# 978-0123869449 Chapter 11 Part 1

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Homework Help

Book Title

Radiative Heat Transfer 3rd Edition

Authors

Michael F. Modest

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CHAPTER 11

11.1 Estimate the eigenfrequency for vibration, νe, for a CO molecule.

290 RADIATIVE HEAT TRANSFER

11.2 A certain gas at 1 bar pressure has a molecular mass of m=10−22 g and a diameter of D=5×10−8cm. At

what temperature would Doppler and collision broadening result in identical broadening widths for a line at

a wavenumber of 4000 cm−1?

T=0.1846 rT

T0

or

T

T0

=8008

0.1846 =43.38

T=4338 K.

CHAPTER 11 291

11.3 Water vapor is known to have spectral lines in the vicinity of λ=1.38 µm. Consider a single, broadened

spectral line centered at λ0=1.33 µm. If the water vapor is at a pressure of 0.1 atm and a temperature of

1000 K, what would you expect to be the main cause for broadening? Over what range of wavenumbers

would you expect the line to be appreciable, i.e., over what range is the absorption coeﬃcient at least 1% of

its value at the line center?

NA

6.023 ×1023/mol =2.989 ×10−23 g.

Then, from equation (11.38) with 0.1 atm =10.132 kPa =10.132 ×104g/cm s

=4.69 ×10−3cm−1since 1 Jg =1 Nmg =1 kg m2/s2g=107(g cm/s)2.

Similarly, for Doppler broadening from equation (11.45)

1.33×10−4cm×2.998×1010 cm/ss2 ln 2×1.3807×10−23 J/K×1000 K

2.989 ×10−23 g

γD=2.12 ×10−2cm−1,

κη0

=0.01 =exp −ln 2 ∆η

γD!2

292 RADIATIVE HEAT TRANSFER

11.4 Compute the half-width for a spectral line of CO2at 2.8µm for both Doppler and collision broadening as a

function of pressure and temperature. Find the temperature as function of pressure for which both broadening

phenomena result in the same half-width. (Note: The eﬀective diameter of the CO2molecule is 4.0×10−8cm).

(4.0×10−8cm)2×105Pa

p

γC=6.00 ×10−2p

2.998×1010 cm/ss2×1.3807×10−23 J/K×100 K

7.305 ×10−23 gln 2 rT

T0

6.00 p

p0rT0

T=0.1928 rT

T0

or

T

T0

=31.1p

p0.

CHAPTER 11 293

11.5 Methane is known to have a vibration-rotation band around 1.7 µm. It is desired to measure the Doppler

half-width of a spectral line in that band at room temperature (T=300 K). In order to make sure that collision

broadening is negligible, the pressure of the CH4is adjusted so that the expected collision half-width is only

1/10 of the Doppler half-width. What is this pressure? (For methane: D=0.381 nm.)

m3=1.75 ×103Pa =0.0175bar

294 RADIATIVE HEAT TRANSFER

11.6 Repeat Problem 11.4 for CO at a spectral location of 4.8µm (Note: The eﬀective diameter of the CO molecule

is 3.4×10−8cm).

(3.4×10−8cm)2×1.0132 ×105Pa

p

γC=5.50 ×10−2p

2.998×1010 cm/ss2×1.3807×10−23 J/K×100 K

4.651 ×10−23 gln 2 rT

T0

or

T

T0

=39.01 p

p0.

CHAPTER 11 295

11.7 A certain gas has two important vibration-rotation bands centered at 4 µm and 10 µm. Measurements of

spectral lines in the 4 µm band (taken at 300 K and 1 bar =105N/m2) indicate a half-width of γη=0.5 cm−1.

Predict the half-width in the 10µm band for the gas at 500K, 3bar. (The diameter of the gas molecules is

known to be between 5 Å <D<40 Å).

