Compress 1

© Srinivas and Auld 2006-2009 --- www.aerodynamics4students.com

Compressible Flow We know that fluids are classified as Incompressible and Compressible fluids. Incompressible fluids do not undergo significant changes in density as they flow. In general, liquids are incompressible; water being an excellent example. In contrast compressible fluids do undergo density changes. Gases are generally compressible;air being the most common compressible fluid we can find. Compressibility of gases leads to many interesting features such as shocks, which are absent for incompressible fluids. Gasdynamics is the discipline that studies the flow of compressible fluids and forms an important branch of Fluid Mechanics. In this book we give a broad introduction to the basics of compressible fluid flow.

Figure 1: Classification of Fluids Though gases are compressible, the density changes they undergo at low speeds may not be considerable. Take air for instance. Fig.2 shows the density changes plotted as a function of Mach Number. Density change is represented as

 0

where

0 is the air density at zero speed (i.e.,

Zero Mach Number). We observe that for Mach numbers up to 0.3, density changes are within about 5% of 0 . So for all practical purposes one can ignore density changes in this region. But as the Mach Number increases beyond 0.3, changes do become appreciable and at a Mach Number of 1, it is 36.5% . it is interesting to note that at a Mach Number of 2, the density changes are as high as 77%. It follows that air flow can be considered incompressible for Mach Numbers below 0.3. Another important difference between incompressible and compressible flows is due to temperature changes. For an incompressible flow temperature is generally constant. But in a compressible flow one will see a significant change in temperature and an exchange between the modes of energy. Consider a flow at a Mach Number of 2. It has two important modes of energy-Kinetic and Internal. At this Mach Number, these are of magnitudes 2.3 x 105 Joules and 2 x 105 Joules. You will recognise that these are of the same order of magnitude. This is in sharp contrast to incompressible flows where only the kinetic energy is important. In addition when the Mach 2 flow is brought to rest as happens at a stagnation point, all the kinetic energy gets converted into internal energy according to the principle of conservation of energy. Consequently the temperature increases at the stagnation point. When the flow Mach number is 2 at a temperature of 200C, the stagnation temperature is as high as 260 0C as indicated in Fig. 3.

Figure 2: Density change as a function of Mach Number

Figure 3: Stagnation Temperature. A direct consequence of these facts is that while calculating compressible flows energy equation has to be considered (not done for incompressible flows). Further, to handle the exchange in modes of energy one has to understand the thermodynamics of the flow. Accordingly we begin with a review of the concepts in thermodynamics. Thermodynamics is a vast subject. Many great books have been written describing the concepts in it and their application. It is not the intention here to give a detailed treatise than it is to review the basic concepts which hep us understand gasdynamics. Reader is referred to exclusive books on thermodynamics for details.

System, Surroundings and Control Volume Concepts in Thermodynamics are developed with the help of systems and control volumes. We define a System as an entity of fixed mass and concentrate on what happens to this fixed mass. Its boundary is not fixed and is allowed to vary depending upon the changes taking place within it. Consider the system sketched, namely water in a container placed on a heater. We are allowed to chose the system as is convenient to us. We could have system as defined in (a) or (b) or (c) as in Fig. 4. Everything outside of a system becomes the Surroundings. Properties of the system are usually measured by noting the changes it makes in the surrounding. For example, temperature of water in system (a) is measured by a the raise of the mercury column in a thermometer which is not a part of the system. Sometimes the system and the surroundings are together called the Universe.

Figure 4 : Definition of a System Control Volume should now be familiar to you. Most of Integral Approach to Fluid Dynamics exploits control volumes, which can be defined as a window in a flow with a fixed boundary. Mass, momentum and energy can cross its boundary. Density, pressure, temperature, etc become properties of a given system. Note that these are all measurable quantities. In addition, these properties also a characterise a system. To define the state of a system (Fig. 5) uniquely we need to specify two properties say (P,T), (P,  ), (T,s) etc., where p, T,  , s are pressure, temperature, density and specific entropy respectively.

