5.3. Hugoniot Relations and the Hydrodynamic Theory of Detonations

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5.3. Hugoniot Relations and the Hydrodynamic Theory of Detonations

If one is to examine the approach to calculating the steady, planar, 1D gaseous detonation velocity, one should consider a system configuration similar to that given in Chapter 4. For the configuration given there, it is important to understand the various velocity symbols used. Here, the appropriate velocities are defined in Figure 5.1. With these velocities, the integrated conservation and static equations are written as

Figure 5.1 Velocities used in analysis of the detonation problem.

$ρ_{1} u_{1} = ρ_{2} u_{2}$

(5.1)

$P_{1} + ρ_{1} u_{1}^{2} = P_{2} + ρ_{2} u_{2}^{2}$

(5.2)

$c_{p} T_{1} + \frac{1}{2} u_{1}^{2} + q = c_{p} T_{2} + \frac{1}{2} u_{2}^{2}$

(5.3)

$P_{1} = ρ_{1} R T_{1} (connects known variables)$

(5.4)

$P_{2} = ρ_{2} R T_{2}$

In this type of representation, all combustion events are collapsed into a discontinuity (the wave). Thus, the unknowns are u₁, u₂, ρ₂, T₂, and P₂. Since there are four equations and five unknowns, an eigenvalue cannot be obtained. Experimentally, it is found that the detonation velocity is uniquely constant for a given mixture. To determine all unknowns, one must know something about the internal structure (rate of reaction), or one must obtain another necessary condition, which is the case for the detonation velocity determination.

5.3.1. Characterization of the Hugoniot Curve and the Uniqueness of the Chapman Jouguet Point

The method of obtaining a unique solution, or the elimination of many of the possible solutions, will be deferred at this point. To establish the argument for the nonexistence of various solutions, it is best to pinpoint or define the various velocities that arise in the problem and then develop certain relationships that will prove convenient.

First, one calculates expressions for the velocities u₁ and u₂. From Eqn (5.1)

$u_{2} = (ρ_{1} / ρ_{2}) u_{1}$

Substituting in Eqn (5.2), one has

$ρ_{1} u_{1}^{2} - (ρ_{1}^{2} / ρ_{2}) u_{1}^{2} = (P_{2} - P_{1})$

Dividing by

$ρ_{1}^{2},$

one obtains

$\begin{matrix} u_{1}^{2} (\frac{1}{ρ_{1}} - \frac{1}{ρ_{2}}) = \frac{P_{2} - P_{1}}{ρ_{1}^{2}}, \\ u_{1}^{2} = \frac{1}{ρ_{1}^{2}} [(P_{2} - P_{1}) / (\frac{1}{ρ_{1}} - \frac{1}{ρ_{2}})] \end{matrix}$

(5.5)

Note that Eqn (5.5) is the equation of the Rayleigh line, which can also be derived without involving any equation of state. Since (ρ₁u₁)² is always a positive value, it follows that if ρ₂ > ρ₁, P₂ > P₁ and vice versa. Since the sound speed c can be written as

$c_{1}^{2} = γ R T_{1} = γ P_{1} (1 / ρ_{1})$

where γ is the ratio of specific heats,

$γ M_{1}^{2} = (\frac{P_{2}}{P_{1}} - 1) / [1 - \frac{(1 / ρ_{2})}{(1 / ρ_{1})}]$

(5.6)

Substituting Eqn (5.5) into Eqn (5.1), one obtains

$u_{2}^{2} = \frac{1}{ρ_{2}^{2}} [(P_{2} - P_{1}) / (\frac{1}{ρ_{1}} - \frac{1}{ρ_{2}})]$

(5.7)

and

$γ M_{2}^{2} = [(1 - \frac{P_{1}}{P_{2}}) / (\frac{(1 / ρ_{1})}{(1 / ρ_{2})} - 1)]$

(5.8)

A relationship called the Hugoniot equation, which is used throughout these developments, is developed as follows. Recall that

$(c_{p} / R) = γ / (γ - 1), c_{p} = R [γ / (γ - 1)]$

Substituting in Eqn (5.3), one obtains

$R [γ / (γ - 1)] T_{1} + \frac{1}{2} u_{1}^{2} + q = R [γ / (γ - 1)] T_{2} + \frac{1}{2} u_{2}^{2}$

Implicit in writing the equation in this form is the assumption that c_p and γ are constant throughout. Since RT = P/ρ, then

$\frac{γ}{γ - 1} (\frac{P_{2}}{ρ_{2}} - \frac{P_{1}}{ρ_{1}}) - \frac{1}{2} (u_{1}^{2} - u_{2}^{2}) = q$

(5.9)

One then obtains from Eqns (5.5) and (5.6)

$\begin{matrix} u_{1}^{2} - u_{2}^{2} = (\frac{1}{ρ_{1}^{2}} - \frac{1}{ρ_{2}^{2}}) [\frac{P_{2} - P_{1}}{(1 / ρ_{1}) - (1 / ρ_{2})}] = \frac{ρ_{2}^{2} - ρ_{1}^{2}}{ρ_{1}^{2} ρ_{2}^{2}} [\frac{P_{2} - P_{1}}{(1 / ρ_{1}) - (1 / ρ_{2})}] \\ = (\frac{1}{ρ_{1}^{2}} - \frac{1}{ρ_{2}^{2}}) [\frac{P_{2} - P_{1}}{(1 / ρ_{1}) - (1 / ρ_{2})}] = (\frac{1}{ρ_{1}} + \frac{1}{ρ_{2}}) (P_{2} - P_{1}) \end{matrix}$

