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SOLUTION - FUNDAMENTALS OF HEAT AND MASS TRANSFER

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PROBLEM 10.59 (Cont.)
Condensation rate, mdot'(kg/s.m)
0.3
0.2
0.1
280
290
300
Surface temperature, Ts(K)
N = 10
N=5
N=2
Clearly, there are significant benefits associated with reducing both Ts and N.
COMMENTS: Note that, since h D,N ∝ N-1/4, the average coefficient decreases with increasing N due
to a corresponding increase in the condensate film thickness. From the result of part (a), the coefficient
for the topmost tube is h D = 6210 W/m2⋅K(10)1/4 = 11,043 W/m2⋅K.
PROBLEM 10.60
KNOWN: Thin-walled concentric tube arrangement for heating deionized water by condensation of
steam.
FIND: Estimates for convection coefficients on both sides of the inner tube. Inner tube wall outlet
temperature. Whether condensation provides fairly uniform inner tube wall temperature approximately
equal to the steam saturation temperature.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible thermal resistance of inner tube wall, (2) Internal flow is fully
developed.
3
PROPERTIES: Deionized water (given): ρ = 982.3 kg/m , cp = 4181 J/kg⋅K, k = 0.643 W/m⋅K, µ =
-6
2
548 × 10 N⋅s/m , Pr = 3.56; Table A-6, Saturated vapor (1 atm): Tsat = 100°C, ρ v = (1/vg) = 0.596
3
kg/m , hfg = 2265 kJ/kg; Table A-6, Saturated water (assume Ts ≈ 75°C, Tf = (75 + 100)°C/2 = 360K):
3
-6
2
ρl = (1/vf) = 967 kg/m , µl = 324 × 10 N⋅s/m , kl = 0.674 W/m⋅K, c p, l = 4203 J/kg⋅K.
ANALYSIS: From an energy balance on the inner tube assuming a constant wall temperature,
(
)
(
hc Tsat − Ts,o = hi Ts,o − Tm,o
)
where h c and hi are, respectively, the heat transfer coefficients for condensation (c) on a horizontal
cylinder and internal (i) flow in a tube.
Condensation. From Eq. 10.40, for the horizontal tube,
1/4
 g ρ ( ρ − ρ ) k 3h ′ 
l l
v l fg 
hc = 0.729 
 µl ( Tsat − Ts ) D 


{
′ = h fg 1 + 0.68c p,l ( Tsat − Ts ) / h fg
where h fg
}
{
h fg
′ = 2265 kJ/kg {1+1.262 ×10 −3 (100 − Ts )}
h ′fg = 2265 kJ/kg 1 + 0.68 × 4203 J / k g ⋅ K (100 − Ts ) /2265 ×103 J/kg
}
hc = 0.729  9.8m/s2 × 967kg/m3 ( 067 − 0.596 ) k g / m3 ( 0.674W/m ⋅ K )3 ×

{
}
1/4
2265 1 + 1.262 ×10 −3 (100 − Ts ) kJ/kg/324 ×10 −6 N ⋅ s / m 2 (100 − Ts ) 0.030 m 

Continued …..
PROBLEM 10.60 (Cont.)
1/4
1 +1.262 ×10 −3 (100 − T ) 
s 
hc = 2.843× 104 
100 − Ts




.
Internal flow. From Eq. 8.6, evaluating properties at Tm , find
Re D =
&
4m
4 × 5 kg/s
=
= 3.872 ×105
6
2
−
πµ D π × 548 ×10 N ⋅ s / m × 0.030 m
and for turbulent flow use the Colburn equation,
hD
0.8 Pr1/3
Nu D = i = 0.023ReD
k
hi =
)
(
0.8
0.023 × 0.643 W / m ⋅ K
3.872 ×105
( 3.56 )1/3 = 2.22 × 104 W / m 2 ⋅ K.
0.03 m
<
Substituting numerical values into the energy balance relation,
(
 1 + 1.262 ×10−3 100 − T
s,o
4
2.843 ×10 

100 − Ts,o

1/4
) 


(100 − Ts,o ) K
(
)
= 2.22 × 104 W / m2 ⋅ K Ts,o − 60 K
and by trial-and-error, find
Ts,o ≈ 75°C.
With this value of Ts, find that
hc = 1.29 × 104 W / m2 ⋅ K
<
which is approximately half that for the internal flow. Hence, the tube wall cannot be at a uniform
temperature. This could only be achieved if h c ? h i .
PROBLEM 10.61
KNOWN: Heat dissipation from multichip module to saturated liquid of prescribed temperature and
properties. Diameter and inlet and outlet water temperatures for a condenser coil.
FIND: (a) Condensation and water flow rates. (b) Tube surface inlet and outlet temperatures. (c)
Coil length.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions since rate of heat transfer from the module is balanced
by rate of heat transfer to coil, (2) Fully developed flow in tube, (3) Negligible changes in potential and
kinetic energy for tube flow.
PROPERTIES: Saturated fluorocarbon (Tsat = 57°C, given): kl = 0.0537 W/m⋅K, c p, l = 1100
3
3
-3
2
J/kg⋅K, h ′fg ≈ h fg = 84,400 J/kg. ρl = 1619.2 kg/m , ρ v = 13.4 kg/m , σ = 8.1 × 10 kg/s , µl = 440
-6
3
× 10 kg/m⋅s, Prl = 9; Table A-6, Water, sat. liquid ( Tm = 300K ) : ρ = 997 kg/m , cp = 4179 J/kg⋅K,
-6
2
µ = 855 × 10 N⋅s/m , k = 0.613 W/m⋅K, Pr = 5.83.
ANALYSIS: (a) With
q = (q ′′ × A )module = 105 W / m 2 ( 0.1m )2 = 10 3 W
the condensation rate is
& con =
m
q
103 W
=
= 0.0118 kg/s
h′fg 84,400 J / k g
<
and the required water flow rate is
& =
m
(
q
c p Tm,o − Tm,i
)
=
1000 W
= 7.98 ×10 −3 kg/s.
4179 J / k g ⋅ K ( 30K )
<
(b) The Reynolds number for flow through the tube is
Re D =
&
4m
4 × 7.98 ×10 −3 k g / s
=
= 1188.
π Dµ π ( 0.01m ) 855 ×10−6 N ⋅ s / m 2
Hence, the flow is laminar. Assuming a uniform wall temperature,
h i = Nu D k / D = 3.66 ( 0.613 W / m ⋅ K/0.01m ) = 224 W / m 2 ⋅ K.
Continued …..
PROBLEM 10.61 (Cont.)
For film condensation on the outer surface, Eq. 10.40 yields
(
)(
)
1/4
 9.8m/s 2 1619.2 k g / m 3 1605.8kg/m3 0.0537 W / m ⋅ K 3 84,400J/kg 
(
)


h o = 0.729 

6
−
440 × 10 k g / m ⋅ s × 0.01m ( Tsat − Ts )


h o = 2150 (57 − Ts )
−1/4
.
From an energy balance on a portion of the tube surface,
h o ( Tsat − Ts ) = h i ( Ts − Tm )
or
3/4
2150 ( 57 − Ts )
= 224 ( Ts − Tm )
At the entrance where ( Tm,i = 285K ) , trial-and-error yields:
<
Ts,i = 50.6°C
and at the exit where ( Tm,o = 315K ) ,
<
Ts,o = 55.4°C
(c) From Eqs. 8.44 and 8.45,
L=
q
hiπ D∆Tlm
where
∆Tl m =
L=
(
( Ts − Tm,i ) − ( Ts − Tm,o ) = 41 −11 = 22.8°C
ln  ( Ts − Tm,i ) / ( Ts − Tm,o ) ln ( 41/11)


1000 W
)
224 W / m2 ⋅ K π ( 0.01m ) 22.8°C
= 6.23m.
<
COMMENTS: Some control over system performance may be exercised by adjusting the water
& ( Tm,o − Tm,i ) is reduced for a prescribed q. The value of h i is increased
flow rate. By increasing m,
substantially if the internal flow is turbulent.
PROBLEM 10.62
KNOWN: Saturated ethylene glycol vapor at 1 atm condensing on a sphere of 100 mm diameter
having surface temperature of 150°C.
FIND: Condensation rate.
SCHEMATIC:
ASSUMPTIONS: (1) Laminar film condensation, (2) Negligible non-condensibles in vapor.
3
PROPERTIES: Table A-5, Saturated ethylene glycol, vapor (1 atm): Tsat = 470K, ρ v ≈ 0 kg/m , hfg
= 812 kJ/kg; Table A-5, Ethylene glycol, liquid (Tf = 423K, but use values at 373K, limit of data in
3
-2
2
table): ρl = 1058.5 kg/m , c p, l = 2742 J/kg⋅K, µl = 0.215 × 10 N⋅s/m , kl = 0.263 W/m⋅K.
ANALYSIS: The condensation rate is given by Eq. 10.33 as
(
)
h L π D2 ( Tsat − Ts )
q
& =
m
=
′
h′fg
hfg
2
where A = π D for the sphere and h ′fg , with Ja = c p, l ∆T / h fg , is given by Eq. 10.26 as
h ′fg = h fg (1 + 0.68Ja ) = 812
kJ 
J

1 + 0.68 × 2742
( 470 − 423) K/812 ×103 J/kg  = 900kJ/kg.

kg 
kg ⋅ K

The average heat transfer coefficient for the sphere follows from Eq. 10.40with C = 0.815,
1/4
g ρ ( ρ − ρ )k 3 h ′ 
v l fg 
l l
hD = 0.815 
 µ l ( Tsat − Ts ) D 


 9 . 8 m / s 2 × 1058.5kg/m 3 (1058.5 − 0 ) k g / m 3 ( 0.263W/m ⋅ K )3× 900 × 10 3 J / k g 
h D = 0.815 

2
−2
0.215 × 10 N ⋅ s / m ( 470 − 423 ) K × 0.100m


1/4
hD = 1674 W / m 2 ⋅ K.
Hence, the condensation rate is
& = 1674W/m 2 ⋅ K ×π ( 0.100m ) 2 ( 470 − 423 ) K/900 ×103 J/kg
m
& = 2.75 ×10 −3 kg/s.
m
<
COMMENTS: Recognize this estimate is likely to be a poor one since properties were not evaluated
at the proper Tf which was beyond the limit of the table.
PROBLEM 10.63
KNOWN: Copper sphere of 10 mm diameter, initially at 50°C, is placed in a large container filled with
saturated steam at 1 atm.
FIND: Time required for sphere to reach equilibrium and the condensate formed during this period.
SCHEMATIC:
ASSUMPTIONS: (1) Laminar film condensation, (2) Negligible non-condensibles in vapor, (3)
Sphere is spacewise isothermal, (4) Sphere experiences heat gain by condensation only.
3
PROPERTIES: Table A-6, Saturated water vapor (1 atm): Tsat = 100°C, ρ v = 0.596 kg/m , hfg =
3
2257 kJ/kg; Table A-6, Water, liquid (Tf ≈ (75 + 100)°C/2 = 360K): ρl = 967.1 kg/m , c p, l = 4203
-6
2
J/kg⋅K, µl = 324 × 10 N⋅s/m , kl = 0.674 W/m⋅K; Table A-1, Copper, pure ( T = 75° C) : ρ sp =
3
8933 kg/m , cp,sp = 389 J/kg⋅K.
ANALYSIS: Using the lumped capacitance approach, an energy balance on the sphere provides,
E& in − E& out = E& st
& h ′fg = h D A s ( Tsat − Ts ) = ρsp c p,sp Vs
m
dTs
.
dt
(1)
Properties of the sphere, ρ sp and cp ,sp ,. Will be evaluated at Ts = ( 50 +100 ) ° C / 2 = 75° C, while water
(liquid) properties will be evaluated at Tf = ( Ts + Tsat ) / 2 = 87.5 °C ≈ 360K. From Eq. 10.26 with Ja =
c p, l ∆T / h fg where ∆T = Tsat - Ts , find
J
kJ
 +


3
× (100 − 75 ) K /2 2 57 × 10 J / k g  = 2328
. (2)
1 0.68 4203



kg 
kg ⋅ K
kg


kJ
h ′fg = h fg (1 + 0.68Ja ) = 2257
To estimate the time required to reach equilibrium, we need to integrate Eq. (1) with appropriate limits.
However, to perform the integration, an appropriate relation for the temperature dependence of hD
needs to be found. Using Eq. 10.40 with C = 0.815,
1/4
 g ρ ( ρ − ρ ) k3 h′ 
v l fg 
l l
hD = 0.815 
.
 µ l ( Tsat − Ts ) D 


Substitute numerical values and find,
1/4
 9 . 8 m / s 2 × 967.1kg/m 3 ( 967.1 − 0.596 ) k g / m 3 ( 0 . 6 7 4 W / m ⋅ K )3 × 2328 × 10 3 J / k g 
h D = 0.815 

−6
2
324 × 10 N ⋅ s / m ( Tsat − Ts ) × 0.010m


−1/4
hD = B ( Tsat − Ts )
where
3/4
B = 30,707W/m 2 ⋅ ( K )
. (3)
Continued …..
PROBLEM 10.63 (Cont.)
1
Substitute Eq. (3) into Eq. (1) for hD and recognize Vs / A s = π D 3 / π D 2 = D / 6 ,
6
dT
−1/4
B ( Tsat − Ts )
( Tsat − Ts ) = ρsp c p,sp ( D / 6 ) s .
dt
(4)
Note that d(Ts) = - d(Tsat – Ts); letting ∆T ≡ Tsat – Ts and separating variables, the energy balance
relation has the form
ρsp c p,sp ( D / 6 ) ∆T d ( ∆T )
t
dt = −
(5)
∆To ∆T3 / 4
0
B
where the limits of integration have been identified, with ∆ To = Tsat − Ti and Ti = Ts(0). Performing
the integration, find
∫
t=−
∫
ρsp cp,sp ( D / 6)
B
⋅
1  1/4

∆T
− ∆ T1/4
o  .
1 − 3 / 4 
Substituting numerical values with the limits, ∆T = 0 and ∆To = 100-50 = 50°C,
t=−
3
8933kg/m × 389J/kg ⋅ K ( 0.010m/6)
30,707 W / m 2 ⋅ K 3 / 4
× 4  01/4 − 501 / 4  K1/4


<
t = 2.0s.
To determine the total amount of condensate formed during this period, perform an energy balance on
a time interval basis,
E in − E out = ∆E = E final − E initial
Ein = ρsp cp,sp V ( Tfinal − Tinitial ]
(6)
where Tfinal = Tsat and Tinitial = Ti = Ts(0). Recognize that
E in = M h ′fg
(7)
where M is the total mass of vapor that condenses. Combining Eqs. (6) and (7),
M=
M=
ρsp cp,sp V
h ′fg
[Tsat − Ti ]
8933kg/m3 × 389J/kg ⋅ K (π / 6)( 0.010m )3
2328 × 103 J/kg
[100 − 50 ] K
M = 3.91 ×10 −5 kg.
<
COMMENTS: The total amount of condensate could have been evaluated from the integral,
t
0
t q
t h D A s ( Tsat − Ts ) dt
dt =
0 h ′fg
0
h ′fg
& =∫
M = ∫ mdt
∫
giving the same result, but with more effort.
PROBLEM 10.64
KNOWN: Saturated steam condensing on the inside of a horizontal pipe.
FIND: Heat transfer coefficient and the condensation rate per unit length of the pipe.
SCHEMATIC:
ASSUMPTIONS: (1) Film condensation with low vapor velocities.
3
PROPERTIES: Table A-6, Saturated water vapor (1.5 bar): Tsat ≈ 385K, ρ v = 0.88 kg/m , hfg =
3
2225 kJ/kg; Table A-6, Saturated water (Tf = (Tsat + Ts)/2 ≈ 380K): ρl = 953.3 kg/m , c p, l = 4226
-6
2
J/kg⋅K, µl = 260 × 10 N⋅s/m , kl = 0.683 W/m⋅K.
ANALYSIS: The condensation rate per unit length follows from Eq. 10.33 with A = π D L and has
the form
&′=
m
&
m
= h D (π D )( Tsat − Ts ) / h ′fg
L
where hD is estimated from the correlation of Eq. 10.42 with Eq. 10.43,
1/4
g ρ ( ρ − ρ )k 3 h ′ 
v l fg 
l l
hD = 0.555 
 µ l ( Tsat − Ts ) D 


where
3
J 3
J
h ′fg = h fg + cp,l ( Tsat − Ts ) = 2225 ×103
+ × 4226
( 385 − 373 ) K
8
kg 8
kg ⋅ K
h ′fg = 2244kJ/kg.
Hence,
1/4
kg
kg
3


2
3
 9.8m/s × 953.3 3 ( 953.3 − 0.88 ) 3 ( 0.683W/m ⋅ K ) 2244 ×10 J/kg 
m
m

hD = 0.555 
6
2
−


260 × 10 N ⋅ s / m ( 385 − 373) K × 0.075m


hD = 7127W/m 2 ⋅ K.
It follows that the condensate rate per unit length of the tube is
& ′ = 7127W/m 2 ⋅ K (π × 0.075m )(385 − 373)K / 2225 ×103 J / k g = 9.06 ×10−3 kg / s ⋅ m.
m
<
PROBLEM 10.65
KNOWN: Horizontal pipe passing through an air space with prescribed temperature and relative
humidity.
FIND: Water condensation rate per unit length of pipe.
SCHEMATIC:
ASSUMPTIONS: (1) Drop-wise condensation, (2) Copper tube approximates well promoted
surface.
PROPERTIES: Table A-6, Water vapor (T∞ = 37°C = 310K): pA,sat = 0.06221 bar; Table A-6,
Water vapor (pA = φ⋅pa,sat = 0.04666 bar): Tsat = 305K = 32°C, hfg = 2426 kJ/kg; Table A-6, Water,
liquid (Tf = (Ts + Tsat )/2 = 297K): c p, l = 4180 J / kg ⋅ K.
ANALYSIS: From Eq. 10.33, the condensate rate per unit length is
&′=
m
q ′ h L (π D )(Tsat − Ts )
=
h′fg
h′fg
where from Eq. 10.26, with Ja = c p, l ( Tsat − T s ) / h fg ,
′ = h fg [1 + 0.68Ja ] = 2426
h fg
kJ
1 + 0.68 × 4180J/kg ⋅ K ( 305 − 288 ) K/2426k J/kg 
kg 
h ′fg = 2474kJ/kg.
Note that Tsat is the saturation temperature of the water vapor in air at 37°C having a relative humidity,
φ = 0.75. That is, Tsat = 305K and Ts = 15°C + 288K. For drop-wise condensation, the correlation
of Eq. 10.44 yields
hdc = 51,104 + 2044Tsat
22° C < Tsat < 100°C
2
where the units of h dc and Tsat are W/m ⋅K and °C.
hdc = 51,104 + 2044 ( 32° C) = 116,510 W / m2 ⋅ K.
Hence, the condensation rate is
& ′ = 116,510 W / m 2 ⋅ K ( π × 0.025m )( 305 − 288 ) K/2474× 10 3 J/kg
m
& ′ = 6.288 ×10−2 kg/s ⋅ m
m
<
COMMENTS: From the result of Problem 10.54 assuming laminar film condensation, the
condensation rate was m& ′film = 4.28 ×10 −3 k g / s ⋅ m which is an order of magnitude less than for the
rate assuming drop-wise condensation.
PROBLEM 10.66
KNOWN: Beverage can at 5°C is placed in a room with ambient air temperature of 32°C and
relative humidity of 75%.
FIND: The condensate rate for (a) drop-wise and (b) film condensation.
SCHEMATIC:
ASSUMPTIONS: (1) Condensation on top and bottom surface of can neglected, (2) Negligible noncondensibles in water vapor-air, and (b) For film condensation, film thickness is small compared to
diameter of can.
PROPERTIES: Table A-6, Water vapor (T∞ = 32°C = 305 K): pA,sat = 0.04712 bar; Water vapor
(pA = φ⋅pA,sat = 0.03534 bar): Tsat ≈ 300 K = 27°C, hfg = 2438 kJ/kg; Water, liquid (Tf = (Ts + Tsat)/2
= 289 K): c p," = 4185 J/kg⋅K.
ANALYSIS: From Eq. 10.33, the condensate rate is
=
m
h (π DL )( Tsat − Ts )
q
=
h ′fg
h ′fg
where from Eq. 10.26, with Ja = c p," (Tsat – Ts)/hfg,
h ′fg = h fg [1 + 0.68 Ja ]
h ′fg = 2438 kJ / kg 1 + 0.68 × 4185 J / kg ⋅ K (300 − 278 ) K / 2438 kJ / kg 
h ′fg = 2501 kJ / kg
Note that Tsat is the saturation temperature of the water vapor in air at 32°C having a relative
humidity of φ∞ = 0.75.
(a) For drop-wise condensation, the correlation of Eq. 10.44 with Tsat = 300 K = 27°C yields
h = h dc = 51,104 + 2044 Tsat
22°C < Tsat ≤ 100°C
2
where the units of h dc are W/m ⋅K and Tsat are °C,
h dc = 51,104 + 2044 × 27 = 106, 292 W / m 2 ⋅ K
Hence, the condensation rate is
= 1.063 × 105 W / m 2 ⋅ K (π × 0.065 m × 0.125 m )( 27 − 5 ) K / 2501 kJ / kg
m
<
= 0.0229 kg / s
m
Continued …..
PROBLEM 10.66 (Cont.)
(b) For film condensation, we used the IHT tool Correlations, Film Condensation, which is based
upon Eqs. 10.37, 10.38 or 10.39 depending upon the flow regime. The code is shown in the
Comments section, and the results are
Reδ = 24, flow is laminar
<
= 0.00136 kg / s
m
Note that the film condensation rate estimate is nearly 20 times less than for drop-wise condensation.
COMMENTS: The IHT code identified in part (b) follows:
/* Results,
NuLbar
0.5093
Part (b) - input variables and rate parameters
Redelta hLbar
mdot
D
L
Ts
24.05
6063
0.001362 0.065
0.125
278
Tsat
300
*/
/* Thermophysical properties evaluated at Tf; hfg at Tsat
Prl
Tf
cpl
h'fg
hfg
kl
mul
nul
7.81 289
4185
2.501E6 2.438E6 0.5964 0.001109 1.11E-6*/
// Other input variables required in the correlation
L = 0.125
b = pi * D
D = 0.065
/* Correlation description: Film condensation (FCO) on a vertical plate (VP). If Redelta<29,
laminar region, Eq 10.37 . If 31<Redelta<1750, wavy-laminar region, Eq 10.38 . If Redelta>=1850,
turbulent region, Eq 10.26, 10.32, 10.33, 10.35, 10.39 . In laminar-wavy and wavy-turbulent transition
regimes, function interpolates between laminar and wavy, and wavy and turbulent correlations. See
Fig 10.15 . */
NuLbar = NuL_bar_FCO_VP(Redelta,Prl)
// Eq 10.37, 38, 39
NuLbar = hLbar * (nul^2 / g)^(1/3) / kl
g = 9.8
// gravitational constant, m/s^2
Ts = 5 + 273
// surface temperature, K
Tsat = 300
// saturation temperature, K
// The liquid properties are evaluated at the film temperature, Tf,
Tf= (Ts + Tsat) / 2
// The condensation and heat rates are
q = hLbar * As * (Tsat - Ts)
// Eq 10.32
As = L * b
// surface Area, m^2
mdot = q / h'fg
// Eq 10.33
h'fg = hfg + 0.68 * cpl * (Tsat - Ts)
// Eq 10.26; hfg evaluated at Tsat
// The Reynolds number based upon film thickness is
Redelta = 4 * mdot / (mul * b)
// Eq 10.35
// Water property functions :T dependence, From Table A.6
// Units: T(K), p(bars);
x=0
// Quality (0=sat liquid or 1=sat vapor)
hfg = hfg_T("Water",Tsat)
// Heat of vaporization, J/kg; evaluated at Tsat
cpl = cp_Tx("Water",Tf,x)
// Specific heat, J/kg·K
mul = mu_Tx("Water",Tf,x)
// Viscosity, N·s/m^2
nul = nu_Tx("Water",Tf,x)
// Kinematic viscosity, m^2/s
kl = k_Tx("Water",Tf,x)
// Thermal conductivity, W/m·K
Prl = Pr_Tx("Water",Tf,x)
// Prandtl number
PROBLEM 10.67
KNOWN: Surface temperature and area of integrated circuits submerged in a dielectric fluid of prescribed
properties. Height and temperature of condenser plates.
FIND: (a) Heat dissipation by an integrated circuit, (b) Condenser surface area needed to balance heat load.
SCHEMATIC:
ASSUMPTIONS: (1) Nucleate pool boiling in liquid, (2) Laminar film condensation of vapor, (3) Negligible heat
loss to surroundings.
3
-4
PROPERTIES: Dielectric fluid (given, Tsat = 50°C): ρl = 1700kg/m , c p, l = 1005J/kg⋅K, µl = 6.80 × 10
2
5
kg/s⋅m, kl = 0.062W/m⋅K, Prl = 11, σ = 0.013 kg/s , h fg = 1.05 × 10 J/kg, Cs,f = 0.004, n = 1.7.
ANALYSIS: (a) For nucleate pool boiling,
1/2
3


