Thermodynamical Equations

The basic energy equation of the airbag can be written as:(1)

d E_{a i r b a g} = - P d V + d H_{i n} = d H_{o u t}

Where,

$E$: Internal energy
$P$: Pressure
$V$: Airbag volume
$H_{i n}$: Incoming enthalpy
$H_{o u t}$: Outgoing enthalpy

When the adiabatic condition is applied and assuming a perfect gas:(2)

P = \frac{(γ - 1) E}{V}

Where, $γ$ is the gas constant. For air, $γ$ = 1.4.

The two equations above allow the current airbag volume to be determined. The energy and pressure can then be found. To know the current airbag volume, derive energy and thus pressure.

Considering a gas such that the constant pressure and the constant volume heat capacities per mass unit (respectively, $c_{p}$ and $c_{v}$ ) vary in temperature $T$ .

The following temperature dependency of the constant pressure heat capacity is assumed:(3)

c_{p} = a + b T + c T^{2}

Where, $a$ , $b$ and $c$ are the constants depending to characteristics of the gas.

The

c_{p}

and

c_{v}

satisfy the Mayer relation:(4)

c_{p} {(T)-c}_{v} (T) = \frac{R}{M}

With

R

is the universal gas constant depending to the unit system (

R = 8.3144 J m o l^{- 1} K^{- 1}

):(5)

\begin{array}{l} c_{v} (T) = \frac{d e}{d T} (T) \\ h (t) = e (t) + \frac{P}{ρ} \end{array}

Where,

$e$: Specific energy
$h$: Specific enthalpy per mass unit of the gas at temperature $T$

You can then obtain:(6)

e (T) = e_{c o l d} + \int_{T_{c o l d}}^{T} c_{v} (T) d T

and(7)

h (T) = e_{c o l d} + \int_{T_{c o l d}}^{T} c_{v} (T) d T + \frac{R}{M} T

Where, the lower index $c o l d$ refers to the reference temperature $T_{c o l d}$ .

Now, assuming an ideal mixture of gas:(8)

P V = n R T

with

n

the total number of moles:(9)

n = \sum_{i} n^{(i)} = \sum_{i} \frac{m^{(i)}}{M^{(i)}}

Where,

$M^{(i)}$: Mass of gas $i$
$M^{(i)}$: Molar weight of gas $i$

It follows:(10)

P = \frac{n R T}{V}

With $n = \sum_{i} \frac{m^{(i)}}{M^{(i)}}$ .