(5.10)

Substituting Eqn (5.10) into Eqn (5.9), one obtains the Hugoniot equation

$\frac{γ}{γ - 1} (\frac{P_{2}}{ρ_{2}} - \frac{P_{1}}{ρ_{1}}) - \frac{1}{2} (P_{2} - P_{1}) (\frac{1}{ρ_{1}} + \frac{1}{ρ_{2}}) = q$

(5.11)

This relationship, of course, will hold for a shock wave when q is set equal to zero. The Hugoniot equation is also written in terms of the enthalpy and internal energy changes. The expression with internal energies is particularly useful in the actual solution for the detonation velocity u₁. If a total enthalpy (sensible plus chemical) in unit mass terms is defined such that

$h \equiv c_{p} T + h^{o}$

where h^o is the heat of formation in the standard state and per unit mass, then a simplification of the Hugoniot equation evolves. Since, by this definition,

$q = h_{1}^{o} - h_{2}^{o}$

Equation (5.3) becomes

$\frac{1}{2} u_{1}^{2} + c_{p} T_{1} + h_{1}^{o} = c_{p} T_{2} + h_{2}^{o} + \frac{1}{2} u_{2}^{2} or \frac{1}{2} (u_{1}^{2} - u_{2}^{2}) = h_{2} - h_{1}$

Or further, from Eqn (5.10), the Hugoniot equation can also be written as

$\frac{1}{2} (P_{2} - P_{1}) [(1 / ρ_{1}) + (1 / ρ_{2})] = h_{2} - h_{1}$

(5.12)

To develop the Hugoniot equation in terms of the internal energy, one proceeds by first writing

$h = e + R T = e + (P / ρ)$

where e is the total internal energy (sensible plus chemical) per unit mass. Substituting for h in Eqn (5.12), one obtains

$\begin{matrix} \frac{1}{2} [(\frac{P_{2}}{ρ_{1}} + \frac{P_{2}}{ρ_{2}}) - (\frac{P_{1}}{ρ_{1}} + \frac{P_{1}}{ρ_{2}})] = e_{2} + \frac{P_{2}}{ρ_{2}} - e_{1} - \frac{P_{1}}{ρ_{1}} \\ \frac{1}{2} (\frac{P_{2}}{ρ_{1}} - \frac{P_{2}}{ρ_{2}} + \frac{P_{1}}{ρ_{1}} - \frac{P_{1}}{ρ_{2}}) = e_{2} - e_{1} \end{matrix}$

Another form of the Hugoniot equation is obtained by factoring:

$\frac{1}{2} (P_{2} + P_{1}) [(1 / ρ_{1}) - (1 / ρ_{2})] = e_{2} - e_{1}$

(5.13)

If the energy equation (Eqn (5.3)) is written in the form

$h_{1} + \frac{1}{2} u_{1}^{2} = h_{2} + \frac{1}{2} u_{2}^{2}$

the Hugoniot relations (Eqns (5.12) and (5.13)) are derivable without the perfect gas and constant c_p and γ assumptions, and thus are valid for shocks and detonations in gases, metals, etc.

There is also interest in the velocity of the burned gases with respect to the tube, since as the wave proceeds into the medium at rest, it is not known whether the burned gases proceed in the direction of the wave (i.e., follow the wave) or proceed away from the wave. From Figure 5.1, it is apparent that this velocity, which is also called the particle velocity (Δu), is

$Δ u = u_{1} - u_{2}$

and from Eqns (5.5) and (5.7)

$Δ u = {[(1 / ρ_{1}) - (1 / ρ_{2})] (P_{2} - P_{1})}^{1 / 2}$

(5.14)

Before proceeding further, it must be established which values of the velocity of the burned gases are valid. Thus, it is now best to make a plot of the Hugoniot equation for an arbitrary q. The Hugoniot equation is essentially a plot of all the possible values of (1/ρ₂, P₂) for a given value of (1/ρ₁, P₁) and a given q. This point (1/ρ₁, P₁), called A, is also plotted on the graph.

The regions of possible solutions are constructed by drawing the tangents to the curve that go through A [(1/ρ_l, P₁)]. Since the form of the Hugoniot equation obtained is a hyperbola, there are two tangents to the curve through A, as shown in Figure 5.2. The tangents and horizontal and vertical lines through the initial condition A divide the Hugoniot curve into five regions, as specified by Roman numerals (I–V). The horizontal and vertical through A are, of course, the lines of constant P and l/ρ. A pressure difference for a final condition can be determined very readily from the Hugoniot relation (Eqn (5.11)) by considering the conditions along the vertical through A, i.e., the condition of constant (1/ρ₁) or constant volume heating:

$\begin{matrix} \frac{γ}{γ - 1} (\frac{P_{2} - P_{1}}{ρ}) - (\frac{P_{2} - P_{1}}{ρ}) = q \\ [(\frac{γ}{γ - 1}) - 1] (\frac{P_{2} - P_{1}}{ρ}) = q, (P_{2} - P_{1}) = ρ (γ - 1) q \end{matrix}$

(5.15)

From Eqn (5.15), it can be considered that the pressure differential generated is proportional to the heat release q. If there is no heat release (q = 0), P₁ = P₂ and the Hugoniot curve would pass through the initial point A. As implied before, the shock Hugoniot curve must pass through A. For different values of q, one obtains a whole family of Hugoniot curves.