 ≈ 6.8 × 10 −4 k g / s ⋅ m 1.05 × 10 5 J / k g
n
 C h Pr 
 s,f fg l 
 g ( ρl − ρv ) 
q s′′ = µ l h fg


σ
1/2
 9 . 8 m / s2 ×1 7 0 0 k g / m3 
×

2
0.013kg/s


c p, l ∆Te
(
)
3


1005 J / kg ⋅ K × 25K
2

 = 84,530W/m
5
1.7 
 0.004 × 1.05 × 10 J / k g × 11 
2
q s = A s q s′′ = 8 4 , 5 3 0 W / m × 25 × 10
−6
<
2
m = 2.11W.
1/4
(b) For laminar film condensation on a vertical surface, Nu L



′ = h fg 1 + 0.68
h fg
 gρ l ( ρl − ρv ) h ′fg L3 
= 0.943 

 µl kl ( Tsat − Ts ) 
cp, l ∆ T 
5
5
 = 1.05 × 10 + 0.68 (1005 J / k g ⋅ K × 35K ) = 1.29 ×10 J / k g

h fg
(
1/4
)

2
3 2
5
3
 9 . 8 m / s 1700kg/m 1.29 × 10 J / k g ( 0.05 m ) 
Nu L ≈ 0.943 

−4
6.8 × 10
k g / s ⋅ m ( 0.062 W / m ⋅ K )( 35K )




= 703
h L = ( k l / L ) Nu L = ( 0.062 W / m ⋅ K / 0 . 0 5 m ) 703 = 872 W / m ⋅ K
2
( )
q c = h LA c ( Tsat − T c ) = 872 W / m ⋅ K ( 35K ) A c = 30,500A c m
2
2
To balance the heat load, qc = Nq s . Hence
Ac =
500 × 2.11 W
30,500 W / m
2
= 0.0346 m
2
2
<
COMMENTS: (1) With A c = 0.0346m and H = 0.05m, the total condenser width is W = A c /H = 692mm. (2) With
& c / b = Γ = q c / h ′fg W = 1055W/1.29 × 10 5 J / k g × 0.692m = 0.0118kg/s ⋅ m, Reδ = 4 Γ / µ l =
m
-4
4(0.0118kg/s⋅m)/6⋅8 × 10
kg/s⋅m = 69.4. Hence condensate film is in the laminar-wavy regime, and a more
accurate estimate of A c would require iteration.
PROBLEM 10.68
KNOWN: Thin-walled thermosyphon. Absorbs heat by boiling saturated water at atmospheric
pressure on boiling section Lb. Rejects heat by condensing vapor into a thick film which falls length of
condensation section Lc back into boiling section.
FIND: (a) Mean surface temperature, Ts,b, of the boiling surface if nucleate boiling flux is 30%
critical flux, (b) Mean surface temperature, Ts,c of condensation section, and (c) Total condensation
& in thermosyphon. Explain how to determine whether film is laminar, wavy-laminar or
flow rate, m,
turbulent.
SCHEMATIC:
ASSUMPTIONS: (1) Laminar film condensation occurs in condensation section which approximates
a vertical plate, (2) Boiling and condensing section are separated by insulated length Li, (3) Top surface
of condensation section is insulated, (4) For condensation, liquid properties evaluated at Tf = 90°C.
PROPERTIES: Table A-6, Saturated water (100°C): ρ l = 1 / v f = 957.9 k g / m 3 , c p, l = 4217
J/kg⋅K, µ l = 279 × 10−6 N ⋅ s / m 2 , Prl = 1.76, hfg = 2257 kJ/kg, σ = 58.9 × 10 N/m; Saturated vapor
-3
3
(100°C): ρ v = 1/vg = 0.5955 kg/m ; Saturated water (90°C): ρ l = 1 / v f = 964.9kg/m 3 , c p, l = 4207
-6
2
J/kg⋅K, µl = 313 × 10 N⋅s/m , k l = 0.676 W / m⋅ K.
ANALYSIS: (a) The heat flux for the boiling section is 30% the critical heat flux which at
atmospheric pressure is
q′′s,b = 0.30q′′max = 0.30× 1.26 × 106 W / m 2 = 3.78 ×105 W / m 2 .
Using the Rohsenow correlation for nucleate boiling with Tsat = 100°C and typical values for the
surface of Cs,f = 0.0130 and n = 1.0, find
3
1/2
 g ( ρl − ρ v )   cp, l Ts,b − Tsat 

q′′s,b = µl h fg 
 
n 
σ

C
h
Pr

 
s,f fg l

(
5
2
3.78 × 10 W / m = 279 × 10
−6
2
)
3
N ⋅ s / m × 2257 × 10 J / k g ×
1/2
 9 . 8 m / s 2 ( 957.9 − 0.5955 ) k g / m3 


−3
58.9 ×10 N / m


 4217 J / kg ⋅ K ( Ts,b − 100 )

3
1.0
 0.013 × 2257 × 10 J/kg1.76
3



Continued …..
PROBLEM 10.68 (Cont.)
<
Ts,b = 114.0 °C.
(b) The heat transferred into the boiling section must be rejected by film condensation,
q c = q b = q′′s,b π D 2 / 4 + π DL b 


2
q c = 3.78 × 105 W / m 2 π ( 0.020m ) / 4 + π ( 0.020m ) × 0.020m  = 592 W.


The mean surface temperature can be determined from the rate equation
(
q c = h Lc ( π DLc ) Tsat − Ts,c
)
where the convection coefficient is determined from Eq. 10.30,
1/4
 g ρ ( ρ − ρ ) k 3 h′ 
v l fg 
l l
hLc = 0.943 
 µ l Tsat − Ts,c Lc 


(
)
1/4
h Lc
 9 . 8 m / s2 × 964.9kg/m3 ( 964.9 − 0.5955 ) k g / m3 ( 0.676W/m ⋅ K )3 2257 × 103 J / k g 

= 0.943 
2
−6


313 × 10 N ⋅ s / m (100 − Ts,c ) 0.040 m
{
(
)
′ = h fg 1 + 0.68c p,l Tsat −T s,c / h fg
where h fg
3
}
{
(
)
3
h ′fg = 2257 ×10 J / k g 1 + 0.68 × 4207 J / k g ⋅ K 100 − Ts,c /2257 × 10 J / k g
(
hLc = 2.517 × 104 100 − Ts,c
Hence,
} ≈ 2257 ×10
3
J/kg.
)−1/4 .
Using the rate equation, now find Ts,c by trial-and-error,
(
592 W = 2.517 × 104 100 − Ts,c
(
9.358 = 100 − Ts,c
) −1/4 (π × 0.020m× 0.040m) (100 − Ts,c ) K
)0.75
<
Ts,c = 80.3 °C.
(c) The condensation rate in the condenser section is
(
)
m
& = qc / h ′fg = 592W/ 2257 ×103 J / k g = 2.623 ×10 −4 k g / s
and from Eq. 10.35,
Reδ =
&
&
4m
4m
4 × 2.623 ×10−4 k g / s
=
=
= 53.3.
µl b µl (π D ) 313 ×10 −6 N ⋅ s / m 2 ( π × 0.020m )
Since 30 < Reδ < 1800, we conclude the film is laminar-wavy.
<
PROBLEM 10.69
KNOWN: Thermosyphon configuration for cooling a computer chip of prescribed size.
FIND: (a) Chip temperature and total power dissipation when chip operates at 90% of critical heat flux,
(b) Required condenser length.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, (2) Saturated liquid/vapor conditions, (3) Negligible heat transfer
from bottom of chip.
PROPERTIES: Fluorocarbon (prescribed): Tsat = 57°C, c p," = 1100 J/kg⋅K, hfg = 84,400 J/kg, ρ" =
1619.2 kg/m3, ρ v = 13.4 kg/m3, σ = 8.1 × 10-3 kg/s2, µ" = 440 × 10-6 kg/m⋅s, Pr" = 9.01, k " = 0.054
W/m⋅K, ν " = µ" ρ" = 0.272 × 10-6 m2/s.
ANALYSIS: (a) With q′′ = 0.9 q′′max and the critical heat flux given by Eq. 10.7, the chip power
dissipation is
1/ 4
 σ g ( ρ" − ρ v ) 
2
q = 0.9Lc × 0.149h fg ρ v 



ρ v2
(
(
)(
)

0.0081kg s 2 9.8 m s 2 1605.8 kg m3
2
3
q = 0.9 ( 0.02 m ) × 0.149 (84, 400 J kg )13.4 kg m 
2

13.4 kg m3

)
(
)
1/ 4





q c = 0.9 4 × 10−4 m 2 1.55 × 105 W m 2 = 55.7 W
<
With operation at q′′ = 1.40 × 105 W/m2 in the nucleate boiling region, Eq. 10.5 yields
T = Tsat +
$
T = 57 C +
Cs,f h fg Pr"n 
c p,"
0.005 (84, 400 J kg )( 9.01)
1.7
1100 J kg ⋅ K
1/ 3



 µ" h fg 
q′′
1/ 6


σ


 g ( ρ" − ρν ) 
5
2

1.40 × 10 W m

−4
 4.4 × 10 kg m ⋅ s × 84, 400 J
1/ 3


kg 


 9.8 m

1/ 6
0.0081 kg s
s
2
(
2
1605.8 kg m
3
)




Continued...
PROBLEM 10.69 (Cont.
<
T = 57$ C + 22.4$ C = 79.4$ C
(b) The power dissipated by the chip must be balanced by the rate of heat transfer from the condensing
section. Hence, with A = πDL, Eq. 10.32 yields the requirement that
hLL =
q
πD Tsat − Ts
=
55.7 W
= 18.5 W m⋅ K
π 0.03 m 32 $ C
$ #
%
To determine h L , we combine Eqs. 10.33 and 10.35 to obtain Re δ = 4q µ " bh ′fg , where b = πD = 0.0942
#
m and h ′fg = h fg + 0.68c p,1 Tsat − Ts = 84,400 J kg + 0.68 1100 J kg ⋅ K 32 $ C = 108,300 J/kg. Hence, Reδ
$
= 4(55.7 W)/4.4 × 10 kg/m⋅s(0.0942 m)108,300 J/kg = 49.6 and the condensate film is in the laminarwavy region. Hence, from Eq. 10.38
-4
hL =
k"
ν" g
2
1/ 3
%
Re δ
0.054 W m⋅ K × 0.409
=
1.22
2
1.08 Re δ − 5.2
0.272 × 10 −6 m 2 s 9.8 m s 2
%
1/ 3
= 1130 W m 2⋅ K
in which case,
L=
18.5 W m⋅ K
= 0.0164 m = 16.4 mm
1130 W m 2⋅ K
<
COMMENTS: The chip operating temperature (T = 79.4°C) is not excessive, and the proposed scheme
provides a compact means of cooling high performance chips.
PROBLEM 10.70
KNOWN: Copper plate, 2m × 2m, in a condenser-boiler section maintained at Ts = 100°C separates
condensing saturated steam and nucleate-pool boiling of saturated liquid X.
FIND: (a) Rates of evaporation and condensation (kg/s) for the two fluids and (b) Saturation temperature
Tsat and pressure p for the steam, assuming that film condensation occurs.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Isothermal copper plate.
PROPERTIES: Fluid-X (Given, 1 atm): Tsat = 80°C, hfg = 700 kJ/kg, portion of boiling curve shown
above for operating condition, ∆Te = Ts − Tsat = (100 − 80)°C = 20°C, q′′s = 5 × 104 W/m2 ; Table A.4,
Water (saturated, 1 atm): Tsat = 100°C, hfg = 2.25 × 106 J/kg; Water (saturated, Tsat): as required in part
(b); Water (saturated, Tf = (Tsat + Ts)/2): as required in part (b).
ANALYSIS: (a) For fluid-X, with ∆Te = Ts − Tsat = (100 − 80)°C = 20 K, the heat flux from the boiling
curve is
q′′s = 50, 000 W m 2
and the heat rate from the copper plate section into liquid-X is
qs = q′′s × As = 50, 000 W m 2 × ( 2 × 2 ) m 2 = 200, 000 W
From an energy balance around liquid-X, the evaporation rate for fluid-X is
X = q s h fg,X = 200, 000 W 700, 000 J kg = 0.286 kg s
m
<
The heat rate into the copper plate section from the steam is qs = 200,000 W, and from an energy balance
around the condensate film, the condensation rate for steam (w)
w = qs h ′fg,w = 200, 000 W 2.25 × 106 J kg = 0.0889 kg s
m
where we are assuming that Tsat,w is only a few degrees above Ts so that h ′fg′ ≈ h fg .
(b) The condensation heat rate, Eq. 10.32 can be expressed as
qs = h L As ( Tsat − Ts )
and assuming laminar film condensation, Eq. 10.30,
1/ 4
 gρ ( ρ − ρ ) k 3 h ′ 
"
"
ν " fg

h L = 0.943 
 µ" k " (Tsat − Ts ) 


Continued...
PROBLEM 10.70 (Cont.)
Recognize that with qs , As and Ts known, this relation can be used to determine Tsat , and from the steam
table, the corresponding psat can be found. The vapor properties (v) are evaluated at Tsat while the liquid
properties ( ) are evaluated at the film temperature Tf = (Tsat + Ts)/2. An iterative solution is required,
beginning by assuming a value for Tsat , evaluate properties and check whether the rate equation returns
the assumed value for Tsat . Using the IHT Correlations Tool, Film Condensation, Vertical Plate for the
laminar region, the results are
Tsat = 381.7 K
psat = 1.367 bar
for which Reδ = 661, so that the flow is wavy-laminar, not laminar. Repeating the analysis but with the
IHT Tool for the laminar, wavy-laminar, turbulent regions, the results with Reδ = 652 are
Tsat = 379.6 K
Psat = 1.27 bar
COMMENTS: A copy of the IHT model for determining Tsat and psat for part (b) is shown below.
// Correlations Tool //Film Condensation, Vertical Plate, laminar, wavy-laminar, turbulent regions
NuLbar = NuL_bar_FCO_VP(Redelta,Prl)
// Eq 10.37, 38, 39
NuLbar = hLbar * (nul^2 / g)^(1/3) / kl
g = 9.8
// Gravitational constant, m/s^2
Ts = 100 + 273
// Surface temperature, K
Tsat = 380
// Saturation temperature, K; explore over range to match q
// The liquid properties are evaluated at the film temperature, Tf,
Tf = Tfluid_avg(Ts,Tsat)
// The condensation and heat rates are
q = hLbar * As * (Tsat - Ts)
// Eq 10.32
As = L * b // Surface Area, m^2
mdot = q / h'fg
// Eq 10.33
h'fg = hfg + 0.68 * cpl * (Tsat - Ts)
// Eq 10.26
// The Reynolds number based upon film thickness is
Redelta = 4 * mdot / (mul * b)
// Eq 10.35
/* Correlation description: Film condensation (FCO) on a vertical plate (VP). If Redelta<29, laminar
region, Eq 10.37 . If 31<Redelta<1750, wavy-laminar region, Eq 10.38 . If Redelta>=1850, turbulent
region, Eq 10.22, 10.32, 10.33, 10.35, 10.39 . In laminar-wavy and wavy-turbulent transition regimes,
function interpolates between laminar and wavy, and wavy and turbulent correlations. See Fig 10.15 . */
// Assigned Variables:
L=2
// Plate height, m
b=2
// Plate width, m
//q = 200000
// Heat rate, W; required heat rate for suitable Tsat
// Properties Tool - Water:
// Water property functions :T dependence, From Table A.6
// Units: T(K), p(bars);
xl = 0
// Quality (0=sat liquid or 1=sat vapor)
pl = psat_T("Water", Tf)
// Saturation pressure, bar
vl = v_Tx("Water",Tf,xl)
// Specific volume, m^3/kg
rhol = rho_Tx("Water",Tf,xl)
// Density, kg/m^3
cpl = cp_Tx("Water",Tf,xl)
// Specific heat, J/kg·K
mul = mu_Tx("Water",Tf,xl)
// Viscosity, N·s/m^2
nul = nu_Tx("Water",Tf,xl)
// Kinematic viscosity, m^2/s
kl = k_Tx("Water",Tf,xl)
// Thermal conductivity, W/m·K
Prl = Pr_Tx("Water",Tf,xl)
// Prandtl number
// Water property functions :T dependence, From Table A.6
// Units: T(K), p(bars);
xv = 1
// Quality (0=sat liquid or 1=sat vapor)
pv = psat_T("Water", Tsat)
// Saturation pressure, bar
vv = v_Tx("Water",Tsat,xv)
// Specific volume, m^3/kg
rhov = rho_Tx("Water",Tsat,xv)
// Density, kg/m^3
hfg = hfg_T("Water",Tsat)
// Heat of vaporization, J/kg
cpv = cp_Tx("Water",Tsat,xv)
// Specific heat, J/kg·K
muv = mu_Tx("Water",Tsat,xv)
// Viscosity, N·s/m^2
nuv = nu_Tx("Water",Tsat,xv)
// Kinematic viscosity, m^2/s
kv = k_Tx("Water",Tsat,xv)
// Thermal conductivity, W/m·K
Prv = Pr_Tx("Water",Tsat,xv)
// Prandtl number
<
PROBLEM 10.71
KNOWN: Thin-walled container filled with a low boiling point liquid (A) at Tsat,A. Outer surface of
container experiences laminar-film condensation with the vapor of a high-boiling point fluid (B).
Laminar film extends from the location of the liquid-A free surface. The heat flux for nucleate pool
boiling in liquid-A along the container wall is given as q′′npb = C(Ts − Tsat)3, where C is a known
empirical constant.
FIND: (a) Expression for the average temperature of the container wall, Ts; assume that the properties of
fluids A and B are known; (b) Heat rate supplied to liquid-A, and (c) Time required to evaporate all the
liquid-A in the container, assuming that initially the container is filled, y = L.
SCHEMATIC:
ASSUMPTIONS: (1) Nucleate pool boiling occurs on the inner surface of the container with liquid-A,
(2) Laminar film condensation occurs on the outer surface of the container with fluid-B over the liquid-A
free surface, y, and (3) Negligible wall thermal resistance.
ANALYSIS: (a) Perform an energy balance on the control surface about the container wall along
locations experiencing boiling (A) and condensation (B) as shown in the schematic above.
E ′′in − E ′′out = 0
(1)
q′′cond − q′′npb = 0
(2)
(
)
(
)
h y (π Dy ) Tsat,B − Ts − (π Dy ) C Ts3 − Tsat,A = 0
(
)
(
h y Tsat,B − Ts = C Ts − Tsat,A
)3
(3)
<
where h y is the average convection coefficient for laminar film condensation over the surface length 0
to y. From Eq. 10.30 and 10.26,
Continued...
PROBLEM 10.71 (Cont.)
1/ 4
 gρ ( ρ − ρ ) k 3h ′ 
"
"
v " fg 
h y = 0.943 
 µ" ( Tsat − Ts ) y 

B
(
h ′fg = h fg,B + 0.68c p,B Tsat,B − Ts
(3)
)
(4)
where the properties are for fluid-B.
(b) The heat flux supplied to liquid-A is, from Eq. (2), q′′cond = q′′npb . Since h y is a function of y, Ts
and, hence, the heat fluxes will be functions of y, the height of liquid A in the container.
(c) To determine the dry-out time, tf, begin with an energy balance on the inside of the container (fluidA). The heat transfer supplied to liquid-A results in an evaporation rate of liquid-A,
q′′npb (π Dy ) −
dM
h fg = 0
dt
(4)
where M is the mass of liquid-A in the container,
(
)
M = ρ",A π D2 4 y
(5)
Substituting Eq. (5) into (4), separating variables and identifying integration limits, find
(
C Ts − Tsat,A
)3 (π Dy ) = dtd  ρ",A (π D2 4 ) y hfg
(
)
ρ",A π D2 4 h fg 0
tf
dy
dt = t f =
3
0
L
Cπ D
Ts − Tsat,A y
∫
∫
(
)
(6)
The definite integral could be numerically evaluated using values for Ts(y) obtained by solving Eq. (3).
PROBLEM 11.1
KNOWN: Initial overall heat transfer coefficient of a fire-tube boiler. Fouling factors following one
year’s application.
FIND: Whether cleaning should be scheduled.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible tube wall conduction resistance, (2) Negligible changes in hc and hh.
ANALYSIS: From Equation 11.1, the overall heat transfer coefficient after one year is
1
1
1
=
+
+ R′′f,i + R′′f,o .
U hi ho
Since the first two terms on the right-hand side correspond to the reciprocal of the initial overall
coefficient,
1
1
=
+ ( 0.0015 + 0.0005) m2 ⋅ K / W = 0.0045 m2 ⋅ K / W
2
U 400 W / m ⋅ K
U = 222 W / m2 ⋅ K.
COMMENTS: Periodic cleaning of the tube inner surfaces is essential to maintaining efficient firetube boiler operations.
PROBLEM 11.2
KNOWN: Type-302 stainless tube with prescribed inner and outer diameters used in a cross-flow heat
exchanger. Prescribed fouling factors and internal water flow conditions.
FIND: (a) Overall coefficient based upon the outer surface, Uo, with air at To =15°C and velocity Vo =
20 m/s in cross-flow; compare thermal resistances due to convection, tube wall conduction and fouling;
(b) Overall coefficient, Uo, with water (rather than air) at To = 15°C and velocity Vo = 1 m/s in crossflow; compare thermal resistances due to convection, tube wall conduction and fouling; (c) For the
water-air conditions of part (a), compute and plot Uo as a function of the air cross-flow velocity for 5 ≤
Vo ≤ 30 m/s for water mean velocities of um,i = 0.2, 0.5 and 1.0 m/s; and (d) For the water-water
conditions of part (b), compute and plot Uo as a function of the water mean velocity for 0.5 ≤ um,i ≤ 2.5
m/s for air cross-flow velocities of Vo = 1, 3 and 8 m/s.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Fully developed internal flow,
PROPERTIES: Table A.1, Stainless steel, AISI 302 (300 K): kw = 15.1 W/m⋅K; Table A.6, Water
( Tm,i = 348 K): ρi = 974.8 kg/m3, µi = 3.746 × 10-4 N⋅s/m2, ki = 0.668 W/m⋅K, Pri = 2.354; Table A.4,
Air (assume Tf ,o = 315K, 1 atm): ko = 0.02737 W/m⋅K, νo = 17.35 × 10-6 m2/s, Pro = 0.705.
ANALYSIS: (a) For the water-air condition, the overall coefficient, Eq. 11.1, based upon the outer area
can be expressed as the sum of the thermal resistances due to convection (cv), tube wall conduction (w)
and fouling (f):
1 Uo Ao = R tot = R cv,i + R f ,i + R w + R f ,o + R cv,o
R cv,i = 1 h i Ai
R cv,o = 1 h o Ao
R f ,i = R ′′f ,i Ai
R f ,o = R ′′f ,o Ao
and from Eq. 3.28,
R w = ln ( Do Di ) ( 2π Lk w )
The convection coefficients can be estimated from appropriate correlations.
Continued...
PROBLEM 11.2 (Cont.)
Estimating h i : For internal flow, characterize the flow evaluating thermophysical properties at Tm,i with
ReD,i =
u m,i Di
νi
=
0.5 m s × 0.022m
3.746 ×10 −4 N ⋅ s m 2 974.8 kg m3
= 28, 625
For the turbulent flow, use the Dittus-Boelter correlation, Eq. 8.60,
0.4
Nu D,i = 0.023Re0.8
D,i Pri
0.8
0.4
Nu D,i = 0.023 ( 28, 625 ) ( 2.354 ) = 119.1
hi = Nu D,i k i Di = 119.1× 0.668 W m 2 ⋅ K 0.022m = 3616 W m 2 ⋅ K
Estimating h o : For external flow, characterize the flow with
V D
20 m s × 0.027m
ReD,o = o o =
= 31,124
νo
17.35 × 10−6 m 2 s
evaluating thermophysical properties at Tf,o = (Ts,o + To)/2 when the
surface temperature is determined from the thermal circuit analysis
result,
(Tm,i − To )
(
R tot = Ts,o − To
)
R cv,o
Assume Tf,o = 315 K, and check later. Using the Churchill-Bernstein
correlation, Eq. 7.57, find
5/8
  Re
D,o 


