Deriving μ and k from One Formula Put Pr Off by 1.98× — the 9/4 Behind a 123.5 K Wall Temperature

Derive μ and k the same way and Pr has to be γ#

Compress air to ten atmospheres and its thermal conductivity barely moves. The density went up tenfold, yet heat flows at the same rate. Kinetic theory explains that in two lines. But if you pull the viscosity out of those same two lines, the Prandtl number — the ratio of momentum diffusivity to thermal diffusivity — gets nailed to $\gamma$. For air that means 1.400. The measured value is 0.707. A number 1.98× too large breaks a wall-temperature prediction, and the factor of $9/4$ that Eucken supplied in 1913 is what pulls it back. The number density slider runs from 1 to 8. crossings and λ move in opposite directions. k̂ barely moves.

While those two bars trade off, k̂ stays put. Only at n = 1 does $\lambda$ outgrow the wall gap and k̂ drop.

What a molecule carries across the tally plane#

Plant an imaginary plane inside the flow. The number of molecules crossing it per unit area and time is

with $n$ the number density and $\bar{c}$ the mean molecular speed. A molecule crossing downward had its last collision one mean free path — the average distance a molecule flies between collisions — above the plane, so it arrives carrying the value from up there. One crossing upward carries the value from $\lambda$ below. Carry momentum and you get viscosity; carry kinetic energy and you get conduction.

Here $\rho = nm$ and $c_v$ is the specific heat at constant volume per unit mass. The two expressions share every leading factor. Only the trailing $c_v$ differs.

Why an eightfold density increase leaves conductivity alone#

The mean free path is inversely proportional to density. For a collision cross section $\sigma$,

so inside $\rho \bar{c} \lambda$ the factors $n$ and $1/n$ erase each other.

Density is gone. More molecules means more crossings, but each one starts closer to the plane and therefore brings a smaller temperature difference with it. That k̂ is tallied from the temperatures the dots actually carried across the yellow plane. n = 1 is where the continuum assumption breaks.

Three routes in one table#

Now divide the two expressions. The Prandtl number is defined as $\Pr = \mu c_p / k$.

Every instance of $\rho$, $\bar{c}$ and $\lambda$ cancels. Molecular diameter, temperature and pressure all drop out of the answer. Only the specific heat ratio survives. This is not an accident of the derivation — it is its structure, because momentum and energy were loaded onto the same carrier in the same way. Viscosity written as one line of molecules is the same spot as the molecular origin of viscosity. No single coefficient is wrong, but two distinct physical processes were tied to one scale.

Rotational degrees of freedom carry no momentum#

$\Pr = \gamma$ fails because $c_v$ was treated as a single lump. For a diatomic molecule, $c_v$ splits into three translational and two rotational degrees of freedom, and only translation feeds momentum transport. A spinning dumbbell may cross the plane, but its rotational energy contributes nothing to the shear stress.

Eucken separated the two parts and gave each its own transport efficiency. Translational energy is correlated with velocity, so faster molecules carry more of it and it picks up a weight of $5/2$. Internal energy is uncorrelated with velocity, so its weight is 1. The result is one line.

For a monatomic gas with $\gamma \to 5/3$ this becomes $1 + 9/4 \cdot 2/3 = 2.5$, exactly $5/2$: with no rotation, only the translational weight remains. Rearranged,

Running all three routes on six gases#

The script computes the three routes side by side and then carries the wrong Pr through to a real calculation: a one-dimensional Couette layer with shear heating. One wall moves at $U$; the opposite wall is adiabatic.

With $u$ linear the source term is constant, and the temperature rise at the adiabatic wall is proportional to $\Pr$ alone.

import math
 
GASES = [
    ("He",  1.667, 0.68),
    ("Ar",  1.667, 0.67),
    ("N2",  1.404, 0.72),
    ("Air", 1.400, 0.707),
    ("CO2", 1.289, 0.77),
    ("H2O", 1.329, 0.98),
]
 
def pr_mean_free_path(gamma):
    """mu = (1/3) rho c_bar lam,  k = (1/3) rho c_bar lam c_v  ->  Pr = c_p/c_v."""
    return gamma
 
def pr_eucken(gamma):
    """k/(mu c_v) = 1 + (9/4)(gamma-1)  ->  Pr = 4 gamma / (9 gamma - 5)."""
    return 4.0 * gamma / (9.0 * gamma - 5.0)
 
