正激波基本控制方程的推导
音速
能量方程的特殊形式
什么情况下流动是可压缩的?
用于计算通过正激波气体特性变化的方
程的详细推导 ; 物理特性变化趋势的讨论
用皮托管测量可压缩流的流动速度
图 8.2 第八章路线图
8.6 CALCULATION OF NORMAL SHOCK-WAVE
PROPERTIES 正激波性质的计算
本节要点只有一个,
? 计算通过正激波的流动特性变化
即
问题, 已知激波前区域 1
的条件，计算激波后区域
2的条件。
TT
1,0
2,0
0,1
0,2
12
1
2
1
2
1
2
2 ???????? p
pss
T
T
p
pM
?
?
2211 uu ?? ?
22222111 upup ?? ???
22
2
2
2
2
1
1
uhuh ???
22 Tch p?
22 RTp ??
复习我们在 8.2节中推导出的正激波基本方程,
(8.2)
(8.6)
(8.10)
(8.49)
(8.50)
Examining the five equations given above,we see that they involve five
unknowns,namely,ρ2,u2,p2,h2,and T2,Hence,Eqs.(8.2),(8.6),(8.10),
(8.49),and (8.50) are sufficient for determining the properties behind a
normal shock wave in a calorically perfect gas,Let us proceed,
2
22
2
1
11
1 u
u
pu
u
p ???
??
12
22
2
11
1 uu
u
p
u
p ???
??
12
2
2
2
1
2
1 uu
u
a
u
a ???
??
)1(2
*)1(
21
222
?
???
? ?
?
?
aua
2
1
22
1 2
1*
2
1 uaa ???? ??
2
2
22
2 2
1*
2
1 uaa ???? ??
(8.6)式除以 (8.2)式,
(8.51)
(8.52) 因为,
?? pa ?
由公式 (8.35),可得,
(8.53)
(8.54)
122
2
2
1
1
2
2
1*
2
1
2
1*
2
1 uuu
u
au
u
a ?????????
?
?
?
?
?
?
?
?
1212
2
12
21
)(2 1*)(2 1 uuuuauuuu ??????? ????
12 1*2 1 2
21
???? ???? auu
21
2* uua ?
将 (8.53),(8.54)式代入 (8.52)式,
整理为,
两边同除以 u2-u1,
(8.55)
21
2* uua ?
(8.55)
Equation (8.55) is called the Prandtl relation and is a useful intermediate
relation for normal shock waves,方程 (8.55)被称为 Prandtl 关系式，
是一个很有用的正激波中间关系式,
**1
21
a
u
a
u? (8.56) (8.55)式还可写成,
由特征马赫数的定义,可得,
** a
uM ?
**1 21 MM?
*
1*
1
2 MM ?
2
2
2
)1(2
)1(*
M
MM
??
??
?
?
1
2
1
2
1
2
2
2
2
)1(2
)1(
)1(2
)1(
?
?
?
?
?
?
?
??
??
??
?
M
M
M
M
?
?
?
?
2/)1(
]2/)1[(1
2
1
2
12
2 ??
???
??
?
M
MM
(8.57)
应用（ 8.48)式,
(8.58)
(8.59)
Equation (8.59) is our first major result for a normal shock wave,
Examine Eq,(8.59) closely; it states that the Mach number behind the
wave,M2,is a function only of the Mach number ahead of the wave,M1,
方程（ 8.59)是我们得到的第一个主要正激波关系式，表明波后马赫
数 M2是波前马赫数 M1的唯一函数,
Moreover,if M1=1,then M2=1,This is the case of an infinitely weak
normal shock wave,defined as a Mach wave,
如果 M1=1，则 M2=1。这种情况对应无限弱的正激波，我们定义
为 马赫波。
Furthermore,if M1>1,then M2<1; i.e.,the Mach number behind the
normal shock wave is subsonic,
如果 M1>1,则 M2<1； 也就是,正激波后的流动是 亚音速 的。
As M1 increases above 1,the normal shock wave becomes stronger,
and M2 becoming progressively less than 1,
当 M1 由 1逐渐增大时，正激波越来越强，激波后马赫数 M2越来越
小（在小于 1的范围内）。
However,in the limit as M1→ ∞,M2 approaches a finite minimum
value,M2→,which for air is 0.378,
然而，当 M1趋于无穷大,M2趋于一有限的最小值 M2→,
对于空气其值为 0.378。
?? 2)1( ?
