1. Basic SI units
Distance = meter
Temperature = Kelvin
1 kilo = 1000 units
1 mega = 1,000,000 units
1 Giga = 1,000,000,000 units (8.8 gig harddrive has 8.8 billion units of storage)
2. Other SI units
Force = kgms^-2 = N = Newton
Pressure = Nm^-2 = Pa= Pascals
Energy = Force × distance = Nm = Joules
Power = J/s = Watt
1 Pascal = 0.01 millibars
Speed of light = c = 2.998 × 10^8 ms^-1
Gas constant = Rd = 287 J K^-1 kg^-1
Standard atmospheric density (0 ƒ C at 1000 mb) = 1.275 kgm^-3
Earth radius = 6.37 × 10^6 m
Average Solar Constant = 1380 Wm^-2
Average sea level pressure = 1013.25 millibars
Earth Coriolis = 7.29 × 10^-5 s^-1
4. Newton's Second Law of Motion
F=mass × acceleration
Interpretation: The sum of the forces acting on a body is directly related to the acceleration of the body. For an individual body, mass can be treated as a constant.
*The five forces in meteorology are pressure gradient, gravity, Coriolis, friction, and centrifugal
Example problem: What is the force of a 4 kilogram mass of air accelerating 10 knots/s?
Answer: First convert knots to m/s. 10 knots is 5 m/s. F=ma, therefore, F = (4kg)(5 m/s^2) =
5. Ideal gas law
Pressure = density × temperature × gas constant
Units = kgm^-3Kkgms^-2mK^-1kg^-1 = kgm^-1s^-2 = Nm^-2 = Pa
Interpretation: The pressure exerted by a mass of air is directly proportional to the density of the air and the temperature. Density is related to volume and mass. If mass is constant: when volume increases, density decreases; when volume decreases, density increases.
*As density in a closed volume increases, pressure increases
*As temperature in a closed volume increases, pressure increases
*If pressure is constant, when density increases temperature decreases and if density decreases temperature increases
Example problem: What is the density of air that has a temperature of 104ƒ F at a pressure of 998 mb?
Answer: First convert pressure to Pascals and temperature to Kelvin.
998 mb = 99,800 Pa; 104ƒ F = 40ƒ C = 313 K
Pressure = (density)(gas const.)(Temperature)
99,800 Pa = (density)(287 J/kg ƒ K)(313 K)
density = 99,800 Pa/(287 J/kg ƒ K)(313 K)
density = 1.11 kgm^-3
Units check = Pa/(Jkg^-1K^-1K)= Nm^-2N^-1m^-1kgK^-1K = kgm^-3
6. Hydrostatic Equation
dp/dz = - density × gravity
Units = kgm^-3ms^-2 = Pa/m = kgm^-2s^-2
Interpretation: The change in pressure with the change in height is directly related to the density of the air. Gravity varies slightly but can be treated as a constant.
*As density increases, the change in pressure with the change in height increases. Therefore, in cold air (more dense air) pressure decreases more rapidly with height than in warm air. This is the reason a deep core polar high will produce high pressure at the surface but a low pressure trough in the upper levels of the atmosphere.
*The negative sign is due to the fact that pressure decreases with height.
Example Problem: What is the average density of air in which the 1000 to 500 millibar thickness is 5,340 geopotential meters?
Answer: First, pressure needs to be converted to Pascals. The change in pressure is 500 mb which is 50,000 Pa.
The change in pressure with height is 50,000 Pa/ 5,340 m = 9.36329588 Pa/m.
Therefore, 9.36329588 Pa/m = -(density)(9.8ms^-2)
density = (9.36329588 Pa/m)/9.8ms^-2) = 0.955 kgm^-3
Units check = kgms^-2m^-2m^-1m^-1s^2 = kgm^-3
7. Virtual Temperature
Tv = T(1 + 0.61w)
T= temperature in Kelvins
w= mixing ratio in kg of moisture per kg of dry air
**Mixing ratio on Skew-T is expressed in g per kg, therefore you must divide Skew-T value by 1,000 to get the w for the virtual temperature equation. If a skew-T mixing ratio is 15 g per kg, w is 15/1,000 or 0.015 kg per kg. Therefore if temperature was 293 K, virtual temperature would be = 293(1 + 0.61 × 0.015) = 295.7 K
Interpretation: Virtual temperature is defined as the temperature that perfectly dry air would need to have to have the same density as the air with moisture. Logically, as the moisture content of the air increases, the virtual temperature increases since an increase of moisture decreases density. Moist air is less dense than dry air. Therefore, air which is perfectly dry needs to have a higher temperature in order to have the same density as the less dense moist air.
You may ask the purpose in needing to know virtual temperature. It is important since it makes meteorological equations MUCH less complicated. If virtual temperature is used, the moisture in the air can be ignored.
*Two air parcels that have the same virtual temperature will also have the same density. By using virtual temperature it can be inferred immediately which parcel is less or more dense.
Example problem: What is the virtual temperature of air that is 50ƒ F and has a mixing ratio of 5g/kg?
