Example

Question:

At a depth of 1000 m in an ocean
(a) What is the absolute pressure?
(b) What is the gauge pressure?
(c) Find the force acting on the window of area \(20\,\mathrm{cm} \times 20\,\mathrm{cm}\) of a submarine at this depth, the interior of which is maintained at sea-level atmospheric pressure.
(The density of sea water is \(1.03 \times 10^3\,\mathrm{kg\,m^{-3}}\), \(g = 10\,\mathrm{m\,s^{-2}}\).)

Solution:

(a) Here \(h = 1000\,\mathrm{m}\), \(\rho = 1.03 \times 10^3\,\mathrm{kg\,m^{-3}}\).
Absolute pressure: \[ P = P_a + \rho g h \] \[ = 1.01 \times 10^5\,\mathrm{Pa} + 1.03 \times 10^3\,\mathrm{kg\,m^{-3}} \times 10\,\mathrm{m\,s^{-2}} \times 1000\,\mathrm{m} \] \[ = 104.01 \times 10^5\,\mathrm{Pa} \approx 104\,\mathrm{atm} \]
(b) Gauge pressure is: \[ P_g = P - P_a = \rho g h \] \[ P_g = 1.03 \times 10^3\,\mathrm{kg\,m^{-3}} \times 10\,\mathrm{m\,s^{-2}} \times 1000\,\mathrm{m} = 103 \times 10^5\,\mathrm{Pa} \approx 103\,\mathrm{atm} \]
(c) The pressure outside the submarine is \(P = P_a + \rho g h\) and the pressure inside is \(P_a\). Hence, the net pressure acting on the window is gauge pressure, \(P_g = \rho g h\). The area of the window is \(A = 0.04\,\mathrm{m}^2\). The force on it is: \[ F = P_g A = 103 \times 10^5\,\mathrm{Pa} \times 0.04\,\mathrm{m}^2 = 4.12 \times 10^5\,\mathrm{N} \]