Figure 5.2 Hugoniot relationship with energy release divided into five regions (I–V) and the shock Hugoniot.

The Hugoniot diagram also defines an angle α_J such that

$\tan α_{J} = \frac{(P_{2} - P_{1})}{(1 / ρ_{1}) - (1 / ρ_{2})}$

From Eqn (5.5), then

$u_{1} = (1 / ρ_{1}) {(\tan α_{J})}^{1 / 2}$

Any other value of α_J obtained, say by taking points along the curve from J to K and drawing a line through A, is positive, and the velocity u₁ is real and possible. However, from K to X, one does not obtain a real velocity due to negative α_J. Thus, region V does not represent real solutions and can be eliminated. A result in this region would require a compression wave to move in the negative direction—an impossible condition.

Regions III and IV give expansion waves, which are the low-velocity waves already classified as deflagrations. That these waves are subsonic can be established from the relative order of magnitude of the numerator and denominator of Eqn (5.8), as has already been shown in Chapter 4.

Regions I and II give compression waves, high velocities, and are the regions of detonation (also as established in Chapter 4).

One can verify that regions I and II give compression waves and regions III and IV give expansion waves by examining the ratio of Δu to u_l obtained by dividing Eqn (5.14) by the square root of Eqn (5.5):

$\frac{Δ u}{u_{1}} = \frac{(1 / ρ_{1}) - (1 / ρ_{2})}{(1 / ρ_{1})} = 1 - \frac{(1 / ρ_{2})}{(1 / ρ_{1})}$

(5.16)

In regions I and II, the detonation branch of the Hugoniot curve, 1/ρ₂ < 1/ρ₁ and the right-hand side of Eqn (5.16) is positive. Thus, in detonations, the hot gases follow the wave. In regions III and IV, the deflagration branch of the Hugoniot curve, 1/ρ₂ > 1/ρ₁ and the right-hand side of Eqn (5.16) is negative. Thus, in deflagrations the hot gases move away from the wave.

At this point in the development, the deflagration and detonation branches of the Hugoniot curve have been characterized, and region V has been eliminated. There are some specific characteristics of the tangency point J that were initially postulated by Chapman [7] in 1889. Chapman established that the slope of the adiabat is exactly the slope through J, that is,

${[\frac{(P_{2} - P_{1})}{(1 / ρ_{1}) - (1 / ρ_{2})}]}_{J} = - {{[\frac{\partial P_{2}}{\partial (1 / ρ_{2})}]}_{s}}_{J}$

(5.17)

The proof of Eqn (5.17) is a very interesting one and is verified in the following development. From thermodynamics, one can write for every point along the Hugoniot curve,

$T_{2} d s_{2} = d e_{2} + P_{2} d (1 / ρ_{2})$

(5.18)

where s is the entropy per unit mass. Differentiating Eqn (5.13), the Hugoniot equation in terms of e is

$d e_{2} = - \frac{1}{2} (P_{1} + P_{2}) d (1 / ρ_{2}) + \frac{1}{2} [(1 / ρ_{1}) - (1 / ρ_{2})] d P_{2}$

since the initial conditions e₁, P₁, and (1/ρ₁) are constant. Substituting this result in Eqn (5.18), one obtains

$\begin{matrix} T_{2} d s_{2} = - \frac{1}{2} (P_{1} + P_{2}) d (1 / ρ_{2}) + \frac{1}{2} [(1 / ρ_{1}) - (1 / ρ_{2})] d P_{2} + P_{2} d (1 / ρ_{2}) \\ = - \frac{1}{2} (P_{1} - P_{2}) d (1 / ρ_{2}) + \frac{1}{2} [(1 / ρ_{1}) - (1 / ρ_{2})] d P_{2} \end{matrix}$

(5.19)

It follows from Eqn (5.19) that, along the Hugoniot curve

$T_{2} {[\frac{d s_{2}}{d (1 / ρ_{2})}]}_{H} = \frac{1}{2} (\frac{1}{ρ_{1}} - \frac{1}{ρ_{2}}) {- \frac{P_{1} - P_{2}}{(1 / ρ_{1}) - (1 / ρ_{2})} + {[\frac{d P_{2}}{d (1 / ρ_{2})}]}_{H}}$

(5.20)

The subscript H is used to emphasize that derivatives are along the Hugoniot curve. Now, somewhere along the Hugoniot curve, the adiabatic curve passing through the same point has the same slope as the Hugoniot curve. There, ds₂ must be zero and Eqn (5.20) becomes

${{[\frac{d P_{2}}{d (1 / ρ_{2})}]}_{H}}_{s} = \frac{(P_{1} - P_{2})}{(1 / ρ_{1}) - (1 / ρ_{2})}$

(5.21)

But notice that the right-hand side of Eqn (5.21) is the value of the tangent that also goes through point A; therefore, the tangency point along the Hugoniot curve is J. Since the order of differentiation on the left-hand side of Eqn (5.21) can be reversed, it is obvious that Eqn (5.17) has been developed.

