The diagram above represents western California’s Sierra Mountains. Air generally moves from west (windward side) to east (leeward side). Assume that a parcel of air is forced to rise up and over the 4000-meter-high mountain (as shown). The DAR is 10°C/1000 m and the MAR is 6°C/1000 m. Assume that condensation begins at 100% relative humidity and that no evaporation takes place as the parcel descends.
The initial temperature of the parcel at sea level is 25°C, and the lifting condensation level (LCL) of the parcel is 2000 meters.
What is the dew point temperature? ______°C
At the dew point temperature how much water vapor can the air hold? That is, what is the maximum specific humidity? ____ grams
At 25°C how much water vapor can the air hold? That is, what is the maximum specific humidity? ____ grams
At 0m the air initially begins with 5.4 grams of water vapor.
What is the initial relative humidity at 0m?
At 4000m the air now only has 2 grams of water vapor and it has lost 3.4 grams of vapor, because the water turned into clouds.
At 0m on the leeward side the air still has 2 grams of water vapor.
How much water vapor can it now hold at 0m on the leeward side? _____ grams
What is the relative humidity at 0m on the leeward side? _______ %
Explain why the parcel is now warmer than it was at sea level on the windward side (what is the source of the heat energy).
Please please help anyone I have no idea and this is due today
To find the dew point temperature, we need to determine the temperature at which saturation occurs. The initial temperature of the parcel at sea level is 25°C, and the lifting condensation level (LCL) of the parcel is 2000 meters.
To calculate the dew point temperature at the LCL, we need to take into account the Dewpoint Depression (DD), which is the difference between the temperature and the dewpoint temperature.
The DD at sea level would be 25°C - 100% relative humidity = 0°C.
Using the Dry Adiabatic Rate (DAR), which is given as 10°C/1000 m, we can calculate how much the temperature drops during the ascent from sea level to the LCL.
In this case, the parcel is forced to rise 2000 meters, so the temperature would decrease by 10°C x 2 = 20°C.
Now, subtract the Dewpoint Depression (DD) from the temperature drop:
20°C - 0°C = 20°C.
Therefore, the dew point temperature at the LCL would be 20°C.
The maximum specific humidity, also known as the maximum amount of water vapor the air can hold, depends on the temperature.
At the dew point temperature of 20°C, the maximum specific humidity is determined using the saturation vapor pressure at that temperature.
To calculate it, we need the saturation vapor pressure values from a table or equation. Suppose the saturation vapor pressure at 20°C is 17.4 mb.
Then, using the maximum specific humidity formula, we can determine the maximum amount of water vapor the air can hold at that temperature:
17.4 mb x 2.87 g/mb = 49.998 g.
So, the maximum specific humidity at the dew point temperature is approximately 50 grams.
To find the initial relative humidity at 0m, we need to calculate the specific humidity at 0m, which is given as 5.4 grams.
The initial relative humidity can be calculated by dividing the specific humidity at 0m by the maximum specific humidity at 0m and multiplying by 100:
(5.4 g / 49.998 g) x 100 = 10.8%.
Now, let's move on to the leeward side. At 4000m, the air has lost 3.4 grams of water vapor due to condensation. Therefore, the specific humidity at 4000m is 2 grams.
To find the maximum specific humidity at the leeward side, we need to determine the temperature at that height. Assuming it's still 25°C, we use the same formula as before:
17.4 mb x 2.87 g/mb = 49.998 g.
So, the maximum specific humidity at the leeward side is also approximately 50 grams.
To find the relative humidity at 0m on the leeward side, we divide the specific humidity (2 g) by the maximum specific humidity (50 g) and multiply by 100:
(2 g / 50 g) x 100 = 4%.
Finally, to explain why the parcel is now warmer on the leeward side than at sea level, it's important to understand the process of adiabatic compression.
When the air parcel descends on the leeward side, it undergoes adiabatic compression, which occurs when a parcel of air is compressed, leading to an increase in its temperature. This compression occurs because the air is descending and experiencing higher pressure as it moves downward.
As the air parcel descends, it compresses, causing the air molecules to collide more frequently and with greater force. This increase in molecular collisions transfers kinetic energy into the parcel, resulting in a rise in temperature. Therefore, the source of the heat energy is the process of adiabatic compression.