week6 - 3.5 Flow past a sphere z θ n R One of the most...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 3.5 Flow past a sphere z θ n R One of the most fundamental problems of fluid mechanics is to understand the flow around obstacles, which stand in the way of a flow. Equivalently (think of Galilean invariance) one can think of a body moving inside a fluid which is at rest (for example an aeroplane). Let us begin studying this problem by considering the perhaps simplest possible body, a sphere of radius R . The sphere is placed inside a uniform stream, which we choose in the direction of the z or symmetry axis: u = U ˆ z . Far from the body, the flow will be uniform. The sphere introduces a perturbation to the flow which decays as r → ∞ , i.e. as one moves away from the body. According to the superposition principle (the Laplace equation is a linear equation), this perturbation is to be added to the potential φ = Uz describing the uniform flow. The simplest such flow we know is a source, with potential φ ∝ 1 /r . However, such a flow cannot describe the physical situation at hand. Far from the sphere, the total mass flow through a closed control surface should be zero, not finite as for a source. Another way of putting it is that the flow field of a source decays too slowly (like 1 /r 2 ) away from the body. This can be helped by taking the derivative of the source solution. Taking the deriva- tive in the μ-direction, one arrives at the potential φ = μ ·∇ 1 r = − μ · x r 3 , which is called the dipole flow (why?). The dipole is oriented in the μ-direction and has strength | μ | . This flow is also a solution of Laplace’s equation, since △ parenleftbigg μ ·∇ 1 r parenrightbigg = μ ·∇△ parenleftbigg 1 r parenrightbigg = 0 . In summary, our ansatz for the potential is φ = Uz + μ · x r 3 . The velocity field is u i = ∂ i φ = Uδ i 3 + ∂ i μ j x j r 3 = Uδ i 3 + μ i r 3 − 3 μ j x j x i r 5 or u = U ˆ z + 1 r 3 parenleftBig μ − 3 μ · x x r 2 parenrightBig . Now we have to satisfy the boundary condition on the surface of the sphere, which is u · n = 0. The normal vector to the sphere is n = x /r . Now u · n = U z r − 2 μ · x r 4 , and the choice μ = UR 3 2 ˆ z will guarantee that this is zero for r = R . In summary, the solution we seek, which satisfies all the boundary conditions is φ = Uz + UR 3 2 z r 3 , (30) and the velocity field is u = U ˆ z + U 2 parenleftbigg R r parenrightbigg 3 parenleftBig ˆ z − 3 z r n parenrightBig . (31) Now we want to calculate the pressure field, and in particular the pressure on the surface of the sphere. This will permit us to calculate the total force on the sphere. On the surface we have u = 3 U 2 parenleftBig ˆ z − z r n parenrightBig , so u 2 = 9 U 2 4 parenleftbigg 1 − z 2 r 2 parenrightbigg ....
View Full Document

This note was uploaded on 01/20/2012 for the course MATH 33200 taught by Professor Eggers during the Fall '11 term at University of Bristol.

Page1 / 10

week6 - 3.5 Flow past a sphere z θ n R One of the most...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online