Weather exam 2 Flashcards
Geosynchronous (GOES)
A. 36,000 km/22,300 mi above Earth
B. Stationary above point on Equator
a. GOES East (75ᵒW)
b. GOES West (135ᵒW)
Geosynchronous (GOES): part 2
A. Advantages:
i. High resolution
ii. 60ᵒN – 60ᵒS
B. Disadvantage:
i. High latitudes
Polar Orbiting (POES):
B. Earth rotates beneath satellite
A. ~840 km/520 mi above Earth
C. Orbital period ~102 min
- Polar Orbiting (POES) part 2
A. Advantage:
B. Disadvantages:
i. High latitudes
ii. Low latitudes
iii. Resolution
i. Multiple times in single image
Visible (VIS)
- Sensor measures reflected VIS
- High resolution
- NO utility at night
Infrared (IR)
- Sensor measures emitted IR
- Lower resolution than VIS
- Utility at night
a. Temperature (S-B)
b. Highest altitude object
Water Vapor (VAP)
- Sensor measures emitted IR
a. Different wavelength of IR - Utility at night
b. Temperature
c. Highest altitude vapor
Base Reflectivity
a. Radar emits microwave pulse
b. Energy scattered back to radar
c. Intensity (dBZ) indicates particle
size
i. Rainfall rate
Doppler effect
a. Motion toward radar
i. Wave compression and increased
frequency
b. Motion away from radar
ii. Wave stretching and decreased
frequency
- Vapor Pressure (e)
a. amount of pressure (in mb) exerted by water vapor in the air
i. measure of the actual vapor content of the air
Initial conditions
vapor pressure = 0
Vapor pressure increasing
vapor pressure > 0
evaporation rate > condensation rate
vapor pressure increases
Vapor Pressure Constant
vapor pressure > 0
evaporation rate = condensation rate
vapor pressure equilibrium
Saturation
- Saturation Vapor Pressure (es)
a. Vapor pressure of the air when the air is saturated
i. Measure of the maximum vapor content of the air
Saturation Vapor Pressure (es) Part 2
- Pressure exerted by the maximum amount of vapor that can
be in the air at a given temperature.
Saturation Vapor Pressure (es) Part 3
- Exponential relationship with T
a. subsaturation (e < es)
b. saturation (e = es)
c. supersaturation (e > es)
Relative Humidity (RH)
Before we define, what is the RH indicated by point “*” on the graph?
a.Explain!
b.T= 80 F and e= 18 mb
C.Hint: what is es at 80 F?
I. If e=18mb & es=36 mb, then RH=50%
Relative Humidity (RH) Part 2
a. the actual vapor pressure of
the air (e) relative to (/) the
vapor pressure exerted by
the air if it were saturated
(es).
b. RH = e/es * 100%
Saturation of subsaturated air
- Add vapor to es
- Cool air to es
- Dew Point Temperature
a. Initially, temperature at which
dew forms
b. Temperature to which air is
cooled to become saturated
i. Td = 61ᵒF
Saturation of Subsaturated Air Part 2
- Add vapor to es
- Cool air to es
a. T is a proxy for the maximum
amount of vapor that can be in
the air…es
a. Td is a proxy for the actual
amount of vapor in the air…e - T ≥ Td
Dew Point Depression Recap
- DPD = T - Td
- Approximate measure of RH
- Magnitude of difference between T
and Td?
a. T=80 F
b. Td=61 F
C. DPD= (80-61) = 19F
Cloud Condensation Nuclei
- Aerosols
a. dust, pollen, smoke, etc.
b. 0.2 – 10 μm
i. concentration decreases as size
increases
c. Hygroscopic
i. Condensation when RH < 100%
d. Hydrophobic
i. No condensation when RH > 100% - Condensation will not occur if nuclei are
not present.
Condensation
- Cool air to Td
- Additional cooling causes air to cool along the es
curve.
a. Air remains saturated as it cools.
b. Vapor content of air decreases - The Earth’s surface cools overlying air.
a. Fog - Rising air causes expansional cooling
a. Clouds
Fog
Cloud at Earth’s surface resulting from
cooling of saturated air.
