Wednesday Mar. 26, 2014

Enough time for a long song "Evil One" from Appleseed Collective before class this afternoon.

The In-class Optional Assignment was returned in class today.  Here are answers to the questions.

I sometimes hide Optional Assignments in the Online Lecture notes.  

There is some suggested reading online that attempts to explain why there is an upper limit to the amount of water vapor that can be found in air and why that upper limit depends on temperature.

A couple of new 1S1P topics are now available online.

A new Optional Assignment is now available and is due next Monday.  Be sure to have the assignment completed before you come to class.


A quick review of the 4 humidity variables introduced on Monday.

humidity variable
mixing ratio
saturation mixing ratio
relative humidity
dew point temperature
"job"
how much water vapor is actually in the air
the maximum amount of water vapor that can be found in air
how close is the air to being filled to capacity with water vapor
1. like mixing ratio it gives an idea of the actual amount of water vapor in the air
2. cool the air to its dew point and RH becomes 100%
units
g/kg
g/kg
%
oF
when and why do the values of these variables change?

this is what we'll be working on today


We're going to work 4 example problems.  It's by doing these problems that you start to understand how these various variables behave and what can cause them to change value.

Example #1
In this example you are given an air temperature of 90 F and a mixing ratio value of 6 g/kg.  You are supposed to determine the mixing ratio and the dew point temperature.



Here's something close to what we ended up with in class.  This would be hard enough to sort out even if you were in class.  So we'll work through this problem in a more detailed, step-by-step manner.



We're given an air temperature of 90 F and a mixing ratio (r) of 6 g/kg.  We're  supposed to find the relative humidity (RH) and the dew point temperature. 
We start by entering this data in the table.

Anytime you know the air's temperature you can look up the saturation mixing ratio value on a chart (such as the one on p. 86 in the ClassNotes); the saturation mixing ratio is 30 g/kg for 90 F air.  90 F air could potentially hold 30 grams of water vapor per kilogram of dry air (it actually contains 6 grams per kilogram in this example). 

Once you know mixing ratio and saturation mixing ratio you can calculate the relative humidity (you divide the mixing ratio by the saturation mixing ratio, 6/30, and multiply the result by 100%).  You ought to be able to work out the ratio 6/30 in your head (6/30 = 1/5 = 0.2).  The RH is 20%. 



The numbers we just figured out are shown on the top line above.

(A) We imagined cooling the air from 90F to 70F, then to 55F, and finally to 45F.

(B) At each step we looked up the saturation mixing ratio and entered it on the chart.  Note that the saturation mixing ratio values decrease as the air is cooling.


(C) The mixing ratio (r) doesn't change as we cool the air.  The only thing that changes r is adding or removing water vapor and we aren't doing either.  This is probably the most difficult concept to grasp.

(D) Note how the relative humidity is increasing as we cool the air.  The air still contains the same amount of water vapor it is just that the air's capacity is decreasing.

Finally at 45 F the RH becomes 100%.  This is kind of a special point.  You have cooled the air until it has become saturated. 
The dew point temperature in this problem is 45 F.

What would happen if we cooled the air further still, below the dew point temperature?



35 F air can't hold the 6 grams of water vapor that 45 F air can.  You can only "fit" 4 grams of water vapor into the 35 F air.  The remaining 2 grams would condense.  If this happened at ground level the ground would get wet with dew.  If it happens above the ground, the water vapor condenses onto small particles in the air and forms fog or a cloud.  Because water vapor is being taken out of the air (the water vapor is turning into water), the mixing ratio will decrease from 6 g/kg to 4 g/kg.  As you cool air below the dew point, the RH stays constant at 100% and the mixing ratio decreases.

In many ways cooling moist air is liking squeezing a moist sponge.



Squeezing the sponge and reducing its volume is like cooling moist air and reducing the saturation mixing ratio.  At first (Path 1 in the figure) when you squeeze the sponge nothing happens, no water drips out.  Eventually you get to a point where the sponge is saturated.  This is like reaching the dew point.  If you squeeze the sponge any further (Path 2) water will begin to drip out of the sponge (water vapor will condense from the air).



Example 2
We're given an air temperature of 90 F and a relative humidity of 50%; we'll try to figure out the mixing ratio and the dew point temperature.  Here's something like what we ended up with in class.


 The problem is worked out in detail below:



First you fill in the air temperature and the RH data that you are given.