Figure 5: State of a System

Properties can be Extensive or Intensive. Extensive properties depend on the mass of the system. On the other hand, Intensive properties are independent of the mass. Volume V , Energy, E, Entropy, S, Enthalpy, H are Extensive properties. Corresponding intensive properties are Specific Volume, v, Specific Energy, e, Specific Entropy, s, and Specific Enthalpy, h, and are obtained by considering extensive properties per unit mass. In other words,

v=

V , m

e=

E , m

s=

S H , h= m m

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Laws of Thermodynamics Thermodynamics centers around a few laws. We will consider them briefly so that the concepts in gasdynamics can be easily developed. Zeroth Law of Thermodynamics This laws helps define Temperature. It states - "Two systems which are in thermal equilibrium with a third system are themselves in thermal equilibrium."

Figure 6: Zeroth Law of Thermodynamics When in thermal equilibrium, we say that the two systems are at the same temperature. In the Figure 6, system A and B are independently in equilibrium with system C. It follows that A and B are themselves in thermal equilibrium and they are at the same temperature. First Law of Thermodynamics The first law of Thermodynamics is a statement of the principle of conservation of energy. It is simply stated as "Energy of a system and surroundings is conserved." Consider a system S. If one adds dq amount of heat per unit mass into the system and the work done by the system is dw per unit mass we have the change in internal energy of the system, du given by,

du−dq−dw where u is Internal Energy. Bringing in Specific Enthalpy defined as

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h=u p v = u

P 

(3)

the statement for the first law can also be written as

dh=dqv.dp

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While writing Eqn. 4 we have included only one form of energy, namely, internal. Other forms such as the kinetic energy have been ignored. Of course, it is possible to account for all the forms of energy. Second Law of Thermodynamics Second Law of Thermodynamics has been a subject of extensive debate and explanation. Its realm ranges from physics, chemistry to biology, life and even philosophy. There are numerous websites and books which discuss these topics. They form an exciting reading in their own right. Our application however is restricted to gasdynamics. The first law is just a statement that energy is conserved during a process. (The term Process stands for the mechanism which changes the state of a system). It does not "worry" about the direction of the process whereas the second law does. It determines the direction of a process. In addition it involves another property - Entropy.

Figure 7: Second Law of Thermodynamics There are numerous statements of the Second Law. Consider a Reversible Process. Suppose a system at state A undergoes changes, say by an addition of heat Q, and attains state B. While doing so the surroundings change from A' to B'. Let us try to bring the state of the system back to A by removing an amount of heat equal to Q. In doing so if we can bring the surroundings also back to state A' then the process is said to be reversible. This is possible only under ideal conditions. In any real process there is friction which dissipates heat. Consequently it is not possible to bring the system back to state A and at the same time, surroundings back to A'. Assuming the process to be reversible the second law defines entropy such that B

s B −s A=∫A

dq T

where s is Specific Entropy. For small changes, the above equation is written as

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T ds=dq Generalising the Equation 6, we have B

s B −s A≥∫A

(6)

dq T

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where an '=' sign is used for reversible processes and > is used for ireversibe processes. Thus with any natural process, entropy of the system and universe increases. In the event the process is reversible entropy remains constant. Such a process is called an Isentropic process. Perfect Gas Law It is well known that a perfect gas obeys

P= R T

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where R is the Gas constant. For any given gas R is given by

R=

ℜ M

(9)

where is known as the Universal Gas Constant with the same value for all gases. Its numerical value is 8313.5 J/kg-mol K. M is the molecular weight of the gas. The following table gives the value of the gas constant (along with other important constants) for some of the gases. Gas

Molecular Weight

Gas Constant, R

CP

J/kg K

J/kg K

CP / CV



Air

28.97

287

1004

1.4

Ammonia

17.03

488.2

2092

1.3

Argon

39.94

2081

519

1.67

Carbon dioxide

44

188.9

845

1.3

Helium

4.003

2077

5200

1.67

Hydrogen

2.016

4124

14350

1.4

Oxygen

32

259.8

916

1.4

Consequences of First Law for a Perfect Gas For a perfect gas internal energy and enthalpy are functions of temperature alone. Hence, u = u(T)

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h = h(T) Specific Heat of a gas depends upon how heat is added - at constant pressure or at constant volume. We have two specific heats, cp, specific heat at constant pressure and cv, specific heat at constant volume. It can be shown that,

