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?
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To calculate the temperatures of the parcel at different elevations and understand the changes in relative humidity, we'll need to follow these steps:
Step 1: Understand the given information and values:
- Initial temperature at sea level: 76.5°F
- Lifting condensation level (LCL): 3000 feet
- Dry Adiabatic Rate (DAR): 5.5°F/1000'
- Saturated Adiabatic Rate (SAR): 3.3°F/1000'
- Condensation begins at 100% relative humidity
- No evaporation takes place as the parcel descends down the lee side of the mountain
Step 2: Calculate the temperature of the parcel at different elevations on the windward side
The temperature change during ascent will vary depending on the elevation. We'll need to consider both the dry adiabatic rate (DAR) and the saturated adiabatic rate (SAR).
(a) To calculate the temperature at 1000 feet, we need to consider the DAR:
Temperature change = DAR * (height change in thousands of feet)
Temperature change = 5.5°F/1000' * 1
Temperature change = 5.5°F
Temperature at 1000 feet = Initial temperature - Temperature change
Temperature at 1000 feet = 76.5°F - 5.5°F = 71.0°F
(b) To calculate the temperature at 3000 feet, we need to consider the DAR until the LCL and then the SAR after the LCL is reached:
Temperature change until LCL = DAR * (LCL in thousands of feet)
Temperature change until LCL = 5.5°F/1000' * 3
Temperature change until LCL = 16.5°F
Temperature change after LCL = SAR * [(Height - LCL) in thousands of feet]
Temperature change after LCL = 3.3°F/1000' * (3000 - 3000)
Temperature change after LCL = 0°F
Temperature at 3000 feet = Initial temperature - Temperature change until LCL - Temperature change after LCL
Temperature at 3000 feet = 76.5°F - 16.5°F - 0°F = 60.0°F
(c) To calculate the temperature at 6000 feet, we need to consider the SAR:
Temperature change = SAR * (height change in thousands of feet)
Temperature change = 3.3°F/1000' * 6
Temperature change = 19.8°F
Temperature at 6000 feet = Initial temperature - Temperature change
Temperature at 6000 feet = 76.5°F - 19.8°F = 56.7°F
Step 3: Calculate the temperature of the parcel after descending down the lee side to sea level
On the lee side, no evaporation takes place, so the temperature change is not affected. The parcel will simply experience the reverse temperature changes it went through during ascent.
(a) To calculate the temperature at sea level on the lee side, we need to consider the temperature change until LCL:
Temperature change until LCL = -DAR * (LCL in thousands of feet)
Temperature change until LCL = -5.5°F/1000' * 3
Temperature change until LCL = -16.5°F
Temperature change after LCL = -SAR * (Height - LCL) in thousands of feet (which is zero in this case)
Temperature at sea level on the lee side = Initial temperature + Temperature change until LCL + Temperature change after LCL
Temperature at sea level on the lee side = 76.5°F + (-16.5°F) + 0°F = 60.0°F
(b) The parcel is now warmer on the lee side than it was at sea level on the windward side because it has descended and compressed. As the parcel descended, the pressure increased, causing the air molecules to compress, which increases the temperature. This is known as the adiabatic compression.
Step 4: Determine the changes in relative humidity
(a) On the windward side, the relative humidity of the parcel is increasing as it rises from sea level to 3000 feet. This is because as the parcel rises, it expands and cools (due to the reduction in air pressure), which causes the relative humidity to increase until it reaches the lifting condensation level (LCL).
(b) The relative humidity is increasing because the temperature is decreasing due to the expansion and cooling of the parcel. As the temperature decreases, the parcel reaches its dew point, which results in the condensation of moisture and an increase in relative humidity.
(a) On the lee side of the mountain, the relative humidity of the parcel is decreasing as it descends from 6000 feet to sea level. This is because the temperature increases during the descent, causing the relative humidity to decrease.
(b) The relative humidity decreases on the lee side because the temperature increases due to adiabatic compression. As the parcel descends, the pressure increases, causing the air molecules to compress and warm up. The increase in temperature decreases the relative humidity of the parcel.