FUNDAMENTALS OF GAS DYNAMICS

Before proceeding further into the question of airfoil characteristics at Mach numbers higher than Mcr, it is necessary to develop some basic relationships relating to compressible subsonic and supersonic flows.

One-DImenslonal Isentroplc Flow

We will begin by considering briefly a reversible, adiabatic flow where the state of the flow is a function only of the position along the flow direction as, for example, a uniform flow through a duct. This simple case illustrates some of the pronounced differences between subsonic and supersonic flows. Ap­plying the momentum theorem to a differential fluid element, Euler’s equation of motion along a streamline was derived in Chapter Two and is again stated here.

VdV + —= 0 P

Also, the continuity equation was derived earlier. For one-dimensional flow through a pipe having a variable cross-sectional area of A,

pAV = constant (5.10)

For an inviscid fluid, the density and pressure are related through the isentropic process

■—r = constant Py

Finally, the properties of the gas are related through the equation of state.

p = pRT (5.12)

Differentiating by parts, Equation 5. JO can be written as

* + £+£-0

p V A

From Chapter Two, the local acoustic velocity is given by

dp

Defining the local Mach number, M, as Via and substituting Equations 5.14 and 5.13 into Equation 5.9 leads to a relationship between и and A.

dV -1 VdA ds 1 — M2 A ds

Since V and A are both positive, we arrive at the surprising result (at least to those who have never seen it) that, for supersonic flow through a duct, an increase in cross-sectional area in the direction of flow will cause the flow to accelerate. Also, Equation 5.15 shows that a Mach number of unity can only occur if dAlds =0 since, for M = 1, dV/ds will be finite only if the cross – sectional area does not change with distance along the duct. This does not mean that M must equal unity when dAlds equals zero but, instead, that dAlds equal to zero is a necessary condition for M = 1.

Consider flow from a reservoir through a converging-diverging nozzle, as pictured in Figure 5.6. Such a nozzle is referred to as a Laval nozzle. If the

reservoir pressure, po, is sufficiently high relative to the exit pressure, pE, the flow will accelerate to Mach 1 at the throat. Beyond the throat, with dA/ds positive, the flow will continue to accelerate, thereby producing a supersonic flow. Downstream of the throat, pressure and density decrease as the velocity increases with the increasing area.

The compressible Bernoulli equation governing one-dimensional isen – tropic flow was derived earlier. In terms of the local acoustic velocity,

V2 a2

— +——- = constant

2 у -1

For this case of flow from a reservoir,

V2. a2 _ ao2 2 у – 1 у — 1

(1+’тім!) (1+izl*p- (.^p-

Dividing this equation through by a2 and using the isentropic relationships among p, p, and T leads to these three quantities as a function of the local Mach number.

These relationships, presented graphically in Figure 5.7, are valid for Mach numbers greater than unity if the flow is shockless. The subject of shock waves will be treated later.

,*2

At the throat the local velocity and the local acoustic velocity are equal. Designating this velocity by a*, Equation 5.16 can be written as

(5.18)

Using Equations 5.11 and 5.14, it follows that

p* Po

2 "p-u

(*y + l)J

= 0.528

Figure 5.7 Isentropic flow. Pressure, density, and temperature as a function of Mach number.

(5.20)

In these equations, the superscript * refers to the throat.

Equation 5.19 shows that the airflow from a reservoir will reach Mach 1 if the reservoir pressure exceeds the exit pressure by a factor of at least 1.894. The mass flow rate, m, through the nozzle will be

m = p*A*a* (5.21)

where A* is the throat area.

