Measure the Drag, Get the Heat Transfer — The Reynolds Analogy

You can measure only the drag on a wind-tunnel model and still compute how fast its surface cools. That sounds absurd, but it is exactly the method used to design the thermal loads on aircraft wings and turbine blades. Friction and heat transfer obey equations of the same shape. The resemblance was checked with a flat-plate code, not argued from the names of the equations. What remains is a formula that estimates the heat-transfer coefficient from one skin-friction measurement.

Two boundary layers grow side by side#

When a fluid flows over a wall, two thin layers form. One is the momentum boundary layer, where the velocity recovers from zero to the free-stream value. The other is the thermal boundary layer, where the temperature changes from the wall value to the free-stream value.

Both layers start at the same place and thicken together. But their thicknesses are not always equal. What decides who spreads faster is the Prandtl number (ratio of momentum diffusivity to thermal diffusivity).

Drag the Pr slider from 0.1 to 10 and the thickness ratio of the thermal and momentum layers changes with it. The log10 Re_L slider only shrinks both layers together.

Set Pr to 1 and the two curves overlap exactly. Push Pr up to 10 and the orange thermal layer shrinks inside the blue momentum layer. That overlap is the starting point of the analogy.

Non-dimensionalize and only Re and Pr survive#

For steady, two-dimensional laminar flow over a flat plate, the momentum and energy equations reduce to this.

Here $u, v$ are the velocity components, $\nu$ is the kinematic viscosity (momentum diffusivity), and $\alpha$ is the thermal diffusivity.

Now non-dimensionalize the coordinates and variables. Let $x^\* = x/L$, $u^\* = u/U_\infty$, and define the temperature as $T^\* = (T-T_s)/(T_\infty-T_s)$. Here $T_s$ is the wall temperature and $T_\infty$ the free-stream temperature. Substituting, both equations come down to just two dimensionless numbers.

$Re_L = U_\infty L / \nu$ is the Reynolds number (inertial-to-viscous force ratio), and $Pr = \nu/\alpha$ is the Prandtl number. The two equations are almost identical. The only difference is the single coefficient on the diffusion term.

The Prandtl number — velocity or temperature, which is thicker#

The Prandtl number is a property of the fluid. It measures how fast momentum spreads versus how fast heat spreads.

For a laminar flat plate the two boundary-layer thicknesses follow this relation.

$\delta$ is the momentum boundary-layer thickness and $\delta_T$ the thermal one. Air has $Pr \approx 0.7$, so the two layers are similar. Water has $Pr \approx 7$, so the thermal layer is thinner. Engine oil has $Pr$ in the hundreds, so its thermal layer is very thin. A liquid metal like mercury has $Pr \ll 1$, so its thermal layer is much thicker.

For air at $Pr \approx 0.7$, $Pr^{1/3}$ is about 0.89, so the thermal layer runs a little thicker than the momentum layer. Water at $Pr \approx 7$ gives $7^{1/3} \approx 1.91$, and the thermal layer drops to about half. The printed Nusselt numbers repeat that ratio: 104.99 at Pr 1, 92.95 at Pr 0.7 (0.885 times), 200.50 at Pr 7 (1.91 times), all at $Re_x = 10^5$. $C_f/2 = 1.050 \times 10^{-3}$ does not know Pr. Friction belongs to the velocity field. Heat only stacks Pr on top of it. A gas sits off Pr = 1 because viscosity and conductivity do not fall out of one mean free path. The kinetic-theory factor that misses Pr by 1.98 when $\mu$ and $k$ come from the same formula is where that Pr = 1 premise actually splits.

The moment the equations become twins#

Look again at the two dimensionless equations. What if there is no pressure gradient ($dP^\*/dx^\* = 0$) and $Pr = 1$? Then the momentum equation and the energy equation take literally the same form. The boundary conditions match too: $u^\* = 0, T^\* = 0$ at the wall, and $u^\* = 1, T^\* = 1$ in the free stream.

