These tutorials supplement the parameter determinations given in lectures 7 and 8. These tutorials are practical application and problem solving. Practice understanding these problems using your laminated Skew-T and erasable marker. Tutorial 1: Mixing Ratio #1 What is the saturation mixing ratio when the temperature is 20 C at 900 mb? answer: 17 g/kg #2 What is the actual mixing ratio when the dewpoint is 10 C at 800 mb? answer: 9.9 g/kg #3 What is the 1000 mb RH (relative humidity) when the 1000 mb temperature is 15 C and the 1000 mb dewpoint is 8 C? answer: mixing ratio = 6.9 g/kg, saturation mixing ratio = 11 g/kg, RH = mixing ratio / saturation mixing ratio = 6.9 / 11 * 100% = 63% #4 What is dewpoint at 1000 mb when 1000 mb temperature is 25 C and RH = 75% answer: saturation mixing ratio = 20.5 g/kg, since RH = 75%, the actual mixing ratio is 20.5*0.75 = 15.4 g/kg. The dewpoint is about 69 F (20.5 C) where the actual mixing ratio is 15.4 g/kg at 1000 mb. Tutorial 2: Vapor pressure Vapor pressure and saturation vapor pressure are found by moving a parcel of air from original pressure level isothermally to the 622 mb pressure surface and reading off the saturation mixing ratio. The units are changed to mb since it is vapor pressure or saturation mixing ratio we are interested in. See 4.4 in plotting book for examples. Finding the vapor pressure and saturation vapor pressure on the Skew-T is finding the graphical solution to the Clausius-Clapeyron equation. Using the C-C equation or determining graphically on Skew-T will result in a similar value of the vapor pressure or saturation vapor pressure parameter (value on Skew-T will not be exact same as C-C equation we have used since L (latent heat) term in C-C equation changes slightly with temperature; we kept L term constant up to this point in course for simplicity). Temperature plugged into C-C equation yields saturation vapor pressure while dewpoint plugged into C-C equation yields vapor pressure. Saturation vapor pressure is the pressure in millibars found graphically on Skew-T by using the temperature. Vapor pressure is the pressure in millibars found graphical on Skew-T by using the dewpoint. Remember from the first half of the course that vapor pressure is related to the mixing ratio by the following formula: w = 0.622*e / (P - e) rearranged to: e = w*p / (0.622 + w) When mixing ratio is known, the above formula can be used to find vapor pressure (or vice versa). The technique of moving the parcel isothermally to 622 on the Skew-T converts mixing ratio to vapor pressure. Example problems using Skew-T: 1. The temperature at 900 mb is 15 C. What is the saturation vapor pressure? answer: plot a point at 900 mb where T = 15 C. Draw a line along the 15 C isotherm until the 622 pressure level is reached. Read off saturation mixing ratio number and give in units of mb. Value is 17.9 mb. 2. The dewpoint at 700 mb is 0 C. What is the vapor pressure? answer: plot a point at 700 mb where T = 0 C. Draw a line along the 0 C isotherm until the 622 pressure level is reached. Read off saturation mixing ratio number and give in units of mb. Value is 6.2 mb. 3. The 1000 mb temperature is 20 C while the dewpoint is 10 C. What is the relative humidity? method 1: find the vapor pressure and saturation vapor pressure of temperature and dewpoint. Vapor pressure is 12.5 mb while saturation vapor pressure is 24.5 mb. RH = e/es = (12.5 mb / 24.5 mb) * 100% = 51% method 2: find the mixing ratio and saturation mixing ratio of temperature and dewpoint at 1000 mb (easier method). Mixing ratio is 7.8 g/kg while saturation mixing ratio is 15 g/kg. RH = w/ws = (7.8 g/kg / 15 g/kg) * 100% = 52% RH is a little more than 50% using each method Tutorial 3: Wet-bulb, wet-bulb potential, wet-bulb zero Finding the wet-bulb on the Skew-T is a more scientific approach than using the 1/3 rule since it accounts for latent heat absorption at any temperature. The 1/3 rule works best for temperatures near freezing. The cooling is closer to 1/2 the dewpoint depression at warm temperatures since warmer air can evaporate much more moisture and thus absorbs much more latent heat than cold air when evaporation occurs. The wet-bulb temperature is the temperature that results after complete evaporational cooling at constant pressure. The wet-bulb potential temperature is found the same as the wet-bulb except the wet-bulb temperature is moved at the WALR to the 