Nu D,o = 0.3 +
1+
1/ 4   282, 000 

 
1 + (0.4 Pr )2 / 3 
 
o


2 1/ 3
0.62 Re1/
D,o Pro
1/ 2
Nu D,o = 0.3 +
0.62 (31,124 )
4/5
(0.705 )1/ 3 1 + 
2 / 3 1/ 4
1 + ( 0.4 0.705 )




5/8
31,124 
 282, 000 


4/5



Nu D,o = 102.6
h o = Nu D,o k o Do = 102.6 × 0.02737 W m ⋅ K 0.027m = 104.0 W m ⋅ K
Using the above values for h i , and h o , and other prescribed values, the thermal resistances and overall
coefficient can be evaluated and are tabulated below.
Rcv,i
(K/W)
0.00436
Rf,i
(K/W)
0.00578
Rw
(K/W)
0.00216
Rf,o
(K/W)
0.00236
Rcv,o
(K/W)
0.1134
Uo
(W/m2⋅K)
92.1
Rtot
(K/W)
0.128
The major thermal resistance is due to outside (air) convection, accounting for 89% of the total
resistance. The other thermal resistances are of similar magnitude, nearly 50 times smaller than Rcv,o.
(b) For the water-water condition, the method of analysis follows that of part (a). For the internal flow,
the estimated convection coefficient is the same as part (a). For an assumed outer film coefficient,
Tf ,o = 292 K, the convection correlation for the outer water flow condition Vo = 1 m/s and To = 15°C,
find
PROBLEM 11.2 (Cont.)
ReD,o = 26, 260
h o = 4914 W m 2 ⋅ K
Nu D,o = 220.6
The thermal resistances and overall coefficient are tabulated below.
Rcv,i
(K/W)
0.00436
Rf,i
(K/W)
0.00579
Rw
(K/W)
0.00216
Rf,o
(K/W)
0.00236
Rcv,o
(K/W)
0.00240
Rtot
(K/W)
0.0171
Uo
(W/m2⋅K)
691
Note that the thermal resistances are of similar magnitude. In contrast with the results for the water-air
condition of part (a), the thermal resistance of the outside convection process, Rcv,o, is nearly 50 times
smaller. The overall coefficient for the water-water condition is 7.5 times greater than that for the waterair condition.
(c) For the water-air condition, using the IHT workspace with the analysis of part (a), Uo was calculated
as a function of the air cross-flow velocity for selected mean water velocities.
Uo (W/m^2.K)
Water (i) - air (o) condition
120
100
80
60
40
5
10
15
20
25
30
Air velocity, Vo (m/s)
Water mean velocity, umi = 0.2 m/s
umi = 0.5 m/s
umi = 1.0 m/s
The effect of increasing the cross-flow air velocity is to increase Uo since the Rcv,o is the dominant
thermal resistance for the system. While increasing the water mean velocity will increase h i , because
Rcv,i << Rcv,o, this increase has only a small effect on Uo.
(d) For the water-water condition, using the IHT workplace with the analysis of part (b), Uo was
calculated as a function of the mean water velocity for selected air cross-flow velocities.
Uo (W/m^2.K)
Water (i) - water (o) condition
1000
900
800
700
600
0.5
1
1.5
2
2.5
Water mean velocity, umi (m/s)
Air velocity, Vo = 1 m/s
Vo = 3 m/s
Vo = 8 m/s
Because the thermal resistances for the convection processes, Rcv,i and Rcv,o, are of similar magnitude
according to the results of part (b), we expect to see Uo significantly increase with increasing water mean
velocity and air cross-flow velocity.
PROBLEM 11.3
KNOWN: Copper tube with prescribed inner and outer diameters used in a shell-and-tube heat
exchanger. Conditions prescribed for internal water flow and steam condensation on external surface.
FIND: (a) Overall heat transfer coefficient based upon the outer surface area, Uo; compare thermal
resistances due to convection, tube wall conduction and condensation, and (b) Compute and plot Uo,
water-side convection coefficient, hi, and steam-side convection coefficient, ho, as a function of the water
i ≤ 0.8 kg/s.
flow rate for the range 0.2 ≤ m
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Fully developed internal flow.
PROPERTIES: Table A.1, Copper, pure (300 K): kw = 401 W/m⋅K; Table A.6, Water (Tm,i = 298 K): µi
= 8.966 × 10-4 N⋅s/m2 , ki = 0.6102 W/m⋅K, Pri = 6.146. Table A.6, Water, (assume Ts,o = 351 K, Tf,o =
362 K): ρ" = 965.7 kg/m3 , cp," = 4205 J/kg⋅K, µ" = 3.172 × 10-4 N⋅s/m2, k " = 0.6751 W/m⋅K; Table
A.6 Water (Tsat = 373 K, 1 atm): ρν = 0.5909 kg/m3 , hfg = 2257 kJ/kg.
ANALYSIS: (a) The overall coefficient, Eq 11.1, based upon the outer surface area can be expressed as
the sum of the thermal resistances due to convection (cv) , tube wall conduction (w, see Eq. 3.28) and
condensation (cnd):
1 Uo Ao = R tot = R cν + R w + R cnd
R cv = 1 h i Ai
R w = n ( Do Di ) ( 2π Lk w )
R cnd = 1 h o Ao
The convection coefficients can be estimated from appropriate correlations.
Estimating hi : For internal flow, characterize the flow using thermophysical properties evaluated at Tm,i
with
ReD,i =
i
4m
4 × 0.2 kg s
=
= 21,847
π Di µi π × 0.013m × 8.966 × 10−4 N s ⋅ m 2
For turbulent flow, use the Dittus-Boelter correlation, Eq. 8.60,
0.8
0.4
0.4
Nu D,i = 0.023 Re0.8
= 140.8
D,i Pri = 0.023 ( 21,847 ) (6.146 )
h i = Nu D,i k i Di = 140.8 × 0.6102 W m ⋅ K 0.013m = 6610 W m 2 ⋅ K
Continued...
PROBLEM 11.3 (Cont.)
Estimating ho : For the horizontal tube, average convection coefficient for film condensation, Eq. 10.40,
is
14
 gρ ( ρ − ρ ) k 3 h ′ 
"
"
ν " fg

h o = 0.729 
 µ" ( Tsat − Ts,o ) Do 


(
)
(
)
h ′fg = h fg + 0.68c p, " Tsat − Ts,o
The vapor (v) properties and hfg are evaluated at Tsat , while the liquid
properties ( ) are evaluated at the film temperature Tf,o = (Ts,o − Tsat )
where the surface temperature is determined from the thermal circuit
analysis result,
(Tm,i − Tsat )
R tot = Ts,o − Tsat
R cnd
Assume Ts,o = 351 K so that Tf,o = 362 K, and check later. Hence,
14
 9.8 m s 2 × 965.7 kg m3 × (965.7 − 0.5909 ) kg m3 × (0.6751W m⋅ K )3 × 2321kJ kg 

h o = 0.729 


3.172 × 10 −4 N ⋅ s m 2 (373 − 351) K × 0.018 m
h o = 11, 005 W m 2 ⋅ K
Using the above values for hi, ho and other prescribed values, the thermal resistances and overall
coefficient can be evaluated and are tabulated below.
Rcv × 103
(K/W)
3.704
Rw × 103
(K/W)
1.292
Rcnd × 103
(K/W)
1.610
Rtot × 103
(K/W)
5.444
Uo
(W/m2 ⋅K)
3249
The largest resistance is that due to convection on the water-side. Interestingly, the wall thermal
resistance for the pure copper, while the smallest for all the process, is still significant relative to that for
the condensation process.
(b) The foregoing relations were entered into the IHT workspace along with the Correlations Tools for
Forced Convection, Internal Flow, Turbulent Flow and for Film Condensation, Horizontal Cylinder with
the appropriate Properties Tools for Air and Water. The coefficients Uo, hi and ho were computed and
plotted as a function of the water flow rate.
25000
20000
hi, ho, Uo (W/m^2.K)
Note that the overall coefficient increases nearly
50% over the range of the water flow rate. The
water-side coefficient increases markedly, by
nearly a factor of 4, with increasing flow rate.
The steam-side coefficient, ho , is larger than hi
by a factor of 2 at the lowest flow rate.
However, ho decreases with increasing water
flow rate since the tube wall temperature, Ts,o ,
decreases causing the water film thickness to
increase with the net effect of reducing ho.
15000
10000
5000
0
0.2
0.3
0.4
0.5
0.6
Water flow rate, mdoti (kg/s)
Water-side, hi (W/m^2.K)
Steam-side, ho (W/m^2.K)
Overall coefficient, Uo (W/m^2.K)
0.7
0.8
PROBLEM 11.4
KNOWN: Dimensions of heat exchanger tube with or without fins. Cold and hot side convection
coefficients.
FIND: Cold side overall heat transfer coefficient without and with fins.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible fouling, (2) Negligible contact resistance between fins and tube wall,
(3) hh is not affected by fins, (4) One-dimensional conduction in fins, (5) Adiabatic fin tip.
ANALYSIS: From Eq. 11.1,
Di ln ( Do / Di )
1
1
Ac
=
+
+
Uc (ηo h ) c
2k
(ηohA )h
Without fins: ηo,c = ηo,h = 1
( 0.02m ) ln ( 26/20 )
1
1
1
20
=
+
+
Uc 8000 W / m 2 ⋅ K
100 W / m ⋅ K
200 W / m 2 ⋅ K 26
(
)
1 / Uc = 1.25 × 10−4 + 5.25 × 10−5 + 3.85 ×10−3 m2 ⋅ K / W = 4.02 × 10−3 m2 ⋅ K / W
Uc = 249 W / m2 ⋅ K.
With fins:
ηo,c = 1, ηo,h = 1 − ( Af / A )(1 − ηf
<
)
Per unit length along the tube axis,
Af = N ( 2Lf + t ) = 16 ( 30 +2 ) mm = 512 mm
A h = A f + (π D o −16t ) = ( 512 + 81.7 − 32 ) mm = 561.7 mm
With
(
m = ( 2h/kt )1/2 = 400 W / m 2 ⋅ K / 5 0 W / m ⋅ K × 0.002m
(
)
)
1/2
= 63.3m− 1
mL f = 63.3m −1 ( 0.015m ) = 0.95
and Eq. 11.4 yields
ηf = tanh ( mLf ) /mLf = 0.739/0.95 = 0.778.
The overall surface efficiency is then
ηo = 1 − ( Af / Ah )(1 − ηf ) = 1 − ( 512/561.7 )(1 − 0.778) = 0.798.
Hence
 2
π ( 20 )
1 
−4 2
=  1.25× 10−4 + 5.25 × 10−5 +
 m ⋅ K / W = 8.78 ×10 m ⋅ K / W

Uc 
0.798 ( 200 ) 561.7 
Uc = 1138 W / m 2 ⋅ K.
<
PROBLEM 11.5
KNOWN: Geometry of finned, annular heat exchanger. Gas-side temperature and convection
coefficient. Water-side flowrate and temperature.
FIND: Heat rate per unit length.
SCHEMATIC:
Do = 60 mm
Di,1 = 24 mm
Di,2 = 30 mm
t = 3 mm = 0.003m
L = (60-30)/2 mm = 0.015m
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) One-dimensional
conduction in strut, (4) Adiabatic outer surface conditions, (5) Negligible gas-side radiation, (6) Fullydeveloped internal flow, (7) Negligible fouling.
-6
2
PROPERTIES: Table A-6, Water (300 K): k = 0.613 W/m⋅K, Pr = 5.83, µ = 855 × 10 N⋅s/m .
ANALYSIS: The heat rate is
(
q = ( UA ) c Tm,h − Tm,c
where
)
1/ ( UA )c = 1/ ( hA )c + R w + 1/ (η ohA) h
Rw =
(
ln Di,2 / Di,1
2π kL
)=
ln ( 30/24 )
= 7.10 ×10 −4 K/W.
2π ( 50 W / m⋅ K) lm
With
Re D =
&
4m
4 × 0.161kg/s
=
= 9990
π Di,1µ π ( 0.024m ) 855 × 10−6 N ⋅ s / m2
internal flow is turbulent and the Dittus-Boelter correlation gives
(
 0.613 W / m ⋅ K  0.023 9990 4 / 5 5.83 0.4 = 1883 W / m 2 ⋅ K
(
) ( )

0.024m


)
h c = k / Di,1 0.023Re 4D/ 5 Pr 0.4 = 
( hA) c−1 =
(
1883 W / m 2 ⋅ K × π × 0.024m
)
−1
= 7.043× 10−3 K/W.
Find the fin efficiency as
ηo = 1 − ( Af / A )(1 − ηf
)
Af = 8 ×2 (L ⋅ w ) = 8 × 2 ( 0.015m ×1m ) = 0.24m 2
A = Af + (π Di,2 − 8t ) w = 0.24m 2 + (π × 0.03m −8 × 0.003m ) = 0.31m 2 .
PROBLEM 11.5 (Cont.)
From Eq. 11.4,
ηf =
tanh ( mL )
mL
where
m = [ 2h/kt ]1/2 =  2 ×100 W / m 2 ⋅ K / 5 0 W / m ⋅ K ( 0.003m )1/2  = 36.5m −1


1/2
mL = ( 2 h / k t )
tanh  ( 2 h / k t )

L = 36.5m −1 × 0.015m = 0.55
1/2 
L = 0.499.

Hence
ηf = 0.800/1.10 = 0.907
ηo = 1 − ( Af / A)(1 − ηf ) = 1 − ( 0.24/0.31)(1 − 0.907 ) = 0.928
(ηohA ) h−1 =
(
0.928 ×100 W / m 2 ⋅ K × 0.31m 2
)
−1
= 0.0347 K / W .
Hence
( UA)c−1 =
( 7.043× 10−3 + 7.1×10−4 + 0.0347) K / W
( UA)c = 23.6 W / K
and
q = 23.6 W / K ( 800 − 300 ) K = 11,800 W
<
for a 1m long section.
COMMENTS: (1) The gas-side resistance is substantially decreased by using the fins (Af >> πDi,2)
and q is increased.
(2) Heat transfer enhancement by the fins could be increased further by using a material of larger k,
but material selection would be limited by the large Tm,h.
PROBLEM 11.6
KNOWN: Condenser arrangement of tube with six longitudinal fins (k = 200 W/m⋅K). Condensing
refrigerant temperature at 45°C flows axially through inner tube while water flows at 0.012 kg/s and
15°C through the six channels formed by the splines.
FIND: Heat removal rate per unit length of the exchanger.
SCHEMATIC:
ASSUMPTIONS: (1) No heat loss/gain to the surroundings, (2) Negligible kinetic and potential
energy changes, (3) Negligible thermal resistance on condensing refrigerant side, hi → ∞, (4) Water
flow is fully developed, (5) Negligible thermal contact between splines and inner tube, (6) Heat
transfer from outer tube negligible.
3
PROPERTIES: Table A-6, Water ( Tc = 15°C = 288 K): ρ = 1000 kg/m , k = 0.595 W/m⋅K, ν = µ/ρ
-6
2
3
-6
2
= 1138 × 10 N⋅s/m /1000 kg/m = 1.138 × 10 m /s, Pr = 8.06; Tube fins (given): k = 200 W/m⋅K.
ANALYSIS: Following the discussion of Section 11.2,
q′ = UA′ ( Th − Tc )
1
1
= R ′h + R ′w + R ′c = R ′w +
UA′
(ηo hA′ )c
where R ′h = 0, due to the negligible thermal resistance on the refrigerant side (h), and
R ′w =
ln ( D2 / D1 )
2π k
=
ln (14 /10 )
2π ( 200 W / m ⋅ K )
= 2.678 × 10−4 m ⋅ K / W.
To estimate the thermal resistance on the water side (c), first evaluate the convection coefficient. The
hydraulic diameter for a passage, where Ac is the cross-sectional area of the passage is
4 π D32 − D22 / 4 − 6 ( D3 − D2 ) t / 2  / 6
4Ac


Dh,c =
=
P
(π D2 − 6t ) / 6 + (π D3 − 6t ) / 6 + 2 ( D3 − D2 ) / 2
)
(
(
)
4 π 502 − 142 / 4 − 6 (50 − 14 ) × 10−6 m 2 / 6


Dh,c =
(14π − 6 × 2 ) / 6 + (50π − 6 × 2 ) / 6 + (50 − 14 ) × 10−3 m
Dh,c =
4 × 2.656 ×10 −4 m 2
6.551×10−2 m
= 0.01622 m.
Hence the Reynolds number is
Continued …..
PROBLEM 11.6 (Cont.)
)
(
(0.012 kg / s / 6 ) / 1000 kg / m3 × 2.656 × 10−4 m 2  × 0.01622m


ReD,c = 
= 107
1.138 × 10−6 m 2 / s
and assuming the flow is fully developed,
h c Dh,c
Nu D,c =
k
= 3.66
h c = 3.66 × 0.595 W / m ⋅ K / 0.01622 = 134 W / m 2 ⋅ K.
The temperature effectiveness of the splines (fins) on the cold side is
ηo = 1 −
A f ,c
Ac
(1 − ηf )
where Af,c and Ac are, respectively, the finned and total (fin plus prime) surface areas, while
ηf =
tanh ( mL )
mL
1/ 2
m = ( 2h c / kt )
ηf =
(
)
(
tanh 25.88 m −1 × 0.018m
25.88 m −1 × 0.018m
Hence
ηo = 1 −
ηo = 1 −
1
ηo hA ′c
=
1/ 2
=  2 ×134 W / m 2 ⋅ K / ( 200 W / m ⋅ K × 0.002m )


= 25.88 m −1
) = 0.4348 = 0.934.
6 ( D3 − D2 )
0.4658
6 ( D3 − D2 ) + (π D2 − 6t )
6 (50 − 14 )
6 (50 − 14 ) + (14π − 6 × 2 )
[1 − ηf ]
(1 − 0.934 ) = 0.943
1
0.943 × 134 W / m ⋅ K [6 ( 50 − 14 ) + (14π − 6 × 2 )] × 10
2
−3
= 3.22 × 10
−2
m⋅K / W
m
and the heat rate is
q′ =
q′ =
Th − Tc
R ′w + 1/ (ηo hA′ )c
( 45 − 15) K
2.678 × 10−4 m ⋅ K / W + 3.22 ×10−2 m ⋅ K / W
= 924 W / m.
<
COMMENTS: (1) The effective length of the fin representing the splines was conservatively
estimated. The heat transfer by conduction through the splines to the outer tube and then by
convection to the water was ignored.
2
(2) Without the splines, find Dh = (D3 –D2) = 36 mm so that hc = 60.5 W/m ⋅K. The heat rate with
A′c = π D2 is
q′ = ( hA′c )( Th − Tc ) = 60.5 W / m 2 ⋅ K ( 0.014π m )( 45 − 15) K = 79 W / m.
The splines enhance the heat transfer rate by a factor of 924/79 = 11.7.
PROBLEM 11.7
KNOWN: Number, inner-and outer diameters, and thermal conductivity of condenser tubes.
Convection coefficient at outer surface. Overall flow rate, inlet temperature and properties of water
flow through the tubes. Flow rate and pressure of condensing steam. Fouling factor for inner surface.
FIND: (a) Overall coefficient based on outer surface area, Uo, without fouling, (b) Overall
coefficient with fouling, (c) Temperature of water leaving the condenser.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible flow work and kinetic and potential energy changes for water flow,
(2) Fully-developed flow in tubes, (3) Negligible effect of fouling on Di.
-4
2
PROPERTIES: Water (Given): cp = 4180 J/kg⋅K, µ = 9.6 × 10 N⋅s/m , k = 0.60 W/m⋅K, Pr = 6.6.
6
Table A-6, Water, saturated vapor (p = 0.0622 bars): Tsat = 310 K, hfg = 2.414 × 10 J/kg.
ANALYSIS: (a) Without fouling, Eq. 11.5 yields
l
l  D  D ln ( Do / Di ) l
=  o + o
+
U o h i  Di 
2 kt
ho
(
)
With Re D = 4 m 1 / π Di µ = 1.60 kg / s / π × 0.025m × 9.6 × 10 −4 N ⋅ s / m 2 = 21, 220, flow in the tubes is
i
turbulent, and from Eq. 8.60
 k
hi = 
 Di

4 / 5 0.4 =  0.60 W / m ⋅ K  0.023 21, 200 4 / 5 6.6 0.4 = 3400 W / m 2 ⋅ K
(
) ( )
 0.023ReDi Pr


 0.025m 

l 
 l  28  0.028 ln ( 28 / 25 )
Uo = 
+
 +
2 × 110
10, 000 
 3400  25 
(3.29 × 10−4 + 1.44 ×10−5 + 10−4 )
−1
−1
W / m2 ⋅ K =
W / m 2 ⋅ K = 2255 W / m 2 ⋅ K
<
(b) With fouling, Eq. 11.5 yields
Uo =  4.43 × 10−4 + ( Do / Di ) R ′′f ,i 


