# 978-0123869449 Chapter 11 Part 3

Document Type

Homework Help

Book Title

Radiative Heat Transfer 3rd Edition

Authors

Michael F. Modest

For full access to CoursePaper, a membership subscription is required.

324 RADIATIVE HEAT TRANSFER

11.32 In a combustor the air-fuel ratio is controlled by measuring the total band absorptance of the fuel (methane)

for its 3.3µm band. The mixtures inlet conditions are 1 atm total pressure, temperature is 400 K, combustor

diameter is L=10 cm, and the design mole fraction for methane is 25%. If the total band absorptance across

the diameter is measured as A3.3=112 cm−1, what is the exact methane mole fraction at that time? Solution

CHAPTER 11 325

11.33 A 1 m thick layer of a mixture of nitrogen and methane (CH4) at T =300 K and p=1 atm has a measured total

emissivity of ǫ=0.010. Estimate the partial pressure of the methane (RCH4=5.128 ×10−6atm m3/g K). It is

known that pCH4≪p.

=0.58292 ×10−8W/m2cm−1K3×28.0 cm−1/(g/m2)

3.3µm band : Ebη0/T3η

T=3020

300 =10.067 cm−1/K×α

2.4µm band : Ebη0/T3η

T=4220

300 =14.067 cm−1/K×α

=0.00002 ×2.9=0.00006 ×10−8W/g K3.

It is clear that, due to the spectral positions of the bands, only the 7.7µm needs to be considered; the other

or

=0.010×5.670×10−8×300 W

0=0.08698, β∗/β∗

0≃0.73,

ω=ω0pT/T0=21.0×√3=36.37 cm−1,

=650.0xCH4g/m2,

where xCH4is the mole fraction of CH4. Now

τ0=α

ωX=28.0 cm−1/(g/m2)

36.37 cm−1650.0xg/m2=500.4xCH4,

500.4=5.90 ×10−3

CHAPTER 11 327

11.34 A mixture of water vapor and nitrogen at a total pressure of 1 atm and a temperature of 300 K is found to have

a total band absorptance of 100 cm−1for the 6.3µm band for a geometric path length of 50 cm. Determine the

partial pressure of the water vapor.

0=0.09427, β∗/β∗

0(T/T0=3) ≃0.60,

ω=ω0pT/T0=56.4×√3=97.69 cm−1,

97.69 cm−1=1.0237;

β=β∗

=365.9xH2Og/m2,

τ0=α

155.9=0.030.

Checking,

11.35 One method of measuring the temperature of a high-temperature gas is to determine the total band absorp-

tance of a vibration-rotation band of the gas under the prevailing conditions. Consider a 20 cm thick layer of

pure methane, CH4, at 1 atm pressure. If the total band absorptance of the 3.3µm band is 587 cm−1, what is

the temperature of the CH4?

Note: Such instruments are generally used only for T>1000 ◦C.

0=0.06973,

0>12. Thus, β=β∗

0(β∗/β∗

0)Pe≥0.06973 ×12 ×1.234 =

RuTL=1.0132×105Pa×16 g/mol

8.3145 J/mol K ×100 K ×0.2 m T0

T=390 ×T0

Tg/m2,

and

τ0=αX/ω =46 ×390 T0

=320 T0

,

10.482 rT0

T=ln "320 T0

T3/2#+1.

CHAPTER 11 329

11.37 A mixture of nitrogen and sulfur dioxide (with 5% SO2by volume) is at 1 atm total pressure. To measure

the temperature of the mixture in a furnace environment (T>1000 K), an instrument is used that measures

total band absorptance for the strong SO2-band centered at ηc=1361 cm−1. For that band it is known that

α=2340 (T0/T) cm−2atm−1,β=0.357 √T/T0Pe,ω=8.8√T/T0cm−1,b=1.28 and n=0.65. What is the

temperature of the mixture if the total band absorptance has been measured as 142 cm−1for a 1 m thick gas

layer?

If we were in the linear regime A∗=τ0, or

T=1329.5T0

T3/2

T

T0

16.136 =82.39 or T=8239 K,

16.136 rT0

T=ln "1329.5T0

T3/2#+1=8.193 +3 ln rT0

T

By trial and error this leads to

1000 42.04 1.14 5.10 4.74 3.82 3.77

1400 25.38 1.35 4.31 4.23 3.37 3.33

1800 17.41 1.53 3.86 3.86 3.04 3.00

2500 10.60 1.80 3.23 3.36 2.61 2.56

330 RADIATIVE HEAT TRANSFER

11.38 The Earth’s pollution with sulfur dioxide (SO2)is determined by measuring the transmission of a light beam

from a satellite. Assuming that the band absorptance of the 7.3 µm band has been measured as 10.0 cm−1, and

that the atmosphere may be approximated as a 10 km thick isothermal layer of nitrogen (with a trace of SO2)

at 0.5 atm and −10◦C, determine the volume fraction of SO2. Use wbmso2 from Appendix F to calculate the

overlap parameter βor use β≃0.357√T/T0Pe.

