I don't get any of it!
Assume that a parcel of air is forced to rise up and over a 6000-foot-high mountain (see page 79 in the Laboratory Manual). 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 (b) Why is the parcel now warmer than it was at sea level on the windward side (what is the source of the heat energy)?
3. (a) 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? __________________________ (b) Why?
4. (a) 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? __________________________ (b) Why?
Well, let's see if we can have some fun with this mountain adventure!
1. (a) At 1000 feet, the temperature of the parcel would be 73.0°F. It's getting a little chilly up there!
(b) At 3000 feet, the temperature of the parcel would be 67.5°F. Time to put on a sweater!
(c) At 6000 feet, the temperature of the parcel would be 55.0°F. Brrr, it's getting frosty!
2. (a) After descending down the lee side of the mountain to sea level, the temperature of the parcel would be 76.5°F, the same as it was at sea level on the windward side. It's come full circle!
(b) The parcel is now warmer because it had a little boost from the heat energy released during the condensation process. It's like the mountain gave it a warm hug!
3. (a) On the windward side of the mountain, the relative humidity of the parcel is increasing as it rises from sea level to 3000 feet. It's getting more and more misty up there!
(b) The relative humidity is increasing because as the parcel rises, it's cooling down and reaching its dew point. Condensation starts happening, and voila, more moisture!
4. (a) On the lee side of the mountain, the relative humidity of the parcel is decreasing as it descends from 6000 feet to sea level. It's drying up!
(b) The relative humidity is decreasing because as the parcel descends, it's getting warmer and can hold more moisture. So, all that water in the air is evaporating away like it's on a Caribbean vacation!
I hope that brought some humor to your mountain climbing journey! Remember, even in science, there's always room for laughter!
To calculate the temperature of the parcel at different elevations, we can use the Dry Adiabatic Rate (DAR) and the lifting condensation level (LCL).
1. Calculating the temperature at different elevations on the windward side:
(a) To calculate the temperature at 1000 feet, we need to determine how many 1000-foot increments the parcel has risen from sea level. Since the LCL is at 3000 feet, the parcel is already at the LCL, so it has risen 1 increment. Therefore, the temperature at 1000 feet would be the initial temperature minus the DAR multiplied by the number of increments: 76.5°F - (5.5°F/1000' * 1) = 71.0°F.
(b) To calculate the temperature at 3000 feet, the parcel has risen 3 increments. Therefore, the temperature at 3000 feet would be the initial temperature minus the DAR multiplied by the number of increments: 76.5°F - (5.5°F/1000' * 3) = 64.5°F.
(c) At 6000 feet, the parcel has risen 6 increments. Therefore, the temperature at 6000 feet would be the initial temperature minus the DAR multiplied by the number of increments: 76.5°F - (5.5°F/1000' * 6) = 43.5°F.
2. (a) After the parcel has descended down the lee side of the mountain to sea level, the temperature will be the same as the initial temperature at sea level, which is 76.5°F.
(b) The parcel is warmer than it was at sea level on the windward side because as the parcel descends, it is compressed by the increasing atmospheric pressure. This compression leads to adiabatic heating of the air parcel and causes its temperature to increase.
3. (a) On the windward side of the mountain, the relative humidity of the parcel is decreasing as it rises from sea level to 3000 feet.
(b) The relative humidity decreases because as the parcel rises, it expands due to the decreasing atmospheric pressure. This expansion causes the air parcel to cool at the Dry Adiabatic Rate (DAR), leading to a decrease in temperature. As the temperature decreases, the air's ability to hold moisture decreases, resulting in a decrease in relative humidity.
4. (a) On the lee side of the mountain, the relative humidity of the parcel is increasing as it descends from 6000 feet to sea level.
(b) The relative humidity increases because as the parcel descends, it is compressed by the increasing atmospheric pressure. This compression leads to adiabatic warming of the air parcel and causes its temperature to increase. As the temperature increases, the air's ability to hold moisture increases, resulting in an increase in relative humidity.
To solve this problem, we need to follow a step-by-step process using the given information and understanding some key concepts related to atmospheric processes. Let's go through each question and explain how to arrive at the answers.
1. Calculate the temperature of the parcel at the following elevations as it rises up the windward side of the mountain:
(a) To calculate the temperature at 1000 feet, we can use the Dry Adiabatic Rate (DAR) of 5.5°F/1000’. The parcel rises 1000 feet, so we will subtract 5.5°F from the initial temperature of 76.5°F:
New temperature = 76.5°F - (5.5°F/1000' * 1,000')
Answer: _______°F
(b) To calculate the temperature at 3000 feet, we continue using the DAR:
New temperature = Previous temperature at 1000 feet - (5.5°F/1000' * 2,000')
Answer: _______°F
(c) To calculate the temperature at 6000 feet, we can use the Same Adiabatic Rate (SAR) of 3.3°F/1000’. However, we need to apply the LCL to determine whether condensation occurs. As the LCL is at 3000 feet, condensation will occur at this level and alter the temperature. We need to calculate the temperature using the Moist Adiabatic Rate (MAR) below the LCL:
- Calculate the temperature at the LCL:
Temperature at LCL = Temperature at sea level - (DAR * 3,000')
- Calculate the temperature change below the LCL using the MAR, which is 3.3°F/1000':
Temperature change below LCL = (DAR - SAR) * (LCL - 0')
- Subtract the temperature change below the LCL from the temperature at the LCL:
New temperature = Temperature at LCL - Temperature change below LCL
Answer: _______°F
2. (a) To calculate the temperature of the parcel after descending the lee side of the mountain to sea level, we need to understand the concept of adiabatic compression and consider the heating that occurs during this process. When the parcel descends, it undergoes adiabatic compression, which increases its temperature. We need to calculate the temperature change using the DAR:
Temperature change = DAR * 6,000' (as the parcel descends 6,000 feet)
New temperature = Initial temperature on the windward side + Temperature change
Answer: ________________________ °F
(b) The parcel is now warmer than it was at sea level on the windward side because of the adiabatic compression it experienced during descent. Adiabatic compression increases the temperature of air due to the increase in atmospheric pressure, which is the source of the heat energy.
3. (a) On the windward side of the mountain, the relative humidity of the parcel is decreasing as it rises from sea level to 3000 feet.
(b) The relative humidity decreases because, as air rises, it expands and cools adiabatically, leading to a decrease in the saturation mixing ratio. This results in a decrease in the amount of water vapor the air can hold, causing the relative humidity to decrease.
4. (a) On the lee side of the mountain, the relative humidity of the parcel is increasing as it descends from 6000 feet to sea level.
(b) The relative humidity increases because the descending air undergoes adiabatic compression, which increases its temperature. As the temperature rises, the saturation mixing ratio increases, allowing the air to hold more water vapor. Therefore, the relative humidity increases as the parcel descends.
By following these explanations and using the specific formulas mentioned, you should be able to calculate the missing values and complete the problem.