Why Is the Calculation of Santa Ana Wind Temperatures Tricky?

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The discussion centers on the calculation of temperature changes in Santa Ana winds as they descend from high elevations to sea level. The key concept is that the air undergoes an adiabatic process, meaning it does not exchange heat with its surroundings, leading to significant temperature increases due to compression. The initial attempt at solving the problem used the ideal gas law incorrectly, resulting in an unrealistic temperature of 180 degrees C. A more accurate approach involves applying the adiabatic relations, which account for changes in pressure, volume, and temperature. The corrected calculation yields a final temperature of approximately 42.45 degrees C, demonstrating the importance of understanding adiabatic processes in meteorology.
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Easy question--but what's wrong?

Homework Statement


When strong winds rapidly carry air down from mountains to a lower elevation, the air has no time to exchange heat with its surroundings. The air is compressed as the pressure rises, and its temperature can increase dramatically. These warm winds are called Chinook winds in the Rocky Mountains and Santa Ana winds in California. Suppose the air temperature high in the mountains behind Los Angeles is 0 degrees C at an elevation where the air pressure is 60 kPa. What will the air temperature be, in C when the Santa Ana winds have carried this air down to an elevation near sea level where the air pressure is 100 kPa ?


Homework Equations





The Attempt at a Solution



This is a pretty straightforward problem, but for some reason I got the answer wrong.

P1/T1=P2/T2

(60kPa)/(273K) = (100kPa)/(T2) T2=455K = 180 degrees C
 
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The important phrase is that the air has no time to exchange heat with its surroundings. What does that signify? What relationship between P and T is true in that case?
 
Hmmm... again I forgot that volume changes, and this is an adiabatic process, and volume decreases

For an adiabatic process: P1V1^gamma = P2V2^gamma

Gamma is the specific heat ratio
gamma= Cp/Cv

since air is a mixture of mostly diatomic gases oxygen & nitrogen gamma = 29/20.8 = 1.40

so (60 kPa)(V1)^1.40 = (100kPa)(V2)^1.40
the ratio of V2/V1 = .694

P1V1/T1 = P2V2/T2

(60kPa)(V1)/(273 K) = (100kPa)(V2)/T2
(.220)*T2 = (V2)/(V1) * (100 kPa)
.220T2= 69.4
T2= 314.45 K = 42.45 degrees C

Is this anywhere near how I should approach this problem?
 
Anybody have any thoughts about this? Thanks.
 
That looks right to me. To save yourself a step or two you can also use the adiabatic relation for P and T directly:

<br /> P^{\gamma -1} T^{-\gamma} = \mbox{ constant}<br />

and there is also one for T and V only. But of course that is essentially what you did.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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