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The hydrostatic equilibrium equation including temperature, T(z)

  1. Nov 9, 2012 #1
    1. The problem statement, all variables and given/known data
    In this problem, you'll model the lower atmosphere of Venus. The atmospheric pressure reaches 1 bar (100 kPa) in the middle of the dense cloud deck, where T ~ 350 K. At the surface, the pressure is 90 bars (9000 kPa). From the surface to the 1 bar level, the temperature T(z) decreases at a rate of dT/dz = -8 K [itex]km^{-1}[/itex], close to the adiabatic lapse rate of -8 K [itex]km^{-1}[/itex].

    (b)Write down the equation for hydrostatic equilibrium, including explicitly the variation T(z).


    2. Relevant equations
    We can estimate the adiabatic lapse rate with the approximation:
    [itex]\frac{dT}{dz}|_{ad}=-\frac{g}{c_p}[/itex]

    The equation for hydrostatic equilibrium:
    [itex]\Delta P = -\rho \Delta z g[/itex]


    3. The attempt at a solution
    Part (a) asks me to confirm the adiabatic lapse rate for Venus, which is simple by plugging in g and cp.

    Part (b) (asking for the hydrostatic equilibrium equation) is what is confusing me. I am specifically having trouble including T(z). I can find an expression for T(z) by integrating the adiabatic lapse rate approximation:

    [itex]∫\frac{dT}{dz} dz|_{ad}=∫-\frac{g}{c_p} dz[/itex]

    [itex]T(z)=T_0-\frac{g}{c_p}z[/itex]

    I called the constant of integration T0 because that makes sense conceptually.

    Now, what I tried was just solving T(z) for g and plugging it into the hydrostatic equilibrium equation because I'm just trying to find a way to include it, but that just yields

    [itex]\Delta P=-\rho \Delta z \frac{c_p}{z}(T_0-T(z))[/itex]

    which seems useless to me, because T(z) would quickly reduce out with a tiny bit of simplification.

    Maybe someone can help by telling me if what I've done so far makes sense and if I should continue, or if I'm just missing something that would make it more clear. Thanks in advance!
     
    Last edited: Nov 9, 2012
  2. jcsd
  3. Nov 9, 2012 #2

    haruspex

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    Just guessing here, but if [itex]\frac{dT}{dz}|_{ad}=-\frac{g}{c_p}[/itex] then can you not write [itex]\Delta T = -\frac{g\Delta z}{c_p} = \frac{g\Delta P}{c_p\rho} [/itex]?
     
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