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            LAPSE RATE PRACTICE SHEET
ANSWERS

 

Let us review lapse rates in general.

 

1.         The normal lapse rate, some times referred to as the environmental lapse rate, is used when you are changing places or only the thermometer is being moved.  This would be used for the situation of determining the temperature at 20,000 feet above sea level, where a plane might be flying.

 

2.         The dry lapse rate is used when the air mass is moving, up or down, when there is no condensation.  Remember that as the air mass moves up or down, the temperature will change because it is an adiabatic temperature change.  That is a change in temperature without the addition or subtraction of heat (energy).  This is caused by the expansion, when the air mass moves up, or the compaction, as the air mass moves down, that occurs with a change in elevation.

 

3.         The wet lapse rate is used when the air mass is moving, up or down, in an area of condensation, such as in a cloud.  The difference between the wet and the dry lapse rate is caused by condensation.  Remember that condensation is a change in state and as condensation occurs latent heat of evaporation is released to the environment.  This release of heat will make the wet lapse rate to be less than the dry lapse rate.

 

The calculation of the lapse rate is really a two step process.  First you must calculate the temperature change over the elevation change.  Then you must calculate the temperature at the new elevation.

 

The lapse rate is given as degrees Fahrenheit/1000 feet or degrees Celcius/1000 meters elevation change.  That means that for every 1000 feet (or 1000 meters) the temperature will change the indicated amount.

 

Example:

 

Lapse Rate = (temperature change/1000 feet) (elevation change)

= (3.5O F/1000’) (5000 feet)

= (3.5 O F) (5)

= 17.5 O F

 

If the beginning temperature was 75 O F then

 

Temperature at 5000 feet higher elevation = 75 O F – 18 O F = 57 O F.

Temperature at 5000 feet lower elevation = 75 O F + 18 O F = 93 O F.

 

LR NORMAL        = 3.5 OF/1000’

LR DRY             = 5.0OF/1000’

LR WET             = 4.5OF/1000’

What is the temperature at each of the indicated elevations if the temperature on the west side of the mountains is 85O F at sea level?

20,000 feet
 
 

 

 

 

 

 

 


West                                                                                                                            East

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 


                                                                                         

 

 

In the diagram above you are presented with several possibilities of lapse rate use.  First we have a plane at 20,000 feet.  What is the temperature at that elevation?  For this example we are simply moving the thermometer from sea level, where the temperature is 85 O F, up to the elevation of the plane.

 

Calculations:

Temperature at the plane at 20,000 feet:

Temperature Change = (degrees Fahrenheit/1000 feet) ( elevation change)

            = (3.5 O F/1000 feet) ( 20,000 feet)

            = (3.5 O F) ( 20)

= 70 O F

 

Temperature at 20,000 feet = Sea Level Temperature – Temperature Change

            = 85 O F – 70 O F

            = 15 O F

 

Now we have the problem of the moving air.  Since there is not cloud in sight we must use the dry adiabatic lapse rate.

 

At 5000 feet:

Temperature Change = (5 F/1000 feet) ( 5000 feet)

            = ( 5 O F) ( 5 )

            = 25 O F

Temperature at 5000 feet = 85 O F – 25 O F

            = 60 O F

 

At 10,000 feet:  You now have a choice.  You can go from 5000 to 10,000 feet or you can start back at sea level.  Either way, unless you have to round your answer at an intermediate step, you will get to the same temperature.  Let us go from 5000 to 10,000 feet.

Temperature Change = (5 O F/1000 feet) ( 5000 feet)

            = (5 O F) ( 5)

            = 25 O F

Temperature at 10,000 feet = 60 O F – 25 O F

            = 35 O F

 

The air mass moves over the mountain range and down the other side, in this case to Lancaster.  Since we have a moving air mass without clouds, we continue to use the dry adiabatic lapse rate.

 

Back at Sea Level:

Temperature Change = (5 O F/1000 feet) ( 10,000 feet)

            = (5 O F) ( 10)

            = 50 O F

Temperature back at sea level = 35 O F + 50 O F

            = 85 O  F

 

Remember that you add because you are going down in elevation.

 

Does it surprise you that the temperature at sea level is the same on both sides of the mountain?  It shouldn’t because we have only moved an air mass.

 

 

 

 

 

 

 

 

 

 

LR NORMAL        = 3.5 OF/1000’

LR DRY             = 5.0OF/1000’

LR WET             = 4.5OF/1000’

What is the temperature at each of the indicated elevations if the temperature on the west side of the mountains is 85O F at sea level?

20,000 feet
 
 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 


                                                                                         

 

 

What is the difference between this diagram and the last one?  We now have a cloud on the side of the San Gabriel Mountains.  Its base is at 5000 feet and its top is at 10,000 feet elevation.  We can say that condensation begins at its base and complete evaporation has occurred at its top.  But the air mass continues on over the mountains.

 

Temperature at the plane at 20,000 feet:

Temperature Change = (degrees Fahrenheit/1000 feet) ( elevation change)

            = (3.5 O F/1000 feet) ( 20,000 feet)

            = (3.5 O F) ( 20)

= 70 O F

 

Temperature at 20,000 feet = Sea Level Temperature – Temperature Change

            = 85 O F – 70 O F

            = 15 O F

 

Temperature at 5000 feet:

Temperature Change = (5 F/1000 feet) ( 5000 feet)

            = ( 5 O F) ( 5 )

            = 25 O F

Temperature at 5000 feet = 85 O F – 25 O F

            = 60 O F

 

Now we have to account for the differences between these diagrams.  Since there is a cloud from 5000 to 10,000 feet, we must use the wet adiabatic lapse rate.  Remember the effect of condensation.

