{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

EAS4101_S11_HW4S - 1 Compare a...

Info iconThis preview shows page 1. Sign up to view the full content.

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

Unformatted text preview: 1. Compare a vortex of strength , to another vortex defined as , where is a rotational speed. Sketch the velocity profiles. Discuss whether or not the flows are well behaved for all . Compute the vorticity fields and compare them. Do the flows possess any strain rate? What is and are viscous forces important? Justify your answer mathematically and physically. How does the circulation of both flows depend on the radius (i.e., evaluate the circulation for ). Solution to problem 3: Velocity profiles: (figures from Power’s notes) • The ideal irrotational vortex, , is infinite at the origin, but well behaved everywhere else. • The ideal rotational vortex, infinity. , goes to infinity as the radius goes to Vorticity fields: • The only possible vorticity field for these flows reduces to . So, , hence the name ideal irroational vortex (IIV). From problem number 1, we know that this flow is irrotational everywhere except at the origin where there is infinite vorticity. • Now, Strain rate and • , hence the name ideal rotational vortex (IRV). : The only possible strain rate for these flow reduces to . So, , so the IIV possesses a non ­zero strain rate. Now, , so and , therefore . So, while the IIV possesses a finite strain rate and shear stress, it is inviscid because . • Now, . Therefore, the IRV possesses zero strain rate and shear stress, and is also inviscid. Physically, this makes sense since the IRV represents pure rotation, therefore zero straining. Circulation: • The circulation is defined as . As per problem 1, , which means that the IIV possesses uniform circulation regardless of the integration path, provided that the singularity is included. • , which means that the circulation for an IRV increases as the square of the radius. Given: Static pressure tap drilled into the surface of a wind tunnel model under two defferent scenarios shown in schematics. (a) A burr is formed prior to the hole deflecting the streamlines upward over the hole. (b) A burr is formed after the hole. Which situation will result in a measurement error to due a pressure increase and which one will result in a decrease? Explain your answer both mathematically and physically Find: Schematics: Solution: An accurate measurement of static pressure requires that none of the fluids kinetic energy be converted into a pressure rise at the point of measurement (tap a and tap b in our case). This requires a smooth hole with no burrs or imperfections. As indicated in the schematics such imperfections can cause the measured pressure to be greater or less than the actual static pressure according to the Euler ­n equation, This indicates that the pressure increases with distance away from the center of the curvature. The positive n direction points towards the “inside” of the curved streamline (a) The first case will have a decrease in pressure. This occurs because the curvature of the streamlines will increase at the point of measurement. (b) The second case will have an increase in pressure. This occurs because the curvature of the streamlines will decrease (convex appearance). Given For An inviscid flow of water over the surfaces shown in the schematics. each case, the depth h of the water at section D ­D is the same and the velocity V of the water is the same. Gravity points downward. The radii of curvature is for the convex and concave case is R. Find: Schematics: *Radius of curvature R for the first two graphs goes from the center of circle to the black dot Solution: (a) Euler’s equation normal to the streamline is written for steady flow as If we then multiply it by we obtain and integrate across the streamline (in the n direction) Plugging in the integral limits the equation becomes Solving the integral Since is the outer limit and is the inner limit then The equation becomes Solving for (b) We begin with the integral (NOTE CHANGE IN COORDINATE SYSTEM) Solving the integral Since is the inner limit and is the outer limit then The equation becomes Solving for (c) We begin with the integral Since there is no curvature in the streamline the equation reduces to Solving the integral Or ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online