KinexiaLEARN PHYSICS

Pressure at Depth — Submarine (Worked Example)

Medium
Problem

A research submarine dives to 300 m in seawater (ρ = 1025 kg/m³). What is the absolute pressure on its hull? (g = 9.8 m/s²)

Given

  • h = 300 m

  • ρ = 1025 kg/m³

  • g = 9.8 m/s²

  • P₀ = 101,325 Pa

Find

The absolute pressure P.

Principle

P = P₀ + ρgh with seawater density — the hull must withstand this crushing load.

Formula

Solution

  1. Substitute: P=101325+(1025)(9.8)(300)P = 101325 + (1025)(9.8)(300)

  2. Gauge part: 3,013,500 Pa3{,}013{,}500\ \mathrm{Pa}

  3. Add: P=101325+3013500=3,114,825 PaP = 101325 + 3013500 = 3{,}114{,}825\ \mathrm{Pa}

3.11 × 10⁶ Pa (3.11 MPa, ≈ 30.7 atm)

Unit check

kg/m³ × m/s² × m = N/m² = Pa, plus Pa. Both terms are pressures — dimensionally consistent.

What this means physically

3.11 MPa — over 30 atmospheres pressing on every square centimetre. Submarine hulls are thick steel cylinders precisely because of this ρgh term.

Common mistake

Using freshwater density. Seawater is 1025 kg/m³ — 2.5% denser, which matters at great depth.