A float valve regulates the flow of oil (sp. gr. 0.8) into a cistern. The spherical float is 15 cm in diameter. A weightless link AOB carries the float at B and a valve at A. The link is hinged at O, with ∠AOB = 135°. OA = 20 cm and OB = 50 cm. When flow is stopped, AO is vertical and the oil surface is 35 cm below the hinge. A force of 9.81 N is required on the valve to stop the flow. Determine the weight of the float.

Float Valve Equilibrium Problem

Problem Statement

A float valve regulates the flow of oil (sp. gr. 0.8) into a cistern. The spherical float is 15 cm in diameter. A weightless link AOB carries the float at B and a valve at A. The link is hinged at O, with ∠AOB = 135°. OA = 20 cm and OB = 50 cm. When flow is stopped, AO is vertical and the oil surface is 35 cm below the hinge. A force of 9.81 N is required on the valve to stop the flow. Determine the weight of the float.

Given Data

  • Sp. gr. of oil = 0.8, \( \rho_{oil} = 800 \, \text{kg/m}^3 \)
  • Float Diameter, D = 15 cm (Radius, R = 7.5 cm = 0.075 m)
  • Valve force, \( P = 9.81 \, \text{N} \)
  • Length OA = 20 cm = 0.2 m
  • Length OB = 50 cm = 0.5 m
  • Vertical distance from hinge O to oil surface = 35 cm = 0.35 m

Diagram of the Float Valve

Diagram of a float valve mechanism in a cistern

Solution

For the system to be in equilibrium, the moments about the hinge O must balance. The clockwise moment from the valve force P must equal the counter-clockwise moment from the net upward force on the float.

1. Geometry and Float Submergence

When AO is vertical, the angle of OB with the vertical is \(135^\circ - 90^\circ = 45^\circ\). Let \(h\) be the depth of the float's center B below the oil surface.

$$ \text{Total vertical distance from O to B} = OB \cos 45^\circ $$ $$ = 0.5 \times \frac{1}{\sqrt{2}} \approx 0.3535 \, \text{m} $$
$$ h = (\text{Vertical distance O to B}) - (\text{Vertical distance O to surface}) $$ $$ h = 0.3535 - 0.35 $$ $$ h = 0.0035 \, \text{m} $$

Since the center of the float is submerged by 0.0035 m (0.35 cm), the total depth of submergence is the radius plus this amount.

$$ \text{Total Submergence} = R + h = 0.075 + 0.0035 = 0.0785 \, \text{m} $$

2. Buoyant Force on the Float (\(F_B\))

The buoyant force is the weight of the displaced oil. The volume displaced is the volume of the submerged portion of the sphere.

$$ V_{displaced} = \frac{\pi (\text{Total Submergence})^2}{3} (3R - \text{Total Submergence}) $$ $$ V_{displaced} = \frac{\pi (0.0785)^2}{3} (3 \times 0.075 - 0.0785) $$ $$ V_{displaced} \approx 0.00645 \times (0.225 - 0.0785) $$ $$ V_{displaced} \approx 0.000945 \, \text{m}^3 $$
$$ F_B = \rho_{oil} \times g \times V_{displaced} $$ $$ F_B = 800 \times 9.81 \times 0.000945 $$ $$ F_B \approx 7.416 \, \text{N} $$

3. Equilibrium of Moments

Let \(W\) be the weight of the float. The net upward force on the float arm is \(F_B - W\). Taking moments about the hinge O:

$$ \text{Moment}_{valve} = \text{Moment}_{float} $$ $$ P \times OA = (F_B - W) \times (\text{Horizontal distance from O to B}) $$ $$ 9.81 \times 0.2 = (7.416 - W) \times (OB \sin 45^\circ) $$ $$ 1.962 = (7.416 - W) \times (0.5 \times 0.7071) $$ $$ 1.962 = (7.416 - W) \times 0.3535 $$ $$ \frac{1.962}{0.3535} = 7.416 - W $$ $$ 5.55 = 7.416 - W $$ $$ W = 7.416 - 5.55 $$ $$ W = 1.866 \, \text{N} $$
Final Result:

The weight of the float is approximately \( 1.87 \, \text{N} \).

Explanation of Concepts

Principle of Moments: The float valve system works like a lever. The problem is solved by balancing the moments about the pivot point (the hinge O). A moment is a turning force (Force × Perpendicular Distance). The clockwise moment from the valve force must equal the counter-clockwise moment from the net force on the float.

Net Force on Float: The float is acted upon by two vertical forces: its own weight (\(W\)) acting downwards and the buoyant force (\(F_B\)) from the displaced oil acting upwards. The net force is the difference between these two (\(F_B - W\)).

Buoyant Force on a Sphere: The buoyant force depends on the volume of the submerged part of the sphere. This volume is calculated using the geometric formula for a spherical cap, which depends on the sphere's radius and the depth of submergence.

Leave a Comment

Your email address will not be published. Required fields are marked *

Scroll to Top