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: Conservation of Momentum, Part VI

Good morning to you all and welcome to Thesis Tuesday on the blog!  Today we'll be discussing the last segment of the conservation of momentum equation: an example using the inviscid reduction of its differential form (i.e. the Bernoulli Equation).

Example: Water flows from the faucet on the first floor of the building shown in the figure below with a maximum velocity of 20 ft/s.  For steady, inviscid flow, determine the maximum water velocity from the basement faucet and from the faucet on the second floor (assume each floor is 12 feet tall).

Given: V1max = 20 ft/s, hf = 12 ft, hgf = 4 ft, gammawater = 62.4 lbf/(ft*ft*ft), rhowater = 1.94 slugs/(ft*ft*ft)

Find: V2max, Vbmax

Assumptions: inviscid, steady, incompressible flow along a streamline, no outside forces, water exits faucet as a free jet

Solution:

Because of our assumptions, we can immediately employ the Bernoulli equation between each set of two points without consideration for the continuity equation, etc.


a) Between floors 1 and 2

p1 + 0.5*rhowater*V1*V1 + gammawater*z1 = p2 + 0.5*rhowater*V2*V2 + gammawater*z2

We can rearrange the equation algebraically to solve for V2 (the velocity from the second floor faucet):

V2 = sqrt((p1 - p2 + 0.5*rhowater*V1*V1 + gammawater*z1 - gammawater*z2)/0.5*rhowater)

Since water from the faucets exits like a free jet, we are essentially saying that the pressure at each exit is atmospheric (0 gage pressure); thus:

p1 = p2 = 0 psi

Since we are given information about the height between the floors:

z1 = 0 ft, z2 = 12 ft

Plugging these and our Given values in to our V2 expression, we find that the maximum velocity from the second floor faucet is: 19.28i ft/s.


The i indicates that this number is an imaginary number; since this number is impossible to achieve physically, we know that water will not flow from the upstairs faucet. 

b) Between floor 1 and the basement 

To determine the velocity from the basement faucet, we can use the manipulated equation above for V2, replacing the subscript 2 with the subscript b (i.e. using the basement values instead of the second floor values):

Vb = sqrt((p1 - pb + 0.5*rhowater*V1*V1 + gammawater*z1 - gammawater*zb)/0.5*rhowater)

Again the free jet assumption tells us:

p1 = pb = 0 psi

And the height between floors information yields:

z1 = 0 ft, zb = -12 ft

Plugging in all our values, we find that the maximum velocity from the basement faucet is: 34.23 ft/s.


That concludes our discussion on momentum.  You should now have the basic tools necessary to work a problem by employing the integral form of the momentum equation, the viscous differential form (Navier Stokes) of the momentum equation, and the inviscid differential form (Bernoulli Equation) of the momentum equation.  Feel free to revisit any of the previous equation derivations or example problems here, and until next time...happy studying!

MrsDrPoe: Conservation of Momentum, Part V

It's Thesis Tuesday on the blog, and today we'll be continuing our examination of the Conservation of Momentum equation, this time focusing on the inviscid reduction of its differential form.

If you recall, the difficulty with the differential momentum equation is that we need to define the stress terms. Previously, we obtained the following results:

d/dx(Sxx) + d/dy(Txy) + d/dz(Txz) = -d/dx(p) + mu*(d/dx(d/dx(u))+d/dy(d/dy(u)) + d/dz(d/dz(u)))

d/dx(Tyx) + d/dy(Syy) + d/dz(Tyz) = -d/dy(p) + mu*(d/dx(d/dx(v))+d/dy(d/dy(v)) + d/dz(d/dz(v)))

d/dx(Tzx) + d/dy(Tzy) + d/dz(Szz) = -d/dz(p) + mu*(d/dx(d/dx(w))+d/dy(d/dy(w)) + d/dz(d/dz(w)))

If we examine the case of inviscid flow, this means we are looking at an area where viscous effects are negligible (mu = 0). Generally this occurs in the far-field, i.e. in a region of the fluid far away from any walls or boundaries. This leaves us with:

d/dx(Sxx) + d/dy(Txy) + d/dz(Txz) = -d/dx(p)

d/dx(Tyx) + d/dy(Syy) + d/dz(Tyz) = -d/dy(p)

d/dx(Tzx) + d/dy(Tzy) + d/dz(Szz) = -d/dz(p)

So our stress tensor terms are ONLY related to changes in pressure. If we replace these new equalities in the differential form of the equation, we end up with a set of equations known as Euler's Equations. It is important to note, however, that this set is only valid for inviscid flows.

