MrsDrPoe: Examples of Lift and Drag

For this Thesis Tuesday here on the blog, I'd like to share an example of the concepts of lift and drag that we have been talking about for the past two weeks.

Example: The average pressure and shear stress acting on the surface of the 1m square flat plate are as indicated i nthe Figure.  Determine the lift and drag generated if shear stress is and is not neglected.

Given: alpha = 7deg, pTa = -1.2 kN/m*m, pBa = 2.3 kN/m*m tauTa = 5.8e-2 kN/m*m, tauBa = 7.6e-2 kN/m*m
Find: D and L with and without shear
Solution:

A = 1 m*m, Theta = 90deg - alpha = 83deg

D = int(p*cos(theta))dA + int(tauw*sin(theta))dA 

Using average values to avoid integration:

DS = (pBa - pTa)*cos(theta)*A + (tauTa + tauBa)*sin(theta)*A = 559.544 N

Dns = (pBa - pTa)*cos(theta)*A = 426.543 N

L = -int(p*sin(theta))dA + int(tauw*cos(theta))dA

Again using average values to avoid integration:

LS = (pBa - pTa)*sin(theta)*A - (tauTa + tauBa)*cos(theta)*A = 3.458e3 N

Lns = (pBa - pTa)*sin(theta)*A = 3.474e3 N 

Next week, get ready to study some internal flows!

MrsDrPoe: Lift

It's once again Thesis Tuesday here on the blog!  Today we'll be discussing another external flow property, lift.

For asymmetrical objects moving through a fluid, there exists a resultant force from pressure and shear stress normal to the upstream velocity, termed lift.  It can be calculated as:

L = -int(p*sin(theta))dA + int(tauw*cos(theta))dA

however, as previously mentioned, these distributions are difficult to determine.  Thus, the lift coefficient is often used:

CL = L/(0.5*rho*U*U*A)

As with the drag coefficient, CL = f(shape, Re, Ma, Fr, epsilon/l).  The Froude number is only important if there is a free surface present.  Surface roughness is often unimportant in therms of lift as well.  The importance of Ma is low except for high-speed subsonic and supersonic flows.  Reynolds number also yields no great impact; however, for high-Re flows, shear stress has little effect on the shape of an object is the key factor in the lift force placed on it.

Bodies designed to generate lift, like airfoils, typically do so by generating a pressure distribution that is different on the top (low) and bottom (high) surface of the body.  For symmetric airfoils to generate lift, they must be moving through the fluid at some angle of attack.  For asymmetric airfoils, there is some non-zero angle of attack for which no lift is generated.  If the angle of attack is too large, however, the boundary layer along the airfoil separates and is unable to reattach to the body.  This event is known as stall, and it is especially dangerous if it occurs for low-flying aircraft.  For calculation of CL, the planform area (A = b*c) is used, where b is the length of the airfoil (into the page), and c is the chord length.  Thus, the lift is the dynamic pressure times the planform area of the wing.  The wing loading or average lift per unit area of the wing (L/A) is also useful for design, as is the aspect ratio (b*b)/A.  From these characteristics, it can be determined that longer wings are more efficient, but more difficult to maneuver in flight.

Since, for most cases, the shear stress is not important for calculating lift, the potential flow solution can be employed to determine this force.  For airfoils at angles of attack not equal to zero, the potential solution alone causes incorrect streamlines at the trailing edge of the airfoil.  This can be corrected with the addition of circulation or clockwise swirl to the solution.  While this may seem random, arbitrary, or inappropriate, this addition has well founded physical and mathematical grounds.


Next week, we'll look at calculating lift and drag.  Happy studying!

MrsDrPoe: Drag

Good morning, and happy Thesis Tuesday!  Today we'll be continuing our look at external flow with a brief discussion on drag.

Drag is a net force in the direction of the flow due to the pressure and shear forces on the surface of the object that is moving through a fluid.  If the pressure distribution and wall shear stress are known, drag can be determined from:

D = int(p*cos(theta))dA + int(tauw*sin(theta))dA

however, there are very few cases for which this data can be found analytically.  Typically, drag is determined from a given drag coefficient, CD:

CD = D/(0.5*rho*U*U*A)

As we will see, CD = f(shape, Re, Ma, Fr, epsilon/l), where epsilon is surface roughness.