 

c p=

∂h ∂T

Then introducing =c p /c v it follows that

P

,

 

cv=

∂u ∂T

(11) v

R −1

c v=

c p =

(12)

R −1

A Calorically Perfect Gas is one for which cp and cv are constants. Accordingly,

u=u T =c v T ,

h=h T =c p T

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Consequences of Second Law for a Perfect Gas We have shown in Eqn. 4 in First Law of Thermodynamics

dh=dqv dp Now assuming a perfect gas and a reversible process we have

dq=c p dT −RT

dP P

T ds=c p T −RT

dP P

14

Integrating between states 1 and 2, we can show that,

s 2−s1 =c p ln

    T2 P −R ln 2 T1 P1

If we assume that the process is isentropic that is adiabatic (implying no heat transfer) and reversible we can show that the above equation leads to  −1 

  

T2 P = 2 T1 P1

 = 2 1

−1

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A familiar form of equation for an isentropic flow is

P =constant 

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Equations of Motion for a Compressible Flow We now write the equations of motion for a compressible flow. Recall that for an incompressible flow one calculates velocity from continuity and other considerations. Pressure is obtained through the Bernoulli Equation. Such a simple approach is not possible for a compressible flow where temperature is not a constant. One needs to solve the energy equation in addition to the continuity and momentum equations. The latter equations have already been derived for incompressible flows. Of course, one has to account for the changes in density. We focus here on the energy equation and briefly outline the other two. We restrict ourselves to an Integral Approach and write the equations for a control volume.

Equations are derived under the following assumptions. • Flow is one-dimensional. • Viscosity and Heat Transfer are neglected. • Behaviour of flow as a consequence of area changes only considered. • The flow is steady. • We consider a one-dimensional control volume as shown.

Figure 8: Control Volume for a Compressible Flow

Continuity Equation For a steady flow it is obvious that the mass flow rates at entry (1) and exit (2) of the control volume must be equal. Hence, 1 u1 A1=1 u 2 A2 (17) Written in a differential form the above equation becomes,

d  u A=0 dx or

d  du dA   =0  u A

(18)

At this stage, it is usual to consider some applications of the above equation. But we note that this equation has always to be solved with momentum and energy equations while calculating any flow. Accordingly, we skip any worked example at this stage.

Momentum Equation The derivation of Momentum Equation closely follows that for incompressible flows. Basically, it equates net force on the control volume to the rate of change of momentum. Defining Pm as the average pressure between the entry (1) and exit (2), (see Fig. 8), we have for a steady flow,

P 1 A1 1 u 1 A1 u1 P m  A2− A1= P 2 A22 u 2 A2 u 2

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For a steady flow through a duct of constant area the momentum equation assumes a simple form,

P 11 u 21=P 22 u 22

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It is to be noted that the above equations can be applied even for the cases where frictional and viscous effects prevail between (1) and (2). But it is necessary that these effects be absent at (1) and (2).

Energy Equation From the first law of Thermodynamics it follows that, for a unit mass,

qwork done = increase in energy

(21)

where q is the heat added. Work done is given by

P 1 v 1−P 2 v 2

(22)

We consider only internal and kinetic energies. Accordingly, we have,

1 2 1 2 change of energy=e 2 u 2−e 1 u1  2 2

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Substituting Eqns 22 and 23 in 21.we have as the energy equation for a gas flow as,

1 2 1 2 qP 1 v 1−P 2 v 2=e 2 u 2−e 1 u1  2 2 Noting that enthalpy,

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h=eP v , we have

1 2 1 2 q=h 2−h1 u 2− u 1 2 2 Considering an adiabatic process, q = 0, we have,

1 2 1 2 h 2 u2 =h1 u 1 2 2

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This equation demands that the states (1) and (2) be in equilibrium, but does allow non-equilibrium conditions between (1) and (2). If the flow is such that equilibrium exists all along the path from (1) to (2) then we have, at any location along 1-2,

1 h u2 =constant 2

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Differentiating the above equation, we have,

dhu du=0

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For a thermally perfect gas i.e., enthalpy, h depends only on temperature, T (h =cpT) the above equation becomes,

c p dT u du=0

Further, for a calorically perfect gas, i.e., cp is constant, we have,

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1 c p T  u 2=constant 2

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Stagnation Conditions What should be the "constant" on the RHS of Eqn. 29 equal to? We have left it as an open question. It appears that stagnation conditions and sonic conditions are good candidates to provide the required constant. We now discuss the consequences of each of these choices.

