Physics

Physics · thermodynamics

Problem

4.1. Table 1 shows air temperatures recorded by a radio sounding balloon at 7:00 am on a November day in Hanoi. The change of temperature with altitude can be approximately described by the formula

T(z)=T(0)-\Lambda z

with different lapse rates

\Lambda

in the three layers

0<\mathrm{z}<96 \mathrm{~m}, 96 \mathrm{~m}<\mathrm{z}<119 \mathrm{~m}

and

119 \mathrm{~m}<\mathrm{z}<215 \mathrm{~m}

. Consider an air parcel with temperature

T_{\text {parcel }}(0)=22^{\circ} \mathrm{C}

ascending from ground. On the basis of the data given in Table 1 and using the above linear approximation, calculate the temperature of the parcel at the altitudes of

96 \mathrm{~m}

and

119 \mathrm{~m}

. Given constants: CHANGE OF AIR TEMPERATURE WITH ALTITUDE, ATMOSPHERIC STABILITY AND AIR POLLUTION Vertical motion of air governs many atmospheric processes, such as the formation of clouds and precipitation and the dispersal of air pollutants. If the atmosphere is stable, vertical motion is restricted and air pollutants tend to be accumulated around the emission site rather than dispersed and diluted. Meanwhile, in an unstable atmosphere, vertical motion of air encourages the vertical dispersal of air pollutants. Therefore, the pollutants' concentrations depend not only on the strength of emission sources but also on the stability of the atmosphere. We shall determine the atmospheric stability by using the concept of air parcel in meteorology and compare the temperature of the air parcel rising or sinking adiabatically in the atmosphere to that of the surrounding air. We will see that in many cases an air parcel containing air pollutants and rising from the ground will come to rest at a certain altitude, called a mixing height. The greater the mixing height, the lower the air pollutant concentration. We will evaluate the mixing height and the concentration of carbon monoxide emitted by motorbikes in the Hanoi metropolitan area for a morning rush hour scenario, in which the vertical mixing is restricted due to a temperature inversion (air temperature increases with altitude) at elevations above

119 \mathrm{~m}

. Let us consider the air as an ideal diatomic gas, with molar mass

\mu=29 \mathrm{~g} / \mathrm{mol}

. Quasi equilibrium adiabatic transformation obey the equation

p V^{\gamma}=

const, where

\gamma=\frac{c_{p}}{c_{V}}

is the ratio between isobaric and isochoric heat capacities of the gas. The student may use the following data if necessary: The universal gas constant is

R=8.31 \mathrm{~J} /(

mol.K). The atmospheric pressure on ground is

p_{0}=101.3 \mathrm{kPa}

The acceleration due to gravity is constant,

g=9.81 \mathrm{~m} / \mathrm{s}^{2}

The molar isobaric heat capacity is

c_{p}=\frac{7}{2} R

for air. The molar isochoric heat capacity is

c_{V}=\frac{5}{2} R

for air. Mathematical hints a.

\int \frac{d x}{A+B x}=\frac{1}{B} \int \frac{d(A+B x)}{A+B x}=\frac{1}{B} \ln (A+B x)

b. The solution of the differential equation

\frac{d x}{d t}+A x=B \quad

(with

\quad A

and

B

constant) is

x(t)=x_{1}(t)+\frac{B}{A}

where

x_{1}(t)

is the solution of the differential equation

\frac{d x}{d t}+A x=0

. c.

\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}=e

1. Change of pressure with altitude. Context question: 1.1. Assume that the temperature of the atmosphere is uniform and equal to

T_{0}

. Write down the expression giving the atmospheric pressure

p

as a function of the altitude

z

. Context answer:

p(z)=p(0) e^{-\frac{\mu g}{R T_{0}} z}

Extra Supplementary Reading Materials: 1.2. Assume that the temperature of the atmosphere varies with the altitude according to the relation

T(z)=T(0)-\Lambda z

where

\Lambda

is a constant, called the temperature lapse rate of the atmosphere (the vertical gradient of temperature is

-\Lambda

). Context question: 1.2.1. Write down the expression giving the atmospheric pressure

p

as a function of the altitude

Z

. Context answer:

p(z)=p(0)\left(1-\frac{\Lambda z}{T(0)}\right)^{\frac{\mu g}{R \Lambda}}

Context question: 1.2.2. A process called free convection occurs when the air density increases with altitude. At which values of

\Lambda

does the free convection occur? Context answer:

0.034

Extra Supplementary Reading Materials: 2. Change of the temperature of an air parcel in vertical motion Consider an air parcel moving upward and downward in the atmosphere. An air parcel is a body of air of sufficient dimension, several meters across, to be treated as an independent thermodynamical entity, yet small enough for its temperature to be considered uniform. The vertical motion of an air parcel can be treated as a quasi adiabatic process, i.e. the exchange of heat with the surrounding air is negligible. If the air parcel rises in the atmosphere, it expands and cools. Conversely, if it moves downward, the increasing outside pressure will compress the air inside the parcel and its temperature will increase. As the size of the parcel is not large, the atmospheric pressure at different points on the parcel boundary can be considered to have the same value

p(z)

