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Chapter 9

Quasi-one-dimensional flow

9.1 Control volume and integral conservation equations

In this chapter we will treat the very general stationary flow of a compressible fluid in achannel without body forces (Gi = 0) shown in figure 9.1 .

Figure 9.1: Control volume enclosing a general stationary channel flow.

Viscous friction imposes a no-slip condition at all solid surfaces. There may be massaddition as well as heat conduction through the wall. The added mass has total enthalpy,htm and stream wise velocity component, Uxm . In addition there may be work done onthe flow by a fan or by the flow on a turbine. The work exchanged with the control volumeis symbolized by a fan assumed to be rotating at a constant angular frequency.

9-1

CHAPTER 9. QUASI-ONE-DIMENSIONAL FLOW 9-2

The appropriate Eulerian-Lagrangian control volume is shown in figure 9.1 . The Eulerianinlet and outlet surfaces of the control volume, A

1

, and A2

are fixed in space as is Aw

which is attached to the non-moving wall of the channel. The Lagrangian surface Af isattached to, and moves with, the surface of the rotating fan. The area where the controlvolume is constantly being stretched and twisted at the corner where A

2

and Af meet isnegligibly small and therefore involves no net contribution to the mass momentum andenergy entering or leaving the control volume.

In chapter 5 we briefly discussed the distinction between steady flow and stationary flow.The flow in the neighborhood of the rotating fan with moving blades is surely unsteadybut as long as the rotation speed of the fan is constant there will be no net gain or lossof mass, momentum or energy inside the control volume. This justifies dropping the timederivative terms in the conservation equations. Another way to look at this is to imaginethat the fan blades are smeared out onto a continuous disk called an actuator disk. Thiswould make the flow truly steady but begs the question as to how such a disk could beconstructed. A third way to justify the use of the steady flow equations is to view any flowvariable as a time average over precisely one blade passage period, or one full fan rotationperiod, or for an averaging time that is so long compared to the blade passage period thata small additional contribution from a portion of a period is negligible.

There are other aspects of the flow that might have to be treated as stationary. Forexample the flow in figure 9.1 could be turbulent in which case unsteady fluctuations inthe flow properties, particularly near the wall, would be characterized by a wide rangeof time scales lacking any obvious periodicity. In this case the flow is stationary as longas the time average of any flow property such as velocity, pressure or density is constant.This brings in some subtle questions as to the length of the time average and specific timeswhen the averaging begins and ends. For our purposes it is su�cient to require that theaveraging is long compared to both the rotation period of the fan and any low frequencyvariations in the turbulent fluctuations.

As a practical matter we are not attempting to treat the case where the flow is being turnedon or o↵ or where the fan is changing speed. These cases would require a full unsteadyanalysis. This is quite do-able but beyond the scope of this course.

9.1.1 Conservation of mass

The integral form of the mass conservation equation in steady flow is evaluated on eachsurface of the control volume.


Z

A(t)⇢�U � UA

�· ndA =

Z

A1

⇢2

U2

dA�Z

A1

⇢1

U1

dA+

Z

Aw

⇢U · ndA+

Z

Af (t)⇢�U � UA

�· ndA = 0

(9.1)

The integral of the mass flux over the wall is simply the total mass added.

Z

Aw

⇢U · ndA = ��m (9.2)

There is no mass addition through the fan surface (the last term in equation (9.1)) and sothe integrated form of the mass equation becomes

Z

A2

⇢2

U2

dA�Z

A1

⇢1

U1

dA = �m. (9.3)

9.1.2 Conservation of x -momentum

Now evaluate the integral form of the momentum equation over the control volume. Notethat U � UA = 0 on the fan surface by the no-slip condition.

Z

A(t)

�⇢U�U � UA

�+ P � �¯⌧

�· ndA

��x

=

Z

A2

(⇢2

U2

U2

+ P2

� ⌧xx2)dA�Z

A1

(⇢1

U1

U1

+ P1

� ⌧xx1)dA+

Z

Aw

�⇢U U + P � �¯⌧

�· ndA

��x

+

Z

Af (t)

�⇢U�U � UA

�+ P � �¯⌧

�· ndA = 0

(9.4)

Over most of the wall the velocity is zero due to the no slip condition. However over theduct through which the added mass enters the control volume there is a contribution tothe x -momentum in the amount

Uxm�m. (9.5)

Any small contributions from the pressure and stress forces acting over the control volumein the neighborhood of the mass injector as well as any non-uniformities in the injector


velocity field have been included in the e↵ective value that might be assigned to Uxm in(9.5). The integrated x -momentum equation now becomes

Z

A(t)

�⇢U�U � UA

�+ P � �¯⌧

�· ndA

��x

=

Z

A2

(⇢2

U2

U2

+ P2

� ⌧xx2)dA�Z

A1

(⇢1

U1

U1

+ P1

� ⌧xx1)dA+

Z

Aw

(P � �¯⌧) · ndA��x

� Uxm�m+ �Fx = 0.

(9.6)

On the fan U = UA and so only the surface pressure and stress contribute an amount Fx

to the momentum flux. The force by the flow on the fan is

�Fx =

Z

Af (t)(P ¯� �¯⌧) · ndA

��x

. (9.7)

If the Reynolds number of the flow is relatively high the major contribution to the totalforce in the x -direction, (9.7), comes from pressure forces integrated over the fan surface.The way to evaluate this force is to sum the stream-wise lift components of all the fanblades. If the sum is negative the fan is adding momentum to the flow. If the sum ispositive, the fan is actually a windmill or turbine taking momentum out of the flow.

9.1.3 Conservation of energy

Next we evaluate the integral form of the energy equation over the control volume.

Z

A(t)

�⇢ (e+ k)

�U � UA

�+ PU �¯⌧ · U + Q

�· ndA =

Z

A2

(⇢2

ht2U2

� ⌧xx2U2

+Q2

)dA�Z

A1

(⇢1

ht1U1

� ⌧xx1U1

+Q1

)dA+

Z

Aw

�⇢htU �¯⌧ · U + Q

�· ndA+

Z

Af (t)

�⇢ (e+ k)

�U � UA

�+ PU �¯⌧ · U + Q

�· ndA = 0

(9.8)


The integral over the wall includes the total enthalpy of the mass injected through thewall.

Z

Aw

�⇢htU �¯⌧ · U + Q

�· ndA = ��Qw � htm�m (9.9)

The last integral in (9.8) is the work done by the flow on the fan.

�W =

Z

Af (t)

�PU �¯⌧ · U

�· ndA (9.10)

Using (9.9) and (9.10) the integrated energy conservation equation is

Z

A2

(⇢2

ht2U2

� ⌧xx2U2

+Q2

)dA�Z

A1

(⇢1

ht1U1

� ⌧xx1U1

+Q1

)dA = �Qw + htm�m� �W.

(9.11)

9.2 Area averaged flow

Every flow variable is a three-dimensional function of space. By averaging the flow acrossthe channel we can produce approximate flow variables that depend only on the stream-wisecoordinate. For example

⇢ (x) =1

A (x)

Z⇢ (x, y, z)dydz. (9.12)

Similarly we can define the following area averages.


T (x) =1

A (x)

ZT (x, y, z)dydz

P (x) =1

A (x)

ZP (x, y, z)dydz

s (x) =1

A (x)

Zs (x, y, z)dydz

U (x) =1

A (x)

ZU (x, y, z)dydz

⌧xx (x) =1

A (x)

Z⌧xx (x, y, z)dydz

Qx (x) =1

A (x)

ZQx (x, y, z)dydz

(9.13)

Thus every variable in the flow can be written as a mean plus a deviation.

⇢ (x, y, z) = ⇢ (x) + ⇢0 (x, y, z)

T (x, y, z) = T (x) + T 0 (x, y, z)

P (x, y, z) = P (x) + P 0 (x, y, z)

s (x, y, z) = s (x) + s0 (x, y, z)

U (x, y, z) = U (x) + U 0 (x, y, z)

⌧xx (x, y, z) = ⌧xx (x) + ⌧ 0xx (x, y, z)

Qx (x, y, z) = Qx (x) +Q0x (x, y, z)

(9.14)

Consider the mass flux integral.

Z

A⇢UdA =

Z

A

�⇢+ ⇢0

� ⇣U + U 0

⌘dA =

Z

A⇢UdA+

Z

A⇢0U 0dA =

⇢ (x) U (x)A (x) + \⇢0U 0A (x)

(9.15)


The terms that are linear in the deviations are zero by the definition of the average. Aslong as spatial correlations of flow variables such as d⇢0U 0 are small and can be neglected,the area average is a useful approximation to the flow. In terms of area averaged variables,the integral equations of motion become

⇢2

U2

A2

� ⇢1

U1

A1

= �m

⇣⇢2

U2

U2

+ P2

� ⌧xx2⌘A

2

�⇣⇢1

U1

U1

+ P1

� ⌧xx1⌘A

1

+

Z

Aw

(P � �¯⌧) · ndA��x

� Uxm�m+ �Fx = 0

⇣⇢2

ht2U2

� ⌧xx2U2

+ Qx2

⌘A

2

�⇣⇢1

ht1U1

� ⌧xx1U1

+ Qx1

⌘A

1

=

�Qw + htm�m� �W.

(9.16)

9.2.1 The traction vector

The pressure-stress integral on the wall in (9.16) (see also (9.7)) needs to be examined alittle further.

Figure 9.2: Traction force on the wall due to pressure and viscous forces.

The unit outward normal to the wall is n = (nx, ny, nz) and the components of the so-calledtraction vector are


T = (P � �¯⌧) · n =

2

4Pnx � ⌧xxnx � ⌧xyny � ⌧xznz

�⌧xynx + Pny � ⌧yyny � ⌧yznz

�⌧zxnx � ⌧zyny + Pnz � ⌧zznz

3

5 . (9.17)

The x -component of the traction vector is indicated in figure 9.2.

If we imagine that the length of the control volume is made very small (�x ! 0) then thepressure and viscous stress variation along the wall of the control volume is small and sothe pressure-viscous stress integral over the surface in (9.16) can be approximated usingaverage values as follows.

Z

Aw

(P � �¯⌧) · ndA��x

=

Z

Aw

(Pnx � ⌧xxnx � ⌧xyny � ⌧xznz)dA ⇠=

✓P1

+ P2

2

◆(A

1

�A2

)�✓⌧xx1 + ⌧xx2

2

◆(A

1

�A2

) + ⌧w⇡

✓D

1

+D2

2

◆�x

(9.18)

where we have used

Z

Aw

nxdA =(A1

�A2

) (9.19)

and

Z

Aw

(⌧xyny + ⌧xznz) dA ⌘ �⌧w⇡

✓D

1

+D2

2

◆�x. (9.20)

This last relation is essentially a definition for the e↵ective wall shear stress ⌧w which isalways taken to be positive. The wetted surface of the channel is defined using the hydraulicdiameter of the channel defined as

D =

✓4A

⇡

◆1/2

. (9.21)

The integrated equations of motion (for small �x) now take the form


⇢2

U2

A2

� ⇢1

U1

A1

= �m

(⇢2

U2

U2

+ P2

� ⌧xx2)A2

� (⇢1

U1

U1

+ P1

� ⌧xx1)A1

�✓P1

+ P2

2

◆(A

1

�A2

) +

✓⌧xx1 + ⌧xx2

2

◆(A

1

�A2

)� ⌧w⇡

✓D

1

+D2

2

◆�x = Uxm�m� �Fx

(⇢2

H2

U2

� ⌧xx2U2

+Q2

)A2

� (⇢1

H1

U1

� ⌧xx1U1

+Q1

)A1

= �Qw + htm�m� �W(9.22)

where the averaging symbol has been dropped for convenience. Now take the limit �x ! 0of each of the equations in (9.22). Each of the terms in the integrated equations becomesa di↵erential.

