Pressure evolution model

In order to devise a pressure evolution model, mass conservation is used. Performing a mass balance on the enclosure volume yields:

˙

me = ˙mgen− ˙mvent (63)

As discussed previously, the vented mass flow rate can be obtained by applying the Bernoulli equation to compute the gas flow velocity. The vented mass flow rate is thus:

˙

mvent= CDρgasvA = CDA(2ρgasP )1/2 (64)

where CD is the discharge coefficient. Using the ideal gas law, the pressure factor can be

replaced by a mass dependent factor:

˙ mvent= CDA  2ρgasmeRT MwV 1/2 (65)

Substituting in the mass balance relation and rearranging the equation yields the fol- lowing result: ˙ me+ CDA  2ρ_gasRT MwV 1/2 m1/2_e − ˙mgen = 0 (66)

Assuming that the combustion rate, temperature and density are constant, this is a non- linear first order ordinary differential equation. This assumption is only necessary in finding analytical solutions to the problem. In the case of numerical solutions, these parameters can be updated as the calculation evolves. Given the finite amount of propellant burning, the

˙

mgen term will be nonzero when 0 ≤ t ≤ tburn. Two cases are thus possible:

˙

me+ CDA

 2ρ_gasRT MwV

1/2

m1/2_e − ˙mgen = 0 when0 ≤ t ≤ tburn (67)

˙ me+ CDA  2ρgasRT MwV 1/2 m1/2_e = 0 when t > tburn (68)

The solution for the first case is difficult, as it requires using the Lambert special func- tion. It is, however, possible to use the differential equation to determine the maximum pressure and the largest rate of pressure change. This analysis will be performed in the next sections.

For the second case, the problem reduces to solving a separable linear first order differ- ential equation. Solving for ˙meand integrating on both sides yield:

me=  k2− 1/2k1(t − tburn) 2 when t > tburn (69) where k2 1 = 2ρgasCD2A2  RT MwV 

and k2 is determined by applying an initial value condi-

tion. For the present case, the initial value condition is that me = mmax when t = tburn,

yielding k2 = m 1/2 max.

By taking a closer look at the expression for k₁2, one can observe that the term in paren- theses can be simplified using the ideal gas law and a known pressure and mass point. Taking the point at which the pressure is at its maximum, one obtains:

k₁2 = 2ρgasCD2A 2Pmax2

m2 max

(70)

where mmaxis the maximum mass of the gases in the enclosure. There is thus a dependency

of k1on the maximum pressure. There is also an expected dependence on the venting area

and sample mass. This last dependence could, however, be more difficult to observe, as the sample masses are modulated with the opening area to keep the maximum pressures in a safe range. The advantage of Eq. (70) is that it is in a form which makes for an easier comparison with empirically calculated values. It also only uses two factors which must be estimated: ρ, the combustion gases density, and CD, the discharge coefficient.

It must be noted that this solution is valid until me(t) reaches a minimum at me(t) = 0.

valid in the interval tburn ≤ t ≤ tburn+ 2k2_k1 . When t ≥ tburn+ 2k2_k1 , the pressure is zero

(which is the atmospheric value here). The evaluation of the maximum mass (and pressure) is carried out next.

4.3.2.1 Maximum pressure

From the previous differential equation, the maximum pressure can be determined by setting ˙me= 0 and solving for me. After some algebra, this yields the following:

me= ˙ m2_genMwV 2C2 DρgasRT A2 (71)

Using the ideal gas law, the maximum pressure is calculated as:

Pmax = ˙ m2 gen 2C2 DρgasA2 (72)

This constitutes the most general form of the solution for the maximum pressure inside a vented enclosure with the assumptions made. Any further simplification would involve finding an appropriate model for the factor ˙mgen. One must, however, be careful in choos-

ing an approximation for ˙mgen, as the model choice will be highly dependent on the case

configuration. The following general result is thus obtained:

Pmax= f _m˙2 gen A2  (73)

A result that is similar to the general form of models found in the literature and dis- cussed previously: Pm = f  r2_S2 A2  (74)

Here, the difference is seen to be strictly in the numerator of the two expressions. These expressions become nearly identical if, for example, it is assumed that the gas generation

rate is simplified as ˙mgen≈ Srρbulk, where ρbulkis the bulk density of the propellant. Such

a mass combustion model would be akin to the case of a propellant in a container with surface area S, burning from the top down. One can observe here that this last statement contains several assumptions: top ignition, constant surface area, laminar vertical propaga- tion through the propellant bed. The problem is that propellant combustion is seldom that ideal. The most general solution therefore remains the best start point for further analysis.

4.3.2.2 Maximum rate of pressure rise

Starting with Eq. (67), the maximum value of the mass rate of change derivative can be obtained by differentiating the equation and setting the second mass derivative equal to zero as follows: ¨ me = 1 2CDA  2ρgasRT MwV 1/2 m−1/2_e m˙e = 0 (75)

It can be seen that the maximum occurs when m(t) = 0, which is at t = 0 in this case. The maximum rate of change of the mass is thus:

˙

mmax = ˙mgen (76)

Transforming the last result to the pressure equivalent yields:

˙ Pmax= ˙ mgenR∆T MwV (77)

Here again, a proper flame propagation model is important in obtaining a precise value for ˙Pmax. The considerations of Chapter 2 are seen to be imperative in making this deter-

4.4

Experimental work