Thus, for case 1 collision broadening dominates. For case 2

γC

= γC

λ2

p2

T1

=23 ×10

296 RADIATIVE HEAT TRANSFER

11.8 It is desired to measure the volume fraction of CO in a hot gas by measuring the transmissivity of a 10 cm long

column, using a blackbody source and a detector responsive around 4.7 µm. The conditions in the column

are 1000 K, 1 atm, and properties for CO around 4.7 µm are known to be S=0.8 cm−2atm−1,γ=0.02 cm−1,

and d=0.05 cm−1. Give an expression relating measured transmissivity to CO volume fraction.

CHAPTER 11 297

11.9 A polyatomic gas has an absorption band in the infrared. For a certain small wavelength range to the

following is known:

Average line-half width: 0.04 cm−1,

Average integrated absorption coeﬃcient: 2.0×10−4cm−1/(g/m2),

Average line spacing: 0.25 cm−1,

The density of the gas at STP is 3 ×10−3g/cm3.

For a 50 cm thick gas layer at 500 K and 1 atm calculate the local mean spectral emissivity using

(a) the Elsasser model,

(b) the statistical model.

Which result can be expected to be more accurate?

(11.78) and (11.79)

Elsasser : W

d=2βL(x)≃2βx"1+πx

25/4#−2/5

=2×0.5027 ×0.711 "1+π×0.711

25/4#−2/5

=0.5265

298 RADIATIVE HEAT TRANSFER

11.10 Consider a gas for which the model is applicable, i.e., ǫη=1−exp(−Wη/d). To predict ǫηfor arbitrary

situations, a band-averaged (or constant) value for γη/dmust be known. Experimentally available are values

for α=R∆η(Sη/d)dηand ǫη=ǫη(η) (for optically thick situations) for given peand T. It is also known that

γη

d≃γη

d0peT0

T1/2

.

Outline how an average value for (γη/d)0can be found.

CHAPTER 11 299

11.11 The following is known for a gas mixture at 600K and 2atm total pressure and in the vicinity of a certain

spectral position: The gas consists of 80% (by volume) N2and 20% of a diatomic absorbing gas with a

molecular weight of 20 g/mol, a mean line half-width γ=0.01 cm−1, a mean line spacing of d=0.1 cm−1,

and a mean line strength of S=8×10−5cm−2/(g/m3). (a) For a gas column 10 cm thick determine the mean

spectral emissivity of the gas. (b) What happens if the pressure is increased to 20atm? (Since no broadening

parameters are known you may assume the eﬀective broadening pressure to be equal to the total pressure).

RuT=(0.2×2 atm)×20 g/mol

8.3144 J/mol K ×600 K ×10.13 N/cm2

atm ×J

102Ncm =162.45 g

m3

(a)

X=162.45 g

m310 cm =1624.5 cm g/m3,

x=8×10−5cm−2/(g/m3)×1624.5 cm(g/m3)

0.1 cm−1. This leads to

β=π, x=8×10−5×16,245

2π×0.1=2.068 (unchanged) and τ=13.012.

Thus:

300 RADIATIVE HEAT TRANSFER

11.12 Repeat Problem 11.11 for a four-atomic gas.

CHAPTER 11 301

11.13 1 kg of a gas mixture at 2000 K and 1 atm occupies a container of 1 m height. The gas consists of 70% nitrogen

(by volume) and of 30% an absorbing species. It is known that, at a certain spectral location, the line half-

width is γ=300 MHz, the mean line spacing is d=2000 MHz, and the line strength is S=100 cm−1MHz. (a)

Calculate the mean spectral emissivity under these conditions. (b) What will happen to the emissivity if the

sealed container is cooled to 300 K?

302 RADIATIVE HEAT TRANSFER

11.14 A 50 cm thick layer of a pure gas is maintained at 1000K and 1 atm. It is known that, at a certain spectral

location, the mean line half-width is γ=0.1 nm, the mean line spacing is d=2 nm, and the mean line strength

is S=0.002 cm−1nm atm−1=2×10−10 atm−1. What is the mean spectral emissivity under these conditions?