1000-mb level after the wet-bulb is determined. Wet-bulb potential standardizes wet-bulb temperature thus allowing for direct comparison of wet bulb temperature at different pressure levels. The wet-bulb zero is the pressure level in the atmosphere where the wet-bulb is equal to 0 C. Example problems: 1. The 900 mb temperature is 21 C while the 900 mb dewpoint is 5 C. What is the wet-bulb temperature? answer: first find the LCL (Lifted Condensation Level). At the LCL, drop the parcel down at the WALR (Wet Adiabatic Lapse Rate) to the original pressure. The LCL is at 710 mb. The temperature is 1.5 C at LCL. Once parcel is dropped to original pressure at WALR, temperature is close to 11.7 C (rounds to 12 C). Wet-bulb is about 12 C. 2. The 900 mb temperature is 21 C while the 900 mb dewpoint is 5 C. What is the wet-bulb potential temperature? answer: at about 11.7 C at 900 mb, continue dropping the parcel at the WALR until the 1000-mb level is reached. Wet-bulb potential is 16 C. **Notice between 500 and 550 mb the wet adiabates are labeled with numbers. Those numbers are the 1000 mb wet-bulb potential temperatures. Notice the wet adiabat we used in above problem is labeled as 16 (same as wet-bulb potential temperature we found by dropping parcel to 1000-mb level). Tutorial 4: Potential temperature, equivalent temperature and THETA-E Definitions: Potential temperature is the temperature a parcel has after being brought dry adiabatically to the 1000 mb level. Potential temperature can be used to standardize temperatures at different pressure levels for comparative purposes. Equivalent temperature: Latent heat of a parcel is released and then brought down at the DALR to original pressure level. Equivalent Potential Temperature (THETA-E): Latent heat of a parcel is released and then brought down at the DALR to the 1000 mb pressure level. THETA-E is often used to analyze instability. Example problems: 1. A parcel of air at 800 mb has a temperature of 15 C. What is its potential temperature? answer: plot a point at 15 C and 800 mb. Bring parcel down at the DALR to 1000 mb. Convert to Kelvins. Potential temperature is 307 K. 2. A parcel of air at 800 mb has a temperature of 5 C and a dewpoint of 0 C. What is its equivalent temperature? answer: plot the 800 mb temperature and dewpoint. find LCL. LCL is at about 740 mb. From top of LCL raise parcel at WALR until slope of wet adiabates becomes parallel to dry adiabates. Next, drop parcel at DALR to original pressure. Equivalent temperature is 19 C or 292 K. 3. A parcel of air at 800 mb has a temperature of 5 C and a dewpoint of 0 C. What is its equivalent potential temperature (THETA-E)? answer: same as previous problem but continue at DALR past the original pressure level to the 1000 mb level. THETA-E is 311 K. Tutorial 5: LCL, CCL, CT LCL= Lifted Condensation Level. This is the pressure level in the atmosphere that a RH of 100% first occurs when air at a particular pressure level is forced lifted. CCL= Convective Condensation Level. This is the pressure level in the atmosphere that a RH of 100% first occurs when air at the surface is heated to a sufficient temperature that allows the air to rise buoyantly on its own. CT= Convective Temperature. This is the surface temperature air must warm to in order for it to buoyantly rise to the CCL. Example Problems: 1. The 1000 mb surface temperature is 25 C while the dewpoint is 18 C. What is the LCL? answer: Draw a line through the dewpoint parallel to the nearest saturation mixing ratio line. Draw a line through the temperature parallel to the nearest dry adiabat. The point of intersection is the LCL. LCL is at about 905 mb. Air forced lifted from surface will result in cloud bases at about 905 mb. 2. The 1000 mb surface temperature is 25 C while the dewpoint is 18 C. The temperature decreases linearly by 20 C between 1000 mb and 700 mb. What is the CCL? answer: First, mark temperature of 25 C at 1000 mb. Since temperature decreases linearly by 20 C between 1000 and 700 mb, the temperature at 700 mb is 5 C. Mark temperature of 5 C at 700 mb. Draw a line connecting the two points. Next, draw a line through the dewpoint parallel to the nearest saturation mixing ratio line until it intersects the temperature line. The point of intersection is the CCL. CCL is at about 840 mb. Cloud bases will be at about 840 mb if air buoyantly rises on its own from the surface. 