⇢2

U2

A2

� ⇢1

U1

A1

) d (⇢UA)

⇢2

U2

U2

� ⇢1

U1

U1

) d�⇢U2A

�

P2

A2

� P1

A1

) d (PA)

⌧xx2A2

� ⌧xx1A1

) d (⌧xxA)

✓P1

+ P2

2

◆(A

1

�A2

) ) PdA

✓⌧xx1 + ⌧xx2

2

◆(A

1

�A2

) ) ⌧xxdA

⌧w⇡

✓D

1

+D2

2

◆�x ) ⌧w⇡Ddx

⇢2

ht2U2

A2

� ⇢1

ht1U1

A1

) d (⇢UhtA)

⌧xx2U2

A2

� ⌧xx1U1

A1

) d (⌧xxUA)

Qx2A2

�Qx1A1

) d (QxA)

(9.23)

Now the equations of motion (9.22) become


d (⇢UA) = �m

d�⇢U2A

�+ d (PA)� d (⌧xxA)� PdA+ ⌧xxdA = �⌧w⇡Ddx+ Uxm�m� �Fx

d (⇢UhtA)� d (⌧xxUA) + d (QxA) = �Qw + htm�m� �W.

(9.24)

Both the momentum and energy equations can be simplified using continuity.

U�m+ ⇢UAdU +AdP �Ad⌧xx = �⌧w⇡Ddx+ Uxm�m� �Fx

ht�m+ ⇢UAdht �⌧xx⇢

�m� ⇢UAd

✓⌧xx⇢

◆+

Qx

⇢U�m+ ⇢UAd

✓Qx

⇢U

◆=

�Qw + htm�m� �W

(9.25)

Finally our quasi-one-dimensional equations of motion are

d (⇢UA) = �m

d (P � ⌧xx) + ⇢UdU = �⌧w

✓⇡Ddx

A

◆+

(Uxm � U) �m

A� �Fx

A

d

✓ht �

⌧xx⇢

+Qx

⇢U

◆=

�Qw

⇢UA� �W

⇢UA+

✓htm �

✓ht �

⌧xx⇢

+Qx

⇢U

◆◆�m

⇢UA.

(9.26)

It is convenient to introduce the friction coe�cient defined as

Cf =⌧w

1

2

⇢U2

. (9.27)

The heat added through the wall and work done per unit mass flow are

�qw =�Qw

⇢UA

�w =�W

⇢UA.

(9.28)


With these definitions, the area-averaged equations of motion take on the fairly conciseform

d (⇢U) =�m

A� ⇢U

dA

A

d (P � ⌧xx) + ⇢UdU = �1

2⇢U2

✓4Cfdx

D

◆+

(Uxm � U) �m

A� �Fx

A

d

✓ht �

⌧xx⇢

+Qx

⇢U

◆= �qw � �w +

✓htm �

✓ht �

⌧xx⇢

+Qx

⇢U

◆◆�m

⇢UA.

(9.29)

The heat added to the flow through the wall per unit mass is �qw and the work done bythe flow per unit mass is �w. The spirit of these equations is that the di↵erential termson the right hand sides of (9.29) such as dA/A , �m , 4Cfdx/D , �qw and �w are assumedto be known or given quantities. They can be thought of as inputs to the flow whereasthe resulting variations in ⇢, U , P and temperature T (recall Qx = �k@T/@x for a linearconducting medium) are the output changes in the flow.

Notice that up to this point we have not invoked an equation of state. Equations (9.29)apply to the steady flow of any continuous medium in the absence of body forces. Bythe same token (9.29) is not a closed system; We have four unknowns but only threeequations.

9.2.2 Example - steady, gravity-free, adiabatic flow of a compressiblefluid in a channel

Figure 9.3: Adiabatic channel flow

The integral form of the energy equation in this case is simply


d

✓ht �

⌧xx⇢

+Qx

⇢U

◆= 0. (9.30)

For typical cases such as the flow of gases at more than a few meters per second, the stressand heat conduction terms on A

1

and A2

are very small and can be neglected. Thus

ht2 = ht1. (9.31)

This is, of course, a result we have seen before. It is extraordinarily useful because there areso many flow situations where an adiabatic approximation can be assumed and where thestagnation enthalpy of a fluid particle is approximately conserved. This applies generally,except for flow near a non- adiabatic wall where significant heat conduction may takeplace and within a shock wave where large stream-wise velocity and temperature gradientsoccur.

9.3 Normal shock waves

Lets consider uni-directional flow in the absence of mass addition, wall friction, and heataddition. The continuity equation in this case is

d (⇢U) = 0. (9.32)

If the fluid is incompressible (9.32) gives the trivial result dU = 0. But if the fluid iscompressible then a non-trivial flow can exist where U and ⇢ vary inversely. The primeexample of such a flow is the shock wave where the properties of the flow change almostdiscontinuously.