−1
(
= 5.55 × 10−4
)
−1
= 1800 W / m 2 ⋅ K
<
(c) The rate at which energy is extracted from the steam equals the rate of heat transfer to the water,
c (T
m
h h fg = m
c p m,o − Tm,i ) , in which case
h h fg
m
10 kg / s × 2.414 × 106 J / kg
Tm,o = Tm,i +
= 15°C +
= 29.4°C
c cp
m
400 kg / s × 4180 J / kg ⋅ K
<
COMMENTS: (1) The largest contribution to the thermal resistance is due to convection at the
, either by increasing m
interior of the tube. To increase Uo, hi could be increased by increasing m
c
1
or decreasing N. (2) Note that Tm,o = 302.4 K < Tsat = 310 K, as must be the case.
PROBLEM 11.8
KNOWN: Diameter and inner and outer convection coefficients of a condenser tube. Thickness, outer
diameter, and pitch of aluminum fins.
FIND: (a) Overall heat transfer coefficient without fins, (b) Effect of fin thickness and pitch on overall
heat transfer coefficient with fins.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible tube wall conduction resistance, (2) Negligible fouling and fin contact
resistance, (3) One-dimensional conduction in fin.
PROPERTIES: Table A.1, Aluminum (T = 300 K): k = 237 W/m⋅K.
ANALYSIS: (a) With no fins, Eq. 11.1 yields
(
U = h i−1 + h o−1
) (
−1
)
= 2 × 10−4 + 0.01
−1
<
W m 2⋅ K = 98.0 W m 2⋅ K
(b) With fins and a unit tube length, Eqs. 11.1 and 11.3 yield
1
1
1
=
+
Uiπ Di h iπ Di ηo h o A′o
(
)
2
and ηo = 1 - ( A′f A′o )(1 − ηf ) . The total fin surface area per unit length is A′f = N′2π roc
− ri2 ,
where the number of fins per unit length is N′ = 1m / S(m) . The total outside surface area per unit length
is A′o = A′f + (1 - N′t )πDi, and the fin efficiency is given by Eq. 3.91 or Fig. 3.19.
For t = 0.0015 m and S = 0.0035 m, roc = (Do/2) + (t/2) = 0.01075 m, N′ ≈ 286, A′f = 0.163 m2/m, and
A′o = (0.163 + 0.018) m2/m = 0.181 m2/m. With roc/ri = 2.15, Lc = 0.00575 m, Ap = 8.625 × 10-6 m2, and
(
2 h kA
L3/
o
p
c
)
1/ 2
= 0.0964, Fig. 3.19 yields ηf ≈ 0.99. Hence, ηo ≈ 1 - (0.163/0.181)(0.01) = 0.99
and
Ui = (1 h i ) + (π Di ηo h o A′o )
−1
Ui =  2 × 10−4 m 2 ⋅ K W + π × 0.01m 0.99 × 100 W m 2⋅ K × 0.181m 2 m 


−1
= 512 W m 2⋅ K
<
We may use the IHT Extended Surface Model (Performance Calculations for a Circular Rectangular Fin
Array) to consider the effect of varying t and S. To maximize N′ , the minimum allowable value of
Continued...
PROBLEM 11.8 (Cont.)
S - t = 1.5 mm should be selected. It is then a matter of choosing between a large number of thin fins or a
smaller number of thicker fins. Calculations were performed for the following options.
t (mm)
1
2
3
4
S (mm)
2.5
3.5
4.5
5.5
N„
400
286
222
182
Ui (W/m2⋅K)
640
512
460
420
Since heat transfer increases with Ui, the best configuration corresponds to t = 1 mm and S = 2.5 mm,
which provides the largest airside surface area.
COMMENTS: The best performance is always associated with a large number of closely spaced fins,
so long as the flow between adjoining fins is sufficient to maintain the convection coefficient.
PROBLEM 11.9
KNOWN: Operating conditions and surface area of a finned-tube, cross-flow exchanger.
FIND: Overall heat transfer coefficient.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible heat loss to surroundings, (2) Negligible kinetic and potential energy
changes, (3) Constant properties, (4) Exhaust gas properties are those of air.
(
)
PROPERTIES: Table A-6, Water Tm = 87° C : cp = 4203J/kg ⋅ K; Table A-4, Air
( Tm ≈ 275° C) :
cp = 1040 J / k g ⋅ K.
ANALYSIS: From the energy balance equations
(
)
& cc p,c Tc,o − Tc,i = 0.5kg/s × 4203J/kg ⋅ K (150 − 25 ) ° C = 2.63 ×105 W
q=m
Th,o = Th,i −
q
2.63× 105 W
= 325°C −
= 198.6°C.
& h c p,h
m
2 k g / s ×1040J/kg ⋅ K
Hence
U = q / A∆Tlm
where
∆Tlm = F∆Tlm,CF .
From Fig. 11.12, with
t − t 150 − 25
T −T
325 −198.6
P= o i =
= 0.42, R = i o =
= 1.01, F = 0.94
Ti − t i 325 − 25
to − ti
150 − 25
∆Tl m,CF =
( 325 − 150 ) − (198.6 − 25)
325 − 150
ln
198.6 − 25
= 174.3°C.
Hence
U=
q
2.63 ×105 W
=
= 160 W / m 2 ⋅ K.
2
AF∆Tlm,CF 10m × 0.94 ×174.3 °C
<
COMMENTS: From the ε - NTU method, Cc = 2102 W/K, Ch = 2080 W/K, (Cmin/Cmax) ≈ 1, qmax =
5
2
6.24 × 10 W and ε = 0.42. Hence, from Fig. 11.18, NTU ≈ 0.75 and U ≈ 156 W/m ⋅K.
PROBLEM 11.10
2
KNOWN: Heat exchanger with two shell passes and eight tube passes having an area 925m ; 45,500
kg/h water is heated from 80°C to 150°C; hot exhaust gases enter at 350°C and exit at 175°C.
FIND: Overall heat transfer coefficient.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible losses to surroundings, (2) Negligible kinetic and potential energy
changes, (3) Constant properties, (4) Exhaust gas properties are approximated as those of atmospheric
air.
(
)
PROPERTIES: Table A-6, Water Tc = ( 80 + 150 )° C / 2 = 388K : c = cp,f = 4236 J/kg⋅K.
ANALYSIS: The overall heat transfer coefficient follows from Eqs. 11.9 and 11.18 written in the
form
U = q / A F∆Tl m,CF
where F is the correction factor for the HXer configuration, Fig. 11.11, and ∆Tlm,CF is the log mean
temperature difference (CF), Eqs. 11.15 and 11.16. From Fig. 11.11, find
T −T
( 350 −175 ) °C = 2.5 P = Tc,o − Tc,i = (150 − 80 ) °C = 0.26
R = h,i h,o =
Tc,o − Tc,i
Th,i − Tc,i ( 350 − 80 ) ° C
(150 − 80 ) °C
find F ≈ 0.97. The log-mean temperature difference, Eqs. 11.15 and 11.17, is
∆Tl m,CF =
( 350 −150 ) °C − (175 − 80 ) °C = 141.1°C.
∆T1 − ∆T2
=
ln ( ∆T1 / ∆T2 ) ln  (350 − 150 ) / (175 − 80 )
From an overall energy balance on the cold fluid (water), the heat rate is
(
& c cc Tc,o − Tc,i
q=m
)
q = 45,500kg/h ×1h/3600s × 4236 J / k g ⋅ K (150 − 80 ) °C = 3.748 × 106 W.
2
Substituting values with A = 925 m , find
U = 3.748× 106 W/925m 2 × 0.97 × 141.1K = 29.6 W / m 2 ⋅ K.
<
COMMENTS: Compare the above result with representative values for air-water exchangers, as
given in Table 11.2. Note that in this exchanger, two shells with eight tube passes, the correction
factor effect is very small, since F = 0.97.
PROBLEM 11.11
KNOWN: Dimensions and thermal conductivity of tubes with or without annular fins. Convection
coefficients associated with condensation and natural convection at the inner and outer surfaces,
respectively.
FIND: (a) Overall heat transfer coefficient Ui for aluminum and copper tubes without fins, (b) Value
of Ui associated with adding aluminum fins.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible fouling and fin contact resistances, (2) One-dimensional
conduction in fins.
ANALYSIS: (a) For unfinned, aluminum tubes of unit length, Eq. 11.5 yields
l
l D ln ( Do / Di ) l  Di 
= + i
+


Ui h i
2k
h o  Do 
0.011ln (13 / 11) l  11  
 l
Ui =
+  
 5000 +
2 × 180
10  13  
−1
(
= 2 × 10
−4
+ 5.1 × 10
−6
+ 846 × 10
-6
−4
2
)
−1
2
= 11.8 W / m ⋅ K
<
-6
For copper the tube conduction resistance is reduced from 5.1 × 10 m ⋅K/W to 2.3 × 10 , but Ui is
essentially unchanged.
Ui = 11.8 W / m 2 ⋅ K
(b) With fins and a unit tube length, Eqs. 11.1 and 11.3 yield
<
π Di
l
l D ln ( Do / Di )
= + i
+
ηo h o A′o
Ui h i
2k
(
)
and ηo = 1 − ( A′f / A ′o )( l − ηf ). The fin surface area is A ′f = N ′2π rfc2 − ro2 and the total outer surface
area is A ′o = A ′f + ( l − N ′t )π D o . With t = 0.001m, rfc = rf + t/2 = (0.0125 + 0.0005)m = 0.0130m and
A ′f = 300 m
−1
( 2π )
(0.0130
2
− 0.0065
2
)m
2
= 0.239m and A ′o = 0.239m + ( l − 0.300 )π ( 0.013m )
=0.268m. With r2c = rf + t/2 = 0.013m, Lc = (rf – ro) + t/2 = 0.0065m, r2c/ro = 2, Ap = Lct = 3.25 ×
1/ 2
-6 2
10 m , and L3c/ 2 ( h o / k A p )
= 0.0685, Fig. 3.19 yields ηf ≈ 0.97. Hence, ηo = l – (0.239/0.268)
(0.03) ≈ 0.973, and
−1
 l

