PPT-03609211-3- Laws of Thermodynamics.pptx mechANICAL

parul univarsity

1. Basic Thermodynamics ( 0360 9211 )

2. Laws of Thermodynamics CHAPTER- 3

3. Zeroth law of thermodynamics and its application The Zeroth Law of Thermodynamics states that if two systems are in thermal equilibrium with a third system, then they are in thermal equilibrium with each other. In other words, if two systems are at the same temperature as a third system, then they are at the same temperature as each other.

4. Application of Zeroth law of Thermodynamics Temperature Scales and Measurement : The Zeroth Law is the basis for establishing temperature scales like Celsius, Fahrenheit, and Kelvin. It allows us to define temperature and create standardized methods for temperature measurement using devices like thermometers. Thermal Equilibrium in Engineering : In engineering, understanding thermal equilibrium is crucial for designing and operating systems effectively. For instance, in the design of heat exchangers, it ensures that fluids reach equilibrium temperatures, maximizing heat transfer efficiency.

5. Calibration of Instruments : The Zeroth Law is essential for calibrating instruments used for temperature measurement, such as thermostats, temperature controllers, and thermal sensors. Calibration ensures accuracy by verifying that instruments reach equilibrium with known reference temperatures. Heating, Ventilation, and Air Conditioning (HVAC) : HVAC systems rely on the Zeroth Law to maintain desired temperatures in buildings and controlled environments. Thermostats use the principle of thermal equilibrium to regulate heating and cooling systems, ensuring comfort and energy efficiency.

6. Thermal Comfort in Architecture : Architects use principles derived from the Zeroth Law to design buildings that provide thermal comfort. By considering factors like insulation, solar orientation, and ventilation, they optimize indoor environments to maintain comfortable temperatures. Food Processing and Preservation : In the food industry, controlling temperature is critical for preserving freshness and safety. The Zeroth Law guides processes such as pasteurization, refrigeration, and freezing to ensure food products remain safe for consumption.

7. Medical Applications : The Zeroth Law is applied in medical diagnostics and treatments. For example, body temperature measurements help diagnose illnesses, monitor patients' health, and regulate medical equipment like incubators and dialysis machines. Environmental Science : Understanding thermal equilibrium is vital in environmental science for studying climate patterns, heat transfer in ecosystems, and temperature fluctuations in oceans and atmosphere.

8. Space Exploration : In space missions, maintaining thermal equilibrium is essential for spacecraft and equipment exposed to extreme temperature variations. Engineers use insulation, radiators, and active temperature control systems to manage thermal conditions. Material Science and Metallurgy : The Zeroth Law is utilized in material science and metallurgy for processes like annealing, tempering, and heat treatment, which involve controlled heating and cooling to alter material properties.

9. Joule's Experiment Joule's experiment, conducted by the English physicist James Joule in the mid-19th century, was crucial in establishing the concept of the mechanical equivalent of heat, which laid the foundation for the First Law of Thermodynamics. The setup and significance of Joule's experiment are as follows:

10. Setup of Joule's Experiment : Joule's experiment involved a simple apparatus consisting of a paddlewheel or turbine immersed in a container of water. The paddlewheel was connected to a system for applying mechanical work, such as weights attached to a pulley system. When the weights were allowed to fall, they exerted a force on the paddlewheel, causing it to rotate and agitate the water . By measuring the increase in temperature of the water and the amount of mechanical work done on the system, Joule sought to establish a relationship between mechanical work and heat transfer.

11. Mathematically his conclusion can be represented as Work ∞ Heat W ∞ Q W= JQ Where J= Joules constant

12. From this observation Joule postulated the  First Law of Thermodynamics   which states that if a closed system undergoes a cyclic process, energies transacted by it in the form of work and heat are equal. Mathematically, (Units: J or kJ) In a cycle involving two processes, the First Law is expressed as  Q 1-2  + Q 2-1  = W 1-2  + W 2-1 For a four process cycle, it is expressed as Q 1-2  + Q 2-3  +    Q 3-4  + Q 4-1  = W 1-2  + W 2-3  +    W 3-4  + W 4-1

13. First law of thermodynamics The First Law of Thermodynamics states that energy can be converted from one form to another with the interaction of heat, work and internal energy, but it  cannot  be created nor destroyed, under any circumstances. Mathematically , this is represented as, Δ U = q + w Where, ΔU = Change in internal energy of the system q = Algebraic sum of heat transfer between system and surroundings W = Work interaction of the system with its surroundings

14. Enthalpy Enthalpy (H) of a substance at any point is quantification of energy content in it, which could be given by summation of internal energy and flow energy. Enthalpy is very useful thermodynamic property for the analysis of engineering systems H = U + PV On unit mass basis, the specific enthalpy could be given as, h = u + pv

15. FIRST LAW OF THERMODYNAMICS APPLIED TO OPEN SYSTEMS Let us take an open system having steady flow. Figure shows steady flow system having inlet at section 1–1, outlet at section 2–2, heat addition Q and work done by the system W .