- Radiation fog
a. Forms during clear calm nights.
b. Earth’s surface cools the overlying air to
Td.
c. Additional cooling causes condensation
to occur.
Advection fog
a. Forms both day and night.
c. Earth’s surface cools the overlying air to
Td.
d. Additional cooling causes condensation
to occur.
b. Air with Td greater than surface T
advected by wind over surface.
e. Often occurs during snowmelt
Steam fog
a. Forms over warm bodies of water.
c. Water warms the overlying air.
d. Water evaporates into the air.
b. Cold air advected by the wind over the
water.
e. Warm moist air rises via convection
es increases
e, Td and RH increase
f. Rising air quickly cooled by surrounding
cold air to Td.
g. Additional cooling causes condensation to
occur.
Why do clouds form?
Lift initially subsaturated (T > Td) air
Cool to Td via expansional cooling
Cloud forms when T < Td
From Clouds to Precipitation (Warm and Cold)
- Warm Clouds: T > 32ᵒF
a. All liquid, no ice
b. Collision and Coalescence
i. Larger droplets collide and merge/coalesce with smaller droplets
ii. Efficiency increases as larger droplet size increases
Cold Clouds: T < 32ᵒF
a. Ice crystals, “supercooled” droplets and vapor coexist in clouds when
-40ᵒF ≤ T ≤ 32ᵒF
i. deposition nuclei
Hexagonal
Rare
b. Cold clouds composed almost entirely supercooled droplets
So, HOW does a cold cloud produce snow???
a. esi
i. sublimation = deposition (slow)
b. esl
i. evaporation = condensation (fast)
- Within a cold cloud, there exists a single
c. For a cloud where T < 32ᵒF, esl > esi
b. e, where esl > esi
a. T, where -40ᵒF ≤ T ≤ 32ᵒF
i. RHl = (e/esl) = 100% saturation
ii. Rhi = (e/esi) > 100% supersaturation
While ice and supercooled droplets coexist in a cold cloud, only ONE can be in equilibrium
- If e = esi
a. RHl < 100%
b. RHi = 100% equilibrium
c. Droplets evaporate - If e = esl
a. RHl = 100% equilibrium
b. RHi > 100%
c. Crystals grow (via deposition)
❄
Bergeron Process (1)
- Given a cloud where
a. -40ᵒF ≤ T ≤ 32ᵒF
b. e = esl
i. RHl = 100%
ii. RHi > 100%
iii. Vapor deposits on crystal
Crystals GROW
e DECREASES
Bergeron Process (2)
- Given a cloud where
a. -40ᵒF ≤ T ≤ 32ᵒF
b. esi < e < esl
i. RHl < 100%
ii. RHi > 100%
e INCREASES
iv. Droplets EVAPORATE
iii. Crystal growth decreases
Bergeron Process (3, 4)
- Given a cloud where
a. -40ᵒF ≤ T ≤ 32ᵒF
b. e = esl (same as 1)
i. RHl = 100%
ii. RHi > 100%
iii. Vapor deposits on crystal
Crystal growth INCREASES
e DECREASES - Repeat Steps 1-3 until ALL droplets EVAPORATE
Bergeron Process (5)
- Given a cloud where only ice crystals exist
a. -40ᵒF ≤ T ≤ 32ᵒF
b. e = esi
Bergeron Process (5)
i. RHi = 100%
deposition = sublimation
crystal growth STOPS
Bergeron Process meaning
…the process by which ice crystals grow
at the expense of supercooled droplets
in a cold cloud
Precipitation associated with cold cloud processes
Snow, freezing rain, sleet
Snow
When air temperature remains below freezing throughout the atmosphere
Sleet
Partly frozen drops refreeze
Freezing Rain
Rain drops become “Supercooled” in cold air and freeze on contact
High
Clockwise and out
Low
Counter-Clockwise and in
Forces that cause surface wind (V)
. Pressure Gradient Force (PGF)
b. Coriolis force (f)
c. Friction (Fr)
Forces and wind represented by vectors
a. Arrow has both direction and magnitude
b. Vectors have both left- and right-hand sides
- Pressure Gradient Force (PGF): ∆p/∆d
a. only force that can act on air at rest
b. directed from high to low pressure
c. points in direction of rate of greatest pressure decrease
d. perpendicular to the isobars
