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I intend to compute the air density, $\rho$ at an altitude of $h=15000ft=4572m$, with an ISA deviation of $\Delta_{ISA}=18^{\circ} F=10\,K$.

Well I first computed the real temperature of the air at $h$ altitude with the following ISA temperature realtions:

$$T_{ISA}[K]=288.15-0.0065\,h[m]=258.432\,K$$ $$T=T_{ISA}+\Delta_{ISA}=268.432\,K$$

I then computed the temperature altitude, i.e. the $ISA$ altitude for a temperature $T$:

$$h_T[m]=\frac{T[K]-288.15}{-0.0065}=3034m$$

With $h_T$ computed I got the air density by the respective ISA relation:

$$\rho[kg/m^3]=1.225\left(1-0.0065\frac{h_T [m]}{288.15}\right)^{4.25588}= 0.906kg/m^3$$

The problem is that this value is different from what I got here ($\rho=0.742 kg/m^3$)

What did I do wrong?

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  • $\begingroup$ Are you comparing different atmosphere models by any chance? $\endgroup$
    – user7241
    Dec 26, 2017 at 19:53

1 Answer 1

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The mistake is subtracting the ISA temperature at sea level from the ISA+18°F temperature at altitude in step two. If your atmosphere is heated up, it needs to be hotter at sea level as well, or your lapse rate in the first step and the denominator is wrong.

First calculate density for ISA conditions. This gives 0.7708 kg/m³.

Then used the ideal gas law to see how a 10 K temperature increase will reduce density at constant pressure. The density ratio is inversely proportional to the temperature ratio, so the result is 0.7421 kg/m³.

When checking your result, input only 10 K into the calculator, not 18 K !!!

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  • $\begingroup$ @ÉlioPereira: Only by assuming a 18 K temperature difference! $\endgroup$ Dec 27, 2017 at 0:38
  • $\begingroup$ But pressure changes with temperature. Why did you consider that a deviation in temperature only affects the density of air? $\endgroup$ Dec 27, 2017 at 0:47
  • $\begingroup$ @ÉlioPereira The ISA deviation is a temperature deviation for a pressure altitude. Your altimeter (which reads pressure) can read 10,000 feet and you can have different outside air temperatures. And here lies the difference between altitude and height. $\endgroup$
    – user14897
    Dec 27, 2017 at 1:20
  • $\begingroup$ I didn't know that. So the 15000ft is a pressure altitude and not an height. Thank you. $\endgroup$ Dec 27, 2017 at 1:24

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