The shaded region in figure 9.4 (the shock) contains gradients in temperature, pressure,density and velocity while the upstream and downstream conditions are perfectly uniform.The thickness of the shock is � . The equations of motion in this case are

d (⇢U) = 0

d�P � ⌧xx + ⇢U2

�= 0

d

✓ht �

⌧xx⇢

+Qx

⇢U

◆= 0.

(9.33)


Figure 9.4: Shock wave schematic

Notice that the continuity equation has been used to make the momentum equation aperfect di↵erential. There are three conserved quantities in this flow.

⇢U = constant1

P � ⌧xx + ⇢U2 = constant2

ht �⌧xx⇢

+Qx

⇢U= constant3

(9.34)

These combinations of flow variables have the same value at every point in the flow includingwithin the shock. Integrate (9.34) between states 1 and 2.

⇢1

U1

= ⇢2

U2

P1

� ⌧xx1 + ⇢1

U1

2 = P2

� ⌧xx2 + ⇢2

U2

2

ht1 �⌧xx1⇢1

+Qx1

⇢1

U1

= ht2 �⌧xx2⇢2

+Qx2

⇢2

U2

(9.35)

Recall that for a Newtonian fluid

⌧xx =

✓4

3µ+ µ⌫

◆@U

@x(9.36)

where µv is the bulk viscosity, and


Qx = �k@T

@x. (9.37)

The flow at 1 and 2 is uniform, i.e., the velocity and temperature gradients are zero andso the equations of motion reduce to the classical shock jump conditions.

⇢1

U1

= ⇢2

U2

P1

+ ⇢1

U1

2 = P2

+ ⇢2

U2

2

ht1 = ht2

(9.38)

It is important to recognize that in deriving (9.38) viscosity and heat conduction werenot neglected. They vanish from the jump conditions because of the uniformity of theupstream and downstream flows. Nevertheless their e↵ects are felt through the mechanismof molecular collision by which changes in the state of the fluid are accomplished withinthe shock. Notice also that we still have not invoked an equation of state. The equations(9.38) govern the propagation of shock waves in any continuous medium, air, water, rock,etc.

9.3.1 The Rankine-Hugoniot relations

A general set of jump relations in which the velocity does not appear can be derived from(9.38). First, use mass and momentum to work out

U1

U2

=P2

� P1

⇢2

� ⇢1

. (9.39)

An alternate form of (9.39) assuming an ideal gas law and constant specific heat is

P2

P1

� 1 = �M1

2

✓1� ⇢

1

⇢2

◆. (9.40)

The energy equation is

✓�

� � 1

◆P1

⇢1

+1

2U1

2 =

✓�

� � 1

◆P2

⇢2

+1

2U2

2. (9.41)

Use continuity to eliminate U1

in (9.41) and solve for U2

.


U2

2 =

✓2�

� � 1

◆P2

⇢1

� P1

⇢2

(⇢2

� ⇢1

) (⇢2

+ ⇢1

)(9.42)

Use (9.42) to replace U2

2 in the momentum jump condition. After some rearrangementthe result is

P2

P1

=

�+1

��1

⇣⇢2

⇢1

⌘� 1

�+1

��1

�⇣⇢2

⇢1

⌘ . (9.43)

Equation (9.43) is the famous Rankine-Hugoniot relation. Another form of (9.43) is

P2

P1

� 1P2

P1

+ 1= �

⇢2

⇢1

� 1⇢2

⇢1

+ 1

!. (9.44)

which can be rearranged to read

P2

� P1

⇢2

� ⇢1

= �

✓P2

+ P1

⇢2

+ ⇢1

◆. (9.45)

9.3.2 Shock property ratios in a calorically perfect ideal gas

For a calorically perfect gas the third jump condition ht1 = ht2 can be written

CpT1

+1

2U1

2 = CpT2

+1

2U2

2. (9.46)

Consider a reference state where the flow speed equals the local speed of sound. The flowvariables at this state will be denoted with a superscript *; Thus U⇤ = a⇤, and ⇢ = ⇢⇤ ,P = P ⇤, T = T ⇤. Each side of (9.46) can be equated with the reference state.

CpT1

+1

2U1

2 =� + 1

2 (� � 1)a⇤2

CpT2

+1

2U2

2 =� + 1

2 (� � 1)a⇤2

(9.47)

Use the ideal gas law to write (9.47) in the form


�

� � 1P1

+1

2⇢2

U2

U1

=� + 1

2 (� � 1)⇢1

a⇤2

�

� � 1P2

+1

2⇢1

U2

U1

=� + 1

2 (� � 1)⇢2

a⇤2.

(9.48)

Subtract the relations in (9.48)

�

� � 1(P

2

� P1

)� 1

2(⇢

2

� ⇢1

)U1

U2

=� + 1

2 (� � 1)a⇤2 (⇢

2

� ⇢1

) . (9.49)

Now use (9.39) to replace P2

� P1

in (9.49) and gather terms. The result is the famousPrandtl relation

U1

U2

= a⇤2. (9.50)

One of the implications of (9.50) is that for a normal shock the flow must be supersonicahead of the shock and subsonic behind the shock. Note that a⇤ is essentially defined by(9.47) and can be determined entirely by values in either region 1 ahead of the shock orregion 2 behind the shock. Prandtl’s relation is the Rosetta stone for generating all of theshock jump relations in terms of the shock Mach number. For example, substitute (9.50)into the first relation in (9.47).

CpT1

+1

2U1

2 =� + 1

2 (� � 1)U1

U2

(9.51)

Divide (9.51) by U1

2. The result is

U2

U1

=2

� + 1

✓�RT

1

U1

2

◆+

✓� � 1

� + 1

◆. (9.52)

The Mach number ahead of the shock is

M1

=U1p

�RT1

(9.53)

and (9.52) becomes the basic relation for the velocity ratio (and density ratio) across theshock in terms of the upstream Mach number.


