Venting Outgoing Mass Determination

Venting, or the expulsion of gas from the airbag, is assumed to be isenthalpic.

The flow is also assumed to be unshocked, coming from a large reservoir and through a small orifice with effective surface area, $T$ .

Conservation of enthalpy leads to velocity, $u$ , at the vent hole. The Bernouilli equation is then written as:

(1)

(airbag) $\frac{γ}{γ - 1} \frac{P}{ρ} = \frac{γ}{γ - 1} \frac{P_{e x t}}{ρ_{v e n t}} + \frac{u^{2}}{2}$ (vent hole)

Applying the adiabatic conditions:(2)

(airbag) $\frac{P}{ρ^{γ}} = \frac{P_{e x t}}{ρ_{v e n t}^{γ}}$ (vent hole)

Therefore, the exit velocity is given by: (3)

u^{2} = \frac{2 γ}{γ - 1} \frac{P}{ρ} (1 - {(\frac{P_{e x t}}{P})}^{\frac{γ - 1}{γ}})

with $ρ = \frac{\sum_{i} m^{(i)}}{V}$ the averaged density of the gas and $γ = \frac{[\sum_{i} m^{(i)} c_{p}^{(i)}] / [\sum_{i} m^{(i)}]}{[\sum_{i} m^{(i)} c_{v}^{(i)}] / [\sum_{i} m^{(i)}]}$ the fraction of massic averages of heat capacities at constant pressure and constant volume.

The mass flow rate is given by:(4)

{\dot{m}}_{o u t} = ρ_{v e n t} A_{v e n t}^{} u = ρ {(\frac{P_{e x t}}{P})}^{1 / γ} A_{v e n t} u

The energy flow rate is given by:(5)

{\dot{E}}_{o u t} = \dot{m} \frac{E}{ρ V} = {(\frac{P_{e x t}}{P})}^{1 / γ} A_{v e n t} u \frac{E}{V}

The total mass flow rate is given by:(6)

d m_{o u t} = ρ {(\frac{P_{e x t}}{P})}^{1 / γ} A_{v e n t} u

Where,

$A_{v e n t}$: Vent hole surface.

The vent hole area or scale factor area,

A_{v e n t}

, can be defined in two ways:

a constant area taking into account a discharge coefficient
a variable area equal to the area of a specified surface multiplied by a discharge coefficient.

Supersonic Outlet Flow

Vent pressure

P_{v e n t}

is equal to external pressure

P_{e x t}

for unshocked flow. For shocked flow,

P_{v e n t}

is equal to critical pressure

P_{c r i t}

and

u

is bounded to critical sound speed:(7)

u^{2} < \frac{2}{γ + 1} c^{2} = \frac{2 * γ}{γ + 1} \frac{P}{ρ}

And,

$P_{c r i t} = P {(\frac{2}{γ + 1})}^{\frac{γ}{γ - 1}}$

$P_{v e n t} = \max (P_{c r i t}, P_{e x t})$

Outgoing Mass per Gas

The mass flow of gas $i$ is $d m^{(i)}_{o u t} = \frac{V^{(i)}}{V} d m_{o u t}$ , where $V^{(i)}$ is the volume occupied by gas $i$ and satisfies:

$V^{(i)} = \frac{n^{(i)}}{n} V$ (from $P V^{(i)} = n^{(i)} R T$ and $P V = [\sum_{i} n^{(i)}] R T$ ).

It comes finally(8)

d m^{(i)}_{o u t} = \frac{n^{(i)}}{\sum_{i} n^{(i)}} d m_{o u t}