print("gas   gamma   Pr_mfp  Pr_euck  Pr_meas   mfp/meas  euck/meas")
for name, g, meas in GASES:
    a, b = pr_mean_free_path(g), pr_eucken(g)
    print("%-5s %5.3f  %6.3f  %7.3f  %7.3f   %8.2f  %9.2f" % (name, g, a, b, meas, a / meas, b / meas))
 
def couette_wall_rise(Pr, U, cp, N=201):
    """d2T/dy2 = -Pr U^2/cp on y in [0,1]; adiabatic at y=0, T=0 at y=1. Thomas sweep."""
    h = 1.0 / (N - 1)
    src = Pr * U * U / cp
    lo = [0.0] * N; di = [0.0] * N; up = [0.0] * N; rhs = [0.0] * N
    di[0], up[0], rhs[0] = -1.0, 1.0, 0.0
    for i in range(1, N - 1):
        lo[i], di[i], up[i] = 1.0, -2.0, 1.0
        rhs[i] = -src * h * h
    lo[N - 1], di[N - 1], rhs[N - 1] = 0.0, 1.0, 0.0
    for i in range(1, N):
        w = lo[i] / di[i - 1]
        di[i] -= w * up[i - 1]
        rhs[i] -= w * rhs[i - 1]
    T = [0.0] * N
    T[N - 1] = rhs[N - 1] / di[N - 1]
    for i in range(N - 2, -1, -1):
        T[i] = (rhs[i] - up[i] * T[i + 1]) / di[i]
    return T
 
U, cp = 600.0, 1005.0
print("\nCouette shear layer, U=%.0f m/s, cp=%.0f J/kg/K, adiabatic wall at y=0" % (U, cp))
print("route            Pr    dT_wall[K]   Pr*U^2/(2cp)[K]")
rise = {}
for label, Pr in (("mean-free-path", pr_mean_free_path(1.4)),
                  ("Eucken", pr_eucken(1.4)),
                  ("measured", 0.707)):
    T = couette_wall_rise(Pr, U, cp)
    rise[label] = T[0]
    print("%-15s %5.3f  %10.1f   %14.1f" % (label, Pr, T[0], Pr * U * U / (2 * cp)))
 
print("\nwall-temperature gap = %.1f K   ratio = %.2f"
      % (rise["mean-free-path"] - rise["measured"], rise["mean-free-path"] / rise["measured"]))
 
print("\nk ~ sqrt(T) extrapolated from 300 K vs measured air conductivity")
k300 = 0.0263
for T_, k_meas in ((400, 0.0338), (600, 0.0469), (800, 0.0573), (1000, 0.0672)):
    k_law = k300 * math.sqrt(T_ / 300.0)
    print("T=%5d K   sqrtT=%.4f   measured=%.4f   error=%+6.1f%%"
          % (T_, k_law, k_meas, 100.0 * (k_law - k_meas) / k_meas))

gas   gamma   Pr_mfp  Pr_euck  Pr_meas   mfp/meas  euck/meas
He    1.667   1.667    0.667    0.680       2.45       0.98
Ar    1.667   1.667    0.667    0.670       2.49       0.99
N2    1.404   1.404    0.735    0.720       1.95       1.02
Air   1.400   1.400    0.737    0.707       1.98       1.04
CO2   1.289   1.289    0.781    0.770       1.67       1.01
H2O   1.329   1.329    0.764    0.980       1.36       0.78
 
Couette shear layer, U=600 m/s, cp=1005 J/kg/K, adiabatic wall at y=0
route            Pr    dT_wall[K]   Pr*U^2/(2cp)[K]
mean-free-path  1.400       249.5            250.7
Eucken          0.737       131.3            132.0
measured        0.707       126.0            126.6
 
wall-temperature gap = 123.5 K   ratio = 1.98
 
k ~ sqrt(T) extrapolated from 300 K vs measured air conductivity
T=  400 K   sqrtT=0.0304   measured=0.0338   error= -10.2%
T=  600 K   sqrtT=0.0372   measured=0.0469   error= -20.7%
T=  800 K   sqrtT=0.0429   measured=0.0573   error= -25.0%
T= 1000 K   sqrtT=0.0480   measured=0.0672   error= -28.5%

The gas buttons and the Eucken toggle draw the three columns as wall-temperature curves. Turn the toggle off and only the Pr = $\gamma$ curve stays beside the measured one.

