Momentum Integral Equation Andrew Ning February 7, 2018 We wish to - - PDF document

Momentum Integral Equation

Andrew Ning February 7, 2018

We wish to combine the integral mass and momentum equations to relate the important variables in the boundary layer. This will allow us to predict skin friction drag and boundary layer development numerically. We consider a general case, and only at the end simplify for incompressible flow. Consider a control volume as shown below through a small slice of the boundary layer. The control volume has a small width ∆x → 0, and it extends until the boundary layer reaches the edge velocity (or using standard boundary layer notation as y/δ → ∞). We will apply a mass and momentum balance to this control volume.

u

u + @u @x∆x

Vye

∆x

τ

Ve

Steady mass balance: The standard equation for a steady mass balance is:

• ρ

V · d A = 0 (1) There is a boundary layer profile coming in on the left, and a different one leaving on the right, and there is also potentially some vertical velocity Vye of unknown magnitude and sign leaving through the top of the control volume as noted on the figure. − h ρudy + h

• ρu + ∂(ρu)

∂x ∆x

• dy + ρeVye∆x = 0

(2) The first integral cancels with part of the second integral, and then ∆x cancels out of the remaining terms leaving: h ∂(ρu) ∂x dy + ρeVye = 0 (3) Thus, we can solve for the unknown vertical velocity at the edge of the boundary layer. Vye = − 1 ρe h ∂(ρu) ∂x dy (4) 1

Steady momentum balance in the x-direction: The general form for a steady momentum balance in the x-direction is:

• ρVx(

V · d A) = −

• pd

Ax + Fx (5) Applying to our situation yields: − h ρu2dy + h

• ρu2 + ∂(ρu2)

∂x ∆x

• dy + VeρeVye∆x =

h pdy − h

• p + ∂p

∂x∆x

• dy − τ∆x

(6) Most of the terms are straightforward. The mass flow rate out of the top is the same, but is multiplied by its x-component of velocity Ve. The pressure on the top surface is unknown, but because it is vertical face it does not contribute any force in the x-direction. The shear stress creates a force on the fluid in the negative x-direction. We can cancel terms just like in the mass balance: h ∂(ρu2) ∂x dy + VeρeVye = − h ∂p ∂xdy − τ (7) Solving for τ and substituting in Vye from the mass balance (Eq. (4)): τ = − h ∂p ∂xdy − h ∂(ρu2) ∂x dy + Ve h ∂(ρu) ∂x dy (8) Euler (outside boundary layer) We can remove the pressure by using the 1D Euler momentum equation in differential form. Because we are

• utside of the boundary layer, the flow is inviscid and so the Euler equation applies:

dp = −ρV dV (9) Thus, dpe dx = −ρeVe dVe dx (10) The pressure derivative in Eq. (8) is actual a total derivative (and not a partial derivative) because according to the boundary layer approximation (dp/dy = 0) so p is only a function of x. Also the value of p is constant across the boundary layer (thus p(x) = pe(x). Thus, we can substitute this equation directly in. Simplifying Substituting this result back into Eq. (8) yields τ = h ρeVe dVe dx dy − h ∂(ρu2) ∂x dy + Ve h ∂(ρu) ∂x dy (11) We can swap out the last term by using the chain rule: ∂(ρuVe) ∂x = ρu∂Ve ∂x + Ve ∂(ρu) ∂x (12) Rearranging: Ve ∂(ρu) ∂x = ∂(ρuVe) ∂x − ρu∂Ve ∂x (13) Substituting this result into Eq. (11) (and noting that Ve does not vary with y by definition and so the derivative is a total derivative) yields: τ = h ρeVe dVe dx dy − h ∂(ρu2) ∂x dy + h ∂(ρuVe) ∂x − ρudVe dx

• dy

(14) 2

We now collect like terms τ = h (ρeVe − ρu)dVe dx dy + h ∂(ρu)(Ve − u) ∂x dy (15) In the first integral we note that dVe/dx is independent of y and can thus be pulled out of the integral. We then multiply and divide that term by ρeVe. For the second integral, we reverse the order of integration and differentiation, and multiply and divide that term by ρeV 2

e

τ = dVe dx ρeVe h

• 1 − ρu

ρeVe

• dy + ∂

∂x

• ρeV 2

e

h ρu ρeVe

• 1 − u

Ve

• dy
• (16)

We see that the first integral is the definition of the displacement thickness δ∗, and the second integral is the definition of the momentum thickness θ. τ = dVe dx ρeVeδ∗ + ∂ ∂x

• ρeV 2

e θ

• (17)

Expanding the second derivative and noting the the partial derivative is a total derivative since all of the quantities do not change in y τ = dVe dx ρeVeδ∗ + ρeV 2

e

dθ dx + 2ρeVe dVe dx θ + dρe dx V 2

e θ

(18) Dividing by ρeV 2

e and collecting like terms:

τ ρeV 2

e

= dVe dx 1 Ve (δ∗ + 2θ) + dθ dx + dρe dx 1 ρe θ (19) We multiply the first term by 1/2 on the top and bottom and use the definition of the local skin friction

• coefficient. For the second term we factor out θ and define a new variable called the shape factor: H = δ∗/θ.

1 2cf = dVe dx θ Ve (H + 2) + dθ dx + dρe dx 1 ρe θ (20) Finally, we would like the relate the derivative of density to the derivative of velocity. If we assume that the flow outside the boundary layer is isentropic, that we can use the isentropic relationship p ργ = constant (21) Taking derivatives yields: dp dx = γp ρ dρ dx (22) Using the definition of the speed of sound gives dp dx = a2 dρ dx (23) We again, make use of Euler’s equation (since the flow can be assumed inviscid outside of the boundary layer), to relate pressure to velocity: dp dx = a2 dρ dx = −ρV dV dx (24) Thus, dρe dx = −ρeM 2

e

1 Ve dVe dx (25) We now substitute this expression into Eq. (20) 1 2cf = dVe dx θ Ve (H + 2) + dθ dx + −M 2

e

1 Ve dVe dx θ (26) 3

We can now combine like terms yielding the final result: 1 2cf = dθ dx + dVe dx θ Ve

• H + 2 − M 2

e

• (27)

This is the Von K´ arm´ an Momentum Integral Equation. We’ve been able to express the mass and momentum balance in a compact equation relating the important quantities in the boundary layer. If the flow is incompressible then that means Me → 0 and we have the incompressible form: 1 2cf = dθ dx + dVe dx θ Ve (H + 2) (28) 4