?? 2)1( ?
2
12
2
1
21
2
1
2
1
1
2 *
* Ma
u
uu
u
u
u ????
?
?
2
1
2
1
2
1
1
2
)1(2
)1(
M
M
u
u
??
???
?
?
?
?
)1()(
1
22
112111
2
22
2
1112 u
uuuuuuupp ??????? ????
)1()1()1(
1
22
1
1
2
2
1
2
1
1
2
1
2
11
1
12
u
uM
u
u
a
u
u
u
p
u
p
pp ??????? ??
?
??
下面我们来推导通过正激波的热力学特性,即,,
的表达式,
12 ?? 12 pp 12 TT
（ 8.61）
(8.62)
(8.63)
?
?
?
?
?
?
?
?????
2
1
2
12
1
1
12
)1(
)1(21
M
MM
p
pp
?
??
)1(
1
21 2
1
1
2 ?
?
?? M
p
p
?
?
???
?
???
?
???
?
???
?
?
2
1
1
2
1
2
?
?
p
p
T
T
2
1
2
12
1
1
2
1
2
)1(
)1(2)1(
1
21
M
MM
h
h
T
T
?
??
??
?
??
? ?
?
???
?
?
?
?
(8.65)
(8.64)
(8.66)
(8.67)
Equations (8.61),(8.65),and (8.67) are important,
Examine them closely,Note that,and
are functions of the upstream Mach number M1 only,
,和 只是上游马赫数 M1 的函数。
The upstream Mach Number M1 is the determining
parameter for changes across a normal shock wave in a
calorically perfect gas,
正激波上游马赫数 M1是确定量热完全气体通过正激
波特性变化的 决定性参数 。
12 ?? 12 pp 12 TT
12 ?? 12 pp 12 TT
3 7 8.02 12lim
1
???
?? ?
?M
M
6
1
1
1
2
1
2
1
2
limlim
lim
11
1
????
?
?
?
?
????
??
T
T
p
p
MM
M ?
?
?
?
If M1=1,then ; i.e.,we have the
case of a normal shock wave of vanishing strength---a Mach
wave,如果 M1=1，那么有 ；即正
激波是无限弱的马赫波。
As M1 increases above 1,
progressively increase above 1,当 M1 大于 1逐渐增加时,
也逐渐沿大于 1的趋势增大。
1121212 ??? TTpp ??
1121212 ??? TTpp ??
121212 an d,,TTpp ??
121212,,TTpp 和??
We have stated earlier that shock waves occur in supersonic flows; a
stationary normal shock such as shown in Fig.8.3 does not occur in
subsonic flow,That is,Eqs,(8.59),(8.61),(8.65),and (8.67),the
upstream Mach number is supersonic,M1≥1,However,on a
mathematical basis,these equations also allow solution for M1≤1,
These equations embody the contiunity,momentum,and energy
equations,which in principle do not care whether the value of M1 is
subsonic or supersonic,Here is an ambiguity which can only be
resolved by second law of thermodynamics,
我们在前面已经指出，激波出现在超音速流中；图 8.3所示的精
制的正激波在亚音速流中不可能出现。也即是，关系式 (8.59),
(8.61),(8.65),和 (8.67)中上游马赫数必须是超音速的；必有 M1≥1。
然而，从纯数学的观点看，这些由连续、动量、能量方程推导出
的关系式在理论上并没有限定 M1 一定要大于 1，即上游流动一定
是超音速的。这种多解性只有借助于热力学第二定律来解决。
1
2
1
2
12 lnln p
pR
T
Tcss
p ???
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
???
)1(
)1(
2
1ln
)1(
)1(2
)1(
1
2
1ln
2
1
2
1
2
12
112
MR
M
M
Mcss
p
?
?
?
?
?
?