If we apply the assumptions of steady, incompressible flow with no outside forces acting on the fluid to Euler's Equations, we get:

u*d/dx(u) + v*d/dy(u) + w*d/dz(u) = gx - (1/rho)*d/dx(p)


u*d/dx(v) + v*d/dy(v) + w*d/dz(v) = gy - (1/rho)*d/dy(p)


u*d/dx(w) + v*d/dy(w) + w*d/dz(w) = gz - (1/rho)*d/dz(p)

Using a vector identity (which we won't discuss in further detail), and applying along a streamline (lines always parallel or tangent to the flow), we can further reduce this equation to:

0.5*V*V + (p/rho) + g*y = constant (along a streamline)

or: 0.5*V1*V1 + (p1/rho) + g*y1 = 0.5*V2*V2 + (p2/rho) + g*y2

This is Bernoulli's Equation, and it is written for two specific points along a streamline.

Restrictions:

Remember that this equation is only applicable to flows of fluids where the assumptions we made to derive the equation (steady, incompressible, inviscid, along a streamline, no outside forces) hold. It is CRUCIAL to realize that if the flow does not meet these qualifications, this simplified equation CANNOT be used!

If our flow cannot be treated as incompressible, but the other assumptions hold, we can modify the Bernoulli Equation so that it can still be used. For compressible gases, we can apply the ideal gas law to obtain:

0.5*V*V + y*g + R*T*ln(p) = constant (along a streamline)

For unsteady flows, we can modify the equation by the addition of a single term integrating the acceleration over a streamline:

0.5*rho*V1*V1 + p1 + rho*g*y1 = rho*int(d/dt(V))ds + 0.5*rho*V2*V2 + p2 + rho*g*y2

Physical Meaning:

The Bernoulli Equation is simply the mathematical statement of the principle "the work done on a particle by all forces acting on the particle is equal to the change of the kinetic energy of the particle." We can better see that if we look at each of the terms in the equation. As the particle moves, we see forces of pressure and gravity acting on it (the z*g and p/rho terms). This is directly connected to the amount of kinetic energy that the particle has (0.5*V*V).

If we multiply each term in the Bernoulli Equation by the density, rho:

0.5*rho*V*V + p + g*rho*y = constant (along a streamline)

we obtain the pressure form of the equation. Here p represents the thermodynamic pressure of the fluid as it flows (this pressure determines the state of a fluid, i.e. whether it is a gas or a liquid). This value is typically termed the static pressure of the fluid, since it could be measured by moving along with the fluid or being static in relation to the fluid. The third term is the hydrostatic pressure, which is associated with the hydrostatic pressure condition. The second term is called the dynamic pressure since it is associated with the velocity of the fluid.

The combination of the first two terms in the equation is the stagnation pressure. This pressure represents the conversion of all the kinetic energy into a pressure rise, which occurs at a stagnation point where the velocity is equal to zero. The streamline leading to a stagnation point is called a stagnation streamline. The sum of all three terms in this form of the Bernoulli Equation is known as the total pressure, which is constant along a streamline.

Often in Mechanical Engineering, the Bernoulli Equation will be written in head form; this form of the equation is particularly useful for pipe flow and can be derived by dividing each term in the original equation by the gravitational constant, g:

(p/gamma) + (0.5*V*V/g) + y = constant (along a streamline)

In this form, the pressure term is called the pressure head, which represents the height of a column of fluid needed to produce the pressure p. The velocity term is called the velocity head, which represents the vertical distance needed for the fluid to fall freely to reach the velocity V from rest. The height term is the elevation head, and it represents the potential energy of the fluid.


