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1 UNIT - I DEFINITIONS AND BASIC CONCEPTS 1.1 What is Thermodynamics? The science, which deals the analysis of various machines by quantity, which involves the transfer of energy into useful work, is called thermodynamics. Many energy conversion devices require the transfer of energy into work. Thermodynamics is applied in various thermal equipments like steam turbine, boiler, condenser, cooling tower, heat exchanger, reciprocating engines, heat pump etc. It is also used in internal combustion engines, turbo jets and rockets. In all these cases, design of thermal equipments essentially requires the in depth knowledge of thermodynamics. Work transfer maybe in any form – Mechanical work, electrical work or from surface tension or magnetic field. Hence the study of work and energy transfers is very important in the field of engineering. 1.2 UNITS SI unit system has three divisions: 1.Base units 2. Derived units 3. Supplementary unit 1.2.1 Base units: The base unit is the first division of SI units in which seven kinds of fundamental quantities are measured. They are given in the table 1.1 Metre (m): Kilogram (Kg Ampere (A Kelvin (K):It Candela (L) : Mole (mol): 1.2.2. Derived units: 1.2.3. Supplementary units: Radian (rad): Steradian (Sr) : 1.3 Important units 1.3.1.Mass and weight: 1.3.2. Pressure: There are various units followed for pressure. They are given below. 1 bar Pa = = 10 5 N / m 2 1N/m2 2 1 Atmospheric pressure Atmospheric pressure 1mm of Hg 1N/m2 = = = = 1.013 x 10 5 N / m 2 760 mm of Hg 1013 x 10 2 / 760 = 760 / 1013 x10 2 = 1ata (atmospheric technical absolute) = 133.3 N / m 2 7.5 x 10 -3 mm of Hg 736 mm of Hg 1.3.2.1. Elements of Measurements of Pressure : There are three elements in measurement of pressure. 1. Atmospheric pressure 2. Gauge pressure 3. Absolute pressure Atmospheric pressure (Patm Gauge pressure (Pg): Absolute pressure: The relation between above three pressures is given below : pabs = patm + pg * Normal temperature and pressure ( NTP ): Normal temperature and pressure refers to the conditions of atmospheric pressure at 760 mm of Hg and 20 ºC Standard temperature and pressure (STP): STP refers that the temperature and pressure of any gas in standard atmospheric pressure are taken at 760 mm of Hg and 0 ºC. 1.3.2.2 Pressure measurement: (i) The Piezometer Tube: (ii) U-Tube manometer: Manometer is commonly used for measuring pressure. It consists of a glass U-tube. Mercury is filled in the U-tube. One end of the manometer is connected to the tank pipe in which the pressure of fluid is to be measured and the other end is open to the atmosphere. The gauge pressure and vacuum pressure can be measured using U tube manometers (Fig 1.4). (iii) Mechanical gauge: 1.3.3.ZEROTH LAW OF THERMODYNAMICS : Temperature: Two scales are used to measure the temperature 1. Centigrade scale 2. Fahrenheit scale Centigrade or Celsius scale: Fahrenheit scale: Absolute temperature: 3 Measurement of Temperature: (i) Gas Filled Thermometer: (ii) Electrical Resistance Thermometer: (iii) Thermocouple: A few thermocouple metal pairs that are commonly used for temperature measurements are Iron – Constantan, Copper – Constantan, Chromel – Alumel, Platinum/Platinum rhodium etc. (iv) Pyrometers: 1.3.4 FORCE: 1.3.5 ENERGY: A capacity to do work is called energy. Energy may be given in two forms: a) Stored energy, b) Transit energy Stored energy: Transit energy: Kinetic energy: Potential energy: Flow energy: Energy is a thermodynamic property of