem 6 Refrigerant-134a enters the compressor of a refrigerator at 140 kPa and -10°C at a rate of 0.3 m3/min and leaves at 1 MPa. The isentropic efficiency of the compressor is 78 percent. The refrigerant enters the throttling valve at 0.95 MPa and 30°C and leaves the evaporator as saturated vapor at -18.5°C. Show the cycle on a T-s diagram with respect to saturation lines, and determine (a) the power input to the compressor, (b) the rate of heat removal from the refrigerated space, and (c) the pressure drop and rate of heat gain in the line between the evaporator and the compressor. Ans: (a) 1.88 kW, (b) 4.99 kW, (c) 1.65 kPa, 0.241 kW 312 Problem 7 Refrigerant-134a enters the compressor of a refrigerator as superheated vapour at 0.14 MPa and -10C at a rate of 0.12 kg/s, and it leaves at 0.7 MPa and 50 C. The refrigerant is cooled in the condenser to 24 C and 0.65 MPa, and it is throttled to 0.15 MPa. Disregarding any heat transfer and pressure drops in the connecting lines between the components, show the cycle on a T-s diagram with respect to the saturation lines, and determine (a) the rate of heat removal from the refrigerated space and the power input to the compressor, (b) the isentropic efficiency of the compressor, and (c) the COP of the refrigerator. Ans: (a) 19.4 kW, 5.06 kW, (b) 82.5%, (c) 3.83 313 Problem 8 A refrigerator uses refrigerant-134a as the working fluid and operates on the ideal vapour-compression refrigeration cycle. The refrigerant enters the evaporator at 120 kPa with a quality of 30 percent and leaves the compressor at 60C. If the compressor consumes 450 W of power, determine (a) the mass flow rate of the refrigerant, (b) the condenser pressure and (c) the COP of the refrigerator. Ans: (a) 0.00727 kg/s, (b) 672 kPa, (c) 2.43 314 Problem 9 A heat pump using refrigerant-134a as a refrigerant operates its condenser at 800 kPa and its evaporator at -1.25C. It operates on the ideal vapour-compression refrigeration cycle, except for the compressor, which has an isentropic efficiency of 85 percent. How much do the compressor irreversibilities reduce this heat pu p s COP as compared to an ideal vapour-compression refrigeration cycle. Ans: 13.1% 315 Problem 10 The liquid leaving the condenser of a 30 kW heat pump using refrigerant-134a as the working fluid is subcooled by 5.4C. The condenser operates at 1 MPa and the evaporator at 0.4 MPa. How does this subcooling change the power required to drive the compressor as compared to an ideal vapour compression refrigeration cycle. Ans: 3.41 kW, 3.25 kW 316 MODULE 9 Ideal Gas Mixtures Composition of a Gas Mixture • The Composition of a gas mixture is a description of the amount of quantity of each constituent present in the mixture • The Composition of a Gas Mixture can be expressed using Molar Analysis or Gravimetric Analysis • Molar Analysis is also known as Volumetric Analysis 318 Mass and Mole Fractions mm   mi k i 1 and n m   ni k mi M ass Fraction, mf i  mm M ole Fraction, y i   mf k i 1 i  1 and ni nm y k i 1 i i 1 1 Average M olecular Weight  M ass of the mixture mm  M oles in the mixture nm m n M    yM n n i m i m i k i 1 i i 319 Mass a d Mole F a tio s… mm mm Mm    mi nm M i 1  mi m M m i mi ni M i Mi   yi mfi  mm nm M m Mm 1 k mfi  i 1 M i 320 Ideal Gas Mixtures • When two or more ideal gases are mixed, the non-reacting mixture also behaves as an ideal gas • Air is an ideal gas even though it is a mixture of nitrogen and oxygen • The prediction of the P-V-T behaviour of gases is based on Dalto s law and A agat s law 321 Dalto s La of Additi e P essu es • The pressure of a gas mixture is equal to the sum of the pressures each gas would exert if it existed alone at the mixture temperature and volume 322 A agat s Law of Additive Volumes • The volume of a gas mixture is equal to the sum of the volumes each gas would occupy if it existed alone at the mixture temperature and pressure 323 Mathematical Representation • Dalto s a d A agat s laws hold exactly for ideal gases but approximately for real gases Pm   Pi Tm ,Vm  k Dalton ' s Law : i 1 Amagat ' s Law : Vm   Vi Tm ,Pm  k i 1 • Pi is called as the Component Pressure or Partial Pressure • Vi is called as the Component Volume or Partial Volume 324 Relation between Partial Pressure and Volume ni RuTm Pi Tm , Vm  Vm ni    yi n R T Pm nm m u m Vm ni RuTm Vi Tm , Pm  Pm ni    yi n R T Vm nm m u m Pm  Pi Vi n   i  yi Pm Vm nm The above result is valid strictly for ideal gases 325 Properties of Gas Mixtures-Extensive Properties U m   U i  mi ui   N i ui k k k i 1 i 1 i 1 k k k i 1 i 1 i 1 H m   H i  mi hi   N i hi S m   Si  mi si   N i si k k k i 1 i 1 i 1 326 Properties of Gas Mixtures-Intensive Properties um   mfi ui k i 1 hm   mfi hi kJ/kg and k i 1 sm   mfi si k i 1 kJ/kg and C p ,m   mfi C p ,i i 1 hm   yi hi kJ/kg  K and Cv ,m   mfi Cv ,i k i 1 kJ/mol k k i 1 u m   yi u i k i 1 kJ/mol sm   yi si kJ/kg  K and kJ/kg  K and kJ/kmol  K k i 1 C v , m   yi C v , i k i 1 C p , m   yi C p , i k i 1 kJ/kmol  K kJ/kmol  K 327 Problem 1 Consider a gas mixture that consists of 3 kg of O2, 5 kg of N2, and 12 kg of CH4. Determine (a) the mass fraction of each component, (b) the mole fraction of each component, and (c) the average molar mass and gas constant of the mixture. Ans: (a) 0.15, 0.25, 0.6 (b) 0.092, 0.175, 0.733 (c) 19.6 kg/kmol, 0.424 kJ/kg-K 328 Problem 2 An insulated rigid tank is divided into two compartments by a partition. One compartment contains 7 kg of O2 at 40 0C and 100 kPa, and the other compartment contains 4 kg of N2 at 20 0C and 150 kPa. Now the partition is removed, and the two gases are allowed to mix. Determine (a) the mixture temperature and (b) the mixture pressure after equilibrium has been established. Take Cv,N2=0.743 kJ/kg-K and Cv,O2=0.658 kJ/kg-K. Ans: (a) 32.2 0C (b) 114.5 kPa 329 Problem 3 A rigid tank contains 8 kmol of O2 and 10 kmol of CO2 gases at 290 K and 150 kPa. Estimate the volume of the tank. Ans: 289 m3 330 Problem 4 A 0.3 m3 rigid tank contains 0.6 kg of N2 and 0.4 kg of O2 at 300 K. Determine the partial pressure of each gas and the total pressure of the mixture. Ans: 178.1 KPa, 103.8 kPa, 282 kPa 331 Thank You