Question 6,7: Rotational Relaxation in a Simple Diatomic Gas (Using Pair S...
Rotational Relaxation in a Simple Diatomic Gas (Using Pair Selection)
In this example, we will simplify the preceding generalized equilibrium distribution function for the case of a simple diatomic gas where pair selection is employed and compare the result to that derived in Section 6.4.2. Also, in this example, we assume that p_{rot} is a constant.
To start, using Eq. 6.64 for rotation,
f(\epsilon _{i};T{i})=\frac{1}{\Gamma (\varsigma _{i}/{2})kT_{i}}\left(\frac{\epsilon _{i}}{kT_{i}} \right)^{{\zeta _{i}}/{2}-1} e^{{-\epsilon_{i}}/{kT_{i}}} =A\epsilon^{{{\zeta _{i}}/{2}-1}}_{i} e^{{-\epsilon _{i}}/{kT_{i}}} (6.64)
we have the following for a specific value of \epsilon_{coll}:
f(\varepsilon _{coll}-\varepsilon _{rot};T_{coll})f(\varepsilon _{rot};T_{coll})P_{rot}
=P_{rot}\frac{1}{\Gamma (\frac{\zeta _{tr}}{2})kT_{coll}} \left(\frac{\varepsilon _{coll}-\varepsilon _{rot}}{kT_{coll}} \right)^{\frac{\zeta _{tr}}{2}-1} e^{-\frac{\varepsilon _{coll}-\varepsilon _{rot}}{kT_{coll}} }\frac{1}{{\Gamma (\frac{\zeta _{tr}}{2})kT_{coll}}} \left(\frac{\varepsilon _{rot}}{kT_{coll}} \right)^{{\frac{\zeta _{tr}}{2}-1}} e^{-\frac{\varepsilon _{rot}}{kT_{coll}} } (6.105)
=P_{rot}\frac{ e^{-\frac{\varepsilon _{rot}}{kT_{coll}} } }{\Gamma (\frac{\zeta _{tr}}{2} )\Gamma (\frac{\zeta _{rot}}{2} )(kT_{coll})^{\frac{\zeta ^{rot}}{2}} (kT_{coll})^{\frac{\zeta ^{rot}}{2}}} (\varepsilon _{coll}-\varepsilon _{rot})^{\frac{\zeta _{tr}}{2}-1 } {\varepsilon _{rot}}^{\frac{\zeta _{tr}}{2}-1 }
=A(\varepsilon_{coll},T_{coll},\zeta _{tr},\zeta _{rot})P_{rot} (\varepsilon_{coll}-\varepsilon _{rot})^{\frac{\zeta_{tr}}{2}-1} (\varepsilon _{rot})^{\frac{\zeta _{tr}}{2}-1}
The maximum value of f (\varepsilon _{coll} −\varepsilon _{rot}; T_{coll})f (ε_{rot}; T_{coll})p_{rot} \text {for a fixed} \varepsilon _{coll} should appear at a specific \varepsilon _{rot},0 where
0=\frac{d}{d\varepsilon _{rot}} \left[f(\varepsilon_{coll}- \varepsilon_{rot};T_{coll})f(\varepsilon _{rot};T_{coll})P_{rot}\right]\mid _{\varepsilon _{rot}=\varepsilon _{rot,0}} (6.106)
which is
\frac{\varepsilon _{rot},0}{\varepsilon _{coll}} =\frac{\zeta _{rot}-2}{\zeta _{rt}+\zeta _{rot}-4} (6.107)
Using Eqs. 6.105 and 6.107, we can write I(ε_{rot}; ε_{coll}, T_{coll}) (appearing in Eq. 6.104), as
I(\varepsilon _{rot};\varepsilon _{coll},T _{coll})=I_{1}(\varepsilon _{rot};\varepsilon _{coll},T _{coll})\times I_{2}(\varepsilon _{rot};\varepsilon _{coll})=I_{1}(\varepsilon _{rot};\varepsilon _{coll},T _{coll})=\frac{f(\varepsilon _{coll}-\varepsilon _{rot},T _{coll})f(\varepsilon _{rot},T _{coll})}{[f(\varepsilon _{coll}-\varepsilon _{rot},T _{coll})f(\varepsilon _{rot},T _{coll})]\mid _{max}}
=\left(\frac{\zeta_{tr}+\zeta_{rot}-4 }{\zeta_{rot}-2} \right)^{\frac{\zeta _{rot}}{2}-1}\left(\frac{\zeta_{tr}+\zeta_{rot}-4 }{\zeta_{tr}-2} \right)^{\frac{\zeta _{tr}}{2}-1 }\left(\frac{\varepsilon_{rot} }{\varepsilon_{coll}} \right)^{\frac{\zeta _{rot}}{2}-1 }\left(1-\frac{\varepsilon_{rot} }{\varepsilon_{coll}} \right)^{\frac{\zeta _{tr}}{2}-1 } (6.108)
First, since p_{rot}|_{max }= p_{rot}, then I_{2}(ε_{rot}; ε_{coll}) = 1. Furthermore, we can clearly see from the above expression, that I_{1} does not explicitly depend on temperature, rather it depends only on ε_{rot}, ε_{coll}, ζ _{tr} and ζ_{rot}.
The distribution function in Eq. 6.108 is general to any selection procedure, as long as \epsilon_{rot} and ζ_{rot} are set appropriately. If pair selection is used, then \epsilon_{rot} corresponds to the combined rotational energy of the collision pair, and ζ_{rot} corresponds to the combined rotational degrees of freedom. With these values (i.e., ζ_{rot} \text {is replaced with} 2ζ_{rot}), Eq. 6.108 becomes equivalent to Eq. 6.76
\frac{P}{P_{max}}=\left\{\frac{\zeta_ {rot}+{1}/{2}-\omega }{{3}/{2}-\omega } \left(\frac{\epsilon_{tr}}{\epsilon _{coll}} \right) \right\}^{{3}/{2}-\omega } =\left\{\frac{\zeta_ {rot}+{1}/{2}-\omega }{\zeta _{rot}-1 } \left(1-\frac{\epsilon_{tr}}{\epsilon _{coll}} \right) \right\}^{\zeta_{rot}-1 } (6.76)
derived in Section 6.4.2. For the special case of a diatomic gas, where ζ_rot = 2, Eq. 6.108 can be further simplified to
I(\varepsilon _{rot};\varepsilon _{coll},T _{coll})=\left(1-\frac{\varepsilon _{rot}}{\varepsilon _{coll}} \right)^{\frac{\zeta _{tr}}{2}-1 } (6.109)