system. The energy is a point function and the value of it depends on the end states only and not on the path by which the change of state occurs in a process. LAW OF CONSERVATION OF ENERGY: Concepts Of Thermodynamics : Thermodynamic system: Boundary Surroundings Boundary System System Gas filled in a vessel Fig 1.9 (a) Fig 1.9 (b) 4 Types of thermodynamic systems: Thermodynamic systems are generally classified in to three as given below: Thermodynamic systems Closed system Open system Isolated system Closed system: A system, which has a fixed mass in it and allows only energy transfer in to the system and out of the system, is referred as closed system. Open system: Isolated system: Thermodynamic Properties: Thermodynamic property is classified in to two types: 1. Intensive property 2. Extensive property Intensive property: Extensive property: State: Change of state: Thermodynamic Process: Path: Thermodynamic cycle: Equilibrium: Mechanical Equilibrium:. Chemical equilibrium: Thermal equilibrium: Thermodynamic equilibrium: Reversible process It is defined as a process, which proceeds in such a manner that the system remains , almost close to an equilibrium state at all time . It is also called quasi static process: It is shown in Fig 1.13. 2 1 Equilibrium states Fig 1.13 5 1.3.6 WORK : Work developed or work output Work done or work input T C work done from turbine (Positive work) work to the compressor (Negative work) In thermodynamics, the work done by the system is considered as positive work transfer. The work done on the system is considered as negative work transfer. There are a few forms of work transfer considered in thermodynamics. They are given below: 1. Displacement work 2. Shaft work 3. Electrical work 4. Paddle wheel work 5. Flow work 6. Work done of a stretching wire 7. Work done in changing the area of surface film 8. Magnetization of paramagnetic solid 9. Free expansion Form of work Compression or expansion work Shaft work Electrical work Flow work Magnetic work Surface tension Paddle wheel work Work in a stretching wire Various forms of work transfer Force Type of displacement Pressure Change in volume Work Torque Electric field pressure Magnetic field Surface tension Torque Rotation Polarization Volume Magnetic moment Peripheral area Change in volume Pdv (flow process) -vdp (non flow process) 2NT EI pv -H.dI -dA -Tdv Stress Strain - AL2 / 2 1.3.7 HEAT : It is a form of energy in transit, which occurs due to the temperature difference between two points in a substance or a body. Heat can be transferred across the boundary only when a system at a higher-level temperature to a system at a lower level temperature. It can never happen in the reverse direction by nature. This contrary will be discussed in the Second law of thermodynamics. Heat is measured in Joules or Kilo Joules. Heat can only be sensed due to the temperature difference. Heat can be transferred by any of the following modes: (i) Conduction (ii) Convection (iii) Radiation 1) Conduction: It is the transfer of heat within a body or more than a body by the physical contact but in the absence of fluid motion. Example for conduction is a metal rod heated up at one end by an external source of heat. Heat is transferred from the source end to the other end of the rod by movement of molecules from one end to the other end. 2) Convection: 6 It is the transfer of heat within a fluid or gas or between two mediums by the motion of molecules from one region to another region. Water heated up in a vessel is an example for convection heat transfer. 