应用我们在第七章推导出的熵增公式，我们可以得到通过正
激波的熵增计算公式,
（ 7.25）
（ 8.68）
From Eq,(8.68),we see that the entropy change s2-s1 across the
shock is a function of M1 only,由方程 (8.68)可以看出，通过正
激波的熵增 s2-s1 只是波前马赫数 M1的函数。
The second law dictates that
s2-s1≥0
In Eq,(8.68),if M1=1,s2=s1,and if M1>1,then s2-s1>0,both
of which obey the second law,However,if M1<1,then
Eq.(8.68) gives s2-s1<0,which is not allowed by the second
law,Consequently,in nature,only cases involving M1≥1 are
valid,i.e.,normal shock waves can occur only in
supersonic flow,
热力学第二定律指出,
s2-s1≥0
由方程 (8.68)可以看出，如果 M1=1,s2=s1 ；如果 M1>1,
then s2-s1>0,两种情况都符合热力学第二定律。然而，
如果 M1<1,则,(8.68) 式的结果为 s2-s1<0，其不符合热
力学第二定律。因此，只有 M1≥1的情况会发生，即,正
激波只能在超音速流动中发生。
Why does the entropy increase across the shock wave? The second law tells
us that it must,but what mechanism does nature use to accomplish this
increase? To answer these questions,recall that a shock wave is a very thin
region ( on the order of 10-5cm ) across which some large changes occur
almost discontinuously,Therefore,within the shock wave itself,large
gradients in velocity and temperature occur; i.e.,the mechanisms of friction
and thermal condition are strong,These are dissipative,irreversible
mechanisms that always increase the entropy,Therefore,the precise
entropy increase predicted by Eq.(8.68) for a given supersonic M1 is
appropriately provided by nature in the form of friction and thermal
conduction within the interior of the shock wave itself,
为什么通过激波会出现熵增？热力学第二定律告诉我们一定会
有熵增，但是这个熵增产生的机理是什么呢？为了回答这些问
题，让我们回忆第七章讨论过的内容：激波是非常薄的，厚度
只有 10-5cm,通过激波流动性质发生剧烈变化，几乎是不连续
的。因此，在激波本身内部，有很大的速度梯度和温度梯度，
即摩擦和热传导的作用是非常强的。这些耗散的、不可逆的机
制总是引起熵增，因此,(8.68)式给出的超音速波前马赫数 M1对
应的精确熵增实际是由于激波本身内部的摩擦与热传导引起的。
总条件 T0与 p0如何变化？
11 ss a ? 22 ss a ?
22
2
2
2
2
1
1
uTcuTc
pp ???
2
2
0
uTcTc
pp ??
2,01,0 TcTc pp ?
2,01,0 TT ?
首先回答 T0如何变化？
（ 8.30)
（ 8.38)
（ 8.39)
能量方程,
总温定义,
Equation (8.39) states that total temperature is constant across
a stationary normal shock wave,
方程（ 8.39）表明,通过静止正激波总温不变。
a
a
a
a
paa p
pR
T
Tcss
1
2
1
2
12 lnln ???
1,0
2,0
1,0
2,0
12 lnln p
p
R
T
T
css p ???
1,0
2,0
12 ln p
p
Rss ???
Rsse
p
p /)(
1,0
2,0 12 ???
总压如何变化？可借助熵增计算公式求出,
(8.70)
(8.71)
(8.72)
(8.73)
From Eq,(8.68),we know that s2-s1>0 for a normal shock
wave,Hence,Eq,(8.73) states that p0,2<p0,1,The total
pressure decreases across a shock wave,The total
pressure ratio p0,2/p0,1 across a normal shock wave is a
function of M1 only,
由公式（ 8.68），我们知道对于正激波 s2-s1>0,因此，
式（ 8.73)表明,p0,2<p0,1,通过正激波总压降低，且
通过正激波总压比 p0,2/p0,1 只是波前马赫数 M1 的函
数。
至此，我们已经全部回答了本节开始提出的问题,
归纳如下,
1,0
2,0
10
20
12
1
2
1
2
1
2
2 ???????? p
p?
T
Tss
T
T
p
pM
,
,
?
?
2/)1(
]2/)1[(1
2
1
2
12
2 ??
???
??
?
M
MM 2
1
2
1
2
1
1
2
)1(2
)1(
M
M
u
u
??
???
?
?
?
? )1(121 21
1
2 ?
??? Mp
p
?
?
2
1
2
12
1
1
2
1
2
)1(
)1(2)1(
1
21
M
MM
h
h
T
T
?
