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

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

Step-by-step explanation:

Given;

specific gravity of oil, γ = 0.75

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

`\gamma =(\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 ↑