Hydrostatic Equilibrium

note on hydrostatic equilibrium
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  Hydrostatic equilibrium Basic ideas The principle of hydrostatic equilibrium is that the pressure at any point in a fluid at rest (whence, “hydrostatic”) is just due to the weight of the overlying fluid.  As pressure is just force per unit area, the pressure at the bottom of a fluid is just the weight of a column of the fluid, one unit of area in cross-section. This principle is simple to apply to incompressible fluids, such as most liquids (e.g., water). [Note that water and other common liquids are not strictly incompressible; but very high pressures are required to change their densities appreciably.] If the fluid is incompressible, so that the density is independent of the pressure, the weight of a column of liquid is just proportional to the height of the liquid above the level where the pressure is measured. In fact, the mass of a unit-area column of height h  and density ρ  is just ρh ; and the weight of the column is its mass times the acceleration of gravity, g . But the weight of the unit-area column is the force it exerts per unit area at its base  —   i.e., the pressure. So P = g ρ h  . Examples For example, the density of water is 1000 kilograms per cubic meter (in SI units), so the weight of a cubic meter of water is 1000 kg times g , the acceleration of gravity (9.8 m/sec 2 ), or 9800 newtons. This force is exerted over 1 m 2 , so the pressure  produced by a 1-meter depth of water is 9800 pascals (the Pa is the SI unit of pressure, equal to 1 newton per square meter). The unit of pressure used in atmospheric work on Earth is the hectopascal; 1 hPa = 100 Pa. So the pressure 1 m below the surface of water (ignoring the pressure exerted  by the atmosphere on top of it) is 98 hPa. Standard atmospheric pressure is 1013.25 hPa, so it takes 1013.25/98 = 10.33 meters of water to produce a pressure of 1 atmosphere. (That's about 34 feet, for those who like obsolete units.) The pressure in the ocean increases by about 1 atmosphere for every 10 meters of depth. The average depth of the ocean is about 4 km, so the pressure on the sea floor is about 400 atmospheres. The homogeneous atmosphere  The density of air, under standard conditions, is about 800 times less than the density of water  —   almost 1.3 kg per cubic meter. So the height of a column of air needed to exert the standard atmospheric pressure of 1013.25 hPa would be about 8 thousand meters (8 km), if   it were all the same density  —   i.e., homogeneous . That height is “the height of the homogeneous atmosphere.”  You can see from the relation above that this height, H , is just P/(gρ) . Even though the atmosphere isn't really homogeneous, this 8-km height is a useful characteristic length that keeps turning up in calculations of atmospheric refraction. (A better name for this concept is  Radau 's term, “reduced height”.)   The real atmosphere Hydrostatic equilibrium is a little more complicated to apply to air, because air is very compressible. The same principle still applies, but we now have to deal with a density that varies with pressure and temperature. Preliminaries: the hydrostatic equation Before delving into the details, let's re-consider the constant-density fluid more mathematically. We need to use a little elementary calculus for this. We'll use the conventional symbol [Greek lowercase rho: ρ] for density. [If your browser doesn't display this correctly, you need to upgrade to a current version.] If P  is the pressure and h  is the height,   dP = −g ρ dh  is the differential equation that expresses hydrostatic equilibrium. [Remember that g  is the local acceleration of gravity, which is needed to convert the element of mass (  ρ dh  ) into the force (i.e., its weight) it adds to the unit area beneath it. The minus sign is there because g  acts in the negative direction along the height scale. We are implicitly assuming that the range of h  is so small compared to the radius of the Earth that g  can be assumed constant.] Its elementary solution is P = −g   ∫  ρ dh  , and if ρ  = const., we can take it outside the integral: P = −g ρ   ∫  dh  or  P = −g ρ   ∫  dh = −g ρ h  . In other words, the pressure is just proportional to the height, h , of the column of fluid, as we already knew. (The minus sign comes from the fact that we measure h  positive upward, but the atmosphere's weight is directed downward.) Equation of state For air, things are not so easy. The density, ρ , depends on both P  and T , the absolute temperature. The equation of state  is the function that tells us the density. For air, it's a very good approximation to use the equation of state for an ideal gas, ρ = μP/(R T)  , where μ  is the (dimensionless) molecular weight  —   about 29 for dry air  —   and R   is the “gas constant” that takes care of the units.  Even though this relation is very simple, it still complicates the integrand of the hydrostatic equation. First of all, it gets P  involved in the integrand, which is no longer a simple function of h . Secondly, it introduces a new independent variable, T . We'd like to express everything inside the integral as a function of a single variable. To do this, we need some additional   relationship among P , ρ , and T , which would allow us to get rid of the second independent variable. Unfortunately, there is no additional physical relationship available, in general. The actual dependence of P , or ρ , or T , on height, is quite variable in the real world. This dependence is what's meant by the phrase “the structure of the atmosphere.” [In the astronomical literature of the 18th and 19th Centuries, it was often called the “constitution” of the atmosphere, which is confusing to a modern reader; “constitution” today means “composition” rather than “structure.”] The structure of the real atmosphere varies considerably from place to place and from time to time. It's often convenient (though somewhat unrealistic) to assume that the structure of the atmosphere is  polytropic; this is explained on the polytropes page.  The homogeneous atmosphere (again) But, even though we can't integrate the hydrostatic equation until some additional information (such as the run of T  with height, or the dependence of P  on ρ ) is available, we can still evaluate the height of the homogeneous  atmosphere. That's just the height the atmosphere would have if it had the same density everywhere (i.e., the  density at the surface of the Earth), and the same pressure at the bottom as the real atmosphere. Given the temperature and pressure at the surface, and the composition of the gas there (which is what determines the mean molecular weight, μ ), we can find the density of air at the surface. Then the height of the homogeneous atmosphere is just H = P/(gρ) = RT/μg   ,  because P  is the weight of a column of gas of height H  and density ρ  in the gravitational field of the Earth, with acceleration g . If you'd rather think in terms of the mass of one molecule, m , then the gas law is ρ  = μP/(NkT) , where N  is Avogadro's number, k   is Boltzmann's constant, and μ  is the (dimensionless) molecular weight. (The molecular mass m  is μ  times the atomic mass unit, u .) Then the height of the homogeneous atmosphere is H = kT/mg . The isothermal atmosphere A notable example that can be integrated is the isothermal atmosphere, with the same temperature T  throughout. Remember  that the hydrostatic equation is   dP = −g ρ dh  ;  but we can use the ideal gas law, ρ = μP/(R T)  , to get rid of ρ .   So, write   dP  as dP = −g μP/(R T) dh  , or (dividing by P )   dP/P = −g μ/(R T) dh  .  Now, integrate this. On the left side, we get ∫   dP/P  , which is just ln  P  ; on the right, T  is constant, so we get some constants times ∫ dh , which is just h . Of course there is a constant of integration; we see that it has to be the value of ln  P  at h  = 0. So: ln  P  = ln  P 0   −   gh μ/(R T)  . Combine the two logarithms to get ln ( P/P 0 ) on the left. Then get rid of the logarithm  by exponentiating both sides: P/P 0  = exp [ −gh μ/(R T) ]  .
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