e. magnitude increases as the pressure gradient increases
i. PGF increases as distance between isobars decreases
Coriolis Force (f)
a. “Apparent” deflection of large-scale motions
b. caused by latitudinal dependence of Earth’s radial velocity (RV)
(2r cos/24hrs; r = 3959 miles (6371 km); = latitude)
c. RV = 1036 mph at equator, 777.8 mph at BG (41.37N), 0 mph at North Pole
Coriolis Force (f) Part 2
a. Always acts 90ᵒ to the right (left) of the wind in Northern (Southern) Hemisphere
b. Increases with wind speed (equals 0 when the wind is calm)
c. For a given V, f = 0 at the equator and increases with latitude
Response of air when both the PGF and f act upon it (sequential)
a. PGF initiates the wind (V)
b. f acts 90ᵒ to the right of V
c. V accelerates and turns to the right to strike a balance between PGF and f
d. Acceleration and turning continue until balance is achieved
Geostrophic Wind (Vg)
a. Wind that results from a balance
between PGF and f
i. Always parallel to isobars
ii. Low pressure to the left
iii. High pressure to the right
iv. Explains 80-85% real wind
Friction (Fr)
a. Causes the following sequence (in order)
i. Vg decreases (becomes V)
iii. PGF becomes dominant force
ii. f decreases
iv. V turns toward PGF
Relationship between upper level and
surface circulations?
a. Surface high and low locations
determined by upper level
circulation pattern.
Constant Altitude:
Column pressure is greater over
warm air than cold air.
Constant Pressure
Warm column height is greater
then cold column height.
500 mb Map
Height varies on constant
pressure (500 mb) surface
Isoheights (solid blue)
3. Isotherms (dashed red)
4. Station models (T, Td, wind)
Block Experiment Recap: Part 1
Rate of pressure decrease with height is larger in cold
air than in warm air.
The height of an upper level pressure surface is
higher in the warm column/lower in the cold
Block Experiment Recap: Part 2
The pressure on an upper level horizontal surface
(levels 1 and 2) is larger in the warm column than the
cold column.
At any upper level, the PGF points from warm
to cold.
Meteorological Application
a. The N-S PGF increases with increasing height.
b. Wind speed increases with increasing height.
i. Not just a result of decreasing friction!
Upper level height and PGF Part 1
Upper level heights decrease with increasing
latitude (colder to the north).
Upper level PGF rules mirror those at the
surface.
a. PGF directed from high to low heights toward
greatest height decrease.
Upper level height and PGF Part 2
PGF perpendicular to the isoheight lines.
c. PGF magnitude increases as distance between
isoheights decreases.
Evolution of upper level wind
a. PGF initiates wind (V).
b. Coriolis force (f) acts 90ᵒ to the right of V.
c. Wind (V) accelerates and turns to the right.
d. Acceleration and turning continue until
balance is achieved.
Evolution of upper level wind Part 2
Wind is geostrophic (Vg).
ii. Upper level wind westerly component
⇾Equator to pole temperature
decrease!!!
⇾Low heights/cold to left of Vg
⇾High heights/warm to right of Vg
- Additional upper level features
a. Ridges:
i. Axis of high heights and warm
column temperatures.
b. Troughs:
i. Axes of low heights and cold
column temperatures.
Additional upper level features
a. Meridional wave pattern
i. High amplitude waves
⇾Cold south in troughs
⇾Warm north in ridges
Additional upper level features part 2
a. Zonal wave pattern
i. Low amplitude waves
⇾Cold air north
⇾Warm air south
Schematic of 3-D of ridges and troughs
Heights slope down toward the north as column
becomes progressively colder.
