Continuing on from last week, we looked at water waves, and I will start with the example that we did last week. We have to make assumptions with this:

• v=0; and the partial derivative with respect to y=0.
• It is inviscid.
• It is incompressible.
• It is irrotational. If it’s incompressible and irrotatinal, this implies that nabla^2*Phi=0.
• g is the only body ‘force’.
• no surface tension.
• the displacement and the velocities are small (ignore products).
• water at the free surface stays there.
• Starts from rest.

We use these equations to find the solution for both finite depth and infinite depth: This is equation 4.1 This is equation 4.6 This is equation 4.5 This is equation 4.3

We can work out the solution by following these steps: We can now imply the boundary conditions: Implying that B=0, we get the solution to be: Dispersion Relation

If we use equation 4.6 we get: this implies that w^2=gk and this is equation 4.9.

Surface Displacement

If we use equation 4.3 we get: Potential

The potential is: Phase Speed

The phase speed is the speed at which the phase of a wave is propagated, the product of the frequency times the wavelength.

In our solution, this is given by: c=w/k.

The following is the example of finite depth:  This example is to do with infinite depth:    This week, going through the examples I’m not sure I fully understand it, so I aim to go back over the examples till I am 100% sure I know where we get the solution from and the stages at each step.

Sources:

Phase speed definition: http://www.thefreedictionary.com/phase+speed

This week we looked more into potential flow as well as doing some examples.

We looked into a Doublet, which was 2-dimensional and steady. The streamline that was given is: To find the isopots of this, we have to find: and: with some manipulation we can get this in the form of x^2+y^2=r^2 As we can see from this, the circle is going to be centered around and with radius .

If we plot the streamlines and isopots on the same graph, we get: We can see these cross at right angles, this seems to be the case for every stream function.

Now we looked at Bernouilli’s equation in potential flow and how to derive it.

This is here: We also looked at another example with water waves.

This week I enjoyed more since we were deriving new equations. I think I will need to go over this week thoroughly so that I understand fully the derivation.

This week we will be going through some examples where we can apply Bernouilli’s equation. We were first looking at a flow around a cylinder. Before we could do anything, we have to make observations of this. We need to see if:

• Flow tangential to cylinder.
• Streamlines = Solid surfaces in inviscid flow.
• Neglect boundary layer effects due to viscosity.

We are given the stream function: If we play around with this equation, getting terms for Ur and Utheta when r=a:  We can now substitute this into Bernouilli’s equation which is: We get: Ps is the stagnation pressure at the point where u=0.

We have also looked at Potential flow this week. This describes the velocity field as the gradient of a scalar function; the velocity potential.

Sources:

Picture and definition. http://en.wikipedia.org/wiki/Potential_flow

Example sheet 4, question 1: