11 replies to this topic

[tip path plane]

Hello everyone I'm new here. I am a student pilot trying to get to grips with meteorology and I was hoping someone may be able to help with a question.
Here goes:
When air reaches saturation vapour pressure it immediately gets warmer, the trouble is I'm not sure why.

This is what I thought may cause this but please post and correct me if I'm wrong.

Saturation will only occur at one temperature "the dewpoint" so if the parcel of air is condensing it will be releasing "latent heat" this heat will leave the parcel and heat up the surrounding atmosphere.

So the parcel was initially rising and cooling at the DALR of 3 C/ 1000ft but became saturated at the dewpoint and released some of this heat and starts to cool at the SALR of an average rate of 1.8 C/ 1000ft.

As the parcel continues to ascend at the SALR the air cools down further and the cold parcel can hold less water vapour and so the SALR becomes closer to the DALR of around 2.2 C/ 1000ft.

Am I on the right track, or completely barking up the wrong tree. Please let me know your opinions as I'm only a beginner and need guidance.

Thanks for your help in advance

[Weather Ship]

At a quick glance you are on the correct track although it's misleading to say it becomes saturated at the dewpoint. Becomes saturated will do. Recommend you read John's guide to reading skew-T diagrams.

http://forum.netweat...kew-t-diagrams/

[tip path plane]

Thanks for that , it was an interesting read. I have another question for you to which I aswered inccorrectly.

the temp at sea level is 16C and the dew point temp is 4C, the condensation level on the lee side of the mountain is at 8000ft what will be the surface temp at the surface on the lee side.

I worked out temp - dew point x 400 = cloud base in feet

16 -4 x 400 = 1200ft cloud base on the windward side. so air reducing at the DALR 3C/1000ft until 1200ft and then at the SALR 1.8/1000ft until the fohn effect kicks in on the lee side when the temp increases back at the DALR again to give a surface temp of 19.6C.
The answer they want is 18.5C, but I dont know how they worked it out.

[Weather Ship]

My brain isn't in gear at the moment but check your condensation level on the windward side, it's too low. Is the height of the mountain not relevant?

Edited by Weather Ship, 12 May 2012 - 11:31 .

[tip path plane]

The height of the mountain is only relevant to show that the SALR changes to DALR at the condensation level on the lee side. Hence the Fohn wind. It would not matter if was 30000ft high as long as it is higher than the condensation level on the lee side.

Does the equation Temperature - Dewpoint x 400= cloud base in feet not apply to the point at which the DALR changes to the SALR on the windward side?

Should it be 16C decreasing at 3C/ 1000ft until 4000ft where the temperature will have decreased to 4C? Is it here where clouds ( visible moisture) will start to form?

Edited by tip path plane, 14 May 2012 - 20:34 .

[Weather Ship]

tip path plane, on 14 May 2012 - 20:15 , said:

The height of the mountain is only relevant to show that the SALR changes to DALR at the condensation level on the lee side. Hence the Fohn wind. It would not matter if was 30000ft high as long as it is higher than the condensation level on the lee side.

Does the equation Temperature - Dewpoint x 400= cloud base in feet not apply to the point at which the DALR changes to the SALR on the windward side?

Should it be 16C decreasing at 3C/ 1000ft until 4000ft where the temperature will have decreased to 4C? Is it here where clouds ( visible moisture) will start to form?

I was only wondering about the height of the mountain as it would affect the temp of the parcel of air before descent but okay forget that.

I'm actaully not familiar with that equation (I thought it a tad more complicated than that) but I assume that's the one you have been given. But yes it would apply and gives a condensation level of 4800 feet which is in the ball park and would be the cloud level. Another thought, were you given the figure for the SALR because often 1.5C, not 1.8C is used.



EDIT

I'm puzzled by this cloud base of 4000ft but did this make it any clearer?

http://www.helitorqu...wtopic&p=156853

Edited by Weather Ship, 15 May 2012 - 09:59 .

[Weather Ship]

I'm not being much help here as this should be relatively simple. Even taking a simplistic view, and using your figures, the condensation level on the windward side is 4800ft and on the lee 8000ft. Thus for 3200ft the air warms at a rate 1.2C higher per 1000ft than it cooled. Whichever way I calculate this I cannot see how the temperature at the surface on the lee side is only 2.5C higher than the windward side. I'm obviously missing something.

Edited by Weather Ship, 15 May 2012 - 10:25 .

[tip path plane]

Thanks for that diagram and the SALR rate of 1.5c average. The average is given to us in the examinations as 1.8c/1000ft or 0.6c/100m for SALR. But thats not set in stone, they like us to use an average of 2C/1000ft up to 36000ft. and ISA 15C at msl 1013.25mb standard atmosphere. Unfortunately the questions vary greatly and it appears that guess work is involved. Temp above 36000ft at -56C and isothermic above.

Another example of the type of question is "What best describes the Tropopause"

A. It is the boundary between the Stratosphere and the Troposphere
B. It separates the Stratosphere from the Troposphere
C. It is an isothermal layer
D. It is a layer at which a temperature inversion may occur

Which would you go for?

I know that for the first part of the stratosphere the temp remains constant so it cant be an inversion as this would indicate temp increase with altitude. I suppose the layer ABOVE the tropopause is isothermic so ruling out D and C. leaving A and B as pretty good descriptions of the tropopause.

I will let you know that I picked the wrong one and lost a mark. Ill post the correct answer later.

With regard to the Temp - dewpoint x 400 formula was given to us students as just a basic one which should give a rough indication of cloud base for us rough pilots " not proper meteorologists"

[Weather Ship]

tip path plane, on 15 May 2012 - 19:40 , said:

Thanks for that diagram and the SALR rate of 1.5c average. The average is given to us in the examinations as 1.8c/1000ft or 0.6c/100m for SALR. But thats not set in stone, they like us to use an average of 2C/1000ft up to 36000ft. and ISA 15C at msl 1013.25mb standard atmosphere. Unfortunately the questions vary greatly and it appears that guess work is involved. Temp above 36000ft at -56C and isothermic above.

Another example of the type of question is "What best describes the Tropopause"

A. It is the boundary between the Stratosphere and the Troposphere
B. It separates the Stratosphere from the Troposphere
C. It is an isothermal layer
D. It is a layer at which a temperature inversion may occur

Which would you go for?

I know that for the first part of the stratosphere the temp remains constant so it cant be an inversion as this would indicate temp increase with altitude. I suppose the layer ABOVE the tropopause is isothermic so ruling out D and C. leaving A and B as pretty good descriptions of the tropopause.

I'm sure if you google that you will know the answer is A. Your last paragraph is not really correct.

Quote

With regard to the Temp - dewpoint x 400 formula was given to us students as just a basic one which should give a rough indication of cloud base for us rough pilots " not proper meteorologists"

Very rough.

Edited by Weather Ship, 15 May 2012 - 22:41 .

[tip path plane]

Well done weather ship, the correct answer is A.

Thanks for your help, Im sitting the exams next week, so it's head back in the books to study the ITCZ and Jet Streams, Ill let you know how the Met turns out when I get the results.

[Weather Ship]

Good luck with that tpp; should keep you busy.

[tip path plane]

Cheers, Weather Ship, thanks again!