Question

Oil of specific gravity 0.75 flows through a smooth contraction in a pipe at a volumetric flow rate of 3.2 cu ft / sec.Find the force required to hold the contraction in place. Indicate the direction of this force.

Answer 1

Answer:

The force required to hold the contraction in place is 665.91 N ↑

Explanation:

Given;

specific gravity of oil, γ = 0.75

Volumetric flow rate, V 3.2 Ft³/s = 0.0906 m³/s

$\gamma =\frac{\rho_o}{\rho_w}$

where;

$\rho_o$ is the density of oil

$\rho_w$ is the density of water = 1000 kg/m³

∴density of oil ($\rho_o$) = γ × density of water($\rho_w$)

                               = 0.75 × 1000 kg/m³

                                = 750kg/m³

Buoyant Force = ρVg

                         = 750 × 0.0906 × 9.8

                         = 665.91 N ↑

This force acts upward or opposite gravitational force.

Therefore, the force required to hold the contraction in place is 665.91 N ↑