251Ch3: space time and motion

8/9/2019 251Ch3: space time and motion

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Chapter 3

Space, Time, and Motion


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(1) Wind Observations

Vectorshave both magnitude anddirection.

Wind is a vector quantity.

The components of wind can beexpressed in the Cartesian coordinates:

x, y, z.


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Wind Instruments:

Anemometer and wind vane

Aerovane

Sonic anemometer


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Anemometer andwind vane.


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Aerovane


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SonicAnemometers


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(2) Plotting Data

Each element has a standard location and format.


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Station V N dd ff TT TdTdTd PPPPP a ppp ww W Cl Cm Ch

Cordoba,

Argentina

11/2 8 25 8 1! 1" 121#2 $ 1! 58 " ! 2 8

o


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Station

%d#

Air

Temp#

&C'

(ew

Point

Temp#

&C'

)eight

&meter*'

Wind

(ire+tion

&degree*'

Wind

Speed

&not*'

-+.rath,

Ala*a#

!2$1

/$ /$0 505 1 2!

O


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(3) Derivatives in Time and Space

%f a graph i* drawn with the al3e* of a 43antit on the

erti+al and timeon the hori6ontal, then drawing a line

tangent to the graph line and determining it* al3e &*lope'

will determine the deriatie of that 43antit with re*pe+t

to time#

The tangent te+hni43e mea*3re* the deriatie how the

43antit +hange* with re*pe+t to time#


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Similarly, if the quantity was drawn on agraph with a spatial direction; e.g., the

x-, y-, or z-direction. Then, the slope ofthe tangent to the line at the location ofinterest determines the derivative of the

quantity with respect to that direction.


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Actually, these derivatives are estimations since it isvery difficult to draw an accurate line tangent to the

parameter line, and we are looking at the changeover a rather large time interval or large x-directioninterval.

Calculus considers the denominator value as

approaching zero.Secondly, these derivatives should be consideredpartial derivativesbecause temperature changes inall three directions as well as time.

If we are considering only the change in the x-direction, we are assuming that there are no changesin the other directions or time.


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(4) Advection

Advection (in meteorology) is the rate ofchange of some property of theatmosphere by the horizontal movement

of air.

The rate of change is a derivative (withrespect to time).


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If at 2:00 pm thecarbon monoxide

concentration was 80ppb at a location 30 nmupstream fromSavannah, Georgia,and the wind were

blowing at 15 knotstoward Savannah,when would theconcentration at

Savannah reach 80ppb with no sources orsinks.


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The 80 ppb air must travel 30 nm and it

is moving at 15 knots. Divide 30 nm by

15 knots (nm/hr) and you get a traveltime of 2 hours. 2:00 pm + 2 hours =4:00 pm.


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Suppose the concentration at Savannahat 2:00 pm were 60 ppb. How rapidly

will the carbon monoxide concentrationchange.

The rate of change = (change in

concentration)/(change in time)


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The change in concentration of carbonmonoxide can be written as:

The subscript is there to indicate we are

only considering the change at Savannahwhich is not moving. The x-direction istoward Savannah.

CO[ ]( )x


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Thus, the rate of change = =

The wind speed (magnitude of thehorizontal wind velocity vector in thedirection of interest) can be written as:

Then,

(80 ppb - 60 ppb)

(2 hours)=

20ppb

2 hrs=

10ppb

hr= 3 x 103 ppb/sec

v = xt

t = x

rv

CO[ ]x

t


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We can then write the rate of change as:

We can get the rate of change with time(derivative with respect to time) from the rateof change with distance (derivative withrespect to distance) if we know the velocity of

the wind.Essentially, it is simply:

CO[ ]( )x

t=

r

v CO[ ]( )

x

x

CO[ ]

t= x

t CO[ ]

x=

r

v CO[ ]

x


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(5) The One-Dimensional Vector Equation

If we consider the change in time and thechange in x as approaching zero, we have theinstantaneous rate of change at a point (e.g.,Savannah).

Writing in calculus form (partial differentialequation since we are only considering thechange in the direction of the wind field (our

x-direction), we have:

CO[ ]

t

x

= u CO[ ]

x

t


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This is a general advection equation

(along the x-coordinate - the west to

east direction). (We are using u thecomponent of the wind along the x-coordinate

One can write such an equation for theadvection along the y-coordinate, or z-coordinate.


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Consider the following analysis of [CO].