U2

U1

=1 +

⇣��1

2

⌘M

1

2

⇣�+1

2

⌘M

1

2

=⇢1

⇢2

(9.54)

All of the important properties of a normal shock wave can be expressed in terms of theupstream Mach number. Using (9.54) and the Rankine-Hugoniot relation (9.43) we canwork out the pressure ratio across the shock.

P2

P1

=

⇣�+1

��1

⌘⇣⇢2

⇢1

⌘� 1

�+1

��1

�⇣⇢2

⇢1

⌘ =

⇣�+1

��1

⌘✓( �+1

2

)M1

2

1+( ��1

2

)M1

2

◆� 1

�+1

��1

�✓

( �+1

2

)M1

2

1+( ��1

2

)M1

2

◆ (9.55)

Simplify (9.55).

P2

P1

=

2��1

M1

2 � 1�+1

��1

(9.56)

The pressure ratio (9.56) is called the shock strength. Note the rapid increase with Machnumber. The temperature ratio is generated using the ideal gas law.

T2

T1

=

✓P2

P1

◆✓⇢1

⇢2

◆=

0

@�M

1

2 �⇣��1

2

⌘

�+1

2

1

A

0

@1 +

⇣��1

2

⌘M

1

2

⇣�+1

2

⌘M

1

2

1

A (9.57)

The downstream Mach number is

✓M

2

M1

◆2

=

✓U2

U1

◆2T

1

T2

. (9.58)

Substitute (9.54) and (9.57).

⌧ij = 2µSij �✓2

3µ� µv

◆�ijSkk

⌧xy = µ

✓@U

@y+

@V

@x

◆⇠= µ

@U

@y

(9.59)


Several factors in (9.59) cancel and the final result for the downstream Mach numberis

M2

2 =

0

@1 +

⇣��1

2

⌘M

1

2

�M1

2 �⇣��1

2

⌘

1

A . (9.60)

The relation (9.60) is plotted below for � = 1.4.

Figure 9.5: Downstream Mach number versus upstream Mach number for a normal shock.

Note that the downstream Mach number has a finite lower limit.

limM

1

!1M

2

=

r� � 1

2�(9.61)

9.3.3 Stagnation pressure ratio across a normal shock wave

The stagnation pressure is determined from the isentropic relations in regions 1 and 2.

Pt1

P1

=

✓1 +

✓� � 1

2

◆M

1

2

◆ ��1

Pt2

P2

=

✓1 +

✓� � 1

2

◆M

2

2

◆ ��1

(9.62)


Divide these two relations.

Pt2

Pt1=

P2

P1

0

@1 +

⇣��1

2

⌘M

2

2

1 +⇣��1

2

⌘M

1

2

1

A

��1

(9.63)

In (9.63) replace P2

/P1

with (9.56) and M2 with (9.60).

Pt2

Pt1=

2��1

M1

2 � 1�+1

��1

!0

BB@

1 +⇣��1

2

⌘✓1+( ��1

2

)M1

2

�M1

2�( ��1

2

)

◆

1 +⇣��1

2

⌘M

1

2

1

CCA

��1

(9.64)

Rearrange (9.64).

Pt2

Pt1=

2��1

M1

2 � 1�+1

��1

!0

B@

⇣�+1

2

⌘2

M1

2

⇣��1

2

⌘⇣2��1

M1

2 � 1⌘⇣

1 +⇣��1

2

⌘M

1

2

⌘

1

CA

��1

(9.65)

Combine factors in (9.65). The stagnation pressure ratio across the shock wave is

Pt2

Pt1=

�+1

��1

2��1

M1

2 � 1

! 1

��1

0

@

⇣�+1

2

⌘M

1

2

1 +⇣��1

2

⌘M

1

2

1

A

��1

. (9.66)

The stagnation pressure ratio across a normal shock wave is plotted in figure 9.6 for twovalues of �.

Two important features of figure 9.6 need to be noted.

1) At Mach numbers close to one the change in stagnation pressure across a normal shockwave is very small.

2) At high Mach numbers the stagnation pressure loss is very large with

limM

1

!1

Pt2

Pt1=

✓� + 1

� � 1

◆ ��1

✓� + 1

2�M1

2

◆ 1

��1

. (9.67)

Recall that we can write the Gibbs equation in terms of the stagnation state of thegas.


Figure 9.6: Stagnation pressure ratio across a normal shock as a function of shock Machnumber.

ds =dhtTt

�RdPt

Pt(9.68)

Integrate (9.68) from state 1 to state 2. Since dht = 0 the entropy change across the waveis determined directly from the change in stagnation pressure.

Pt2

Pt1= e

�⇣

s2

�s1

R

⌘

(9.69)

The connection (9.69) between the stagnation pressure and entropy in an adiabatic flowis an extremely important one that we see often when we solve problems in compressiblechannel flow.

The equations for the flow properties across the shock are the fundamental relations usedto generate the shock tables.

9.3.4 Example - stagnation at a leading edge in supersonic flow

The figure below shows a supersonic flow of Helium (atomic weight equals 4) over theleading edge of a thick flat plate at a free stream Mach number M

1

= 2.0.

The temperature of the free stream is 300K and the pressure is one atmosphere.


Figure 9.7: Supersonic flow of helium over a leading edge.

1) Determine the energy per unit mass of a fluid element located at points 1 (free stream), 2(just behind the shock) and 3 (at the stagnation point on the body). State the assumptionsused to solve the problem. Express your answer in Joules/kg.