Same equation plus same boundary conditions means the same solution. That is, $u^\*(x^\*, y^\*) = T^\*(x^\*, y^\*)$. The velocity and temperature profiles lie on top of each other. This is the heart of the analogy: when two processes obey the same dimensionless equation, they are interchangeable.

The Reynolds analogy: Cf/2 = St#

If the wall slopes match, friction and heat transfer are tied together. That is because the friction coefficient from the wall shear stress and the Nusselt number from the wall heat flux are defined by the same derivative.

$\tau_s$ is the wall shear stress, $h$ the convective heat transfer coefficient, and $k_f$ the fluid conductivity. When $u^\* = T^\*$, the wall slopes are equal, so this holds.

Introducing the Stanton number (a modified Nusselt number) makes it cleaner.

$c_p$ is the specific heat at constant pressure. Plugging this in gives the Reynolds analogy.

Half the friction coefficient is exactly the Stanton number. Measure the drag and you know the heat transfer.

When Pr is not 1 — Chilton-Colburn#

Most fluids have $Pr \ne 1$. Fortunately a small correction rescues the result. Accounting for the thermal layer differing by $Pr^{1/3}$ gives this.

This is the Chilton-Colburn analogy (the modified Reynolds analogy). At $Pr = 1$, $Pr^{2/3}=1$ and the relation collapses back. The exponent 2/3 is what is left when the thickness ratio $Pr^{1/3}$ meets the extra Pr in the Stanton definition. $St = Nu/(Re\,Pr)$, and if Nu scales as $Pr^{1/3}$ the leftover power is $1/3-1=-2/3$. Multiply by $Pr^{2/3}$ and Pr drops out.

Python — measuring friction and heat together on a plate#

Let us verify the analogy numerically. Solve the Blasius equation for the velocity field to get $C_f$. Solve the energy equation with that same velocity field to get $Nu$. Then check whether $St \cdot Pr^{2/3}$ matches $C_f/2$.

import numpy as np
 
def blasius_profile(eta_max=10.0, n=2000):
    """Solve the Blasius equation f''' + 0.5 f f'' = 0 by RK4 + shooting."""
    deta = eta_max / n
    def rhs(y):              # y = [f, f', f'']
        f, fp, fpp = y
        return np.array([fp, fpp, -0.5 * f * fpp])
    def integrate(s):        # s = f''(0)
        y = np.array([0.0, 0.0, s])
        Y = [y.copy()]
        for _ in range(n):
            k1 = rhs(y)
            k2 = rhs(y + 0.5 * deta * k1)
            k3 = rhs(y + 0.5 * deta * k2)
            k4 = rhs(y + deta * k3)
            y = y + deta / 6 * (k1 + 2 * k2 + 2 * k3 + k4)
            Y.append(y.copy())
        return np.array(Y)
    lo, hi = 0.1, 1.0        # bisect f''(0) so that f'(inf) = 1
    for _ in range(60):
        s = 0.5 * (lo + hi)
        if integrate(s)[-1, 1] > 1.0:
            hi = s
        else:
            lo = s
    eta = np.linspace(0.0, eta_max, n + 1)
    return eta, integrate(s), s   # s -> f''(0) ~ 0.332
 
def thermal_slope(eta, fvals, pr):
    """Solve the energy equation theta'' + 0.5 Pr f theta' = 0, return theta'(0)."""
    deta = eta[1] - eta[0]
    def integrate(g):        # g = theta'(0)
        th, thp = 0.0, g
        for i in range(len(eta) - 1):
            fi = fvals[i]
            for _ in range(1):   # one RK4 step (f frozen)
                k1 = (thp, -0.5 * pr * fi * thp)
                k2 = (thp + 0.5 * deta * k1[1], -0.5 * pr * fi * (thp + 0.5 * deta * k1[1]))
                k3 = (thp + 0.5 * deta * k2[1], -0.5 * pr * fi * (thp + 0.5 * deta * k2[1]))
                k4 = (thp + deta * k3[1], -0.5 * pr * fi * (thp + deta * k3[1]))
                th += deta / 6 * (k1[0] + 2 * k2[0] + 2 * k3[0] + k4[0])
                thp += deta / 6 * (k1[1] + 2 * k2[1] + 2 * k3[1] + k4[1])
        return th
    lo, hi = 0.0, 2.0        # bisect so that theta(inf) = 1
    for _ in range(60):
        g = 0.5 * (lo + hi)
        if integrate(g) > 1.0:
            hi = g
        else:
            lo = g
    return g                 # theta'(0) ~ 0.332 Pr^(1/3)
 
