Metfi 120 (introduction to Weather and Climato) Orline - Unit 2 Review Adiabatic processes in the mountaln. The diagram below shows the lapse rates that air parcel will encounter while it is forced to move over a mountain. If an alr parcel descends through the atmosphere the air will heat at the dry adiabatic rate as it will necessarily be unsaturated. In summary, while rising air cools at varying rates, descending air heats at a uniform rate. 16. Using the table below, fill in the blanks for temperature and dew-point temperature at various helghts on the windward and leeward sides of the mountain. Assume that the WALR is 0.5C/100m. (Note: temperature and dew-paint drop together above LCL when a saturated air parcel rises and water vapor condonses. In addition, start from the 500m at windward side 1000m1500m2000m Peak 2000m at leeward side 1500m1000m500m Sea Level at leeward side.) 17. Compare to those of on the windward side, how do air temperature and dew point on the leeward side? 18. From the previous question, Why? MEIR120 (introduction to Weather and CImate) Onine - Unit2 Review Here is a challenging question. Assume that the wet adiabatic lapse rate (WALR) is 0.6C/100m. Fill the blanks in the boxes in the figure. Here is how to get answers. 1. At the LCL, temperature (2C) and dew point (2C) are the same. Until air parcel reaches LCL, temperature falls at the DALR (1.0C/100m). At the LCL, temperature falls 25C from the surface (27C2C=25C). It means air parcel rises 2500m. Thus LCL is 2500m. 2. Below the LCL, dew point changes at the Dew Point Lapse Rate (DPLR, 0.2C/100m ), it means that dew point has changed 5.0C(0.2C/100m2500m). Thus dew point at the surface is 7C/100m 3. Above the LCL, both temperature and dew point fall at the WALR (0.6C/100m). Atthe peak, dew point is 1C. It is 3C cooler than the LCL. It means air parcel rises another 500m(0.6C/100m500m=3C). Thus, peak elevation is 3000m(LCL+500m=2500m+500m) 4. At the leeward side, temperature and dew point rise at the DALR (1.0C/100m) and DPLR (0.2C/100m), respectively. Elevation has changed 2500m(3000m500m). Thus temperature and dew point rise 25C/100m and 5C. Thus, temperature will be 24C(1C+25C) and dew point will be 4C(1C+5C). Chapter 6: Precipitation and its formation Steps in the formation of precipitation according to the Bergeron process. The Bergeron process relies on the fact that cloud droplets do not freeze until they reach a temperature below the freezing point, and even then only in the presence of freezing nuclei (solid particles that have a crystal form similar to that of ice). Because freezing nuclel are much less abundant than condensation nuclei, many clouds exist in the liquid state while at temperatures well below 0C. These are supercooled clouds. The freezing nuclei present promote the formation of a few scattered ice crystals. Since lce crystals are more efficient absorbers of water vapor, they consume the "excess" water vapor, which lowers the relative humid.