Assume that a parcel of air is forcd to rise up and over a 6000 foot high mountain.The initial temp of the parcel at sea level is 76.5 F, and the LCL of the parcel is 3000 feet. The DAR is 5.5 * F/1000m and the SAR is 3.3*c/1000m. Assume that condensation begins at 100% relative humidity and that no evaporation takes place as the parcel descends.Indicate calculated temperatures to one decimal place.

Question

Assume that a parcel of air is forcd to rise up and over a 6000 foot high mountain.The initial temp of the parcel at sea level is 76.5 F, and the LCL of the parcel is 3000 feet. The DAR is 5.5 * F/1000m and the SAR is 3.3*c/1000m. Assume that condensation begins at 100% relative humidity and that no evaporation takes place as the parcel descends.Indicate calculated temperatures to one decimal place.

Answer 1

Please xplain what you mean by LCL, DAR and SAR. Why are you using degrees F for one and degrees C for the other?

Answer 2

To calculate the temperatures at different altitudes as the parcel of air rises and descends over the mountain, we will use the Dry Adiabatic Rate (DAR) and Saturated Adiabatic Rate (SAR), as well as the initial temperature at sea level.

1. Start with the given initial temperature: 76.5°F.

2. As the parcel of air rises above sea level, it cools down at the Dry Adiabatic Rate (DAR) of 5.5°F/1000 feet or 9.8°C/1000 meters.

The temperature at the Lifting Condensation Level (LCL) of 3000 feet can be calculated as follows:

76.5°F - DAR * (Altitude / 1000)
= 76.5°F - 5.5°F/1000ft * (3000ft / 1000)
= 76.5°F - 5.5°F * 3
= 76.5°F - 16.5°F
= 60.0°F

So, the temperature at the LCL is 60.0°F.

3. As the parcel continues to rise above the LCL, it cools down at the Saturated Adiabatic Rate (SAR) of 3.3°C/1000 meters.

To calculate the temperature at the top of the mountain (6000 feet or 1828.8 meters), we need to convert the units:

Altitude = 6000 feet * 1 meter/3.28084 feet
= 1828.8 meters

Temperature at the summit = LCL temperature - SAR * (Altitude / 1000)
= 60.0°F - 3.3°C/1000m * (1828.8m / 1000)
= 60.0°F - 6.03°C
≈ 60.0°F - 10.86°F
≈ 49.14°F

So, the temperature at the summit is approximately 49.1°F.

4. As the parcel descends from the summit back to sea level, there is no evaporation, and thus the temperature remains the same.

Therefore, the temperature at sea level after the parcel has descended remains 76.5°F.

To summarize:
- Initial temperature at sea level: 76.5°F
- Temperature at the LCL (3000 feet): 60.0°F
- Temperature at the summit (6000 feet): 49.1°F
- Temperature at sea level after descent: 76.5°F