???
?
?
??
? ?
???? ?
?
?
?
??
?
??
? ?
?????
?
?
?
?
?
??
??
?
??
? ?
???? )1()1(
21ln
)1(
)1(2)1(
1
21ln 2
12
1
2
12
112 MRM
MMcss
p ?
?
?
?
?
?
2,01,0 TT ?
Rsse
p
p /)(
1,0
2,0 12 ???
这些关系式在附录 B中以列表形式给出。
In summary,we have now verified the qualitative changes across a
normal shock wave as sketched in Fig.7.4b and as originally
discussed in Sec,7.6,
??
?
??
? ?
??
? 1)(
1
2 /)1(
1
1,0
2
12
1
??
? p
pau
8.7 MEASUREMENT OF VELOCITY IN A COMPRESSIBLE
FLOW 可压缩流动的速度测量
8.7.1 Subsonic Compressible Flow 亚音速可压缩流
??
?
??
? ?
??
? 1)(
1
2 /)1(
1
1,02
1
??
? p
pM
)1/(2
1
1
1,0 )
2
11( ???? ??? M
p
p （ 8.42）
（ 8.74）
（ 8.75）
??
?
??
? ?
??
? 1)(
1
2 /)1(
1
1,0
2
12
1
??
? p
pau (8.75)
From Eq,(8.75),we see that,unlike incompressible
flow,a knowledge of p0,1 and p1 is not sufficient to
obtain u1; we also need the freestream speed of sound,
a1,
从（ 8.75）式可以看到,与不可压缩流不同，只
知道 p0,1 和 p1 还不足以得到速度 u1；我们还需要
知道自由流的音速,a1。
8.7.2 Supersonic Flow 超音速流
Fig,8.8 A Pitot tube in supersonic flow
A fluid element moving along streamline cde will first decelerated
nonisentropically to a subsonic velocity at point b just behind the
shock, Then it is is isentropically compressed to zero velocity at
point e,As a result,the pressure at point e is not the total pressure
of the freestream but rather the total pressure behind a normal
shock wave,p0,2,This is the Pitot pressure read at the end of the
tube,
沿流线 cde 的流体微团首先非等熵地在 d点减速为亚音速，然
后被等熵地在 e点压缩为驻点速度零。因此,e点的压强不是
自由流的总压而是正激波后的总压 p0,2。 这是 皮托管测得的总
压。
Keep in mind that because of the entropy increase across the shock,
there is a loss in total pressure across the shock,p0,2<p0,1,However,
knowing p0,2 and the freestream static pressure p1 is still sufficient to
calculate the freestream Mach number M1,as follows,
一定要记住的是：由于通过激波引起熵增，所以通过激波会有
总压损失,p0,2<p0,1。然而，知道 p0,2 和自由来流静压 p1，仍足
以使我们计算出自由来流马赫数 M1，方法如下,
1
2
2
2,0
1
2,0
p
p
p
p
p
p ? (8.76)
)1/(2
2
2
2,0 )
2
11( ???? ??? M
p
p (8.77)
2/)1(
]2/)1[(1
2
1
2
12
2 ??
???
??
?
M
MM
(8.78)
1
21
)1(24
)1( 21
)1/(
2
1
2
1
2
1
2,0
?
??
???
?
???
?
??
??
?
?
??
??
?
??
M
M
M
p
p
)1(
1
21 2
1
1
2 ?
?
?? M
p
p
?
?
(8.79)
(8.80)
Equation (8.80) is called the Rayleigh Pitot tube formula,It relates the
Pitot pressure p0,2 and the freestream static pressure p1 to the
freestream Mach number M1,Equation (8.80) gives M1 as an implicit
function of p0,2/p1 and allows the calculation of M1 from known p0,2/p1,
For convenience in making calculations,the ratio p0,2/p1 is tabulated
versus M1 in App.B,
(8.80)式被称为雷利皮托管公式。它将皮托管测得的总压 p0,2 和自
由来流静压 p1与自由来流马赫数 M1 联系起来了。 (8.80)式中 M1为
p0,2/p1 的隐式函数，可以由 p0,2/p1 的值计算出 M1 。为方便应用,
附录 B给出了 p0,2/p1 随 M1的变化表。
音速
能量方程的特殊形式
什么情况下流动是可压缩的?