3. The 1000 mb surface temperature is 25 C while the dewpoint is 18 C. The temperature decreases linearly by 20 C between 1000 mb and 700 mb. What is the CT? answer: From CCL, drop parcel along dry adiabat to the surface (1000 mb in this case). CT is about 30 C. Once surface temperature warms to 30 C, convective cloud bases will theoretically begin to develop even in the absence of significant lift. Tutorial 6: LFC, EL LFC = Level of Free Convection. This is the pressure level in which a parcel of air first becomes equal to the temperature of the surrounding environment. This is the pressure level at the bottom of the most significant CAPE region on a sounding. A LFC will only be present if instability exists in the troposphere. EL = Equilibrium Level. This is the pressure level in which a parcel of air becomes equal again to the surrounding environmental temperature after parcel rises through CAPE region. This is the pressure level at the top of the most significant CAPE region on a sounding. An EL will only be present if instability exists in the troposphere. Example Problem: The surface temperature is 22 C while the dewpoint is 18 C. The surface is at 1000 mb. The lapse rate of temperature cools at the WALR between 1000 and 800 mb. The temperature then cools linearly by 35 C between 800 and 500 mb. The temperature is isothermal (constant temperature with height) between 500 and 350 mb. What is the LFC and EL resulting from forced lift (use LCL)? answer: First, plot the points 22 C and 18 C at 1000 mb. From the 22 C temperature, parallel a wet adiabat up to 800 mb. The temperature at 800 mb is about 14 C. Since the temperature drops 35 C from 800 to 500 mb, the temperature at 500 mb is -21 C. Plot -21 at 500 mb. Draw a line connecting the 800 mb and 500 mb points. From 500 mb, parallel the -21 C isotherm to the 350 mb level. When finding LFC and EL, use a surface based lifted parcel. Find the LCL. The LCL is at 950 mb. From the LCL, parallel the wet adiabat until it has intersected the temperature curve twice. The first intersection is the LFC. The LFC is located at about 730 mb. The second intersection is the EL. The EL is located at about 405 mb. CAPE and thus instability exist between 730 and 405 mb. Tutorial 7: Dewpoint depression Dewpoint depression: the difference in temperature between the actual temperature and dewpoint. The formula is temperature minus dewpoint. Thus, dewpoint depression can never be negative. Example problems: 1. The 850 mb temperature is 20 C with a dewpoint of 7 C. What is the dewpoint depression? answer: 20 - 7 = 13 units of C **dewpoint depression is a change in temperature and not an actual temperature. 2. The 1000 mb temperature is 20 C with a dewpoint of 7 C. What is the dewpoint depression after this air is forced lifted to the 900 mb level? answer: Draw a line parallel to saturation mixing ratio through the dewpoint until it reaches 900 mb level. Dewpoint at 900 mb is 6 C. Draw a line parallel to dry adiabat through the temperature until it reaches 900 mb level. Temperature at 900 mb is 11 C. The dewpoint depression at 900 mb is 11 C - 6 C = 5 units of C Tutorial 8: LI, TT, KI LI= Lifted Index. Comparison in temperature between actual 500 mb temperature and temperature of a parcel of air, force lifted from the surface to the 500 mb level. Calculated as: Te500 - Tp500 TT= Total Totals thermodynamic index. Calculated as: (T850 - T500) + (Td850 - T500) KI= The K thermodynamic index. Calculated as: (T850 - T500) + (Td850 - Tdd700) Examples problems: 1. What is the LI if the surface temperature is 30 C with a dewpoint of 21 C. The surface is at 1000 mb. The 500 mb actual temperature is -7 C (negative 7)? answer: Plot 1000 mb temperature and dewpoint. Find LCL. LCL is at 880 mb. From LCL, raise parcel at WALR to the 500 mb level. Parcel temperature is about -3 (negative 3) at 500 mb. Tp = -3, Te = -7 Te - Tp = -7 - (-3) = -4 (negative 4) LI is -4 (negative 4) 2. What is TT index if the temperature is 15 C while the dewpoint is 11 C at 850 mb? The 500 mb environmental temperature is -9 (negative 9). T850 = 15, Td850= 11, T500= -9 (negative 9) (T850 - T500) + (Td850 - T500) (15 + 9) + (11 + 9) = 24 + 20 = 44 3. What is KI if the temperature is 18 C while the dewpoint is 13 C at 850 mb? The 500 mb environmental temperature is -7 (negative 7) while the 700 mb dewpoint depression is 6 units of Celsius. (T850 - T500) + (Td850 - Tdd700) (18 + 7) + (13 - 6) = 25 + 7 = 32 |