Chapter 3 The First Law of Thermodynamics: Closed System

1. Chapter 3 The First Law of Thermodynamics: Closed System

2. 3-1 Introduction To The First Law of Thermodynamics 3-1-1 Conservation of energy principle Energy can be neither created nor destroyed;it can only change forms 3-1-2 The First Law of Thermodynamics Neither heat nor work can be destroyed;they can only change from one to another, that is:

3. 3-1-2 The shortcomings of Q=W • Can’t be employed in engineering calculation • Can’t show the quality difference between heat and work In engineering area we would rather use a formula like this: The net energy transferred from the system The net energy transferred to the system - The net change in the total energy of the system =

4. 3-2 Work Work is energy in transition 3-2-1Definition: work is the energy transfer associated with a force acting through a distance Denoted by W ----------kJ work on a unit-mass basis is denoted by w w----------kJ/kg work done per unit time is called power power is denoted as

5. 3-2-2Positive and negative Since work is the energy transferred between system and its boundary, then we define that: work done by a system is positive; and work done on a system is negative

6. 3-3 Mechanical forms of Work 3-3-1Moving boundary work

7. work done per unit: P-v Chart

8. The condition of the formula Is that: Reversible Process A process that not only system itself but also system and surrounding keeps equilibrium System undergoes a reversible process

9. 3-3-2Gravitational work

10. 3-3-3Accelerational work

11. 3-3 Heat Transfer Heat is energy in transition 3-3-1Definition: Heat is defined as the form of energy that is transferred between two systems due to temperature difference . denote as Q ----------kJ heat transferred per unit mass of a system is denoted as q----------kJ/kg We define heat absorbed by a system is positive

12. 3-3-2 Historical Background 3-3-3 Modes of Heat Transfer: Conduction:

13. Convection: Radiation:

14. 3-3-4 Thermodynamic calculation of Heat 1. Q=mCΔT 2. Consider: P------the source to do work dV-----the indication to show if work has been done the source to lead to heat transfer is T, then there should be:

15. What is dx here? dx-----the indication to show if heat has been transferred Consider: We define that x is called entropy and denoted asS The unit of S is kJ/K Specific entropy is denoted as s The unit of s is: kJ/kg.K

16. Also needs the condition of reversible process! T-s chart

17. 3-4 The First Law of Thermodynamics 3-4-1 Modeling

18. 1. The net energy transfer to the system: Win , Qin 2. The net energy transfer from the system: Wout , Qout 3. The total Energy ofthe system: E

19. 3-4-2 The First-LawRelation (Qin＋Win) - (Qout＋Wout) = ΔE (Qin - Qout) + (Win - Wout) = ΔE Consider the algebraic value of Q and W (Qin - ∑Qout) - (∑Wout - ∑Win) =ΔE Q - W = ΔE Q = ΔE + W

20. 3-4-3 Other Forms of the First-LawRelation 1. Differential Form: δQ = dE + δW As to a system without macroscopic form energy δQ = dU + δW On a unit-mass basis δq = de + δw δq = du + δw

21. 2. Reversible Process δQ = dE + PdV or δQ = dU + PdV On a unit-mass basis δq = de + pdv δq = du + pdv

22. 3. Cycle δq = du + δw ∮δq = ∮du + ∮δw since∮du = 0 then∮δ q = ∮δw if ∮δ q = 0 ∮δw =0 This can illustrate that the first kind of perpetual motion machine can’t be produced

23. 4.Reversible Process under a Constant Pressure δQ = dU + PdV Since p=const δQ = dU + d(pV) 5.Isolated System dE = 0

24. 3-5 Specific Heats 3-5-1 Definition of specific heat The energy required to raise the temperature of the unit mass of a substance by one degree Then q=CT or δq=CdT 3-5-2 Specific heat at constant volume The specific heat at constant volume Cv can be viewed as the energy required to raise the temperature of unit mass of substance by 1 degree as the volume is maintained constant. At constant volume , δq = du CvdT=du

25. 3-5-3 Specific heat at constant pressure The specific heat at constant volume Cp can be viewed as the energy required to raise the temperature of unit mass of substance by 1 degree as the volume is maintained constant pressure. Similarly, at constant pressure δq = du+ δ w=du+Pdv=du+d(Pv)=dh CpdT=dh

26. 3-5-4 Specific Heats of Ideal-Gas A: Specific heat at constant volume Since there are no attraction among molecules of ideal-gas,then: u= f (T )

27. B: Specific heat at constant pressure Since:u= f (T ) h=u+pv =f (T ) +RT =f ’( T )

28. 3-6The internal energy, enthalpy of Ideal-Gas 2-7-1 Internal energy and enthalpy

29. We define that u =0 while T =0, then: Obviously, h =0 while T =0, then: Meanwhile:

30. We define k =Cp / Cv

31. This Chapter is over Thank you!