Answer: First change mixing ratio of kg/kg. 5g/kg =0.005kg/kg.
Second, convert temperature to Kelvins. 50ƒ F = 10ƒ C = 283.0 K
The mixing ratio is UNITLESS since kg/kg cancels.
Tv = 283.0 K(1 + (0.61 × 0.005)) = 283.9 K
8. Hypsometric Equation
Z2 - Z1 =change in height = Rd × Tv × 1/gravity × natural log (Pressure at base / Pressure of top)
UNITS = meters = kgms^-2mK^-1kg^-1Km^-1s^2 = m
Rd= gas constant = 287 J K^-1 kg^-1
Tv= Average virtual temperature in Kelvins between the two pressure levels (slightly higher than actual temp)
Natural log = The LN button on scientific calculator
Pressure at base and pressure of interest = suppose you wanted to know the change in height between 700 and 850 mb, the pressure at base would be 850 mb while the pressure of top would be 700 mb. The two pressures are divided by each other then the natural log is taken of this value. LN (850/700) = 0.194156014
Gravity= can be treated as a constant 9.81 ms^-2
Interpretation: The change in height is directly related to temperature. As temperature (almost the same value as virtual temperature) increases, the change in height (thickness between the two pressure levels) increases. As temperature decreases, the change in height decreases. Therefore, warm air advection causes expansion of air while cold air advection causes compression of air.
Example problem: What is the 1000 to 500 millibar thickness of dry air that has an average temperature of 5ƒ C?
Answer: Since the air is dry Tv = T = 278 K
Z = (278 K)(1/9.8 ms^-2)(287 JK^-1kg^-1) LN (1000/500) = 5,643 geopotential meters
Units check: Kkgms^-2mK^-1kg^-1m^-1s^2 = m
9. The dry adiabatic lapse rate
The change in temperature with height of a parcel of air if relative humidity is less than 100%
dT/dz = g/cp
Units = ms^-2J^-1kgK = ms^-2kg^-1m^-1s^2m^-1kgK = Km^-1
g = gravity 9.81 ms^-2
cp = 1004 Jkg^-1K^-1
Interpretation: The dry adiabatic lapse rate is a direct function of gravity. Since gravity is basically a constant, the dry adiabatic lapse rate is basically a constant.
Example problem: What is the dry adiabatic lapse rate on the planet Venus? How does this compare to the dry adiabatic lapse rate on Earth? The gravity on Venus is 0.904 that of earth. Assume the atmosphere of Venus is pure CO2 (it is actually 96%). The cp of C02 is 840 Jkg^-1K^-1.
Answer: First find gravity on Venus = 9.8ms^-2(0.904) = 8.87ms^-2
dT/dz = 8.87ms^-2/840 Jkg^-1K^-1 = 10.6 ƒ K/km = 10.6ƒ C/km
A rising parcel of dry air on Venus cools at about the same rate as on Earth
10. Relative Humidity
RH = 100 × w/ws or 100 × E/Es
UNITS = no units
w = mixing ratio, E= vapor pressure
ws= saturation mixing ratio, Es= saturation vapor pressure
*Relative humidity is NOT the dewpoint divided by the temperature
*RH can be found be using either a Skew-T, the vapor pressure/ mixing ratio as a function of temperature graph, or solving equations.
*Skew-T method for RH at any pressure level. 1. Find the mixing ratio which goes through the dewpoint (this is w) at the pressure level of interest 2. Find the mixing ratio which goes through the temperature (this is ws) at the pressure level of interest. 3. Divide w by ws and multiple by 100. A graph of temperature/dewpoint and mixing ratio can also be used. (One can be found on page 79 of the Intro of Meteorology book, The Atmosphere seventh edition by Lutgens and Tarbuck)
*Vapor pressure method. 1. Find the graph that displays vapor pressure as a function of temperature. 2. Find the vapor pressure using Temperature then find vapor pressure using dewpoint temperature. 3. Divide dewpoint vapor pressure by temperature vapor pressure and multiple by 100. Can also be done by hand using the Clausius-Clapeyron equation covered in this essay.
Example problem: What is the relative humidity when the temperature is 70ƒ F and the dewpoint is 50ƒ F?
Answer: First convert to Celsius scale. Temp = 70ƒ F = 21.1ƒ C
Dewpoint = 50ƒ = 10ƒ C; the saturation mixing ratio when the temperature is 10ƒ C = 7.8 g/kg;
the saturation mixing ratio when the temperature is 21.1ƒ C = 16.0 g/kg;
Therefore the relative humidity is the saturation mixing ratio of the dewpoint divided by the saturation mixing ratio of the temperature.
7.8/16.0 = 49 % Relative Humidity
11. The Stefan-Boltzman Law
Blackbody energy emission = constant × T^4
Units = Wm^-2
Constant = 5.67 × 10^-8 Wm^-2K^-4
T= temperature in Kelvins
T^4 = T × T × T × T
Interpretation: As temperature increases, the energy emitted from a blackbody increases exponentially (very rapidly with increasing temperature).