Constant from Stagnation Conditions Stagnation conditions are reached when the flow is brought to rest,i.e., u = 0. Temperature, pressure, density, entropy and enthalpy become equal to "Stagnation Temperature" , T0, "Stagnation Pressure", P0 , "Stagnation Density", 0 , "Stagnation Entropy", s0 and "Stagnation Enthalpy, h0. These are also known as "Total" conditions. This is in contrast to incompressible flows where we have only the Stagnation Pressure. Rewrite Eqn. 26 as,

1 h u2 =h0 2

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Recalling that h = cpT for a calorically perfect gas, we have,

1 c p T  u 2=c p T 0 2

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The constant we have arrived at is h0 or cpT0. It is to be noted that there does not have to be a stagnation point in a flow in order to use the above equations. Stagnation or Total conditions are only reference conditions. Further, it is apparent that there can be only one stagnation condition for a given flow. Such a statement is to be qualified and is true only for isentropic flows. In a non-isentropic flow every point can have its own stagnation conditions, meaning if the flow is brought to rest locally at every point, one can have a series of stagnation points. In an adiabatic flow (a flow where heat is not added or taken away) the stagnation or total temperature,T0 does not change. This is true even in presence of a shock as we will see later. But the total pressure p0 can change from point to point. Consider again the control volume shown in Fig. 8. As long as the flow is adiabatic, we have,

T 0 1=T 0 2

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To deduce the conditions for total (stagnation) pressure we consider the Second Law of Thermodynamics,

 s0 2− s0 1≥0

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For a perfect gas the above equation becomes,

R ln Since

T 0 1=T 0 2 , we have,

   

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 P 0 1 ≥0  P 0 2

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 P 0 1 T 0 2 c p ≥0  P 0 2 T 0 1

"Equals" sign applies when the flow is isentropic and "Greater Than " sign applies for any nonisentropic flow. Thus for any natural process involving dissipation total pressure drops. It is preserved for an isentropic flow. A very good example of a non-isentropic flow is that of a shock. Across a shock there is a reduction of total pressure.

Area-Velocity Relation We are all used to the trends of an incompressible flow where velocity changes inversely with area changes - as the area offered to the flow increases, velocity decreases and vice versa. This seems to be the "commonsense". But a compressible flow at supersonic speeds does beat this commonsense. Let us see how. Consider the Continuity Equation, Eqn. 18 , which reads,

d  du dA   =0  u A Consider also the Euler Equation (derived before for incompressible flows)

u du

dP =0 

Rewriting the equation,

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u du=

−dP −dP d  d = =−a 2   d 

Where we have brought in speed of sound, which is given by Mach Number, M the above equation becomes,

a=  dp/d  . Further introducing

d du =−M 2  u

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Upon substituting this in the continuity equation 18 we have

du −dA/ A = u 1−M 2

(38)

Studying Eqn 38 and Fig. 9 one can observe the following,

•

For incompressible flows, M ≈0 . As the area of cross section for a flow decreases velocity increases and vice versa.

•

For subsonic flows, M<1 , the behaviour resembles that for incompressible flows.

•

For supersonic flows,M > 1 , as the area decreases, velocity also decreases, and as the area increases, velocity also increases. We can explain this behaviour like this. In response to an area change all the static properties change. At subsonic speeds changes in density are smaller. The velocity decreases when there is an increased area offered (and vice versa). But in case of a supersonic flow with increasing area density decreases at a faster rate than velocity. In order to preserve continuity velocity now increases (and decreases when area is reduced). vice versa). Not apparent from the above equation is another important property. If the geometry of the flow involves a throat, then mathematically it can shown that if a sonic point occurs in the flow, it occurs only at the Throat. But the converse - The flow is always sonic at throat , is not true.

