Assume that a parcel of air is forced to rise up and over a 6000-foot high mountain (shown below). 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 the condensation begins at 100% relative humidity and that no evaporation takes place as the parcel descends. Indicate calculated temperatures to one decimal place.
1. Calculate the temp of the parcel at the following elevations as it rises up the windward side of the mountain
A 1000' ________________
B 3000' __________________
C 6000'__________________
2 after the parcel of air has descended dpwn the lee side of the mountain to sea level what is the temperature of the parcel?____________
why?_______________
5.One the winward side of the mountain, should the relative humidity of the parcel change as it rises from 3000 to 6000 feet?__________________
why_____________________
6. As the air rises up the windward side of the mountain
a what is the capacity (saturation mixing rate) of the rising air at 3000 feet?_____________________________g/kg
what is the capacity of the air at 6000 feet?________________________________g/kg
7. what is the capacity of the air after it descended back down to sea level on the lee side of the moutain? ____________
8. Assuming no water vapo is added as the parcel descends down the lee side of the mountain to sea, is the water vapor content (the mixing ration) of the parcel higher or lower than before it began to rise over the mountain?______________________
why_____________________
what is the lifting condensation level of this parcel now? ___________________________feet
To answer these questions, we'll need to use the given information and follow some steps.
Step 1: Determine the temperature at each elevation on the windward side of the mountain.
Given: Initial temperature at sea level = 76.5°F, DAR = 5.5°F/1000', LCL = 3000'
A) 1000 feet:
To find the temperature at 1000 feet, we need to calculate the temperature lapse rate (TLR) between sea level and the LCL.
TLR = DAR + SAR
= 5.5 + 3.3
= 8.8°F/1000'
Change in temperature from sea level to 1000 feet = (TLR * elevation change)/1000
= (8.8 * 1000)/1000
= 8.8°F
Temperature at 1000 feet = Initial temperature - change in temperature
= 76.5 - 8.8
= 67.7°F
B) 3000 feet:
The temperature at the LCL is the same as the calculated temperature at 1000 feet (67.7°F). From the LCL to 3000 feet, the temperature will remain constant.
Temperature at 3000 feet = 67.7°F
C) 6000 feet:
To find the temperature at 6000 feet, we need to calculate the temperature lapse rate from the LCL to 6000 feet.
TLR = DAR
= 5.5°F/1000'
Change in temperature from LCL to 6000 feet = (TLR * elevation change)/1000
= (5.5 * 3000)/1000
= 16.5°F
Temperature at 6000 feet = Temperature at LCL - change in temperature
= 67.7 - 16.5
= 51.2°F
Step 2: Determine the temperature of the parcel after descending the lee side of the mountain to sea level.
The temperature remains constant during descent, so the temperature at sea level after descending the lee side will be the same as the temperature at 6000 feet: 51.2°F.
Step 3: Determine if the relative humidity changes as the parcel rises from 3000 to 6000 feet on the windward side of the mountain.
No information is given about changes in relative humidity, so we assume it remains constant. Therefore, the relative humidity of the parcel will not change as it rises from 3000 to 6000 feet.
Step 4: Calculate the saturation mixing ratio (capacity) of the rising air at 3000 and 6000 feet.
The saturation mixing ratio can be calculated using the temperature values.
Capacity (g/kg) = 6.122 * e^(17.67 * T / (T + 243.5))
where T is the temperature in degrees Celsius.
a) At 3000 feet:
Temperature at 3000 feet = 67.7°F
Converting to Celsius: T = (67.7 - 32) * (5/9) = 19.8°C
Capacity at 3000 feet = 6.122 * e^(17.67 * 19.8 / (19.8 + 243.5))
= 6.122 * e^(350.9466 / 263.3)
≈ 16.6 g/kg
b) At 6000 feet:
Temperature at 6000 feet = 51.2°F
Converting to Celsius: T = (51.2 - 32) * (5/9) = 10.7°C
Capacity at 6000 feet = 6.122 * e^(17.67 * 10.7 / (10.7 + 243.5))
= 6.122 * e^(189.0669 / 254.2)
≈ 11.1 g/kg
Step 5: Determine the capacity of the air after descending back down to sea level on the lee side of the mountain.
The capacity remains constant during descent, so the capacity at sea level will be the same as the capacity at 6000 feet: 11.1 g/kg.
Step 6: Determine if the water vapor content (mixing ratio) of the parcel is higher or lower after descending down the lee side.
No water vapor is added or removed during descent, so the water vapor content (mixing ratio) remains the same. Therefore, it is the same as the capacity of the air after descending: 11.1 g/kg.
Step 7: Calculate the new lifting condensation level (LCL) after the parcel descends down the lee side.
The LCL can be calculated using the temperature at sea level after descending: 51.2°F
LCL = (T - 56) / (DAR - SAR)
= (51.2 - 56) / (5.5 - 3.3)
≈ 300 feet
The new LCL after descending is approximately 300 feet.