用于计算通过正激波气体特性变化的方
程的详细推导 ; 物理特性变化趋势的讨论
用皮托管测量可压缩流的流动速度
图 8.2 第八章路线图
8.6 CALCULATION OF NORMAL SHOCK-WAVE
PROPERTIES 正激波性质的计算
本节要点只有一个,
? 计算通过正激波的流动特性变化
即
问题, 已知激波前区域 1
的条件，计算激波后区域
2的条件。
TT
1,0
2,0
0,1
0,2
12
1
2
1
2
1
2
2 ???????? p
pss
T
T
p
pM
?
?
2211 uu ?? ?
22222111 upup ?? ???
22
2
2
2
2
1
1
uhuh ???
22 Tch p?
22 RTp ??
复习我们在 8.2节中推导出的正激波基本方程,
(8.2)
(8.6)
(8.10)
(8.49)
(8.50)
Examining the five equations given above,we see that they involve five
unknowns,namely,ρ2,u2,p2,h2,and T2,Hence,Eqs.(8.2),(8.6),(8.10),
(8.49),and (8.50) are sufficient for determining the properties behind a
normal shock wave in a calorically perfect gas,Let us proceed,
2
22
2
1
11
1 u
u
pu
u
p ???
??
12
22
2
11
1 uu
u
p
u
p ???
??
12
2
2
2
1
2
1 uu
u
a
u
a ???
??
)1(2
*)1(
21
222
?
???
? ?
?
?
aua
2
1
22
1 2
1*
2
1 uaa ???? ??
2
2
22
2 2
1*
2
1 uaa ???? ??
(8.6)式除以 (8.2)式,
(8.51)
(8.52) 因为,
?? pa ?
由公式 (8.35),可得,
(8.53)
(8.54)
122
2
2
1
1
2
2
1*
2
1
2
1*
2
1 uuu
u
au
u
a ?????????
?
?
?
?
?
?
?
?
1212
2
12
21
)(2 1*)(2 1 uuuuauuuu ??????? ????
12 1*2 1 2
21
???? ???? auu
21
2* uua ?
将 (8.53),(8.54)式代入 (8.52)式,
整理为,
两边同除以 u2-u1,
(8.55)
21
2* uua ?
(8.55)
Equation (8.55) is called the Prandtl relation and is a useful intermediate
relation for normal shock waves,方程 (8.55)被称为 Prandtl 关系式，
是一个很有用的正激波中间关系式,
**1
21
a
u
a
u? (8.56) (8.55)式还可写成,
由特征马赫数的定义,可得,
** a
uM ?
**1 21 MM?
*
1*
1
2 MM ?
2
2
2
)1(2
)1(*
M
MM
??
??
?
?
1
2
1
2
1
2
2
2
2
)1(2
)1(
)1(2
)1(
?
?
?
?
?
?
?
??
??
??
?
M
M
M
M
?
?
?
?
2/)1(
]2/)1[(1
2
1
2
12
2 ??
???
??
?
M
MM
(8.57)
应用（ 8.48)式,
(8.58)
(8.59)
Equation (8.59) is our first major result for a normal shock wave,
Examine Eq,(8.59) closely; it states that the Mach number behind the
wave,M2,is a function only of the Mach number ahead of the wave,M1,
方程（ 8.59)是我们得到的第一个主要正激波关系式，表明波后马赫
数 M2是波前马赫数 M1的唯一函数,
Moreover,if M1=1,then M2=1,This is the case of an infinitely weak
normal shock wave,defined as a Mach wave,
如果 M1=1，则 M2=1。这种情况对应无限弱的正激波，我们定义
为 马赫波。
Furthermore,if M1>1,then M2<1; i.e.,the Mach number behind the
normal shock wave is subsonic,
如果 M1>1,则 M2<1； 也就是,正激波后的流动是 亚音速 的。
As M1 increases above 1,the normal shock wave becomes stronger,
and M2 becoming progressively less than 1,
当 M1 由 1逐渐增大时，正激波越来越强，激波后马赫数 M2越来越
小（在小于 1的范围内）。
However,in the limit as M1→ ∞,M2 approaches a finite minimum
value,M2→,which for air is 0.378,
然而，当 M1趋于无穷大,M2趋于一有限的最小值 M2→,
对于空气其值为 0.378。
?? 2)1( ?
