Engineering 41: Thermofluid Mechanics
Professor Fred Orthlieb
Laboratory # 1
Prepared by
September 21, 2004
It can be shown that the cross sections of a uniformly rotating fluid within a vertical cylindrical container are a family of parabolas centered on the rotational axis. If z and r represent the vertical and radial coordinates, and the rotational axis is the cylinder centerline, constant pressure surfaces are given by
z = w^2 / (2g) * r^2 + z0 (1)
where z0 is the height of the fluid at the center of rotation.
Rewriting the above expression, one can determine the angular rate of rotation given by
w = sqrt( 2g / r^2 * (z- z0 ) ) (2)
In lab, we took cross sections of the spinning fluid at two different speeds—“slow” and “fast”—and were able to mark off the silhouettes. Using Microsoft Excel, the points were plotted on a scatterplot, and a parabolic regression equation was used to test the goodness of fit. The graphs are shown below
Note how well the regression lines characterize the data set as evident in the high correlation coefficient, R 2 . The constant pressure surfaces of the rotating fluid are indeed a family of parabolas.
Using equation (2) and plugging in the appropriate values, we obtain angular speeds, which are shown below in the table.
Speed |
Theoretical ? [rad/sec] |
Measured ? [rad/sec] |
Slow |
1.2879 |
.8812 |
Fast |
4.4793 |
1.1877 |
It appears that at lower speeds, the theoretical and measured angular speeds are relatively closer than at higher speeds. This can be accounted for by friction of the air fluid above the spinning fluid. At low speeds, the friction from the air has less effect on the surface of the spinning fluid than at high speeds, causing a wide discrepancy between the measured and calculated angular speeds.
Note that the result does not depend on the fluid density at all. The theory applies to all fluids.
The Rayleigh problem solution of a fluid in rotation can be given in terms of the Complementary Error Function:
u(z, t) = Uerfc(?) = 1 – erf(?)
Rewriting the above with the Complementary Error function substituted with its linear approximation for small ?,
(Courtesy of Aron Dobos)
We can test how well the Rayleigh problem solution applies by considering various z values and exploring the time needed for the fluid to reach 95% of its final velocity value.
Consider z = 1 cm. Noting that the kinematic viscosity [v = µ/?] of water at room temperature is about 10 -6 m 2 /second, the time is 12732 seconds or 212 minutes. And considering z = 7 cm, the time increases to 623887 seconds or 173 hours! Our fluid was at an even higher z-level, and the solution suggests an even longer time for the fluid to reach final speed. Clearly something is unaccounted for when we apply the Rayleigh problem solution, considering we saw that the fluid took roughly 6 minutes to speed up to its final speed.
The secondary flow in the spinning fluid could possibly be the cause of the discrepancy. The Rayleigh problem solution seems to not account for the interaction of the fluid, which can be seen in the diagram shown in the following section. Every fluid layer is thus subject to speeding up rather than having to wait for the bottom layer to speed up the above layers one by one.
The fluid at the z = 0 layer started spinning first since it is in contact with the container. The layer above that started spinning due to the first layer, and this chain reaction continued until the entire fluid was spinning as a solid body. The fluid did not all spin together instantly. This created multiple layers in the fluid spinning at different speeds, with the bottom layer spinning fastest and the top the slowest initially. Similarly, the layers near the tank walls started spinning with the tank first, while the other radial layers soon followed. Over time, the speeds began to match, but interesting events occurred during the process.
Various marking objects were placed in the rotating fluid including party glitter, colored dyes, and potassium permanganate. The glitter tended to surface, while the dyes and the potassium permanganate drifted toward the bottom of the tank at various speeds. As a result, we were able to witness the interaction of the objects with the fluid layers.
On the bottom of the tank the fluid wanted to force the markers to the outside of the tank, while the top most layer forced them to the inside. This is most evident at the higher speed, due to the increasing parabolic nature of the fluid surfaces as gravity wants to bring the objects downward. The phenomena can be described by the fact that a secondary flow is occurring as shown below.
Imagine the parabolic shape of the fluid in the tank. Gravity can be seen as the agent responsible for the inward flow at the top layer since markers gravitate downward to the center of the tank from a high fluid elevation to a lower one. The agent responsible for the fluid flow outward at the bottom layer is the tangential acceleration due to the spinning tank. A situation of an object on a merry-go-round is an appropriate analogy—the object will fly off unless it holds on tight to the handlebars.
In the layers between the top and bottom one, we observed that the sinking markers produced a slant path as shown in the above diagram. This justifies the existence of the secondary flow in the rotating fluid.
Looking at the markers from a radial perspective, the objects seemed to have spun faster when closest to the tank and slower on the inside. This makes sense in that the inner layers are the last to rotate uniformly with the tank.
Recalling that the acceleration a of a fluid particle of mass m is due to the sum of all applied forces—in this case, pressure and viscous forces. In other words,
ma = f pressure + f viscosity
The bottom layer has higher pressure than other layers of the fluid, which explains why the markers at the bottom layer were spinning more forcefully and faster.
When the tank was suddenly stopped from spinning, the fluid initially spun violently due to the friction it encountered on the tank walls. But it kept on spinning, with the bottom layer the first to slow down, causing the upper layers to follow suit soon after.
Cubes of various materials with specific weights less than that of water will float in different orientations on an undisturbed water surface as shown in the below diagram.
As long as the buoyant force is equal to the weight of the object submerged, the object will float. Depending on the density of the cube in relation to that of water, more or less of the cube's volume will be submerged. An object with a density close to half that of water will sit nicely on top of the fluid, and prefer orientation A. But as the density increases, the object begins to sink, still preferring the same orientation. At a critical density, the object begins to tilt and prefers orientation B with a pyramidal volume above surface.
The principle of a floating object is that it is stable at a position if it returns to that position after a small perturbation. Consider the diagram below where C is the center of buoyancy of the cube. The stability of an object depends on the difference between its metacenter (M)—determined by the intersection of the buoyant force before and after the tilt—and its center of gravity (G). The height GM is called the metacentric height as shown in the below diagram. For stability, GM must be greater than zero. As GM increases, the stability (restoring force after a slight disturbance) increases.
GM is greatest when the cube is not submerged, but begins to decrease as the block sinks. That explains why the cube prefers A when it sits on the surface, and prefers B when it is mostly submerged. Again, the amount of submersion depends on the relative density of the object to that of the fluid, in our case water at room-temperature.
After solid-body rotation of the water has been established, the cubes were placed in it. It appeared that the orientations of the cubes in still water were maintained in the spinning fluid, though they were now oriented normally to the parabolic shape of the fluid.