Example problem: How much energy is emitted from lead if it's temperature is 400 K? Assume the lead is a blackbody
Answer: E = (5.67×10^-8Wm^-2K^-4)(400 K)^4 = 1,452 Wm^-2
12. Potential temperature (aka Poisson's Equation)
PT = T(1000/P)^Rd/cp = T(1000/P)^0.286
T = temperature in Kelvins
P = pressure in millibars
Rd = gas constant for dry air
Cp = constant pressure process
Interpretation: Potential temperature is the temperature a parcel of air will have if raised or lowered to the 1000-millibar level. Potential temperature is the same for a parcel of air, as it rises or sinks, assuming adiabatic conditions.
*Potential temperature decreasing with height is an indication of atmospheric instability
Example Problem: What is the potential temperature of an air parcel at 500 mb that has a temperature of 0ƒ C.
Answer: PT = (273ƒ K)(2)^0.286 = 333ƒ K
13. Equivalent potential temperature
Theta-E = T(1000/P)^0.286 + 3w = PT + 3w
T = Temperature in Kelvins
P = Pressure in millibars
w = Mixing ratio in grams per kilogram
PT = Potential temperature
Interpretation: Theta-E is a function of moisture (via w) and temperature (via potential temperature). Theta-E increases by either increasing moisture content of the air or increasing the temperature. Areas with higher Theta-E have a greater potential for positive buoyancy.
Example problem: What is the Theta-E of an air parcel at 850 mb that has a temperature of 16ƒ C and is saturated?
Answer: First find the mixing ratio. Since the air is saturated the saturation mixing ratio will be w = 11.56 g/kg.
Next convert temperature to Kelvins, 16ƒ C = 289 K
Theta-E = 289(1000/850)^0.286 + 3(11.56) = 337 K
14. Clausius-Clapeyron equation
LN(Es/6.11) = (L/Rv )(1/273 - 1/T)
Es = Saturation vapor pressure
L = Latent heat of vaporization = 2.453 × 10^6 J/kg
Rv = Gas constant for moist air = 461 J/kg
T = Temperature in Kelvins
Interpretation: This equation gives the relationship between saturation vapor pressure and the temperature in Kelvins. This equation is used also to calculate relative humidity and other moisture variables.
Example problem: What is the saturation vapor pressure when the temperature is 30ƒ C?
Answer: Convert temperature to Kelvins, 30ƒ C = 303 K
LN(Es/6.11) = (2.453×10^6 J/kg/461 J/kg)(1/273 - 1/303)
LN(Es/6.11) = (5,321.041215)(0.003663004 - 0.00330033)
LN (Es/6.11) = 1.929801333
Es/6.11 = e^1.929801333
Es = (e^1.929801333)(6.11) = 42.1 mb
15. Relationship between vapor pressure and mixing ratio
Two forms of the equation
w = (0.622 × e)/(P - e) or e = (w × p)/(.622 + w)
w = mixing ratio in kilograms per kilogram
e = vapor pressure in millibars
P = pressure in millibars
Interpretation: If mixing ratio or vapor pressure is known, that can be converted into either the mixing ratio or the vapor pressure.
16. Fluid temperature equilibrium
(c1×m1× (TF - T1)) + (c2×m2× (TF - T2)) + Ö.. = 0
c1 = specific heat capacity in cal/g/ƒ C (a constant for the material, water has a c1 of 1.00)
m1 = mass in grams, can be in other units if c1, c2 have same mass or heat units
TF = Final temperature in Kelvins or Celsius
T1 = Temperature of fluid 1 in Kelvins or Celsius
Ö.. = denotes there can be more than 2 fluids
Interpretation: Fluids mixing together that have temperature differences will come to an equilibrium temperature after being mixed. Two masses of water mixed together which have different temperatures will result in a new equilibrium temperature. Most materials have a lower c1 than water (e.g. Aluminum = 0.2 , copper = 0.095)
Example problem: 3 kg of water at 5ƒ C mixing with 10 kg of water at 15ƒ C, what is the new temperature after the two water samples have mixed together?
Answer: The heat capacity of water = 1 = both c terms
(1)3kg (TF - 5ƒ C) + (1)10kg (TF - 15ƒ C) = 0
3TF - 15 + 10TF -150 = 0
13TF = 165
TF = 12.7ƒ C
17. Latent heat release/absorption
Latent heat of vaporization or condensation = 2.5 × 10^6 J/kg
Latent heat of freezing or melting = 3.34 × 10^5 J/kg
Latent heat of deposition or sublimation = 2.83 × 10^6 J/kg
*The latent heat of deposition and sublimation is found by adding the previous two terms together
These processes release latent heat (warm surrounding air) : Condensation, freezing, deposition
These processes absorb latent heat (chill surrounding air) : Vaporization, melting, sublimation
Amount of latent heat release/absorption in Joules = Latent heat constant (given above) × Mass in kilograms
Example problem: A storm produces 1,000,000 kg of water. How much latent heat of condensation was released to produce this rain?
Answer: (2.5 × 10^6 J/kg)(1,000,000 kg) = 2.5×10^12 Joules