Friction drag, Df, is the portion of drag due to the viscous shear stress on the object (the second term in the drag equation above).  Typically, the surface of an object rotated in any fashion has parts parallel and perpendicular to the flow, but if we examine the case of a very thin flat plate, it can be seen that there is no friction drag force on the plate rotated perpendicularly to the flow.  The friction drag on a flat plate can be determined by:

Df = 0.5*rho*U*U*b*l*CDf

Pressure drag, Dp, is the portion of drag due to the pressure (or normal stresses) on an object (the first term in the drag equation above).  The pressure drag can be found by:

Dp = 0.5*rho*U*U*A*CDp

where:  CDp = int(Cp*cos(theta))dA/A

Here, Cp is a pressure coefficient:

Cp = (p - po)/(0.5*rho*U*U)

where po is a reference pressure, which does not influence the drag (the difference is important).


In most cases, the net effects of friction drag and pressure drag are considered instead of examining each type of drag individually as in the second equation above.  As mentioned before, drag is influenced by many different aspects of the flow.


One of the most important contributors to the drag coefficient is the shape of the submerged body.  Obviously, objects range in size from streamlined to blunt; more blunt objects (l/D -> 0, D >> l) result in larger drag coefficients.  For extremely thin, streamlined bodies (l/D -> infinity, D << l) such as thin airfoils, the bodies are essentially tread as flat plates.


Reynolds number also affects the drag coefficient.  At very low Re (<1), inertial effects are very small and CD = 2*C/Re, where C is a constant dependent upon size.  For moderate Re, CD = Re^-0.5.  For very high Re, CD increases for streamlined bodies and decreases for blunt bodies.  For extremely blunt bodies, there CD depends little upon Reynolds number.


For sufficiently large object velocities (Ma > 0.5), compressibility effects become important and the drag coefficient becomes a function of Mach number (Ma = U/c).  Sharp-pointed bodies develop their maximum drag coefficient around Ma = 1 (sonic flow), while that for blunt bodies increases with Ma far above Ma = 1.


For streamlined bodies, drag increases with increased surface roughness; however, for blunt bodies the opposite is true.


The Froude number (U/sqrt(g*l)) is the ratio of free-stream speed to a typical wave speed on the interface of two fluids, such as the surface of the ocean.  Wave drag, Dw can be a complex function of the Froude number and body shape:

CDw = Dw/(0.5*rho*U*U*l*l)

Often approximate drag calculations for a complex body can be obtained by examining the body as a collection of various parts.  The drag on each simpler part can be calculated and added together to determine the overall drag on the body.




So that's drag!  Until next week, happy studying!

MrsDrPoe: Momentum Integral Example

As promised last week, we'll look at an example problem using the momentum integral boundary layer equation.

Example: If the boundary layer velocity profile is approximated at u = U*y/delta in the boundary layer and u = U in the free stream, and the displacement thickness is measured at 6.7 cm, determine the boundary layer and momentum thicknesses.

Given: u(y) = U*y/delta, u(y) = U if y > delta, delta* = 6.7 cm

Find: boundary layer thickness, momentum thickness
Solution:

Choose some Y in the free stream:

delta* = int(1 - (u(y)/Uinfinity))dy|0,Y 

= int(1 - (u(y)/Uinfinity))dy|0,delta + int(1 - (u(y)/Uinfinity))dy|delta,Y 

= int(1 - (U*y/delta)/Uinfinity)dy|0,delta + int(0)dy|delta,Y 

= int(1 - (y/delta))dy|0,delta 

= (delta - (delta*delta)/(2*delta)) 

= delta - delta/2 = delta/2

delta = 2*delta* = 13.4 cm

Theta = int((u(y)/Uinfinity)*(1 - u(y)/Uinfinity))dy|0,Y

= int((u(y)/Uinfinity)*(1 - u(y)/Uinfinity))dy|0,delta + int((u(y)/Uinfinity)*(1 - u(y)/Uinfinity))dy|delta,Y

= int((u(y)/Uinfinity)*(1 - u(y)/Uinfinity))dy|0,delta + int(0)dy|delta,Y

=int((y/delta)*(1 - y/delta))dy|0,delta

= (delta*delta)/(2*delta) - (delta*delta*delta)/(3*delta*delta)

= delta/6

Theta = delta/6 = 2.233 cm

Not too bad, huh?  Until next week, happy studying!

MrsDrPoe: Momentum Integral Boundary Layer Equation

Happy Thesis Tuesday to all!  Today we'll be looking at another case of external flow- the momentum integral boundary layer equation.

