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

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

p = ρ2gz2 + ρ1gz1

= 900 × 9.81 × 1 + 1000 × 9.81 × 3

= 8829 + 29430 = 38259 N/m2