0.011ln (13 /11)
π × 0.011
Ui = 
+
+

360
0.973 × 10 × 0.244 
 5000
(
= 2 ×10−4 + 5.1× 10−6 + 145 × 10−4
)
−1
= 68.0 W / m 2 ⋅ K
<
COMMENTS: There is significant advantage to installing fins on the outer surface, which has a
much smaller convection2coefficient. The thermal resistance at the outer surface has been reduced
from 0.0846 to 0.0145 m ⋅K/W and could be reduced further by increasing Df and/or N ′. However,
the spacing between adjoining fins must not be so small as to restrict buoyancy driven flow in the
associated air space.
PROBLEM 11.12
KNOWN: Properties and flow rates for the hot and cold fluid to a heat exchanger.
FIND: Which fluid limits the heat transfer rate of the exchanger?
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, and (3) Negligible losses to
the surroundings and kinetic and potential energy changes.
ANALYSIS: The properties and flow rates for the hot and cold fluid to the heat exchanger are
tabulated below.
3
Density, kg/m
Specific heat, J/kg⋅K
Thermal conductivity, W/m⋅K
2
Viscosity, N⋅s/m
3
Flow rate, m /h
Cold fluid
Hot fluid
997
4179
0.613
-4
8.55 × 10
14
1247
2564
0.287
-4
1.68 × 10
16
The fluid which limits the heat transfer rate of the exchanger is the minimum fluid,
C min = bm
& ⋅ cgmin. For the hot and cold fluids, find
Ch = m
& h c h = 16 m 3 / h × 1247 kg / m 3 × 2564 J / kg ⋅ K × b1h / 3600sg = 14.21 kW / K
Cc = m
& c c c = 14 m3 / h × 997 kg / m 3 × 4179 J / kg ⋅ K × b1h / 3600sg = 16.20 kW / K
Hence, the hot fluid is the minimum fluid,
C min = C h
<
For any exchanger, the heat rate is q = ε qmax, where ε depends upon the exchanger type. The
maximum heat rate is qmax = Cmin (Th,i - Tc,i). Hence, it is the conditions for the minimum fluid that
limit the performance of the exchanger.
PROBLEM 11.13
KNOWN: Process (hot) fluid having a specific heat of 3500 J/kg⋅K and flowing at 2 kg/s is to be
cooled from 80°C to 50°C with chilled-water (cold fluid) supplied at 2.5 ks/g and 15°C assuming an
2
overall heat transfer coefficient of 2000 W/m ⋅K.
FIND: The required heat transfer areas for the following heat exchanger configurations; (a)
Concentric tube (CT) - parallel flow, (b) CT - counterflow, (c) Shell and tube, one-shell pass and 2 tube
passes; (d) Cross flow, single pass, both fluids unmixed. Use the IHT Tools | Heat Exchanger models
as your solution tool.
SCHEMATIC:
Th,i = 80oC
.
.
T h,i = 80oC
Th,o = 50oC
Th,o = 50oC
Tc,o
Tc,o
T c,i = 15oC
.
Tc,i = 15oC
.
U = 2000 W/m2-K
(a) Parallel flow hxer
(b) Counterflow hxer
ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible losses to the surroundings and kinetic
and potential energy changes, (3) Overall heat transfer coefficient remains constant with different
configurations, and (4) Constant properties.
ANALYSIS: The IHT Tools | Heat Exchanger models are based upon the effectiveness-NTU
method and suited for design-type problems. The table below summarizes the results of our analysis
using the IHT models including model equations, figures, and the required heat transfer area. The cold
fluid outlet temperature for all configurations is Tc,o = 35.1°C. The IHT code for the concentric tube,
parallel flow heat exchanger is provided in the Comments.
Heat exchanger type
(a)
(b)
(c)
(d)
CT -Parallel flow
CT -Counterflow
Shell and tube (1 - sp, 2 - tp)
Crossflow (1 - p, unmixed)
Eqs.
11.29b
11.30b
11.31b
11.33
Figs
11.14
11.15
11.10, 16
11.12, 18
2
A(m )
3.09
2.64
2.83
2.84
COMMENTS: (1) Referring to the tabulated results, note that for the concentric tube exchangers,
the area required for parallel flow is 17% larger than for counterflow. Under what circumstances
would you choose to use the PF arrangement if the area has to be significantly larger?
(2) The shell-tube and crossflow exchangers require nearly the same heat transfer area. What are
other factors that might influence your decision to select one type over the other for an application?
(3) Based upon area considerations only, the CF arrangement requires the smallest heat transfer area.
2
What practical issues need to be considered in making a CF heat exchanger with a 2.6 m area?
Continued …..
PROBLEM 11.13 (Cont.)
(4) The IHT code used for the concentric tube, parallel flow heat exchanger is shown below. Note the
use of the water property function, cp_Tx, and the intrinsic function, Tfluidavg, to provide the specific
heat at the mean water (cold fluid) temperature.
/” Results - energy balance only
Cc
Ch
1.045E4
7000
Tco
35.1
/” Results of sizing
A
CR
NTU
3.87
0.6699 0.882
*/
eps
0.4615
cc
4180
q
2.1E5
Tci
15
Thi
80
Tho
50
// Design conditions
Thi = 80
Tho = 50
mdoth = 2
ch = 3500
mdotc = 2.5
Tci = 15
U = 2000
// For the parallel-flow, concentric-tube heat exchanger,
// For the parallel-flow, concentric-tube heat exchanger,
NTU = -ln(1 - eps * (1 + Cr))/(1 + Cr)
// Eq 11.29b
// where the heat-capacity ratio is
Cr = Cmin/Cmax
// and the number of transfer units, NTU, is
NTU = U * A/Cmin
// Eq 11.25
// The effectiveness is defined as
eps = q/qmax
qmax = Cmin * (Thi - Tci)
// Eq 11.20
// See Tables 11.3 and 11.4 and Fig 11.14
// Energy balances
q = Cc * (Tco - Tci)
q = Ch * (Thi - Tho)
Cc = mdotc * cc
Ch = mdoth * ch
Cmin = Ch
Cmax = Cc
// Water property functions: T dependence, From Table A.6
// Units: T(K), p(bars):
xc = 0
// Quality (0=sat liquid or 1-sat vapor)
cc = cp_Tx(“Water”, Tcm,xc)
// Specific heat, J/kg⋅K
Tcm = Tfluid_avg(Tci, Tco)
// Mean temperature; K; intrinsic function
ch
3500*/
PROBLEM 11.14
KNOWN: A shell and tube Hxer (two shells, four tube passes) heats 10,000 kg/h of pressurized
water from 35°C to 120°C with 5,000 kg/h water entering at 300°C.
FIND: Required heat transfer area, As.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible heat loss to surroundings, (2) Negligible kinetic and potential energy
changes, (3) Constant properties.
PROPERTIES: Table A-6, Water ( Tc = 350K ) : cp = 4195 J/kg⋅K; Table A-6, Water (Assume Th,o
≈ 150°C, Th ≈ 500 K): cp = 4660 J/kg⋅K.
ANALYSIS: The rate equation, Eq. 11.14, can be written in the form
As = q / U∆Tlm
(1)
and from Eq. 11.18,
∆Tl m = F∆Tl m,CF
∆Tl m,CF =
where
∆T1 − ∆T2
.
ln ( ∆T1 / ∆T2 )
(2,3)
From an energy balance on the cold fluid, the heat rate is
(
)
& c cp,c Tc,o − Tc,i =
q=m
10,000 k g / h
J
× 4195
(120 − 35 ) K = 9.905 ×105 W.
3600s/h
kg ⋅ K
From an energy balance on the hot fluid, the outlet temperature is
& h c p,h = 300°C − 9.905 ×105 W /
Th,o = Th,i − q / m
5000 kg
J
× 4660
= 147°C.
3600 s
kg ⋅ K
From Fig. 11.11, determine F from values of P and R, where P = (120 – 35)°C/(300 – 35)°C = 0.32, R
= (300 – 147)°C/(120-35)°C = 1.8, and F ≈ 0.97. The log-mean temperature difference based upon a
CF arrangement follows from Eq. (3); find
( 300 −120 )
= 143.3K.
(147 − 35)
<
As = 9.905 ×10 5 W/1500W/m 2 ⋅ K × 0.97 × 143.3K = 4.75m 2
<
∆Tl m =  ( 300 − 120 ) − (147 − 35 ) K / ln
COMMENTS: (1) Check Th ≈ 500 K used in property determination; Th = (300 + 147)°C/2 = 497 K.
(2) Using the NTU-ε method, determine first the capacity rate ratio, Cmin/Cmax = 0.56. Then
ε≡
q
q max
=
(
) = 1 × (120 − 35) °C = 0.57.
Cmin (Th,i − Tc,i ) 0.56 ( 300 − 35) ° C
Cmax Tc,o − Tc,i
2
From Fig. 11.17, find that NTU = AU/Cmin ≈ 1.1 giving As = 4.7 m .
PROBLEM 11.15
KNOWN: The shell and tube Hxer (two shells, four tube passes) of Problem 11.14, known to have an
2
area 4.75m , provides 95°C water at the cold outlet (rather than 120°C) after several years of
operation. Flow rates and inlet temperatures of the fluids remain the same.
FIND: The fouling factor, Rf.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible heat loss to surroundings, (2) Negligible kinetic and potential energy
changes, (3) Constant properties, (4) Thermal resistance for the clean condition is R ′′t = (1500
2
-1
W/m ⋅K) .
PROPERTIES: Table A-6, Water ( Tc ≈ 338 K): cp = 4187 J/kg⋅K; Table A-6, Water (Assume Th,o
≈ 190°C, Th ≈ 520 K): cp = 4840 J/kg⋅K.
ANALYSIS: The overall heat transfer coefficient can be expresses as
U = 1/ ( R′′t + Rf′′ )
R′′f = 1 / U − Rt′′
or
(1)
where R ′′t is the thermal resistance for the clean condition and R ′′f , the fouling factor, represents the
additional resistance due to fouling of the surface. The rate equation, Eq. 11.14 with Eq. 11.18, has the
form,
U = q / As F∆ Tlm,CF
∆Tlm,CF = ( ∆T1 − ∆ T2 ) / ln ( ∆T1 / ∆ T2 ) .
(2)
From energy balances on the cold and hot fluids, find
(
)
& c cp,c Tc,o − Tc,i = (10,000/3600 k g / s ) 4187 J / k g ⋅ K ( 95 − 35 ) K = 6.978 × 105 W
q=m
& h cp,h = 300°C − 6.978 × 10 5 W / ( 5000/3600kg/s × 4840J/kg ⋅ K ) = 196.2°C.
Th,o = T h,i − q / m
The factor, F, follows from values of P and R as given by Fig. 11.11 with
P = ( 95 − 35 ) / ( 300 − 35 ) = 0.23
R = (300 − 196 ) / (120 − 35 ) = 1.22
giving F ≈ 1. Based upon CF arrangement,
∆Tl m,CF =  (300 − 95 ) − (196 − 35)  ° C / ln ( 300 − 95) / (196.2 − 35 ) = 182K.
Using Eq. (2), find now the overall heat transfer coefficient as
U = 6.978 ×105 W/4.75m2 × 1× 182K = 806 W / m 2 ⋅ K.
From Eq. (1), the fouling factor is
R ′′f =
1
806 W / m 2 ⋅K
−
1
1500 W / m 2 ⋅ K
= 5.74 ×10 −4 m 2 ⋅ K/W.
<
COMMENTS: Note that the effect of fouling is to nearly double (Uclean/Ufouled = 1500/806 ≈ 1.9)
the resistance to heat transfer. Note also the assumption for Th,o used for property evaluation is
satisfactory.
PROBLEM 11.16
KNOWN: Inner tube diameter (D = 0.02 m) and fluid inlet and outlet temperatures corresponding to
design conditions for a concentric tube heat exchanger. Overall heat transfer coefficient (U = 500
W/m2⋅K) and desired heat rate (q = 3000 W). Cold fluid outlet temperature after three years of operation.
FIND: (a) Required heat exchanger length, (b) Heat rate, hot fluid outlet temperature, overall heat
transfer coefficient, and fouling factor after three years.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible heat loss to the surroundings and kinetic and potential energy changes,
(2) Negligible tube wall conduction resistance, (3) Constant properties.
ANALYSIS: (a) The tube length needed to achieve the prescribed conditions may be obtained from Eqs.
11.14 and 11.15 where ∆T1 = Th,i - Tc,o = 80°C and ∆T2 = Th,o - Tc,i = 120°C. Hence, ∆T1m = (120 80)°C/ln(120/80) = 98.7°C and
L=
q
3000 W
=
= 0.968 m
(π D ) U∆T1m (π × 0.02 m ) 500 W m2⋅ K × 98.7$ C
<
(b) With q = Cc(Tc,o - Tc,i), the following ratio may be formed in terms of the design and 3 year
conditions.
(
(
)
)
Cc Tc,o − Tc,i
q
60$ C
=
=
= 1.333
$
q3 Cc Tc,o − Tc,i
45
C
3
Hence,
<
q3 = q 1.33 = 3000 W 1.333 = 2250 W
Having determined the ratio of heat rates, it follows that
(
(
) = 20$ C = 1.333
)3 160$ C − Th,o(3)
Ch Th,i − Th,o
q
=
q3 Ch Th,i − Th,o
Hence,
<
Th,o(3) = 160$ C − 20$ C 1.333 = 145$ C
With ∆Tlm,3 = (125 − 95 ) ln (125 95 ) = 109.3$ C ,
U3 =
q3
2250 W
=
= 338 W m 2⋅ K
$
DL
T
π
∆
(
) 1m,3 π (0.02 m ) 0.968 m 109.3 C
(
)
<
Continued...
PROBLEM 11.16 (Cont.)
With U = (1 h i ) + (1 h o )
R ′′f ,c =
−1
and U3 = (1 h i ) + (1 h o ) + R ′′f ,c 
−1
,
1
1  1
1  2
−4 2
− =
−
 m ⋅ K W = 9.59 × 10 m ⋅ K W
U3 U  338 500 
COMMENTS: Over time fouling will always contribute to a degradation of heat exchanger
performance. In practice it is desirable to remove fluid contaminants and to implement a regular
maintenance (cleaning) procedure.
<
PROBLEM 11.17
KNOWN: Counterflow, concentric tube heat exchanger of Example 11.1; maintaining the outlet oil
temperature of 60°C, but with variable rate of cooling water, all other conditions remaining the same.
FIND: (a) Calculate and plot the required exchanger tube length L and water outlet temperature Tc,o
for the cooling water flow rate in the range 0.15 to 0.3 kg/s, and (b) Calculate U as a function of the
water flow rate assuming the water properties are independent of temperature; justify using a constant
value of U for the part (a) calculations.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible losses to the surroundings and kinetic
and potential energy changes, (3) Overall heat transfer coefficient independent of water flow rate for
this range, and (4) Constant properties.
PROPERTIES: Table A-6, Water Tc = 35$ C = 308 K : c p = 4178 J / kg ⋅ K, µ = 725 × 10-6
"
'
%
N ⋅ s / m2 , k = 0.625 W / m ⋅ K, Pr = 4.85, Table A-4, Unused engine oil Th = 353 K :
c p = 2131 J / kg ⋅ K.
ANALYSIS: (a) The NTU-ε method will be used to calculate the tube length L and water outlet
temperature Tc,o using this system of equations in the IHT workspace:
NTU relation, CF hxer, Eq. 11.30b
NTU =
$
ε −1
1
n
ε Cr − 1
Cr − 1
C r = C max / C min
$
(1, 2)
NTU = U ⋅ A / C min
(3)
A = π Di ⋅ L
(4)
Capacity rates, find minimum fluid
h c h = 01
Ch = m
. kg / s × 2131 J / kg ⋅ K = 213.1 W / K
$
c cc = 015
Cc = m
. to 0.30 kg / s × 4178 J / kg ⋅ K = 626.7 − 1253 W / K
(5)
C min = C h
(6)
Effectiveness and maximum heat rate, Eqs. 11.19 and 11.20
ε = q / q max
(7)
!
&
!
q max = C min Th,i − Tc,i = Cc Th,i − Tc,i
&
(8)
Continued …..
PROBLEM 11.17 (Cont.)
!
q = C h Th,i − Th,o
&
(9)
With the foregoing equations and the parameters specified in the schematic, the results are plotted in
the graphs below.
Th e effe ct of w a te r flo w ra te o n ou tle t te m p e ra tu re
Th e e ffe ct o f w a te r flo w ra te o n re q u ire d e xch a n g e r le n g th
44
70
68
E xch a n g e r le n g th , L (m )
Te m p e ra ture , Tco (C )
42
40
66
64
38
62
36
60
0 .15
0 .2
0 .25
0 .3
0 .1 5
0 .2
W a te r flo w ra te , m o d tc (kg /s )
0 .2 5
0 .3
W a te r flo w ra te , m d o tc (kg /s )
(b) The overall coefficient can be written in terms of the inner (cold) and outer (hot) side convection
coefficients,
U = 1/ 1 / h i + 1 / ho
$
(10)
2
From Example 11.1, ho = 38.4 W/m ⋅K, and hi will vary with the flow rate from Eq. 8.60 as
!
i/m
i,b
h i = h i,b m
&0.8
(11)
i = 0.2 kg / s. From these equations, the results
where the subscript b denotes the base case when m
are tabulated.
$
c kg / s
m
015
0.20
0.25
0.30
h i W / m2 ⋅ K
"
1787
2250
2690
3112
'
h o W / m2 ⋅ K
"
38.4
38.4
38.4
38.4
'
U W / m2 ⋅ K
"
'
37.6
37.8
37.9
37.9
Note that while hi varies nearly 50%, there is a negligible effect on the value of U.
COMMENTS: Note from the graphical results, that by doubling the flow rate (from 0.15 to 0.30
kg/s), the required length of the exchanger can be decreased by approximately 6%. Increasing the
flow rate is not a good strategy for reducing the length of the exchanger. However, any increase in
the hot-side (oil) convection coefficient would provide a proportional decrease in the length.
PROBLEM 11.18
2
KNOWN: Concentric tube heat exchanger with area of 50 m with operating conditions as shown on
the schematic.
FIND: (a) Outlet temperature of the hot fluid; (b) Whether the exchanger is operating in counterflow
or parallel flow; or can’t tell from information provided; (c) Overall heat transfer coefficient; (d)
Effectiveness of the exchanger; and (e) Effectiveness of the exchanger if its length is made very long
SCHEMATIC:
ASSUMPTIONS: (1) Negligible heat loss to surroundings, (2) Negligible kinetic and potential
energy changes, and (3) Constant properties.
ANALYSIS: From overall energy balances on the hot and cold fluids, find the hot fluid outlet
temperature
(
)
(
q = Cc Tc,o − Tc,i = Ch Th,i − Th,o
(
)
3000 W / K (54 − 30 ) K = 6000 60 − Th,o
(1)
)
Th,o = 48°C
<
(b) HXer must be operating in counterflow (CF) since Th,o < Tc,o. See schematic for temperature
distribution.
2
(c) From the rate equation with A = 50 m , with Eq. (1) for q,
(
)
q = Cc Tc,o − Tc,i = UA∆T"m
∆T"m =
(2)
(60 − 54 ) K − ( 48 − 30 ) K 10.9 C
∆T1 − ∆T2
=
=
°
m ( ∆T1 / ∆T2 )
n (6 /18 )
(3)
3000 W / K (54 − 30 ) K = U × 50 m 2 × 10.9 K
U = 132 W / m 2 ⋅ K
<
(d) The effectiveness, from Eq. 11.20, with the cold fluid as the minimum fluid, Cc = Cmin,
ε=
q
q max
=
(
) = (54 − 30 ) K = 0.8
Cmin ( Th,i − Tc,i ) (60 − 30 ) K
Cc Tc,o − Tc,i
<
(e) For a very long CF HXer, the outlet of the minimum fluid, Cmin = Cc, will approach Th,i. That is,
(
)
q → Cmin Tc,o − Tc,i → q max
ε =1
<
PROBLEM 11.19
KNOWN: Specifications for a water-to-water heat exchanger as shown in the schematic including
the flow rate, and inlet and outlet temperatures.
FIND: (a) Design a heat exchanger to meet the specifications; that is, size the heat exchanger, and (b)
Evaluate your design by identifying what features and configurations could be explored with your
customer in order to develop more complete, detailed specifications.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible heat loss to surroundings, (2) Negligible kinetic and potential
energy changes, (3) Tube walls have negligible thermal resistance, (4) Flow is fully developed, and
(5) Constant properties.
ANALYSIS: (a) Referring to the schematic above and using the rate equation, we can determine the
value of the UA product required to satisfy the design requirements. Sizing the heat exchanger
involves determining the heat transfer area, A (tube diameter, length and number), and the associated
overall convection coefficient, U, such that U × A satisfies the required UA product. Our approach
has five steps: (1) Calculate the UA product: Select a configuration and calculate the required UA
product; (2) Estimate the area, A: Assume a range for the overall coefficient, calculate the area and
consider suitable tube diameter(s); (3) Estimate the overall coefficient, U: For selected tube
diameter(s), use correlations to estimate hot- and cold-side convection coefficients and the overall
coefficient; (4) Evaluate first-pass design: Check whether the A and U values (U × A) from Steps 2
and 3 satisfy the required UA product; if not, then (5) Repeat the analysis: Iterate on different values
for area parameters until a satisfactory match is made, (U × A) = UA.
To perform the analysis, IHT models and tools will be used for the effectiveness-NTU method
relations, internal flow convection correlations, and thermophysical properties. See the Comments
section for details.
Step 1 Calculate the required UA. For the initial design, select a concentric tube, counterflow heat
exchanger. Calculate UA using the following set of equations, Eqs. 11.30a,
ε=
1 − exp  − NTU (1 − Cr )
(1)
1 − Cr exp  − NTU (1 − Cr )
NTU = UA / Cmin
Cr = C min / Cmax
ε = q / q max
q max = Cmin Th,i − Tc,i
(
(2,3)
)
(4,5)
c , and cp is evaluated at the average mean temperature of the fluid, T
where C = m
m = (Tm,i +
p
Tm,o)/2. Substituting numerical values, find
ε = 0.464
NTU = 0.8523 q = 2.934 ×106 W
Th,o = 65.0°C
Continued …..
PROBLEM 11.19 (Cont.)
UA = 9.62 × 104 W / K
<
Step 2 Estimate the area, A. From Table 11.2, the typical range of U for water-to-water exchangers is
4
2
2
850 – 1700 W/m ⋅K. With UA = 9.619 × 10 W/K, the range for A is 57 – 113 m , where
A = π Di LN
(6)
where L and N are the length and number of tubes, respectively. Consider these values of Di with L =
10 m to describe the exchanger:
2
Case
1
2
Di (mm)
25
50
L (m)
10
10
N
73-146
36-72
A (m )
57-113
57-113
3
75
10
24-48
57-113
<
Step 3 Estimate the overall coefficient, U. With the inner (hot) and outer (cold) fluids in the
concentric tube arrangement, the overall coefficient is
1/ U = 1/ hi + 1/ ho
(7)
and the h are estimated using the Dittus-Boelter correlation assuming fully developed turbulent flow.
Coefficient, hot side, h i . For flow in the inner tube,
ReDi =
h,i
4m
π Di µ h
h =m
hi ⋅ N
m
(8,9)
and the correlation, Eq. 8.60 with n = 0.3, is
h D
Nu D = i i = 0.037 Re4Di/ 5 Pr 0.3
k
(10)
where properties are evaluated at the average mean temperature, Th = ( Thi + Tho ) / 2.
Coefficient, cold side, h o . For flow in the annular space, Do – Di, the above relations apply where
the characteristic dimension is the hydraulic diameter,
Dh,o = 4 Ac,o / Po
)
(
Ac,o = π Do2 − Di2 / 4
Po = π ( Do + Di )
(11-13)
To determine the outer diameter Do, require that the inner and outer fluid flow areas are the same, that
is,
π Di2 / 4 = π Do2 − Di2 / 4
Ac,i = Ac,o
(14,15)
(
)
Summary of the convection coefficient calculations. The results of the analysis with L = 10 m are
summarized below.
Continued …..
PROBLEM 11.19 (Cont.)
Case
1a
2a
3a
Di
(mm)
N
25
50
75
73
36
24
A
2
(m )
57
57
57
hi
2
(W/m ⋅K)
4795
2424
1616
ho
2
(W/m ⋅K)
4877
2465
1644
U
2
(W/m ⋅K)
2418
1222
814
U×A
W/K
5
1.39 ×10
4
6.91 x 10
4
4.61 × 10
For all these cases, the Reynolds numbers are above 10,000 and turbulent flow occurs.
Step 4 Evaluate first-pass design. The required UA product value determined in step 1 is UA = 9.62
4
× 10 W/K. By comparison with the results in the above table, note that the U × A values for cases 1a
and 2a are, respectively, larger and smaller than that required. In this first-pass design trial we have
identified the range of Di and N (with L = 10 m) that could satisfy the exchanger specifications. A
strategy can now be developed in Step 5 to iterate the analysis on values for Di and N, as well as with
different L, to identify a combination that will meet specifications.
(b) What information could have been provided by the customer to simplify the analysis for design of
the exchanger? Looking back at the analysis, recognize that we had to assume the exchanger
configuration (type) and overall length. Will knowledge of the customer’s installation provide any
insight? While no consideration was given in our analysis to pumping power limitations, that would
affect the flow velocities, and hence selection of tube diameter.
COMMENTS: The IHT workspace with the relations for step 3 analysis is shown below, including
summary of key correlation parameters. The set of equations is quite stiff so that good initial guesses
are required to make the initial solve.
/* Results, Step 3 - Di = 25 mm, N = 73, L = 10 m
A
Do
U
UA
Di
L
57.33 0.03536 2418
1.386E5 0.025
10
ReDi
ReDo
hDi
hDo
5.384E4
1.352E4 4795
4877
*/
/* Results, Step 3 - Di = 50 mm, N = 36, L = 10 m
A
Do
U
UA
Di
L
56.55 0.07071 1222
6.912E4 0.05
10
ReDi
ReDo
hDi
hDo
5.459E4
1.371E4 2424
2465
*/
/* Results, Step 3 - Di = 75 mm, N = 24, L = 10 m
A
Do
U
UA
Di
L
56.55 0.1061 814.8
4.608E4 0.075
10
ReDi
ReDo
hDi
hDo
5.459E4
1.371E4 1616
1644
*/
// Input variables
//Di = 0.050
Di = 0.025
//Di = 0.075
//N = 36
N = 73
//N = 24
L = 10
mdoth = 28
Thi_C = 90
Tho_C = 65.0
mdotc = 27
Tci_C = 34
Tco_C = 60
N
73
N
36
N
24
// From Step 1
Continued …..
PROBLEM 11.19 (Cont.)
// Flow rate and number of tubes, inside parameters (hot)
mdoth = N * umi * rhoi * Aci
Aci = pi * Di^2 /4
1 / U = 1 / hDi + 1/ hDo
UA = U * A
A = pi * Di * L * N
// Flow rate, outside parameters (cold)
mdotc = rhoo * Aco * umo * N
Aco = Aci
// Make cross-sectional areas of equal size
Aco = pi * (Do^2 - Di^2) / 4
Dho = 4 * Aco / P
// hydraulic diameter
P = pi * ( Di + Do)
// wetted perimeter of the annular space
// Inside coefficient, hot fluid
NuDi = NuD_bar_IF_T_FD(ReDi,Pri,n) // Eq 8.60
n = 0.3 // n = 0.4 or 0.3 for Tsi>Tmi or Tsi<Tmi
NuDi = hDi * Di / ki
ReDi = umi * Di / nui
/* Evaluate properties at the fluid average mean temperature, Tmi. */
Tmi = Tfluid_avg(Thi,Tho)
//Tmi = 310
// Outside coefficient, cold fluid
NuDo = NuD_bar_IF_T_FD(ReDo,Pro,nn) // Eq 8.60
nn = 0.4 // n = 0.4 or 0.3 for Tsi>Tmi or Tsi<Tmi
NuDo= hDo * Dho / ko
ReDo= umo * Dho/ nuo
/* Evaluate properties at the fluid average mean temperature, Tmo. */
Tmo = Tfluid_avg(Tci,Tco)
//Tmo = 310
// Water property functions :T dependence, From Table A.6
// Units: T(K), p(bars);
x=0
// Quality (0=sat liquid or 1=sat vapor)
rhoi = rho_Tx("Water",Tmi,x)
// Density, kg/m^3
nui = nu_Tx("Water",Tmi,x)
// Kinematic viscosity, m^2/s
ki = k_Tx("Water",Tmi,x)
// Thermal conductivity, W/m·K
Pri = Pr_Tx("Water",Tmi,x)
// Prandtl number
rhoo = rho_Tx("Water",Tmo,x)
// Density, kg/m^3
nuo = nu_Tx("Water",Tmo,x)
// Kinematic viscosity, m^2/s
ko = k_Tx("Water",Tmo,x)
// Thermal conductivity, W/m·K
Pro = Pr_Tx("Water",Tmo,x)
//Prandtl number
// Conversions
Thi_C = Thi - 273
Tho_C = Tho - 273
Tci_C = Tci - 273
Tco_C = Tco - 273
PROBLEM 11.20
KNOWN: Counterflow concentric tube heat exchanger.
FIND: (a) Total heat transfer rate and outlet temperature of the water and (b) Required length.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible heat loss to surroundings, (2) Negligible kinetic and potential energy
changes, (3) Negligible thermal resistance due to tube wall thickness.
PROPERTIES: (given):
3
ρ (kg/m )
Water
1000
Oil
800
cp (J/kg⋅K)
4200
1900
2
ν (m /s)
-7
7 × 10
-5
1 × 10
k (W/m⋅K)
0.64
0.134
Pr
4.7
140
ANALYSIS: (a) With the outlet temperature, Tc,o = 60°C, from an overall energy balance on the hot
(oil) fluid, find
(
)
& h c h Th,i − Th,o = 0.1kg/s ×1900 J / k g ⋅ K (100 − 60 ) ° C = 7600 W.
q=m
<
From an energy balance on the cold (water) fluid, find
& c cc = 30°C + 7600 W / 0 . 1 k g / s × 4200 J / k g ⋅ K = 48.1°C.
Tc,o = Tc,i + q / m
<
(b) Using the LMTD method, the length of the CF heat exchanger follows from
q = UA∆Tlm,CF = U ( π DL ) ∆Tlm,CF
where
∆Tlm,CF =
L = q / U ( π D ) ∆Tlm,CF
( 60 − 30 ) °C − (100 −48.1) °C
∆T1 − ∆T2
=
= 40.0°C
ln ( ∆T1 / ∆T2 )
ln (30/51.9 )
L = 7600 W / 6 0 W / m 2 ⋅ K ( π × 0.025m ) × 40.0 °C = 40.3m.
<
COMMENTS: Using the ε-NTU method, find Cmin = Ch = 190 W/K and Cmax = Cc = 420 W/K.
Hence
(
)
q max = Cmin Th,i − Tc,i = 190 W / K(100 − 30 ) K = 13,300 W
and ε=q/qmax = 0.571. With Cr = Cmin/Cmax = 0.452 and using Eq. 11.30b,
NTU =
 ε −1 
UA
1
1
0.571 −1


ln 
ln 
=
= 1.00
=
Cmin Cr − 1  ε Cr − 1  0.452 − 1  0.571× 0.452 − 1 
so that with A = πDL, find L = 40.3 m.
PROBLEM 11.21
KNOWN: Counterflow, concentric tube heat exchanger undergoing test after service for an extended
2
period of time; surface area of 5 m and design value for the overall heat transfer coefficient of
2
Ud = 38 W/m ⋅K.
FIND: Fouling factor, if any, based upon the test results of engine oil flowing at 0.1 kg/s cooled from
110°C to 66°C by water supplied at 25°C and a flow rate of 0.2 kg/s.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible losses to the surroundings and kinetic
and potential energy changes, (3) Constant properties.
PROPERTIES: Table A-5, Engine oil ( Th = 361 K): c = 2166 J/kg⋅K; Table A-6, Water
(Tc = 304 K, assuming Tc,o
)
= 36$C : c = 4178 J/kg⋅K.
ANALYSIS: For the CF conditions shown in the Schematic, find the heat rate, q, from an energy
balance on the hot fluid (oil); the cold fluid outlet temperature, Tc,o, from an energy balance on the
cold fluid (water); the overall coefficient U from the rate equation; and a fouling factor, R, by
comparison with the design value, Ud.
Energy balance on hot fluid
h c h Th,i − Th,o = 01
q=m
. kg / s × 2166 J / kg ⋅ K 110 − 66 K = 9530 W
!
&
$
Energy balance on the cold fluid
!
&
ccc Tc,o − Tc,i ,
q=m
find
Tc,o = 36.4$ C
Rate equation
q = UA∆T"n,CF
$
∆T"n,CF =
$
!Th,i − Tc,o & − !Th,o − Tc,i & = 110 − 36.4$ C − 66 − 25$ C = 55.7 C
n 73.6 / 41.0
n !Th,i − Tc,o & / !Th,o − Tc,i &
$
9530 W = U × 5 m2 × 55.7$ C
U = 34.2 W / m2 ⋅ K
Overall resistance including fouling factor
U = 1/ 1/ Ud + R ′′f
34.2 W / m2 ⋅ K = 1/ 1/ 38 W / m2 ⋅ K + R ′′f
R ′′f = 0.0029 m2 ⋅ K / W
<
PROBLEM 11.22
KNOWN: Prescribed flow rates and inlet temperatures for hot and cold water; UA value for a shelland-tube heat exchanger (one shell and two tube passes).
FIND: Outlet temperature of the hot water.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible heat loss to surroundings, (2) Constant properties, (3) Negligible
kinetic and potential energy changes.
PROPERTIES: Table A-6, Water ( Tc = (20 + 60)/2 = 40°C ≈ 310 K): cc = cp,f = 4178 J/kg⋅K;
Water ( Th = (80 +60)/2=70°C≈340 K): ch = cp,f = 4188 J/kg⋅K.
ANALYSIS: From an energy balance on the hot fluid, the outlet temperature is
& h c h.
Th,o = Th,i − q / m
(1)
The heat rate can be written in terms of the effectiveness and qmax as
(
q = ε q max = ε Cmin Th,i − Tc,i
)
(2)
where for this HXer, the cold fluid is the minimum fluid giving
(
& ) Th,i − Tc,i
q max = ( mc
c
)
q max = ( 5000/3600 ) kg/s × 4178 J / k g ⋅ K (80 − 20 ) °C = 348.2 kW.
The effectiveness can be determined from Figure 11.16 with
NTU =
UA
11,600 W / K
=
= 2.0
Cmin (5000/3600 ) k g / s × 4178J/kg ⋅ K
giving, ε = 0.7 for Cr = Cmin/Cmax = (5,000 × 4178)/(10,000 × 4188) = 0.499. Combining Eqs. (1) and
(2), find
Th,o = 80°C − 0.7 × 348.2 × 103 W / (10,000/3600 ) k g / s × 4188 J / k g ⋅ K
(
Th,o = ( 80 − 21.0 ) ° C = 59°C.
)
<
& )c (Tc,o – Tc,i), find that Tc,o =
COMMENTS: (1) From an energy balance on the cold fluid, q = ( mc
62°C. For evaluating properties at average mean temperatures, we should use Th = (59 + 80)/2 =
70°C = 343 K and Tc = (20 + 62)/2 = 41°C = 314 K. Note from above that we have indeed assumed
reasonable temperatures at which to obtain specific heats.
(2) We could have also used Eq. 11.31a to evaluate ε using Cr = 0.5 and NTU = 2 to obtain ε = 0.693.
PROBLEM 11.23
KNOWN: Flow rates and inlet temperatures for automobile radiator configured as a cross-flow heat
exchanger with both fluids unmixed. Overall heat transfer coefficient.
FIND: (a) Area required to achieve hot fluid (water) outlet temperature, Tm,o = 330 K, and (b) Outlet
temperatures, Th,o and Tc,o, as a function of the overall coefficient for the range, 200 ≤ U ≤ 400 W/m2⋅K
with the surface area A found in part (a) with all other heat transfer conditions remaining the same as for
part (a).
SCHEMATIC:
ASSUMPTIONS: (1) Negligible heat loss to surroundings and kinetic and potential energy changes, (2)
Constant properties.
(
)
PROPERTIES: Table A.6, Water ( Th = 365 K): cp,h = 4209 J/kg⋅K; Table A.4, Air Tc ≈ 310 K : cp,c
= 1007 J/kg⋅K.
ANALYSIS: (a) The required heat transfer rate is
(
)
h cp,h Th,i − Th,o = 0.05 kg s ( 4209 J kg ⋅ K ) 70 K = 14, 732 W .
q=m
Using the ε -NTU method,
Cmin = Ch = 210.45 W/K
Cmax = Cc = 755.25 W K .
Hence, Cmin/Cmax = 0.279 and
(
)
q max = Cmin Th,i − Tc,i = 210.45 W K (100 K) = 21, 045 W
ε = q q max = 14, 732 W 21, 045 W = 0.700 .
Figure 11.18 yields NTU ≈ 1.5, hence,
A = NTU ( Cmin U ) = 1.5 × 210.45 W K
(200 W m2 ⋅ K ) =1.58 m2 .
<
(b) Using the IHT Heat Exchanger Tool for Cross-flow with both fluids unmixed arrangement and the
Properties Tool for Air and Water, a model was generated to solve part (a) evaluating the efficiency
using Eq. 11.33. The following results were obtained:
A = 1.516 m 2
NTU = 1.441
Tc,o = 319.5K
Using the model but assigning A = 1.516 m2, the outlet temperature Th,o and Tc,o were calculated as a
function of U and the results plotted below.
Continued...
PROBLEM 11.23 (Cont.)
Outlet temperatures, Tco or Tho (K)
350
With a higher U, the outlet temperature of the
hot fluid (water) decreases. A benefit is
enhanced heat removal from the engine block
and a cooler operating temperature. If it is
desired to cool the engine with water at 330 K,
the heat exchanger surface area and, hence its
volume in the engine component could be
reduced.
300
200
300
Overall coefficient, U (W/^2.K)
Cold fluid (air), Tco (K)
Hot fluid (water), Tho (K)
COMMENTS: (1) For the results of part (a), the air outlet temperature is
Tc,o = Tc,i + q Cc = 300 K + (14, 732 W 755.25 W K ) = 319.5 K .
(2) For the conditions of part (a), using the LMTD approach, ∆Tlm = 51.2 K, R = 0.279 and P = 0.7.
Hence, Fig. 11.12 yields F ≈ 0.95 and
(
)
A = q FU∆Tlm = (14, 732W ) 0.95 200 W m 2 ⋅ K 51.2K  = 1.51m 2 .