Defining the boundary layer height as the height from the plate where the velocity reached 99% of the free stream velocity is appealing, but often this is difficult to physically measure.  Instead, we can measure the displacement thickness, delta*, which is how far the free stream streamlines are shifted (or displaced) due to the boundary layer.  We can derive an expression for the displacement thickness by examining conservation of mass in a control volume that has a top control surface along a displaced streamline:

d/dt(int(rho)dV) + int(rho*(V.n))dA = 0

We will assume that the flow is steady and 1D with the flow crossing through two areas of the control surface with unit width (y(1) and H(1)) such that:

int(rho*u)dy|0,y - int(rho*Uinfinity)dy|0,H = 0

Assuming the flow is also incompressible, we can divide the rho out of both integrals:

int(u)dy|0,y - int(Uinfinity)dy|0,H = 0

We know that the free stream velocity is a constant and that y can be written as H + delta* and u can be written as Uinfinity + u - Uinfinity:

Uinfinity*H = int(Uinfinity + u - Uinfinity)dy|0,y = int(Uinfinity)dy|0,H+delta* + int(u-Uinfinity)dy|0,y = Uinfinity*H + Uinfinity*delta* + int(u - Uinfinity)dy|0,y

Further simplification yields:

Uinfinity*delta* = int(Uinfinity - u)dy|0,y or delta* = int(1 -(u/Uinfinity))dy|0,y

As previously mentioned, this value is much easier to measure than delta.

Another boundary layer thickness definition commonly employed is the momentum thickness (Theta), which is related to the amount of momentum "lost" due to drag in the boundary layer.  We can derive an expression for this thickness by applying the x-momentum equation to the control volume previously defined:

d/dt(int(rho*V)dV) + int(rho*V*(V.n))dA = Sum(F)

becomes:

int(rho*V*u)dA - int(rho*V*Uinfinity)dA = Fdrag

Taking the x-momentum equation:

int(rho*u*u)dA - int(rho*Uinfinity*Uinfinity)dA = -Fdrag

and applying the flow through an area of unit width with Fdrag = rho*Uinfinity*Uinfinity*Theta, then:

-rho*Uinfinity*Uinfinity = int(rho*u*u)dy|0,y - int(rho*Uinfinity*Uinfinity)dy|0,H = int(rho*u*u)dy|0,y - Uinfinity*int(rho*Uinfinity)dy|0,H

Employing the continuity equation:

-rho*Uinfinity*Uinfinity*Theta = int(rho*u*u)dy|0,y - Uinfinity*int(rho*u)dy|0,y

We can divide both sides by -rho*Uinfinity*Uinfinity (assuming an incompressible fluid) to obtain:

Theta = -int((u*u)/(Uinfinity*Uinfinity))dy|0,y + int(u/Uinfinity)dy|0,y = int((u/Uinfinity) - (u*u)/(Uinfinity*Uinfinity))dy|0,y

or finally:

Theta = int((u/Uinfinity)*(1 - (u/Uinfinity)))dy|0,y

This momentum thickness is also more easily measured than delta.  For most cases, Theta < delta* < delta.  These integrals are valid for any y in the free stream.  It can also be noted that the shear stress on a flat plate can be written in terms of the momentum integral:

tauw = rho*U*U*d/dx(Theta)

Next week, we'll look at a quick example of these equations in action, but until then, happy studying!

MrsDrPoe: Blasius Boundary Layer Example

It's another Thesis Tuesday on the blog, and as promised, I'll be giving you an example of the Blasius Boundary Layer Solution.

Problem: A viscous fluid with a known density (rho = 1300 kg/m^3) flows past a flat plate such that the boundary layer thickness at a distance 1.3 m from the leading edge is 12 mm.  Determine the boundary layer thickenss, shear stress and friction coefficient at a distance of 0.2 m from the leading edge if the free stream velocity is 1.5 m/s.

Given: xk = 1.3 m, deltak = 12 mm, rho = 1300 kg/m^3, U = 1.5 m/s
Find: delta, tauw, and Cf at 0.2 m
Solution:

Rek = (5*xk/deltak)^2 = 2.934x10^5

nu = U*xk/Rek = 6.646x10^-6 m*m/s

mu = nu*rho = 8.64x10^-3 N*s/m*m

Rex = U*x/nu =  45,140

delta = 5*x/sqrt(Rex) =4.707 mm

tauw = 0.3332*U^(3/2)*sqrt(rho*mu/x) = 4.571 Pa

Cf = 0.664/sqrt(Rex) = 3.125x10^-3

Pretty simple, huh?  Until next time...keep studying!

MrsDrPoe: The Blasius Boundary Layer Solution

It's Thesis Tuesday on the blog, and today we'll be looking at the Blasius Boundary Layer solution.