, with

z

- the altitude of the parcel center. The temperature in the parcel is uniform and equals to

T_{\text {parcel }}(z)

, which is generally different from the temperature of the surrounding air

T(z)

. In parts 2.1 and 2.2, we do not make any assumption about the form of

T(z)

. Context question: 2.1. The change of the parcel temperature

T_{\text {parcel }}

with altitude is defined by

\frac{d T_{\text {parcel }}}{d z}=-G

. Derive the expression of

G\left(T, T_{\text {parcel }}\right)

. Context answer:

\frac{\gamma-1}{\gamma} \frac{\mu g}{R} \frac{T_{\text {parcel }}}{T}

Extra Supplementary Reading Materials: 2.2. Consider a special atmospheric condition in which at any altitude

z

the temperature

T

of the atmosphere equals to that of the parcel

T_{\text {parcel }}, T(z)=T_{\text {parcel }}(z)

. We use

\Gamma

to denote the value of

G

when

T=T_{\text {parcel }}

, that is

\Gamma=-\frac{d T_{\text {parcel }}}{d z}

(with

\left.T=T_{\text {parcel }}\right) . \Gamma

is called dry adiabatic lapse rate. Context question: 2.2.1. Derive the expression of

\Gamma

Context answer:

\Gamma=\frac{\mu g}{c_{p}}

Context question: 2.2.2. Calculate the numerical value of

\Gamma

. Context answer:

0.00978

Context question: 2.2.3. Derive the expression of the atmospheric temperature

T(z)

as a function of the altitude. Context answer:

T(z)=T(0)-\Gamma z

Context question: 2.3. Assume that the atmospheric temperature depends on altitude according to the relation

T(z)=T(0)-\Lambda z

, where

\Lambda

is a constant. Find the dependence of the parcel temperature

T_{\text {parcel }}(z)

on altitude

z

. Context answer:

T_{\text {parcel }}(z)=T_{\text {parcel }}(0)\left(\frac{T(0)-\Lambda z}{T(0)}\right)^{\frac{\Gamma}{\Lambda}}

Context question: 2.4. Write down the approximate expression of

T_{\text {parcel }}(z)

when

|\Lambda z|<<T(0)

and

T(0) \approx T_{\text {parcel }}(0)

. Context answer:

T_{\text {parcel }}(0)-\Gamma z

Extra Supplementary Reading Materials: 3. The atmospheric stability. In this part, we assume that

T

changes linearly with altitude. Context question: 3.1. Consider an air parcel initially in equilibrium with its surrounding air at altitude

z_{0}

, i.e. it has the same temperature

T\left(z_{0}\right)

as that of the surrounding air. If the parcel is moved slightly up and down (e.g. by atmospheric turbulence), one of the three following cases may occur: - The air parcel finds its way back to the original altitude

z_{0}

, the equilibrium of the parcel is stable. The atmosphere is said to be stable. - The parcel keeps moving in the original direction, the equilibrium of the parcel is unstable. The atmosphere is unstable. - The air parcel remains at its new position, the equilibrium of the parcel is indifferent. The atmosphere is said to be neutral. What is the condition on

\Lambda

for the atmosphere to be stable, unstable or neutral? Context answer:

\Lambda>\Gamma

is unstable

\Lambda<\Gamma

is stable

\Lambda=\Gamma

is neutral Context question: 3.2. A parcel has its temperature on ground

T_{\text {parcel }}(0)

higher than the temperature

T(0)

of the surrounding air. The buoyancy force will make the parcel rise. Derive the expression for the maximal altitude the parcel can reach in the case of a stable atmosphere in terms of

\Lambda

and

\Gamma

. Context answer:

h=\frac{1}{\Lambda}\left[T(0)-\left(\frac{(T(0))^{\Gamma}}{\left(T_{\text {parcel }}(0)\right)^{\Lambda}}\right)^{\frac{1}{\Gamma-\Lambda}}\right]

Extra Supplementary Reading Materials: 4. The mixing height Table 1 Data recorded by a radio sounding balloon at 7:00 am on a November day in Hanoi.

Altitude, m	Temperature, ${ }^{\circ} \mathrm{C}$
5	21.5
60	20.6
64	20.5
69	20.5
75	20.4
81	20.3
90	20.2
96	20.1
102	20.1
109	20.1
113	20.1
119	20.1
128	20.2
136	20.3
145	20.4
153	20.5
159	20.6
168	20.8
178	21.0
189	21.5
202	21.8
215	22.0
225	22.1
234	22.2
246	22.3
257	22.3

Related theory

thermodynamics

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