Equation (5.17) is useful in developing another important condition at point J. The velocity of sound in the burned gas can be written as

$c_{2}^{2} = {(\frac{\partial P_{2}}{\partial ρ_{2}})}_{s} = - \frac{1}{ρ_{2}^{2}} {[\frac{\partial P_{2}}{\partial (1 / ρ_{2})}]}_{s}$

(5.22)

Cross-multiplying and comparing with Eqn (5.17), one obtains

$ρ_{2}^{2} c_{2}^{2} = - {[\frac{\partial P_{2}}{\partial (1 / ρ_{2})}]}_{s} = {[\frac{(P_{2} - P_{1})}{(1 / ρ_{1}) - (1 / ρ_{2})}]}_{J}$

${[c_{2}^{2}]}_{J} = \frac{1}{ρ_{2}^{2}} {[\frac{(P_{2} - P_{1})}{(1 / ρ_{1}) - (1 / ρ_{2})}]}_{J} = {[u_{2}^{2}]}_{J}$

Therefore

${[u_{2}]}_{J} = {[c_{2}]}_{J} or {[M_{2}]}_{J} = 1$

Thus, the important result is obtained that at J the velocity of the burned gases (u₂) is equal to the speed of sound in the burned gases. Furthermore, an identical analysis would show, as well, that

${[M_{2}]}_{Y} = 1$

Recall that the velocity of the burned gas with respect to the tube (Δu) is written as

$Δ u = u_{1} - u_{2}$

or at J

$u_{1} = Δ u + u_{2}, u_{1} = Δ u + c_{2}$

(5.23)

Thus, at J, the velocity of the unburned gases moving into the wave—i.e., the detonation velocity—equals the velocity of sound in the gases behind the detonation wave plus the mass velocity of these gases (the velocity of the burned gases with respect to the tube). It will be shown presently that this solution at J is the only solution that can exist along the detonation branch of the Hugoniot curve for actual experimental conditions.

Although the complete solution of u₁ at J will not be attempted at this point, it can be shown readily that the detonation velocity has a simple expression now that u₂ and c₂ have been shown to be equal. The conservation of mass equation is rewritten to show that

$ρ_{1} u_{1} = ρ_{2} u_{2} = ρ_{2} c_{2} or u_{1} = \frac{ρ_{2}}{ρ_{1}} c_{2} = \frac{(1 / ρ_{1})}{(1 / ρ_{2})} c_{2}$

(5.24)

Then from Eqn (5.22) for c₂, it follows that

$u_{1} = \frac{(1 / ρ_{1})}{(1 / ρ_{2})} (1 / ρ_{2}) {- {[\frac{\partial P_{2}}{\partial (1 / ρ_{2})}]}_{s}}^{1 / 2} = (\frac{1}{ρ_{1}}) {- {[\frac{\partial P_{2}}{\partial (1 / ρ_{2})}]}_{s}}^{1 / 2}$

(5.25)

With the condition that u₂ = c₂ at J, it is possible to characterize the different branches of the Hugoniot curve in the following manner:

Region I: Strong detonation since P₂ > P_J (supersonic flow to subsonic)

Region II: Weak detonation since P₂ < P_J (supersonic flow to supersonic)

Region III: Weak deflagration since P₂ > P_Y (subsonic flow to subsonic)

Region IV: Strong deflagration since P₂ < P_Y (1/ρ₂ > 1/ρ_Y) (subsonic flow to supersonic)

At points above J, P₂ > P_J; thus, u₂ < u_2,J. Since the temperature increases somewhat at higher pressures, c₂ > c_2,J [c = (γRT)^1/2]. More exactly, it is shown in the next section that above J, c₂ > u₂. Thus, M₂ above J must be less than 1. Similar arguments for points between J and K reveal M₂ > M_2,J and hence supersonic flow behind the wave. At points past Y, 1/ρ₂ > 1/ρ₁, or the velocities are greater than u_2,Y. Also past Y, the sound speed is about equal to the value at Y. Thus, past Y, M₂ > 1. A similar argument shows that M₂ < 1 between X and Y. Then, past Y, the density decreases; therefore, the heat addition prescribes that there be supersonic outflow. But, in a constant area duct, it is not possible to have heat addition and proceed past the sonic condition. Thus, region IV is not a physically possible region of solutions and is ruled out.

Region III (weak deflagration) encompasses the laminar flame solutions that were treated in Chapter 4.

There is no condition by which one can rule out strong detonation; however, Chapman stated that in this region only velocities corresponding to J are valid. Jouguet [8] gave the following analysis.

If the final values of P and 1/ρ correspond to a point on the Hugoniot curve higher than the point J, it can be shown (next section) that the velocity of sound in the burned gases is greater than the velocity of the detonation wave relative to the burned gases because, above J, c₂ is shown to be greater than u₂. (It can also be shown that the entropy is a minimum at J and that M_J is greater than values above J.) Consequently, if a rarefaction wave due to any reason whatsoever starts behind the wave, it will catch up with the detonation front; u₁ − Δu = u₂. The rarefaction will then reduce the pressure and cause the final value of P₂ and 1/ρ₂ to drop and move down the curve toward J. Thus, points above J are not stable. Heat losses, turbulence, friction, etc., can start the rarefaction. At the point J, the velocity of the detonation wave is equal to the velocity of sound in the burned gases plus the mass velocity of these gases, so that the rarefaction will not overtake it; thus, J corresponds to a “self-sustained” detonation. The point and conditions at J are referred to as the C−J results.