Often when discussing boundary layer flow over a flat plate, the Blasius boundary layer solution is examined.  This solution is based on the fact that, even though the boundary layer profile changes along the plate, its general shape remains the same.  To capture this relationship, Blasius defined a similarity variable, eta = y*sqrt(Uinfinity/(2*nu*x)).  We can define a stream function for the flow based on the volumetric flow rate between streamlines: psi = sqrt(2*nu*Uinfinity*x)*f(eta).  Using our stream function/velocity relations and the product rule, we can see that:

u = d/dy(psi) = Uinfinity*f'(eta)

v = -d/dx(psi) = sqrt((nu*Uinfinity)/(4*x))*(eta*f'(eta) - f(eta)).

Since we now have our velocity components in terms of x and y, we can plug these expressions into the boundary layer momentum equation to obtain (with MUCH manipulation): 

f''' + f*f'' = 0.

The boundary conditions for this differential equation are f = f' = 0 at eta = 0 and f' -> 1 as eta -> infinity.  Check out the figure here.

From the computed solution, it can be seen that u = 0.99*Uinfinity when eta = 5.0.  This gives us: delta = 5*sqrt(nu*x/Uinfinity) = 5*x/sqrt(Rex), where Rex is the local Reynolds number at any x-location along the plate (Rex = Uinfinity*x/nu).  Furthermore, since we now know the velocity profile in the boundary layer, we can calculate the shear stress at any x-location along the plate by tauw = 0.332*Uinfinity^(3/2)*sqrt(rho*mu/x).  We can also determine the friction coefficient at a particular x-location Cf = 0.664/sqrt(Rex).

Next week we'll look at an example problem involving these equations.  Happy studying!

MrsDrPoe: Boundary Layer Equations

It's once again Thesis Tuesday on the blog, and today we'll look some more at external flow.

One of the most important considerations of external flow is this region around the body in the flow that is affected by viscous stresses.  This region, called a boundary layer, possesses a large velocity gradient as the velocity transitions from 0 at the wall (from the no-slip condition) to 0.99*Uinfinity at the top of the boundary layer.  Since tau = mu*d/dy(u), we know that it is because of this velocity gradient that viscous stresses are important!  Typically, the height (distance from the surface to the top of the boundary layer) is denoted delta.

Fluid flow in the boundary layer is governed by a special reduced form of the governing equations.  We can find these equations using simple scaling analysis.  We will begin with the continuity equation for flow over a flat plate:

d/dt(rho) + d/dx(rho*u) + d/dy(rho*v) + d/dz(rho*w) = 0

Reducing for incompressible, 2D flow:

d/dx(u) + d/dy(v) = 0

We can say that du scales with Uinfinity (Uinfinity is a representative x-velocity), dx scales with L, dv scales with vs, and dy scales with delta.   For continuity to be satisfied, these two scaled terms must be of the same magnitude.  Thus, setting them equal, we find that vs = U*delta/L.

Next we will look at the Navier-Stokes y-momentum equation (so we've made assumptions of Newtonian Fluid, steady, constant properties, laminar, incompressible) for 2D flow:

rho*(u*d/dx(v) + v*d/dy(v)) = -d/dy(p) + v*(d/dx(d/dx(v)) + d/dy(d/dy(v)))

Applying our scaling arguments as before with the exception of p scaling with rho*U*U: rho*U*vs/L, rho*vs*vs/delta, rho*U*U/delta, v*vs/(L*L), v*vs/(delta*delta).  If we substitute in the value for vs that we found from the continuity equation:

rho*U*(1/L)*(U*delta/L), rho*(U*delta/L)*(1/delta)*(U*delta/L), rho*U*U/delta, v*(1/(L*L))*(U*delta/L), v*(1/delta)*(U*delta/L)

We can make these terms dimensionless by multiplying both sides by delta/(U*U*rho): (delta*deta)/(L*L), (delta*delta)/(L*L), 1, (delta*delta)/(L*L), 1/Re, 1/Re.  If we then take the limit as delta/L (the boundary layer is VERY thin) -> 0 and Re -> infinity, we see that the only term that does not go to zero is the pressure term, d/dy(p).  From these arguments, we can see that d/dy(p) = 0, or the pressure does not vary significantly in the direction normal to the wall.

