PPL Met question

[Pilotpaul787]

Hi all, I recently passed my Airlaw exam, and have now been studying for met. Iv just finished the chapter on clouds in the book, and used the confuser to test my knowledge on the chapter. I did fine on most of the questions but one of them that came up puzzled me and I was wondering
if someone could give me a hand. The question was;

Given a surface temperature of 21 degrees C and a dew point of 7 degrees at approximately what height will the cloud base of a cumulus cloud be
a) 560 feet
b) 56,000 feet
c) 6500 feet
d) 5600 feet
D - 5600 feet is the correct answer, why is this?

Thanks in advance,
Paul.

[PAPI-74]

Cloud base is 5600

(OAT-Dpt) x 400

(21-7) x 400

14 x 400 = 5600

Why?

Air cools at the Dry Adiabatic Lapse Rate until saturated at it Due Piont.
OAT is 21
-3degC per 1000 '
Reaches DP of 7 deg at 5600' and condensed out.

It then cools at the Saturated LR of 1.5 / 1000' and the tops are another story altogether, subject to inversions etc...

[FREDAcheck]

There's another issue, as 5600 feet corresponds to 2.5C per thousand feet. The rule I remember is that the difference between temp and dew point drops by (on average) 1C per 400 feet, i.e. 2.5C per thousand. Temp drops by an average of 3C per thousand feet (rising unsaturated air), but dew point also drops by 0.5C. Something to do with density, but I can't remember exactly.

Edit: I found this reference in http://www.abc15.com/content/weather...cloudbase.aspx :

Dew Point, along with Temperature, can tell you the estimated height of a cloud's base. How? When air rises in a convective current, it cools at the rate of 5.4° F [3C] /1,000 ft, and its dew point decreases 1° F [0.55C] /1,000 ft. Therefore, the temperature and dew point are converging at 4.4° F [2.5C] /1,000 ft. Since clouds form when the temperature/dew point spread is 0°, you can use that information to estimate the base of a cumulus cloud. Here's how it works: Subtract the surface dew point from the surface temperature to give you the temperature/dew point spread. Then divide that number by 4.4° F. Finally, multiply this result by 1,000 ft. This total will be your estimate of the cloud base's height above ground level (AGL).

Or, multiply temp / dew point diff in deg C by 400, to give cloud base in feet.

[niknak]

PAPIs answer in theory is correct, but the SALR and the DALR depend on several variables, not least of which, is the atmosphere stable or unstable, but you don't need to concern yourself with that to any great extent other than looking out of the window.
If you see cumulus or similar cloud, its unstable, if you see strato cu its stable.

All you need to know is, that in the "Standard Atmosphere" ie the assumed day to day weather conditions of any given day at any given time of year, air temperature cools at 2.5 deg C for every 1000ft.
Therefore 21 deg C - 7 deg C = 14, divided by 2.5 deg C = 5.6 (5600ft).

[PAPI-74]

Perfectly corect if you want to go that far.
The DP is a temp to which a volume of air must be cooled at a constant temp.
As you say, the presure changes as the air is lifted, changing the DP temp slightly at altitude. That is as much as I can remember, but I know that for PPL-ATPL it isn't considdered.
I am making a guess but as Press reduces, so does temp (unless there is an inversion). If press reduces it would aid condesation and increase the temp the cloud would form, lowering the cloud base (beware of low press days) but it can only be a few feet....at a guess.

[Pilotpaul787]

Thanks everyone for your posts I understand now

[IO540]

You have to give the answer the CAA is looking for (temp-dp * 400ft) and not the correct answer which is "it depends on many things and any formula is at best a very rough guess"

It's a long time since I saw a day where the cloudbase was as per the formula.

[bookworm]

It's a long time since I saw a day where the cloudbase was as per the formula.

A couple from recent METARs with exactly that pair:

KCEF 291557Z 01010KT 7SM FEW012 21/07 A3025=
METAR UKLL 291500Z 25002MPS 220V290 9999 SCT040 21/07 Q1022 31090070 NOSIG=

or if you fancy an example in the other direction:

UAUU 291700Z 31003MPS CAVOK 09/07 Q1020 330///70 NOSIG RMK QFE749/0999=
EFUT 291650Z AUTO 08011KT CAVOK 13/12 Q1027=

It's absurd to set a question like that.

a) 560 feet
b) 56,000 feet
c) 5600 feet

is not unreasonable as a set of choices, but to give students the impression that a 14 point spread allows determination of the dewpoint to the nearest 1000 ft is just bad science.

[rustle]

Originally Posted by bookworm

...but to give students the impression that a 14 point spread allows determination of the dewpoint to the nearest 1000 ft is just bad science.

I would imagine the 6500 or 5600 choice was to identify the dyslexics rather than the mathematicians