Using Equations 5.18 and 5.20, this becomes

m = 0.579p0a0A* (5.22)

Observe that this is the maximum mass flow rate that can be obtained from a given reservoir independent of the exit pressure. For example, consider two tanks connected together through a nozzle having a throat area of 1 m2, as pictured in Figure 5.6. Assume that the air in both tanks is at standard sea level conditions. We will now begin to lower the pressure in one tank, causing air to flow from the other tank into the one with the vacuum. As the pressure in the vacuum tank is gradually reduced, the mass flow through the pipe will increase continuously, assuming the volume in the other tank, or reservoir, is sufficiently large so that its pressure and density does not change

significantly. The pressure drop along the pipe resulting from skin friction will exactly equal the pressure difference between the two tanks. However, when the pressure in the vacuum tank is reduced to 53.5 kN/m2 (from Equation 5.19), a value of M = 1 occurs at the throat. The nozzle, or flow, is then said to be “choked,” since a further reduction in the pressure downstream of the throat will not result in any further increase in the mass flow. From Equation 5.22, this critical mass flow will equal 242.3 kg/s.

If we assume that the flow beyond the throat is still isentropic, Equations 5.17, 5.18, 5.20, and 5.21 can be combined to give

(A2 y- (2Іу+1)(у+ту~1)

UV 2 [l-(plpo)(y~"ly](plpo)2ly

This is known as St. Venant’s equation.

Substituting for the local pressure ratio in terms of Mach number, this can also be written as

Since p, p, and T are related through the adiabatic process and the equation of state, it follows from the foregoing that p, p, and T are all uniquely related to their corresponding reservoir values by the ratio of A to A*. This obviously raises some problems since, in the example of Figure 5.6, the pressure at the exit into the vacuum tank does not necessarily have to match the pressure from Equation 5.23 corresponding to the area of the duct at its connection to the vacuum tank. Some nonisentropic mechanism must exist that will allow the pressure to adjust to exit conditions. This leads us to the concept of a shock wave.

Normal Shock Waves

across which the flow properties p, p, T, and V change discontinuously. We will now examine the equations governing the flow to see if such a standing wave is possible and to determine the relationships between the upstream and downstream fluid properties. To begin, the equations governing the con­servation of mass and momentum must hold.

continuity
PiVi = p2V2 = m	(5.25)
momentum
і-p2 = m(V2- V,)	(5.26)

In addition, the equation of state,(Equation 2.1, must also hold. It is repeated here for convenience. ‘ 1

equation of state p = pRT

A fourth relationship, which has not been used as yet, is the energy equation.

energy

V,2 V->2

CpTt + Ц – = CPT2 + Ц – = CpTo

This equation, derived in Reference 5.6, applies to adiabatic flows where no heat is added to the flow. Cp is the specific heat at constant pressure. The product CPT is the enthalpy of the flow per unit mass. Thus Equation 5.28 states that the sum of the enthalpy and kinetic energy per unit mass of an adiabatic flow remains constant. Cp, R, y, and the specific heat at constant volume, C„, are all interrelated.

R = Cp — Cv y = ~C

V-‘t)

Ry

(y — l)

If Equation 5.29 is substituted into the energy equation (Equation 5.28), it is interesting to note that one obtains the compressible Bernoulli equation (Equation 5.16). Thus Equation 5.16 and the energy equation are equivalent for isentropic flow. However, across the shock wave the flow is not a reversible, adiabatic process, so the changes in state are not related by Equation 5.11.

In order to see how p, p, T, and M change across a normal shock wave, we begin by substituting Equation 5.25 into Equation 5.26 so that,

Pi + PiVi2 = p2 + pV22

Since a2 = yplp, it follows that

Pi + ypiM2 = p2 + УР2М22

P2^i + yM2 p2 1 + yM-2

Manipulating Equation 5.28 in a similar manner, it follows that

/02V 1 + [(7 ~ 1)/2]M^ _ P2P1 T2 W 1 + [(-y – l)/2]M2i pip2 Ti

Thus,

P2 p2l + l(y-W]Mi2 Pi p, l + [(y-l)/2]M25

Next, Equation 5.25 is written as

£lMi£i = t л Р2М2С12 u

Substituting Equations 5.30, 5.31, and 5.32 into the preceding equations leads to an implicit relationship for M2 as a function of Mt.

