1 / 9

# C ONVECTIVE A VAILABLE P OTENTIAL E NERGY - PowerPoint PPT Presentation

C ONVECTIVE A VAILABLE P OTENTIAL E NERGY. WHAT IS CAPE?. CAPE is defined as the amount of energy a parcel of air would have if lifted a certain distance vertically through the atmosphere.

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about ' C ONVECTIVE A VAILABLE P OTENTIAL E NERGY' - henry-barlow

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

C ONVECTIVE

A VAILABLE

P OTENTIAL

E NERGY

CAPE is defined as the amount of energy a parcel of air would have if lifted a certain distance vertically through the atmosphere.

A positive number for CAPE indicates instability in the atmosphere. This occurs when warm moist air is underneath dry cool air therefore causing the warm air to rise and create clouds and eventually storms. CAPE is measured in J/kg.

Higher CAPE values result in stronger instability and therefore can cause extremely strong thunderstorms.

< 300 Very weak convection

300-1000 Weak

1000-2500 Moderate

2500-3000 Strong

3000+ Very Strong

On May 3rd, 1999 a F-5 tornado ripped through Oklahoma City. That day CAPE values reached up to 5,885 J/kg. The background image shown here is that tornado.

Here is how CAPE is calculated. First meteorologists send up weather balloons in the atmosphere to obtain data about certain aspects of the upper atmosphere. From this we can calculate CAPE because we know the temperature of the upper air compared to the air on the surface. With this we can graph the two aspects out, take the area under the curve and calculate CAPE.

In matematical terms CAPE=

This is an image of some data that a weather balloon has collected. The green line represents the temperature of a parcel of air that would be risen from the surface and the red line represents the actual temperature. To find CAPE you must find the area under the curve between these two lines. NB is the level of neutral buoyancy, represented by EL on this sounding. LFC is the level of free convection where a parcel will begin rising. This is the level where instability occurs. Other factors contribute so the actual integration is not just

(Tparcel- Tenv)dh.

dh

LFC

In order to solve this integral we must first know the equations of both the temperature of the parcel and the temperature of the environment with a change in height.

First I am going to take points from the graph and form a regression. I only need to take points between the NB and LFC since that is the bounds of our integral.

(Pressure, Temperature)

We perform a regression on each data set to find the equation. The red line is going to be broken up into two regressions since there is such a strong difference in the slope at 550 millibars. The green line is a smooth curve therefore we don’t have to worry about it as much.

Regression website:

http://www.xuru.org/rt/TOC.asp

T parcel

y = 9.660956158·10-8 x3 - 3.630009144·10-4 x2 + 3.890413407·10-1 x - 123.1159255

Regression Results

T environment

y = -6.008450238·10-11 x5 + 1.112450615·10-7 x4 - 8.009029231·10-5 x3 + 2.778012636·10-2 x2 - 4.441126835 x + 199.5174406 from 550 to 196 Mb

y = 68.08439949 ln(x) - 438.5814911 from 656 to 550 mb

Now using our equation we can determine CAPE equation. The red line is going to be broken up into two regressions since there is such a strong difference in the slope at 550

-

-

-6.008450238·10-11 x5 + 1.112450615·10-7 x4 - 8.009029231·10-5 x3 + 2.778012636·10-2 x2 - 4.441126835 x + 199.5174406

9.660956158·10-8 x3 - 3.630009144·10-4 x2 + 3.890413407·10-1 x - 123.1159255

9.660956158·10-8 x3 - 3.630009144·10-4 x2 + 3.890413407·10-1 x - 123.1159255

68.08439949 ln(x) - 438.5814911

550

196

dh

dh

656

550

68.08439949 ln(x) - 438.5814911

-6.008450238·10-11 x5 + 1.112450615·10-7 x4 - 8.009029231·10-5 x3 + 2.778012636·10-2 x2 - 4.441126835 x + 199.5174406

+

I used my calculator to solve this ugly integral

=

1393.2 J/kg

How did we do? equation. The red line is going to be broken up into two regressions since there is such a strong difference in the slope at 550

Image of Iowa City tornado on April 13, 2006 during a lightning flash

High powered computers compute these complex calculations, thank God for computers!

If we go back to the original graph that was provided, on the right hand side you can see a number of calculations that are given. Within these calculations there is CAPE.

The CAPE value given here is 1357 J/kg

Using the percent error equation:

We get a percent error value of 2.6676 %

All things considered that is not too bad

This is my hand equation. The red line is going to be broken up into two regressions since there is such a strong difference in the slope at 550

THE END

NOTICE THE DATE! This is the same day as the Iowa City tornado