Solution

The energy per unit mass of a flowing gas is the sum of internal energy and kinetic energyper unit mass, e + k .

a) Assume the gas is calorically perfect - constant heat capacities.

b) Assume the flow is adiabatic from station 1 to station 3.

c) Assume the body is adiabatic.

For Helium the number of degrees of freedom equals 3 and at the conditions of the freestream we have the following values.


R =8314.472

4= 2078.62m2/sec2 �K

Cv =3

2R = 3117.93m2/sec2 �K

Cp =3 + 2

2R = 5196.55m2/sec2 �K

� = 5/3

a =p�RT =

r5

32078.62 (300) = 1019.46m2/sec2

U1

= 2 (1019.46) = 2038.92m2/sec2

e1

+ k1

= CvT1

+1

2U1

2 = 3117.93 (300) + 0.5(2038.92)2 = 935379 + 2078597

e1

+ k1

= 3013976 J/kg

(9.70)

The stagnation temperature of the free stream is determined from

Tt

T= 1 +

✓� � 1

2

◆M2. (9.71)

Thus

Tt1 = 300

✓1 +

4

3

◆= 700K (9.72)

Across a normal shock at Mach 2 the temperature ratio is

T2

T1

=

⇣1 +

⇣��1

2

⌘M

1

2

⌘⇣�M

1

2 �⇣��1

2

⌘⌘

⇣�+1

2

⌘2

M1

2

(9.73)

which gives


T2

T1

=

�1 + 4

3

� �5

3

(4)��1

3

��4

3

�2

(4)=

�7

3

� �19

3

��4

3

� �16

3

� =7

4

✓19

16

◆= 2.078. (9.74)

Assume the flow is adiabatic from the free stream to the stagnation point.

h1

+1

2U1

2 = h2

+1

2U2

2 = h3

+1

2U3

2 (9.75)

We can rewrite this equation as follows.

RT1

+ (e1

+ k1

) = RT2

+ (e2

+ k2

) = RT3

+ (e3

+ k3

) (9.76)

The temperatures at stations 1, 2 and 3 are respectively

T1

= 300K

T2

= 2.078 (300) = 623.44K

T3

= Tt1 = 700K.

(9.77)

Now

e1

+ k1

= 3.014⇥ 106 J/kg

(e2

+ k2

) = (e1

+ k1

)�R (T2

� T1

) = 3.014⇥ 106 � 2078.62 (623.44� 300)= 3.014⇥ 106 � 0.6723⇥ 106 = 2.3417⇥ 106 J/kg

(e3

+ k3

) = (e1

+ k1

)�R (T3

� T1

) = 3.014⇥ 106 � 2078.62 (700� 300)= 3.014⇥ 106 � 0.8314⇥ 106 = 2.1826⇥ 106 J/kg.

(9.78)

The energy of a fluid element decreases considerably across the shock and then decreasesfurther to the stagnation point.

2) Describe the mechanism by which the energy of the fluid element changes as it movesfrom station 1 to station 3.

The work done by the pressure and viscous normal force field on the fluid element is themechanism by which the energy decreases in moving from station 1 to station 3. The flowenergy decreases across the shock wave through a combination of pressure and viscous


normal stress forces of roughly equal magnitude that act to compress the fluid elementincreasing its internal energy while decelerating it and reducing its kinetic energy. The lossof kinetic energy dominates the increase in internal energy.

Between stations 2 and 3 the flow further decelerates as the pressure increases toward thestagnation point. Viscous normal forces also act in region 2 to 3 but because the streamwisevelocity gradients are small (compared to the shock) viscous forces are generally muchsmaller than the pressure forces.

9.4 Shock wave thickness

Let’s use the jump conditions and our knowledge of entropy generation to estimate thethickness of a shock wave.

Figure 9.8: Linear model for temperature and pressure variation within a shock.

The integral form of the entropy transport equation, derived in Chapter 7, over an Euleriancontrol volume is

d

dt

Z

V⇢sdV +

Z

A

✓⇢Us� k

TrT

◆· ndA =

Z

V

✓⌥+ �

T

◆dV (9.79)

where for a Newtonian, linear, heat conducting medium

� = 2µ

✓Sij �

1

3�ijSkk

◆✓Sij �

1

3�ijSkk

◆+ µv (SiiSkk) (9.80)

and

⌥ =k

T

✓@T

@xj

@T

@xj

◆. (9.81)


The stress tensor is

⌧ij = 2µSij � ((2/3)µ� µv) �ijSkk. (9.82)

For the unidirectional flow within the shock wave the stress tensor reduces to

⌧ij =

2

6666664

✓4

3µ+ µv

◆dU

dx0 0

0

✓�2

3µ+ µv

◆dU

dx0

0 0

✓�2

3µ+ µv

◆dU

dx

3

7777775. (9.83)

Note that the viscous normal stresses in the y and z directions are not zero since Skk =r · U = dU/dx. The modified rate-of-strain tensor that appears in the dissipation is

Sij �1

3�ijSkk =

2

66664

2

3

dU

dx0 0

0 �1

3

dU

dx0

0 0 �1

3

dU

dx

3

77775(9.84)

and therefore the kinetic energy dissipation within the shock is

� =

✓4

3µ+ µv

◆✓dU

dx

◆2

. (9.85)

The temperature dissipation is

⌥ =k

T

✓dT

dx

◆2

. (9.86)

Now we integrate the entropy transport equation over the control volume indicated in thefigure 9.8. Since the flow is steady

d

dt

Z

V⇢sdV = 0. (9.87)

The integrated entropy equation becomes


✓⇢UsA� k

T

dT

dxA

◆

2

�✓⇢UsA� k

T

dT

dxA

◆

1

=

✓Z �

0

✓⌥+ �

T

◆dx

◆A. (9.88)

The areas cancel on both sides of the equation and the temperature gradients at stations1 and 2 are zero. The product ⇢U is constant through the wave.

