Thermodynamics

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Thermodynamics

  1. 1. Thermodynamics
  2. 2. THERMODYNAMICSThermodynamics isthe study of energyrelationships thatinvolve heat,mechanical work,and other aspects ofenergy and heattransfer. Central Heating
  3. 3. A THERMODYNAMIC SYSTEM• A system is a closed environment in which heat transfer can take place. (For example, the gas, walls, and cylinder of an automobile engine.) Work done on gas or work done by gas
  4. 4. INTERNAL ENERGY OF SYSTEM• The internal energy U of a system is the total of all kinds of energy possessed by the particles that make up the system. Usually the internal energy consists of the sum of the potential and kinetic energies of the working gas molecules.
  5. 5. TWO WAYS TO INCREASE THE INTERNAL ENERGY, U. + UWORK DONE HEAT PUT INTO ON A GAS A SYSTEM (Positive) (Positive)
  6. 6. TWO WAYS TO DECREASE THE INTERNAL ENERGY, U. Wout Qout - U Decrease hot hot WORK DONE BY HEAT LEAVES A EXPANDING GAS: SYSTEM W is positive Q is negative
  7. 7. THERMODYNAMIC STATEThe STATE of a thermodynamicsystem is determined by fourfactors: • Absolute Pressure P in Pascals • Temperature T in Kelvins • Volume V in cubic meters • Number of moles, n, of working gas
  8. 8. THE FIRST LAW OF THERMODYAMICS:• The net heat put into a system is equal to the change in internal energy of the system plus the work done BY the system. Q= U+W final - initial)• Conversely, the work done ON a system is equal to the change in internal energy plus the heat lost in the process.
  9. 9. SIGN CONVENTIONS FOR FIRST LAW +Wout +Qin• Heat Q input is positive U• Work BY a gas is positive -Win• Work ON a gas is negative U• Heat OUT is negative -Qout Q= U+W final - initial)
  10. 10. APPLICATION OF FIRST LAW OF THERMODYNAMICSExample 1: In the figure, thegas absorbs 400 J of heat and Wout =120 Jat the same time does 120 Jof work on the piston. Whatis the change in internalenergy of the system? Qin 400 JApply First Law: Q= U+W
  11. 11. Example 1 (Cont.): Apply First LawQ is positive: +400 J (Heat IN) Wout =120 JW is positive: +120 J (Work OUT) Qin Q= U+W 400 J U=Q-W U=Q-W = (+400 J) - (+120 J) U = +280 J = +280 J
  12. 12. FOUR THERMODYNAMIC PROCESSES:• Isochoric Process: V = 0, W = 0• Isobaric Process: P=0• Isothermal Process: T = 0, U = 0• Adiabatic Process: Q=0 Q= U+W
  13. 13. HEAT ENGINES A heat engine is any Hot Res. TH device which throughQhot Wout a cyclic process: Engine • Absorbs heat QhotQcold • Performs work Wout Cold Res. TC • Rejects heat Qcold
  14. 14. THE SECOND LAW OF THERMODYNAMICS Hot Res. TH It is impossible to construct anQhot Wout engine that, operating in a cycle, produces no effect other Engine than the extraction of heat from a reservoir and theQcold performance of an equivalent amount of work.Cold Res. TC Not only can you not win (1st law); you can’t even break even (2nd law)!
  15. 15. THE SECOND LAW OF THERMODYNAMICS Hot Res. TH Hot Res. TH400 J 100 J 400 J 400 J Engine Engine300 J Cold Res. TC Cold Res. TC• A possible engine. • An IMPOSSIBLE engine.
  16. 16. EFFICIENCY OF AN ENGINE The efficiency of a heat engineHot Res. TH is the ratio of the net workQH W done W to the heat input QH. Engine W QH- QC e= =QC QH QH Cold Res. TC QC e=1- QH
  17. 17. EFFICIENCY EXAMPLE An engine absorbs 800 J and Hot Res. TH wastes 600 J every cycle. What800 J W is the efficiency? QC Engine e=1-600 J QH Cold Res. TC 600 J e=1- e = 25% 800 JQuestion: How many joules of work is done?
  18. 18. REFRIGERATORS A refrigerator is an engine operating in reverse: Hot Res. TH Work is done on gas Qhot Win extracting heat from cold reservoir and depositing Engine heat into hot reservoir.Qcold Win + Qcold = Qhot Cold Res. TC WIN = Qhot - Qcold
  19. 19. THE SECOND LAW FOR REFRIGERATORS It is impossible to construct a Hot Res. TH refrigerator that absorbs heatQhot from a cold reservoir and deposits equal heat to a hot Engine reservoir with W = 0. QcoldCold Res. TC If this were possible, we could establish perpetual motion!

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