1 / 103

111302 Aero Engineering Thermodynamics by Mr.Suresh Chandra Khandai

111302 Aero Engineering Thermodynamics by Mr.Suresh Chandra Khandai. Unit - I. Thermodynamic Systems, States and Processes. Objectives are to: define thermodynamics systems and states of systems explain how processes affect such systems

mattox
Download Presentation

111302 Aero Engineering Thermodynamics by Mr.Suresh Chandra Khandai

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. 111302 Aero Engineering ThermodynamicsbyMr.Suresh Chandra Khandai

  2. Unit - I

  3. Thermodynamic Systems, States and Processes Objectives are to: • define thermodynamics systems and states of systems • explain how processes affect such systems • apply the above thermodynamic terms and ideas to the laws of thermodynamics

  4. Internal Energy of a Classical ideal gas • “Classical” means Equipartition Principle applies: each molecule has average energy ½ kT per in thermal equilibrium. At room temperature, for most gases: • monatomic gas (He, Ne, Ar, …) • 3 translational modes (x, y, z) diatomic molecules (N2, O2, CO, …) 3 translational modes (x, y, z) + 2 rotational modes (wx, wy)

  5. Internal Energy of a Gas A pressurized gas bottle (V = 0.05 m3), contains helium gas (an ideal monatomic gas) at a pressure p = 1×107 Pa and temperature T = 300 K. What is the internal thermal energy of this gas?

  6. WORK done by the system on the environment Changing the Internal Energy • Uis a “state” function --- depends uniquely on the state of the system in terms of p, V, T etc. • (e.g. For a classical ideal gas, U = NkT) • There are two ways to change the internal energy of a system: Wby = -Won HEAT is the transfer of thermal energy into the system from the surroundings Q Thermal reservoir Work and Heat are process energies, not state functions.

  7. Increase in volume, dV +dV Positive Work (Work is done by the gas) -dV Negative Work (Work is done on the gas) Work Done by An Expanding Gas The expands slowly enough to maintain thermodynamic equilibrium.

  8. +dV Positive Work (Work is done by the gas) -dV Negative Work (Work is done on the gas) A Historical Convention Energy leaves the system and goes to the environment. Energy enters the system from the environment.

  9. Total Work Done To evaluate the integral, we must know how the pressure depends (functionally) on the volume.

  10. Work depends on the path taken in “PV space.” The precise path serves to describe the kind of process that took place. Pressure as a Function of Volume Work is the area under the curve of a PV-diagram.

  11. Different Thermodynamic Paths The work done depends on the initial and final states and the path taken between these states.

  12. p p p V V V Work done by a Gas • When a gas expands, it does work on its environment • Consider a piston with cross-sectional area A filled with gas. For a small displacement dx, the work done by the gas is: • We generally assume quasi-static processes (slow enough that p and T are well defined at all times): This is just the area under the p-V curve dx dWby = F dx = pA dx = p (A dx)= p dV Note that the amount of work needed to take the system from one state to another is not unique! It dependson the path taken.

  13. Up to mid-1800’s heat was considered a substance -- a “caloric fluid” that could be stored in an object and transferred between objects. After 1850, kinetic theory. • A more recent and still common misconception is that heat is the quantity of thermal energy in an object. • The term Heat (Q) is properly used to describe energy in transit, thermal energy transferred into or out of a system from a thermal reservoir … (like cash transfers into and out of your bank account) Q U What is Heat? • Q is not a “state” function --- the heat depends on the process, not just on the initial and final states of the system • Sign of Q : Q > 0 system gains thermal energy • Q < 0 system loses thermal energy

  14. An Extraordinary Fact The work done depends on the initial and final states and the path taken between these states. BUT, the quantity Q - W does not depend on the path taken; it depends only on the initial and final states. Only Q - W has this property. Q, W, Q + W, Q - 2W, etc. do not. So we give Q - W a name: the internal energy.

  15. The First Law of Thermodynamics (FLT) -- Heat and work are forms of energy transfer and energy is conserved. U = Q + Won change in total internal energy work done on the system heat added to system State Function Process Functions or U = Q - Wby

  16. 1st Law of Thermodynamics • statement of energy conservation for a thermodynamic system • internal energy U is a state variable • W, Q process dependent

  17. The First Law of Thermodynamics What this means: The internal energy of a system tends to increase if energy is added via heat (Q) and decrease via work (W) done by the system. . . . and increase via work (W) done on the system.

  18. Isoprocesses • apply 1st law of thermodynamics to closed system of an ideal gas • isoprocess is one in which one of the thermodynamic (state) variables are kept constant • use pV diagram to visualise process

  19. Isobaric Process • process in which pressure is kept constant

  20. Isochoric Process • process in which volume is kept constant

  21. Isothermal Process • process in which temperature is held constant

  22. Thermodynamic processes of an ideal gas( FLT: DU = Q - Wby ) 2 p Q Q 1 Temperature changes FLT: V p • Isobaric (constantpressure) 1 2 p FLT: Temperature and volume change V • Isochoric (constant volume)

  23. 1 2 p Q Thermal Reservoir T FLT: V Volume and pressure change ( FLT: DU = Q - Wby ) • Isothermal (constant temperature)

  24. The First Law Of Thermodynamics §2-1.The central point of first law §2-2. Internal energy and total energy §2-3.The equation of the first law §2-4.The first law for closed system §2-5.The first law for open system §2-6.Application of the energy equation

  25. §2-1.The central point of first law 1.Expression In a cyclic process, the algebraic sum of the work transfers is proportional to the algebraic sum of the heat transfers. Energy can be neither created nor destroyed; it can only change forms. The first law of thermodynamics is simply a statement of energy principle.