CHAPTER 11 331

11.39 To determine the average atmospheric temperature on a distant planet, the total band absorptance for the

3.3µm CH4band has been measured as A3.3=100 cm−1. It is known from other measurements that methane

is a trace element in the atmosphere (which contains mostly nitrogen and whose total pressure is 2 atm), and

that the absorption path length for methane on that planet, for which A3.3was measured, is 4.14 g/m2. What

is the temperature?

0=0.06973, β∗/β∗

0from Fig. 11-25,

n=0.8,b=1.3,Pe=h2+0.3pCH4)i0.8≃20.8=1.7411.

Thus,

τ0=α

1.7857 t−1/2=2s3.4007 t−1/20.1214 β∗

β∗

0−0.1214 β∗

β∗

0

,

or

0.1214x2−1.2851x+1.7857 =0,where x= β∗

332 RADIATIVE HEAT TRANSFER

11.40 Using the exponential wide band model, evaluate the total emissivity of a 1 m thick layer of a nitrogen–water

vapor mixture at 2 atm and 400 K if the water vapor content by volume is (a) 0.01%, (b) 1%, or (c) 100%.

Compare with Leckner’s model using subroutine totemiss.

A=499.6 cm−1.

A=1275.5 cm−1.

6.3µm band:

0)×β∗

0=0.58

|{z}

112.8×1097 xH2O=400.7xH2O.

CHAPTER 11 333

(a)xH2O=10−4:β=0.110, τ0=0.0401 < β (linear regime),

A∗=τ0,A=ωτ0=4.5 cm−1.

=0.01497 A∗

71 +0.03782 A∗

6.3

(a)xH2O=10−4ǫ=0.01497 ×0.3286 +0.03782 ×0.0401 =0.0064

pco2=0.d0

tg=400.d0

xl=1.d2

do i=1,3

ph2o=1.d1**(2*i-6)*ptot

call totemiss (ph2o,pco2,ptot,tg,xl,epsh2o,epsco2,epstot)

334 RADIATIVE HEAT TRANSFER

11.41 Using the exponential wide band model, evaluate the total emissivity of a 1 m thick layer of a nitrogen–CO2

mixture at 0.75 atm and 600 K if the CO2content by volume is (a) 0.01%, (b) 1%, or (c) 100%. Compare with

Leckner’s model using subroutine totemiss.

The 9.4 µm and 10.4 µm bands are negligibly weak (only important for CO2laser generation). The 2.7µm

band is relatively weak and, combined with its spectral position, is almost negligible, but will be included

here. Therefore, only the 2.7µm, 4.3µm and 15 µm bands will contribute to the total emissivity. The partial

0)×β∗

0=2.82

|{z}

Fig. 9-13

A=43.1 cm−1.

4.3µm band:

α=α0=110.0 cm−1/(g/m2),

27.4×670 xCO2=2686.4xCO2.

CHAPTER 11 335

2.7 µm band:

α=(α/α0)

| {z }

α0=1.044 ×4.0=4.2 cm−1/(g/m2),

57.6×670 xCO2=48.9xCO2.

(a)xCO2=10−4:β=0.313, τ0=5×10−3< β, (linear regime),

A∗=τ0,A=0.29 cm−1.

(b)xCO2=10−2:β=0.314, τ0=0.49 <1/β, ( sqrt regime)

5.670×10−8×600 A∗

5.670×10−8×600 A∗

+0.1327 ×10−857.6

5.670 ×10−8×600 A∗

=0.01189 A∗

15 +0.0612 A∗

4.3+0.00225 A∗

2.7

(a)xCO2=10−4ǫ=0.01189 ×0.041 +0.00612 ×0.269 +0.00225 ×0.005 =0.0021

xl=1.d2

do i=1,3

pco2=1.d1**(2*i-6)*ptot

call totemiss (ph2o,pco2,ptot,tg,xl,epsh2o,epsco2,epstot)

write(*,100) pco2,epstot

CHAPTER 11 337

11.42 Evaluate the Planck-mean absorption coeﬃcients for the two gases in Problems 11.40 and 11.41, based on the

data given in Table 11.3. Compare the results with Fig. 11-31.

6.3µm band : α=α0=41.2 cm−1/(g/m2),

η/T=1600 cm−1/400 K =4 cm−1/K,

5.670 ×10−8W/m2K4×400 K

κP=6.7×10−2(g/m2)−1.

=5.415 ×10−3g

J×105Pa

1 bar =541.5g

m3bar,

κpP = ρ

p!κρP=541.5g

m3bar ×6.7×10−2(g/m2)−1=0.363 cm−1bar−1.

4.3µm band : α=α0=110.0 cm−1/(g/m2),

2.7µm band : α=α0×α

α0

=4.0 cm−1/(g/m2)×1.1=4.4 cm−1/(g/m2),

5.670 ×10−8W/m2K4×600 K

κP=3.2×10−2(g/m2)−1.

=8.820 ×10−3g

J×105Pa

1 bar =882 g

m3bar,

## Trusted by Thousands of

Students

Here are what students say about us.

###### Resources

###### Company

Copyright ©2021 All rights reserved. | CoursePaper is not sponsored or endorsed by any college or university.