(3) The IHT workspace with the model to generate the above plot is shown below. Note that it is
necessary to enter the overall energy balances on the fluids from the keyboard.
// Heat Exchanger Tool - Cross-flow with both fluids unmixed:
// For the cross-flow, single-pass heat exchanger with both fluids unmixed,
eps = 1 - exp((1 / Cr) * (NTU^0.22) * (exp(-Cr * NTU^0.78) - 1))
// Eq 11.33
// where the heat-capacity ratio is
Cr = Cmin / Cmax
// and the number of transfer units, NTU, is
NTU = U * A / Cmin
// Eq 11.25
// The effectiveness is defined as
eps = q / qmax
qmax = Cmin * (Thi - Tci)
// Eq 11.20
// See Tables 11.3 and 11.4 and Fig 11.18
// Overall Energy Balances on Fluids:
q = mdoth * cph * (Thi - Tho)
q = mdotc * cpc * (Tco - Tci)
// Assigned Variables:
Cmin = Ch
// Capacity rate, minimum fluid, W/K
Ch = mdoth * cph
// Capacity rate, hot fluid, W/K
mdoth = 0.05
// Flow rate, hot fluid, kg/s
Thi = 400
// Inlet temperature, hot fluid, K
Tho = 330
// Outlet temperature, hot fluid, K; specified for part (a)
Cmax = Cc
// Capacity rate, maximum fluid, W/K
Cc = mdotc * cpc
// Capacity rate, cold fluid, W/K
mdotc = 0.75
// Flow rate, cold fluid, kg/s
Tci = 300
// Inlet temperature, cold fluid, K
U = 200
// Overall coefficient, W/m^2.K
// Properties Tool - Water (h)
// Water property functions :T dependence, From Table A.6
// Units: T(K), p(bars);
xh = 0
// Quality (0=sat liquid or 1=sat vapor)
rhoh = rho_Tx("Water",Tmh,xh)
// Density, kg/m^3
cph = cp_Tx("Water",Tmh,xh)
// Specific heat, J/kg·K
Tmh = Tfluid_avg(Thi,Tho )
// Properties Tool - Air(c)
// Air property functions : From Table A.4
// Units: T(K); 1 atm pressure
rhoc = rho_T("Air",Tmc)
// Density, kg/m^3
cpc = cp_T("Air",Tmc)
// Specific heat, J/kg·K
Tmc = Tfluid_avg(Tci,Tco)
400
PROBLEM 11.24
KNOWN: Flowrates and inlet temperatures of a cross-flow heat exchanger with both fluids unmixed.
Total surface area and overall heat transfer coefficient for clean surfaces. Fouling resistance associated
with extended operation.
FIND: (a) Fluid outlet temperatures, (b) Effect of fouling, (c) Effect of UA on air outlet temperature.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible heat loss to surroundings and negligible kinetic and potential energy
changes, (2) Constant properties, (3) Negligible tube wall resistance.
PROPERTIES: Air and gas (given): cp = 1040 J/kg⋅K.
ANALYSIS: (a) With Cmin = Ch = 1 kg/s × 1040 J/kg⋅K = 1040 W/K and Cmax = Cc = 5 kg/s × 1040
J/kg⋅K = 5200 W/K, Cmin/Cmax = 0.2. Hence, NTU = UA/Cmin = 35 W/m2⋅K(25 m2)/1040 W/K = 0.841
and Fig. 11.18 yields ε ≈ 0.57. With Cmin(Th,i - Tc,i) = 1040 W/K(500 K) = 520,000 W = qmax, Eqs.
(11.21) and (11.22) yield
Th,o = Th,i − ε q max Ch = 800 K − 0.56 (520, 000 W ) 1040 W K = 520 K
Tc,o = Tc,i + ε q max Cc = 300 K + 0.56 (520, 000 W ) 5200 W K = 356 K
(b) With fouling, the overall heat transfer coefficient is reduced to
(
U f = U −1 + R ′′f
)
−1
= ( 0.029 + 0.004 ) m 2 ⋅ K W 

−1

<
<
= 30.3 W m 2⋅ K
This 13.4% reduction in performance is large enough to justify cleaning of the tubes.
<
(c) Using the Heat Exchangers option from the IHT Toolpad to explore the effect of UA, we obtain the
following result.
Air outlet temperature, Tco(K)
400
380
360
340
320
300
500
900
1300
1700
2100
2500
Heat transfer parameter, UA(W/K)
The heat rate, and hence the air outlet temperature, increases with increasing UA, with Tc,o approaching a
maximum outlet temperature of 400 K as UA → ∞ and ε → 1.
COMMENTS: Note that, for conditions of part (a), Eq. 11.33 yields a value of ε = 0.538, which reveals
the level of approximation associated with reading ε from Fig. 11.18.
PROBLEM 11.25
KNOWN: Cooling milk from a dairy operation to a safe-to-store temperature, Th,o ≤ 13°C, using
ground water in a counterflow concentric tube heat exchanger with a 50-mm diameter inner pipe and
2
overall heat transfer coefficient of 1000 W/m ⋅K.
FIND: (a) The UA product required for the chilling process and the length L of the exchanger, (b)
The outlet temperature of the ground water, and (c) the milk outlet temperatures for the cases when
the water flow rate is halved and doubled, using the UA product found in part (a)
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible heat loss to surroundings and kinetic
and potential energy changes, and (3) Constant properties.
PROPERTIES: Table A-6, Water Tc = 287 K, assume Tc,o = 18$ C : ρ = 1000 kg / m3 ,
4
9
c p = 4187 J / kg ⋅ K; Milk (given): ρ = 1030 kg / m3 , c p = 3860 J / kg ⋅ K.
ANALYSIS: (a) Using the effectiveness-NTU method, determine the capacity rates and the
minimum fluid.
Hot fluid, milk:
= 1030 kg / m3 × 250 liter / h × 10-3 m3 / liter × 1 h / 3600 s = 0.0715 kg / s
h = ρ h∀
m
h
h c h = 0.0715 kg / s × 3860 J / kg ⋅ K = 276 W / K
Ch = m
Cold fluid, water:
c cc = 1000 kg / m3 × 0.72 / 3600 m3 / s × 4187 J / kg ⋅ K = 837 W / K
Cc = m
4
9
It follows that Cmin = Ch. The effectiveness of the exchanger from Eqs. 11.19 and 11.21 is
ε=
q
q max
=
3
8 = 138.6 − 136K = 0.895
C min 3Th,i − Tc,i 8 138.6 -106K
C h Th,i − Th,o
(1)
The NTU can be calculated from Eq. 10.30b, where Cr = Cmin/Cmax = 0.330,
NTU =
ε −l
l
n
ε Cr − 1
Cr − 1
NTU =
l
0.895 − l
n
= 2.842
0.330 − 1
0.895 × 0.330 − l
(2)
Continued …..
PROBLEM 11.25 (Cont.)
From Eq. 11.25, find UA
<
UA = NTU ⋅ C min = 2.842 × 276 W / K = 785 W / K
and the exchanger tube length with A = π DL is
L = UA / π DU = 785 W / K / π 0.050 m × 1000 W / m2 ⋅ K = 5.0 m
<
(b) The water outlet temperature, Tc,o, can be calculated from the heat rates,
3
8
3
C h Th,i − Th,o = Cc Tc,o − Tc,i
1
8
6
(3)
3
8
276 W / K 38.6 − 13 K = 837 W / K Tc,o − 10 K
Tc,o = 18.4$ C
<
(c) Using the foregoing Eqs. (1 - 3) in the IHT workspace, the hot fluid (milk) outlet temperatures are
evaluated with UA = 785 W/K for different water flow rates. The results, including the hot fluid
outlet temperatures, are compared to the base case, part (a).
Case
Cc (W/K)
1, halved flow rate
Base, part (a)
2, doubled flow rate
419
837
1675
Tc,o (°C)
Th,o (°C)
14.9
13
12.3
25.6
18.4
14.3
COMMENTS: (1) From the results table in part (c), note that if the water flow rate is halved, the
milk will not be properly chilled, since Tc,o = 14.9°C > 13°C. Doubling the water flow rate reduces
the outlet milk temperature by less than 1°C.
(2) From the results table, note that the water outlet temperature changes are substantially larger than
those of the milk with changes in the water flow rate. Why is this so? What operational advantage is
achieved using the heat exchanger under the present conditions?
(3) The water thermophysical properties were evaluated at the average cold fluid temperature,
3
8
Tc = Tc,i + Tc,o / 2. We assumed an outlet temperature of 18°C, which as the results show, was a
good choice. Because the water properties are not highly temperature dependent, it was acceptable to
use the same values for the calculations of part (c). You could, of course, use the properties function
in IHT that will automatically use the appropriate values.
PROBLEM 11.26
KNOWN: Flow rate, inlet temperatures and overall heat transfer coefficient for a regenerator.
Desired regenerator effectiveness. Cost of natural gas.
FIND: (a) Heat transfer area required for regenerator and corresponding heat recovery rate and outlet
temperatures, (b) Annual energy and fuel cost savings.
SCHEMATIC:
ASSUMPTIONS: (a) Negligible heat loss to surroundings, (b) Constant properties.
PROPERTIES: Table A-6, water ( Tm ≈ 310K ) : c p = 4178 J / kg ⋅ K.
ANALYSIS: (a) With Cr = 1 and ε = 0.50 for one shell and two tube passes, Eq. 11.31c yields E =
1.414. With Cmin = 5 kg/s × 4178 J/kg⋅K = 20,890 W/K, Eq. 11.31b then yields
ln ( E − 1) / ( E + l )
C
20,890 W / K ln (0.171)
A = − min 
=−
= 13.05 m 2
1/
2
2
U
2000 W / m ⋅ K 1.414
l + C2r
)
(
<
With ε = 0.50, the heat recovery rate is then
(
)
q = ε Cmin Th,i − Tc,i = 679, 000 W
<
and the outlet temperatures are
Tc,o = Tc,i +
q
679, 000 W
= 5°C +
= 37.5°C
Cc
20,890 W / K
<
Th,o = Th,i −
q
679, 000 W
= 70°C −
= 37.5°C
Ch
20,890 W / K
<
(b) The amount of energy recovered for continuous operation over 365 days is
∆E = 679, 000 W × 365d / yr × 24 h / d × 3600s / h = 2.14 × 1013 J / yr
The annual fuel savings SA is then
SA =
∆E × Cng
η
2.14 ×107 MJ / yr × $0.0075 / MJ
=
= $178, 000 / yr
0.9
<
COMMENTS: (1) With Cc = Ch, the temperature changes are the same for the two fluids, (2) A
larger effectiveness and hence a smaller value of A can be achieved with a counterflow exchanger
(compare Figs. 11.15 and 11.16 for Cr = 1), (c) The savings are significant and well worth the cost of
the heat exchanger. An additional benefit is that, with Th,o reduced from 70 to 37.5°C, less energy is
consumed by the refrigeration system used to restore it to 5°C.
PROBLEM 11.27
KNOWN: Heat exchanger in car operating between warm radiator fluid and cooler outside air.
−0.2
& air
Effectiveness of heater is ε ~ m
since water flow rate is large compared to that of the air. For
low-speed fan condition, heat warms outdoor air from 0°C to 30°C.
& air to be doubled
FIND: (a) Increase in heat added to car for high-speed fan condition causing m
while inlet temperatures remain the same, and (b) Air outlet temperature for medium-speed fan
condition where air flow rate increases 50% and heat transfer increases 20%.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible heat losses from heat exchanger to surroundings, (2) Th,i and Tc,i
remain fixed for all fan-speed conditions, (3) Water flow rate is much larger than that of air.
ANALYSIS: (a) Assuming the flow rate of the water is much larger than that of air,
& air cp,c.
Cmin = Cc = m
Hence, the heat rate can be written as
(
)
(
)
& airc p,air Th,i − Tc,i .
q = ε q max = ε Cmin Th,i − Tc,i = ε ⋅ m
Taking the ratio of the heat rates for the high and low speed fan conditions, find
& 0.8
m
& air )
air hi
qhi ( ε m
0.8
hi =
=
= 2 = 1.74
& air )
qlo ( ε m
& 0.8
m
lo
air lo
( )
( )
<
0.2
& −air
where we have used ε ~ m
and recognized that for the high speed fan condition, the air flow rate
is doubled. Hence the heat rate is increased by 74%.
(b) Considering the medium and low speed conditions, it was observed that,
( m& air ) med
( m& air ) lo
qmed
= 1.2
qlo
= 1.5.
To find the outlet air temperature for the medium speed condition,
(
)
&

qmed  m airc p,c Tc,o − Tc,i  med
=
qlo
m
& c
T
T 
 air p,c c,o − c,i  lo
(
1.2 =
(
& airc p,c Tc,o − 0° C
1.5 m
& air cp,c ( 30 − 0° C)
m
)
)
Tc,o = 24°C.
<
PROBLEM 11.28
KNOWN: Counterflow heat exchanger formed by two brazed tubes with prescribed hot and cold
fluid inlet temperatures and flow rates.
FIND: Outlet temperature of the air.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible loss/gain from tubes to surroundings, (2) Negligible changes in
kinetic and potential energy, (3) Flow in tubes is fully developed since L/Dh = 40 m/0.030m = 1333.
-6
2
PROPERTIES: Table A-6, Water ( Th = 335 K): ch = cp,h = 4186 J/kg⋅K, µ = 453 × 10 N⋅s/m , k
-7
2
= 0.656 W/m⋅K, Pr = 2.88; Table A-4, Air (300 K): cc = cp,c = 1007 J/kg⋅K, µ = 184.6 × 10 N⋅s/m , k
= 0.0263 W/m⋅K, Pr = 0.707; Table A-1, Nickel ( T = (23 + 85)°C/2 = 327 K): k = 88 W/m⋅K.
ANALYSIS: Using the NTU - ε method, from Eq. 11.30a,
ε=
1 − exp  − NTU (1 − Cr ) 
NTU = U A / Cmin
1 − C r exp − NTU (1− Cr ) 
C r =C min / C max.
(1,2,3)
Estimate UA from a model of the tubes and flows, and determine the outlet temperature from the
expression
(
)
(
)
ε = Cc Tc,o − Tc,i / C min T h,i − Tc,i .
Water-side:
Re D =
(4)
&h
4m
4 × 0.04 k g / s
=
= 11,243.
π Dµ π × 0.010m × 453× 10−6 N ⋅ s / m 2
The flow is turbulent and since fully developed, use the Dittus-Boelter correlation,
0.3 = 0.023 11,243
Nu h = h h D / k = 0.023Re0.8
(
)
D Pr
0.8
( 2.88 ) 0.3 = 54.99
hh = 54.99 × 0.656 W / m ⋅K / 0.01m = 3,607 W / m 2 ⋅ K.
Air-side:
Re D =
&c
4m
4 × 0.120 k g / s
=
= 275,890.
π Dµ π × 0.030m × 184.6 × 10−7 N ⋅ s / m 2
The flow is turbulent and since fully developed, again use the correlation
0.8 Pr 0.4 = 0.023 275,890 0.8 0.707 0.4 = 450.9
Nu c = hc D / K = 0.023ReD
(
) (
)
hc = 450.9 × 0.0263 W / m ⋅ K/0.030m = 395.3 W / m 2 ⋅ K.
Overall coefficient: From Eq. 11.1, considering the temperature effectiveness of the tube walls and
the thermal conductance across the brazed region,
Continued …..
PROBLEM 11.28 (Cont.)
1
1
1
1
=
+
+
UA (ηo hA ) h K′t L (ηo hA ) c
(5)
where ηo needs to be evaluated for each of the tubes.
2
Water-side temperature effectiveness: A h = π Dh L = π ( 0.010m ) 40m = 1.257m
(
ηo,h = ηf,h = tanh ( mLh ) /mL h
(
m = hh P / k A
m = 3607 W / m 2 ⋅ K / 8 8 W / m ⋅ K × 0.002m
-1
)
1/2
)
1/2
= ( h h /kt )1/2
= 143.2m −1
-1
and with Lh = 0.5 πDh, ηo,h = tanh(143.2 m × 0.5 π × 0.010m)/143.2 m × 0.5 π × 0.010 m = 0.435.
2
Ac = πDcL = π(0.030m)40m = 3.770 m
Air-side temperature effectiveness:
(
ηo,c = ηf,c = tanh ( mL c ) /mLc m = 395.3 W / m2 ⋅ K / 8 8 W / m ⋅ K × 0.002m
-1
)
1/2
= 47.39m−1
-1
and with Lc = 0.5πDc, ηo,c = tanh(47.39 m × 0.5 π × 0.030m)/47.39 m × 0.5 π × 0.030m = 0.438.
Hence, the overall heat transfer coefficient using Eq. (5) is
1
UA
1
=
2
0.435 × 3607 W / m ⋅ K × 1.257m
2
+
1
100 W / m⋅ K( 40m )
UA =  5.070 ×10−4 + 2.50 ×10−4 +1.533 × 10−3 


1
+
−1
2
2
0.438 × 395.3 W / m ⋅ K × 3.770m
W / K = 437 W / K .
Evaluating now the heat exchanger effectiveness from Eq. (1) with
}
& h ch = 0.040kg/s × 4186 J / k g ⋅ K = 167.4 W / K ← Cmax
Ch = m
& ccc = 0.120kg/s × 1007 J / k g ⋅ K = 120.8 W / K ← Cmin Cr = Cmin / Cmax = 0.722
Cc = m
NTU =
1 − exp  −3.62 (1 − 0.722 ) 
UA
437 W / K
=
= 3.62 ε =
= 0.862
C min 120.8 W / K
1 − 0.722 exp  −3.62 (1 − 0.722 ) 
and finally from Eq. (4) with Cmin = Cc,
0.862 =
(
Cc Tc,o − 23 °C
Cc (85 − 23 ) °C
)
Tc,o = 76.4 °C
<
COMMENTS: (1) Using overall energy balances, the water outlet temperature is
(
)
Th,o = Th,i + ( Cc / C h ) Tc,o − Tc,i = 85°C − 0.722 ( 76.4 − 23) ° C = 46.4°C.
(2) To initially evaluate the properties, we assumed that Th ≈ 335 K and Tc ≈ 300 K. From the
calculated values of Th,o and Tc,o, more appropriate estimates of Th and Tc are 338 K and 322 K,
respectively. We conclude that proper thermophysical properties were used for water but that the
estimates could be improved for air.
PROBLEM 11.29
= 0.003 kg/s. Cold
KNOWN: Twin-tube counterflow heat exchanger with balanced flow rates, m
airstream enters at 280 K and must be heated to 340 K. Maximum allowable pressure drop of cold
airstream is 10 kPa.
FIND: (a) Tube diameter D and length L which satisfies the heat transfer and pressure drop
requirements, and (b) Compute and plot the cold stream outlet temperature Tc,o, the heat rate q, and
pressure drop ∆p as a function of the balanced flow rate from 0.002 to 0.004 kg/s.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible heat loss to surroundings, (3) Average
pressure of the airstreams is 1 atm, (4) Tube walls act as fins with 100% efficiency, (4) Fully developed
flow.
PROPERTIES: Table A.4, Air ( Tm = 310 K, 1 atm) : ρ = 1.128 kg/m3, cp = 1007 J/kg⋅K, µ = 18.93 ×
10-6 m2 / s, k = 0.0270 W/m⋅K, Pr = 0.7056.
ANALYSIS: (a) The heat exchanger diameter D and length L can be specified through two analyses: (1)
heat transfer based upon the effectiveness-NTU method to meet the cold air heating requirement and (2)
pressure drop calculation to meet the requirement of 10 kPa. The heat transfer analysis begins by
determining the effectiveness from Eq. 11.22, since Cmin = Cmax and Cr = 1,
ε=
q
q max
=
(
) = (340 − 280 ) K = 0.750
C ( Th,i − Tc,i ) (360 − 280 ) K
C Tc,o − Tc,i
From Table 11.4, Eq. 11.29b for Cr = 1,
0.750
ε
NTU =
=
=3
1 − ε 1 − 0.750
where NTU, following its definition, Eq. 11.25, is
UA
NTU =
Cmin
with
Cmin = mc
p = 0.003 kg s × 1007 J kg ⋅ K = 3.021K W
(1)
(2)
(3)
(4)
Continued...
PROBLEM 11.29 (Cont.)
and 1 UA represents the thermal resistance between the two fluids at Tm,h and Tm,c as illustrated in the
above-right schematic. Since the tube walls are isothermal, it follows that
1 UA = 1 hc A + 1 h h A
(5)
and since the flow conditions are nearly identical hc = h h so that
U = 0.5h
(6)
where the heat transfer area is
A = π DL
(7)
Hence, Eq. (3) can now be expressed as
3=
0.5h (π DL )
3.021K W
hDL = 5.7697
(8)
Assuming an average mean temperature Tm,c = 310 K , characterize the flow with
Re D =
4m
π Dµ
=
4 × 0.003 kg s
=
π × D × 18.93 × 10−6 m 2 s
201.78
(9)
D
and assuming the flow is both turbulent and fully developed using the Dittus-Boelter correlation, Eq.
8.57, with n = 0.4,
hD
0.4
= 0.023 Re0.8
Nu D =
D Pr
k
hD = 0.023 × 0.0270 W m ⋅ K ( 201.78 D )
0.8
(0.7056 )0.4
hD1.8 = 0.0377
The pressure drop for fully developed flow, Eq. 8.22a, is
ρ u 2m
L
2D
2
/(ρπD /4) so that
where the mean velocity is um = m
∆p = f
)
(
2
ρπ D
4ρ m
L
∆p = f
∆p =
2D
2
( 0.003 kg s )2
=
2
π2
f
(10)
(11)
m2 ρ 2 L
D5
(1.128 kg m3 ) L = 2.3206 ×10−6 f L D−5
(12)
f
D5
π2
Recall that the pressure drop requirement is ∆p = 10 kPa = 104 N/m2 , so that Eq. (12) can be rewritten
as
fLD −5 = 4.3092 × 1010
(13)
Continued...
PROBLEM 11.29 (Cont.)
For the Reynolds number range, 3000 ≤ ReD ≤ 5 ×106 , Eq. 8.21 provides an estimate for the friction
factor,
f = ( 0.790"n ( ReD ) − 1.64 )
−2
f = ( 0.790"n ( 201.78 D ) − 1.46 )
−2
(14)
In the foregoing analysis, there are 4 unknowns (D, L, f, h ) and 4 equations (8, 10, 13, 14). Using the
IHT workspace, find
D = 8.96 mm
L = 3.52 m
h = 182.9 W m 2 ⋅ K
f = 0.02538
For this configuration, ReD = 22,520 so the flow is turbulent and since L/D = 3.52/0.00896 = 390 >> 10,
the fully developed assumption is reasonable.
(b) The foregoing analysis entered into the IHT workspace was used to determine Tc,o , q and ∆p as a
.
function of the balanced flow rate, m
250
16
200
q (W)
deltap (kPa)
20
12
150
8
4
0.002
100
0.003
0.004
Flow rate, mdot (kg/s)
0.002
0.003
0.004
Flow rate, modt (kg/s)
Tco (K)
345
340
335
0.002
0.003
0.004
Flow rate, mdot (kg/s)
The outlet temperature of the cold air, Tc,o , is nearly insensitive to the flow rate. It follows that the heat
plot above. The
rate, q, must be nearly proportional to the flow rate as can be seen in the q vs. m
pressure drop varies with the mean velocity squared.
PROBLEM 11.30
KNOWN: Dimensions and thermal conductivity of twin-tube, counterflow heat exchanger. Contact
resistance between tubes. Air inlet conditions for one tube and pressure of saturated steam in other
tube.
FIND: Air outlet temperature and condensation rate.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible heat exchange with surroundings, (2) Negligible kinetic and
potential energy and flow work changes, (3) Fully developed air flow, (4) Negligible fouling, (5)
Constant properties.
(
)
PROPERTIES: Table A-4, air Tc ≈ 325 K, p = 5atm : cp = 1008 J/kg⋅K, µ = 196.4 × 10
-7
2
N⋅s/m , k = 0.0281 W/m⋅K, Pr = 0.703. Table A-6, sat. steam (p = 2.455 bar): Th,i = Th,o = 400 K,
hfg = 2183 kJ/kg.
ANALYSIS: With Cmax → ∞, Cr = 0 and Eqs. 11.22 and 11.36a yield
Tc,o − Tc,i
= 1 − exp ( − NTU )
(1)
R′
1
1
1
=
+ t+
UA (ηo hA )c L (ηo hA )h
(2)
ε=
Th,i − Tc,i
From Eq. 11.1,
/ π Di µ = 0.12 kg / s / π (0.05m )196.4 × 10−7 N ⋅ s / m 2 = 38,900, the air flow is
With Re D = 4m
turbulent and the Dittus-Boelter correlation yields
 k 
4 / 5 0.4
 0.023 Re D Pr
D
 i
h c ≈ h fD = 
 0.0281 W / m ⋅ K  0.023 (38, 900 )4 / 5 ( 0.703 )0.4 = 52.7 W / m 2 ⋅ K

