Would someone please help me with this? Or at least tell me the formula I should be using? I've been trying to get help with this for four days now. I'm VERY frustrated.
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
To answer the questions, we need to understand the concept of the Dry Adiabatic Rate (DAR) and Saturated Adiabatic Rate (SAR). The DAR is the rate at which unsaturated air cools as it rises, which is given as 5.5°F per 1000 feet of elevation. The SAR is the rate at which saturated air cools as it rises and condenses, which is given as 3.3°F per 1000 feet of elevation.
Now, let's calculate the temperature at each given elevation:
1. To calculate the temperature of the parcel at different elevations as it rises up the windward side of the mountain, we need to consider whether the air is saturated or unsaturated.
(a) At 1000 feet:
Since this elevation is below the Lifting Condensation Level (LCL), the air is unsaturated. Therefore, we will use the DAR.
Formula: Temperature = Initial Temperature - (DAR * (Elevation / 1000))
Substituting the values:
Temperature = 76.5°F - (5.5°F/1000' * (1000'/1000))
Temperature = 76.5°F - (5.5°F)
Temperature = 71°F
(b) At 3000 feet:
At this elevation, the air is still unsaturated, so we use the DAR.
Temperature = 76.5°F - (5.5°F/1000' * (3000'/1000))
Temperature = 76.5°F - (16.5°F)
Temperature = 60°F
(c) At 6000 feet:
At this elevation, the air is saturated, so we use the SAR.
Temperature = 60°F - (3.3°F/1000' * (3000'/1000))
Temperature = 60°F - (9.9°F)
Temperature = 50.1°F
2. After the parcel of air has descended down the lee side of the mountain to sea level, we need to calculate the temperature at sea level.
Since the air is still saturated after descending, the temperature remains the same as the temperature at 6000 feet, which was 50.1°F.
Therefore, the temperature of the parcel at sea level after descending down the lee side of the mountain is 50.1°F.