(1 nm =10−9m.)

d=0.040

CHAPTER 11 303

11.15 The following data for a diatomic gas at 300 K and 1 atm are known: The mean line spacing is 0.6 cm−1and the

mean line half-width is 0.03 cm−1; the mean line strength (=integrated absorption coeﬃcient) is 0.8 cm−2atm−1

(based on a pressure absorption coeﬃcient). Calculate the mean spectral emissivity for a path length of 1 cm.

In what band approximation is the optical condition?

Table 11.1,

β=πγ

d=π×0.03

0.6=π

20 =0.157,X=pL =1 cm atm,

x=SX

2πγ =0.8 cm−2atm−11 cm atm

2π×0.03 cm−1=4.244,

304 RADIATIVE HEAT TRANSFER

11.16 The average narrow band transmissivity of a homogeneous gas mixture has, at a certain wavenumber η, been

measured as 0.70 for a length of 10cm, and as 0.58 for a length of 20 cm. What is the expected transmissivity

for a gas column of 30 cm length, assuming the Malkmus model to hold?

β=(ln ¯

t1)2−(L1/L2)(ln ¯

t2)2

ln ¯

t1−(L1/L2) ln ¯

t2

=(ln 0.7)2−0.5(ln 0.58)2

ln 0.7−0.5 ln 0.58 =0.2508

This formula applies for any two lengths. Thus,

CHAPTER 11 305

11.17 1 kg of a gas mixture at 2000 K and 1 atm occupies a container of 1 m height. The gas consists of 70% nitrogen

(by volume) and 30% of an absorbing species. It is known that, at a certain spectral location, the nitrogen-

broadening line half-width at STP (1atm and 300K) is γn0=0.05 cm−1, the self-broadening line half-width is

γa0=0.02 cm−1, the mean line spacing is d=0.4 cm−1, and the density and mean line strength (for the given

mixture conditions) are ρ=0.800 kg/m3and ¯

S=4×10−3cm−1/(g/m2), respectively. Under these conditions

collision broadening is expected to dominate.

(a) Calculate the mean spectral emissivity based on the height of the container.

(b) What will happen to the emissivity if the sealed container is cooled to 300 K at constant pressure (with

fixed container cross-section and sinking top end)?

Note: the mean line intensity is directly proportional to the number of molecules of the absorbing gas and

otherwise constant. The line half-width is given by

γ=[γn0pn+γa0pa]rT0

T(pin atm,T0=300 K),

where pnand paare partial pressures of nitrogen and absorbing species.

Solution

306 RADIATIVE HEAT TRANSFER

11.18 A certain gas is known to behave almost according to the rigid-rotor/harmonic-oscillator model, resulting in

gradually changing line strengths (with wavenumber) and somewhat irregular line spacing. Calculate the

local mean emissivity for a 1 m thick layer of the gas at 0.1 atm pressure. In the wavelength range of interest, it

is known that the integrated absorption coeﬃcient is equal to 0.80 cm−2atm−1, the line half-width is 0.04 cm−1

and the average line spacing is 0.40 cm−1.

CHAPTER 11 307

11.19 A narrow band of a certain absorbing gas contains a single spectral line of Lorentz shape at its center. For a

narrow band width of ∆η=10γ, determine the corresponding reordered kvs. 1distribution.

Hint: this can be achieved without a lot of math.

(η›η0)/

›5 ›4 ›3 ›2 ›1 0 1 2 3 4 5

0

0.2

0.6

0.8

0 0.25 0.5 0.75 1

0

0.2

0.6

0.8

Masochists can, of course, also solve this problem mathematically:

Diﬀerentiating

dη

dκη

=∓γkmax

2κ2

η,skmax

κη−1,

308 RADIATIVE HEAT TRANSFER

1=

kmin

f(k)dk or 1 −1=

k

f(k)dk =

k

kmaxdk

10k2√kmax/k−1.

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