ATM OCN (Meteorology) 100

Answers for Homework 3 
ATMOSPHERIC HUMIDITY

Fall 1999

Date Due: Wednesday, 20 October 1999

The total maximum points were 50. Point distribution for each question noted below. 

PART A. SLING PSYCHROMETER

Part A

(6 pts. or 2 pts each)

This question involves reading (hopefully, correctly) the Psychrometric Tables. Note that the vertical columns in the Psychrometric Tables contain the Wet bulb depression, defined as the difference between the dry bulb (or air) temperature and the wet bulb temperature.

 

Wet bulb depression =  Relative Humidity =  Dewpoint temperature =

(15 deg C - 12 deg C) = 3 deg C  70%  9.6 deg C (10 deg C is acceptable)

PART B. OTHER HUMIDITY MEASURES

a. If the air temperature were 76°F and the dewpoint were 56°F, what is the relative humidity?

b. If air temperature were 62°F and the relative humidity were 80%, what is the dewpoint?

c. Which condition a. or b. would have more water vapor? Please explain!

Part B

a. (4 pts.)

 

Relative humidity = 49.93% (or 50% is acceptable)  The relative humidity is determined from RH = e/es x 100. Since the air temperature is 76 deg F, you can determine that the saturation vapor pressure es at that temperature is 30.64 mb from the table. Using the dewpoint temperature of 56 deg, the tabular entry would indicate an actual vapor pressure of 15.30 mb.  Hence, RH = (15.30 mb / 30.64 mb) x 100 = 49.93%.

b. (4 pts.)

 

Dewpoint = 56 deg F  In this case from the air temperature = 62 deg F, the saturation vapor pressure es = 18.96 mb can determined from the Table. Since the relative humidity is known (80%), you can solve for the actual vapor pressure, by rearranging the equation from above. e = es x RH/100% or e = 18.96 mb x [80% / 100%] = 15.17 mb. Now enter the table to find the dewpoint corresponding to this vapor pressure - or 55.8 deg F (using a simple interpolation), which can be rounded up to 56 deg F.

 

Both cases, a and b, have the same water vapor content,  since the dewpoint in each case is 56 deg; equivalently, both have a vapor pressure of approximately 15.3 mb. This problem is meant to illustrate that while the relative humidity in b is higher (80% as compared with 50% in a), the air temperature is lower, but the dewpoint is essentially the same. NOTE: Suggestion that b is more humid because of a higher relative humidity is incorrect!

2. ADIABATIC PROCESSES - THE MOUNTAIN BARRIER

PART A. The initial ascent 
(Clearly show your work for partial credit!)

An air parcel is part of a Pacific maritime air mass moving toward the West Coast. The parcel, located just above the ocean surface, has an initial air temperature of 10°C and a dewpoint temperature of 5°C. This air parcel is forced to ascend the Coastal Range, with a 2000 m elevation.

A.1. How far up the west (windward) slope of the mountain range would the air have to be lifted in order for it to become saturated? (Assume no change in moisture content of the air, no physical phase changes and that the dewpoint remained constant).

[Please include units!]

A.2. What is the air temperature at this level?

A.3. What is the dewpoint temperature at this level? 

---

Part A (6 pts. or 2 pts. each)

A.1.

 

The height of the cloud base, where the lifted air just becomes saturated is 500 meters (1/2 km).  Assume that the air parcel cools at a dry adiabatic lapse rate of 10 deg C per 1000 meters and that the surface dewpoint remains constant. Since the parcel only needs to cool by 5 deg C (= 10 deg - 5 deg C) to become saturated, only a 500 meter lift is needed.

A.2

 

Air temperature = 5 deg C  See above discussion; using the dry adiabatic lapse rate over 500 meters, the air parcel cools by 5 deg C, yielding 10 - 5 = 5 deg C.

A.3

 

Dewpoint = 5 deg C  For this simple model, the dewpoint is assumed to remain constant for a dry adiabatic ascent (See above discussion). {In reality, the dewpoint does decrease when lifted by a rate of about 0.2 deg C per 1000 meters).

---

PART B. To the top

The air parcel continues up the 2000 meter mountain, condensing out moisture as clouds as it rises and all condensate falls to the ground as precipitation.

B.1. What is the air parcel temperature at the crest of the mountain (2000 m above sea level)?

[Assume the average moist adiabatic lapse rate of 7°C per 1 km]

B.2. What is the dewpoint of the parcel at the crest of the mountain? 

---

Part B (6 pts. or 3 pts. each)

B.1.

 

Air temperature = -5.5 deg C  Since the air parcel is now saturated, the air parcel will cool at the saturation adiabatic lapse rate (7 deg C per 1000 meters) over (2000 m - 500 m) = 1500 meters. In other words, the parcel cools by 10.5 Celsius degrees, meaning that 5 C - 10.5 C = -5.5 deg C.

B.2

 

Dewpoint = -5.5 deg C  When the saturated air is lifted, condensation takes place and the dewpoint simultaneously cools along with the air temperature, since the parcel will remain saturated.

---

PART C: The descent.

The air parcel now moves down the east slope of the mountain, to a valley that has an elevation essentially at sea level.

C.1. What is the air parcel temperature in this valley?

[Assume no physical phase changes in any residual moisture or clouds]

C.2. What is the dewpoint temperature of the parcel in this valley? 

---

Part C (6 pts. or 3 pts. each)

C.1.

 

Air temperature = 14.5 deg C  Using the dry adiabatic lapse rate, causing an air parcel to sink over 2000 meters, will heat the parcel by 20 C degrees, from -5.5 deg C to 14.5 deg C.

C.2

 

Dewpoint = -5.5 deg C  For this simple model, since the dewpoint is assumed to remain constant for a dry adiabatic descent (see above discussion), the dewpoint will remain the same as air at the top of the mountain.

---

PART D: Comparison.

D.1. How has the parcel temperature changed from start to the end?

D.2. How has the parcel dewpoint changed from start to the end?

D.3 How has the relative humidity changed from beginning to end?

---

Part. D (9 pts. or 3 pts. each)

D.1.

 

Air temperature = 4.5 C deg warmer Parcel started at 10 deg C and ended at 14.5 deg C for a net warming of +4.5 C deg.

D.2
 

Dewpoint = 10.5 C deg less Parcel dewpoint started at 5 deg C and ended at -5.5 deg C, for a net change of -10.5 C deg. C.

 D.3
 

Relative humidity is less at the end Since the air temperature increased and the dewpoint decreased, the temperature-dewpoint spread increased, which means that the relative humidity decreased. (This qualitative answer was sufficient)  For a rigorous proof, the Relative humidity of the parcel initially was approximately 71% (since T = 10 deg C = 50 deg F, which yields a saturation vapor pressure of 12.3 mb and a dewpoint of 5 deg C = 41 deg F yields an actual vapor pressure of 8.7 mb)  The relative humidity at the end was 24.5% (since T = 14.5 deg C = 58.1 deg F or saturation vapor pressure of 1.5 mb and a dewpoint of -5.5 deg C = 22.1 deg F has an actual vapor pressure of 4.05 mb).

URL Address: aos100/homework/f9hmk03a.htm