I am investigating on the terminal velocity of steel balls moving in water. I used the balls with different diameters, such as 3.17mm, 6.02mm, etc. And I used a full-filled 1L graduated cylinder. I dropped the ball below the water surface. And I calculate the terminal velocity by a program called Logger Pro.
Theoretically, the terminal velocity of steel ball with diameter 3.17mm in water is about 37m/s. But in my experiment, it is only 0.7m/s. The difference is really huge.
And also, I got a graph of terminal velocity as a function of square of radius of steel ball. It is a straight line, but it does not go through origin.
Can anyone help me with that?
Answer
Stokes Law is not going to apply in this situation because the water flow around the ball will be turbulent not laminar. The way to see this is to calculate the Reynold's number. For a sphere this is approximately given by:
$$ Re \approx \frac{\rho_wdv}{\mu} $$
If we feed in $\rho_w = 1000$ kg/m$^3$, $d = 0.00317$ m, $v = 37$ m/s and $\mu = 0.001$ Pa.s then we get $Re \approx 117290$. A Reynold's number of greater than one means the flow is turbulent, and since Re for a $3.17$ mm ball travelling at $37$ m/s is far greater than one we conclude that the flow is turbulent and Stoke's law doesn't apply.
In a turbulent regime the viscous drag is given by:
$$ F_d = \tfrac{1}{2}\rho C_d A v^2 $$
where $\rho$ is the density of the fluid (i.e. water), $A$ is the frontal area of the object, $v$ is the velocity and $C_d$ is a dimensionless parameter called the drag coefficient. The value of $C_d$ depends on the Reynold's number (so it isn't really a constant) but it's generally of order one.
No comments:
Post a Comment