3) Radiation : It is the transfer of heat from hot body to cold body by movement particles in the space or between gas mediums. 1.3.7.1 Heat may be classified into two forms : 1. Sensible heat 2. Latent heat 1. Sensible heat: 2. Latent heat : Three types of latent heat are: a) Latent heat of melting or fusion b) Latent heat of freezing or sublimation c) Latent heat of vaporization or evaporation 1.3.7.2 Specific heat capacity: The values of specific heat for heating and cooling the gas may be taken in two ways : 1) Specific heat at constant pressure (Cp) 2) Specific heat at constant volume (Cv) . It is to be noted that the Cp value is always greater than Cv. Specific heat at constant pressure ( Cp ): Specific heat at constant pressure ( Cv ): 1.3.7.2 Molar specific heats of gases : 1.4 FIRST LAW OF THERMODYNAMICS: [Qalgebraic] = [Walgebraic ] cycleQ = cycleW Qb Qa System Wa Boundary Wb Fig 1.22 1.4.1 Perpetual motion machine of first kind (PMM – I ): 7 1.4.2 VARIOUS THERMODYNAMIC PROCESSES : Non-Flow Process Fig 1.24 Reversible Non-flow process Irreversible Non-Flow process Flow Process Fig 1.25 Steady flow Process Unsteady flow process REVERSIBLE NON-FLOW PROCESS Constant volume Process Isochori c Constant Pressure Constant Temperature Process Process Isobari c Isothermal Hyperbolic ReversibleAdi abatic Process Process Isentropi c Fig 1.26 1.4.2.1 Constant volume process: 1. P-V-T relationship: We know that the general gas equation is PV = m RT 2 P/T = Constant P/T = P1/T1 = P2/T2 P 1 V 1.4.2.2 Constant pressure process: 1. P-V-T relationship: Polytropic Process 8 We know that the general gas equation is PV = m RT V/T = Constant V/T = V1/T1 = V2/T2 p 1 2 WORK TRANSFER W1-2 V 1.4.2.3. Constant Temperature process: 1. P-V-T relationship: We know that the general gas equation is PV = m RT As the temperature is constant during the process the gas equation becomes pV = constant P V Fig 1.29 1.4.2.4 Reversible Adiabatic process: 1. P-V-T relationship: We know that the general gas equation is PV = m RT pV = Constant pV = P1V1 = P2V2 pV = Constant P Work transfer W1-2 V 1.4.2.5. Polytropic process : 1. P-V-T relationship: We know that the general gas equation is PV = m RT pV n = Constant pV n = P1V1n = p2V2n pV n = Constant 9 It is important to note that the sign conversions for heat and work should be carefully followed in case of heating or expansion and cooling or compression of gases. The expansion of gases is considered to be positive work and cooling of gas is considered as negative work. Similarly the expansion or heating of gas in which positive heat transfer is considered whereas cooling or compression of gas is negative heat transfer. Flow processes or open systems : Steady flow energy equation ( SFEE ) As per energy balance at both the sides of inlet and outlet Sum of all energies at inlet = Sum of all energies at outlet Heat transfer + [ Kinetic energy + Potential energy + flow energy + Internal energy ] inlet = Work transfer + [ Kinetic energy + Potential energy + flow energy + Internal energy ] outlet Applications of steady flow processes : 1. Boiler 2. Evaporator 3. Condenser 4. Heat exchanger 5. Steam nozzle 6.Turbine 7. Reciprocating compressor 8. Rotary compressor 1) Boiler : Boiler is a closed vessel in which steam is generated by adding heat to water. Boiler is assumed as a steady flow heat absorbing system. Assumptions made : 1. No work transfer (W = 0) 2. Change in kinetic energy is negligible 3. Change in potential energy is negligible 2)Evaporator: 