Transcript Chapter2 The First Law of Thermodynamics

Chapter 3
The First Law of
Thermodynamics:
Closed System
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:
Q W
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 to the system
The net energy transferred
from the system
= The net change in the total energy
of the system
Work is
3-2 Work energy
in
3-2-1 Definition:
transitio
work is the energy transfer associated
n 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 W
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 Mechanical forms of Work
3-3-1 Moving boundary work
W   12 Fds
  12 PAds

2
1
dV
PA
  12 PdV
A
work done per unit:
w   12 Pdv
P-v Chart
or
w  Pdv
Reversible Process
A process that not only system itself but also system
and surrounding keeps equilibrium
The condition of the formula w 

2
1
System undergoes a reversible process
Pdv Is that:
3-3-2 Gravitational work
W   12 Fds
  12 mgdz
 mg  12 dz
 mg( z2  z1 )
3-3-3 Accelerational work
dV
F  ma  m
dt
ds
V
 ds  Vdt
dt
W   Fds  
2
1
2
1
(1)
(2)
1
 dV 
2
2

m
(
V

V
m
V
dt


2
1 )
2
 dt 
3-3
Heat
Heat
is energy
3-3-1 Definition:in transition
Transfer
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
Q
q
m
We define heat absorbed by a system is positive
3-3-2 Historical Background
3-3-3 Modes of Heat Transfer:
Conduction:
Qcond
T
 kA
x
Convection:
Qconv  hA(Ts  Tf )
Radiation:
Qrad  AT 4
3-3-4 Thermodynamic calculation of Heat
1. Q=mCΔT
2. Consider:
W  PdV
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?
Q  Tdx
dx-----the indication to show if heat has been transferred
Consider:
dx 
Q
T
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
Since
q  Tds
then
q   12Tds
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 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
 u 

 T  v
Cv dT=du  Cv  
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
 h 

C

 
p
Cp dT=dh
 T  p
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 )
 du 
Cv   
 dT  v
B: Specific heat at constant pressure
Since:
u= f (T )
h=u+pv
= f (T ) +RT
=f ’( T )
 dh 
Cp   
 dT  p
3-6 The internal energy, enthalpy of Ideal-Gas
2-7-1 Internal energy and enthalpy
 du 
Cv   
 dT  v
 du 
Cp   
 dT  p
du  Cv dT
dh  C pdT
u  Cv T
h  C p T
We define that u =0 while T =0, then:
u  C vT
h  u  pv
h  u  RT
Obviously, h =0 while T =0, then:
h  C pT
Meanwhile:
h  u  RT
C pT  CvT  RT
C p  Cv  R
We define k =Cp / Cv
C p  C v  R

 k  C p C v
k
Cp 
R
k 1
R
Cv 
k 1
This Chapter is over
Thank you!