Finally, we will apply scaling arguments to the Navier-Stokes x-momentum equation (again already reduced for 2D flow): rho*(u*d/dx(u) + v*d/dy(u)) = -d/dx(p) + v*(d/dx(d/dx(u)) + d/dy(d/dy(u))).  The terms become: rho*U*U/L, rho*vs*U/delta, rho*U*U/L, v*U/L, v*U/delta.  Substituting our expression for vs: rho*U*U/L, rho*(U*delta/L)*U/delta, rho*U*U/L, v*U/L, v*U/delta and non-dimensionalizing by multiplying by L/(U*U*rho): 1, 1, 1, 1/Re, (L*L/delta*delta), 1/Re.  If we again take the limit as delta/L -> 0 and Re -> infinity, we can see that the fourth term goes to zero; however, it is unclear what the last term becomes.  We can examine three possibilities: a) it's less that 1, b) it's greater than 1, or c) it's equal to 1.  

For case (a), scaling allows us to eliminate this term (since it is insignificant compared to the terms that reduced to 1) leaving: rho*(u*d/dx(v) + v*d/dy(v)) = -d/dx(p).  This equation is true for potential flow (irrotational, no viscous stresses), but since we definitely have these stresses present in the boundary layer, this option cannot be true.  

For case (b), the fifth term is the most significant term in the momentum equation so it would become: 0 = v*d/dy(d/dy(u)).  While this equation accounts for viscous stresses, it is purely diffusive and incorrect (we can see that d/dy(u) changes along the plate in the x-direction and is not constant).  

So case (c) must be correct, which means our boundary layer x-momentum equation can be written: rho*(u*d/dx(u) + v*d/dy(u)) = -d/dx(p) + v*d/dy(d/dy(u)).

A few miscellaneous notes:

Our full set of boundary layer equations consists of: 

d/dx(u) + d/dx(v) = 0

 

rho*(u*d/dx(u) + v*d/dy(u)) = -d/dx(p) + v*d/dy(d/dy(u))

 

0 = d/dy(p)

The boundary conditions for these equations are: u(x,0) = 0 (no-slip), v(x,0) = 0 (no-slip), u(x,delta) = Uinfinity, and u(x0,y) = uin(y) (starting velocity profile).

Incidentally, for our scaling: delta/L scales with 1/sqrt(Re).  Also, just outside the boundary layer, Bernoulli's equation applies.

MrsDrPoe: Characteristics of Flow Past an Object

Happy Thesis Tuesday to you all!  Today we'll be starting a series on external flow by examining some of this flow's general characteristics.

Flows in which an object is completely submerged in a fluid are termed external flows; however, flows such as those around buildings are also considered external flows, even though buildings aren't completely submerged.  We can consider cases where the object or body is stationary and fluid is flowing around it, where the object is moving through a stationary fluid, or some combination of the two.  For all of these situations, if we fix our coordinate system with the body, we can analyze these scenarios as if fluid were flowing over a stationary body.  We will consider the velocity upstream of the body (Uinfinity) to be constant with respect to time and space.  The bodies in the flow can be classified using one of two systems:

1a) 2D object extending infinitely in a third direction
1b) axi-symmetric bodies formed by rotating a cross-section about an axis of symmetry
1c) 3D bodies

2a) streamlined bodies
2b) blunt bodies

As you can imagine, flow past an object is influenced by both the fluid properties and the size and shape of the object.  These characteristics are typically grouped in dimensionless parameters; those used most often for external flows are Reynolds number, Mach number, and the Froude number.

We can examine some general differences in flows for a range of Reynolds number values by taking a closer look at flows over a flat plate.  Here we consider a plate of length, l, in the same fluid with the same viscosity and density, but we will continue to increase the upstream velocity, Uinfinity, so that Re = 0.1, 10, and 10^7.  As the Reynolds number increases, the region around the plate where the viscous forces/stresses are important shrinks considerably, causing outer streamlines to be deflected from the plate less and less.  The wake region behind the plate also shrinks as the Reynolds number increases.  It should be noted that flows with Re < 1 are dominated by viscous effects while flows with Re > 1 are dominated by inertia.

If we look at flow over a cylinder, we can make some more generalizations about how flow over blunt bodies is affected by changes in Reynolds number.  For low Reynolds number flows (0.1), again we see large deflection of the streamlines far away from the body.  These streamlines appear to be symmetric about the center of the cylinder (the stagnation streamline) as we saw when we investigated the velocity potential function.  As the Reynolds number increases (50, 10^5), we see this area affected by the viscous stresses again shrinking.  We can also see that the flow separates from the surface of the clyinder, creating a recirculation bubble or wake behind the cylinder.

Next week we'll continue looking at external flows, focusing on the region affected by the viscous forces.  Until then, happy studying!