第4章 理想气体的热力过程
Chapter 4 Thermodynamic processes of Ideal Gas
4.1基本热力过程
Basic thermodynamic processes--p.134,p.313
4.2多变过程
Polytropic process ---P.135
4.3活塞式压气机的理论压缩过程
Theoretical compression process ---P.336
4.4多级压缩中间冷却
Multistage Compression with inter-cooling-- P.337~P343
1.本章的研究任务
The task of this chapter
A. It is to determine the change in the properties of working
fluid during a process;
(在于确定在一个过程中工质状态参数的变化量)
B. It is to determine the amount of heat, work interaction
between the system and its surroundings during a process.
(确定在一个过程中系统与外界所交换的功量和热量的多少)
2. 本章研究的对象 The object of this chapter
Reversible processes of Ideal Gases in closed systems
(闭口系统中理想气体的可逆过程)
3. 所采取的步骤 The procedures adopted
Pv=RT
u  cv T
q  cv dT  pdv
Tds  cv dT  pdv
p1v1 p2 v2

T1
T2
h  c p T
q  c p dT  vdp
Tds  c p dT  vdp
cp
cv
R
ds  dT  dv
T
v
R
ds  dT  dp
T
p
T2
v2
s  cv ln  R ln
T1
v1
T2
p2
s  c p ln  R ln
T1
p1
w  pdv
wt  vdp
s  c p ln
q  Tds
v2
p
 R ln 2
v1
p1
4.1 Basic Thermodynamic Process
(4.1 基本热力过程)
1.等容过程
Isochoric-----Constant Volume process
(1) 过程方程 (Process equation)
For a constant volume process, the addition or
removal of heat will lead to a change in the
temperature and pressure of the gas, as shown on
the two graphs above.
Substitute
vC
into
p1 p2

T1 T2
p1v1 p2 v2

T1
T2
(2)内能、焓及熵的变化量
The change in Internal Energy, Enthalpy
and Entropy
q  cv dT  pdv
du  q  cv dT
dh  c p dT
The amount of heat added to a closed
system during a constant volume process
equals to the increase in internal energy.
定容过程中加入闭口系统的热量等于系统的内能的增
加量
Entropy Change
To find the Entropy change, start with the expression
derived from the first law, replacing dU using the
definition of specific heat at constant volume and
using the definition of entropy
Tds  cv dT
T2
s  cv ln
T1
(3) Work done and Heat Transferred(功量和热量)
Applying the first law of thermodynamics to the process
dU  Q  W
Replacing δW with the reversible work
dU  Q  PdV
since the volume is constant dV = 0
W  0
Q  dU
q  du
using the definition of the specific heat at
constant volume
to replace dU in the first law.
Q  mCV dT
T2
Q  m  CV dT
T1
The technical work done
p2
wt    vdp  v( p1  p 2 )
p1
2.等压过程 Isobaric-----Constant Pressure

(1) 过程方程 Process equation
For a constant pressure process, the
addition or removal of heat will lead to a
change in the temperature and volume of
the gas, as shown on the two graphs
above.
pC
Substitute
into
p1v1 p2 v2

T1
T2
v1 v2

T1 T2
(2)内能、焓及熵的变化量
The change in Internal Energy, Enthalpy
and Entropy
du  cv dT
q  cv dT  pdv
dh  c p dT
q  cv dT  d ( pv)
q  dh
The amount of heat added to a closed
system during a constant pressure process
equals to the increase in enthalpy.
（3） Work done and Heat Transferred
q  cv dT  pdv
q  cv dT  d ( pv)
q  dh
The amount of heat added to a closed
system during a constant pressure process
equals to the increase in enthalpy.
Q  dH
T2
Q  m  C P dT
T1
During a constant pressure process, heat
is added or removed and the temperature
and volume changes. The volume at the
end of the process can be found using the
ideal gas law and the work calculated from
p2
wt    vdp  0
p1
3. 等温压缩和膨胀
Isothermal Compression and Expansion
(1)过程方程 (Process equation)
For a constant temperature process, the addition
or removal of heat will lead to a change in the
volume and pressure of the gas, as shown on the
two graphs above.
Substitute
T C
p1v1 p2 v2