?? 2)1( ?
2
12
2
1
21
2
1
2
1
1
2 *
* Ma
u
uu
u
u
u ????
?
?
2
1
2
1
2
1
1
2
)1(2
)1(
M
M
u
u
??
???
?
?
?
?
)1()(
1
22
112111
2
22
2
1112 u
uuuuuuupp ??????? ????
)1()1()1(
1
22
1
1
2
2
1
2
1
1
2
1
2
11
1
12
u
uM
u
u
a
u
u
u
p
u
p
pp ??????? ??
?
??
下面我们来推导通过正激波的热力学特性,即,,
的表达式,
12 ?? 12 pp 12 TT
（ 8.61）
(8.62)
(8.63)
?
?
?
?
?
?
?
?????
2
1
2
12
1
1
12
)1(
)1(21
M
MM
p
pp
?
??
)1(
1
21 2
1
1
2 ?
?
?? M
p
p
?
?
???
?
???
?
???
?
???
?
?
2
1
1
2
1
2
?
?
p
p
T
T
2
1
2
12
1
1
2
1
2
)1(
)1(2)1(
1
21
M
MM
h
h
T
T
?
??
??
?
??
? ?
?
???
?
?
?
?
(8.65)
(8.64)
(8.66)
(8.67)
Equations (8.61),(8.65),and (8.67) are important,
Examine them closely,Note that,and
are functions of the upstream Mach number M1 only,
,和 只是上游马赫数 M1 的函数。
The upstream Mach Number M1 is the determining
parameter for changes across a normal shock wave in a
calorically perfect gas,
正激波上游马赫数 M1是确定量热完全气体通过正激
波特性变化的 决定性参数 。
12 ?? 12 pp 12 TT
12 ?? 12 pp 12 TT
3 7 8.02 12lim
1
???
?? ?
?M
M
6
1
1
1
2
1
2
1
2
limlim
lim
11
1
????
?
?
?
?
????
??
T
T
p
p
MM
M ?
?
?
?
If M1=1,then ; i.e.,we have the
case of a normal shock wave of vanishing strength---a Mach
wave,如果 M1=1，那么有 ；即正
激波是无限弱的马赫波。
As M1 increases above 1,
progressively increase above 1,当 M1 大于 1逐渐增加时,
也逐渐沿大于 1的趋势增大。
1121212 ??? TTpp ??
1121212 ??? TTpp ??
121212 an d,,TTpp ??
121212,,TTpp 和??
We have stated earlier that shock waves occur in supersonic flows; a
stationary normal shock such as shown in Fig.8.3 does not occur in
subsonic flow,That is,Eqs,(8.59),(8.61),(8.65),and (8.67),the
upstream Mach number is supersonic,M1≥1,However,on a
mathematical basis,these equations also allow solution for M1≤1,
These equations embody the contiunity,momentum,and energy
equations,which in principle do not care whether the value of M1 is
subsonic or supersonic,Here is an ambiguity which can only be
resolved by second law of thermodynamics,
我们在前面已经指出，激波出现在超音速流中；图 8.3所示的精
制的正激波在亚音速流中不可能出现。也即是，关系式 (8.59),
(8.61),(8.65),和 (8.67)中上游马赫数必须是超音速的；必有 M1≥1。
然而，从纯数学的观点看，这些由连续、动量、能量方程推导出
的关系式在理论上并没有限定 M1 一定要大于 1，即上游流动一定
是超音速的。这种多解性只有借助于热力学第二定律来解决。
1
2
1
2
12 lnln p
pR
T
Tcss
p ???
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
???
)1(
)1(
2
1ln
)1(
)1(2
)1(
1
2
1ln
2
1
2
1
2
12
112
MR
M
M
Mcss
p
?
?
?
?
?
?