(A) since you know the air's temperature you can look up the saturation mixing ratio (30 g/kg). 

(B)  Then you might be able to figure out the mixing ratio in your head.  Air that is filled to 50% of its capacity could hold up to 30 g/kg.  Half of 30 is 15, that is the mixing ratio.  Or you can substitute into the relative humidity formula and solve for the mixing ratio.  The details of that calculation are shown above at B.



Finally you imagine cooling the air.  Notice how the saturation mixing ratio decreases, the mixing ratio stays constant, and the relative humidity increases as the air is cooled.   In this example the RH reached 100% when the air had cooled to 70 F.  That is the dew point temperature.



We can use results from humidity problems #1 and #2 to learn and understand a useful rule.



In the first example the difference between the air and dew point temperatures was large (45 F) and the RH was low (20%).

In the 2nd problem the difference between the air and dew point temperatures was smaller (20 F) and the RH was higher (50%). 

The easiest way to remember this rule is to remember the case where there is no difference between the air and dew point temperatures.  The RH then would be 100%.



Example 3



You're given the the mixing ratio = 10.5 g/kg and a relative humidity of 50%.   
You need to figure out the air temperature and the dew point temperature.  Here's the play by play solution to the question:

(1) The air contains 10.5 g/kg of water vapor.  This is 50% (half) of what the air could potentially hold.  So the air's capacity, the saturation mixing ratio must be 21 g/kg (you can either do this in your head or use the RH equation following the steps shown above). 

(2) Once you know the saturation mixing ratio you can look up the air temperature in a table (80 F air has a saturation mixing ratio of 21 g/kg)

(3) Then you imagine cooling the air until the RH becomes 100%.  This occurs at 60 F.  The dew point is 60 F.


Example 4
Probably the most difficult problem of the bunch. 

But one of the things we said about dew point is that it has the same job as mixing ratio - it gives you an idea of the actual amount of water vapor in the air.  This problem will show that if you know the dew point, you can quickly figure out the mixing ratio.  Knowing the dew point is equivalent to knowing the mixing ratio.


Here's what we ended up with in class, we were given the air temperature and the dew point temperature.  We were supposed to figure out the mixing ratio and the relative humidity. 


We enter the two temperatures onto a chart and look up the saturation mixing ratio for each.


We ignore the fact that we don't know the mixing ratio.  We do know that if we cool the 90 F air to 50 F the RH will become 100%.  So on the 50 F row, we can set the mixing ratio equal to the value of the saturation mixing ratio at 50 F, 7.5 g/kg.



Remember back to the three earlier examples.  When we cooled air to the the dew point, the mixing ratio didn't change.  So the mixing ratio must have been 7.5 all along.   Once we know the mixing ratio in the 90 F air it is a simple matter to calculate the relative humidity, 25%.


The figure below is on p. 87 in the photocopied ClassNotes.  It explains how you can dry moist air. 



At Point 1 we start with some 90 F air with a relative humidity of 25%, fairly dry air.   These are the same numbers in Example Problem #4.  We imagine cooling this air to the dew point temperature, 50 F.  While doing that the mixing ratio, r, would stay constant.  Relative humidity would increase and eventually reach 100%.  A cloud would form (Pt. 2 in the figure above). 

Then we continue to cool the air below the dew point, to 30 F.  Air that is cooled below the dew point finds itself with more water vapor than it can contain.  The excess moisture must condense (we will assume it falls out of the air as rain or snow).  Mixing ratio will decrease, the relative humidity will remain 100%.  When air reaches 30 F it contains 3 g/kg, less than half the moisture that it originally did (7.5 g/kg).  The air is being warmed back up to 90 F along Path 4.  As it warms the mixing ratio remains constant.  At Point 5, the air now has a RH of only 10%.

Drying moist air is very similar to wringing moisture from a wet sponge.



You start to squeeze the sponge and it gets smaller.  That's like cooling the air and reducing the saturation mixing ratio, the air's capacity for water vapor.  At first squeezing the sponge doesn't cause anything to happen (that's like cooling the air, the mixing ratio stays constant as long as the air doesn't lose any water vapor).  Eventually water will start to drop from the sponge (with air this is what happens when you reach the dew point and continue to cool the air below the dew point).  Then you let go of the sponge and let it expand back to its original shape and size (the air warms back to its original temperature).  The sponge (and the air) will be drier than when you started.