Thus, it appears that solutions in region I are possible, but only in the transient state, since external effects quickly break down this state. Some investigators have claimed to have measured strong detonations in the transient state. There also exist standing detonations that are strong. Overdriven detonations have been generated by pistons, and some investigators have observed oblique detonations that are overdriven.

The argument used to exclude points on the Hugoniot curve below J is based on the structure of the wave. If a solution in region II were possible, there would be an equation that would give results in both region I and region II. The broken line in Figure 5.2 representing this equation would go through A and some point, say Z, in region I and another point, say W, in region II. Both Z and W must correspond to the same detonation velocity. The same line would cross the shock Hugoniot curve at point X′. It follows in Section E, the structure of the detonation is a shock wave followed by chemical reaction. Thus, to detail the structure of the detonation wave on Figure 5.2, the pressure could rise from A to X′, and then be reduced along the broken line to Z as there is chemical energy release. To proceed to the weak detonation solution at W, there would have to be further energy release. However, all the energy is expended for the initial mixture at point Z. Hence, it is physically impossible to reach the solution given by W as long as the structure requires a shock wave followed by chemical energy release. Therefore, the condition of tangency at J provides the additional condition necessary to specify the detonation velocity uniquely. Thus, the physically possible solutions represented by the Hugoniot curve are only those shown in Figure 5.3.

Figure 5.3 The only physical possible steady-state results along the Hugoniot—the point J and region III. The broken line represents transient conditions.

5.3.2. Determination of the Speed of Sound in the Burned Gases for Conditions above the C−J Point

5.3.2.1. Behavior of the Entropy along the Hugoniot Curve

Equation (5.18) was written as

$T_{2} {[\frac{d s_{2}}{d (1 / ρ_{2})}]}_{H} = \frac{1}{2} (\frac{1}{ρ_{1}} - \frac{1}{ρ_{2}}) {{[\frac{d P_{2}}{d (1 / ρ_{2})}]}_{H} - \frac{P_{1} - P_{2}}{(1 / ρ_{1}) - (1 / ρ_{2})}}$

with the further consequence that [ds₂/d(1/ρ₂)] = 0 at points Y and J (the latter is the C−J point for the detonation condition).

Differentiating again and taking into account the fact that

$[d s_{2} / d (1 / ρ_{2})] = 0$

at point J, one obtains

${[\frac{d^{2} s}{d {(1 / ρ_{2})}^{2}}]}_{H at J or Y} = \frac{(1 / ρ_{1}) - (1 / ρ_{2})}{2 T_{2}} [\frac{d^{2} P_{2}}{d {(1 / ρ_{2})}^{2}}]$

(5.26)

Now [d²P₂/d(1/ρ₂)²] > 0 everywhere, since the Hugoniot curve has its concavity directed toward the positive ordinates (see formal proof later).

Therefore, at point J, [(1/ρ_l) − (1/ρ₂)] > 0, hence the entropy is minimum at J. At point Y, [(1/ρ_l) − (1/ρ₂)] < 0, hence s₂ goes through a maximum.

When q = 0, the Hugoniot curve represents an adiabatic shock. Point 1(P₁, ρ_l) is then on the curve and Y and J are 1. Then [(1/ρ_l) − (1/ρ₂)] = 0, and the classical result of the shock theory is found; that is, the shock Hugoniot curve osculates the adiabat at the point representing the conditions before the shock.

Along the detonation branch of the Hugoniot curve, the variation of the entropy is as given in Figure 5.4. For the adiabatic shock, the entropy variation is as shown in Figure 5.5.

5.3.2.2. The concavity of the Hugoniot Curve

Solving for P₂ in the Hugoniot relation, one obtains

$P_{2} = \frac{a + b (1 / ρ_{2})}{c + d (1 / ρ_{2})}$

(5.27)

where

$a = q + \frac{γ + 1}{2 (γ - 1)} \frac{P_{1}}{ρ_{1}}, b = - \frac{1}{2} P_{1}, c = - \frac{1}{2} ρ_{1}^{- 1}, d = \frac{γ + 1}{2 (γ - 1)}$

Figure 5.4 Variation of entropy along the Hugoniot.

Figure 5.5 Entropy variation for an adiabatic shock.

From this equation for the pressure, it is obvious that the Hugoniot curve is a hyperbola. Its asymptotes are the lines

$\frac{1}{ρ_{2}} = (\frac{γ - 1}{γ + 1}) (\frac{1}{ρ_{1}}) > 0, P_{2} = - \frac{γ - 1}{γ + 1} P_{1} < 0$

The slope is

${[\frac{d P_{2}}{d (1 / ρ_{2})}]}_{H} = \frac{b c - a d}{{[c + d (1 / ρ_{2})]}^{2}}$

(5.28)

where

$b c - a d = - [\frac{γ + 1}{2 (γ - 1)} q + \frac{P_{1}}{ρ_{1}} \frac{γ}{{(γ - 1)}^{2}}] < 0$

(5.29)

since q > 0, P₁ > 0, and ρ_l > 0. A complete plot of the Hugoniot curves with its asymptotes would be as shown in Figure 5.6. From Figure 5.6, it is seen, as in earlier figures, that the part of the hyperbola representing the strong detonation branch has its concavity directed upward. It is also possible to determine directly the sign of

Figure 5.6 Asymptotes to the Hugoniot curves.

