Chapter 3 The First Law of Thermodynamics: Closed System

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

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 change in the total energy of the system The net energy transferred to the system - The net energy transferred from the system =

3-2 Work Definition: work is the energy transfer associated with a force acting through a distance Work is energy in transitio n 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

3-2-2 Positive 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

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

work done per unit: P-v Chart

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

3-3-2 Gravitational work

3-3-3 Accelerational work

3-3 Heat Transfer Definition: Heat is defined as the form of energy that is transferred between two systems due to temperature difference. Heat is energy in transition 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

3-3-3 Modes of Heat Transfer : Conduction: Historical Background

Convection: Radiation:

3-3-4 Thermodynamic calculation of Heat 1. Q=mC Δ T 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:

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

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

3-4 The First Law of Thermodynamics Modeling

1. The net energy transfer to the system: W in, Q in 2. The net energy transfer from the system: W out, Q out 3. The total Energy of the system: E

3-4-2 The First-Law Relation (Q in ＋ W in ) - (Q out ＋ W out ) = ΔE (Q in - Q out ) + (W in - W out ) = ΔE Consider the algebraic value of Q and W (Q in - ∑Q out ) - (∑W out - ∑Win) =ΔE Q - W = ΔE Q = ΔE + W

3-4-3 Other Forms of the First-Law Relation 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

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

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

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

3-5 Specific Heats 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 Specific heat at constant volume The specific heat at constant volume C v 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 C v dT=du

3-5-3 Specific heat at constant pressure The specific heat at constant volume Cp 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 Cp Cp dT=dh

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

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

3-6 The internal energy, enthalpy of Ideal-Gas Internal energy and enthalpy

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

We define k =C p / C v

This Chapter is over Thank you!