⇢U (s2

� s1

) =

Z �

0

4

3

µ+ µv

T

!✓dU

dx

◆2

+k

T 2

✓dT

dx

◆2

!dx (9.89)

In order to model the integral on the right-hand-side of (9.89) we make the following linearapproximations for the gradients inside the wave and for the mean temperature.

dU

dx⇠=

U2

� U1

�

dT

dx⇠=

T2

� T1

�

T ⇠=T2

+ T1

2

(9.90)

We assume that the viscosity and thermal conductivity are evaluated at the mean temper-ature. The control volume balance of entropy now becomes

⇢U (s2

� s1

)

✓T2

+ T1

2

◆� =

✓4

3µ+ µv

◆(U

2

� U1

)2 + 2k(T

2

� T1

)2

(T2

+ T1

)

!(9.91)

which can be expressed as

⇢U (s2

� s1

) � =

0

BBBB@

2�4

3

µ+ µv�

U1

2

T1

⇣U2

U1

� 1⌘2

+ 4k

⇣T2

T1

�1

⌘2

⇣T2

T1

+1

⌘

⇣T2

T1

+ 1⌘

1

CCCCA. (9.92)

Use U2

/U1

= ⇢1

/⇢2

and move the entropy term to the right-hand-side.


⇢U�

µ=

0

BBBB@

2 (� � 1)⇣4

3

+ µv

µ

⌘M

1

2

⇣⇢1

⇢2

� 1⌘2

+ 4 kµCp

⇣T2

T1

�1

⌘2

⇣T2

T1

+1

⌘

1

�

⇣T2

T1

+ 1⌘⇣

s2

�s1

Cv

⌘

1

CCCCA(9.93)

Recall that the Prandtl number is defined as

Pr =µCp

k. (9.94)

The entropy jump across the shock can be expressed in terms of the density and tempera-ture ratio using Gibbs’ equation.

s2

� s1

Cv= ln

T2

T1

✓⇢1

⇢2

◆(��1)

!(9.95)

Finally

⇢U�

µ=

0

BBBB@

2� (� � 1)⇣4

3

+ µv

µ

⌘M

1

2

⇣⇢1

⇢2

� 1⌘2

+ 4 �Pr

⇣T2

T1

�1

⌘2

⇣T2

T1

+1

⌘

⇣T2

T1

+ 1⌘ln

✓T2

T1

⇣⇢1

⇢2

⌘(��1)

◆

1

CCCCA. (9.96)

The right hand side of (9.96) is expressed in terms of ratios across the shock and canbe written entirely in terms of the upstream Mach number using the shock jump rela-tions.

⇢1

⇢2

=(� � 1)M

1

2 + 2

(� + 1)M1

2

T2

T1

=

�2�M

1

2 � (� � 1)� �

(� � 1)M1

2 + 2�

(� + 1)2M1

2

(9.97)

The left hand side of (9.96) can be interpreted as the shock Reynolds number where thecharacteristic length is taken to be the shock thickness. In summary

⇢U�

µ= F

✓M

1

2, �, Pr,µv

µ

◆. (9.98)


The function F�M

1

2, �, Pr, µv/µ�in (9.96) is plotted in figure 9.9 for two values of the

ratio of bulk to shear viscosity µv/µ.

Figure 9.9: Shock Reynolds number as a function of Mach number.

The results of our model indicate that the Reynolds number of a shock wave is small anddecreases with increasing Mach number, a�rming the strongly viscous nature of the flowthrough the wave. Note also that, at a given Mach number, a gas with a large bulk viscositywill generate a much thicker shock wave.

Results from kinetic theory relating the viscosity to the mean free path are provided inAppendix A. To a good approximation µ = ⇢a� where � is the mean free path and a is thespeed of sound. Equation (9.98) can be written

⇢U�

⇢a�⇠= F

✓M

1

2, �, Pr,µv

µ

◆. (9.99)

In the spirit of the approximations used earlier let the Mach number appearing on the lefthand side of (9.99) be approximated as

U

a=

✓U1

+ U2

2

◆✓2

a1

+ a2

◆= M

1

0

@⇢1

⇢2

+ 1q

T2

T1

+ 1

1

A . (9.100)


Now the ratio of shock thickness to the mean free path in the gas (the inverse of the shockKnudsen number) can be estimated.

�

�⇠=

F⇣M

1

2, �, Pr,µv

µ

⌘

M1

0

@

qT2

T1

+ 1⇢1

⇢2

+ 1

1

A (9.101)

This function is plotted in figure 9.10.

Figure 9.10: Shock thickness number as a function of Mach number.

We can see from this final result that in fact a shock is extremely thin, on the order of afew mean free paths thick. It is important that the model gives a shock thickness greaterthan the mean free path so as to be consistent with the fact that the shock is viscous andthat a molecule experiences at least several collisions as it passes through the shock. It isthe increased randomization of the molecular motion due to these collisions that leads tothe entropy rise across the shock imposed by the shock jump conditions.

This result also serves as a warning that the model, based on the Navier-Stokes equations,has its limitations! Notice that for Mach numbers above about 2, the shock is a very smallnumber of mean free paths thick. This is inconsistent with the assumption that the gas isin local thermodynamic equilibrium. The equilibrium state of the gas, with well definedvalues of temperature (to which the viscosity and thermal conductivity are related) andpressure, can only be established through collisions and four or five collisions is barelyenough.


The state of the gas inside a very strong shock wave is far from thermodynamic equilibriumand a proper theory for the flow requires a much more sophisticated treatment. One isforced to a much more complex analysis based on the Boltzmann equation which describesthe space-time evolution of the velocity probability density function and explicitly accountsfor molecular collisions.