МЛ/І + К7-І)/2JM,1 M2Vl + [(y-m]M2i

——– гтда———————— ГТтМ?——– – f(M)

Obviously, one solution of the above is M2 = M,, in which case p2 = pi, p2 = pu and T2 = T, so that there is no discontinuity in the flow and the solution is trivial. The other solution is apparent from Figure 5.9, where f(M)

0 0.4 0.8 1.2 1.6 2.0

Mach number, M

Figure 5.9 Function to determine conditions across a normal shock wave.

is presented as a. function of M. It is seen that the same value for /(M) is obtained from two different values of M, one greater and one less than unity. For example, if Mi is equal to 2.0, Equation 5.33 would be satisfied by an M2 value of approximately 0.57. One might, of course, say that a value of Mi equal to 0.57 with M2 equal to 2.0 would also satisfy Equation 5.33 which, indeed, would be the case. However, it can be argued on the basis of the second law of thermodynamics that the flow ahead of the shock wave must be supersonic (see Ref. 5.6, p. 234). An entropy loss, in violation of the second law, will occur if Mi is less than unity. Therefore, in Figure 5.9, M, is greater than unity and M2 is the value of M less than unity for which

, , ; /(М2) = f(Mi)

l’*

Notice from Equation 5.33 that /(M) approaches a value of (у ~ 1 )mly, or approximately 0.319, as M,-»°°. Thus, behind a normal shock wave, the Mach number has a lower limit of approximately 0.38.

Since there is an entropy gain across the normal shock wave, a loss occurs in the total, or reservoir, pressure as the flow passes through the wave. With some algebraic manipulation of the energy equation and application of the isentropic relationships before and after the wave (but not across it), the following equation can be obtained.

Pb = P2 Г1 + [(у – l)/2]МЛ»-‘1 Po, Pi ІЛ + [(y – 1)/2]M,2J

Poj and po, are the reservoir pressures behind and ahead of the wave, respectively.

Equations 5.30 to 5.34 are unwieldy to use because of their implicit nature. However, after some algebraic manipulation, they can all be reduced to explicit functions of Mx. These can be found in Reference 5.7 and are repeated here. Following the lead of Reference 5.7, the second form of each equation is for у = 7/5.

Figure 5.10 presents these relationships graphically and can be used for approximate calculations.

Let us now return to the problem of flow in the duct illustrated by Figure 5.8. Suppose the pressure in the vacuum tank is lowered to a value of 80 kN/m2. Furthermore, let us assume that the area of the duct entering the second tank is large so that, as the air enters the tank, its velocity, and hence dynamic pressure, is low. Thus, 80 kN/m2 would represent approximately the reservoir pressure downstream of a normal shock in the duct. The upstream reservoir pressure is equal to the standard sea level value of 101.3 kN/m2. Thus,

— = 0.790

P o,

From Figure 5.10, the Mach number, Mb just upstream of the normal shock wave equals 1.85. In addition,

— = 3.85 Pi

^ = 2.43 Pi

M2 = 0.605

Figure 5.11 A detached shock wave ahead of a blunt-nosed shape traveling at supersonic speed.

Thus, from Equation 5.24

A.,.50

Therefore, we predict for the pressure in the vacuum tank of 80 kN/m2 that a normal shock wave must be positioned in the duct at a location where the duct area is one and a half times greater than the throat area.

This supersonic flow through a duct that must ultimately come to rest in the vacuum is directly comparable to a blunt-nosed body, or airfoil, traveling at supersonic speeds. Figure 5.11 depicts a supersonic airfoil with a rounded leading edge traveling at a Mach number of 1.85. Since the flow must come to rest at the stagnation point on the nose, it obviously must be subsonic for some extent ahead of the nose. The result is a shock wave that is normal to the flow in the vicinity of the nose. As in the case of the duct flow, immediately behind this wave, the flow is subsonic with a Mach number of 0.605. The shock wave, positioned away from the nose some small distance, is referred to as a “detached” shock wave, since it is detached from the surface.