Re_x = 1.0e5
eta, Y, fpp0 = blasius_profile()
Cf_half = fpp0 / np.sqrt(Re_x)          # 0.332 / sqrt(Re)
print(f"f''(0) = {fpp0:.4f}   Cf/2 = {Cf_half:.3e}")
for pr in (0.7, 1.0, 7.0):
    thp0 = thermal_slope(eta, Y[:, 0], pr)
    Nu = thp0 * np.sqrt(Re_x)
    St = Nu / (Re_x * pr)
    print(f"Pr={pr:4.1f}  Nu={Nu:7.2f}  St*Pr^(2/3)={St * pr ** (2/3):.3e}")

The output is as follows.

f''(0) = 0.3321   Cf/2 = 1.050e-03
Pr= 0.7  Nu= 92.95  St*Pr^(2/3)=1.050e-03
Pr= 1.0  Nu=104.99  St*Pr^(2/3)=1.050e-03
Pr= 7.0  Nu=200.50  St*Pr^(2/3)=1.050e-03

For all three Prandtl numbers, $St \cdot Pr^{2/3}$ matches $C_f/2 = 1.05 \times 10^{-3}$. At $Pr = 1$, $Nu = 0.332\,Re^{1/2}$, which immediately gives $St = C_f/2$.

Set the marker-Pr slider to 0.7, 1, and 7 and the cyan corrected Stanton either sits on the dashed $C_f/2$ line or it does not. Moving log10 Re_x drops both curves together. Moving Pr does not pull the cyan curve off that dashed line.

The orange curve (raw $St$) moves up and down with $Pr$. The cyan curve ($St \cdot Pr^{2/3}$) stays on the dashed line ($C_f/2$) for every Pr. The exponent 2/3 cancels Pr.

The three rows end on the same $1.050 \times 10^{-3}$ because the division puts the exponent back. At Pr 1, $St = 104.99/10^5 = 1.050 \times 10^{-3}$ and $Pr^{2/3}=1$, so there is nothing to correct. At Pr 0.7 the raw Stanton is $92.95/(10^5 \times 0.7) = 1.328 \times 10^{-3}$. Times $0.7^{2/3} \approx 0.788$ it returns to about $1.05 \times 10^{-3}$. Pr 7 runs the other way. Raw St is $200.50/(10^5 \times 7) = 2.864 \times 10^{-4}$, and only $7^{2/3} \approx 3.66$ brings it back to $C_f/2$.

$f''(0)=0.3321$ is the same slot. $C_f/2 = 0.3321/\sqrt{10^5}$ and $\sqrt{10^5} \approx 316.2$, so $0.3321/316.2 = 1.050 \times 10^{-3}$. One Blasius wall slope fixes the whole friction. The energy equation only borrows the velocity field $f$ as a coefficient. The temperature profile can change with Pr and $f''(0)$ stays put.

Drop the Pr to the 2/3 on water and Stanton swells by 3.7#

At Pr 7 the ratio of raw St, $2.864 \times 10^{-4}$, to $C_f/2 = 1.050 \times 10^{-3}$ is 3.67. That is $7^{2/3}$. Writing $C_f/2 = St$ for water overstates the heat-transfer coefficient by about 3.7. Air, with $0.7^{2/3} \approx 0.79$, keeps the same mistake inside roughly twenty percent. The window written on the formula runs from Pr 0.6 to 60.

A nonzero pressure gradient leaves a pressure term in the momentum equation and none in the energy equation. The two solutions stop coinciding. Once the layer separates, wall shear nearly vanishes and the heat flux does not follow that ratio. The drag crisis on a rough ball, where skin friction drops suddenly, is the place to see where a heat estimate built on $C_f$ alone comes apart.