The latest version of this IPython notebook is available at https://github.com/jckantor/Airbag-Design-for-Cargo-Airdrop under the MIT License.

J.C. Kantor ([email protected])

Venting Mass Flowrate

Mass Flow from an Orifice

The mass flow through the airbag vent orifice is given by

$$\dot{m} = C_DA_o \rho_o v_o$$

where the subscript $o$ refers to conditions at the orifice. We begin by constructing an estimate for $\rho_o v_o$ using Bernoulli's equation for compressible flow on a streamline, the pressure-density relationship for the isentropic expansion of an ideal gas, and an assumption regarding pressure at the orifice.

In [5]:
#Initializations
from IPython.core.display import HTML
HTML(open("../styles/custom.css", "r").read())
Out[5]:
In [6]:
%matplotlib inline

import numpy as np
import matplotlib.pyplot as plt

import math
pi = math.pi

from pint import UnitRegistry
ur = UnitRegistry()

Coefficient of Discharge

Based on Browning (1963)

$$C_D = 0.9 - 0.3\frac{P_a}{P}$$

Flow under Unchoked Conditions

Bernoulli's Equation

Bernoulli's equation for the behavior of a compressible fluid along a streamline

$$\frac{v_o^2}{2} + \frac{\gamma}{\gamma-1} \frac{P_o}{\rho_o} = \frac{\gamma}{\gamma-1} \frac{P}{\rho}$$

which, for a streamline starting deep within the airbag, gives for $v_o$

$$v_o = \sqrt{\frac{2\gamma}{\gamma -1}\left(\frac{P}{\rho} - \frac{P_o}{\rho_o}\right)}$$

then for $\dot{m} = C_D A_o\rho_ov_o$

$$\dot{m} = C_D A_o \sqrt{\frac{2\gamma\rho_o^2}{\gamma -1}\left(\frac{P}{\rho} - \frac{P_o}{\rho_o}\right)}$$

Adiabatic Expansion

For the adiabatic expansion of an ideal gas,

$$\frac{\rho_o}{\rho} = \left(\frac{P_o}{P}\right)^{\frac{1}{\gamma}}$$

Substituting into the expression for $\rho_ov_o$,

$$\dot{m} = C_D A_o \sqrt{\frac{2\gamma\rho^2}{\gamma -1}\left(\frac{P_o}{P}\right)^\frac{2}{\gamma}\left(\frac{P}{\rho} - \frac{P_o}{\rho\left(\frac{P_o}{P}\right)^\frac{1}{\gamma}}\right)}$$

After rearrangement

$$\dot{m} = C_D A_o \sqrt{\frac{2\gamma\rho P}{\gamma -1}\left[\left(\frac{P_o}{P}\right)^\frac{2}{\gamma} - \left(\frac{P_o}{P}\right)^{1+\frac{1}{\gamma}}\right]}$$

Evaluating at the Orifice

Assuming the pressure at the orifice is equal to the surrounding ambient pressure, $P_o = P_a$. With that assumption we have

$$\dot{m} = C_D A_o \sqrt{\frac{2\gamma\rho P}{\gamma -1}\left[\left(\frac{P_a}{P}\right)^\frac{2}{\gamma} - \left(\frac{P_a}{P}\right)^{1+\frac{1}{\gamma}}\right]}$$

This expression is found in Handbook of Chemical Hazard Analysis Procedures, Appendix B, Federal Emergency Management Agency, U.S. Dept. of Transportation, and U.S. Environmental Protection Agency, 1989. The relevant section is on page B-4 located on the page 391 of the pdf file.

In [7]:
# ideal gas parameters for air
gamma = 1.4
R = 8.314 * ur.J/(ur.degK*ur.mol)
MW = 28.966 * ur.grams/ur.mol

# ambient conditions
Pa = 1.0 * ur.atm
Ta = ur.Quantity(20,ur.degC).to(ur.degK)

def Qunchoke(P,Pa):
    Pr = Pa/P
    rho = MW*P/(R*Ta)
    b = 2.0*(gamma/(gamma-1))*rho*P
    Q = np.sqrt(b*(Pr**(2.0/gamma)-Pr**((1.0+1.0/gamma))))
    return Q.to(ur.kg/ur.m**2/ur.sec)

P = np.linspace(1.0,3.0,100) * ur.atm
Qu = [Qunchoke(p,Pa) for p in P]

plt.plot(P,[q.magnitude for q in Qu])
plt.title('Unchoked Flow from an Orifice, CD = 1')
plt.ylabel(Qu[0].units)
plt.xlabel(P.units)
Out[7]:
<matplotlib.text.Text at 0x102b341d0>

Choked Flow

When the flow is choked, flow through the orific depends only on the upstream pressure $P$

$$\dot{m} = C_D A \sqrt{\gamma \rho P \left(\frac{2}{\gamma + 1}\right)^\left(\frac{\gamma + 1}{\gamma - 1}\right)}$$

Parameter Description Units
$\dot{m}$ mass flow rate kg/s
$C_d$ discharge coefficient dimensionless
$A$ area of discharge m**2
$\gamma$ ratio of heat capacities $\frac{C_p}{C_v}$ dimensionless
$\rho$ gas density kg/m**3
$P$ pressure Pa
$T$ temperature K
In [10]:
# ideal gas parameters for air
gamma = 1.4
R = 8.314 * ur.J/(ur.degK*ur.mol)
MW = 28.966 * ur.grams/ur.mol

# ambient conditions
Pa = 1.0 * ur.atm
Ta = ur.Quantity(20,ur.degC).to(ur.degK)

def Qchoke(P):
    rho = MW*P/(R*Ta)
    a = gamma*P*rho
    Q = np.sqrt(a*(2.0/(gamma+1))**((gamma+1)/(gamma-1)))
    return Q.to(ur.kg/ur.m**2/ur.sec)

