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I Hydrost Pressure Integration Problem (Environmental Physics)

  1. Mar 10, 2016 #1
    Hey guys!

    I'm working on a problem for which i somehow just can't figure out what i'm doing wrong.

    This is the problem:
    I think i've figured it out, but somehow i think i make an error with the maths.

    First of let's determine the equation for the temperature (in Kelvin):
    $$T(z) = 293.15-0.0098z$$
    For the hydrostatic pressure we can use $$\frac{dp}{dz}=-\rho g$$ with the ideal gas law given as $$\rho = \frac{p}{R_d T(z)}$$.
    Combining it all and putting all the constants into C:
    $$\frac{dp}{dz}=-\frac{p g}{R_d \times (293.15-0.0098z)}=C\frac{p}{293.15-0.0098z}$$
    $$\int_{p_0}^p \frac{1}{\tilde{p}}d\tilde{p}=C\int\frac{1}{293.15-0.0098z}dz$$
    $$\ln{p}-\ln{p_0}=\ln{\frac{p}{p_0}}=C\times\left(-\frac{1}{0.0098}\right) \ln{(293.15-0.0098z)} \\ =\ln{\left[ (293.15-0.0098z)^\frac{g}{0.0098 R_d} \right] }$$
    Taking the exponential of this form will give us finally:
    $$p\approx p_0 (293.15-0.0098z)^{3.4874}$$

    It's weird that the hydrostatic air pressure is increasing with height right? Since from the hydrostatic pressure theorem it should decrease. Can anyone point out to me what i'm doing wrong exactly.
     
  2. jcsd
  3. Mar 10, 2016 #2

    BvU

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    Check the bounds and the result for ##\int dz##.
     
  4. Mar 10, 2016 #3
    Thanks! If i'm correct, i'll get:
    $$
    \ln{p}-\ln{p_0}=\ln{\frac{p}{p_0}}=C\times\left(-\frac{1}{0.0098}\right) \ln{(293.15-0.0098z)} \\ =\ln{\left[ (293.15-0.0098z)^\frac{g}{0.0098 R_d} \right] }-\ln{\left[ (293.15-0.0098z_0)^\frac{g}{0.0098 R_d} \right] }
    $$

    with z0=0 this becomes:

    $$
    \ln{p}-\ln{p_0}=\ln{\frac{p}{p_0}}=C\times\left(-\frac{1}{0.0098}\right) \ln{(293.15-0.0098z)} \\ =\ln{\left[ (293.15-0.0098z)^\frac{g}{0.0098 R_d} \right] }-\ln{\left[ (293.15)^\frac{g}{0.0098 R_d} \right] } \\
    =\ln{\left[ \left( \frac{293.15-0.0098z}{293.15}\right)^\frac{g}{0.0098 R_d} \right] }
    $$

    $$
    p\approx p_0 \left(\frac{293.15-0.0098z}{293.15}\right)^{3.4874}
    $$

    And that solves all. Great! One of those small mistakes which you just don't notice when you're working too long on the same thing haha. Thanks so much!
     
  5. Mar 10, 2016 #4

    BvU

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    Equation (40) here !
     
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