Question 2.15: A U-tube with a radius of 10 in and containing mercury to a ...

Convert the angular velocity to radians per second:

From Table 2.1 we find for mercury that \gamma = 846 lbf/ft^3 and hence ρ = 846/32.2 = 26.3 slugs/ft^3. At this high rotation rate, the free surface will slant upward at a fierce angle [about 84°; check this from Eq. (2.47)], but the tubing is so thin that the free surface will remain at approximately the same 30-in height, point B. Placing our origin of coordinates at this height, we can calculate the constant C in Eq. (2.45) from the condition p_B = 2116 lbf/ft^2 at (r, z) = (10 in, 0):

or

C = 2116 – 3245 = -1129 lbf/ft^2

p = const  –  \gamma z + \frac{1}{2}\rho r^2 \Omega^2               (2.45)

z = \frac{p_0 – p_1}{\gamma} + \frac{r^2 \Omega^2}{2g} = a + br^2                   (2.47)

We then obtain p_A by evaluating Eq. (2.46) at (r, z) = (0, -30 in):

p = p_0  –  \gamma z + \frac{1}{2}\rho r^2 \Omega^2               (2.46)

This is less than atmospheric pressure, and we can see why if we follow the free-surface paraboloid down from point B along the dashed line in the figure. It will cross the horizontal portion of the U-tube (where p will be atmospheric) and fall below point A. From Fig. 2.23 the actual drop from point B will be

h = \frac{\Omega^2 R^2}{2g} = \frac{(18.85)^2(\frac{10}{12})^2}{2(32.2)} = 3.83 ft = 46 in

Thus p_A is about 16 inHg below atmospheric pressure, or about \frac{16}{12}(846) = 1128  lbf/ft^2 below p_a = 2116  lbf/ft^2, which checks with the answer above. When the tube is at rest,

Hence rotation has reduced the pressure at point A by 77 percent. Further rotation can reduce p_A to near-zero pressure, and cavitation can occur.