assume that a parcel of air is forced to rise up and over a 6000 foot high mountain. The intial temperature of the parcel at sea level is 76.5 F, and the lifting condensation level (lvl) of the parcel is 3000 feet. The DAR is 5.5 F/1000 and the SAR is 3.3 F/1000. Assume that conscends. Indicate calculated temperatures to one decimal place.
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. After the parcel of air has descended down the lee side of the mountain to sea level, what is the temperature of the parcel?
Why is the parcel now warmer than it was at sea level on the windward side (what is the source of the heat energy)?
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?
Why?
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?
why?
To calculate the temperature of the parcel at different elevations as it rises up the windward side of the mountain, we need to follow the dry adiabatic rate (DAR) and the lifting condensation level (LCL) until the parcel reaches the desired elevation.
1. (a) At 1000 feet, the parcel has risen from sea level to this elevation. To calculate the temperature, we need to determine the change in temperature for the given elevation using the DAR. The DAR is given as 5.5 F/1000. So, we calculate:
Change in temperature = DAR * (elevation change/1000)
= 5.5 F/1000 * (1000 ft/1000)
= 5.5 F
Temperature at 1000 ft = Initial temperature at sea level - Change in temperature
= 76.5 F - 5.5 F
= 71.0 F
Therefore, the temperature of the parcel at 1000 feet is 71.0°F.
1. (b) Similarly, to calculate the temperature at 3000 feet, we continue to use the DAR:
Change in temperature = DAR * (elevation change/1000)
= 5.5 F/1000 * (3000 ft/1000)
= 16.5 F
Temperature at 3000 ft = Initial temperature at sea level - Change in temperature
= 76.5 F - 16.5 F
= 60.0 F
Therefore, the temperature of the parcel at 3000 feet is 60.0°F.
1. (c) Finally, at 6000 feet, we again use the DAR:
Change in temperature = DAR * (elevation change/1000)
= 5.5 F/1000 * (6000 ft/1000)
= 33.0 F
Temperature at 6000 ft = Initial temperature at sea level - Change in temperature
= 76.5 F - 33.0 F
= 43.5 F
Therefore, the temperature of the parcel at 6000 feet is 43.5°F.
2. After the parcel of air descends down the lee side of the mountain to sea level, we need to consider the sinking adiabatic rate (SAR) to calculate the temperature. The SAR is given as 3.3 F/1000.
Change in temperature = SAR * (elevation change/1000)
= 3.3 F/1000 * (6000 ft/1000)
= 19.8 F
Temperature at sea level (after descending) = Temperature at 6000 ft + Change in temperature
= 43.5 F + 19.8 F
= 63.3 F
Therefore, the temperature of the parcel after descending to sea level on the lee side is 63.3°F.
The parcel is now warmer on the lee side because as air descends, it is compressed, which causes it to warm up. This compression occurs due to the increase in atmospheric pressure when the air sinks, resulting in an increase in temperature. This process is known as adiabatic compression heating.
On the windward side of the mountain, the relative humidity of the parcel is decreasing as it rises from sea level to 3000 feet. This is because as the parcel rises, it expands due to the decrease in atmospheric pressure. As the air expands, its capacity to hold water vapor increases, causing the relative humidity to decrease. Eventually, when the parcel reaches the lifting condensation level (LCL), the air cools at the moist adiabatic rate (MAR) until it reaches saturation and condenses into clouds.
On the lee side of the mountain, the relative humidity of the parcel is increasing as it descends from 6000 feet to sea level. As the parcel descends, it gets compressed, leading to an increase in temperature. The increased temperature causes the parcel's capacity to hold water vapor to increase, resulting in a decrease in relative humidity. However, since no new moisture is added to the parcel during descent, the water vapor content remains constant, which makes the relative humidity increase.