3) Condenser : 4) Turbine : 5) Rotary compressor: 6) Reciprocating compressor : 7) Steam nozzle 10 SOLVED PROBLEMS IN FIRST LAW OF THERMODYNAMICS , NON FLOW PROCESSES AND FLOW PROCESSES Summary : First law of thermodynamics applied to A ) Thermodynamic cycle : Q = W B ) During a change of state in a process : Q = W + U Table: 1.5 Type of P-V-T Work Change Heat Change P-V Process relationship transfer in transfer in diagram internal enthalp energy y 1 Constant P volume P/T = C 2 process 0 m Cv T m Cv T m Cp V P/T = P1/T1 T ( Isochoric = P2/T2 ) Constant 1 2 pressure V/T = C P process V/T = V1/T1 P (V2- V1) m Cp m Cv T m Cp T = T V (Isobaric) V2/T2 Constant Temperatu re process pV = C pV = p1V1 = pV2 2.3 P1V1 log10 (V2/V1 ) 0 2.3 P1V1 log10 (V2/V1 ) P 0 Reversible adiabatic process 1 pV = C pV = p1V1 = pV2 2.3 P1V1 log10 (V2/V1 ) 0 2.3 P1V1 log10 (V2/V1 ) P =C pV = p1V T 2 P 2 T = pV m Cv T 0 2 V P [ P2V2 P1V1 ] / (1) 1 1 2 m Cp T V 2 PVn = C PVn = p1V 1 n = pV 2 n [ P2V2 P1V1 ] / (n1) m Cv T [ P2V2 P1V1 ] / (n1) + mCv ( T2 – T1) 1 m Cp T 2 1 m Cv T pV T 1 P V 1 Polytropic process 2 P 1 1 (Isotherma l) Hyperboli c process P-T Diagram P 2 V P 2 T 11 Problem 1: A system undergoes a thermodynamic cycle . Four processes are involved in the cycle. The values of work transfer , heat transfer and internal energy in each process are tabulated in the table . Find rate of work. Check the work and heat transfer follows first law. Thermodynamic process A-B B-C C-D D-A Work transfer KJ / min 250 -800 -600 --- Change in internal energy kJ / min -400 100 Heat transfer KJ / min 300 -200 50 Process A- B : Q A- B = 300 = U A- B W A- B + U A- B 250 + U = 50 kJ/min Process B - C : Q B- C = Q B- C Q B- C W B - C+ U B-C = -800 +400 = -400 kJ/min Process C - D : Q C-D = 200 = U = W C - D + U C - D -600 +U C - D 800 kJ/min Process D - A : Q D-A = 50 = W = W D - A + U D - A W +100 -50 kJ/min Thermodynamic cycle : Q = W Qalgebric = Walgebric Q A-B + Q B-C + Q C-D + QD-A = W A-B +W B-C + W C-D +W D-A 300 –400+200 + 50 = 250 +400 - 600 +100 CONSTANT VOLUME PROCESS: 12 Problem 2: A cylindrical vessel has a diameter of 1.1 m contains 1kg mole of nitrogen at 100 º C. If he gas is cooled to 60 o Cat constant volume , calculate the Work transfer , change in internal energy , change in enthalpy and heat transfer. The ratio of the specific heats is 1.4 and 1 kg of mole nitrogen is 28. Assume that the stroke of the cylinder is equal to diameter of the cylinder. 4. Change in enthalpy H = m cp (T2 -T1 ) kJ H = 1 x 1.036 x (333-373 ) H = 1 x 1.036 x (-40) H = -41.44 kJ/mole CONSTANT PRESSURE PROCESS: Problem 3: 0.5 kg of a perfect gas is heated from 100 º to 300 º C at a constant pressure of 2.8 bar.. Determine the work transfer , change in internal energy , change in enthalpy , heat transfer. Take gas constant R = 0.28 kJ/kg K. Cp = 1 kJ/kg K & Cv = 0.72 kJ/kg K Given: m = T1 = 0.5 kg; 100 + 273 = 373 K; p T2 = = 2.8 bar = 2.8 x 10 5 N/m2; 300 + 273 = 573 K; CONSTANT TEMPERATURE PROCESS: Problem 4: A certain quantity of a perfect gas is heated in a reversible isothermal process from 1 bar and 40 o C to 10 bar. Find the work done per kg of gas . Take R = 287 J/Kg K. T2 P T1 T Given: T = p1 = 40 + 273 = 313 K; 1 bar = 1 x 10 5 N/m2; m = 1 kg; p2 = 10 bar = 10 x 10 5 N/ m2; POLYTROPIC PROCESS: Problem 5: 1 kg of air at a pressure of 7 bar and a temperature of 363 K undergoes a reversible polytropic process which may be represented by p v 1. 