METEOROLOGIST JEFF HABY
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 SkewT 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 SkewT is finding the graphical solution to
the ClausiusClapeyron equation. Using the CC equation or determining graphically on SkewT will
result in a similar value of the vapor pressure or saturation vapor pressure parameter (value on
SkewT will not be exact same as CC equation we have used since L (latent heat) term in CC equation
changes slightly with temperature; we kept L term constant up to this point in course for simplicity).
Temperature plugged into CC equation yields saturation vapor pressure while dewpoint plugged
into CC equation yields vapor pressure.
Saturation vapor pressure is the pressure in millibars found graphically on SkewT by using the temperature.
Vapor pressure is the pressure in millibars found graphical on SkewT 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 SkewT converts mixing ratio to vapor pressure.
Example problems using SkewT:
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: Wetbulb, wetbulb potential, wetbulb zero
Finding the wetbulb on the SkewT 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 wetbulb temperature is the temperature that results after complete evaporational cooling at constant pressure.
The wetbulb potential temperature is found the same as the wetbulb except the wetbulb temperature
is moved at the WALR
to the 1000mb level after the wetbulb is determined. Wetbulb potential standardizes wetbulb
temperature thus allowing for direct comparison of wet bulb temperature at
different pressure levels.
The wetbulb zero is the pressure level in the atmosphere where the wetbulb 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 wetbulb 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).
Wetbulb is about 12 C.
2. The 900 mb temperature is 21 C while the 900 mb dewpoint is 5 C. What is the wetbulb
potential temperature?
answer: at about 11.7 C at 900 mb, continue dropping the parcel at the WALR until the 1000mb level is reached.
Wetbulb potential is 16 C.
**Notice between 500 and 550 mb the wet adiabates are labeled with numbers. Those numbers are the 1000 mb
wetbulb potential temperatures. Notice the wet adiabat we used in above problem is labeled as 16
(same as wetbulb potential temperature we found by dropping parcel to 1000mb level).
Tutorial 4: Potential temperature, equivalent temperature and THETAE
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 (THETAE): Latent heat of a parcel is released and then brought down at
the DALR to the 1000 mb pressure level. THETAE 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 (THETAE)?
answer: same as previous problem but continue at DALR past the original pressure level to
the 1000 mb level. THETAE 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


