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Deriving Equations for Atmospheric Pressure and Density

The pressure and density of the atmosphere as used in the Barometric Formula depend on the altitude, the amount and composition of the atmosphere and the temperature profile. Atmospheric pressure, density and temperature are related via the

Ideal Gas Law

(1)

$P = R_\mathrm{s} \cdot \rho \cdot T \qquad \Leftrightarrow \qquad \rho = { P \over R_\mathrm{s} \cdot T }$

Ideal Gas Law

where^'

$P$ ^'	=^'	^'air pressure
$\rho$ ^'	=^'	^'air density
$T$ ^'	=^'	^'absolute temperature
$R_\mathrm{s}$ ^'	=^'	^'Specific Gas Constant; dry air = 287.058 J/(kg·K)

We can not measure the density of the atmosphere (mass per volume element) at a certain altitude directly, but we can measure the temperature profile using balloons and then apply physical laws to derive the pressure and density gradient as follows:

The atmospheric pressure as a function of altitude can be derived by investigating the pressures acting on a small volume element. Using calculus we can then derive the pressure gradient for the whole atmosphere. Using the temperature profile and the ideal gas law we can also derive the density gradient from the pressure and temperature gradient.

determining the pressures on all sides of a small volume element of air
creating an emirical simplified model for the temperature profile of the atmosphere
deriving the atmospheric pressure profile as a function of altitude and temperature profile using calculus
deriving the density profile from the pressure and temperature profile using the ideal gas law

Pressure Differential

To derive the pressure of the atmosphere as a function of altitude, we need to calculate how the pressure changes with altitude. To do this, lets regard an ifinitesimal small cube of atmosphere and calculate the pressures acting at the top and bottom area of the cube. The forces acting on the sides of the cube cancel each other, so they are irrelevant for the calculation of the vertical pressure gradient.

The downward force $F(h+\mathrm{d} h)$ acting on the area $\mathrm{d} A$ at the top of the cube is the weight $W_\mathrm{col}$ of the column of atmosphere above the cube. The downward force $F(h)$ acting at the bottom of the cube is the force at the top plus the weight $\mathrm{d} W$ of the cube. So we have the pressures at the top and bottom of the cube respectively:

(2)	$P(h+\mathrm{d} h) = { F(h+\mathrm{d} h) \over \mathrm{d} A } = { W_\mathrm{col} \over \mathrm{d} A }$
(3)	$P(h) = { F(h+\mathrm{d} h) + \mathrm{d} W \over \mathrm{d} A } = { W_\mathrm{col} + \mathrm{d} W \over \mathrm{d} A }$

The pressure difference between the top and bottom of the small cube is:

(4)

$\mathrm{d} P = P(h+\mathrm{d} h) - P(h) = { W_\mathrm{col} \over \mathrm{d} A } - { W_\mathrm{col} + \mathrm{d} W \over \mathrm{d} A } = - { \mathrm{d} W \over \mathrm{d} A }$

with

$\mathrm{d} W = g \cdot \rho \cdot \color{blue}{\mathrm{d} V} = g \cdot \rho \cdot \color{blue}{\mathrm{d} A \cdot \mathrm{d} h}$

where^'

$\mathrm{d} W$ ^'	=^'	^'weight of the cube
$g$ ^'	=^'	^'gravitational acceleration
$\rho$ ^'	=^'	^'density of the air in the cube
$\mathrm{d} V$ ^'	=^'	^'volume of the cube
$\mathrm{d} A$ ^'	=^'	^'base and top area of the cube
$\mathrm{d} h$ ^'	=^'	^'height of the cube

Inserting $\mathrm{d} W$ into the equation for $\mathrm{d} P$ we get:

(5)	$\mathrm{d} P = - { \mathrm{d} W \over \mathrm{d} A } = -{ g \cdot \rho \cdot \color{red}{\mathrm{d} A} \cdot \mathrm{d} h \over \color{red}{\mathrm{d} A} }$

So the vertical pressure differential acting on a small volume element of density $\rho$ under the gravitational acceleration g is:

(6)

$\mathrm{d} P(h) = -g \cdot \rho(h) \cdot \mathrm{d} h$

where^'

$\mathrm{d} P(h)$ ^'	=^'	^'air pressure differential at altitude h
$g$ ^'	=^'	^'gravitational acceleration
$\rho(h)$ ^'	=^'	^'air density at altitude h
$\mathrm{d} h$ ^'	=^'	^'height differential

Using the ideal gas law (1) we can replace the density in (6) by the pressure to get the following differential equation:

(7)

$\mathrm{d} P(h) = -{ g \cdot P(h) \over R_\mathrm{s} \cdot T(h) } \cdot \mathrm{d} h$

where^'

$\mathrm{d} P(h)$ ^'	=^'	^'air pressure differenctial at altitude h
$g$ ^'	=^'	^'gravitational acceleration
$P(h)$ ^'	=^'	^'air pressure at altitude h
$T(h)$ ^'	=^'	^'absolute air temperature at altitude h in Kelvin
$R_\mathrm{s}$ ^'	=^'	^'287.058 J/(kg·K) = Specific Gas Constant for dry air
$\mathrm{d} h$ ^'	=^'	^'height differential

Solving Nummerically

The differential equation (7) can be solved for the pressure gradient nummerically. In this case g and T can be arbitrary functions or measurements. Nummerically we solve the following equation iteratively:

(8)	$P_{i+1}(h_{i+1}) = P_i(h_i) + \Delta P_i(h_i)$
with	$\Delta P_i(h_i) = -{ g_i(h_i) \cdot P_i(h_i) \over R_\mathrm{s} \cdot T_i(h_i) } \cdot \Delta h$
and	$h_{i+1} = h_i + \Delta h$

We begin with some start conditions, which could be measurements, for P₀(h₀), g₀(h₀), T₀(h₀). After each iteration we get the new pressure for the next height increment and we may have different g and T values. There are more complicated ways for nummerical integrations, that yield more accuracy.

Analytical Solutions

If we make the simplifying assumptions that g is constant for the height range we want to solve for and T is constant or a linear function of h, we can solve the differential equation (9) analytically.

(9)	$\mathrm{d} P = -{ g \cdot P \over R_\mathrm{s} \cdot T } \cdot \mathrm{d} h$

We can bring the pressure P from the right side of the equal sign to the left side, so we have all pressure terms on the left, which simplifies solving the differential equation:

(10)

${ \mathrm{d} P \over P } = -{ g \over R_\mathrm{s} \cdot T } \cdot \mathrm{d} h$

Note: Pressure and temperature are commonly dependent on altitude. The temperature has to be measured empirically. It can also be constant over some altitude range. The gravitational acceleration can be considered constant below a certain altitude range.

In the International Standard Atmosphere model, used in aviation and other fields, the empirically measured temperature gradient is divided into some ranges and approximated either as constant (isotherm) or linearly dependent on altitude.

Pressure Isotherm

Lets first derive the pressure equation for the simpler case of constant temperature.

We can solve (10) by integrating both sides of the equal sign. The integration ∫ 1/P dP = ln(P) is applied to the left side, the integration of the right side is straight forward A · ∫ dh = A·h:

(11)	$\int_{ P_\mathrm{ref} }^{ P(h) } { 1 \over P } \ \mathrm{d} P = -{ g \over R_\mathrm{s} \cdot T } \cdot \int_{ h_\mathrm{ref} }^{ h } \ \mathrm{d} h$
(12)	$\big[ \, \ln\left( P \right) \, \big]_{P_\mathrm{ref}}^{P(h)} = -{ g \over R_\mathrm{s} \cdot T } \cdot \big[ h \big]_{ h_\mathrm{ref} }^{ h }$

Applying the limits:

(13)	$\ln\left[ P(h) \right] - \ln\left[ P_\mathrm{ref} \right] = \ln\left( { P(h) \over P_\mathrm{ref} } \right) = -{ g \over R_\mathrm{s} \cdot T } \cdot ( h - h_\mathrm{ref} )$

To get rid of the logarithms we apply e^x to both sides, because e^ln(X) = X:

(14)	${ P(h) \over P_\mathrm{ref} } = \mathrm{exp} \left[ - { g \over R_\mathrm{s} \cdot T } \cdot ( h - h_\mathrm{ref} ) \right]$

Bringing the constant $P_\mathrm{ref}$ to the right hand side we get finally the formula for pressure as a function of altitude for constant temperature:

(15)

$P(h) = P_\mathrm{ref} \cdot \mathrm{exp} \left[ -{ g \over R_\mathrm{s} \cdot T } \cdot ( h - h_\mathrm{ref} ) \right]$

where^'

$P(h)$ ^'	=^'	^'pressure at altitude h above sea level
$P_\mathrm{ref}$ ^'	=^'	^'pressure at the reference altitude h_ref above sea level
$g$ ^'	=^'	^'9.80665 m/s² = gravitational acceleration
$R_\mathrm{s}$ ^'	=^'	^'287.058 J/(kg·K) = Specific Gas Constant for dry air
$T$ ^'	=^'	^'constant absolute temperature in the range from h_ref to h
$h$ ^'	=^'	^'altitude above sea level
$h_\mathrm{ref}$ ^'	=^'	^'reference altitude above sea level

This formula makes the simplification that the gravitational acceleration g is constant in the altitude range h_ref ... h.

Pressure with Linear Temperature Gradient

We start again with the differential equation derived at (10), repeated here:

(16)	${ \mathrm{d} P \over P } = -{ g \over R_\mathrm{s} \cdot T } \cdot \mathrm{d} h$

But this time temperature T is not constant, but a linear function of altitude h, so that T(h_ref) = T_ref:

(17)

$T(h) = \alpha \cdot (h - h_\mathrm{ref}) + T_\mathrm{ref}$

Inserting (17) into (16) we get the new differential equation:

(18)	${ \mathrm{d} P \over P } = -{ g \over R_\mathrm{s} } \cdot { 1 \over \alpha \cdot (h - h_\mathrm{ref}) + T_\mathrm{ref} } \cdot \mathrm{d} h$

We can solve (18) by integrating both sides of the equal sign. The general integral ∫ (1/X) dX = ln(X) is applied to both sides:

(19)	$\int_{ P_\mathrm{ref} }^{ P(h) } { 1 \over P } \ \mathrm{d} P = -{ g \over R_\mathrm{s} } \cdot \int_{ h_\mathrm{ref} }^{ h } { 1 \over \alpha \cdot (h - h_\mathrm{ref}) + T_\mathrm{ref} } \ \mathrm{d} h$
(20)	$\big[ \, \ln\left( P \right) \, \big]_{P_\mathrm{ref}}^{P(h)} = -{ g \over R_\mathrm{s} } \cdot \left[ { 1 \over \alpha } \cdot \ln\left( \alpha \cdot (h - h_\mathrm{ref}) + T_\mathrm{ref} \right) \right]_{ h_\mathrm{ref} }^{ h }$

Putting in the limits yields:

(21)	$\ln\left[ P(h) \right] - \ln\left[ P_\mathrm{ref} \right] = -{ g \over R_\mathrm{s} \cdot \alpha } \cdot \left[ \ln\left( \alpha \cdot ( h - h_\mathrm{ref} ) + T_\mathrm{ref} \right) - \ln\left( T_\mathrm{ref} \right) \right]$

Applying ln(A) − ln(B) = ln(A/B) on both sides yields:

(22)	$\ln\left( { P(h) \over P_\mathrm{ref} } \right) = - \beta \cdot \ln \left[ { \alpha \cdot ( h - h_\mathrm{ref} ) + T_\mathrm{ref} \over T_\mathrm{ref} } \right]$
with	$\beta = { g \over R_\mathrm{s} \cdot \alpha }$

To get rid of the logarithms we apply e^x to both sides and because exp(−β · ln(X)) = exp(ln(X)·(−β)) = exp(ln(X))^−β = X^−β we get:

(23)	${ P(h) \over P_\mathrm{ref} } = \mathrm{exp}\left\{ -\beta \cdot \ln \left[ { \alpha \cdot ( h - h_\mathrm{ref} ) + T_\mathrm{ref} \over T_\mathrm{ref} } \right] \right\} = \left( { \alpha \cdot ( h - h_\mathrm{ref} ) + T_\mathrm{ref} \over T_\mathrm{ref} } \right)^{ -\beta }$

Bringing $P_\mathrm{ref}$ to the right hand side we finally get our formula for pressure as function of altitude in the case of a linear temperature gradient:

(24)

$P(h) = P_\mathrm{ref} \cdot \left( { \alpha \cdot ( h - h_\mathrm{ref} ) + T_\mathrm{ref} \over T_\mathrm{ref} } \right)^{ -\beta } = P_\mathrm{ref} \cdot \left( 1 + { \alpha \cdot ( h - h_\mathrm{ref} ) \over T_\mathrm{ref} } \right)^{ -\beta }$

with

$\beta = { g \over R_\mathrm{s} \cdot \alpha }$

where^'

$P(h)$ ^'	=^'	^'pressure at altitude h above sea level
$P_\mathrm{ref}$ ^'	=^'	^'pressure at reference altitude h_ref above sea level
$T_\mathrm{ref}$ ^'	=^'	^'temperature at reference altitude h_ref
$\alpha$ ^'	=^'	^'temperature gradient (negative lapse rate)
$h$ ^'	=^'	^'altitude above sea level
$h_\mathrm{ref}$ ^'	=^'	^'reference altitude above sea level
$g$ ^'	=^'	^'9.80665 m/s² = mean gravitational acceleration at sea level
$R_\mathrm{s}$ ^'	=^'	^'287.058 J/(kg·K) = Specific Gas Constant for dry air

Density Isotherm

Using the ideal gas law (1) we can derive the equation for the density in an altitude range of constant air temperature T from the equation for Pressure Isotherm:

(25)	$\rho(h) = { P(h) \over R_\mathrm{s} \cdot T } = { P_\mathrm{ref} \over R_\mathrm{s} \cdot T } \cdot \mathrm{exp} \left[ -{ g \over R_\mathrm{s} \cdot T } \cdot ( h - h_\mathrm{ref} ) \right]$

And because $P_\mathrm{ref}/(R_\mathrm{s} \cdot T)$ is the reference density $\rho_\mathrm{ref}$ we get finally:

(26)

$\rho(h) = \rho_\mathrm{ref} \cdot \mathrm{exp} \left[ -{ g \over R_\mathrm{s} \cdot T } \cdot ( h - h_\mathrm{ref} ) \right]$

where^'

$\rho(h)$ ^'	=^'	^'air density at altitude h
$\rho_\mathrm{ref}$ ^'	=^'	^'air density at reference altitude h_ref, at which the constand temperature range starts
$T$ ^'	=^'	^'constant absolute temperature in the range from h_ref to h
$h$ ^'	=^'	^'altitude above sea level
$h_\mathrm{ref}$ ^'	=^'	^'reference altitude above sea level
$g$ ^'	=^'	^'9.80665 m/s² = mean gravitational acceleration at sea level
$R_\mathrm{s}$ ^'	=^'	^'287.058 J/(kg·K) = Specific Gas Constant for dry air

Density with a Linear Temerature Gradient

Using the ideal gas law (1) we can derive the equation for the density in an altitude range with a linear temperture gradient (17) from the equation for Pressure with a Linear Temerature Gradient:

(27)

$\rho(h) = { P(h) \over R_\mathrm{s} \cdot T(h) } = { P_\mathrm{ref} \cdot { \left( 1 + { \alpha \cdot ( h - h_\mathrm{ref} ) \over T_\mathrm{ref} } \right) }^{ -\beta } \over R_\mathrm{s} \cdot T_\mathrm{ref} \cdot \left( 1 + { \alpha \cdot ( h - h_\mathrm{ref} ) \over T_\mathrm{ref} } \right) } = \rho_\mathrm{ref} \cdot \left( 1 + { \alpha \cdot ( h - h_\mathrm{ref} ) \over T_\mathrm{ref} } \right)^{ -\beta } \cdot \left( 1 + { \alpha \cdot ( h - h_\mathrm{ref} ) \over T_\mathrm{ref} } \right)^{ -1 }$

We can combine the 2 equal expressions in parenthesis to get finally:

(28)

$\rho(h) = \rho_\mathrm{ref} \cdot \left( 1 + { \alpha \cdot ( h - h_\mathrm{ref} ) \over T_\mathrm{ref} } \right)^{ -\beta - 1 }$

with

$\beta = { g \over R_\mathrm{s} \cdot \alpha }$

where^'

$\rho(h)$ ^'	=^'	^'density at altitude h above sea level in a range with a linear temperature gradient
$\rho_\mathrm{ref}$ ^'	=^'	^'density at reference altitude h_ref above sea level
$\alpha$ ^'	=^'	^'temperature gradient (negative lapse rate)
$T_\mathrm{ref}$ ^'	=^'	^'temperature at reference altitude h_ref
$h$ ^'	=^'	^'altitude above sea level
$h_\mathrm{ref}$ ^'	=^'	^'reference altitude above sea level
$g$ ^'	=^'	^'9.80665 m/s² = mean gravitational acceleration at sea level
$R_\mathrm{s}$ ^'	=^'	^'287.058 J/(kg·K) = Specific Gas Constant for dry air

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WaBis

walter.bislins.ch

Walter Bislin's Blog-En

Deriving Equations for Atmospheric Pressure and Density

Ideal Gas Law

Pressure Differential

Solving Nummerically

Analytical Solutions

Pressure Isotherm

Pressure with Linear Temperature Gradient

Density Isotherm

Density with a Linear Temerature Gradient

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