Next week we'll look at some very simple examples of how to employ the Bernoulli Equation. Stay tuned!

MrsDrPoe: Conservation of Momentum, Part IV

Thesis Tuesday is upon us yet again, ladies and gents! As promised, today we will be continuing our discussion of the Conservation of Momentum Equation with an example of how to solve the Navier Stokes equations.

We'll start with fluid flowing between two parallel plates, separated by H meters and infinite in the z-direction. (Note: infinite plates do not exist, but this consideration allows us to approximate the flow as 2D instead of 3D). We will consider the pressure driven flow case where both plates are stationary (I), as well as the Couette Flow case where the upper plate moves in the +x-direction with some velocity U m/s (II).

Given: gap height = H meters
Find: the flow field description (u, v, w, and p)
Assumptions: steady, laminar, incompressible, infinite plates (w = 0, d/dz(...) = 0), constant properties, ignore edge effects, Newtonian fluid, fully developed (d/dx(...) = 0)
Solution:

Step 1: We must first apply the continuity equation.

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

Since the flow is steady (or because it's incompressible), the first term is equal to zero. For fully developed, laminar flow, d/dx(rho*u) = 0; for infinite plates, d/dz(rho*w) = 0. Because the fluid is incompressible, rho can be pulled out of the derivative in the remaining term. Thus, we are left with:

rho*d/dy(v) = 0 or d/dy(v) = 0

Since the derivative of v is equal to zero, that means that v must be constant (found by integrating both sides of the above equation):

v = const

But we know that because we are dealing with a viscous fluid, the no-slip condition holds. This tells us that the fluid adjacent to either plate has the same velocity as the plate; therefore, since the bottom plate is fixed in both cases, u = v = w = 0 at this location. Putting this boundary information with the result from the continuity equation, v = 0 throughout the domain.

Step 2: Next, we simplify the Navier Stokes equations.

x-momentum:

rho*[d/dt(u) + u*d/dx(u) + v*d/dy(u) + w*d/dz(u)] = -d/dx(p) + rho*gx + mu*[d/dx(d/dx(u)) + d/dy(d/dy(u)) + d/dz(d/dz(u))]

d/dt(u) = 0 since the flow is steady; d/dx(u) = 0 since the flow is fully developed; d/dy(u) = d/dz(u) = 0 because the flow is laminar; thus, the entire left side of the equation equals zero for this case. On the right-hand side, rho*gx = 0 because gravity only acts in the downward (-y) direction; d/dx(d/dx(u)) = 0 also because the flow is fully developed; d/dz(d/dz(u)) = 0 since we are dealing with infinite plates. Rearranging the two terms that are left, we have:

d/dx(p) = mu*d/dy(d/dy(u))

y-momentum:

rho*[d/dt(v) + u*d/dx(v) + v*d/dy(v) + w*d/dz(v)] = -d/dy(p) + rho*gy + mu*[d/dx(d/dx(v)) + d/dy(d/dy(v)) + d/dz(d/dz(v))]

From the result of our continuity equation examination, we know that all the terms containing a v must be equal to zero. We can quickly see that we are only left with two terms:

d/dy(p) = rho*gy

When both sides of this equation are integrated, we obtain the hydrostatic pressure distribution.

z-momentum:

rho*[d/dt(w) + u*d/dx(w) + v*d/dy(w) + w*d/dz(w)] = -d/dz(p) + rho*gz + mu*[d/dx(d/dx(w)) + d/dy(d/dy(w)) + d/dz(d/dz(w))]

Because our flow is laminar, all terms containing a w or d/dz(...) are equal to zero. Also, we know that the term rho*gz = 0, since gravity is only acting in the downward (-y) direction. This equation quickly reduces to:

0 = 0

Step 3: Then we integrate the reduced Navier-Stokes equations and solve for the unknowns.

z-momentum:

