Tank Test

By using data from a tank test output, it is possible to obtain the temperature and the mass flow of the gas supplied, which can be used as input to Radioss.

With a tank test it is possible to measure the pressure at the injection point or in the middle of the tank, the two values are equal so the pressure variation is well known. Also, the amount of gas supplied and the characteristics of the gas in the test are known.

For temperature, it is more difficult to get accurate test data because the temperature gauges are often not accurate enough. Therefore, the temperature from the tank test may be wrong.

The following cases take into account that you may or may not know the injected temperature and the temperature in the tank.

In case the temperature at the injector and in the tank is not known:

Initial and injected gas composition
Knowing the molecular weight ( $M W_{i}$ ) of each elements and the molar fraction ( $X_{i}$ ), it is possible to define the molar weight of the gas ( $M W$ ):(1)
$M W = \sum X_{i} \cdot M W_{i}$

The average heat capacity, per mass unit of a mixture of gases is given by the Amagat-Leduc equation:(2)
$C_{p} (T) = \frac{\sum m_{i} \cdot C_{p a}}{\sum m_{i}} + \frac{\sum m_{i} \cdot C_{p b}}{\sum m_{i}} T + \frac{\sum m_{i} \cdot C_{p c}}{\sum m_{i}} T^{2}$

Using the previous equation, the heat capacity coefficients ( $C_{p} (T)$ ) of the initial and injected mixture may be defined.

Knowing the characteristics of the injected gas, the initial gas, and the mixture, it is possible to find the mass flow and the temperature for the inflator. The following basic equations are used to carry out the analysis.

The perfect gas equation of state is:(3)
$P V = n R T, n = \frac{m}{M W}$
with $R = 8.314 \frac{J}{m o l e \cdot K}$
And the adiabatic equation:(4)
$H = c o n s t .$
with $H$ being the total enthalpy of the system (inflator + tank).

From the conservation of the energy, the basic energy equation of the tank test can be written as:(5)

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

Here $d H_{o u t}$ =0, since the tank test is adiabatic. The constant volume of the tank test means $d V$ =0.

Therefore, Equation 5 is summarized as:(6)

U_{f} - U_{0} = d H_{i n}

$\Leftrightarrow \int_{0}^{T_{m i x}} m_{(i n + a i r)} \cdot C_{V (i n + a i r)} d T - \int_{0}^{T_{a i r}} m_{(a i r)} \cdot C_{V (a i r)} d T = \int_{0}^{T_{i n}} m_{(i n)} \cdot C_{P (i n)} d T$

$\begin{array}{l} \Leftrightarrow m_{(m i x)} T_{m i x} (C_{p a (m i x)} + C_{p b (m i x)} \frac{T_{m i x}}{2} + C_{p c (m i x)} \frac{T_{m i x}^{2}}{3} - \frac{R}{M W_{m i x}}) - m_{(a i r)} T_{0} \\ (C_{p a (a i r)} + C_{p b (a i r)} \frac{T_{0}}{2} + C_{p c (a i r)} \frac{T_{0}^{2}}{3} - \frac{R}{M W_{a i r}}) = m_{(i n)} T_{i n} (C_{p a (i n)} + C_{p b (i n)} \frac{T_{i n}}{2} + C_{p c (i n)} \frac{T_{i n}^{2}}{3}) \end{array}$

Inflator Temperature

In Equation 6, the unknown variable is only $T_{i n i}$ .

The other variables are known or could be determined using Equation 2 and Equation 3:

$m_{(i n)} m_{(a i r)} = M W_{a i r} \frac{P_{0} V}{R T_{0}}$ and $n_{(m i x)} = \frac{m_{(i n)}}{M W_{i n}} + \frac{m_{(a i r)}}{M W_{a i r}}$
for i = a to c: $C_{p i}_{(i n)}, C_{p i}_{(a i r)}, C_{p i}_{(m i x)}$ are calculated with Equation 2, and
$T_{m i x} = \frac{P_{t a n k} V_{t a n k}}{n_{(m i x)} R}$

Therefore, Equation 6 finds the temperature at the injector $T_{i n}$ of the injected gas by iterating on $T_{i n}$ . First, the temperature is guessed and six iterations are sufficient to converge to the solution.

Mass Flow

Knowing the evolution of the pressure versus time at the top of the tank test, it is possible to determine the mass flow rate with:(7)

\dot{m} = \frac{\partial m}{\partial P} \frac{\partial P}{\partial t} ≅ \frac{Δ M}{Δ P} \frac{\partial P}{\partial t}

With,

$Δ P$: Total pressure variation during the experiment
$Δ M$: Total injected mass

Equation 7 may be written if the variation of mass versus the variation of the pressure is a function strictly growing, which is the case.

Inflator Gas Velocity

Since the pressure is quickly uniform, the following equation may be written:

P_{i n} = P (t)

, knowing

T_{i n}

the density may be expressed as a function of

T_{i n}

and

P (t)

:(8)

ρ_{i n} = \frac{P (t)}{\frac{T_{i n} R}{M W_{i n}}}

Besides,

V_{i n} (t) = \frac{\frac{d m (t)}{d t}}{S ρ_{i n} (t)}

, the sound speed into the gas is equal to:(9)

c_{i n}^{2} = γ \frac{R}{M W_{i n}} T_{i n}

and if

V_{i n} > c_{i n}

, then

V_{i n} = c_{i n}

the velocity is major by the sound speed velocity.

If the injector temperature is known:
$\frac{d m}{d t} = \frac{v}{γ (γ - 1) C_{V} T_{i n}} \frac{d P}{d t}$
If the temperature in the tank is known:
$\frac{d m}{d t} = \frac{V}{γ (γ - 1) C_{V} T^{2}} (T \frac{d P}{d t} - P \frac{d T}{d t})$ or $m = \frac{P V}{(γ - 1) C_{V} T}$

$T_{i n} = \frac{T^{2}}{γ} (\frac{\frac{d P}{d t}}{T \frac{d P}{d t} - P \frac{d T}{d t}}) or T_{i n} = \frac{T}{γ} + \frac{m}{γ} \frac{\frac{d T}{d t}}{\frac{d m}{d t}}$
If the temperature in the tank is constant:
$T_{i n} = \frac{T}{γ}$