Physics

Physics · thermodynamics

Problem

5.1. Derive the differential equation determining the CO pollutant concentration C(t)C(t) as a function of time. 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 m119 \mathrm{~m}. Let us consider the air as an ideal diatomic gas, with molar mass μ=29 g/mol\mu=29 \mathrm{~g} / \mathrm{mol}. Quasi equilibrium adiabatic transformation obey the equation pVγ=p V^{\gamma}= const, where γ=cpcV\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 J/(R=8.31 \mathrm{~J} /( mol.K). The atmospheric pressure on ground is p0=101.3kPap_{0}=101.3 \mathrm{kPa} The acceleration due to gravity is constant, g=9.81 m/s2g=9.81 \mathrm{~m} / \mathrm{s}^{2} The molar isobaric heat capacity is cp=72Rc_{p}=\frac{7}{2} R for air. The molar isochoric heat capacity is cV=52Rc_{V}=\frac{5}{2} R for air. Mathematical hints a. ∫dxA+Bx=1B∫d(A+Bx)A+Bx=1Bln⁡(A+Bx)\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 dxdt+Ax=B\frac{d x}{d t}+A x=B \quad (with A\quad A and BB constant) is x(t)=x1(t)+BAx(t)=x_{1}(t)+\frac{B}{A} where x1(t)x_{1}(t) is the solution of the differential equation dxdt+Ax=0\frac{d x}{d t}+A x=0. c. lim⁡x→∞(1+1x)x=e\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 T0T_{0}. Write down the expression giving the atmospheric pressure pp as a function of the altitude zz. Context answer: p(z)=p(0)e−μgRT0zp(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)−ΛzT(z)=T(0)-\Lambda z where Λ\Lambda is a constant, called the temperature lapse rate of the atmosphere (the vertical gradient of temperature is −Λ-\Lambda ). Context question: 1.2.1. Write down the expression giving the atmospheric pressure pp as a function of the altitude ZZ. Context answer: p(z)=p(0)(1−ΛzT(0))μgRΛp(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 Λ\Lambda does the free convection occur? Context answer: 0.0340.034 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)p(z), with zz - the altitude of the parcel center. The temperature in the parcel is uniform and equals to Tparcel (z)T_{\text {parcel }}(z), which is generally different from the temperature of the surrounding air T(z)T(z). In parts 2.1 and 2.2, we do not make any assumption about the form of T(z)T(z). Context question: 2.1. The change of the parcel temperature Tparcel T_{\text {parcel }} with altitude is defined by dTparcel dz=−G\frac{d T_{\text {parcel }}}{d z}=-G. Derive the expression of G(T,Tparcel )G\left(T, T_{\text {parcel }}\right). Context answer: γ−1γμgRTparcel T\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 zz the temperature TT of the atmosphere equals to that of the parcel Tparcel ,T(z)=Tparcel (z)T_{\text {parcel }}, T(z)=T_{\text {parcel }}(z). We use Γ\Gamma to denote the value of GG when T=Tparcel T=T_{\text {parcel }}, that is Γ=−dTparcel dz\Gamma=-\frac{d T_{\text {parcel }}}{d z} (with T=Tparcel ).Γ\left.T=T_{\text {parcel }}\right) . \Gamma is called dry adiabatic lapse rate. Context question: 2.2.1. Derive the expression of Γ\Gamma Context answer: Γ=μgcp\Gamma=\frac{\mu g}{c_{p}} Context question: 2.2.2. Calculate the numerical value of Γ\Gamma. Context answer: 0.009780.00978 Context question: 2.2.3. Derive the expression of the atmospheric temperature T(z)T(z) as a function of the altitude. Context answer: T(z)=T(0)−ΓzT(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)−ΛzT(z)=T(0)-\Lambda z, where Λ\Lambda is a constant. Find the dependence of the parcel temperature Tparcel (z)T_{\text {parcel }}(z) on altitude zz. Context answer: Tparcel (z)=Tparcel (0)(T(0)−ΛzT(0))ΓΛT_{\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 Tparcel (z)T_{\text {parcel }}(z) when ∣Λz∣<<T(0)|\Lambda z|<<T(0) and T(0)≈Tparcel (0)T(0) \approx T_{\text {parcel }}(0). Context answer: Tparcel (0)−ΓzT_{\text {parcel }}(0)-\Gamma z Extra Supplementary Reading Materials: 3. The atmospheric stability. In this part, we assume that TT changes linearly with altitude. Context question: 3.1. Consider an air parcel initially in equilibrium with its surrounding air at altitude z0z_{0}, i.e. it has the same temperature T(z0)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 z0z_{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 Λ\Lambda for the atmosphere to be stable, unstable or neutral? Context answer: Λ>Γ\Lambda>\Gamma is unstable Λ<Γ\Lambda<\Gamma is stable Λ=Γ\Lambda=\Gamma is neutral Context question: 3.2. A parcel has its temperature on ground Tparcel (0)T_{\text {parcel }}(0) higher than the temperature T(0)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 Λ\Lambda and Γ\Gamma. Context answer: h=1Λ[T(0)−((T(0))Γ(Tparcel (0))Λ)1Γ−Λ]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, mTemperature, ∘C{ }^{\circ} \mathrm{C}
521.5
6020.6
6420.5
6920.5
7520.4
8120.3
9020.2
9620.1
10220.1
10920.1
11320.1
11920.1
12820.2
13620.3
14520.4
15320.5
15920.6
16820.8
17821.0
18921.5
20221.8
21522.0
22522.1
23422.2
24622.3
25722.3
Context question: 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)−ΛzT(z)=T(0)-\Lambda z with different lapse rates Λ\Lambda in the three layers 0<z<96 m,96 m<z<119 m0<\mathrm{z}<96 \mathrm{~m}, 96 \mathrm{~m}<\mathrm{z}<119 \mathrm{~m} and 119 m<z<215 m119 \mathrm{~m}<\mathrm{z}<215 \mathrm{~m}. Consider an air parcel with temperature Tparcel (0)=22∘CT_{\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 m96 \mathrm{~m} and 119 m119 \mathrm{~m}. Context answer: 294.04294.04, 293.81293.81 Context question: 4.2. Determine the maximal elevation HH the parcel can reach, and the temperature Tparcel (H)T_{\text {parcel }}(H) of the parcel. HH is called the mixing height. Air pollutants emitted from ground can mix with the air in the atmosphere (e.g. by wind, turbulence and dispersion) and become diluted within this layer. Context answer: 142, 293.6 Extra Supplementary Reading Materials: 5. Estimation of carbon monoxide (CO) pollution during a morning motorbike rush hour in Hanoi. Hanoi metropolitan area can be approximated by a rectangle with base dimensions LL and WW as shown in the figure, with one side taken along the south-west bank of the Red River. <image_1> It is estimated that during the morning rush hour, from 7:00 am to 8:00 am, there are 8×1058 \times 10^{5} motorbikes on the road, each running on average 5 km5 \mathrm{~km} and emitting 12 g12 \mathrm{~g} of CO per kilometer. The amount of CO pollutant is approximately considered as emitted uniformly in time, at a constant rate MM during the rush hour. At the same time, the clean north-east wind blows perpendicularly to the Red River (i.e. perpendicularly to the sides LL of the rectangle) with velocity uu, passes the city with the same velocity, and carries a part of the CO-polluted air out of the city atmosphere. Also, we use the following rough approximate model: - The CO spreads quickly throughout the entire volume of the mixing layer above the Hanoi metropolitan area, so that the concentration C(t)C(t) of CO\mathrm{CO} at time tt can be assumed to be constant throughout that rectangular box of dimensions L,WL, W and HH. - The upwind air entering the box is clean and no pollution is assumed to be lost from the box through the sides parallel to the wind. - Before 7:00 am, the CO\mathrm{CO} concentration in the atmosphere is negligible.

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