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Pressure: Question 5

Syllabus 1.8

Structured Extended 11 marks

A research submarine is cruising at a depth of 150 m150\text{ m} below the surface of the sea. Take the density of sea water as 1030 kg/m31030\text{ kg/m}^3, the gravitational field strength as g=10 N/kgg = 10\text{ N/kg}, and atmospheric pressure at the sea surface as 1.0×105 Pa1.0\times10^{5}\text{ Pa}.

(a) Calculate the increase in pressure, due to the sea water alone, at this depth. [2]

(b) Calculate the total pressure acting on the outside of the submarine's hull at this depth, including the atmosphere. [2]

(c) The submarine has a circular observation window of radius 0.20 m0.20\text{ m}. Calculate the area of this window. [2]

(d) Calculate the total force exerted on the outside of the window by the water and the atmosphere. [3]

(e) Inside the submarine, the air is kept at close to normal atmospheric pressure at all times, regardless of depth. Explain, in terms of the pressure difference across the window, why this window must be far stronger than a window of the same size in a building at the surface. [2]

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Worked solution

Part (a): Pressure increase due to the sea water

Use Δp=ρgΔh\Delta p = \rho g \Delta h with ρ=1030 kg/m3\rho = 1030\text{ kg/m}^3, g=10 N/kgg = 10\text{ N/kg} and Δh=150 m\Delta h = 150\text{ m}:

Δp=1030×10×150=1 545 000 Pa=1.545×106 Pa\Delta p = 1030 \times 10 \times 150 = 1\,545\,000\text{ Pa} = 1.545\times10^{6}\text{ Pa}

Part (b): Total pressure on the hull

Add the atmospheric pressure pushing down at the sea surface:

ptotal=patmosphere+Δp=1.0×105 Pa+1.545×106 Pa=1.645×106 Pap_{\text{total}} = p_{\text{atmosphere}} + \Delta p = 1.0\times10^{5}\text{ Pa} + 1.545\times10^{6}\text{ Pa} = 1.645\times10^{6}\text{ Pa}

Part (c): Area of the circular window

The window is a circle of radius r=0.20 mr = 0.20\text{ m}:

A=πr2=π×(0.20 m)2=π×0.0400 m2=0.126 m2 (3 s.f.)A = \pi r^2 = \pi \times (0.20\text{ m})^2 = \pi \times 0.0400\text{ m}^2 = 0.126\text{ m}^2 \text{ (3 s.f.)}

Part (d): Total force on the window

Rearrange p=F/Ap = F/A to make force the subject: F=pAF = pA. Use the total pressure from part (b) and the area from part (c):

F=ptotal×A=1.645×106 Pa×0.126 m2≈2.07×105 NF = p_{\text{total}} \times A = 1.645\times10^{6}\text{ Pa} \times 0.126\text{ m}^2 \approx 2.07\times10^{5}\text{ N}

This is an enormous force, equivalent to the weight of over 20 tonnes, pressing on a single window just 0.40 m0.40\text{ m} across.

Part (e): Why the window must be so strong

Inside the cabin, the air pressure is kept close to normal atmospheric pressure (1.0×105 Pa1.0\times10^{5}\text{ Pa}) no matter how deep the submarine goes. Outside, at 150 m150\text{ m} depth, the pressure is 1.645×106 Pa1.645\times10^{6}\text{ Pa}, more than 1616 times greater. This creates a very large pressure difference across the window, and therefore a huge net force pushing inward on it (as calculated in part (d)). A window in a building at the surface has almost no pressure difference across it, since the air pressure is nearly the same on both sides, so it experiences essentially no net force from pressure. The submarine window must therefore be far thicker and stronger than an ordinary window, to withstand this large inward force without cracking or being pushed into the cabin.

Final answers

  • (a) Pressure increase due to the water == 1.545×106 Pa1.545\times10^{6}\text{ Pa}
  • (b) Total pressure on the hull == 1.645×106 Pa1.645\times10^{6}\text{ Pa}
  • (c) Window area == 0.126 m20.126\text{ m}^2
  • (d) Total force on the window == 2.07×105 N2.07\times10^{5}\text{ N}
  • (e) The window must be far stronger because of the huge pressure difference between the high outside pressure and the near-atmospheric inside pressure, producing a large net inward force that a surface window never experiences