Cave DiverAn average of 8 persons a year drown while exploring spring-fed underwater caves in Florida. In one cave a local dive club posted warning signs and installed a mercury barometer. The schematic shows a diver's view of the barometer and its location relative to the spring surface. Calculate the depth of the diver beneath the surface of the spring.

♦  By inspection we know the barometer reading and the vapor pressure of mercury in the barometer (almost zero). The pressure of air at the surface is atmospheric. The depth of the diver, H, is the unknown. To solve, using the hydrostatic equation, we imagine a path through fluids (air, water and mercury) between the two known pressures. Our path can start at either place but must end at the other. We write the pressure changes that occur at a point that moves through fluids from the first point (the spring surface) to last point, in the mercury vapor. Later such equations will be written on one line. Here we write it repeatedly as the terms are explained.

Begin in air at the surface of the spring where we assume logically:

p_air,surface = p_atm

Moving toward the mercury vapor of the barometer requires crossing the air/water interface for which there is no pressure change. At the spring surface in water:

p_{water,surface} = p_atm

In water, we imagine our point to procede downward to the point (A) at depth H. As the point moves there, its pressure will increase by the column of water of height, H:

p_{water,depth H} = p_atm + ρ_waterg_o H

At depth H, the "point" that traces our path must move horizontally, with no changes of pressure, to arrive just above the exposed mercury of the barometer attached to the cave wall. Then imagine the point to move downward (across the inferface) into mercury (with no pressure change) then upward in mercury and across the liquid/vapor interface at the top of the barometer. (Since the liquid-to-vapor interface of the mercury is not flat there is a pressure change across the interface. It is small and negligible to this calculation). Now in the mercury vapor, the pressure of the "point" becomes that of mercury vapor, (B).

p_{Hg vapor} = p_atm + ρ_water gH - ρ_Hgg_o(36.4/12)ft

However, p_{Hg vapor} is nearly zero. Thus,

0 = p_atm + ρ_watergH - ρ_Hgg_o(36.4/12)ft

Since the original sketch presented English Engineering Units are used.

The above is A single equation with one unknown, H.

While the above approach is clear, some persons skilled with these calculations write the equation on one line. This technique will be illustrated in the next examples.