${[\frac{d^{2} P_{2}}{d {(1 / ρ_{2})}^{2}}]}_{H}$

By differentiating Eqn (5.28), one obtains

$\frac{d^{2} P_{2}}{d {(1 / ρ_{2})}^{2}} = \frac{2 d (a d - b c)}{{[c + d (1 / ρ_{2})]}^{3}}$

Now, d > 0, ad − bc > 0 (Eqn (5.29)), and

$c + d (\frac{1}{ρ_{2}}) = \frac{1}{2} [\frac{γ + 1}{γ - 1} (\frac{1}{ρ_{2}}) - (\frac{1}{ρ_{1}})] > 0$

The solutions lie on the part of the hyperbola situated on the right-hand side of the asymptote.

$(1 / ρ_{2}) = [(γ - 1) / (γ + 1)] (1 / ρ_{1})$

Hence

$[d^{2} P_{2} / d {(1 / ρ_{2})}^{2}] > 0$

5.3.2.3. The Burned Gas Speed

Here,

$d s = {(\frac{\partial s}{\partial (1 / ρ)})}_{P} d (\frac{1}{ρ}) + {(\frac{\partial s}{\partial P})}_{1 / ρ} d P$

(5.30)

Since ds = 0 for the adiabat, Eqn (5.30) becomes

$0 = {[\frac{\partial s}{\partial (1 / ρ)}]}_{P} + {(\frac{\partial s}{\partial P})}_{1 / ρ} {[\frac{\partial P}{\partial (1 / ρ)}]}_{s}$

(5.31)

Differentiating Eqn (5.30) along the Hugoniot curve, one obtains.

${[\frac{d s}{d (1 / ρ)}]}_{H} = {[\frac{\partial s}{\partial (1 / ρ)}]}_{P} + {(\frac{\partial s}{\partial P})}_{1 / ρ} {[\frac{d P}{d (1 / ρ)}]}_{H}$

(5.32)

Subtracting and transposing Eqns (5.31) and (5.32), one has

${[\frac{d P}{d (1 / ρ)}]}_{H} - {[\frac{\partial P}{\partial (1 / ρ)}]}_{s} = \frac{{[d s / d (1 / ρ)]}_{H}}{{(\partial s / \partial P)}_{1 / ρ}}$

(5.33)

A thermodynamic expression for the enthalpy is

$d h = T d s + d P / ρ$

(5.34)

With the conditions of constant c_p and an ideal gas, the expressions

$d h = c_{p} d T, T = P / R ρ, c_{p} = [γ / (γ - 1)] R$

are developed and substituted in

$d h = (\frac{\partial h}{\partial P}) d P + \frac{\partial h}{\partial (1 / ρ)} d (\frac{1}{ρ})$

to obtain

$d h = (\frac{γ}{γ - 1}) R [(\frac{1}{R ρ}) d P + \frac{P}{R} d (\frac{1}{ρ})]$

(5.35)

Combining Eqns (5.34) and (5.35) gives

$\begin{matrix} d s = \frac{1}{T} [(\frac{γ}{γ - 1}) \frac{1}{ρ} d P - \frac{d P}{ρ} + (\frac{γ}{γ - 1}) P d (\frac{1}{ρ})] \\ = \frac{1}{T} [(\frac{1}{γ - 1}) \frac{d P}{ρ} + (\frac{γ}{γ - 1}) ρ R T d (\frac{1}{ρ})] \\ = \frac{d P}{(γ - 1) ρ T} + (\frac{γ}{γ - 1}) ρ R d (\frac{1}{ρ}) \end{matrix}$

Therefore

${(\partial s / \partial P)}_{1 / ρ} = 1 / (γ - 1) ρ T$

(5.36)

Then substituting in the values of Eqn (5.36) into Eqn (5.33), one obtains

${[\frac{\partial P}{\partial (1 / ρ)}]}_{H} - {[\frac{\partial P}{\partial (1 / ρ)}]}_{s} = (γ - 1) ρ T {[\frac{\partial s}{\partial (1 / ρ)}]}_{H}$

(5.37)

Equation (5.20) may be written as

${[\frac{\partial P_{2}}{\partial (1 / ρ_{2})}]}_{H} - \frac{P_{1} - P_{2}}{(1 / ρ_{1}) - (1 / ρ_{2})} = \frac{2 T_{2}}{(1 / ρ_{1}) - (1 / ρ_{2})} {[\frac{\partial s_{2}}{\partial (1 / ρ_{2})}]}_{H}$

(5.38)

Combining Eqns (5.37) and (5.38) gives

$\begin{matrix} {[- \frac{\partial P_{2}}{\partial (1 / ρ_{2})}]}_{s} - \frac{P_{2} - P_{1}}{(1 / ρ_{1}) - (1 / ρ_{2})} \\ = {[\frac{\partial s_{2}}{\partial (1 / ρ_{2})}]}_{H} [- \frac{2 T_{2}}{(1 / ρ_{1}) - (1 / ρ_{2})} + (γ - 1) ρ_{2} T_{2}] \end{matrix}$

$ρ_{2}^{2} c_{2}^{2} - ρ_{2}^{2} u_{2}^{2} = \frac{P_{2}}{R} [γ - \frac{1 + (ρ_{1} / ρ_{2})}{1 - (ρ_{1} / ρ_{2})}] {[\frac{\partial s_{2}}{\partial (1 / ρ_{2})}]}_{H}$

