Assume that a parcel of air is forced to rise up and over a 6000-foot-high mountain. The initial temperature of the parcel at sea level is 76.5°F, and the lifting condensation level (LCL) of the parcel is 3000 feet. The DAR is 5.5°F/1000’ and the SAR is 3.3°F/1000’. Assume that condensation begins at 100% relative humidity and that no evaporation takes place as the parcel descends. Indicate calculated temperatures to one decimal point.
1. Calculate the temperature of the parcel at the following elevations as it rises up the windward side of the mountain:
(a) 1000’_______°F
(b) 3000’ ______ °F
(c) 6000’ ______ °F
2. (a) After the parcel of air has descended down the lee side of the mountain to sea level, what is the temperature of the parcel?
________________________ °F
(b) Why is the parcel now warmer than it was at sea level on the windward side (what is the source of the heat energy)?
3. (a) On the windward side of the mountain, is the relative humidity of the parcel increasing or decreasing as it rises from sea level to 3000 feet?
(b) Why?
4. (a) On the lee side of the mountain, is the relative humidity of the parcel increasing or decreasing as it descends from 6000 feet to sea level?
(b) Why?
To answer the given questions, we need to apply the principles of adiabatic lapse rates, lifting condensation level, and the changes in relative humidity as the parcel of air rises up and over the mountain. Let's solve the questions step by step:
1. To calculate the temperature of the parcel at different elevations on the windward side of the mountain, we'll use the dry adiabatic rate (DAR) until the lifting condensation level (LCL) is reached, and then switch to the saturated adiabatic rate (SAR).
(a) At 1000 feet:
We start from sea level where the initial temperature is 76.5°F. At this elevation, we apply the DAR of 5.5°F/1000' to calculate the temperature change:
Temperature at 1000' = 76.5°F - (5.5°F/1000' * 1) = 71°F
(b) At 3000 feet (the LCL):
At the LCL, the parcel has reached its dew point and condensation begins. From this point onward, we need to use the SAR of 3.3°F/1000' to calculate the temperature change:
Temperature at 3000' = 71°F - (3.3°F/1000' * 2) = 64.4°F
(c) At 6000 feet:
Using the SAR again, we calculate the temperature change:
Temperature at 6000' = 64.4°F - (3.3°F/1000' * 3) = 54.5°F
Therefore, the calculated temperatures are:
(a) 1000 feet: 71.0°F
(b) 3000 feet: 64.4°F
(c) 6000 feet: 54.5°F
2. (a) After descending down the lee side of the mountain to sea level, the temperature of the parcel is the same as the initial temperature at sea level, which is 76.5°F.
(b) The parcel is warmer on the lee side compared to the windward side because, during its descent, it experienced compression, which resulted in an increase in temperature due to the conversion of potential energy to heat. This process is called adiabatic compression.
3. (a) On the windward side, the relative humidity of the parcel is increasing.
(b) As the parcel rises from sea level to 3000 feet, the temperature decreases according to the adiabatic lapse rates. According to the Clausius-Clapeyron equation, cold air holds less moisture than warm air. As the temperature decreases, the air becomes cooler, causing the relative humidity to increase until it reaches 100% at the lifting condensation level (LCL).
4. (a) On the lee side of the mountain, the relative humidity of the parcel is decreasing.
(b) As the parcel descends from 6000 feet to sea level, it encounters adiabatic warming or compression. The compression caused by the downward movement of air increases the temperature of the parcel. As the temperature increases, the parcel's capacity to hold moisture also increases, leading to a decrease in relative humidity.
Please note that these calculations are based on simplified assumptions and may not accurately represent real-world atmospheric conditions, but the principles and concepts used are applicable to understand the given scenario.