Applications of Hydrostatic Equilibrium

When we study some thing we look for its application in our daily day life. The hydrostatic equilibrium has wide applications in engineering & other fields of science.

Some Applications of  Hydrostatic Equilibrium are here

1. Barometric Equation
2. Manometers
3. Centrifuge & Decanters

Barometric Equation

Barometric Equation states that Pressure(P) decreases with increase in Height(h)

It is  given as follow:

P=Cexp(−MgRTh).

Derivation Considering an ideal gas with density  ρ is compressedPPP through pressure  P then according to ideal gas law: PV=mMRT,⇒P=mVMRT=ρMRT. Here T is the absolute temperature, R is the universal gas constant equal to 8.314JK⋅mol, M is the molar mass, which is for air equal to 0.029kgmol. It follows from here that the density is given by the formula ρ=MPRT. Putting this into the differential relation for dP gives: dP=−ρgdh=−MPRTgdh,⇒dPP=−MgRTdh. We obtain a differential equation describing the gas pressure P as a function of the altitude h. Integrating gives the equation: ∫dPP=−∫MgRTdh,⇒lnP=−MgRTh+lnC. Getting rid of the logarithms, we obtain the so-called barometric formula P=Cexp(−MgRTh). The constant of integration C can be determined from the initial condition P(h=0)=P0, where P0 is the average sea level atmospheric pressure. Thus, dependency of the barometric pressure on the altitude is given by the formula P=P0exp(−MgRTh). Substituting the known constant values (see Figure 2 above), we find that the dependence P(h) (in kilopascals) is expressed by the formula: P(h)=101.325exp(−0.02896⋅9.8078.3143⋅288.15h)=101.325exp(−0.00012h)[kPa], where the height h above sea level is expressed in meters. If the pressure is given in millimeters of mercury (mmHg), the barometric formula is written in the form: P(h)=760exp(−0.00012h)[mmHg]. In case when the height h is given in feet, and pressure in inches of mercury (inHg), this formula is written as P(h)=29.92exp(−0.00039h)[inHg]. The barometric formula is often used for estimating the air pressure under different conditions, although it gives slightly higher values compared with the real ones.

Example 1

Determine at what altitude the air pressure is twice less than on the sea level? Solution. To estimate the altitude, we use the barometric formula P(h)=P0exp(−0.00012h). When h=0, the pressure P(h) is equal to the average atmospheric sea level pressure P0. At a certain altitude H, the pressure is twice less: P(H)=P02=P0exp(−0.00012H). It follows from here that exp(−0.00012H)=12. Taking logarithms of both sides, we find the altitude H: ln12=−0.00012H,⇒ln2=0.00012H,⇒H=ln20.00012≈5780m.

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