16. Consider flow of fluid through a generalized open system as shown in figure. The working fluid enter the system at section 1 and leave the system at section 2 and passing at a steady rate. Let, m= mass flowing through the control volume, kg/s Q= heat entering the control volume, kJ W= work transferred from control volume, kJ C1,C2= velocity of fluid at entrance and exit, m/s p1,p2 = pressure of fluid at entrance and exit, N/m2 u1,u2 =internal energy per kg of fluid at entrance and exit, kJ/kg v1,v2 = specific volume of fluid at entrance and exit m3/kg h1,h2 = enthalpy of fluid at entrance and exit, kJ/kg w= work transferred from the control volume per kg of fluid, kJ/kg z1,z2 = elevation of entrance section and exit section, m.

17. The energy balance required in open system for flow process may be written as follows

18. For steady flow process, increase of stored energy within the control volume is zero.

19. Condition for Steady flow: Mass flow rate should be constant with time Fluid properties at any point within system and at boundaries of the system should remain constant with time. Energy flow rates across the boundaries of the system should remains constant with time.

20. This equation is called steady flow energy equation(SFEE).

21. Limitation of First law of Thermodynamics: How much quantity of one form of energy can be converted in to other form? i.e. information about quantity of energy to be converted For given two forms of energy, which form can be converted in to other form? i.e. information about direction of energy conversion process. Which types of situation is necessary form energy conversion process i.e. information about possibility of energy conversion process.

22. Image source : Google Second law of Thermodynamics Heat Source and Heat Sink Thermal Energy Reservoir : Hypothetical body with relatively very large Thermal Energy Capacity ( mass x Sp. Heat ) that can supply or absorb finite amount of Heat without undergoing change in Temperature. e.g. ocean, lake, atmosphere, two-phase system, industrial furnace, etc.

23. Image source : Google Reservoir that supplies Energy in form of Heat is known as SOURCE . Source Heat Reservoir that absorbs Energy in form of Heat is known as SINK . Sink Heat

24. Image source : Google Heat Engine In heat engine the heat supplied to the engine is converted into useful work. If Q h  is the heat supplied to the engine and Q c  is the heat rejected from the engine, the net work done by the engine is given by: 𝑊 = Q h   − Q c The energy efficiency of a heat engine is given by where, W = Net work transfer from the engine, and Q 1 = Heat transfer to engine.

25. Image source : Google Heat Pump and Refrigerator Refrigerators, air conditioners, and heat pumps all use work to transfer heat from a cold object to a hot object: Q h = 𝑊 + Q c

26. Image source : Google The coefficient of performance, (COP) of refrigerators is, The COP for a heat pump is

27. Image source : Google Heat engine efficiency (COP) of refrigerators COP for a heat pump Relation between Heat pump and Refrigerator

28. Second Law of Thermodynamics Kelvin-Planck statement Clausius statement

29. 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. The Kelvin-Planck statement of the second law of thermodynamics states that no heat engine can produce a net amount of work while exchanging heat with a single reservoir only. In other words, the maximum possible efficiency is less than 100 percent. Fig.4.8 Heat Engine

30. 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”. Fig.4.9 Refrigerator

31. Carnot Cycle The cycle was first suggested by a French engineer Sadi Carnot in 1824 which works on reversible cycle and is known as Carnot cycle . The assumptions made for describing the working of the Carnot engine are as follows : ( i ) The piston moving in a cylinder does not develop any friction during motion. ( ii ) The walls of piston and cylinder are considered as perfect insulators of heat. ( iii ) The cylinder head is so arranged that it can be a perfect heat conductor or perfect heat insulator. ( iv ) The transfer of heat does not affect the temperature of source or sink. ( v ) Working medium is a perfect gas and has constant specific heat. ( vi ) Compression and expansion are reversible.

