Wed., Mar. 22 notes

Wednesday Mar. 22, 2006

The Experiment #3 reports and the revised Expt. 2 reports were due today.

All of the Experiment #4 materials have been checked out. There are now plenty of Expt. 3 materials available, however. If you have not done a book, scientific paper, or experiment report you should check out an Experiment #3 kit and perform that experiment.

All of the Assignment #2 1S1P reports have been graded.

Saturation mixing ratio values for air with different temperatures. Data are given in table and graph form. The beakers illustrate the relative amounts of water vapor that air can contain. We referred to this data often while working the humidity sample problems that follow.

Here is the first sample problem that we worked in class. You'll have a heck of a time unscrambling this if you're seeing it for the first time. The series of steps that we followed are retraced below:

We were given some starting information: an air temperature of 90 F and a mixing ratio value of 6 g/kg. We are supposed to find the relative humidity and the dew point temperature.

We start by entering the data we were given in the table. Once you know the air's temperature you can look up the saturation mixing ratio value; it 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.

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 doesn't change as we cool the air. The only thing that changes r is adding or removing water vapor and we are doing neither.

(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 capacity is decreasing.

Finally at 45 F the RH becomes 100%. The dew point temperature in this problem is 45 F.

What would happen if you cooled the air 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. Now because water vapor is being taken out of the air (and turned into water), the mixing ratio will decrease from 6 to 4.

Here is the second sample problem. Given an air temperature of 90 F and a relative humidity of 50% you are supposed to figure out the mixing ratio (15 g/kg) and the dew point temperature (70 F). The problem is worked out in detail below.

First you fill in the air temperature and the RH data that you are given. Since you know the air's temperature you can look up the saturation mixing ratio (30 g/kg). Then you can substitute into the relative humidity formula and solve for the mixing ratio (15 g/kg).

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

Something new you can learn from Examples 1 and 2. In the first example the RH was low and the difference between the air and dew point temperatures was large. In the second example, the RH was higher and the difference between the air and dew point temperatures was smaller. If there were no difference between the air and dew point temperature the RH would be 100%.

In the 3rd problem we started with RH=50% and r=10.5 g/kg.

The air contains 10.5 g/kg of water vapor, this is 50% of what the air could potentially hold. So the air's capacity, the saturation mixing ratio must be 21 g/kg. Once you know the saturation mixing ratio you can look up the air temperature in a table. Then you imagine cooling the air until the RH becomes 100%. This occurs at 60 F. The dew point is 60 F.

In the last example problem we start with an air temperature of 90 F and a dew point temperature of 50 F.
The first thing you can do is look up the saturation mixing ratio for 90 F and 50 F air.

Then we know that if we cool the 90 F air to 50 F the RH will become 100%. Since we know the saturation mixing ratio value at 50 F is 7.5 g/kg we can say the mixing ratio is 7.5 g/kg.

Now that we know the mixing ratio we can compute the relative humidity.

Dew point temperature and mixing ratio both give you an idea of the actual amount of water vapor in the air. If you know the mixing ratio you can determine the dew point temperature. If you know the dew point temperature you can determine the mixing ratio (that's what we did in this problem).

90 F air with 25% relative humidity is cooled to 50 F the dew point. The relative humidity reaches 100% at that point. Then the air is cooled some more, to 30 F. The 30 F air can't hold the 7.5 g/kg of water vapor that was found in the 50 F air. Some of the moisture must condense. Next the 30 F air now containing only 3 g/kg of water vapor is warmed to 90 F, the starting temperature. The air now has a RH of only 10%. The air contains less than half the moisture it did originally.

This figure was not shown in class. Cooling moist air below the dew point is kind of like squeezing out or wringing out a wet sponge. You start to squeeze the sponge until water starts to drop out and then squeeze it some more. Then you let go of the sponge and let it expand back to its orignal shape and size. The sponge will be drier than when you started.

These two figures show where this kind of thing can occur. In the winter cold air is brought inside your house or apartment and warmed. Imagine 30 F air with a RH of 100 % brought inside and warmed to 70 F. The RH will decrease to 20%.

The air in an airplane comes from outside the plane. The air outside the plane is very cold (-50 F perhaps) and contains very little water vapor. When brought inside and warmed to a comfortable temperature the RH of the air in the plane will be very close 0%. Actually the ventilation system in the plane will add moisture to the air so that it doesn't get that dry.