l ' I '

I l ~ ft

I l

I I !

11-1

Chapter 11

INTRODUCTION TO COMPRESSIBLE FLOW

In Chapter 2 we briefly discussed the two most important questions we must ask before analyzing a fluid flow: whether or not the flow is viscous, and whether or not the flow is compressible. We subsequently considered incompressible, inviscid flows (Chapter 6) and incompressible, viscous flows (Chapters 8 and 9). We are now ready to study flows that experience compressibility effects. Because this is an introductory text, our focus will be mainly on one-dimensional compressible, inviscid flows, although we will also review some important compressible, viscous flow phenomena. After our consideration of one-dimensional flows, we will introduce some basic concepts of two-dimensional steady compressible flows.

We first need to establish what we mean by a "compressible" flow. This is a flow in which there are significant or noticeable changes in fluid density. Just as in­viscid fluids do not actually exist, so incompressible fluids do not actually exist For example, in this text we have treated water as an incompressible fluid, although in fact the density of seawater increases by 1 % for each mile or so of depth. Hence, whether or not a given flow can be treated as incompressible is a judgment call: Liq­uid flows will almost always be considered incompressible (exceptions include phe­nomena such as the "water hanuner" effect in pipes), but gas flows could easily be ei­ther incompressible or compressible. As we will see (in Example Problem 11.5), our judgment will be guided by the rule of thumb that flows for which the Mach number Mis less than about 0.3 can be considered incompressible.

The consequences of compressibility are not limited simply to density changes. Density changes mean that we can have sigruficant compression or expansion work on a gas, so the thermodynamic state of the fluid will change, meaning that in general all properties-temperature, internal energy, entropy, and so on-can change. In particular, density changes create a mechanism (just as viscosity did) for exchange of energy between "mechanical" energies (kinetic, potential, and "pressure") and the thermal internal energy. For this reason, we begin with a review of the thermodynam­ics needed to study compressible flow.

REVIEW OF THERMODYNAMICS

The pressure, density, and temperature of a substance may be related by an equation of state. Although many substances are complex in behavior, experience shows that most gases of engineering interest, at moderate pressure and temperature, are well represented by the ideal gas equation of state,

p = pRT (11.1)

589

590 CHAPTER 11 I INTRODUCTION TO COMPRESSIBLE FLOW

where Risa unique constant for each gas;1 R is given by

R = R,, Mm

where R" is the universal gas constant, R" = 8314 N · m!(kgmole · K) = 1544 ft . !bf/ (lbmole · 0 R) and Mm is the molecular mass of the gas. Although the ideal gas equa­tion is derived using a model that has the unrealistic assumptions that the gas mole­cules (a) take up zero volume (i.e., they are point masses) and (b) do not interact With one another, many real gases conform to Eq. 11.1, especially ifthe pressure is "low" enough and/or temperature "high" enough (see, e.g., [l-3]). For example, at room temperature, as long as the pressure is less than about 30 atm, Eq. 11. I models the air density to better than 1 percent accuracy; similarly, Eq. 11.1 is accurate for air at 1 attn for temperatures that are greater than about -130°C (140 K).

The ideal gas has other features that are useful. In general, the internal energy of a simple substance may be expressed as a function of any two independent proper­ties, e.g., u = u(v, T), where v = l/p is the specific volume. Then

du=(du) dT+(ilu) dv iJT v dv T

The specific heat at constant volume is defined as cv = ( au/aT)"' so that

du= Cv dT+(ilu) dv iJv T

In particular, for an ideal gas it can be shown (see, e.g., Chapter 11 of [I]) that the in­ternal energy, u, is a function of temperature only, so (au/av),= 0, and

(11.2)

for an ideal gas. This means that internal energy and temperature changes may be re­lated if Cv is known. Furthermore, siuce u = u(T), then from Eq. 11.2, Cv = Cv (T).

The enthalpy of any substance is defined as h = u + pl p. For an ideal gas, p = pRT, and so h = u + RT. Since u = u(T) for an ideal gas, h also must be a function of temperature alone.

We can obtain a relation between h and T by recalling once again that for a simple substance any property can be. expressed as a function of any two other independent properties [I], e.g., h = h(v, T) as we did for u, or h = h(p, T). We choose the latter in order to develop a useful relation,

dh = (~) dT + (ilh) dp iJT P ilp T

Since the specific heat at constant pressure is defined as cP = ( ahlaT)p,

dh = cp dT + (~; )T dp

We have shown that for an ideal gas his a function of T only. Consequently, ( ah/ap), =

Oand

dh = cPdT (11.3)

1For air, R = 287 N · m/(kg · K) = 53.3 ft · lbf/(lbm · 0 R).

11-1 REVIEW Of THERMODYNAMICS 591

Since h is a function of Talone, Eq. 11.3 requires that cP be a function of T only for an ideal gas.

Although specific heats for an ideal gas are functions of temperature, their dif­ference is a constant for each gas. To see this, from

h = u +RT

we can write

dh =du+ RdT

Combining this with Eq. 11.2 and Eq.11.3, we can write

dh = cP dT =du + RdT = cv dT + R dT

Then

Cp - Cv = R (11.4)

This result may seem a bit odd, but it means simply that although the specific heats of an ideal gas may vary with temperature, they do so at the same rate, so their differ­ence is always constant.

