A U-tube shown in figure is filled with a liquid of specific gravity 1.25 to a height of 15cm in both the limbs. It is rotated about a vertical axis 15cm from one limb and 30cm from the other. If the speed of rotation is 60rpm, find the difference in the liquid levels in the two limbs. Also find the pressure at points M and N at the base of U-tube.

Problem Statement

A U-tube is filled with a liquid of specific gravity 1.25 to a height of 15 cm in both limbs. It is rotated about a vertical axis 15 cm from one limb and 30 cm from the other. If the speed of rotation is 60 rpm, find:

1. The difference in the liquid levels in the two limbs after rotation
2. The pressure at points M and N at the base of the U-tube

Given Data

Solution

1. Convert units to SI and calculate angular velocity

   \( r_1 = 0.3 \text{ m} \)
   \( r_2 = 0.15 \text{ m} \)
   \( N = 60 \text{ rpm} \)
   
   \( \omega = \frac{2\pi N}{60} = \frac{2\pi \times 60}{60} = 2\pi = 6.28 \text{ rad/s} \)

2. Calculate rise in liquid levels using parabolic equation

   For a rotating fluid, the surface forms a paraboloid of revolution. The elevation at any radius is:

   \( z = \frac{\omega^2 r^2}{2g} \)
   
   At left limb (r₁ = 0.3 m):
   \( z_1 = \frac{\omega^2 r_1^2}{2g} = \frac{(6.28)^2 \times (0.3)^2}{2 \times 9.81} = \frac{3.55}{19.62} = 0.181 \text{ m} \)
   
   At right limb (r₂ = 0.15 m):
   \( z_2 = \frac{\omega^2 r_2^2}{2g} = \frac{(6.28)^2 \times (0.15)^2}{2 \times 9.81} = \frac{0.887}{19.62} = 0.045 \text{ m} \)

3. Calculate the difference in liquid levels

   \( \text{Difference} = z_1 – z_2 = 0.181 – 0.045 = 0.136 \text{ m} = 13.6 \text{ cm} \)
   The difference in liquid levels in the two limbs is 13.6 cm.

4. Calculate the actual heights after rotation

   We need to account for mass conservation. The sum of the heights changes due to rotation, but the total volume remains constant.

   \( \text{Sum of elevations due to rotation} = z_1 + z_2 = 0.181 + 0.045 = 0.226 \text{ m} \)
   
   \( \text{Initial total height} = 0.15 + 0.15 = 0.3 \text{ m} \)
   
   \( \text{Difference} = 0.3 – 0.226 = 0.074 \text{ m} \)

   This difference is equally distributed between the two limbs:

   \( \text{Adjustment per limb} = \frac{0.074}{2} = 0.037 \text{ m} \)
   
   \( \text{Actual level in left limb} = 0.15 + 0.181 – 0.037 = 0.294 \text{ m} \)
   \( \text{Actual level in right limb} = 0.15 + 0.045 – 0.037 = 0.158 \text{ m} \)

5. Calculate pressures at points M and N

   The pressure at any point in the fluid is calculated using:

   \( P = \rho g h = SG \times \rho_{water} \times g \times h \)
   
   \( \text{Pressure at N} = 1.25 \times 9810 \times 0.218 = 2673 \text{ Pa} \)
   \( \text{Pressure at M} = 1.25 \times 9810 \times 0.082 = 1005.5 \text{ Pa} \)
   The pressure at point N is 2673 Pa.
   The pressure at point M is 1005.5 Pa.

Discussion

When the U-tube rotates, the centrifugal force causes the liquid to rise higher in the limb that is farther from the axis of rotation (left limb in this case) and lower in the limb closer to the axis (right limb). The parabolic surface equation derived from the balance of pressure gradient and centrifugal force allows us to calculate the exact difference in heights.

Applications

This principle of fluid behavior under rotation has several practical applications:

• Centrifugal pumps and compressors
• Rotating machinery with fluid cooling systems
• Laboratory centrifuges for separation processes
• Design of rotating vessels in chemical processing
• Analysis of fluid behavior in rotating spacecraft