•

Figure 9: Response of Subsonic and Supersonic Flows to Area Changes

Isentropic Relations For an isentropic flow all the static properties such as P ,  , T and s when expressed as a ratio of their stagnation values become functions of Mach Number, M and  alone. This can be shown as below. Recall the energy equation,

1 2 c p T  u =c p T 0 2 Eliminating T using the equation for speed of sound (still not proved),

a2 a2 1  u 2= 0 −1 2 −1

(39) 2 a = R T , we have

(40)

where a0 is the stagnation speed of sound. Multiplying throughout by

−1/a 2 yields, a 20 T 0 −1 2 = =1 M 2 T 2 a

Thus we have a relationship which connects temperature ratio with Mach Number. Assuming isentropy and using the relation, P=constant   (see Eqn. 16, we can derive expressions for pressure and density as,

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P0 −1 2 = 1 M P 2 0 −1 2 = 1 M  2

 

 

 −1

1 −1

(42) (43)

The relations just developed prove very useful in calculating isentropic flows. Once Mach Number is known it is easy now to calculate pressure, density and temperature as ratios of their stagnation values.

Software and Data There are also data tables and calculation scripts available for isentropic compressible flow properties.

• • • • •

http://www.aoe.vt.edu/aoe3114/calc.html : Aerodynamics Calculator. (web script) http://s6.aeromech.usyd.edu.au/aero/info/compressible.exe : Isentropic Flows (MS windows executable). http://s6.aeromech.usyd.edu.au/aero/compressible/ : Isentropic Flows ( web script ) http://s6.aeromech.usyd.edu.au/aero/info/ctable1.txt : Isentropic Flow (Subsonic) (Tabulated Data) http://s6.aeromech.usyd.edu.au/aero/info/ctable2.txt : Isentropic Flow (Supersonic) (Tabulated Data)

Sonic Point as Reference The preceding relations were arrived at with stagnation point as the reference. As stated before, it is also possible to choose sonic point, the position where M = 1 as the reference. At this point let u = u* and a = a * . Since M = 1 , we have u* = a *. As a consequence the energy equation 40, Isentropic Relations) , becomes, 2

2

*2

*2

u a u a 1 1 * 2  =  = a 2 −1 2 −1 2 −1

(44)

Comparing with the energy equation, Eqn 40 we obtain, *2

*

a 2 T = = 2 a 0 1 T 0 As a result for air with =1.4

(45)

we have

*

*

T =0.833 , T0

a =0.913 a0

P* 2 = P0 1

 −1

* 2 = 0 1

1 −1

   

=0.528

=0.634

(46)

It may be pointed out a sonic point need not be present in the flow for the above equations to be applicable.

Mass Flow Rate Now derive an equation for mass flow rate in terms of Mach Number of flow. Denote the area in an isentropic flow where the Mach Number becomes 1, as A*. We have for mass flow rate, For an isentropic flow *

*

m= A u= A u ˙ It follows by noting that

M =1 ,

*

u *=a *

A  * a* * 0 a * = = *  u 0  u A *

 , 

Substituting for terms such as 2

 

0 

and

 

a

*

etc and simplifying one obtains,

A 1 2 −1 2 = 2 1 M * 2 A M 1



1 −1

This is a very useful relation in Gasdynamics, connecting the local area and local Mach Number. Tables in Appendix also list this function, i.e., A/A* as a function of Mach Number. It helps one to determine changes in Mach Number as area changes.

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Figure 10: Mach Number as a function of area. Figure 10 shows the area function A/A* plotted as a function of Mach Number. We again see what was found under "Area Velocity" rule before. The difference is that now we have a relationship between area and Mach Number. For subsonic flows Mach Number increases as the area decreases and it decreases as the area increases. While with supersonic flows, Mach Number decreases as area decreases and it increases as area increases.

Equations of Motion in absence of Area Changes We can now gather the equations that we have derived for mass, momentum and energy. If we ignore any area change, these become,

1 u1 =2 u 2 2

2

P 11 u 1=P 22 u 2

1 1 h 1 u 21=h2 u 22 2 2

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