To get the rate of

change of [CO]

(partial derivative

of [CO] with

respect to time) at

Savannah, weneed the partial

derivative of [CO]

with respect to x

and the average

wind speed.otice that concentrations at Savannah are less than the! are

"pwind, so the rate of change of [CO] over time sho"ld #e positive.


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$"st as on a graph, to

get the change in

concentration of [CO],

pic% two points on

either side of the point

of interest and get thedifference #etween

those val"es. &n this

case, '*+ .

ow, divide #! the distance #etween those points, -nm. This will

#e the slope of the graph line at Savannah which will #e the rate

of change of [CO] with respect to distance at Savannah.50ppb 70ppb

30nm=20ppb

30nm= 0.667ppb/nm


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The wind speed is everywhere 15 knots (nm/hr) sothe average wind speed is 15 nm/hr.

Then, the rate of change of [CO] with time is:

Similarly, the equation can be set up for any spatially-varying atmospheric variable; such as, temperature:

CO[ ]

t

x= u

CO[ ]

x

t=15nm/hr 0.667ppb/nm( ) = 10ppb/hr

Tt

= u Tx


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(6) Equations on the Brain

To understand the equation and how itrelates to the atmosphere:

Say it in words.

See if it makes sense if each variable orderivative, in turn, is zero.

See if the signs make sense.

See it it makes sense if certain variables get

larger or smaller.Make up a concrete example and workthrough it.


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For the advection equation.

The equations states that the rate ofchange of temperature with respect to timeat a particular location is equal to thenegative of the wind speed times the rateof variation of temperature in the direction

toward which the wind is blowing.

T

t= u

T

x


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For various terms set to zero.

If wind speed is zero, no air is being

transported, so the wind is not changingthe temperature, so the advection is zero.

If is zero, then the temperature

is uniform along the x-axis, so theair blowing in is the same temperature asthe air blowing out so advection is zero.

T

t= 0

T

x

T

t= u 0

T

x


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For this situation,remember, theequation relates toadvectionat that

instant, at a particularlocation. Here theadvection is zerobecause the change intemperature at that

instant is zero.You would have to average u and T/x over a muchlarger distance to get a non-zero value, not just locally.


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Does the sign makesense?If this were analyzedtemperature, orient x to point

toward where the wind is

blowing, then u will always be

positive.

The sign of the temperature

change with time then will be determined by the sign of thedownwind variation with temperature.

If T/x is positive, (T down - T up), then if you graphed T versus

x, the slope would be positive.Warmer temperatures would be downwind.

If T/ x is negative, then a slope of T vs x would have coldertemperatures downwind (slope would be negative) as we have.


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Check the magnitude.Suppose the average wind is 5 m/s towardeast and temperatures are colderupstream, warmer downstream.

Then, if the wind were stronger, the changewould occur faster and temperatures woulddrop faster - greater change in temperaturewith time - greater advection.

If the temperature change with distance issmall, the temperature change over timewould be small.


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Check with numbers.

Assume a wind speed of 5 m/s andtemperatures are colder by 5oK over adistance of 100km upstream.

Then,with a wind speed of 5 m/s, how long

will it take the colder air to travel 100km?

So, temperature should drop at a rate of

5o

K every 2 x 104

seconds. This is:

100km

5mm

=1x10

5km

5ms

= 2x104 sec

T

t=

5oK

(2x104s)

=2.5x10oKs


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Check equation with numbers.

Assume a wind speed of 5 m/s andtemperatures are colder by 5oK over a

distance of 100km upstream.Then,

and

T

x= 5

oK100km

= 5oK

100000m= 5.0 10

5 oKm

T

t= u

T

x= 5ms

50K

1x105m

= 25x10

5

o

Ks = 2.5x104

o

Ks


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The differences inthe horizontal

scale is simplydue to the speedof the wind.


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(8) Phase Speed

The examples have been using wind as thecause of the advection, but the same conceptcan be applied to anything that is movingregardless of the cause.

As long as you can tell how fast it is moving(e.g., a cold front), the speed can be used inthe advection equation to determine thechange over time.

The time record of observed meteorologicalvariables at a particular space (location) can beconverted directly into a depiction of thehorizontal structure of the phenomena.


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The front is moving at 20knots and it passed NorthPlatte two hours ago, so itshould be 40 miles pastNorth Platte, at location B.

If temperature at North

Platte two hours ago was52oF and it is moving towardDodge City at 20 knots, the52oF air should arrive inDodge City in 5 hours after

being in North Platte, or inanother 3 hours.


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Questions:Do: 1, 2, 3, 4, 5

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251Ch3: space time and motion