(5.39)

Since the asymptote is given by

$1 / ρ_{2} = [(γ - 1) / (γ + 1)] (1 / ρ_{1})$

values of (1/ρ₂) on the right-hand side of the asymptote must be

$1 / ρ_{2} > [(γ - 1) / (γ + 1)] (1 / ρ_{1})$

which leads to

$[γ - \frac{1 + (ρ_{1} / ρ_{2})}{1 - (ρ_{1} / ρ_{2})}] < 0$

Since also

$[\partial s / \partial (1 / ρ_{2})] < 0,$

the right-hand side of Eqn (5.39) is the product of two negative numbers, or a positive number. If the right-hand side of Eqn (5.39) is positive, c₂ must be greater than u₂; that is,

$c_{2} > u_{2}$

5.3.3. Calculation of the Detonation Velocity

With the background provided, it is now possible to calculate the detonation velocity for an explosive mixture at given initial conditions. Equation (5.25), developed earlier,

$u_{1} = (\frac{1}{ρ_{1}}) {[- \frac{d P_{2}}{d (1 / ρ_{2})}]}_{s}^{1 / 2}$

(5.25)

shows the strong importance of density of the initial gas mixture, which is reflected more properly in the molecular weight of the products, as will be derived later.

For ideal gases, the adiabatic expansion law is

$P υ^{γ} = constant = P_{2} {(1 / ρ_{2})}^{γ_{2}}$

Differentiating this expression, one obtains

${(\frac{1}{ρ_{2}})}^{γ_{2}} d P_{2} + P_{2} {(1 / ρ_{2})}^{γ_{2} - 1} γ_{2} d (1 / ρ_{2}) = 0$

which gives

$- {[\frac{d P_{2}}{d (1 / ρ_{2})}]}_{s} = \frac{P_{2}}{(1 / ρ_{2})} γ_{2}$

(5.40)

Substituting Eqn (5.40) into Eqn (5.25), one obtains

$u_{1} = \frac{(1 / ρ_{1})}{(1 / ρ_{2})} {[γ_{2} P_{2} (1 / ρ_{2})]}^{1 / 2} = \frac{(1 / ρ_{1})}{(1 / ρ_{2})} {(γ_{2} R T_{2})}^{1 / 2}$

If one defines

$μ = (1 / ρ_{1}) / (1 / ρ_{2})$

then

$u_{1} = μ {(γ_{2} R T_{2})}^{1 / 2}$

(5.41)

Rearranging Eqn (5.5), it is possible to write

$P_{2} - P_{1} = u_{1}^{2} \frac{(1 / ρ_{1}) - (1 / ρ_{2})}{{(1 / ρ_{1})}^{2}}$

Substituting for

$u_{1}^{2}$

from above, one obtains

$(P_{2} - P_{1}) \frac{(1 / ρ_{2})}{γ_{2} P_{2}} = (\frac{1}{ρ_{1}} - \frac{1}{ρ_{2}})$

(5.42)

Now Eqn (5.12) was

$e_{2} - e_{1} = \frac{1}{2} (P_{2} + P_{1}) [(1 / ρ_{1}) - (1 / ρ_{2})]$

Substituting Eqn (5.43) into Eqn (5.13), one has

$e_{2} - e_{1} = \frac{1}{2} (P_{2} + P_{1}) \frac{(P_{2} - P_{1}) (1 / ρ_{2})}{γ_{2} P_{2}}$

$e_{2} - e_{1} = \frac{1}{2} \frac{(P_{2}^{2} - P_{1}^{2}) (1 / ρ_{2})}{γ_{2} P_{2}}$

Since

$P_{2}^{2} > P_{1}^{2}$

$e_{2} - e_{1} = \frac{1}{2} \frac{P_{2}^{2} (1 / ρ_{2})}{γ_{2} P_{2}} = \frac{1}{2} \frac{P_{2} (1 / ρ_{2})}{γ_{2}}$

As established, all expressions are in mass units; therefore, the gas constant R is not the universal gas constant. Indeed, it should now be written R₂ to indicate this condition. Thus

$e_{2} - e_{1} = \frac{1}{2} \frac{P_{2} (1 / ρ_{2})}{γ_{2}} = \frac{1}{2} \frac{R_{2} T_{2}}{γ_{2}}$

(5.43)

As presented earlier, e is the sum of the sensible internal energy plus the internal energy of formation. Equation (5.43) is the one to be solved in order to obtain T₂, and hence u₁. However, it is more convenient to solve this expression on a molar basis, because the available thermodynamic data and stoichiometric equations are in molar terms.

Equation (5.43) may be written in terms of the universal gas constant R′ as

$e_{2} - e_{1} = \frac{1}{2} (R^{'} / M W_{2}) (T_{2} / γ_{2})$

(5.44)

where MW₂ is the average molecular weight of the products. The gas constant R used throughout this chapter must be the engineering gas constant since all the equations developed are in terms of unit mass, not moles. R′ specifies the universal gas constant. If one multiplies through Eqn (5.44) with MW₁, the average molecular weight of the reactants, one obtains

$(M W_{1} / M W_{2}) e_{2} (M W_{2}) - (M W_{1}) e_{1} = \frac{1}{2} (R^{'} T_{2} / γ_{2}) (M W_{1} / M W_{2})$

$n_{2} E_{2} - E_{1} = \frac{1}{2} (n_{2} R^{'} T_{2} / γ_{2})$

(5.45)

where the E's are the total internal energies per mole of all reactants or products and n₂ is (MW₁/MW₂), which is the number of moles of the product per mole of reactant. Usually, one has more than one product and one reactant; thus, the E's are the molar sums.