The ratio of specific heats is defined as

(11.5)

Using the definition of k, we can solve Eq. 11.4 for either cP or cv in terms of k and R. Thus,

kR c =--p k -1

(ll.6a)

and R

Cv=--k-1

(ll.6b)

Although the specific heats of an ideal gas may vary with temperature, for moderate temperature ranges they vary only slightly, and can be treated as constant, so

(l l.7a)

(ll.7b)

Data for Mm, cP, cv, R, and k for common gases are given in Table A.6 of Appendix A.

We will find the property entropy to be extremely useful in analyzing compressi­ble flows. State diagrams, particularly the temperature-entropy (Ts) diagram, are valuable aids in the physical interpretation of analytical results. Since we shall make extensive use of Ts diagrams in solving compressible flow problems, let us review briefly some useful relationships involving the property entropy [l-3].

Entropy is defined by the equation

ilS= f oQ or Jrev T

dS = (oQ) T rev

(11.8)

where the subscript signifies reversible.

592 CHAPTER 11 I INTRODUCTION TO COMPRESSIBLE FLOW

The inequality of Clausius, deduced from the second law, states that

As a consequence of the second law, we can write

dS?. /JQ T

or TdS?.oQ

For reversible processes, the equality holds, and

(reversible process)

The inequality holds for irreversible processes, and

Tds > /JQ (irreversible process) m

For an adiabatic process, BQ/m = 0. Thus

ds = 0 (reversible adiabatic process)

and

ds > 0 (irreversible adiabatic process)

(l 1.9a)

(1 l.9b)

(l l.9c)

(l l.9d)

(ll.9e)

Thus a process that is reversible and adiabatic is also isentropic; the entropy remains constant during the process. Inequality 11.9e shows that entropy must increase for an adiabatic process that is irreversible.

Equations 11.9 show that any two of the restrictions-reversible, adiabatic, or isentropic-imply the third. For example, a process that is isentropic and reversible must also be adiabatic.

A useful relationship among properties (p, v, T, s, u) can be obtained by consid­ering the first and second laws together. The result is the Gibbs, or T ds, equation

Tds =du+ pdv (11.lOa)

This is a differential relationship among properties, valid for any process between any two equilibrium states. Although it is derived from the first and second laws, it is, in itself, a statement of neither.

An alternative form of Eq. 11.lOa can be obtained by substituting

du = d (h - pv) = dh - pdv - v dp

to obtain

Tds=dh-vdp (11.lOb)

For an ideal gas, entropy change can be evaluated from the T ds equations as

dup dT dv ds = -+-dv = Cv-+R-

T T T v

dhv dT dp ds=---dp=c --R-

T T P T p

For constant specific heats, these equations can be integrated to yield

11-1 REVIEW OF THERMODYNAMICS 593

(11.l la)

(11.llb)

and also

P2 V2 s2 - si = cv In - + cP In -Pi Vi

(11.llc)

(Equation 11.llc can be obtained from either Eq. 11.lla or 11.llb using Eq. 11.4 and the ideal gas equation, Eq. 11.1, written in the form pv = RT, to eliminate T.)

Example Problem 11.1 shows use of the above governing equations (the T ds equations) to evaluate property changes during a process.

For an ideal gas with constant specific heats, we can use Eqs. 11.11 to obtain re­lations valid for an isentropic process. From Eq. 11.1 la

Then, using Eqs. 11.4 and 11.5,

( ~i )(Vvzi )R/cv -- 0 or T k-i "' k-i T k-i ,, 2v2 = 1iVi = v = constant

where states 1 and 2 are arbitrary states of the isentropic process. Using v = lip,

Tvk-i = i"-i =constant p

(l l.12a)

We can apply a similar process to Eqs. 11.11 b and 11.1 lc, respectively, and obtain the following useful relations:

1-k Tp T = constant

pvk = ~ = constant p

Equations 11.12 are for an ideal gas undergoing an isentropic process.

(l l.12b)

(ll.12c)

Qualitative information that is useful in drawing state diagrams also can be ob­tained from the T ds equations. To complete our review of the thermodynamic funda­mentals, we evaluate the slopes of lines of constant pressure and of constant volume on the Ts diagram in Example Problem 11.2.

EXAMPLE 11.1 Property Changes in Compressible Duel flow Air flows through a long duct of constant area at 0.15 kg/s. A short section of the duct is cooled by liquid nitrogen that surrounds the duct. The rate of heat loss in this sec­tion is 15.0 kJ/s from the air. The absolute pressure, temperature, and velocity entering the cooled section are 188 kPa, 440 K, and 210 mis, respectively. At the outlet, the ab­solute pressure and temperature are 213 kPa and 351 K. Compute the duct cross­sectional area and the changes in enthalpy, internal energy, and entropy for this flow.

-·

594 ·· CHAPTER 11 I INTRODUCTION TO COMPRESSIBLE FLOW 11-1 REVIEW OF THERMODYNAMICS 594

EXAMPLE PROBLEM 11.1

GIVEN: Air flows steadily through a short section of constant-area duct that is cooled by liquid nitrogen.