0.05m


=
As shown on the inset, each tube wall may be modelled as two fins, each of length Lf ≈ π Di/2 =
2
0.0785m. The total surface area for heat transfer is At = π DiL = 0.785 m = Ac, which is equivalent
to the surface area of the fins. With NAf = At from Eq. 3.102, ηo = ηf. Because the outer surface of
1/2
the tube is insulated, a wall thickness of 2t must be used in evaluating ηf. With m = (2h/k × 2t) =
1/2
-1
1/2
2
(h/kt) = [52.7 W/m ⋅K/(60 W/m⋅K × 0.004m)] = 14.8 m , Lc = Lf for an adiabatic tip, and mLf =
1.163, Eq. 3.89 yields
ηf =
tanh mLf 0.821
=
= 0.706 = ηo,c
mLf
1.163
Continued …..
PROBLEM 11.30 (Cont.)
Similarly, for the steam tube, m = (h/kt)
and mLf = 11.33. Hence,
ηf =
1/2
2
= [5,000 W/m ⋅K/(60 W/m⋅K × 0.004m)]
1/2
-1
= 144.3 m
tanh mLf
1.00
=
= 0.088 = ηo,h
mLf
11.33
Substituting into Eq. (2),
1
0.01
1


UA = 
+
+
5
0.088 × 5000 × 0.785 
 0.706 × 52.7 × 0.785
(
−1
W
W
= 25.6
K
K
)
cp = 0.03 kg / s × 1008 J / kg ⋅ K = 30.2 W / K, NTU = UA/Cmin =
Hence, with Cmin = m
c
0.847 and ε = 1 – exp (-NTU) = 0.571. From Eq. (1), the air outlet temperature is then
(
)
Tc,o = Tc,i + ε Th,i − Tc,i = 17°C + 0.571(127 − 17 ) °C = 79.8°C
<
The rate of heat transfer to the air is
(
)
cp Tc,o − Tc,i = 0.03 kg / s × 1008 J / kg ⋅ K × 62.8°C = 1900 W
q=m
and the rate of condensation is
cond =
m
q
1900 W
=
= 8.70 × 10−4 kg / s
h fg 2.183 ×106 J / kg
<
COMMENTS: (1) With Tc = 321.4 K, the initial estimate of 325K is reasonable and iteration on
the property values is not necessary, (2) The major contribution to the total thermal resistance
is due to air-side convection, (3) The foregoing results are independent of air pressure.
PROBLEM 11.31
KNOWN: Tube inner and outer diameters and longitudinal and transverse pitches for a cross-flow heat
exchanger. Number of tubes in transverse plane. Water and gas flow rates and inlet temperatures.
Water outlet temperature.
FIND: (a) Gas outlet temperature and number of longitudinal tube rows, (b) Effect of gas flowrate and
inlet temperature on fluid outlet temperatures.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible heat loss to surroundings, (2) Negligible kinetic and potential energy
changes, (3) Constant properties, (4) Negligible fouling.
PROPERTIES: Table A.6, Water ( Tc = 320 K: cp = 4180 J/kg⋅K, µ = 577 × 10-6 N⋅s/m2, kf = 0.640
W/m⋅K, Pr = 3.77; Table A.4, Air ( Th ≈ 550 K): cp = 1040 J/kg⋅K, µ = 288.4 × 10-7 N⋅s/m2, k = 0.0439
W/m⋅K, Pr = 0.683, ρ = 0.633 kg/m3.
ANALYSIS: (a) The required heat transfer rate is
(
)
7
c
q=m
c p,c Tc,o − Tc,i = 50 kg s ( 4180 J kg ⋅ K ) 60 K = 1.254 × 10 W .
h cp,h ,
Hence, with Th,o = Th,i - q m
<
Th,o = 700 K − 1.254 × 107 W ( 40 kg s × 1040 J kg ⋅ K ) = 398.6 K
Use the ε - NTU method to compute the hot side HX surface area, AH. To calculate Uh, we must find hh.
For the tube bank, SD = 44.7 mm > (ST + D)/2 = 30 mm. Hence, with ρVmax = [ST (ST − Do )] ρ V =
[ST (ST − Do )]( m h
WH ) ,
ρ Vmax = ( 40 20 )  40 kg s

( 2 ×1.2 ) m 2  = 33.3 kg
s ⋅ m2
Re D,max = ( ρ Vmax Do ) µ = 33.3 kg s ⋅ m 2 (0.02 m ) 288.4 × 10−7 N ⋅ s m 2 = 23,116 .


From the Zhukauskas correlation, with (Pr/Prs) ≈ 1, and Table 7.7,
0.36
Nu D = 0.35 Re0.6
= 0.35 ( 23,116 )
D Pr
where it is assumed that NL > 20. Hence,
0.6
(0.683)0.36 = 127
h h = Nu D ( k Do ) = 127 ( 0.0439 W m⋅ K 0.02 m ) = 279 W m 2⋅ K .
From Eq. 11.1,
Continued...
PROBLEM 11.31 (Cont.)
1
Uh
1
Uh
=
1
20 0.02 m ln ( 20 15 )
1
1 Do Do ln ( Do Di ) 1
=
+
+
+
+
60 W m⋅ K
h c Di
2k
hh
3000 W m 2⋅ K 15
279 W m2⋅ K
)
(
= 4.44 × 10−4 + 9.59 × 10−5 + 3.58 × 10−3 m 2 ⋅ K W = 4.12 × 10−3 m 2 ⋅ K W
U h = 243 W m2⋅ K .
With Ch = Cmin = 4.160 × 104 W/K and Cc = Cmax = 2.09 × 105 W/K, Cmin/Cmax = 0.199 and qmax = Cmin(Th,i
- Tc,i) = 4.16 × 104 W/K(410 K) = 1.71 × 107 W. Hence, ε = (q/qmax) = (1.254 × 107 W/1.71 × 107 W) =
0.735. With Cmin mixed and Cmax unmixed, Eq. 11.35b gives NTU = 1.54 and
(
)
A h = NTU ( Cmin U h ) = 1.54 4.160 × 104 W K 243 W m 2⋅ K = 264 m 2 .
Hence, N L =
Ah
(π Do W ) NT
264 m 2
=
<
= 70
π ( 0.02 ) 2 ( 30 ) m
(b) Using the IHT Correlations, Heat Exchangers and Properties Toolpads to perform the parametric
calculations, we obtain the following results for NL = 90.
2
400
340
Gas outlet temperature, Tho(K)
Water outlet temperature, Tco(K)
350
330
320
310
300
20
24
28
32
36
40
380
360
340
320
20
24
Gas flowrate, mdoth (kg/s)
28
32
36
40
Gas flowrate, mdoth (kg/s)
Thi = 700 K
Thi = 600 K
Thi = 500 K
Thi = 700 K
Thi = 600 K
Thi = 500 K
h , q, and hence, Tc,o, increases with increasing m
h , as well as
Since hh, and hence Uh, increases with m
a
, the proportionality is not linear (q α m
, where a <
with increasing Th,i. Although q increases with m
h
h
, in which case Th,o must increase. From the above
1) and (Th,i - Th,o) must decrease with increasing m
h
results, it is clear that operation is restricted to m
h ≥ 40 kg/s and Th,i ≥ 700 K, if corrosion of the heat
exchanger surfaces is to be avoided.
COMMENTS: To check the presumed value of hc = 3000 W/m2⋅K, compute
4 (m
4 (50 kg s ) 70 × 30
c N)
=
= 3500 .
Re D =
π Di µ
π ( 0.015 m ) 577 × 10−6 N ⋅ s m 2
Hence, Nu D = 0.023 Re 4D/ 5 Pr 0.4 = 0.023 (3500 )
4/5
(3.77 )0.4 = 26.8
h c = ( k D ) Nu D = ( 0.640 W m⋅ K 0.015 m ) 26.8 = 1142 W m 2⋅ K .
Hence, the cold side convection coefficient has been overestimated and the calculations should be
repeated using the smaller value of hc.
PROBLEM 11.32
KNOWN: Single pass, cross-flow heat exchanger with hot exhaust gases (mixed) to heat water
(unmixed)
FIND: Required surface area.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible heat loss to surroundings, (2) Negligible kinetic and potential energy
changes, (3) Exhaust gas properties assumed to be those of air.
PROPERTIES: Table A-6, Water ( Tc = (80 + 30)°C/2 = 328 K): cp = 4184 J/kg⋅K; Table A-4, Air
(1 atm, Th = (100 + 225)°C/2 = 436 K): cp = 1019 J/kg⋅K.
ANALYSIS: The rate equation for the heat exchanger follows from Eqs. 11.14 and 11.18. The area
is given as
A = q / U∆Tlm = q / U F∆Tlm,CF
(1)
where F is determined from Fig. 11.13 using
P=
80 − 30
225 − 100
= 0.26 and R =
= 2.50 giving F ≈ 0.92.
225 − 30
80 − 30
(2)
From an energy balance on the cold fluid, find
(
)
& c cc Tc,o − Tc,i = 3
q=m
kg
J
× 4184
( 80 − 30 ) K = 627,600 W.
s
kg ⋅ K
(3)
From Eq. 11.15, the LMTD for counter-flow conditions is
∆Tl m,CF =
( 225 − 80 ) − (100 − 30)
∆T1 − ∆T2
=
° C = 103.0°C.
ln ( ∆T1 / ∆T2 )
ln (145/70 )
(4)
Substituting numerical values resulting from Eqs. (2-4) into Eq. (1), find the required surface area to be
A = 627,600 W / 2 0 0 W / m 2 ⋅K × 0.92 ×103.0K = 33.1m 2.
<
COMMENTS: Note that the properties of the exhaust gases were not needed in this method of
analysis. If the ε-NTU method were used, find first Ch/Cc = 0.40 with Cmin = Ch = 5021 W/K. From
Eqs. 11.19 and 11.20, with Ch = Cmin, ε = q/qmax = (Th,i – Th,o)/(Th,i – Tc,i) = (225 – 100)/(225 – 30)
= 0.64. Using Fig. 11.19 with Cmin/Cmax = 0.4 and ε = 0.64, find NTU = UA/Cmin ≈ 1.4. Hence,
A = NTU ⋅ Cmin / U ≈ 1.4 × 5021W/K/200 W / m 2 ⋅ K = 35.2m 2 .
Note agreement with above result.
PROBLEM 11.33
KNOWN: Concentric tube heat exchanger operating in parallel flow (PF) conditions with a thin-walled
separator tube of 100-mm diameter; fluid conditions as specified.
FIND: (a) Required length for the exchanger; (b) Convection coefficient for water flow, assumed to be
fully developed; (c) Compute and plot the heat transfer rate, q, and fluid inlet temperatures, Th,o and Tc,o,
as a function of the tube length for 60 ≤ L ≤ 400 m with the PF arrangement and overall coefficient
!U = 200 W m ⋅ K& , inlet temperatures (T
2
h,i
= 225°C and Tc,i = 30°C), and fluid flow rates from Problem
11.23; (d) Reduction in required length relative to the value found in part (a) if the exchanger were
operated in the counterflow (CF) arrangement; and (e) Compute and plot the effectiveness and fluid
outlet temperatures as a function of tube length for 60 ≤ L ≤ 400 m for the CF arrangement of part (c).
SCHEMATIC:
ASSUMPTIONS: (1) No losses to surroundings, (2) Negligible kinetic and potential energy changes,
(3) Separation tube has negligible thermal resistance, (4) Water flow is fully developed, (5) Constant
properties, (6) Exhaust gas properties are those of atmospheric air.
PROPERTIES: Table A-4, Hot fluid, Air (1 atm, T = (225 +100)°C /2 = 436 K): cp = 1019 J/kg⋅K;
Table A-6, Cold fluid, Water T = (30 + 80)°C /2 ≈ 328 K): ρ = 1/vf = 985.4 kg/m3, cp = 4183 J/kg⋅K, k =
0.648 W/m⋅K, µ = 505 × 10-6 N⋅s/m2, Pr = 3.58.
ANALYSIS: (a) From the rate equation, Eq. 11.14, with A = πDL, the length of the exchanger is
L = q U ⋅ π D ⋅ ∆T"n,PF .
(1)
The heat rate follows from an energy balance on the cold fluid, using Eq. 11.7, find
(
)
3
c
q=m
c c Tc,o − Tc,i = 3 kg s × 4183 J kg ⋅ K (80 − 30 ) K = 627.5 × 10 W .
h for later use.
Using an energy balance on the hot fluid, find m
(
)
3
h = q c h Th,i − Th,o = 627.5 × 10 W 1019 J kg ⋅ K ( 225 − 100 ) K = 4.93 kg s
m
For parallel flow, Eqs. 11.15 and 11.16,
∆T"m,PF =
∆T1 − ∆T2
"n∆T1
∆T2
(2)
( 225 − 30 )$ C − (100 − 80 )$ C 76.8$ C
.
=
=
"n ( 225 − 30 ) (100 − 80 )
Substituting numerical values into Eq. (1), find
L = 627.5 × 103 W 200 W m 2 ⋅ K (π × 0.1m ) 76.8K = 130m .
<
Continued...
PROBLEM 11.33 (Cont.)
(b) Considering the water flow within the separator tube, from Eq. 8.6,
(
)
−6
π Dµ = 4 × 3 kg s π × 0.1m × 505 × 10
Re D = 4m
N s ⋅ m 2 = 75, 638 .
Since ReD > 2300, the flow is turbulent and since flow is assumed to be fully developed, use the DittusBoelter correlation with n = 0.4 for heating,
0.8
0.4
Nu D = 0.023 Re0.8
= 0.023 ( 75, 638 )
D Pr
h = Nu D
k
D
(3.58 )0.4 = 306.4
= 306.4 × 0.648 W m ⋅ K ( 0.1m ) = 1985 W m 2 ⋅ K .
<
(c) Using the IHT Heat Exchanger Tool, Concentric Tube, Parallel Flow, Effectiveness relation, and the
Properties Tool for Water and Air, a model was developed for the PF arrangement. With U = 200
W/m2⋅K and prescribed inlet temperatures, Th,i = 225°C and Tc,i = 30°C, the outlet temperatures, Th,o and
Tc,o and heat rate, q, were computed as a function of tube length L.
Temperature (C) or Heat rate (W*10^-4)
Parallel flow arrangement
140
115
90
65
40
60
120
180
240
300
360
Tube length, L (m)
Cold outlet temperature, Tco (C)
Hot outlet temperature, Tho (C)
Heat rate, q*10^-4 (W)
As the tube length increases, the outlet temperatures approach one another and eventually reach Th,o =
Tc,o = 85.6°C.
(d) If the exchanger as for part (a) is operated in
counterflow (rather than parallel flow), the log
mean temperature difference is
∆T"m,CF =
∆T"m,CF =
∆Ti − ∆T2
"n∆T1
∆T2
( 225 − 80 ) − (100 − 30 )
= 103.0$ C .
"n ( 225 − 80 ) 100 − 30
Using Eq. (1), the required length is
L = 627.5 × 103 W 200 W m 2 ⋅ K × π × 0.1m × 103.0 K = 97 m .
The reduction in required length of CF relative to PF operation is
Continued...
PROBLEM 11.33 (Cont.)
<
∆L = ( L PF − LCF ) L PF = (103 − 97 ) 103 = 5.8%
(e) Using the IHT Heat Exchanger Tool, Concentric Tube, Counterflow, Effectiveness relation, and the
Properties Tool for Water and Air, a model was developed for the CF arrangement. For the same
conditions as part (c), but CF rather than PF, the effectiveness and fluid outlet temperatures were
computed as a function of tube length L.
Counterflow arrangement
Tho, Tco (C) or eps*100
140
120
100
80
60
40
20
60
120
180
240
300
360
Tube length, L (m)
Cold outlet temperature, Tco (C)
Hot outlet temperature, Tho (C)
Effectiveness, eps*100
Note that as the length increases, the effectiveness tends toward unity, and the hot fluid outlet
temperature tends toward Tc,i = 30°C. Remember the heat rate for an infinitely long CF heat exchanger is
qmax and the minimum fluid (hot in our case) experiences the temperature change, Th,i - Tc,i.
COMMENTS: (1) As anticipated, the required length for CF operations was less than for PF operation.
(2) Note that U is substantially less than hi implying that the gas-side coefficient must be the controlling
thermal resistance.
PROBLEM 11.34
KNOWN: Cross-flow heat exchanger (both fluids unmixed) cools blood to induce body hypothermia
using ice-water as the coolant.
(liter/min), (c) Surface area of
FIND: (a) Heat transfer rate from the blood, (b) Water flow rate, ∀
c
the exchanger, and (d) Calculate and plot the blood and water outlet temperatures as a function of the
≤ 4 liter/min, assuming all other parameters remain
water flow rate for the range, 2 ≤ ∀
unchanged.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible losses to the surroundings and kinetic
and potential energy changes, (3) Overall heat transfer coefficient remains constant with water flow
rate changes, and (4) Constant properties.
2
7
PROPERTIES: Table A-6, Water Tc = 280K , ρ = 1000 kg / m3 , c = 4198 J / kg ⋅ K. Blood
(given): ρ = 1050 kg / m3 , c = 3740 J / kg ⋅ K.
ANALYSIS: (a) The heat transfer rate from the blood is calculated from an energy balance on the
hot fluid,
= 1050 kg / m3 × 5 liter / min × 1 min / 60 s × 10−3 m3 / liter = 0.0875 kg / s
h = ρ h∀
m
h
1
3
6
1
8
6
h c h Th,i − Th,o = 0.0875 kg / s × 3740 J / kg ⋅ K 37 − 25 K = 3927 W
q=m
< (1)
(b) From an energy balance on the cold fluid, find the coolant water flow rate,
3
c cc Tc,o − Tc,i
q=m
8
(2)
c × 4198 J/kg ⋅ K (15 − 0 ) K
3927 W = m
c = 0.0624 kg/s
m
3
3
3
=m
∀
c c / ρ c = 0.0624 kg / s / 1000 kg / m × 10 liter / m × 60 s / min = 3.74 liter / min
<
(c) The surface area can be determined using the effectiveness-NTU method. The capacity rates for
the exchanger are
h c h = 327 W / K
Ch = m
c c c = 262 W / K
Cc = m
C min = Cc
(3, 4, 5)
From Eq. 11.19 and 11.20, the maximum heat rate and effectiveness are
Continued …..
PROBLEM 11.34 (Cont.)
3
1
8
6
q max = C min Th,i − Tc,i = 262 W / K 37 − 0 K = 9694 W
(6)
ε = q / q max = 3927 / 9694 = 0.405
(7)
For the cross flow exchanger, with both fluids unmixed, substitute numerical values into Eq. 11.33 to
find the number of transfer units, NTU, where C r = C min / C max .
1
!
6
L"$#
J
ε = 1 − exp 1/ C r NTU0.22 exp − C r NTU0.78 − 1
(8)
NTU = 0.691
From Eq. 11.25, find the surface area, A.
NTU = UA / C min
A = 0.691 × 262 W / K / 750 W / m2 ⋅ K = 0.241 m2
<
(d) Using the foregoing equations in the IHT workspace, the blood and water outlet temperatures, Th,o
and Tc,o, respectively, are calculated and plotted as a function of the water flow rate, all other
parameters remaining unchanged.
Temperature, Tco and Tho (C)
Outlet temperatures for blood flow rate 5 liter/min
28
26
24
22
20
18
16
14
2
2.5
3
3.5
4
Water flow rate, Qc (liter/min)
Water outlet temperature, Tco
Blood outlet temperature, Tho
From the graph, note that with increasing water flow rate, both the blood and water outlet
temperatures decrease. However, the effect of the water flow rate is greater on the water outlet
temperature. This is an advantage for this application, since it is desirable to have the blood outlet
temperature relatively insensitive to changes in the water flow rate. That is, if there are pressure
changes on the water supply line or a slight miss-setting of the water flow rate controller, the outlet
blood temperature will not change markedly.
PROBLEM 11.35
KNOWN: Steam at 0.14 bar condensing in a shell and tube HXer (one shell, two tube passes consisting
of 130 brass tubes off length 2 m, Di = 13.4 mm, Do = 15.9 mm). Cooling water enters at 20°C with a
mean velocity 1.25 m/s. Heat transfer convection coefficient for condensation on outer tube surface is ho
= 13,500 W/m2 ⋅K.
FIND: (a) Overall heat transfer coefficient, U, for the HXer, outlet temperature of cooling water, Tc,o,
h ; and (b) Compute and plot Tc,o and m
h as a function of the
and condensation rate of the steam m
c ≤ 30 kg/s with all other conditions remaining the same, but accounting for
water flow rate 10 ≤ m
changes in U.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible heat loss to surroundings, (2) Negligible kinetic and potential energy
changes, (3) Constant properties, (4) Fully developed water flow in tubes.
PROPERTIES: Table A-6, Steam (0.14 bar): Tsat = Th = 327 K, hfg = 2373 kJ/kg, cp = 1898 J/kg⋅K;
Table A-6, Water (Assume Tc,o ≈ 44°C or Tc ≈ 305 K): vf = 1.005 × 10-3 m3/kg , cp = 4178 J/kg⋅K,
µf = 769 × 10-6 N⋅s/m2 , kf = 0.620 W/m⋅K, Prf = 5.2; Table A-1, Brass - 70/30 (Evaluate at T = (Th +
Tc )/2 = 316 K): k = 114 W/m⋅K.
ANALYSIS: (a) The overall heat transfer coefficient based upon the outside tube area follows from Eq.
11.5,
−1
 1 ro ro  ro  1 
+ "n +    .
Uo = 
ri  ri  h i 
 ho k
(1)
The value for hi can be estimated from an appropriate internal flow correlation. First determine the
nature of the flow within the tubes. From Eq. 8.1,
−3 3
Di (1.005 × 10 m kg )
=
−1
Re D = ρ u m
i
µ
× 1.25m s × 13.4 × 10 −3 m
769 × 10−6 N ⋅ s m 2
= 21, 673 .
The water flow is turbulent and fully developed (L/Di = 2 m /13.4 × 10-3 m = 150 > 10). The DittusBoelter correlation with n = 0.4 is appropriate,
0.8
0.4
Nu D = hi Di k f = 0.023 Re0.8
D Prf = 0.023 × ( 21, 673 )
(5.2 )0.4 = 130.9
Continued...
PROBLEM 11.35 (Cont.)
k
0.620W m ⋅ K
× 130.9 = 6057 W m 2 ⋅ K .
h i = f Nu D =
−
3
Di
13.4 × 10 m
Substituting numerical values into Eq. (1), the overall heat transfer coefficient is
(
)