0 = 0 tells us nothing

y-momentum:

p = -rho*gy*y + fp(x)
essentially the hydrostatic pressure distribution, the integration constant in this case, fp(x), is constant in the y direction, but may be a function of x

x-momentum:

(1/mu)*d/dx(p) = d/dy(d/dy(u)) rearranging

(1/mu)*d/dx(p)*y + const1 = d/dy(u) integrating once

(0.5/mu)*d/dx(p)*y*y + const2 = u integrating twice

We now have an expression for each of the variables necessary for a full flow field description (v = 0, w = 0, p = ..., u = ...). To completely finish the problem, we must examine our given boundary conditions to determine

Step 4 (I): We need two boundary conditions since there are two constants of integration that need to be determined. For this case, both plates are stationary; therefore, the no-slip condition tells us that u = 0 at y = 0 and u = 0 at y = H (if we consider the bottom plate to be located at y = 0). Applying the first yields:

const2 = 0

Applying the second gives:

const1 = -(0.5/mu)*d/dx(p)*H

So: u = (0.5/mu)*d/dx(p)*(y*y - H*y)

Step 4 (II): We again need two boundary conditions since there are two constants of integration that need to be determined. For this case, the bottom plate is stationary and the top plate is moving with constant velocity, U; therefore, the no-slip condition tells us that u = 0 at y = 0 and u = U at y = H (if we consider the bottom plate to be located at y = 0). Applying the first yields:

const2 = 0

Applying the second gives:

const1 = -(0.5/mu)*d/dx(p)*H + U/H

So: u = (0.5/mu)*d/dx(p)*(y*y - H*y) + (U*y)/H


We can plot the velocity for both of these cases verses y for specified values of d/dx(p), U, mu, and H; this will give us the velocity distribution of the flow between the plates. Here you can see an animation of velocity profiles for various values of U and d/dx(p). If you'd like to play with this program yourself, you can download it (for free I think) from Wolfram Mathematics.

And that's the Navier Stokes equations in a nutshell! Like I mentioned last week, a lot of assumptions need to be made in order to obtain a version of these equations that we can work with; however, once these have been made, the resulting forms of the equations can be solved rather simply.

If you're feeling extra adventurous, try to go through the procedure again by yourself, examining what the velocity profile looks like for the case where the top plate is moving and bottom one is stationary or one plate is moving in the +x-direction and one is moving in the -x-direction or both plates are moving in the +x-direction at the same speed and at different speeds.

MrsDrPoe: Conservation of Momentum, Part III

Once again Thesis Tuesday is upon us, so we'll be continuing our discussion of the Conservation of Momentum equation. Today we'll change gears a bit, moving toward examination of the differential form of the momentum equation. When I taught this introductory fluid mechanics class, this form of the equations was the most loved and most hated at the same time by my students- loved because it was much simpler to work with and hated because I always made them start from the full form of the (inviscid or viscous) equation and tell me why they ended up with the final result.

Recall that the linear momentum equation is actually a vector equation that is composed of 3 scalar equations (for a 3D problem). The differential (scalar) form of the linear momentum equation is:

d/dt(rho*u) + u*d/dx(rho*u) + v*d/dy(rho*u) + w*d/dz(rho*u) = rho*gx + d/dx(Sxx) + d/dy(Txy) + d/dz(Txz) + sum(Fx)

d/dt(rho*v) + u*d/dx(rho*v) + v*d/dy(rho*v) + w*d/dz(rho*v) = rho*gy + d/dx(Tyx) + d/dy(Syy) + d/dz(Tyz) + sum(Fy)

d/dt(rho*w) + u*d/dx(rho*w) + v*d/dy(rho*w) + w*d/dz(rho*w) = rho*gz + d/dx(Tzx) + d/dy(Tzy) + d/dz(Szz) + sum(Fz)

We've discussed each of the components in these equations before except for the S and T terms. These terms are ones from the stress tensor/matrix- S represents the normal stresses on a fluid due primarily to pressure forces; T represents the shear stresses on a fluid due primarily to the viscosity of the fluid. These stress terms reduce to different forms depending on whether we are dealing with a viscous or an inviscid fluid. Today we'll examine the viscous fluid equations.