应用我们在第七章推导出的熵增公式，我们可以得到通过正
激波的熵增计算公式,
（ 7.25）
（ 8.68）
From Eq,(8.68),we see that the entropy change s2-s1 across the
shock is a function of M1 only,由方程 (8.68)可以看出，通过正
激波的熵增 s2-s1 只是波前马赫数 M1的函数。
The second law dictates that
s2-s1≥0
In Eq,(8.68),if M1=1,s2=s1,and if M1>1,then s2-s1>0,both
of which obey the second law,However,if M1<1,then
Eq.(8.68) gives s2-s1<0,which is not allowed by the second
law,Consequently,in nature,only cases involving M1≥1 are
valid,i.e.,normal shock waves can occur only in
supersonic flow,
热力学第二定律指出,
s2-s1≥0
由方程 (8.68)可以看出，如果 M1=1,s2=s1 ；如果 M1>1,
then s2-s1>0,两种情况都符合热力学第二定律。然而，
如果 M1<1,则,(8.68) 式的结果为 s2-s1<0，其不符合热
力学第二定律。因此，只有 M1≥1的情况会发生，即,正
激波只能在超音速流动中发生。
Why does the entropy increase across the shock wave? The second law tells
us that it must,but what mechanism does nature use to accomplish this
increase? To answer these questions,recall that a shock wave is a very thin
region ( on the order of 10-5cm ) across which some large changes occur
almost discontinuously,Therefore,within the shock wave itself,large
gradients in velocity and temperature occur; i.e.,the mechanisms of friction
and thermal condition are strong,These are dissipative,irreversible
mechanisms that always increase the entropy,Therefore,the precise
entropy increase predicted by Eq.(8.68) for a given supersonic M1 is
appropriately provided by nature in the form of friction and thermal
conduction within the interior of the shock wave itself,
为什么通过激波会出现熵增？热力学第二定律告诉我们一定会
有熵增，但是这个熵增产生的机理是什么呢？为了回答这些问
题，让我们回忆第七章讨论过的内容：激波是非常薄的，厚度
只有 10-5cm,通过激波流动性质发生剧烈变化，几乎是不连续
的。因此，在激波本身内部，有很大的速度梯度和温度梯度，
即摩擦和热传导的作用是非常强的。这些耗散的、不可逆的机
制总是引起熵增，因此,(8.68)式给出的超音速波前马赫数 M1对
应的精确熵增实际是由于激波本身内部的摩擦与热传导引起的。
总条件 T0与 p0如何变化？
11 ss a ? 22 ss a ?
22
2
2
2
2
1
1
uTcuTc
pp ???
2
2
0
uTcTc
pp ??
2,01,0 TcTc pp ?
2,01,0 TT ?
首先回答 T0如何变化？
（ 8.30)
（ 8.38)
（ 8.39)
能量方程,
总温定义,
Equation (8.39) states that total temperature is constant across
a stationary normal shock wave,
方程（ 8.39）表明,通过静止正激波总温不变。
a
a
a
a
paa p
pR
T
Tcss
1
2
1
2
12 lnln ???
1,0
2,0
1,0
2,0
12 lnln p
p
R
T
T
css p ???
1,0
2,0
12 ln p
p
Rss ???
Rsse
p
p /)(
1,0
2,0 12 ???
总压如何变化？可借助熵增计算公式求出,
(8.70)
(8.71)
(8.72)
(8.73)
From Eq,(8.68),we know that s2-s1>0 for a normal shock
wave,Hence,Eq,(8.73) states that p0,2<p0,1,The total
pressure decreases across a shock wave,The total
pressure ratio p0,2/p0,1 across a normal shock wave is a
function of M1 only,
由公式（ 8.68），我们知道对于正激波 s2-s1>0,因此，
式（ 8.73)表明,p0,2<p0,1,通过正激波总压降低，且
通过正激波总压比 p0,2/p0,1 只是波前马赫数 M1 的函
数。
至此，我们已经全部回答了本节开始提出的问题,
归纳如下,
1,0
2,0
10
20
12
1
2
1
2
1
2
2 ???????? p
p?
T
Tss
T
T
p
pM
,
,
?
?
2/)1(
]2/)1[(1
2
1
2
12
2 ??
???
??
?
M
MM 2
1
2
1
2
1
1
2
)1(2
)1(
M
M
u
u
??
???
?
?
?
? )1(121 21
1
2 ?
??? Mp
p
?
?
2
1
2
12
1
1
2
1
2
)1(
)1(2)1(
1
21
M
MM
h
h
T
T
?