Ti= 440K P1 = 188 kPa(abs) V1 = 2!0m/s

I

Flow-~ I

1

FIND: (a) Duct area. (b) dh. (c) du. (d) ds.

SOLUTION:

I 1 " :;:::·.·····'.

Q<O

The duct area may be found from the continuity equation.

= 0(1)

Governing equation: AJcvpdV+ L,rv·iiA=o Assumptions:(!) Steady flow.

(2) Uniform flow at each section. (3) Ideal gas with constant specific heats.

Then

or

'

~cv

2

since A = A 1 = A2 =constant. Using the ideal gas relation,p = pRT, we find

= ...£L = 1.88x10s ~ x kg . K x _I_ = l.49kg!m' PI R1) m 2 287 N · m 440 K

From continuity,

T2 = 351 K P2 = 213 kPa (abs)

(4.12)

A - m _O.l5kg m3

s _ 479 I0-4 2 A ---- -x---X---. x m P1VI s l.49kg 210 m '-----------"

For an ideal gas, the change in enthalpy is

dh = h, - h1 = JT' cp dT = cp(T2 -1)) T,

(l l.7b)

dh = l .OO --1<.:!_ x (351 - 440) K = - 89.0 kJ/kg M kg·K <-------------'~

Also, the change in internal energy is

du= u2 - u1 = JT' cv dT = cv(T2 -1)) T,

(l l.7a)

du= 0

·717 --1<.:!_ x (351- 440) K = -63.8 kJ/kg du

kg·K <------------'=C

11-1 REVIEW OF THERMODYNAMICS 595

The entropy change may be obtained from Eq. 11.11 b,

As = s2 - s1 = cP ln Tz - R In Pz T1 P1

= 1.00 _I'.!_ x Jn (351)- 0.287 _I'.!_ x lo (2.13 x 105

)

kg · K 440 kg · K 1.88 x 105

t>.s = - 0.262 kJ/(kg · K)<-----------------~6.s

We see that entropy may decrease for a nonadiabatic process in which the gas is cooled.

This problem illustrates the use of the governing equations for computing property changes of an ideal gas during a process.

EXAMPLE 11.2 Conslanl-Property Lines cm a Ts Diagram For an ideal gas, find the equations for lines of (a) constant volume and (b) constant pressure in the Ts plane.

EXAMPLE PROBLEM 11 .2

FIND: Equations for lines of (a) constant volume and (b) constant pressure in the Ts plane for an ideal gas.

SOLUTION: (a) We are interested in the relation between T ands with the volume v held constant. This suggests use of Eq. 11.!la,

=O

+Rloi We relabel this equation so that state 1 is now reference state 0, and state 2 is an arbitrary state,

T s-s0 = cvln- or

To

Hence, we conclude that constant volume lines in the Ts plane are exponential.

(I)

(b) We are interested in the relation between T ands \Vith the pressure p held constant. This suggests use of Eq. 11.l lb, and using a similar approach to case (a), we find

s-so

T = T0 ecp (2)

Hence, we conclude that constant pressure lines in the Ts plane are also exponential. What about the slope of these curves? Because cP > Cv for all gases, we can see that the exponential,

and therefore the slope, of the constant pressure curve, Eq. 2, is smaller than that for the constant volume curve, Eq. 1. This is shown in the sketch below:

,•

596 ,

CHAPTER 11 I INTRODUCTION TO COMPRESSIBLE FLOW

Constant volume

Constant pressure --------

1 ncreasing p

It \

J----_:J------1-----

Decreasing ¥

Entropy

This Example Problem illustrates use of governing equations to explore relations among properties. '

11-2 PROPAGATION OF SOI.IND WAVES

Speed o! Sound

Supersonic and subsonic are familiar terms; they refer to speeds that are. respectively. greater than and less than the speed of sound. The speed of sound (which is a pres­sure wave of infinitesimal strength) is therefore an important characteristic parameter for compressible flow. We have previously (Chapters 2 and 7) introduced the Mach number, M = Vic, the ratio of the local flow speed to the local speed of sound, as an important nondimensional parameter characterizing compressible flows. Before studying compressible flows let us determine the speed of sound in any medium (gas, liquid, or solid). As we do so, it is worth keeping in mind the question "Would a liq­uid, such as water, behave this way, or just a gas, such as air?"

Consider propagation of a sound wave of infinitesimal strength into an undis­turbed medium, as shown in Fig. II.la. We are interested in relating the speed of wave propagation, c, to fluid property changes across the wave. If pressure and den­sity in the undisturbed medium ahead of the wave are denoted by p and p, passage of the wave will cause them to undergo infinitesimal changes to become p + dp and p + dp. Since the wave propagates into a stationary fluid, the velocity ahead of the wave, V"' is zero. The magnitude of the velocity behind the wave, V, + dV" then will be simply dVx; in Fig. I I.la, the direction of the motion behind the wave has been as­sumed to the left. 2

2 The same final result is obtained regardless of the direction initially assumed for motion behind the wave (see Problem 11.19).