The viscous form of the momentum equations is achieved by applying Stokes' postulates:

1) Each stress component is a linear function of strain rate (Newtonian)
2) The properties of a fluid are isotropic (not varying with direction)
3) when all strain rates are equal to zero, hydrostatic pressure is the only stress present

and Stokes' assumption:

lambda = -(2/3)*mu, where lambda is the bulk viscosity (related to the volumetric dilatation of the fluid) and mu is the viscosity

These allow the stress terms in the equations above to be rewritten as:

d/dx(Sxx) + d/dy(Txy) + d/dz(Txz) = -d/dx(p) + mu*(d/dx(d/dx(u))+d/dy(d/dy(u)) + d/dz(d/dz(u)))

d/dx(Tyx) + d/dy(Syy) + d/dz(Tyz) = -d/dy(p) + mu*(d/dx(d/dx(v))+d/dy(d/dy(v)) + d/dz(d/dz(v)))

d/dx(Tzx) + d/dy(Tzy) + d/dz(Szz) = -d/dz(p) + mu*(d/dx(d/dx(w))+d/dy(d/dy(w)) + d/dz(d/dz(w)))

Placing these expressions in the initial differential form of the momentum equation gives us the Navier Stokes Equations. What these scalar equations physically represents is:

Left side- "mass times acceleration" of a fluid particle (per unit volume)

Right side-
First term- force of gravity of a fluid particle per unit volume; if positive, this terms acts to accelerate the particle in the +x-direction

Second term- pressure difference across the fluid particle; if negative, than this term acts to accelerate the particle in the +x-direction

Third-fifth terms- viscous stresses acting on the fluid particle; if sum is positive, than this group acts to accelerate the particle in the +x-direction

Sixth term- any other forces (support, etc.) per unit volume acting on the fluid; this term is often neglected

By combining these equations with the continuity equation, we can provide a complete mathematical description of the flow of an incompressible, Newtonian fluid: 4 equations allows us to solve for 4 possible unknowns (p, u, v, and w). If we have exactly four unknowns, our problem is well-defined. Fewer and our problem is over-defined; more (unknown external forces, etc.) and our problem is ill-posed and we cannot solve it unless more information is obtained.

These equations may seem a bit intimidating, but remember, we can always employ simplifying assumptions in order to obtain a more manageable set. In fact, that's what we'll look at next week- application of the Navier Stokes Equations. So until next time, happy studying!

MrsDrPoe: Conservation of Momentum, Part II

It's that time again folks! For this Thesis Tuesday, we will continue our discussion on the conservation of momentum equation by looking at a brief example.

Consider water flowing through a 20 deg bend shown below at a rate of 0.025 m3/s. The flow is frictionless, gravitational effects are negligible, and the pressure at section 1 is 150 kPa and section 2 is 74.5 kPa. Determine the x and y components of force required to hold the bend in place.


Alright...let's organize the problem from the given information in the statement, the original figure (not shown), and a fluid property table:

Given: theta = 20 deg, Q1 = 0.025 m3/s, P1 = 150 kPa, P2 = 74.5 kPa, rho = 999 kg/m3, D1 = 0.1 m, D2 = 0.05 m
Find: Fx, Fy
Assume: steady, laminar, incompressible, uniform flow, frictionless, neglect gravity

Now onto the solution (employing the process we outlined last week):

Step 1: We must first choose a CV, keeping in mind the criteria for the "best" CV. In the figure above, the CV we will use is represented by the dashed line- note that it is perpendicular to flow into and out of the bend and that it cuts through the bend support.

Step 2: Next, we must apply the continuity equation:

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

For steady (d/dt(...) = 0), incompressible (rho = constant), uniform (V = constant) flow with 2 in/out surfaces:

rho*int(V1.n1)dA1 + rho*int(V2.n2)dA2 = rho*|V2|*A2 - rho*|V1|*A1 = 0

Since we know the volumetric flowrate (Q1) and bend diameter at 1, we can find the velocity at 1:

|V1|= Q1/(0.25*pi*D1*D1)

Similarly, since we know the diameters at 1 and 2, we can determine the areas:

A1 = 0.25*pi*D1*D1 (etc.)