1 = constant. The final pressure is 1.4 bar. Evaluate a) the final specific volume, temperature . b) the work done and the heat transfer during the process. Assume R = 287 J/Kg K and = 1.4. Given: T1 = p1 = R = 363 K; 7 bar = 7 x 10 5 N/m2; 287 J/Kg K; m p2 = = = 1Kg; 1.4; 1.4 bar = 1.4 x 10 5 N/m2; 13 n = 1.1. Problem 6: A steam turbine in a steam power plant receives steam at 40 bar. The specific enthalpy and specific volume of steam at inlet of turbine are 3214 kJ/kg and 0.07m3/kg, respectively. Steam after expanding on turbine has the pressure at exit of 35 bar. The enthalpy and volume of steam are 3202.6 kJ/Kg and 0.084 m3/Kg, respectively. If the steam is carried in a 0.2m diameter pipe the heat loss in the pipeline is 8.5 kJ/Kg. Calculate the mass flow rate of steam. Given: p1 = v1 = h2 = Q = 40 bar; 0.073 m3/kg; 3202.6KJ/kg; 8.5 kJ/kg. h1 p2 v2 = = = 3214 kJ/kg; 35 bar; 0.084 m3/Kg; Problem 7: Air enters a compressor with a velocity of 60m/s, pressure of 100 kpa, temperature of 40 C and leaves the compressor with a velocity of 90m/s, pressure of 500 kpa, temperature of 120 C . Consider the system is adiabatic. Find the power of the motor for a mass rate of flow of 40 kg/min. Write the assumptions made. Given: V1 = T1 = p2 = m = 60m/s; p1 = 313 K; V2 500 kpa = 500 x 1000 N/ m2; T2 40 kg/min=0.66 kg/s. 100 kpa=100 x 1000 N/ m2; = 90m/s; = 393 K; Problem 8: Air is compressed from 100 kpa and at 22 C to a pressure of 1Mpa while being cooled at the rate of 16kJ/kg by circulating the water through the compressor casing. The volume flow rate of air inlet condition is 150m3/min and the power input to the compressor is 500Kw. Determine Mass flow rate. b) Temperature of air exit. Neglect the datum head. Given: p1 = q = w = Problem 9: 100 kpa; 16 kJ/s; -500 kw; p2 v1 T1 = = = 1Mpa; 150 m3/min =2.5m3/s; 22 + 273 = 295K. 14 Steam enters an adiabatic turbine at 10Mpa and 500 oC at the rate of 3 kg/s and leaves at 50Kpa. If the power output of the turbine is 2Mw, determine the temperature of steam at exit of turbine. Neglect the kinetic energy changes. Given: P1 = m = w = 10Mpa = 10 x 10 6 N/m2; 3 kg/s; p2 2Mw = 2 x 10 3Kw. T1 = = 500 + 273 = 773 K; 50 kpa = 50 x 10 3 N/m2; Problem 10: A steady flow system in which 1 kg/s of air at 20ºC and consumes a power of 15 kw. The inlet and outlet velocities of air are 100 m/s and 150 m/s respectively. Find the exit of air temperature, assuming adiabatic conditions. Take the cp of air is 1.005 kJ/Kg K. Given: T1 = V1 = 20 + 273 = 293 K; 100 m/s; w V2 = = 15 kw; 150 m/s; Problem 11: In an heat exchanger, oil is cooled from 90oC to 30oC. Water is used to cool the oil. Inlet temperature of water is 25oC and the outlet temperature is 70oC. The enthalpies of oil and water at inlet and outlet conditions are given below. Water : hinlet = 593.64 kJ/kg; houtlet = 699.77 kJ/kg; Oil : hoilin = 748.19 kJ/kg; hoilout = 605.43 kJ/kg; Calculate the mass flow rate of water required to cool the oil for cooling 2.78 kg/s of oil ? Given: Tw1 = 25 + 273=298 K; Tw2 = 90 + 273=363 K; Toil1 = 70 + 273=343 K; Toil2 = 30 + 273=303 K; moil = 2.78 kg/s. OBJECTIVE TYPE QUESTIONS : 1.Which one is not the extensive property of a thermodynamic system ? (i) Mass (ii)Total volume (iii) Total internal energy (iv) Density 2.One bar in SI unit is equal to (a) 1x10 5 N/m2 (b) 1x10 4 N/m2 (c) 1x10 3 N/m2 (d) 1 N/m2 3. Unit of power is (a) Watt (b)Joule (c) Joule-metre (d) None of the above 4. The law that states that heat and work are mutually convertible is known as (a) Zeroth law of thermodynamics (b) first law of thermodynamics (c) Second law of thermodynamics (d) None of the above