The Standard Atmosphere
The "Standard Atmosphere" is a hypothetical vertical distribution of atmospheric properties which, by international agreement, is roughly representative of year-round, mid-latitude conditions. Typical usages include altimeter calibrations and aircraft design and performance calculations. It should be recognized that actual conditions may vary considerably from this standard.
The most recent definition is the "US Standard Atmosphere, 1976" developed jointly by NOAA, NASA, and the USAF. It is an idealized, steady state representation of the earth's atmosphere from the surface to 1000 km, as it is assumed to exist during a period of moderate solar activity. The 1976 model is identical with the earlier 1962 standard up to 51 km, and with the International Civil Aviation Organization (ICAO) standard up to 32 km.
Up to 86 km, the model assumes a constant mean molecular weight, and comprises of a series of six layers, each defined by a linear temperature gradient (lapse rate). (The assumption of linearity conveniently avoids the need for numerical integration in the computation of properties.) The bottom layer, with a negative lapse rate, represents the earth's troposphere, a region where most clouds form, and with generally turbulent conditions. Higher layers form part of the earth's stratosphere, where winds may be high, but turbulence is generally low.
The model is derived by assuming a constant value for g (gravitational acceleration). Strictly speaking, altitudes in this model should therefore be referred to as "geopotential altitudes" rather than "geometric altitudes" (physical height above mean sea level). The relationship between these altitudes is given by:
hgeometric = hgeopotential x Rearth / (Rearth - hgeopotential)
where Rearth is the earth's effective radius. The difference is small, with geometric altitude and geopotential altitude differing from by less than 0.5% at 30 km (~100,000 ft).
The standard is defined in terms of the International System of Units (SI). The air is assumed to be dry and to obey the perfect gas law and hydrostatic equation, which, taken together, relate temperature, pressure and density with geopotential altitude.
It should also be noted that since the standard atmosphere model does not include humidity, and since water has a lower molecular weight than air, its presence produces a lower density. Under extreme circumstances, this can amount to as much as a 3% reduction, but typically is less than 1% and may be neglected.
The following symbols are used to define the relationships between variables in the model. Subscript n indicates conditions at the base of the nth layer (or at the top of the (n-1)th layer) or refers to the constant lapse rate in the nth layer The first layer is considered to be layer 0, hence subscript 0 indicates standard, sea level conditions, or lapse rate in the bottom layer.
h = Pressure/Geopotential Altitude
T = Temperature
p = Pressure
ρ = Density
θ = T/T0 (Temperature Ratio)
δ = p/p0 (Pressure Ratio)
σ = ρ/ρ0 (Density Ratio)
μ = Dynamic Viscosity
ν = μ/ρ = Kinematic Viscosity
(Note: = indicates an exact value; ≅ indicates an approximate value correct to 6 significant figures)
Fundamental Constants:
Absolute Zero = -273.15°C (= -459.67°F)Earth:
Rearth = 6356.766 km (≈ 3950 statute miles)Gas Properties:
g0 = 9.80665 m/sec2 (≅ 32.1740 ft/sec2) (≅ gravitational acceleration at 45° latitude)
R* = 8.31432 J/mole·K (universal gas constant)Viscosity (empirical constants):
M = 0.0289644 kg/mole (mean molecular mass of air)
R = R*/M ≅ 287.053 J/kg·K (≅ 1716.56 ft2/sec2·R) (gas constant for air)
γ= 1.40 (ratio of specific heat capacities of air, cp/cv)
Sea Level Conditions (by definition):S = 110.4 K (Sutherland constant)
β = 1.458 x 10-6 kg/s·m·R1/2
T0 = 15.0°C (= 59.0°F)Hence:
p0 = 101,325 N/m2 (= 760mm Hg, ≅ 2116.22 lbs/ft2)
ρ0 ≅ 1.22500 kg/m3 (≅ 0.00237689 slugs/ft3)Definition of Layers in Model:
μ0 ≅ 1.78938 x 10-5 kg/m·sec (≅ 3.73720 x 10-7 slugs/ft·sec)
ν0 ≅ 1.46072 x 10-5 m2/sec (≅ 1.57231 x 10-4 ft2/sec)
Layer
|
Base Geopotential Altitude, hn (km) |
Base Geopotential Altitude, hn (ft)
|
Lapse Rate, λn (K/km)
|
Type
|
0
|
0
|
0
|
-6.5
|
|
1
|
11
|
36,089.2
|
0
|
Isothermal
|
2
|
20
|
65,616.8
|
+1.0
|
Inversion
|
3
|
32
|
104,986.9
|
+2.8
|
Inversion
|
4
|
47
|
154,199.5
|
0
|
Isothermal
|
5
|
51
|
167,322.8
|
-2.8
|
|
6
|
71
|
232,939.6
|
-2.0
|
|
7
|
84.8520
|
278,385.8
|
-
|
A positive lapse rate (λ > 0) means temperature increases with height. The temperature at the base of layer n is given by:
Tn = Tn-1 + (hn - hn-1)λn-1
or:
&thetan/&thetan-1 = 1 + (hn - hn-1)λn-1/Tn-1
Ideal gas law:
p = ρRT
and hence: σ = δ / θ
Hydrostatic equilibrium:
dp/dh = -gρ
In this model, viscosity is a function of temperature only, given by the following empirical relationship, valid at all altitudes:
μ = β·T3/2 / (T+S)
Combining the gas law and hydrostatic equations:
dp/p = (-g/R) dh/T
For the case of a constant lapse rate, λ
T = Tn + (h-hn)λn
and
dT/dh = λn
Substituting:
∫ dp/p = -g/(λnR) ∫ dT/T
Integrating:
loge(p/pn) = loge(T/Tn)-g/λnR
or
p/pn = (1 + (h-hn)λn/Tn)-g/λnR
For isothermal layers (λ=0), and using the approximation loge(1+ε)→ε as ε→0, this becomes:
p/pn = e-(h-hn)g/RTn
These solutions lead to the following model equations:
h ≤ h1θ = 1 + hλ0/T0h1 < h ≤ h2 (Isothermal)
δ = θ (-g/λ0R)
σ= (1 + hλ0/T0) (-g/λ0R)-1
hence:θ1 = 1 + h1λ0/T0
δ1 = θ1(-g/λ0R)
θ = θ1h2 < h ≤ h3 (Inversion)
δ = δ1 e-(h-h1)g/RT1
σ= (δ1/θ1) e-(h-h1)g/RT1
hence:θ2 = θ1
δ2 = δ1 e-(h2-h1)g/RT1
θ = θ2 + (h-h2)λ2/T0h3 < h ≤ h4 (Inversion)
δ = δ2 (θ/θ2)(-g/λ2R)
σ= δ2 (θ2 + (h-h2)λ2/T0) (-g/λ2R)-1 (1/θ2)(-g/λ2R)