9.5 Problems

Problem 1 - Heat in the amount of 106 J/kg is added to a compressible flow of helium ina diverging channel. The heat is distributed so that the area averaged velocities at stations1 and 2 are the same.

Figure 9.11: Heat added to a diverging channel flow.

The temperature at station 1 is 1000K and the area ratio is A2

/A1

= 2. Determine T2

/T1

,⇢2

/⇢1

, P2

/P1

, and (s2

� s1

) /Cp.

Problem 2 - Recall Problem 5.3. Consider steady flow in one dimension where U =(U (x) , 0, 0) and all velocity gradients are zero except

A11

=@U

@x. (9.102)

Work out the components of the Newtonian viscous stress tensor ⌧ij . Note the role ofthe bulk viscosity. Inside a normal shock wave the velocity gradient can be as high as1010 sec�1. Using values for Air at 300K and one atmosphere, estimate the magnitude ofthe viscous normal stress inside a shock wave. Express your answer in atmospheres.

Problem 3 - Consider a normal shock wave in helium with Mach number M1

= 3. Thetemperature of the upstream gas is 300K and the pressure is 105N/m2.

1) Determine the stagnation temperature in region 2 as measured by an observer at restwith respect to the upstream gas. This is an observer that sees the shock wave propagating


Figure 9.12: Normal shock wave in helium

to the left at Mach 3.

2) Determine the stagnation pressure in region 2 as measured by an observer at rest withrespect to the upstream gas.

Problem 4 - The figure below shows supersonic flow of carbon dioxide,� = 4/3, past acylindrical bullet at a free stream Mach number, M

1

= 2.77.

Figure 9.13: Supersonic flow of carbon dioxide past a bullet. See Van Dyke page 163.


a) Determine the temperature, pressure and Mach number of the gas on the centerline justdownstream of the shock wave.

b) Estimate the temperature and pressure at the stagnation point on the upstream face ofthe cylinder.

c) Determine the entropy increase across the shock wave.

d) Estimate the thickness of the shock wave.

e) Estimate the acceleration of the fluid element as it traverses the shock wave. Expressyour answer in m/sec2.


Problem 5 - Estimate the thickness of the shock wave in Helium discussed in section9.4.

Problem 6 - The sketch below shows supersonic flow of air, � = 1.4, past a sphere at afree stream Mach number, M

1

= 1.53. See Van Dyke page 164.

Figure 9.14: Supersonic flow past a sphere.

a) Compare each of the following properties of the gas; stagnation enthalpy, ht, stagnationpressure, Pt and entropy per unit mass, s at locations 1, 2 and 3 identified in the figure.State the assumptions needed to make your comparisons.

b) What can you say about the same properties of the gas at station 4? How certain isyour answer? Why?

c) Determine the Mach number at station 2.

d) Is the energy per unit mass (internal plus kinetic) of a gas particle at 1 and 2 the same?Prove your answer.

Problem 7 - We often encounter practical situations involving weak shock waves wherethe Mach number upstream of the wave is very close to one.

Figure 9.15: Weak normal shock

Let the Mach number ahead of the wave be M1

= 1 + " where " < 1. Derive the weakshock jump relations M

2

⇠= 1� " and


U2

� U1

a1

⇠= � 4

� + 1"

T2

� T1

T1

⇠=?

P2

� P1

P1

⇠=?

Pt2 � Pt1

Pt1

⇠= �16

3

�

(� + 1)2"3.

(9.103)

The last result in (9.103) is extremely important in that it shows that the stagnation lossacross a weak shock is extremely small indeed. This fact is exploited in the design ofsupersonic inlets. Note that first and second order terms in " have cancelled. I suggest youuse symbol manipulation software such as Mathematica to derive this result.

Problem 8 - The photo below shows a rifle bullet moving in air at a Mach number of 1.1.The air temperature is 300K. On the centerline the flow from the left passes through anormal shock wave and then stagnates on the nose of the bullet.

Figure 9.16: Supersonic rifle bullet.

a) Determine the temperature, pressure and density change across the wave.

b) Compare the temperature, pressure and density of the gas at the nose of the bullet tovalues in the freestream.

c) Evaluate the entropy change.

d) State any assumptions used.

Problem 9 - Use the weak shock theory developed in problem 7 to estimate the thickness ofthe shock wave depicted in Problem 8. Develop an expression for estimating the thicknessof a weak shock wave � in terms of " and �.


Problem 10 - The figure below shows supersonic flow of air over a model of a re-entrybody at a free stream Mach number, M

1

= 2.

Figure 9.17: Supersonic flow over a re-entry body.


1) Determine the stagnation temperature and pressure of a fluid element located at stations1 (free stream), 2 (just behind the shock) and 3 (at the stagnation point on the body).State the assumptions used to solve the problem. Express your answers in degrees Kelvinand atmospheres.

2) What can you say about the state of the gas at point 4?

3) Refer the stagnation temperatures at 1, 2 and 3 to an observer at rest with respect tothe upstream gas. To such an observer the body is moving to the left at a Mach numberof 2.0.

Problem 11 - The photo below shows the flow of Helium gas past a sphere at a Machnumber of 1.05. The pressure is one atmosphere (1.01325⇥105N/m2) and the temperatureis 300K. The viscosity of Helium at that temperature is µ

1

= 1.98 ⇥ 10�5 kg/m� sec.Consider a fluid element that passes through the shock on the flow centerline.

Figure 9.18: Slightly supersonic flow past a sphere.

Estimate the acceleration of the fluid element as it traverses the shock wave. Express youranswer in m/sec2.

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