A parcel of air is being forced up the windward side of a 10,000-foot-high mount

Question

A parcel of air is being forced up the windward side of a 10,000-foot-high mountain. The parcel has a temperature of 86.5 degree F at 0 feet in elevation. If the LCL is located at 8,000 feet, what is the temperature of this parcel of air at 1,000 feet? What is the temperature of the parcel of air, detailed in question 1, at the LCL (8,000 feet)? What is the temperature of the parcel of air, detailed in question 1, at the top of the mountain (10,000 feet)? After the parcel of air has descended down the leeward side of the mountain to sea level, what is the temperature of the parcel? What is the source of heat energy that has caused the parcel of air to be warmer on the leeward side of the mountain and how much has the temperature increased as a result? Using Table 13-1 (Page 66) of the laboratory manual, what is the capacity (saturated mixing ratio) of the parcel of air at 2,000 feet on the windward side of the mountain (estimate the value as needed)? Assuming the mixing ratio of the parcel of air at 2,000 feet is 5.6 g/kg what it is the relative humidity of the parcel in percent? Assuming the mixing ratio of the parcel of air on the leeward side of the mountain is 4.5 g/kg and that no additional water vapor is added to parcel of air, does the relative humidity increase or decrease as it descends the mountain? What explains the increase or decrease in relative humidity identified in Question 8?

Explanation / Answer

Answer:

The lifted condensation level is the point where relative humidity becomes 100 %. And it can be calculated as follows;

Height= 125*(T-Td)

where, T= temperature of air, Td= temperature at dew point.

Calculate first dew point temperature then find temperature at different heights by keeping dew point constant.

Example: 125( 86.5- Td) = 8000

Td= 22.5 F

Now, for height 1000 ft

1000= 125(T-22.5)

T= 30.5 F

Likewise above, we can calculate the temperature at 8000 ft and 10,000 ft.

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