Flame speed correlation¶
Equation 8.33 In Turns $$S_L = S_{L,ref}\left(\frac{T_u}{T_{u,ref}}\right)^{\gamma}\left(\frac{P}{P_{ref}}\right)^{\beta}(1-2.1Y_{dil}).$$
$$T_{ref}=298\, K,$$ $$P_{ref}=1\, atm,$$
$$S_{L_ref} = B_M + B_2(\Phi-\Phi_M)^2,$$
$$\gamma = 2.18 - 0.8(\Phi - 1),$$
$$\beta = -0.16 + 0.22(\Phi-1),$$
where $B_M$, $B_2$, and $\Phi_M$ are constants for fuels: methanol, propane, isooctane, and gasoline. $Y_{dil}$ is the mass fraction of diluent in the air-fuel mixture (to account for, e.g., recirculated products.
In [6]:
def SL_corr(fuel, ϕ, P, Tu, Ydil) :
Pref = 1 # atm
Turef = 298.15 # K
ϕM = {"methanol":1.11, "propane":1.08, "isooctane":1.13, "gasoline":1.13}
BM = {"methanol":36.92,"propane":34.22,"isooctane":26.32,"gasoline":27.58} # cm/s
B2 = {"methanol":-140.51,"propane":-138.64, "isooctane":-84.72, "gasoline":-78.34} # cm/s
gamma = 2.18 - 0.8*(ϕ-1)
beta = -0.16 + 0.22*(ϕ-1)
SLref = BM[fuel] + B2[fuel]*(ϕ-ϕM[fuel])**2.0
return SLref*(Tu/Turef)**gamma * (P/Pref)**beta * (1-2.1*Ydil)
In [7]:
import numpy as np
import matplotlib.pyplot as plt
%matplotlib inline
fuel = "gasoline"
P = 18.38 # atm
Tu = 685.0 # K
Ydil = 0.0
ϕ = np.linspace(0.5,2.0,100)
SL = SL_corr(fuel, ϕ, P, Tu, Ydil)
igood = np.where(SL>0.0)
#print("S_L = %g cm/s" %(SL))
plt.rc('font', size=14)
plt.plot(phi[igood],SL[igood])
plt.xlabel(r'$\phi$')
plt.ylabel(r'S$_L$ (cm/s)');
In [ ]: