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

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:

WQ

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 Work3-2-1 Definition:

work is the energy transfer associated with a force acting through a distance

W

Work is energy in transition

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

FdsW 21

PAds 21

A

dVPA 2

1 PdV 21

3-3 Mechanical forms of Work

Pdvw 21

work done per unit:

P-v Chart

Pdvw or

Reversible Process

A process that not only system itself but also system and surrounding keeps equilibrium

Pdvw 21The condition of the formula Is that:

System undergoes a reversible process

FdsW 21

mgdz 21

dzmg 21

)( 12 zzmg

3-3-2 Gravitational work

maF )1(dt

dm

V

dt

dsV )2(dtds V

3-3-3 Accelerational work

FdsW 21 dt

dt

dm V

V

2

1)(

2

1 21

22 VV m

3-3 Heat Transfer3-3-1 Definition:

Heat is defined as the form of energy that is transferred between two systems due to temperature difference .

Heat is energy in transition

m

Qq

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:

3-3-2 Historical Background

x

TkAQcond

Convection:

)( fsconv TThAQ

Radiation:

4ATQrad

3-3-4 Thermodynamic calculation of Heat

1. Q=mCΔT

2. Consider:

PdVW

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:

TdxQ

What is dx here?

dx-----the indication to show if heat has been transferred

Consider:

T

Qdx

We define that x is called entropy and denoted as SThe unit of S is kJ/K Specific entropy is denoted as s The unit of s is: kJ/kg.K

Tdsq Since

T-s chart

Tdsq 21then

Also needs the condition of reversible process!

3-4 The First Law of Thermodynamics

3-4-1 Modeling

1. The net energy transfer to the system:

Win , Qin

2. The net energy transfer from the system:

Wout , Qout

3. The total Energy of the system:

E

3-4-2 The First-Law Relation

(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

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 Heats3-5-1 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

Cv dT=duv

v T

uC

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

Cp dT=dhp

p T

hC

A: Specific heat at constant volume

Since there are no attraction among molecules of ideal-gas,then:

u= f (T )

vv dT

duC

3-5-4 Specific Heats of Ideal-Gas

B: Specific heat at constant pressure

Since: u= f (T )

h=u+pv

= f (T ) +RT

=f ’( T )

pp dT

dhC

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

2-7-1 Internal energy and enthalpy

vv dT

duC

dTCdu v

TCu v

pp dT

duC

dTCdh p

TCh p

We define that u =0 while T =0, then: TCu v

pvuh

RTuh Obviously, h =0 while T =0, then:

TCh p

Meanwhile:

RTuh

RTTCTC vp

RCC vp

We define k =Cp / Cv

vp

vp

CC

RCC

k

RC p 1

k

k

1

k

RvC

This Chapter is over

Thank you!