T1
T2
into
p1v1  p2v2
(2) 内能、焓及熵的变化量
The change in Internal Energy, Enthalpy
and Entropy
du  cv dT  0
q  cv dT  pdv
Tds  pdv
v2
s  R ln
v1
dh  c p dT  0
q  pdv
pdv
ds 
T
p1
s  R ln
p2
Rdv
ds 
v
(3)功量和热量
Work Done and Heat Transferred
In an isothermal process, the temperature is
constant. Applying the first law of
thermodynamics to this closed process
q  cv dT  pdv
For an ideal gas, the internal energy is a
function of temperature only, and since the
temperature is constant, then dU is zero and
q  w  pdv
using the ideal gas law and integrating
between the start and end of the process
V2
dV
V2
p1
Q  W  mRT 
 mRT ln
 mRT ln
V
V1
p2
V1
During an isothermal process, the work
done by the system is equal to the heat
added to the system, and all the work is
technically usable.
p2
p2
p1
RT
wt    vdp   
dp  RT ln
p
p2
p1
p1
4.绝热过程 Adiabatic Process
(1) Process equation (过程方程)
Quasi-static, adiabatic process for an ideal gas
δq = cvdT + pdv
and δq = cpdT – vdp
then cvdT = -pdv and cp dT =vdp
therefore c
dv
dp
v dp

cv
p dv
p
then
v2
p2
k ln  ln
0
v1
p1
or
or
k

v k
p
p 2 v2
1
k
p1v1
finally, we arrive at the very useful expression
from which it can also be shown that
pv k  C
When the temperatures at the start and end of the
process are known, the pressure is calculated from
k
p2
T2 k 1
( )
p1
T1
T2
v1 k 1
( )
T1
v2
(2) The change in Internal Energy, Enthalpy
and Entropy (内能、焓及熵的变化量）
Entropy Change
There is no heat transfer to or from the gas and the
process is reversible so that
(3) Work Done and Heat Transferred(功量和热量)
When the temperatures at the start and end of the
process are known, the work done is calculated
from
w  pdv  cv (T2  T1 )
2
2
1
1
w   pdv  
pv k
1
dv 
( p1v1  p2 v2 )
k
v
k 1
R(T1  T2 )

k 1
Q0
wt  kw
wt  vdp
2
2
1
1
wt   pdv   d ( pv)
R(T1  T2 )

 R(T2  T1 )
k 1
k

R(T1  T2 )
k 1
5. 多变过程
Polytropic Process
 Many processes can be approximated by
the law:
 where,
P Pressure,
v Volume,
n an index depending on the process type.
Polytropic processes are internally reversible. Some examples
are vapors and ideal gas in many non-flow processes, such as:
 n=0, results in P=constant i.e. isobaric
process.
 n=infinity, results in v=constant i.e.
isochoric process.
 n=1, results in P v=constant, which is an
isothermal process for a perfect gas.
 n=k, which is a reversible adiabatic
process for a perfect gas.
 Some polytropic processes are shown in
figure below:
The initial state of working fluid is shown by point 0 on the P-V diagram. The
polytropic state changes are:
•0 to 1= constant pressure heating,
•0 to 2= constant volume heating,
•0 to 3= reversible adiabatic compression,
•0 to 4= isothermal compression,
•0 to 5= constant pressure cooling,
•0 to 6= constant volume cooling,
•0 to 7= reversible adiabatic expansion,
•0 to 8= isothermal expansion.
When the temperatures at the start and end of
the process are known, the pressure is
calculated from
n
p2
T2 n 1
( )
p1
T1
T2
v1 n 1
( )
T1
v2
(2)内能、焓及熵的变化量
The change in Internal Energy, Enthalpy
and Entropy
Entropy Change
There is no heat transfer to or from the gas and the
process is reversible so that
(3) Work Done and Heat Transferred(功量和热量)
When the temperatures at the start and end of the
process are known, the work done is calculated
from
wt  vdp
w  pdv
2
2
n
2
2
pv
1
w   pdv   n dv 
( p1v1  p2 v2 ) wt   pdv   d ( pv)
v
n 1
1
1
1
1
R(T1  T2 )
R(T1  T2 )

 R(T2  T1 )

