Unlock document.
This document is partially blurred.
Unlock all pages and 1 million more documents.
Get Access
466 RADIATIVE HEAT TRANSFER
20.8 The new planet in an adjacent solar system recently found by Penn State (and other) researchers has been
determined to have an atmosphere consisting of nitrogen with 1% by volume NO. The planet’s surface has
an emittance of (ǫ=0.5), and a temperature of Ts=900 K. The atmosphere’s total pressure is known to be
p(z)=pse−z/L(surface pressure ps=5 bar, characteristic length L=10 km). Assuming radiative equilibrium
prevails, what is the heat loss from the planet? You may assume that for NO line broadening is unaffected by
temperature.
(a) To make a coarse approximation, replace the atmosphere by a constant pressure (ps) layer of a thick-
ness that would contain the correct total pressure path length. Evaluate radiative properties as if the
atmosphere’s temperature were constant at Ts.
(b) The problem is to be solved by the P1approximation combined with the box model. Find the appropriate
absorption coefficient(s) and other necessary parameters. You may assume that the spectral width of
bands for NO is unaffected by altitude (evaluate at surface conditions). Set up equation(s) and boundary
condition(s).
(c) Determine the heat loss from the planet.
(d) What would change if an infinitely thick atmosphere with exponentially decaying pressure were con-
sidered?
RnTs
8.3144kJ/kmol K900K =20.0g
m3
The box model parameters are found from equations (20.33) and (20.34) as
κ∆ηe=αρNO =9.0×20.0=180.0cm−1m−1
A(Lm)= ∆ηe(1 −e−κLm)≃∆ηe,
where Lmis the mean beam length from Table 20.1, Lm=1.76L=17.6 km. From Table 9.2:
m2
0.2435
q=14,070 W
468 RADIATIVE HEAT TRANSFER
20.9 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 box model together with
(a) the P1approximation as well as (b) the exact formulation.
=20.9 cm−1/(g/m2)×0.2×105Pa ×28 kg/kmol ×1.76m
114 cm−1×8 314 J/kmol K ×2000 K
p
p0
=10.87 p
p0
10.87 exp 8.817
c−1=228 Exact
126 P1,
i.e., the 7% error of theP1approximation translates into an 82% error in the pressure prediction!
CHAPTER 20 469
20.10 The coal particles of Problem 12.3 are burnt in a long cylindrical combustion chamber of R=1 m radius. The
combustor walls are gray and diffuse, with ǫw=0.8, and are at 800 K. Since it is well stirred, combustion
results in uniform heat generation throughout of ˙
Q′′′ =720 kW/m3.
(a) Determine the maximum temperature in the combustor, using the P1/differential approximation, assum-
ing radiation is the only mode of heat transfer (use κ=4.5 m−1and σs=0.5 m−1if results of Problem
12.3 are not available).
(b) How will the answer change if, instead, the combustion gas is responsible for the radiation with
κλ=10 cm−1,4µm<λ<5µm
0,elsewhere ;σs=0?
(c) What if both are present?
Solution
part c needs work
470 RADIATIVE HEAT TRANSFER
20.11 Consider a sphere of very hot molecular gas of radius 50 cm. The gas has a single vibration-rotation band
at η0=3000 cm−1, is suspended magnetically in a vacuum within a large cold container and is initially at a
uniform temperature T1=3000 K. For this gas (ρaα)(T)=500 cm−2, ω(T)=100 √T/100 K cm−1and β≫1.
These properties imply that the absorption coefficient may be determined from
κη=κ0e−2|η−η0|/ω, κ0=ρaα
ω
and the band absorptance from
A(s)=ωA∗=ω[E1(κ0s)+ln(κ0s)+γE], γE=0.577216.
Using the stepwise-gray model together with the P1-approximation and neglecting conduction and convec-
tion, specify the total heat loss per unit time from the entire sphere at time t=0. Outline the solution
procedure for times t>0.
Hint: Solve the governing equation by introducing a new dependent variable 1(τ)=τ(4πIb−G).
2
√3cosh √3τR+1−2
3τRsinh √3τR
and
Ebη
√3coth √3×0.196×50+1−2
3×0.196×50
Q=2.62 ×106W.
For the transient case, neglecting conduction and convection reduces overall energy, equation (10.72), to
∂T
∂
472 RADIATIVE HEAT TRANSFER
20.15 An infinitely long cylinder of radius R=10 cm is bounded by a wall that is isothermal at Tw=1500 K and
has a gray emittance of ǫ=0.3. Inside the cylinder there is uniform heat generation of ˙
Q′′′ =38,136 W/m3.
The cylinder is filled with a mixture of combustion gases at p=1 atm, containing 10% by volume CO2and
20% water vapor. Assuming the gas to be well-stirred (i.e., isothermal) determine the gas temperature using
the weighted sum of gray gases approach, using the data of Table 20.2. This problem will require an iteration
and, thus, is most conveniently solved on a computer.
(a) Set up all necessary equations and explain the procedure. You may use the exact relations of Section
14.6 or the P1-approximation.
(b) Write a small computer code to find the gas temperature.
Note for the P1-approximation: The solution to the ODE
1
r
d
dr rd f
dr !−ν2f=0
is
f(r)=C1I0(νr)+C2K0(νr),
where I0and K0are modified Bessel functions. Note also that K0(0) → ∞ and I′
0(x)=I1(x).
a3=0.065 −0.0270 ×10−3T, κ3=239p,
where p=pCO2 +pH2O =0.3 atm. The total flux is then calculated from
q(R)=
3
X
4σhak(T)T4−ak(Tw)T4
wi
√3I0(√3τRk).I1(√3τRk)+22
kp(3)=(/0.89,15.5,239./)
!
! set constants
qw=0.5d0*R*Qs ! wall flux, W/m2
pco2=0.1 ! atm
ph2o=2*pco2
Tmax=2.*Tw
100 Tv=.5*(Tmin+Tmax)
qc=0.d0
DO i=1,3
av=a0(i)+a1(i)*Tv
aw=a0(i)+a1(i)*Tw
end
474 RADIATIVE HEAT TRANSFER
20.20 Repeat Problem 20.19 using the WSGG approach together with the correlation of Truelove.
Solution
