Convective Available Potential Energy (CAPE) Calculation Procedure

The procedure is relatively simple and is best done on a spreadsheet program (for example, Excel). Of course, if you are unfamiliar with spreadsheet programs, then using an hand calculator or doing it by hand will work (but is very tedious).

We will start with a schematic sounding (Fig. 1)--Remember. this is only an example, and does not correspont to a real case. Only the portion of the sounding from the LFC to the EL is shown. Remember, the temperatures should be in Kelvin. So add 273o to convert from Centigrade. Figure 1: Portion of schematic sounding from the LFC at 800 mb to the EL above the tropopause at 200 mb. Elevations of isobaric levels are shown. Procedure to estimate CAPE is summarized at bottom, as discussed in text.

The base of the portion of the sounding considered is indicated by the line at the 800 mb level. The thick solid line represents the ELR and the thin solid curve represents the wet adiabatic ascent curve of the rising air parcel. Note that the area of the polygon bounded by the two curves would be the CAPE.

To estimate the CAPE for a given layer, say the 650 mb to 500 mb layer, one would need to estimate the average value obtained from Equation 1 for the whole layer. A simple approximation of that would be to evaluate Equation 1 at the 500 mb level and at the 650 mb level. Add the result and average by dividing by 2. Multiplying the result by g (9.8 m s-2) and the thickness of the layer (in this case, 1500 m) would yield an estimate of the mean CAPE for the layer.

This procedure would need to be followed for each layer considered. In this example, there would be four values of layer CAPE estimates that you would need to add to estimate the total CAPE for the case.

In the case of this exercise, you will use layers 1500 m (1.5 km) thick. You'll be making the (bad) assumption that the conditions at each level represents conditions for a layer 1500 m thick centered at each level. Clearly, that is not estimating conditions correctly, but it is a start.

Procedure Detail for Homework 2 Assignment Using a Different Sounding (KOUN, 12 UTC 3 May 1999)

Step 1: After you have examined the sounding and determined the LFC (or CCL) and EL, you should tabulate the environmental temperature and air parcel temperature (remember, in Kelvin) for each pressure level from the LFC (or CCL) to the EL. Fill in the Table below. Pressure "Layer" Number Temp Parcel Temp Env Temp Diff Temp Quotient* 700 1 600 2 500 3 400 4 300 5 250 6

Table 1: Tally Table for CAPE Calculations

*Temperature Quotient = (Temp Diff/TempEnv)

Step 2: Using the information from Table 1, you will need to compute the values needed to

solve Equation 1 at each level. I suggest having a second table like the one given as Table 2.

 Layer Delta z gravity Temp Quotient* Layer CAPE** 1 1500 m 9.8 m s-2 2 1500 m 9.8 m s-2 3 1500 m 9.8 m s-2 4 1500 m 9.8 m s-2 5 1500 m 9.8 m s-2 6 1500 m 9.8 m s-2 Total CAPE***

Table 2: Computational Results (from Equation 1 and Table 1)

*Should really be Layer Average Quotient but what we will simply use is the values from Table 1

** Layer Cape = Layer Average Quotient X gravity X 1500 m

***Total CAPE = Add all the values above

Once you have complete Table 2, you simply sum the values in the fifth column to obtain your estimate of CAPE for the sounding. It's going to be an underestimate because we used values for levels instead of layer averages. You can see how a computer could do this better, by computing layer averages for layers that were very small.

I have completed the Table for the KOUN sounding for 12 UTC 3 May 1999, a day in which an EF-5 tornado devastated Moore, OK. I've actually combined both tables into one below.

 Pressure Elevation Temp Env Temp Parcel Diff ∆z Quotient Layer Avg Q Layer CAPE 700 3000 277 277 0 1500 0 600 4500 267 271 4 1500 0.014981273 0.014981273 220.2247191 500 6000 260 261 1 1500 0.003846154 0.011336791 166.6508211 400 7500 246 250 4 1500 0.016260163 0.01818324 267.293621 300 9000 227 231 4 1500 0.017621145 0.025751227 378.5430321 250 10500 218 218 0 1500 0 0.008810573 129.5154185 CAPE 1032.712193 w= 45.44694034