32. In a Carnot cycle, the system executing the cycle undergoes a series of four internally reversible processes: two adiabatic processes alternated with two isothermal processes. Fig.4.11 Carnot cycle on P-V and T-S diagram

33. Four stages of Carnot cycle : Stage 1. (Process 1-2). Hot energy source is applied. Heat Q 1 is taken in whilst the fluid expands isothermally and reversibly at constant high temperature T 1. Fig.4.12 stage-1

34. Stage 2. (Process 2-3). The cylinder becomes a perfect insulator so that no heat flow takes place. The fluid expands adiabatically and reversibly whilst temperature falls from T 1 to T 2. Fig.4.13 stage-2

35. Stage 3. (Process 3-4). Cold energy source is applied. Heat Q 2 flows from the fluid whilst it is compressed isothermally and reversibly at constant lower temperature T 2. Fig.4.14 stage-3

36. Stage 4. (Process 4-1). Cylinder head becomes a perfect insulator so that no heat flow occurs. The compression is continued adiabatically and reversibly during which temperature is raised from T 2 to T 1. Fig.4.15 stage-4

37. The work delivered from the system during the cycle is represented by the enclosed area of the cycle. Again for a closed cycle, according to first law of the thermodynamics the work obtained is equal to the difference between the heat supplied by the source ( Q 1) and the heat rejected to the sink ( Q 2). ∴ W = Q 1 – Q 2 Also, thermal efficiency,

38. The Carnot cycle cannot be performed in practice because of the following reasons : 1. It is impossible to perform a frictionless process. 2. It is impossible to transfer the heat without temperature potential. 3. Isothermal process can be achieved only if the piston moves very slowly to allow heat transfer so that the temperature remains constant. 4. Adiabatic process can be achieved only if the piston moves as fast as possible so that the heat transfer is negligible due to very short time available. The isothermal and adiabatic processes take place during the same stroke therefore the piston has to move very slowly for part of the stroke and it has to move very fast during remaining stroke. This variation of motion of the piston during the same stroke is not possible.

39. Numerical Exp.1 A heat engine receives heat at the rate of 1500 kJ/min and gives an output of 8.2 kW. Determine : The thermal efficiency ; (ii) The rate of heat rejection. Solution. Heat received by the heat engine, Q1 = 1500 kJ/min =1500/60 = 25 kJ/s Work output, W = 8.2 kW = 8.2 kJ/s. Thermal efficiency, η th = W/Q1 = 8.2/25 =0.328 = 32.8% (ii) Rate of heat rejection, Q2 = Q1 – W = 25 – 8.2 = 16.8 kJ/s Hence, the rate of heat rejection = 16.8 kJ/s.

40. Exp.2 During a process a system receives 30 kJ of heat from a reservoir and does 60 kJ of work. Is it possible to reach initial state by an adiabatic process ? Solution. Heat received by the system = 30 kJ Work done = 60 kJ Process 1-2 : By first law of thermodynamics, Q1–2 = (U2 – U1) + W1–2 30 = (U2 – U1) + 60 ∴ (U2 – U1) = – 30 kJ. Process 2-1 : By first law of thermodynamics, Q2–1 = (U1 – U2) + W2–1 ∴ 0 = 30 + W2–1 ∴ W2–1 = – 30 kJ. Thus 30 kJ work has to be done on the system to restore it to original state, by adiabatic process.

42. Hence transfer rate in condenser = Q1 According to the first law Q1 = Q2 + W = 12000 + 0.75 × 60 × 60 = 14700 kJ/h Hence, heat transfer rate = 14700 kJ/h.

43. Exp.4 A domestic food refrigerator maintains a temperature of – 12°C. The ambient air temperature is 35°C. If heat leaks into the freezer at the continuous rate of 2 kJ/s determine the least power necessary to pump this heat out continuously. Solution. Freezer temperature, T2 = – 12 + 273 = 261 K Ambient air temperature, T1 = 35 + 273 = 308 K Rate of heat leakage into the freezer = 2 kJ/s Least power required to pump the heat : The refrigerator cycle removes heat from the freezer at the same rate at which heat leaks into it as shown in Fig.

44.   For minimum power requirement   = ( 2/261) × 308 = 2.36 kJ/s W = Q1 – Q2 = 2.36 – 2 = 0.36 kJ/s = 0.36 kW Hence, least power required to pump the heat continuously = 0.36 kW.

45. Exp.5 A house requires 2 × 105 kJ/h for heating in winter. Heat pump is used to absorb heat from cold air outside in winter and send heat to the house. Work required to operate the heat pump is 3 × 104 kJ/h. Determine : (i) Heat abstracted from outside ; (ii) Co-efficient of performance. Solution. ( i ) Heat requirement of the house, Q 1 (or heat rejected) = 2 × 105 kJ/h Work required to operate the heat pump, W = 3 × 104 kJ/h Now, Q 1 = W + Q 2 where Q 2 is the heat abstracted from outside. ∴ 2 × 105 = 3 × 104 + Q 2 Thus Q 2 = 2 × 105 – 3 × 104 = 200000 – 30000 = 170000 kJ/h Hence, heat abstracted from outside = 170000 kJ/h.

46.   Hence, co-efficient of performance = 6.66. (ii) (C.O.P.)heat pump

47. Image source : Google

48. Image source : Google

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