Far-Field Force Theory of Steady Flow

Abstract

This chapter studies the lift and drag experienced by a body in a viscous, compressible and steady flow. By a rigorous linear far-field theory and the Helmholtz decomposition of velocity field, we prove that the KJ lift formula for 2D inviscid potential flow, Filon’s drag formula for 2D incompressible viscous flow, and Goldstein’s lift and drag formulas for 3D incompressible viscous flow are universally true for the whole field of viscous compressible flow in a wide range of Mach number, from subsonic to supersonic flows. Thus, the steady lift and drag are always exactly determined by the values of vector circulation $\varvec{\varGamma }_\phi$ due to the longitudinal velocity and inflow $Q_\psi$ due to the transverse velocity, respectively, no matter how complicated the near-field viscous flow surrounding the body might be. We call this result the unified force theorem (UF theorem for short). However, velocity potentials are not directly testable either experimentally or computationally, and hence neither is the UF theorem. Thus, a testable version of it is also derived, which holds in the linear far field. We call it the testable unified force formula (TUF formula for short). Due to its linear dependence on the vorticity, TUF formula is also valid for statistically stationary flow, including time-averaged turbulent flow. For 2D flow, some careful RANS simulations of the flow over a RAE-2822 airfoil with angle of attack $\alpha = 2.31^\circ$ and $5.0^\circ$, Reynolds number $Re = 6.5\times 10^6$, and incoming flow Mach number $M\in [0.1,2.0]$ is performed to examine the validity of the TUF formula. The computed Mach-number dependence of L and D and its underlying physics, as well as the physical implication of the theorem, are also addressed. These results strongly support and enrich the UF theorem.

Appendix: The Calculations of Circulation and Inflow

The circulation due to the longitudinal velocity and the inflow due to the transversal velocity are given by (3.3.27) and (3.3.28), respectively,

$$\begin{aligned} \varvec{\varGamma }_\phi= & {} - \frac{1}{\rho _0} \int _S (\varvec{n}\times \nabla ) (\varvec{F}\cdot \nabla G_\psi ) \text {d} S, \end{aligned}$$

(3.6.1a)

$$\begin{aligned} Q_\psi= & {} \frac{1}{\rho _0} \int _S (\varvec{n}\times \nabla )\cdot (\varvec{F}\times \nabla G_\psi ) \text {d} S, \end{aligned}$$

(3.6.1b)

with

$$\begin{aligned} \nabla G_\psi= & {} \frac{e^{-k(r-x)}}{4 \pi U r} \left( 1, -\frac{y}{r-x}, -\frac{z}{r-x} \right) . \end{aligned}$$

(3.6.2)

In fact, due to the exponential factor $e^{-k(r-x)}$ in (3.6.2), (3.6.1b) can be reduced to a wake-plane integral with $\varvec{n}= \varvec{e}_x$,

$$\begin{aligned} Q_\psi= & {} \frac{1}{\rho _0} \int _W (-\partial _z \varvec{e}_y + \partial _y \varvec{e}_z) \cdot (\varvec{F}\times \nabla G_\psi ) \text {d} S, \end{aligned}$$

(3.6.3)

which can more or less simplify our analysis.

Since (3.6.1) are linearly dependent on $\varvec{F}$, we can estimate their results by assigning $\varvec{F}$ with a specific value. Suppose $\varvec{F}= D \varvec{e}_x$, then $\varvec{\varGamma }_\phi \equiv \mathbf 0$ since $\partial G_\psi /\partial x$ is regular due to (3.6.2). However, (3.6.3) reduces to

$$\begin{aligned} Q_\psi= & {} - \frac{D}{4\pi \rho _0 U} \int _S \left\{ \frac{\partial }{\partial y} \left[ \frac{y e^{-k(r-x)}}{r(r-x)} \right] + \frac{\partial }{\partial z} \left[ \frac{z e^{-k(r-x)}}{r(r-x)} \right] \right\} \text {d} S \nonumber \\= & {} \frac{D}{4\pi \rho _0 U} \int _W \frac{kr^2 + krx + x}{r^3} e^{-k(r-x)} \text {d} S =\frac{D}{\rho _0 U}. \end{aligned}$$

(3.6.4)

Due to the symmetry of y and z in (3.6.2), for the lift or side force case we only need to consider $\varvec{F}= L \varvec{e}_z$ in (3.6.1a) or (3.6.3). In this case, (3.6.3) reduces to

$$\begin{aligned} Q_\psi= & {} \frac{L}{4\pi \rho _0 U} \int _W \frac{\partial }{\partial z} \left[ \frac{e^{-k(r-x)}}{r} \right] \text {d} y \text {d} z =0. \end{aligned}$$

(3.6.5)

However, (3.6.1a) needs more algebra, which can be simplified by letting S be a sphere surface with $\varvec{n}= \varvec{e}_r$. Thus, in this situation (3.6.1a) reduces to

$$\begin{aligned} \varvec{\varGamma }_\phi = (0, \varGamma _{\phi y}, \varGamma _{\phi z}), \end{aligned}$$

(3.6.6)

where

$$\begin{aligned} \varGamma _{\phi y}= & {} \frac{L}{4\pi \rho _0 U} \int _S (\varvec{e}_y \times \varvec{e}_r) \cdot \nabla \left[ \frac{ze^{-k(r-x)}}{r(r-x)} \right] \text {d} S \nonumber \\= & {} \frac{L}{4\pi \rho _0 U} \int _S \left( \frac{z}{r} \frac{\partial }{\partial x} - \frac{x}{r}\frac{\partial }{\partial z} \right) \frac{ze^{-k(r-x)}}{r(r-x)} \text {d} S \nonumber \\= & {} \frac{L}{4\pi \rho _0 U} \int _S \frac{x^2+z^2 - r x + kz^2(r-x)}{r^2 (r-x)^2} e^{-k(r-x)} \text {d} S\nonumber \\= & {} \frac{L}{4\pi \rho _0 U} \int _0^{\pi } \!\!\!\! \int _0^{2\pi } \frac{\cos ^2\theta - \cos \theta + \sin ^2\theta \sin ^2\varphi [1 + kr (1-\cos \theta )]}{(1-\cos \theta )^2 e^{kr(1-\cos \theta )}} \sin \theta \text {d} \theta \text {d} \varphi \nonumber \\= & {} \frac{L}{4 \rho _0 U} \int _{-1}^{1} \frac{2t^2 + (1-t^2) - 2t + kr (1-t^2)(1-t)}{(1-t)^2 e^{kr(1-t)}} \text {d} t \nonumber \\= & {} \frac{L}{4 \rho _0 U} \int _{-1}^{1} [1 + kr (1+t)] e^{-kr(1-t)} \text {d} t = \frac{L}{2 \rho _0 U}, \end{aligned}$$

(3.6.7)

and

$$\begin{aligned} \varGamma _{\phi z}= & {} \frac{L}{4\pi \rho _0 U} \int _S (\varvec{e}_z \times \varvec{e}_r) \cdot \nabla \left[ \frac{ze^{-k(r-x)}}{r(r-x)} \right] \text {d}S \nonumber \\= & {} \frac{L}{4\pi \rho _0 U} \int _S \left( -\frac{y}{r} \frac{\partial }{\partial x} + \frac{x}{r}\frac{\partial }{\partial y} \right) \frac{ze^{-k(r-x)}}{r(r-x)} \text {d}S \nonumber \\= & {} -\frac{L}{4\pi \rho _0 U} \int _S \frac{1+k(r-x)}{r^2 (r-x)^2} y z e^{-k(r-x)} \text {d}S =0. \end{aligned}$$

(3.6.8)