???
?
?
??
? ?
???? ?
?
?
?
??
?
??
? ?
?????
?
?
?
?
?
??
??
?
??
? ?
???? )1()1(
21ln
)1(
)1(2)1(
1
21ln 2
12
1
2
12
112 MRM
MMcss
p ?
?
?
?
?
?
2,01,0 TT ?
Rsse
p
p /)(
1,0
2,0 12 ???
这些关系式在附录 B中以列表形式给出。
In summary,we have now verified the qualitative changes across a
normal shock wave as sketched in Fig.7.4b and as originally
discussed in Sec,7.6,
??
?
??
? ?
??
? 1)(
1
2 /)1(
1
1,0
2
12
1
??
? p
pau
8.7 MEASUREMENT OF VELOCITY IN A COMPRESSIBLE
FLOW 可压缩流动的速度测量
8.7.1 Subsonic Compressible Flow 亚音速可压缩流
??
?
??
? ?
??
? 1)(
1
2 /)1(
1
1,02
1
??
? p
pM
)1/(2
1
1
1,0 )
2
11( ???? ??? M
p
p （ 8.42）
（ 8.74）
（ 8.75）
??
?
??
? ?
??
? 1)(
1
2 /)1(
1
1,0
2
12
1
??
? p
pau (8.75)
From Eq,(8.75),we see that,unlike incompressible
flow,a knowledge of p0,1 and p1 is not sufficient to
obtain u1; we also need the freestream speed of sound,
a1,
从（ 8.75）式可以看到,与不可压缩流不同，只
知道 p0,1 和 p1 还不足以得到速度 u1；我们还需要
知道自由流的音速,a1。
8.7.2 Supersonic Flow 超音速流
Fig,8.8 A Pitot tube in supersonic flow
A fluid element moving along streamline cde will first decelerated
nonisentropically to a subsonic velocity at point b just behind the
shock, Then it is is isentropically compressed to zero velocity at
point e,As a result,the pressure at point e is not the total pressure
of the freestream but rather the total pressure behind a normal
shock wave,p0,2,This is the Pitot pressure read at the end of the
tube,
沿流线 cde 的流体微团首先非等熵地在 d点减速为亚音速，然
后被等熵地在 e点压缩为驻点速度零。因此,e点的压强不是
自由流的总压而是正激波后的总压 p0,2。 这是 皮托管测得的总
压。
Keep in mind that because of the entropy increase across the shock,
there is a loss in total pressure across the shock,p0,2<p0,1,However,
knowing p0,2 and the freestream static pressure p1 is still sufficient to
calculate the freestream Mach number M1,as follows,
一定要记住的是：由于通过激波引起熵增，所以通过激波会有
总压损失,p0,2<p0,1。然而，知道 p0,2 和自由来流静压 p1，仍足
以使我们计算出自由来流马赫数 M1，方法如下,
1
2
2
2,0
1
2,0
p
p
p
p
p
p ? (8.76)
)1/(2
2
2
2,0 )
2
11( ???? ??? M
p
p (8.77)
2/)1(
]2/)1[(1
2
1
2
12
2 ??
???
??
?
M
MM
(8.78)
1
21
)1(24
)1( 21
)1/(
2
1
2
1
2
1
2,0
?
??
???
?
???
?
??
??
?
?
??
??
?
??
M
M
M
p
p
)1(
1
21 2
1
1
2 ?
?
?? M
p
p
?
?
(8.79)
(8.80)
Equation (8.80) is called the Rayleigh Pitot tube formula,It relates the
Pitot pressure p0,2 and the freestream static pressure p1 to the
freestream Mach number M1,Equation (8.80) gives M1 as an implicit
function of p0,2/p1 and allows the calculation of M1 from known p0,2/p1,
For convenience in making calculations,the ratio p0,2/p1 is tabulated
versus M1 in App.B,
(8.80)式被称为雷利皮托管公式。它将皮托管测得的总压 p0,2 和自
由来流静压 p1与自由来流马赫数 M1 联系起来了。 (8.80)式中 M1为
p0,2/p1 的隐式函数，可以由 p0,2/p1 的值计算出 M1 。为方便应用,
附录 B给出了 p0,2/p1 随 M1的变化表。