P = np.linspace(1.0,3.0,100) * ur.atm
Qc = [Qchoke(p) for p in P]

plt.plot(P,[q.magnitude for q in Qc],P,[q.magnitude for q in Qc])
plt.title('Choked Flow from an Orifice, CD = 1')
plt.ylabel(Qc[0].units)
plt.xlabel(P.units)
Out[10]:
<matplotlib.text.Text at 0x10877ba50>

Critical Pressure Ratio

The critical pressure ratio is the ratio of upstream to downstream pressure of a compressible gas at which leads to choked flow. Given the ratio of heat capacities

$$\gamma = \frac{C_p}{C_v}$$

the critical ratio is

$$\left(\frac{P_2}{P_1}\right)_{critical} = \left(\frac{\gamma+1}{2}\right)^\left({\frac{\gamma}{\gamma-1}}\right)$$

In [3]:
gamma = dict()

gamma['air (dry)'] = 1.4
gamma['argon'] = 1.67
gamma['butane'] = 1.096
gamma['carbon dioxide'] = 1.30
gamma['chlorine'] = 1.33
gamma['helium'] = 1.660
gamma['hydrogen'] = 1.41
gamma['nitrogen'] = 1.404
gamma['oxygen'] = 1.400
gamma['propane'] = 1.15

print "{0:<15s}   {1:6s}  {2:6s}".format("Species","CR", "1/CR")

for s in sorted(gamma.keys()):
    r = ((gamma[s]+1)/2)**(gamma[s]/(gamma[s]-1))
    print "{0:<15s}   {1:6.4f}  {2:6.4f}".format(s, r, 1.0/r)


def isChoked(Phi, Plo = 101325.0, gamma = 1.4):
    r = ((gamma+1)/2)**(gamma/(gamma-1))
    return Phi > r*Plo
Species           CR      1/CR  
air (dry)         1.8929  0.5283
argon             2.0548  0.4867
butane            1.7079  0.5855
carbon dioxide    1.8324  0.5457
chlorine          1.8506  0.5404
helium            2.0488  0.4881
hydrogen          1.8990  0.5266
nitrogen          1.8953  0.5276
oxygen            1.8929  0.5283
propane           1.7410  0.5744

Procedure to Calculate the Airbag Vent Mass Flowrate

Given the initial airbag pressure $P_i$ and absolute temperature $T_i$, the ambient pressure $P_a$, and gas constant $\gamma$, and orifice vent area $A_o$:

Step 1. Calculate the intial air density as

$$\rho_i = \frac{M P_i}{R T_i}$$

For adiabatic (isentropic) compression, subsequent values of the airbag pressure, temperature and air density satisfy

$$ \frac{P}{P_i} = \left(\frac{\rho}{\rho_i}\right)^\gamma = \left(\frac{T}{T_i}\right)^\frac{\gamma}{\gamma-1}$$

Step 2. Given the ambient pressure $P_a$, compute the discharge coefficient

$$C_D = 0.9 - 0.3\frac{P_a}{P}$$

Step 3. Compute the critical pressure ratio

$$r = \left(\frac{\gamma+1}{2}\right)^\left({\frac{\gamma}{\gamma-1}}\right)$$

Step 4. Compute the mass flow for either unchoked or choked flow conditions

$$ \dot{m} = \begin{cases} C_D A_o \sqrt{\frac{2\gamma\rho P}{\gamma -1}\left[\left(\frac{P_a}{P}\right)^\frac{2}{\gamma} - \left(\frac{P_a}{P}\right)^{1+\frac{1}{\gamma}}\right]} & \mbox{if } P \leq r P_a \\ C_D A_o \sqrt{\gamma \rho P \left(\frac{2}{\gamma + 1}\right)^\left(\frac{\gamma + 1}{\gamma - 1}\right)} & \mbox{otherwise} \end{cases} $$

In [34]:
from pint import UnitRegistry
ur = UnitRegistry()

# ideal gas parameters for air
gamma = 1.4
R = 8.314 * ur.J/(ur.degK*ur.mol)
MW = 28.966 * ur.grams/ur.mol

# ambient conditions
Pa = 1.0 * ur.atm
Ta = ur.Quantity(20,ur.degC).to(ur.degK)

def CD(P, Pa = 1.0 * ur.atm):
    return 0.9 - 0.3*Pa/P

def Qflow(T, P, Pa = 1.0 * ur.atm):
    rho = MW*P/(R*T)
    r = ((gamma+1)/2)**(gamma/(gamma-1))
    if P <= r*Pa:
        Pr = Pa/P
        b = 2.0*(gamma/(gamma-1))*rho*P
        Q = np.sqrt(b*(Pr**(2.0/gamma)-Pr**((1.0+1.0/gamma))))
    else:
        a = gamma*P*rho
        Q = np.sqrt(a*(2.0/(gamma+1))**((gamma+1)/(gamma-1)))
    return Q.to(ur.kg/ur.m**2/ur.sec)

def Mflow(T, P, Pa, A = 1.0*ur.m**2):
    return (CD(P,Pa)*A*Qflow(T,P,Pa)).to_base_units()

P = np.linspace(1.0,2.5,100) * ur.atm
M = [Mflow(Ta,p,Pa,A = 100*ur.cm**2) for p in P]

plt.plot(P,[m.magnitude for m in M])
plt.title('Mass Flow from an 100 cm**2 Orifice')
plt.ylabel(M[0].units)
plt.xlabel(P.units)
Out[34]:
<matplotlib.text.Text at 0x10c610f50>