Thus, we can solve for the velocity at surface 2:

|V2|= |V1|*D1*D1/(D2*D2) = 12.732 m/s

Step 3: Then we look at the external forces acting on the bend. In this case, we have two pressure forces and the support force of the bend (which we would not "see" if our CV did not intersect the support apparatus):

Fp1 = (p1*A1 0 0)
Fp2 = (Fp2x Fp2y 0) = (-p2*cos(theta)*A2 p2*sin(theta)*A2 0)
Fs = (Fsx Fsy Fsz)

Step 4: Next, we must look at the left side of the linear momentum equation for steady, uniform, incompressible flow with 2 surfaces:

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

rho*int(V1*(V1.n1))dA1 + rho*int(V2*(V2.n2))dA2

-rho*int(V1*|V1|)dA1 + rho*int(V2*|V2|)dA2

-rho*V1*|V1|*A1 + rho*V2*|V2|*A2

Step 5: Finally, we combine the right and left sides of the momentum equation, splitting it into scalar components (remember: this equation is a vector equation!), and solve for the unknown support forces:

Vector Eqn:

-rho*V1*|V1|*A1 + rho*V2*|V2|*A2 = Fs + Fp1 + Fp2

X-Momentum:

-rho*|V1|*|V1|*A1 + rho*|V2|*|V2|*cos(theta)*A2 = Fsx + rho*A1 -p2*cos(theta)*A2

Fsx = -821.322 N

Y-Momentum:

-rho*|V2|*|V2|*sin(theta)*A2 = Fsy + p2*sin(theta)*A2

Fsy = -158.79 N

Z-Momentum:

0 = Fsz

While it may seem a bit complicated at first, I can assure you that the more problems you work with this outlined process, the easier it becomes. Please be sure to comment with any questions you may have, and stay tuned for more momentum madness next week!

MrsDrPoe: Conservation of Momentum, Part I

It's another Thesis Tuesday on the blog! Today we'll be looking at the second conservation equation that governs fluid mechanics: Conservation of Momentum (AKA Newton's 2nd Law or the Linear Momentum Equation).

In Lagrangian terms, this equation simply states that the sum of the forces acting on an object is equal to the mass of the object times its acceleration (or the time rate of change of its linear momentum). Recall that we wish to employ an Eulerian methodology; therefore, we must use Reynolds Transport Theorem to convert this equation.

For this case, our extensive (mass dependent) property is linear momentum, and our intensive (mass independent) property is linear momentum per unit mass, or density times velocity. Combining the result from the Reynolds Transport Theorem with the Lagrangian form of the conservation of momentum equation, we get the sum of the forces acting on the control volume is equal to:

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

Here, F is a force, rho is the density, V is the velocity, and n is the unit outward normal vector.

The first thing we should notice about this equation is that it is a vector equation and can thus be written as 1, 2, or 3 equations depending on how many dimensions our problem has. For a 3D problem:

sum(Fx) = d/dt(int(rho*u)dV) + int(rho*u*(V.n))dA

sum(Fy) = d/dt(int(rho*v)dV) + int(rho*v*(V.n))dA

sum(Fz) = d/dt(int(rho*w)dV) + int(rho*w*(V.n))dA

The second thing we should notice is the similarity between these equations and the conservation of mass equation. Each one has something equal to a term tracking how our intensive property changes with time in the control volume plus a term tracking how much of the intensive property is crossing the control surface (into or out of the control volume).

We will employ the following methodology to a fixed, non-deforming control volume:

1) Choose the best CV, establishing all bounds and known conditions
2) Apply the conservation of mass equation
3) Determine all forces acting on the CV
4) Evaluate the right side of the conservation of momentum equation in terms of unknown and known variables
5) Combine both sides of the conservation of momentum equation and solve for the unknowns

Now, let that stew, and tune in next Tuesday for some fun examples.