n 1
n 1
n

R(T1  T2 )
wt  nw
n 1
q  cv dT  pdv
R(T1  T2 )
q  cv (T2  T1 ) 
n 1
R
R

(T2  T1 ) 
(T2  T1 )
k 1
n 1
nk

R(T2  T1 )
n 1
29
4.3活塞式压气机的工作过程
Working Process of gas piston compressor
1. Theoretical processes of single-staged
piston compressor
(理想单级活塞式压气机的工作原理)
（1）b-1:charge stroke (吸气冲程)，
吸气量增大，吸入气体的状态（P,T)不变;
(2) 1- 2:compression stroke( 压缩冲程)，
吸气量不变，压力升高.
(3) 2- a: discharge process (排气过程),
气缸内的气体减少，气体状态不变化
let the initial pressure of the gas is p1
(令气体初态压力为 p1 )
the final pressure is p 2 (终态压力为 p2 )，
pressure ratio is
p2
p1
(升压比)
2. Analysis on theoretical work consumed by
these processes (压气机的理论耗功量分析)
 取活塞右行一次的吸气量为控制质量，并忽略动能
差及位能差进行分析。
吸气过程：气体推动活塞移动，做推挤功 p1v1 ；
2
压缩过程：气体向活塞做膨胀功
 pdv
1
排气过程：气体向活塞做推挤功
 p2v2
压缩过程功:指1- 2，闭口系的膨胀功，是可逆过程；
压气机耗功：指a- 1- 2- b,全过程是开口系，包括流动功。
气体在全过程中所做的总功为：
2
2
1
1
w  p1v1   pdv  p 2 v 2    vdp
3.三种不同压缩过程的比较
(Comparison of three kinds of compression
processes)
Adiabatic process (绝热过程)：
压缩过程进行的很快，热量来不及释放；
Isothermal process (等温过程)：压缩过程中，理想
的冷却条件，压缩产生的热量可及时排出；
Polytropic process (多变过程)：采用了一定的冷却措
施，但压缩期间, 温度仍继续升高。
(1) Work consumption (耗功量）
If it is isothermal compression, then
若压缩过程为可逆定温过程，则：
p
p
w c  w t,T  p1v1ln
2
p1
 RgT1 ln
2
p1
If it is isentropic compression, then
若压缩过程为可逆绝热压缩，则:
w c  w t,s
 k ( p2v2  p1v1 )  k RgT1[( p2 )
k 1
k 1
p2
k 1
k
 1]
If it is polytropic compression, then
若压缩过程为可逆多变过程，则 :
w c  w t,n 
p2
n
n
( p 2 v 2  p1v1 ) 
R g T1 [( )
n 1
n 1
p1
Wt,T  Wt,n  Wt , s
（2） Discharge Temperature (排气温度）
T2,T  T2,n  T2, s
n 1
n
 1]
4.余隙容积的影响
The influence of residual volume of inter-space.
p
2”
2’
2
1
V
（1）The influence on discharge volume( 对排气量的影响）
Residual Volume (余隙容积）：V3
We note that if the pressure ratio is too high, then it can not work
normally.
(压缩压力不能太高，压力升高，效率下降，在极限情况下容积效率
可达零 ）
Volume efficiency is defined as
Let
V3
c
V1  V3
v 
V1  V4
V3 V4
1
(  1)
V1  V3
V1  V3 V3
1
then
p
v  1  c[( 2 ) n  1]
p1
The discharge volume will decrease .
(余隙存在将使排气量减少）
（2） The influence on compression work（对耗功量的
影响）
n 1
n 1
p3 n
p2 n
n
n
wc 
p1V1 [( )  1] 
p 4V4 [( )  1]
n 1
p1
n 1
p4
p
n

p1 (V1  V4 )[( 2 )
n 1
p1
n 1
n
p
n
 1] 
p1V [( 2 )
n 1
p1
n 1
n
There is no influence on compression
work.(对耗功量无影响)
 1]
4.4 Multistage Compression with inter-cooling
（多级压缩、中间冷却）
Multistage compression with intercooler is especially attractive when a
gas is to be compressed to very high pressures.
1. two stage compression （两级压缩）
intercooler
Assumed that the compression process 1-2 and
3-4 are with the same polytropic index n
(假设1-2和3-4的多变指数相等。)
n
p2
wc 
RgT1[( )
n 1
p1
Assumed ( 假设)
T
3
n 1
n
n
p4
 1] 
Rg T3[( )
n 1
p3
n 1
n
 1]
 T1
To minimize the total compression work, the
optimal pressure ratio can be determined by
(为了使压缩过程的耗功量最小,可以采用如下方法确
定最佳中间压力）
We obtain
dwc
0
dp2
p2 p4

p1 p3
That is, to minimize the compression work during
two stage compression, the pressure ratio across
each stage of the compressor must be the same.
（各级压缩比都相等时多级压缩最省功。）
When this condition is satisfied, the compression
work at each stage becomes identical.
(当满足上述条件时，每一级压缩的耗功量都相等）
Wcomp,  Wcomp,
2. Z-Stage Compression (n级压缩）
Optimal compression ration is
p2
 
p1
z
p z 1
p1
3. Isentropic Efficiency and Isothermal Efficiency
(等熵效率和等温效率）
Isentropic conpressor work Wc，s
s 

Actual compressor wor
Wc,n
 Isentropic efficiency is defined as the ratio of the
work input is required to raise the pressure of a gas
to a specified value in an isentropic manner to the
actual work input.
Isothermal Efficiency (等温效率）
Isothermal conpressor work Wc，s
T 

Actual compressor wor
Wc,n
 Isothermal efficiency is defined as the ratio of the work
input is required to raise the pressure of a gas to a
specified value in a reversible isothermal manner to
the actual work input.
Exercise (练习)
 Draw the following polytropic process of
air on p-v and T-s diagram.
(1) Pressure rises, temperature increases
and with heat rejection
(2) Working medium expands and the
temperature drops with heat rejection
at the same time
Exercise (练习)
(3)Expansion process with n=1.6,please judge
the sign of q, w, u
q, w,with
u n=1.3,please
(4)Compression process
judge the sign of
Draw the following process on the T-s
Diagram.