Temperature at 10,000 feet:

Temperature change = ( 4.5 F/1000 feet) ( 5000 feet)

            = (4.5 O F) (5)

            = 22.5 O F

Temperature at 10,000 feet = 60 O F – 23 O F

            = 37 O F

 

At the top of the cloud evaporation is complete and therefore as the air mass descends into Lancaster, we will use the dry adiabatic lapse rate again.

 

Back at Sea Level:

Temperature Change = (5 O F/1000 feet) ( 10,000 feet)

            = (5 O F) ( 10)

            = 50 O F

Temperature back at sea level = 37 O F + 50 O F

            = 87 O  F

 

Remember that you add because you are going down in elevation.

 

Why should it be hotter in Lancaster?  Since part of the temperature change on our side of the mountain occurred in the cloud, the rate of change was less than the rate of change used going down slope into Lancaster.

 

LR NORMAL        = 3.5 OF/1000’

LR DRY             = 5.0OF/1000’

LR WET             = 4.5OF/1000’

What is the temperature at each of the indicated elevations if the temperature on the west side of the mountains is 85O F at sea level?  If the HR is 75% at sea level on the west side of the mountain, what is the relative humidity at the other locations?

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 


                                                                                         

 

 

Since this is the same diagram as the preceding one our temperatures will be the same.  What has changed is that I am asking for relative humidity calculations to be made as well.  To make these calculations we will need not only the temperature but the chart for determining the capacity at different temperatures.  That chart follows this exercise.

 

Calculations:

At sea level:  The temperature is 85 O F and the HR is 75%.  What is the content?  This is important because that content is what the air mass will carry with it until it is changed!

 

HR = content/capacity  

OR 

content = ( HR ) ( capacity )

 

            = ( .75 ) ( 11.5 grains/cubic foot)  This is taken from the chart at the end of this exercise.

            = 8.6 grains/cubic foot

 

At 5000 feet:  The temperature is 60 O F.  What is the relative humidity?

By observation we see that a cloud has formed.  This means that condensation is taking place.  For condensation to take place the relative humidity must be 100%.  Therefore the real question is what is the capacity at the temperature at this location?  We know the temperature, so we can refer to the Capacity Chart and see that the capacity is 5.5 grains/cubic foot at 60 O F.  Since the relative humidity is 100%, the content must equal the capacity and be 5.5 grains/cubic foot.

 

The difference between the content determined above at sea level and that determined at 5000 feet is due to the condensation you see, the cloud.

 

 

At 10,000 feet:  By inspection we can see that in the cloud the relative humidity is 100% and therefore the content = capacity.  So the question is what is the capacity at 10,000 feet.  The temperature is 37 O F and from the capacity chart the capacity at 37 O F is approximately 2.5 grains per cubic foot.  Therefore the content is also 2.5 grains per cubic foot.

 

Also occurring here is evaporation.  Immediately above the cloud evaporation has equaled condensation so that all of the moisture has been converted back to water vapor.  Therefore the content has returned to the 8.6 grains per cubic foot that was determined above.

 

Back at sea level:  As the air mass travels back down to sea level the temperature is going up.  This will cause the capacity to increase and therefore the relative humidity will decrease.

At Sea Level:  The temperature is 87 O F and the content is 8.6 grains per cubic foot, while the capacity is 13 grains per cubic foot.

HR = 8.6/13 = 66%

 

No surprise that with the higher temperature that the relative humidity will be lower.  But what if thee was rain on our side.  How does that affect the numbers?
LR NORMAL        = 3.5 OF/1000’

LR DRY             = 5.0OF/1000’

LR WET             = 4.5OF/1000’

What is the temperature at each of the indicated elevations if the temperature on the west side of the mountains is 85O F at sea level?  If the HR is 75% at sea level on the west side of the mountain, what is the relative humidity at the other locations?

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 


                                                                                         

 

Now we have rain.  What does the rain do?  It effectively removes water from the system.  For the purposes of this exercise we will ignore the replacement by evaporation that occurs.

 

Will the rain change our temperatures?  No.

 

Will the rain change our relative humidity?  Yes because due to the rain we are removing water from the system.  Therefore at the top of the cloud, instead of evaporation returning the content of water vapor back to its original value, the rain has removed that water vapor and the content will equal the capacity at that temperature.

 

At 10,000 feet:  there is 100 % relative humidity and the temperature is 37 O F, determined previously.  From the Capacity chart we can see that the capacity at that temperature is approximately 2.5 grains per cubic foot.  Therefore the content will be 2.5 grains per cubic foot.

 

Back down at Sea Level:  Coming down the mountain to Lancaster, we have a temperature increase, already calculated to be 87 O F.  However, we have a much lower content and therefore the relative humidity will be lower as well.

 

HR = content/capacity

At 87 O F, the capacity is 13 grains per cubic foot.

Therefore HR = 2.5/13 = 19%

 

What these calculations demonstrate are the Rain Shadow Effect that we talked about in our discussion of deserts and the geologic history of North America.

 

CAPACITY CHART