  26. §2-1.The central point of first law 2.Central point The energy conservation law is used to conservation between work and heat. Perpetual motion machines of the first kind.(PMM1) Heat: see chapter 1; Work: see chapter 1;

  27. §2-2.Internal Energy 1.Definition: Internal energy is all kinds of micro-energy in system. 2. Internal energy is property It include: • Kinetic energy of molecule (translational kinetic, vibration, rotational energy) • Potential energy • Chemical energy • Nuclear energy

  28. §2-2.Internal Energy 3.The symbol u: specific internal energy, unit –J/kg, kJ/kg ; U: total internal energy, unit – J, kJ; 4.Total energy of system E=Ek+Ep+U Ek=mcf2/2 Ep=mgz ΔE=ΔEk+ΔEp+ΔU per unit mass: e=ek+ep+u Δe=Δek+Δep+Δu

  29. §2-3. The equation of the first law 1. The equation ( inlet energy of system) – (outlet energy of system) = (the change of the total energy of the system) Ein-Eout=ΔEsystem

  30. Q W §2-4.The first law in closed system 1. The equation Ein-Eout=ΔEsystem

  31. §2-4.The first law in closed system Q-W=ΔEsystem=ΔU Q=ΔU+W Per unit mass: q= Δu+w dq=du+dw If the process is reversible, then: dq=du+pdv This is the first equation of the first law. Here q, w, Δu is algebraic.

  32. §2-4.The first law in closed system The only way of the heat change to mechanical energy is expansion of working fluid.

  33. §2-5. The first law in open system 1. Stead flow For stead flow, the following conditions are fulfilled: • The matter of system is flowing steadily, so that the flow rate across any section of the flow has the same value; • The state of the matter at any point remains constant; • Q, W flow remains constant;

  34. §2-5. The first law in open system 2. Flow work Wflow=pfΔs=pV wflow=pv p V

  35. §2-5. The first law in open system 3. 技术功 “ Wt” are expansion work and the change of flow work for open system. 4. 轴功 “ Ws” is “ Wt” and the change of kinetic and potential energy of fluid for open system.

  36. §2-5. The first law in open system 5. Enthalpy for flow fluid energy: U+pV +mcf2/2+mgz H =U+pV unit: J, kJ For Per unit mass: h=u+pv unit: J/kg, kJ/kg

  37. Q W §2-5. The first law in open system 6. Energy equation for steady flow open system , mcf12/2, mgz1 U1+p1V1 H1 U2+p2V2 H2 , mcf22/2, mgz2

  38. §2-5. The first law in open system

  39. §2-5. The first law in open system Per unit mass:

  40. §2-5. The first law in open system If neglect kinetic energy and potential energy , then: If the process is reversible, then: This is the second equation of the first law.

  41. Q W §2-5. The first law in open system 7. Energy equation for the open system Inlet flows Out flows 1 1 Open system 2 2 …… … … i j

  42. §2-5. The first law in open system Energy equation for the open system

  43. Q Wi §2-6. Application of The Energy Equation 1. Engine a). Turbines energy equation: Ein-Eout=Esystem=0 Wi=H2-H1 wi=h2-h1 , mcf12/2, mgz1 =0 U1+p1V1 H1 Q≈0 U2+p2V2 H2 mcf22/2, mgz2 =0

  44. H2 H1 Wt §2-6. Application of The Energy Equation 1. Engine b). Cylinder engine energy equation: Wt=H2-H1+Q=(U+pV) 2-(U+pV) 1 +Q Ek1, Ep1≈0 Q Ek1, Ep1≈0

  45. H2 H1 Wc Q≈0 §2-6. Application of The Energy Equation 2. Compressors Energy equation: Wc=- Wt =H2-H1 Ek1, Ep1≈0 Ek1, Ep1≈0

  46. §2-6. Application of The Energy Equation 3. Mixing chambers Energy equation: m1h1 + m2h2 -m3h3=0 Mixing water: m3h3 hot water: m2h2 Cold water: m1h1

  47. m3h3 m5h5 m2h2 m1h1 m4h4 m6h6 §2-6. Application of The Energy Equation 4. Heat exchangers Energy equation: (m1h1 + m2h2 + m3h3)-(m4h4 + m5h5 + m6h6)= 0

  48. h2 §2-6. Application of The Energy Equation 5. Throttling valves Energy equation: h1 -h2 =0 h1

  49. Unit - II Air Cycles

  50. OTTO CYCLE

More Related