Boundary Conditions -

In today’s lesson, we will be discussing the following topics:

1. Introduction to Boundary Conditions

2. Common Boundary Conditions

3. Example 1 – Using Boundary Conditions to find temperature as a function of radius

4. Example 2 – Using Boundary Conditions to determine unknown constants

video coming soon

Introduction to Boundary Conditions

The three main equations of fluid motion that were derived in the first chapter are given below. These equations contain a total of five unknowns: density (rho), V, p, û, and T.

Knowing in mathematics that we need as many equations as the number of unknowns to solve a system of equations, we would need five equations here in order to solve for the five unknowns. We have three equations, but we will find two more depending on the system and situation we are trying to solve the variables for. This is where knowing more information through the boundary conditions can come into play when solving for a variable.

So how do we solve a system using boundary conditions? Here are some rules:

(a) If the flow is unsteady, there must be a given initial condition for each variable (at time = 0)

(b) After the initial time, there must be other information given about each variable (at time > 0)

Common Boundary Conditions

There are some common boundary conditions that come up. Boundary can specify a liquid-gas, solid-gas or solid-liquid. Note that when we say fluid, it can refer to both a liquid or a gas.

Let us review these common boundary conditions:

(1) Solid Wall

When fluid particles are against a solid wall (these are the boundary layer), the velocity and temperature of the fluid at that boundary is the same as that of the wall, since there is no slip or temperature change between the wall and the fluid. Note that there is one exception for a rare type of gas flow where this condition is not true.

Note that the above term boundary layer means a thin layer of fluid which is “sticking to” a solid surface. This is where the velocity of the fluid right at the wall is zero because of this boundary condition.

Using this knowledge, the two equations below will be true:

Another term you should be familiar with is thickness of the boundary layer. This is a term that refers to the distance away from the solid wall (or the height above the surface) where the velocity reaches 99% of the freestream (or approach) velocity. These velocities are those that are “free,” or far away from the surface where the velocity is not influenced by the solid wall. This is the symbol that defines the boundary layer thickness:

(2) Inlet or Outlet

When fluid particles are at the inlet or outlet of the system being analyzed, the velocity, pressure, and temperature should be given. These are usually at +/- infinity, which simulates a body immersed in an infinite amount of fluid.

(3) Liquid-Gas Interface

At a liquid-gas interface (otherwise known as a free surface), the following conditions apply:

Vertical Velocity: The vertical velocity, w, must be equal on both the liquid and gas sides of the interface (position), which is the sideways n. The vertical velocity of the liquid is equal to the vertical velocity of the gas. This is then equal to the derivative of the interface (position) with respect to time (since velocity is displacement or position divided by time). This then further splits into the derivative of the velocity components in each direction. It gives the following equation:

In many cases, the equation can be simplified without the components:

This basically means that the velocity tangent, as well as normal to the interface, will be continuous. These are known as the Kinematic Boundary Conditions. The kinematic conditions are only concerned with relating the motion of the interface to the liquid and gas velocities at the surface.

The normal vs tangential velocity flow vectors between liquid-gas interface

The Dynamic Boundary Condition are those that are concerned with the force balance at the free surface. These conditions require that the stress (tau) is continuous across the free surface that separates the gas and liquid. This means that the traction exerted by the liquid onto the gas is the same as the traction exerted by the gas onto the liquid.

Mechanical Equilibrium – Viscous Shear Stress: The shear stress tau (slanted T) of the liquid and the gas need to balance. This gives the following equations:

Pressure: The pressures must balance at the interface, to give the following equation:

This equation can be rewritten as the following:

In many cases, for example open-channel flow, the pressure can be reduced to the following. Here, the pressure of the liquid and gas is about equal to atmospheric pressure:

Heat Transfer: The heat transfer, q, must balance at the interface. Neglecting radiation, we get the the following equation.

Replacing q with an equivalent equation gives:

This image below shows how the three common boundary conditions can be applied to a system:

Example 1 – Using Boundary Conditions to find temperature as a function of radius

We are being asked to find T(r). In the equations for incompressible flow with constant properties, the energy equation is the only one with a temperature (T) in it. Therefore, we are using this equation. The equation given in the question is a version of the simplified energy equation below (4.75).

Follow the steps below with explanations in the caption to get the answer. Note that you will need to know general derivative and integral rules in order to do this:

Cancel the left side of the equation after setting Vr = 0, as Vr = 0 is given in the question

Simplify the equation to move the term with viscosity over to one side of the equation

Replace Vz with the equation given in the question

Take the derivative of the term where you replaced Vz, with respect to r

Take the derivative of this part of the equation as shown above, with respect to r. Keep U and R constant. Note that the derivative of the first U term alone is 0.

Square the answer once you get the derivative. There was a squared on the outside of the brackets from earlier. Make sure to add in the viscosity u in the equation at the end, as it was there in the beginning before we took the derivative

Move the (k/r) to the other side of the equation, then integrate both sides with respect to r (add dr to the end)

On the left side of the equation, taking the derivative (d/dr) and then integrating with respect to dr cancels out. On the right side of the equation, combine the *r^2* and r to make *r^2*. Integrating this side adds r^4/4, as highlighted in red

Simplify the equation. Note that you need to add a C1 term on the right side after integrating.

Bring the r on the left side of the equation to the right. Then integrate both sides with respect to dr again.

Dividing the right side by r means it goes into both terms, including the C1. This made it *C1/r,* and integrating it using integration rules makes it *C1* ln(r).* (This is just a generic integral rule if the variable is in the denominator). You will also need to add a C2 to the end after integrating for the second time

Simplify the equation to get the final answer for T. Now we have to make it only a function of r *(T = T(r))*. This means we need to find the constants C1 and C2.

Now we can apply the boundary condition of a solid-fluid interface, since the liquid (fluid) is inside a pipe (solid). These were the conditions at that interface, as mentioned above:

Using a radius of zero (r = 0), this would mean that we are right at the edge of the pipe. This is how the picture would look:

If we use r = 0 in the final equation for T that we found, it would have an ln(0) in one of the terms. ln(0) gives infinity, which would not work unless C1 is also 0. Thus, we will take C1 to be zero. Now in the question, it was said that the radius of the pipe is R. In the energy equation, r is radius. So we can use r = R since the question told us that the radius variable should be equal to R.

Simplify the equation by cancelling out C1 and replacing r with R. Both the R^4 then cancel out, and the fractions are multiplied so they can be combined as I have done.

It is given in the question that T = Tw, so replacing that as well as isolating C2 on one side of the equation gives us the answer for C2.

Using the final equation for T that was found earlier (before we found C1 and C2), then substituting in C1 and C2 (C1 = 0 and C2 is the equation we just found in the last step) gives this final equation circled in red. The equation right above the circled one is also correct, but in the second step here I have just factored out a common term.

Example 2 – Using Boundary Conditions to determine unknown constants

For this example, we are doing almost the same thing as the question above. However, we don’t need to solve for anything further aside from C1 and C2. In Example 1, we solved for C1 and C2, then substituted them back in to find T(r). This one will be easier! Follow the steps to get the answer:

(1) Assume steady and incompressible flow, as we usually do.

(2) Pick conditions of y = 0 and y = δ. This is because we were given the information regarding y = δ, and we know that at y = 0, the velocity is also 0 because of the solid-fluid interface. Substituting these into the equation above, we get the following: