Hydrost Pressure Integration Problem (Environmental Physics)

[MrGianmarcoD]

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:

During daytime the vertical temperature profile of the atmosphere is observed to obey the dry adiabatic lapse rate (-9.8K/km). At the surface the temperature is 20°C, and the air pressure is 1.018·10⁵ Pa. Use the hydrostatic equation and the gas law to compute (numerically) the vertical profile of the pressure. What are the pressure values at 0.5 and 1 km?

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.

 

[BvU]

Check the bounds and the result for $\int dz$.

 

[MrGianmarcoD]

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!

 

[BvU]

Equation (40) here !