hence:θ3 = θ2 + (h3-h2)λ2/T0
δ3 = δ2 (θ3/θ2)(-g/λ2R)
θ = θ3 + (h-h3)λ3/T0h4 < h ≤ h5 (Isothermal)
δ = δ3 (θ/θ3)(-g/λ3R)
σ= δ3 (θ3 + (h-h3)λ3/T0) (-g/λ3R)-1 (1/θ3)(-g/λ3R)
hence:θ4 = θ3 + (h4-h3)λ3/T0
δ4 = δ3 (θ4/θ3)(-g/λ3R)
θ = θ4h5 < h ≤ h6
δ = δ4 e-(h-h4)g/RT4
σ= (δ4/θ4) e-(h-h4)g/RT4
hence:θ5 = θ4
δ5 = δ4 e-(h5-h4)g/RT4
θ = θ5 + (h-h5)λ5/T0h6 < h ≤ h7
δ = δ5 (θ/θ5)(-g/λ5R)
σ= δ5 (θ5 + (h-h5)λ5/T0) (-g/λ5R)-1 (1/θ5)(-g/λ5R)
hence:θ6 = θ5 + (h6-h5)λ5/T0
δ6 = δ5 (θ6/θ5)(-g/λ5R)
θ = θ6 + (h-h6)λ6/T0
δ = δ6 (θ/θ6)(-g/λ6R)
σ= δ6 (θ6 + (h-h6)λ6/T0) (-g/λ6R)-1 (1/θ6)(-g/λ6R)
hence:θ7 = θ6 + (h7-h6)λ6/T0
δ7 = δ6 (θ7/θ6)(-g/λ6R)
(where h is in feet)
Layer n
|
θn
|
δn
|
σn = δn / θn
|
0
|
1
|
1
|
1
|
1
|
0.751865
|
2.23361E-01
|
2.97076E-01
|
2
|
0.751865
|
5.40330E-02
|
7.18652E-02
|
3
|
0.793510
|
8.56668E-03
|
1.07959E-02
|
4
|
0.939268
|
1.09456E-03
|
1.16533E-03
|
5
|
0.939268
|
6.60635E-04
|
7.03351E-04
|
6
|
0.744925
|
3.90468E-05
|
5.24172E-05
|
7
|
0.648780
|
3.68501E-06
|
5.67991E-06
|
h ≤ 36,089
θ = 1 - h / 145,44236,089 < h ≤ 65,617 (Isothermal)
δ = (1 - h / 145,442)5.255876
σ = (1 - h / 145,442)4.255876
θ = 0.75186565,617 < h ≤ 104,987 (Inversion)
δ = 0.223361 e-(h-36,089)/20,806
σ = 0.297076 e-(h-36,089)/20,806
θ = 0.682457 + h / 945,374104,987< h ≤ 154,199 (Inversion)
δ = (0.988626 + h / 652,600)-34.16320
σ = (0.978261 + h / 659,515)-35.16320
θ = 0.482561 + h / 337,634154,199 < h ≤ 167,323 (Isothermal)
δ = (0.898309 + h / 181,373)-12.20114
σ = (0.857003 + h / 190,115)-13.20114
θ = 0.939268167,323 < h ≤ 232,940
δ = 0.00109456 e-(h-154,199)/25,992
σ = 0.00116533 e-(h-154,199)/25,992
θ = 1.434843 - h / 337,634232,940 < h ≤ 278,386
δ = (0.838263 - h / 577,922)+12.20114
σ = (0.798990 - h / 606,330)+11.20114
θ = 1.237723 - h / 472,687
δ = (0.917131 - h / 637,919)+17.08160
σ = (0.900194 - h / 649,922)+16.08160
When actual conditions vary from those given by the standard atmosphere model, the expressions given above, do not apply. Two definitions need to be introduced at this point:
The pressure altitude is calculated by inverting the appropriate expression for pressure ratio, δ. Density altitude is calculated by inverting the appropriate expression for σ as a function of geopotential, h. Thus, for non-standard temperatures, the density altitude, hd, is given by:
σ ≥ σ1hd = (σ-1/(g/λ0R+1) - 1) T0/λ0σ 1 > σ ≥ σ2 (Isothermal)
hd = (-RT1/g) loge(σθ1/δ1) + h1σ 2 > σ ≥ σ3 (Inversion)
hd = ( (σ/δ2) θ2(-g/λ2R) )-1/(g/λ2R+1) - θ2) T0/λ2 + h2σ 3 > σ ≥ σ4 (Inversion)
hd = ( (σ/δ3) θ3(-g/λ3R) )-1/(g/λ3R+1) - θ3) T0/λ3 + h3σ 4 > σ ≥ σ5 (Isothermal)
hd = (-RT4/g) loge(σθ4/δ4) + h4σ 5 > σ ≥ σ6
hd = ( (σ/δ5) θ5(-g/λ5R) )-1/(g/λ5R+1) - θ5) T0/λ5 + h5σ 6 > σ ≥ σ7
hd = ( (σ/δ6) θ6(-g/λ6R) )-1/(g/λ6R+1) - θ6) T0/λ6 + h6
Page last updated on November 23, 2014
Copyright © 1991-2014, Graham Gyatt. All Rights Reserved.