Question

An open tank contains water up to a depth of 3m and above it an oil of sp.gr 0.9 for a depth of 1m. Find pressure intensity 

1. at the interface of two liquids 

2. at the bottom of the tank.

Answer 1

Height of water, z₁ = 3 m

Height of oil, z₂ = 1 m

Density of oil

= Sp.gr of oil × Density of water

= 0.9 × 1000 kg/m³ = 9000 kg/m³

=ρ₂

Density of water = ρ₁= 1000 kg/m³

1. At the interface, i.e., at A,

P= ρ₂g × z₂ = 900 × 9.81 × 1

= 8829 N/m²

2. At the bottom i.e., at B,

p = ρ₂gz₂ + ρ₁gz₁

= 900 × 9.81 × 1 + 1000 × 9.81 × 3

= 8829 + 29430 = 38259 N/m²