Constructing an adiabatic graph

Graphing adiabatic temperature change.

Now that you know how to calculate the condensation level, let's look at how to graph the adiabatic temperature change experienced by a parcel of air and its resulting stability. We'll use a new example for this discussion.

The first step is to plot the values of the environmental lapse rate. I have not provided the data here as there's no need to know what they are for this example. See graph. For this example:

Ground level (altitude 0 meters) air temperature of parcel = 15^o C
Dew Point Temperature of the air parcel = 10^o C

Recall that we need three points to construct the adiabatic temperature change graph:

Point 1 - Ground level air temperature
Point 2 - The condensation level
Point 3 - The end point (highest elevation for the problem)

Point 1 is given above, 15^o C at ground level so it can be plotted right away.

Point 2 is the condensation level. I'll use the same procedure as I did earlier to find condensation level. First determine the amount of temperature change (cooling) that takes place between the surface and the condensation level (altitude where the dew point is reached by adiabatic cooling). Given that the parcel temperature is 15^o C at the surface and at the condensation level it will be 10^o C (the dew point), the air rises and cools by 5^o C (i.e., 15^o C - 10^o C).

Take the amount of temperature change (5^o)and multiply it by the inverted DAR like we did before to get the condensation level in meters. You should arrive at a value of 500 meters. Now plot this point on your graph, that is, plot 10^o C at 500 meters. This, again, is your condensation level. Connect point 1 with point 2 using a straight line. Because we used the dry adiabatic rate to compute this portion of the graph, label the line DAR. (See graph).

Point 3 will be the temperature of the air at the highest point of our graph which in this problem is 3,000 meters. There's a three - step process to calculate point 3:

Step 1. Calculate the change in elevation between points 2 and 3.

The change in elevation is equal to the starting elevation minus the ending elevation. The starting elevation is 500 meters because we have brought the air up to that altitude already. The ending elevation is 3,000 meters so,

Change in elevation = 500m - 3,000m = -2,500 m

Step 2. Calculate the amount of temperature change between points 2 and 3.

To calculate the amount of temperature change, or amount of cooling that will take place between points 2 and 3 we'll take the change in elevation and multiply it by the saturated adiabatic rate (SAR) of .6^oC/100m. Why the SAR? We use the SAR because the air is rising upwards from the condensation level where it reached its dew point temperature. When it reaches dew point the air is fully saturated. So,

Amount of temperature change = -2,500 m X .6^oC/100m = -15^o C

What does the -15^oC mean? It means that the air will rise and cool by 15^oC going from 500m to 3,000m.

Step 3. The last step is to calculate the new temperature.

To calculate the new temperature, that is the temperature at 3,000 meters, take the starting temperature and add the amount of temperature change found in Step 2 above. The starting temperature is the 10^o C because we started at 500 meters and rose upwards to 3,000 meters. So,

10^o C + (-15^o C) = - 5^o C

Now we're ready to finish the graph. Plot - 5^o C at 3,000 meters on the graph and connect this point to the dew point temperature at the condensation level. Because I used the SAR to compute this new temperature, I'll label this portion of the graph "SAR". (See graph) and I have a completed graph of the adiabatic temperature change.

Finally, draw and label a line that indicates the level of condensation. This is the altitude of the dew point temperature. Also label where the air is unstable and stable. The air is unstable from the surface up to 2400 meters because the parcel is warmer and less dense than the surrounding air between these two elevations. From 2400 meters to 3000 meters the air is stable because the parcel is colder than the surrounding air between these two levels. A cloud will start to form at the condensation level and grow upwards through that portion of the graph where the air is unstable. Once it reaches the elevation where the air is stable it will likely stop growing as it will resist uplift. (See graph)

What would happen to a parcel of air lifted to an elevation of 2800 meters and then released? (See graph).

Air density changes

Two things fundamentally determine the air's density, its temperature and the air pressure. Under the same external pressure (same elevation), colder air is more dense than warm air. As the air rises into the lower pressure environment aloft it will expand, (and yes, cool). But because of the expansion, it is less dense aloft even though it's colder that it is at the surface. Let's see how this relates to our example.

If I compare the density of my parcel to that of the surrounding air at the surface, the parcel will be less dense. Why? We are comparing the parcel to the surrounding air at the same elevation, hence the pressure is the same and thus temperature will determine the density. Because the parcel is warmer than the surrounding air at the surface, it is less dense (and will likely begin to rise).

Now given what was said above, compare the density of the parcel at the surface to its density at 3,000 meters (See graph). Where is it less dense?