DOC PREVIEW
UW ATMS 101 - Study Guide

This preview shows page 1 out of 4 pages.

Save
View full document
Premium Document
Do you want full access? Go Premium and unlock all 4 pages.
Access to all documents
Download any document
Ad free experience

Unformatted text preview:

Homework Assignment 3 solutions and discussion 1 Contouring exercise On the back of this page is a simplified map of sea level pressure as reported by weather stations across North America at 10 p m Eastern Time on 27 January 1999 A real weather map would contain many more stations and other data in addition to pressure Draw isobars lines of constant pressure on the figure from 1012 mb up to 1028 mb contouring every 4 mb i e 1012 1016 1020 1024 and 1028 mb Label high pressure and low pressure centers with H and L respectively Hints Use a pencil first then smooth the contours as you trace them in ink The 1020 mb isobar is probably the easiest contour line to start your contouring with Remember the guidelines listed on the back of the sheet handed out in quiz section Bonus Winds blow counterclockwise around low pressure centers and clockwise around high pressure centers parallel to the isobars we will explain this phenomenon in class later The strength of the wind is proportional to the pressure gradient which is represented on your drawing by how closely the isobars are spaced Using this information draw a few sample wind arrows on your figure make the length of the arrow proportional to the wind speed See the solution in class or posted outside the TA office Also make sure you understand all the rules for contouring Especially label your contours do not extend your contours too far from where you have data and make sure all the values on one side of a isobar are greater than its value while values to the other side of the contour are less 2 a I have 6 kg of dry air in a rigid box The volume of the box is 5 m3 Using the formula for density mass volume calculate the density of the air in the box density mass volume 6 kg 5 m3 1 2 kg m3 b Now I add some air 1 8 kg to the box without changing the volume Calculate the new air density in the box density 6 0 kg 1 8 kg 5 m3 7 8 kg 5 m3 1 56 kg m3 c Now consider a balloon containing the same mass and with the same volume as the box in part a Initially the density inside the balloon is the same as inside the box Now we raise the balloon several hundred meters What happens to the balloon Why If the air pressure outside the balloon is less than the air pressure inside the balloon the walls of the balloon will experience a net outward force and the balloon will quickly expand If the balloon expands the air density inside the balloon will decrease because the mass of air inside the balloon does not change but the volume of the balloon increases mass volume We also know that the air inside the balloon will cool during the expansion process because expansion of the balloon s walls requires work read energy which comes from the internal kinetic energy of the molecules of air inside the balloon So if the density of the air inside the balloon decreases and the temperature of the air inside the balloon also decreases we know from the ideal gas law that the pressure inside the balloon must also decrease as it expands p RT Air pressure always decreases with height in the atmosphere So as the balloon rises in the atmosphere it experiences less pressure from the outside and the balloon must expand and cool until the pressure inside the balloon equals the pressure outside the balloon At any given height the pressure inside the balloon must equal the pressure outside the balloon or the balloon will quickly expand or contract until this condition is met Note that the reason the balloon expands is because pressure decreases with height NOT because density decreases with height The density inside the balloon and outside the balloon may be very different but as long as the pressure inside the balloon and the pressure outside the balloon are the same the balloon will remain the same size d If the volume of the balloon increases to 5 8 m3 in part c but the mass of air inside the balloon stays the same what is the new density of air inside the balloon Do you think this is greater than less than or equal to the density of air outside the balloon at this height How about the pressure inside the balloon versus outside the balloon density 6 kg 5 8 m3 1 03 kg m3 We have already noted that the pressure inside the balloon must be the same as the pressure outside the balloon Figuring out the air density inside the balloon versus the air density outside the balloon is a little trickier and you actually do not know for sure which is higher unless you also know the temperature both inside the balloon and outside the balloon But here is a good argument for why the density is probably greater inside the balloon recall that the environmental lapse rate in the troposphere is about 6 5 C km on average but the dry adiabatic lapse rate is about 10 C km So if the balloon rises say 1 km the air inside the balloon will be 10 C colder than it was at the surface assuming that the air inside the balloon is relatively dry while the temperature outside the balloon is probably only about 6 5 C colder than the surface air Since the pressure inside and outside the balloon is the same then the colder air the air inside the balloon will be less dense than the warmer air the air outside the balloon You can work this out for yourself using the ideal gas law for both gases p inside inside R T inside p outside outside R T outside since p inside p outside inside R T inside outside R T outside inside outside T outside T inside note that if T outside is greater than T inside then inside will be larger than outside You certainly did not have to do this calculation to get credit for this problem but this is a good illustration of how density pressure and temperature are related e Do you think the air inside the balloon at this height is colder or warmer than the air inside the box on the ground in part a How do you know From the argument in part c we know that the balloon must have cooled as it rose and expanded Again the work of expanding the walls of the balloon requires energy which comes from the internal kinetic energy temperature of the air inside the balloon Since the box did not rise it is at the same temperature as it started So the temperature of the air inside the balloon is colder than the temperature of the air in the box Make sure that you make the difference between adiabatic cooling which is taking place inside the balloon and the temperature decrease of the environment outside the balloon The …


View Full Document

UW ATMS 101 - Study Guide

Download Study Guide
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view Study Guide and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view Study Guide and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?