CHAPTER 20
20.1 A long, cylindrical furnace bounded by a cold, black wall of 1 m radius contains pure CO2that is isothermal
at 1700 K and at a pressure of patm. Using the mean-beam-length method, determine the nondimensional
wall heat flux Ψ = qw/σT4as a function of pressure. Plot Ψvs. p(actual calculations for p=0.001, 0.01, 0.1,
and 1.0 should suffice).
0.1 0.1053 0.996 0.105 47.36 4.955
1 0.1537 1.000 0.154 47.36 7.277
This solution is also given, together with a plot, as part (iii) of the solution to Problem 20.5 below.
454
CHAPTER 20 455
20.2 A high-pressure isothermal mixture (p>40 atm) of 80% N2and 20% CO at 2000 K is contained between two
large, parallel, cold black plates, spaced 1m apart. If the radiative flux to each wall may not exceed 100 kW/m2,
what is the maximum pressure the gas mixture may be raised to? Use the mean-beam-length method.
σT4A∗
4.7(Lm),
or
A∗(Lm)=qmax
ωEbη0
From Table 11.3, for the 4.7µm CO-band:
=10.87 p
p0
10.87 exp (8.817 −1)=228
456 RADIATIVE HEAT TRANSFER
20.3 An isothermal mixture of N2and soot (m=2.5−0.15i) at 2000 K is contained between two large, parallel,
cold black plates, spaced 1m apart. If the radiative flux to each wall may not exceed 100 kW/m2, what is the
maximum volume fraction of soot, fv, allowed? Use the mean-beam-length method.
C2
1.4388 cm K fv=3,213 fvcm −1
With Lm=176 cm, κmLm=5.655 ×105fv. Inverting the equation for ǫmax,
and
fv=1
5.655 ×105ln 1
1−0.1102 =2.07 ×10−7.
CHAPTER 20 457
20.4 Two parallel, infinite, black plates at constant temperatures T1and T2are separated by a nongray medium of
geometrical thickness d=10 cm that is at radiative equilibrium. The absorption characteristics of the medium
are such that they can be approximated by
κλ=κ=1 cm−13µm<λ<7µm,
0 elsewhere.
Calculate the nondimensional heat flux, q/σ(T4
1−T4
2), for a number of T2(T2=500 K, 750 K, 1000 K, 1500 K,
and 2000 K) and T1=300 K by
(a) the differential approximation, using a gray gas with Planck-mean absorption coefficient κP,
(b) the nongray differential approximation.
For the evaluation of κPyou may use Tm=(T1+T2)/2. Plot, compare, and discuss your results.
750 0.66370 0.11030 9928 17,481 0.501 525 0.41876 0.01782 0.401 0.250
1500 0.92366 0.56429 103,155 286,585 0.683 900 0.76180 0.20535 0.556 0.193
2000 0.96285 0.73778 204,184 906,741 0.801 1150 0.85823 0.37237 0.486 0.215
For very low T2the λT-values for the band are much less than (λT)max =2898 µmK, i.e., where the emissive
CHAPTER 20 459
20.5 A cold-walled cylindrical furnace of 1 m radius contains pure CO2that is isothermal at 1700 K and at a
pressure of patm. Using the (i) gray and (ii) nongray differential approximation with single band strength
κ, determine the nondimensional wall heat flux Ψ = qw/σT4as a function of pressure. Plot Ψvs. p(actual
calculations for p=0.001, 0.01, 0.1, and 1.0 should suffice; for simplification, you may assume that band width
is not a function of p).
κ2.7=4.0×1.61 ×315.5/459.0=4.4m−1
52.4=333.8cm−1
κ15 =19 ×315.5/333.8=18.0m−1
Applying the P-1 approximation, equations (20.48) through (20.48), to a one-dimensional, solid cylinder leads
to:
1
r
d
dr(rqη)=κη(4Ebη−Gη)
dr2+1
r
dr −3κ2
3κη
dr .
This is a modified Bessel’s equation (with constant non-homogeneous term), and
Gη=4Ebη+C1I0√3κηr+C2K0√3κηr.
Here, C2=0 since K0(0) → ∞,and C1is found from the boundary condition [using the fact that dI0(ax)/dx =
aI1(ax)]. Thus,
dGη
I0√3κηR+2
√3I1√3κηR
,
and
qwη=1
2Gη(R)=Ebη2
1−
I0√3κηR
(20.5-C)
Ψn1,a=
i=1
σT4
φ(κiR),
The total emissivity for CO2may be calculated from eq. (9.89),
ǫCO2=
3
X
EbηiAi
σT4,
C1=3.7418D-8 ! W CMˆ4/Mˆ2
T=1.7D3 ! K
DENS=3.155D2 ! G/Mˆ3 AT 1ATM
TFAC=DSQRT(T/1.D2)
XLM=1.88D0
PTOT=1.0132D0*P
PH2O=0.D0
PCO2=PTOT
c
c calculate Planck-mean kappa and total emissivity from wideband model
c
PMABS= DENS*(PLANCK(ETAC(1),T)*ALFA(1)+PLANCK(ETAC(2),T)*ALFA(2)
c
c calculate total emissivity from Leckner’s model
c
CALL TOTEMISS(PH2O,PCO2,PTOT,T,XLM_LECK,EPSH2O,EPSCO2,EPSTOT)
c
c qw for gray:
ARGGR=RT3*PMABS
QWGR=FI01(ARGGR)
c
QWMBB=EPSTOT
WRITE(*,10) P,QWGR,QWNGA,QWNGB,QWMBA,QWMBB
ENDDO
ENDDO
462 RADIATIVE HEAT TRANSFER
ENDIF
ELSE
IF(T.LE.B) THEN
ASTAR=T
ELSEIF(T.LT.1.D0/B) THEN
RETURN
END
c***********************************************************************
Discussion: as seen from the figure all 3 box models go to the same, correct optically-thin limit (for p→0);
the mean-beam length method underpredicts this limit by 6% (Lm/Lo=0.94 for cylinders); mean-beam length
+wide band model is significantly lower at p=10−3because the extremely strong 4.3µm band of CO2
10-3 10-2 10-1 100
p (atm)
10-2
10-1
100
qw/σT4
Gray box
Nongray box (single κ)
Nongray box (individual κ)
Mean-beam + wideband model
Mean-beam + Leckner’s model
Comparison of nongray box models and mean-beam-length models for Problem 20.5.
464 RADIATIVE HEAT TRANSFER
20.7 An infinitely long cylinder of radius R=10 cm is bounded by a cold black wall. Inside the cylinder there
is uniform heat generation of ˙
Q′′′ =38,136 W/m3. Estimate wall heat fluxes and temperature distributions
using the P1approximation if
(a) the medium has a band at λ=4µm of width ∆λ=1µm; across the band it has a constant absorption
coefficient such that κR=100.
(b) the medium is gray with an “appropriately” chosen κP, say by evaluating κPat the volume-averaged
temperature Tav, that is
T4
av =1
VZV
T4dV.
σ
T5
av
5.670 ×10−8=0.0775; κPR=7.75
1
T4(0) =1.6815 ×1010 1
T(0) =585 K,T(R)=371 K.