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Adiabatic Wind

by EmBista
Tags: adiabatic, altitude, chinook, pressure, temp
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EmBista
#1
May28-12, 03:12 AM
P: 25
1. The problem statement, all variables and given/known data
Adiabatic wind. The normal airflow over the Rocky Mountains is west to east. The air loses much of its moisture content and is chilled as it climbs the western side of the mountains. When it descends on the eastern side, the increase in pressure toward lower altitudes causes the temperature to increase. The flow, then called a chinook wind, can rapidly raise the air temperature at the base of the mountains. Assume that the air pressur p depends on altitude y according to [itex]p=p_0e[/itex]-ay where [itex]p_0=1.00atm[/itex] and [itex]a=1.16*10[/itex]-4m-1. Also assume that the ratio of the molar specific heats is: [itex]γ=4/3[/itex]. A parcel of air with an initial temperature of -5.00℃ descends adiabatically from [itex]y_1=4267m[/itex] to [itex]y=1567m[/itex]. What is its temperature at the end of the descent?


2. Relevant equations
pVγ=a constant
TVγ-1=a constant


3. The attempt at a solution
I really don't know if I went right here.. but here it goes.

[itex]p_1=1*e^{-a*4267}[/itex]
[itex]p_1=0.609587975[/itex]
[itex]p_2=1*e^{-a*4267}[/itex]
[itex]p_2=0.833791423[/itex]

[itex]p_1V_1^γ=p_2V_2^γ[/itex]
[itex]V_2^γ=(p_1/p_2)[/itex] as [itex]v_1^γ=1[/itex]
[itex]ln(V_2)=(ln(p_1/p_2))/γ[/itex]
[itex]e^{ln(V_2)}=e^{ln(p_1/p_2)/γ}[/itex]
plug in all the numbers and [itex]V_2=0.790455348[/itex]

[itex]T_2V_2^{γ-1}=T_1V_1^{γ-1}[/itex]
[itex]V_1=1[/itex] [itex]V_2=0.790455348[/itex] [itex]T_1=-5[/itex] [itex]T_2=?[/itex]

[itex]T_2=\frac{T_1V_1^{γ-1}}{V_2^{γ-1}}[/itex] as [itex]V_1^{γ-1}=1[/itex]
[itex]T_2=\frac{T_1}{V_2^{γ-1}}[/itex]
[itex]T_2=\frac{-5}{0.790455348^{4/3-1}}[/itex]
[itex]T_2=-5.40767883[/itex]

This answer doesn't make sense because the temperature is supposed to INCREASE.

So can anybody tell me where I went wrong? Maybe tell me I used the wrong formulas..


Thank you :)
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gneill
#2
May28-12, 07:53 AM
Mentor
P: 11,625
Try it using the Kelvin scale.
EmBista
#3
May28-12, 07:01 PM
P: 25
That solved it :) wow i'm an idiot


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