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
1(a) 71
1(b) 60
1(c) 51
2(a) 84
To calculate the temperature of the parcel at different elevations, you will need to use the Dry Adiabatic Rate (DAR) and the Saturated Adiabatic Rate (SAR). The DAR and the SAR are rates at which the temperature of a parcel of air changes as it rises or descends.
1. To calculate the temperature of the parcel at different elevations on the windward side of the mountain, you can use the following formula:
Temperature = Initial Temperature + (Rate of Change * (Elevation - LCL))
(a) For an elevation of 1000 feet:
Temperature = 76.5°F + (5.5°F/1000' * (1000' - 3000'))
Temperature = 76.5°F + (5.5°F/1000' * -2000')
Temperature = 76.5°F - 11°F = 65.5°F
(b) For an elevation of 3000 feet (the LCL):
Temperature will be the initial temperature at sea level, which is 76.5°F.
(c) For an elevation of 6000 feet:
Temperature = 76.5°F + (5.5°F/1000' * (6000' - 3000'))
Temperature = 76.5°F + (5.5°F/1000' * 3000')
Temperature = 76.5°F + 16.5°F = 93°F
2. To calculate the temperature after the parcel descends down the lee side of the mountain to sea level, you can use the same formula:
Temperature = Initial Temperature + (Rate of Change * Elevation)
(a) For an elevation of 6000 feet:
Temperature = 93°F + (3.3°F/1000' * 6000')
Temperature = 93°F + (3.3°F/1000' * 6000')
Temperature = 93°F + 19.8°F = 112.8°F
So, after the parcel of air descends down the lee side of the mountain to sea level, the temperature of the parcel will be approximately 112.8°F.