Focus on isoheight curvature
a. Clockwise circulation around highs/
ridges
b. Counterclockwise circulation around
lows/troughs
Rotation in a carnival ride
a. Centrifugal or Centripetal force???
i. Motion is tangent to the ride.
ii. Ride pushes person IN along the circle.
iii. Centrifugal force (brown) is apparent/fake.
iv. Centripetal force (white) is real!
⇾Centripetal force ALWAYS directed inward
toward middle of circle.
If the isoheights are curved
a. Centripetal force (Ce) vector points toward
the center of the circle.
Ridges
Ce acts in the same direction as the
Coriolis force (f).
i. f becomes large relative to the PGF.
⇾V increases and becomes faster
than geostrophic, or
“supergeo-
strophic”(SUP).
Troughs
Ce acts in the same direction as the PGF.
i. f becomes small relative to the PGF.
⇾V decreases and becomes slower
than geostrophic, or
“subgeostrophic”
(SUB).
Gradient Wind (Vgr)
a. Results from a balance between PGF,
Coriolis force (f) and centripetal force
(Ce).
i. Supergeostrophic (SUP) in ridges.
ii. Subgeostrophic (SUB) in troughs.
Gradient Wind (Vgr) part 2
a. Supergeostrophic (SUP) in ridges.
b. Subgeostrophic (SUB) in troughs.G
c. Geostrophic (G) in areas of straight flow
between ridges and troughs.
d. So, Vgr accelerates and decelerates as
air moves through the wave pattern.
Greatest deceleration of V in 500 mb wave
Downstream of ridge
Where the winds are geostrophic (G)
I. Mass converges (CON)
Ii. Column weight increases
Iii. Surface High pressure
Iv. Sinking air
Surface high pressure located beneath region of 500 mb CON
Sinking mid-tropospheric air
I. compressional warming
Ii. relative humidity decrease
Iii. relatively cloud-free near high center
Greatest acceleration of V in 500 mb wave
a. Downstream of trough
b. Where the winds are geostrophic (G)
i. Mass diverges (DIV)
ii. Column weight decreases
iii. Surface LOW pressure
iv. Rising Air
Surface low pressure located beneath
region of 500 mb DIV
a. Rising mid-tropospheric air
ii. relative humidity increase
iii. clouds and precipitation near low
center
Atmospheric Stability and Vertical Motions:
Shape indicates “stability” and intensity of vertical motion.
Stability
determined by the response of an object displaced from its initial position
Stable
Object displaced from initial position returns to
that position
Unstable
Object displaced from initial position
accelerates away from that position
Neutral
Object displaced from initial position
remains where displaced
Parcel
small hypothetical volume
Parcel Part 2
Defined by 3 interactions with surrounding environment
1. flexible boundaries
2. no MASS exchange with surrounding environment
3. no ENERGY exchange with surrounding environment …ADIABATIC
Expansional Cooling as Parcels Rise
ENVIRONMENT
(TE = PE + KE = constant)
Parcel part 3
Expansional Cooling as Parcels Rise
Rising Parcel
Dry adiabatic lapse rate (d) = 10C/km (Expansional cooling
➣Moist adiabatic lapse rate (m) = 6C/km (Expansional Cooling - Latent Heat Release)
➣Environmental lapse rate (e) = ?C/km
CP
Sounding
Vertical temperature profile of environment (i.e., 𝚪e)
Rate of temperature change in the
environment is NOT constant
How to determine atmospheric stability for a given
sounding:
Part 1
if the temperature of the parcel air is colder than that of the environment then
parcel more dense than environment
parcel sinks to surface
STABLE
How to determine atmospheric stability for a given
sounding:
Part 2
If the temperature of the parcel air is warmer than that of the environment then
UNSTABLE
How to determine atmospheric stability for a given
sounding:
Part 3
if the temperature of the parcel air is equal to that of the environment then
parcel and environment densities are identical
parcel remains at 1 km
NEUTRAL