Pick He and the gap between the two parabolas at the wall end opens widest. A larger $\gamma$ costs more when the rotational degrees of freedom are left out.

A factor of 9/4 decided 123.5 K of wall temperature#

The numerical and analytical columns agree to within 1 K — the $O(h^2)$ error on a 201-point grid. The third column is the temperature that split. At $U = 600$ m/s the adiabatic wall sits at 249.5 K above the moving wall if Pr is taken as $\gamma$, and 126.0 K above it with the measured value. That is a 123.5 K spread. In a supersonic wall heat flux prediction, that gap changes the material you select.

The Eucken column lands within 2–4% of measurement for every monatomic and diatomic gas listed. Splitting one dumbbell into translation and rotation turned a 2.45× error into 2%.

For air, $\gamma=1.400$ in $4\gamma/(9\gamma-5)$ has numerator $4\times1.400=5.600$ and denominator $9\times1.400-5=7.600$. The quotient is $0.7368$, which matches the table's Eucken value $0.737$ at the third decimal. The same relation as $1+(9/4)(\gamma-1)$ gives $9/4\times0.400=0.900$, and $1.400/1.900=0.737$. Helium has $\gamma=1.667$, so the mean-free-path route is $1.667$, Eucken is $0.667$, and the measurement is $0.680$. Then $1.667/0.680=2.45$, the table's mfp/meas. Eucken over measurement is $0.667/0.680=0.98$.

On the Couette side, $U=600$ and $c_p=1005$ give $U^2/(2c_p)=360000/2010=179.1$ K. Times $1.400$ that is $250.7$ K, times $0.707$ it is $126.6$ K, times $0.737$ it is $132.0$ K. The 201-point numerical values are $249.5$, $126.0$, and $131.3$. They sit $1.2$ K, $0.6$ K, and $0.7$ K off the closed form, inside the stated 1 K. And $249.5-126.0=123.5$, while $249.5/126.0=1.98$. Push $k=0.0263$ at $300$ K by $\sqrt{1000/300}$ and $\sqrt{3.333}\times0.0263$ lands near $0.0480$. Against the measured $0.0672$, $(0.0480-0.0672)/0.0672$ is about $-28.6\%$, the printed $-28.5\%$ inside rounding. For water, $0.764/0.980=0.780$, the table's euck/meas = 0.78, about 22% low.

Why the √T law falls 28.5% short at 1000 K#

The same derivation also hands you a temperature dependence. If $\bar{c} \propto \sqrt{T}$ and $\sigma$ is constant, then

The last output block extrapolates that law from 300 K. It reads 10.2% low at 400 K and 28.5% low at 1000 K. The fact that the sign never flips is the clue: the constant-$\sigma$ assumption is what broke. Faster molecules graze past one another more closely, so the effective cross section shrinks. This is the slot Chapman–Enskog theory fills with the $\Omega^{(2,2)*}$ collision integral, and that practical codes hand to the constant $S$ in Sutherland's law. Tying wall friction and wall heat with one ratio is what the Reynolds analogy does when it reads heat transfer off drag alone. Feed that ratio a Pr that is 1.98× too large and the heat flux moves with it.

Where Eucken misses H₂O by 22%#

H₂O is the one row with euck/meas = 0.78, 22% low. Water is a polar molecule and dipole interactions dominate its collisions. Rotational relaxation is far slower than translational, so the assumption that internal energy rides along uncorrelated with velocity does not hold. The modified Eucken relation multiplies the internal contribution by its own $\rho D_{\text{int}}/\mu$ to capture this. In a combustion calculation rich in steam, holding Pr constant loads that discrepancy straight into the wall heat flux.

What you actually sign when you type Pr = 0.71#

The line Pr = 0.71 is not a constant declaration. It asserts that momentum and energy diffuse at different rates even though the same molecules carry both, and that the ratio comes from two rotational degrees of freedom. Three conditions break that assertion: a monatomic gas (it falls to 0.67), a polar polyatomic one (it rises to 0.98), and a rarefied gas where $\lambda$ outgrows the cell (n = 1 in the first simulation). If your calculation lives in one of those three, changing a single constant is not the end of the job.