Presently to solve for T₂, first assume a T₂ and estimate ρ₂ and MW₂, which do not vary substantially for burned gas mixtures. For these approximations, it is possible to determine 1/ρ₂ and P₂.

If Eqn (5.43) is multiplied by (P₁ + P₂)

$(P_{1} + P_{1}) {(1 / ρ_{1}) - (1 / ρ_{2})} = (P_{2}^{2} - P_{1}^{2}) (1 / ρ_{2}) / γ_{2} P_{2}$

Again

$P_{2}^{2} ≫ P_{1}^{2}$

, so that

$\begin{matrix} \frac{P_{1}}{ρ_{1}} - \frac{P_{1}}{ρ_{2}} + \frac{P_{2}}{ρ_{1}} - \frac{P_{2}}{ρ_{2}} = \frac{P_{2} (1 / ρ_{2})}{γ_{2}} \\ \frac{P_{2}}{ρ_{1}} - \frac{P_{1}}{ρ_{2}} = \frac{P_{2} (1 / ρ_{2})}{γ_{2}} + \frac{P_{2}}{ρ_{2}} - \frac{P_{1}}{ρ_{1}} \end{matrix}$

$\begin{matrix} \frac{P_{2} ρ_{2}}{ρ_{2} ρ_{1}} - \frac{P_{1} ρ_{1}}{ρ_{1} ρ_{2}} = \frac{R_{2} T_{2}}{γ_{2}} + R_{2} T_{2} - R_{1} T_{1} \\ R_{2} T_{2} [\frac{(1 / ρ_{1})}{(1 / ρ_{2})}] - R_{1} T_{1} [\frac{(1 / ρ_{2})}{(1 / ρ_{1})}] = R_{2} T_{2} - R_{1} T_{1} \end{matrix}$

In terms of μ

$R_{2} T_{2} μ - R_{1} T_{1} (1 / μ) = [(1 / γ_{2}) + 1] R_{2} T_{2} - R_{1} T_{1}$

which gives

$μ^{2} - [(1 / γ_{2}) + 1 - (R_{1} T_{1} / R_{2} T_{2})] μ - (R_{1} T_{1} / R_{2} T_{2}) = 0$

(5.46)

This quadratic equation can be solved for μ; thus, for the initial condition (1/ρ_l), (1/ρ₂) is known. P₂ is then determined from the ratio of the state equations at 2 and 1:

$P_{2} = μ (R_{2} T_{2} / R_{1} T_{1}) P_{1} = μ (\frac{M W_{1} T_{2}}{M W_{2} T_{1}}) P_{1}$

(5.47)

For the assumed T₂, P₂ is known. Then it is possible to determine the equilibrium composition of the burned gas mixture in the same fashion as described in Chapter 1. For this mixture and temperature, both sides of Eqn (5.43) or (5.44) are deduced. If the correct T₂ was assumed, both sides of the equation will be equal. If not, reiterate the procedure until T₂ is found. The correct γ₂ and MW₂ will be determined readily. For the correct values, u₁ is determined from Eqn (5.41) written as

$u_{1} = μ {(\frac{γ_{2} R^{'} T_{2}}{M W_{2}})}^{1 / 2}$

(5.48)

The physical significance of Eqn (5.48) is that the detonation velocity is proportional to (T₂/MW₂)^1/2; thus, it will not maximize at the stoichiometric mixture, but at one that is more likely to be fuel rich.

The solution is simpler if the assumption P₂ > P₁ is made. Then, from Eqn (5.42)

$(\frac{1}{ρ_{1}} - \frac{1}{ρ_{2}}) = \frac{1}{γ_{2}} (\frac{1}{ρ_{2}}), (\frac{1}{ρ_{2}}) = \frac{γ_{2}}{1 + γ_{2}} (\frac{1}{ρ_{1}}), μ = \frac{γ_{2} + 1}{γ_{2}}$

Since one can usually make an excellent guess of γ₂, one obtains μ immediately and, thus, P₂. Furthermore, μ does not vary significantly for most detonation mixtures, particularly when the oxidizer is air. It is a number close to 1.8, which means, as Eqn (5.24) specifies, that the detonation velocity is 1.8 times the sound speed in the burned gases.

Gordon and McBride [12] present a more detailed computational scheme and the associated computational program.

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Table of Contents for 5.3. Hugoniot Relations and the Hydrodynamic Theory of Detonations

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5.3. Hugoniot Relations and the Hydrodynamic Theory of Detonations

5.3.1. Characterization of the Hugoniot Curve and the Uniqueness of the Chapman Jouguet Point

5.3.2. Determination of the Speed of Sound in the Burned Gases for Conditions above the C−J Point

5.3.2.1. Behavior of the Entropy along the Hugoniot Curve

5.3.2.2. The concavity of the Hugoniot Curve

5.3.2.3. The Burned Gas Speed

5.3.3. Calculation of the Detonation Velocity

Table of Contents for
5.3. Hugoniot Relations and the Hydrodynamic Theory of Detonations