15.9 × 10−3 m 2 15.9 15.9
1
1


"n
+
+
×
Uo =
13, 500 W m 2 ⋅ K
2
115 W m ⋅ K
13.4 13.4 6057 W m ⋅ K 


U o = 7.407 × 10−5 + 1.183 × 10−5 + 19.590 × 10−5 

−1

−1
<
W m 2 ⋅ K = 3549 W m 2 ⋅ K .
To find the outlet temperature of the water, we’ll employ the ε − NTU method. From an energy balance
on the cold fluid,
Tc,o = Tc,i + q Cc
(3)
where the heat rate can be expressed as
(
q = ε q max
)
q max = C min Th,i − Th,o .
(4,5)
The minimum capacity rate is that of the cold water since Ch → ∞ . Evaluating, find
(
)c = 22.8 kg s × 4178 J kg ⋅ K = 95, 270 W K .
Cmin = Cc = mc
p
where
2
3
m
c = ( ρ Au m ) N = 995.0 kg/m × π 4 ( 0.0134m ) × 1.25 m s × 130 = 22.8 kg s
To determine ε, use Fig. 11.16 (one shell and any multiple of tube passes) with
NTU =
Uo Ao
Cmin
=
3549 W m 2 ⋅ K (π 0.0159m × 2m × 130 × 2 )
95, 270 W K
= 0.968
where 130 and 2 represent the number of tubes and passes, respectively, to find ε ≈ 0.62. Combining
Eqs. (4) and (5) into Eq. (3), find
(
Tc,o = Tc,i + ε Cmin Th,i − Tc,i
)
<
Cc = 20$ C + 0.62 (327 − 293) K = 41.1$ C .
The condensation rate of the steam is given by
m
h = q h fg
(6)
where the heat rate can be determined from Eq. (3) with Tc,o ,
(
m
h = Cc Tc,o − Tc,i
)
h fg = 95, 270 W K ( 41.1 − 20.0 ) K 2373 × 103 J kg ⋅ K = 0.85 kg s .
<
(b) Using the IHT Heat Exchanger Tool, All Exchangers, Cr = 0, and the Properties Tool for Water, a
model was developed and the cold outlet temperature and condensation rate were computed and plotted.
Continued...
PROBLEM 11.35 (Cont.)
Tco (C) and mdoth*10^-1 (kg/s)
50
40
30
20
10
0
10
14
18
22
26
30
Cold flow rate, mdotc (kg/s)
Cold outlet temperature, Tco (C)
Condensation rate, mdoth*10^-1 (kg/s)
With increasing cold flow rate, the cold outlet temperature decreases as expected. The condensation rate
h are nearly linear with the cold flow rate.
increases with increasing cold flow rate. Note that Tc,o and m
COMMENTS: For part (a) analysis, note that the assumption Tc,o ≈ 44°C used for evaluation of the cold
h =
fluid properties was reasonable. Using the IHT model of part (b), we found Tc,o = 40.2°C and m
0.812 kg /s.
PROBLEM 11.36
KNOWN: Shell-and-tube (one shell, two tube passes) heat exchanger design. Water flow rate and inlet
temperature. Steam pressure and convection coefficient.
c , for the range, 5 ≤ m
c ≤
FIND: (a) Water outlet temperature, Tc,o; (b) Tc,o as a function of flow rate, m
20 kg/s, with all other conditions remaining the same, but accounting for changes in the overall
coefficient, U; and (c) Plot Tc,o on the same graph considering fouling factors of R′′f = 0.0002 and
0.0005 m2⋅K/W
SCHEMATIC:
ASSUMPTIONS: (1) Negligible heat loss to surroundings and kinetic and potential energy changes, (2)
Negligible wall conduction and fouling resistances, (3) Constant properties.
PROPERTIES: Table A-6, Sat. water (p = 1.0133 bar): Tsat = T = 373.1 K; ( Tc ≈ 320 K): cp = 4180
J/kg⋅K, µ = 577 × 10-6 N⋅s/m2 , k = 0.640 W/m⋅K, Pr = 3.77.
ANALYSIS: Using the NTU-effectiveness method, calculate for U by finding hi . With
(
)
−6
π Dµ = [4 (10 kg s ) 100 ] π ( 0.02m ) 577 × 10
Re D = 4m
N ⋅ s m 2  = 11, 033

(1)
and using the Dittus-Boelter correlation,
Nu D = 0.023 Re 4D/ 5 Pr 0.4 = 0.023 (11, 033)
4/5
(3.77 )0.4 = 67.05
(2)
h i = ( k D ) Nu D = (0.640 W m ⋅ K 0.02m ) 67.05 = 2146 W m 2 ⋅ K .
From Eq. 11.5
1 U = 1 h i + 1 h o = [(1 10, 000 ) + (1 2146 )] m 2 ⋅ K W = 5.66 × 10−4 m 2 ⋅ K W
(3)
U = 1766 W m 2 ⋅ K .
The heat transfer surface area, capacity rates and NTU are
A = N (π D ) 2L = 100 (π 0.02m ) 2 × 2m = 25.1m 2
Cmin = Cc = 10 kg s ( 4180 J kg ⋅ K ) = 41,800 W K
NTU = UA Cmin = 1766 W m 2 ⋅ K × 25.1 m 2 41,800 W K = 1.06
From Eq. 11.36a
Continued...
PROBLEM 11.36 (Cont.)
ε = 1 − exp ( − NTU ) = 1 − exp ( −1.06 ) = 0.654 .
(4)
With
(
)
q max = Cmin Th,i − Tc,i = 41,800 W K (373.15 − 290 ) K = 3.48 × 106 W
(5)
)
(
q = ε q max = 0.654 3.48 × 106 W = 2.27 × 106 W
find
(
)
Tc,o = Tc,i + ( q Cc ) = 290 K + 2.27 × 106 W 41,800W K = 344.4K .
(6)
<
(b,c) Using the IHT Heat Exchanger Tool, All Exchangers, Cr = 0, the Properties Tool for Water and the
Correlation Tool, Forced Convection, Internal Flow, for Turbulent, fully developed conditions, a model
was developed following the foregoing analysis to compute and plot the outlet temperature Tc,o as a
c . The expression for the overall coefficient, Eq.(1), was modified
function of the cold fluid flow rate, m
to include the fouling factor,
1 U = 1 h i + R ′′f + 1 h o .
Outlet temperature, Tco (K)
360
350
340
330
320
310
5
8
11
14
17
20
Water flow rate, mdotc (kg/s)
Fouling factor, R''f = 0
R''f = 0.0002 m^2.K/W
R''f = 0.0005 m^2.K/W
The effect of increasing the cold flow rate is to decrease the outlet temperature. The effect of the fouling
resistance is to decrease the outlet temperature as well.
COMMENTS: (1) For the part (a) analysis, Tc = 317 K and the initial guess of 320 K was reasonably
good.
(2) In the anlysis of parts (b,c), ReD,c is as low as 4880, below the turbulent range (10,000) and above the
laminar range (2300). We chose to treat the flow as turbulent.
PROBLEM 11.37
KNOWN: Saturated steam at 110°C condensing in a shell and tube heat exchanger (one shell pass, 2,
4, tube passes) with a UA value of 2.5 kW/K; cooling water enters at 40°C.
FIND: Cooling water flow rate required to maintain a heat rate of 150 kW; and (b) Calculate and plot
the water flow rate required to provide heat rates over the range 130 to 160 kW, assuming that UA is
independent of flow rate. Comment on the validity of the assumption.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible heat loss to surroundings, (2) Negligible kinetic and potential
energy changes, (3) UA independent of flow rate, and (4) Constant properties.
PROPERTIES: Table A-6, Water (Tm,c = (Tc,i + Tc,o)/2 = 49.5°C = 322.5 K): cp,c = 4181 J/kg⋅K.
ANALYSIS: (a) For the shell-tube heat exchanger with any multiple of two-tube passes, from Eq.
11.36a with Cr = 0, using Eqs. 11.20 and 11.23,
ε = 1 − exp ( − NTU )
NTU = UA / Cmin
ε = q / q max
q max = Cc Th,i − Tc,i
(
(1,2)
)
(3,4)
c c p,c ,
By combining the equations with Cmin = Cc = m

q
UA 
= 1 − exp  −

 
c cp,c Th,i − Tc,i
m
 mc cp,c 
(
)
(5)
Substituting numerical values, and solving using IHT find
<
c = 1.89 kg / s
m
The specific heat of the cold fluid, cp,c, is evaluated at the average of the mean inlet and outlet
temperatures, Tm,c = (Tc,i + Tc,o)/2, with Tc,o determined from the energy balance equation,
(
)
cp,c Tc,o − Tc,i .
q=m
(6)
(b) Solving the above system of equations in the IHT workspace, the graph below illustrates the water
flow rate required to provide a range of heat rates.
Continued …..
PROBLEM 11.37 (Cont.)
Water flow rate required for specified heat rate
Water flow rate, mdotc (kg/s)
4
3
2
1
0
130
140
150
160
Heat rate, q (kW)
COMMENTS: (1) The assumption that UA is constant with flow rate is a poor one. Because the
heat transfer coefficient for condensation is so high, the overall coefficient is controlled by the waterside coefficient. Presuming the flow is turbulent, from the Dittus-Boelter correlation, we’d expect
U
0.8
mc . Over the range of the graph above, U will vary by approximately a factor of (3.5/1)
0.8
2.7.
(2) If we considered UA to vary with the cold water flow rate as just described, make a sketch of
m
c vs. q and compare it to the graph above.
=
PROBLEM 11.38
KNOWN: Temperature, convection coefficient and condensation rate of saturated steam. Tube
diameter for shell-and-tube heat exchanger with one shell pass and two tube passes. Velocity and
inlet and maximum allowable exit temperatures of cooling water.
FIND: (a) Minimum number of tubes and tube length per pass, (b) Effect of tube-side heat transfer
enhancement on tube length.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible heat exchange with surroundings, (2) Negligible flow work and
kinetic and potential energy changes, (3) Negligible tube wall conduction and fouling resistance, (4)
Constant properties, (5) Fully developed internal flow throughout.
6
PROPERTIES: Table A-6, Sat. water (340 K): hfg = 2.342 × 10 J/kg; Sat. water
(Tc = 22.5°C ≈ 295 K ) :
3
-6
2
ρ=998 kg/m , cp = 4181 J/kg⋅K, µ = 959 × 10 N⋅s/m , k = 0.606 W/m⋅K,
Pr = 6.62.
ANALYSIS: (a) The required heat rate and the maximum allowable temperature rise of the water
determine the minimum allowable flow rate. That is, with
cond h fg = 2.73 kg / s × 2.342 × 106 J / kg = 6.39 × 106 W
q = qcond = m
c,min =
m
q
(
cp,c Tc,o − Tc,i
)
=
6.39 × 106 W
= 101.9 kg / s
4181 J / kg ⋅ K (15°C )
2
3
2
c,1 =ρumπ D /4 = 998 kg/m × 0.5 m/s × π (0.019m) /4 =
With a specified flow rate per tube of m
0.141 kg/s, the minimum number of tubes is
N min =
c,min
m
c,1
m
=
101.9 kg / s
= 720
0.141 kg / s
<
To determine the corresponding tube length, we must first find the required heat transfer surface area.
3
-6
2
With ReD = ρumD/µ = 998 kg/m (0.5 m/s) 0.019m/959 × 10 N⋅s/m = 9,886, the Dittus-Boelter
equation yields
4/5
hi = ( k / D ) 0.023 Re D
Pr
0.4
4/5
= ( 0.606 W / m ⋅ K / 0.019m ) 0.023 (9886 )
(6.62 )0.4 = 2454 W / m 2 ⋅ K
Continued …..
PROBLEM 11.38 (Cont.)
−1
Hence, U =  h i−1 + h o−1  = 1970 W / m 2 ⋅ K


5
c cp,c = 101.9 kg/s × 4181 J/kg⋅K = 4.26 × 10 W/K, qmax = Cmin (Th,i – Tc,i)
With Cr = 0, Cmin = m
5
7
= 4.26 × 10 W/K (340 – 288) K = 2.215 × 10 W and ε = q/qmax = 0.289, Eq. 11.36b yields NTU =
– ln (1 – ε) = – ln (1 – 0.289) = 0.341. Hence the tube length per pass is
L=
A
NTU × Cmin
0.341× 4.26 × 105 W / K
=
=
= 0.858m
2Nπ D
2Nπ DU
2 × 720 × π (0.019m )1970 W / m 2 ⋅ K
<
2
(b) If the tube-side convection coefficient is doubled, h i = 4908 W / m 2 ⋅ K and U = 3292 W/m ⋅K.
Since q, Cr, Cmin, qmax and hence ε are unchanged, the number of transfer units is still NTU = 0.341.
Hence, the tube length per pass is now
NTU × Cmin
0.341× 4.26 × 105 W / K
L=
=
= 0.513m
2 Nπ DU
2 × 720 × π ( 0.019m ) 3292 W / m 2 ⋅ K
COMMENTS: Heat transfer enhancement for the flow with the smallest convection coefficient
significantly reduces the size of the heat exchanger.
<
PROBLEM 11.39
KNOWN: Pressure and initial flow rate of water vapor. Water inlet and outlet temperatures. Initial
and final overall heat transfer coefficients.
FIND: (a) Surface area for initial U and water flow rate, (b) Vapor flow rate for final U.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible heat loss to surroundings, (2) Negligible wall conduction resistance.
PROPERTIES: Table A-6, Sat. water ( Tc = 310 K): cp,c = 4178 J/kg⋅K; (p = 0.51 bars): Tsat = 355
K, hfg = 2304 kJ/kg.
ANALYSIS: (a) The required heat transfer rate is
& h h fg = 1.5 k g / s 2.304 ×106 J / k g = 3.46 ×106 W
q=m
(
)
and the corresponding heat capacity rate for the water is
(
)
Cc = Cmin = q / Tc,o − Tc,i = 3.46 ×106 W / 4 0 K = 86,400 W / K.
Hence,
(
)
ε = q / Cmin  Th,i − Tc,i  = 3.46 ×106 W/86,400 W / K ( 65 K ) = 0.62.
Since Cmin/Cmax = 0, Eq. 11.36b yields
NTU = − ln (1 − ε ) = − ln ( 1− 0.62 ) = 0.97
and
(
)
A = NTU ( Cmin / U ) = 0.97 86,400 W / K / 2 0 0 0 W / m 2 ⋅ K = 41.9m 2
<
& c = Cc / c p,c = 86,400 W / K / 4 1 7 8 J / k g ⋅ K = 20.7 kg/s.
m
<
(b) Using the final overall heat transfer coefficient, find
NTU = UA/Cmin = 1000 W / m 2 ⋅ K 41.9m 2 /86,400 W / K = 0.485.
(
)
Since Cmin/Cmax = 0, Eq. 11.36a yields
ε = 1 − exp ( − NTU ) = 1 − exp ( −0.485 ) = 0.384.
Hence,
(
)
q = ε Cmin Th,i − Tc,i = 0.384 ( 86,400 W / K ) 65 K = 2.16 ×10 6 W
& h = q / h fg = 2.16 ×106 W/2.304 ×106 J / k g = 0.936 kg/s.
m
<
& h represents a significant loss in turbine power.
COMMENTS: The significant reduction (38%) in m
Periodic cleaning of condenser surfaces should be employed to minimize the adverse effects of fouling.
PROBLEM 11.40
KNOWN: Two-fluid heat exchanger with prescribed inlet and outlet temperatures of the two fluids.
FIND: (a) Whether exchanger is operating in parallel or counter flow, (b) Effectiveness of the
exchanger when Cc = Cmin.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible heat loss to the surroundings, (2) Negligible kinetic and potential
energy changes, (3) Cold fluid is minimum fluid.
ANALYSIS: (a) To determine whether operation is PF or CF, consider the temperature distributions.
From the distributions we note that PF or CF operation is possible.
(b) The effectiveness of the exchanger follows from Eq. 11.20,
ε = q / q max
(1)
where from Eq. 11.19,
(
)
q max = Cmin Th,i − Tc,i .
(2)
Since Cmin = Cc and performing an energy balance on the cold fluid, Eq. (1) with Eq. (2) becomes
(
)
(
) (
)(
ε = Cc Tc,o − Tc,i / C min T h,i − Tc,i = Tc,o − Tc,i / Th,i − Tc,i
)
ε = ( 30 − 15 ) °C / ( 65 − 15 ) °C = 0.30.
COMMENTS: If Tc,o were greater than Th,o, parallel-flow operation would not be possible.
<
PROBLEM 11.41
KNOWN: Concentric tube heat exchanger.
FIND: Length of the exchanger.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible heat loss to surroundings, (2) Negligible kinetic and potential energy
changes, (3) Constant properties.
PROPERTIES: Table A-6, Water ( Tc = (35 + 95)°C/2 = 338 K): cp,c = 4188 J/kg⋅K
ANALYSIS: From the rate equation, Eq. 11.14, with Ao = πDoL,
L = q / U oπ D o∆T lm.
The heat rate, q, can be evaluated from an energy balance on the cold fluid,
(
)
& c cc Tc,o − Tc,i =
q=m
225 k g / h
× 4188J/kg ⋅ K ( 95 − 35 ) K =15,705 W.
3600 s / h
In order to evaluate ∆Tlm , we need to know whether the exchanger is operating in CF or PF. From
an energy balance on the hot fluid, find
& h c h = 210°C − 15,705 W /
Th,o = Th,i − q / m
225 k g / h
J
× 2095
= 90.1°C.
3600s/h
kg ⋅ K
Since Th,o < Tc,o it follows that Hxer operation must be CF. From Eq. 11.15,
∆Tl m,CF =
( 210 − 95 ) − ( 90.1 − 35)
∆T1 − ∆T2
=
°C = 81.4°C.
ln ( ∆T1 / ∆T2 )
ln (115/55.1)
Substituting numerical values, the HXer length is
2
L = 15,705 W / 5 5 0 W / m ⋅ Kπ ( 0.10m ) × 81.4K = 1.12m.
<
COMMENTS: The ε-NTU method could also be used. It would be necessary to perform the hot
fluid energy balance to determine if CF operation existed. The capacity rate ratio is Cmin/Cmax = 0.50.
From Eqs. 11.19 and 11.20 with q evaluated from an energy balance on the hot fluid,
Th,i − Th,o 210 − 90.1
=
= 0.69.
ε=
Th,i − Tc,i
210 − 35
From Fig. 11.15, find NTU ≈ 1.5 giving
L = NTU ⋅ Cmin / Uo π Do ≈ 1.5 ×130.94
Note the good agreement in both methods.
W
W
/550
⋅ π ( 0.10m ) ≈ 1.14m.
K
m2 ⋅ K
PROBLEM 11.42
KNOWN: A very long, concentric tube heat exchanger having hot and cold water inlet temperatures,
85°C and 15°C, respectively; flow rate of hot water is twice that of the cold water.
FIND: Outlet temperatures for counterflow and parallel flow operation.
SCHEMATIC:
ASSUMPTIONS: (1) Equivalent hot and cold water specific heats, (2) Negligible kinetic and
potential energy changes, (3) No heat loss to surroundings.
ANALYSIS: The heat rate for a concentric tube heat
exchanger with very large surface area operating in the
counterflow mode is
(
q = q max = Cmin Th,i − Tc,i
)
where Cmin = Cc. From an energy balance on the hot fluid,
(
)
q = Ch Th,i − T h,o .
Combining the above relations and rearranging, find
(
)
(
)
C
C
Th,o = − min Th,i − Tc,i + Th,i = − c Th,i − Tc,i + Th,i .
Ch
Ch
Substituting numerical values,
Th,o = −
1
(85 −15 ) °C + 85°C =50 °C.
2
<
For parallel flow operation, the hot and cold outlet
temperatures will be equal; that is, Tc,o = Th,o.
Hence,
(
)
(
)
Cc Tc,o − Tc,i = Ch Th,i −T h,o .
Setting Tc,o = Th,o and rearranging,

  C 
C
Th,o =  Th,i + c Tc,i  /  1 + c 
Ch

  Ch 
1
1
Th,o =  85 + ×15  °C /  1 +  = 61.7°C.
2


 2
COMMENTS: Note that while ε = 1 for CF operation, for PF operation find ε = q/qmax = 0.67.
<
PROBLEM 11.43
KNOWN: Saturation temperature and condensation rate of refrigerant. Frontal area of condenser
and dependence of overall coefficient on inlet velocity. Operational range of the air inlet temperature.
FIND: (a) Required heat exchanger area and air outlet temperature for prescribed air inlet velocity
and temperature, (b) Variation in air velocity needed to achieve prescribed condensation rate.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible heat loss to surroundings, (2) Constant properties.
5
3
PROPERTIES: Given (R-12): hfg = 1.35 × 10 J/kg. Table A-4, air (Tc,i = 303 K): ρc = 1.17 kg/m ,
cp,c = 1007 J/kg⋅K.
ANALYSIS: (a) With m c = ρc VA fr = 1.17 kg / m3 × 2 m / s × 0.25 m 2 = 0.585 kg / s,
c
C min = m
c p,c = 589 W / K. Hence, from Eq. (11.19), with Th,i = Tsat,
(
)
q max = Cmin Th,i − Tc,i = 589 W / K ( 45 − 30 ) K = 8,836 W
5
and with q = m
h h fg = 0.015 kg / s × 1.35 × 10 J / kg = 2025 W
ε=
q
q max
=
2025
= 0.229
8836
From Eq. 11.36b we then obtain (for Cr = 0),
C
C
589 W / K
A = min NTU = − min ln ( l − ε ) = −
ln (0.771) = 3.067 m 2
2
U
U
50 W / m ⋅ K
<
With q = Cmin (Tc,o – Tc,i), the outlet temperature is
Tc,o = Tc,i +
q
Cmin
= 30°C +
2
2025 W
= 33.4°C
589 W / K
2
(b) With q = 2025 W, A = 3.06 m and U = 50 W/m ⋅K (V/2)
solved to obtain V as a function of Tc,i.
<
0.7
, the foregoing equations may be
Continued…..
PROBLEM 11.43 (Cont.)
Air in le t ve lo city, m /s
6
5
4
3
2
1
0
27
28
29
30
31
32
33
34
35
36
37
38
Air in le t te m p e ra tu re , C
With increasing Tc,i, the driving potential for heat transfer, Th,i – Tc,i, decreases and a larger value of
U, and hence V, is needed to maintain the required heat rate. For 27 ≤ Tc,i ≤ 38°C, 1.56 ≤ V ≤ 5.66
2
m/s and 42.1 ≤ U ≤ 103.6 W/m ⋅K.
COMMENTS: The variation of V with Tc,i is nonlinear, and, in principle, V → ∞ as Tc,i → Tsat.
PROBLEM 11.44
KNOWN: Conditions of oil and water for heat exchanger, one shell with 4 tube passes.
FIND: Length of exchanger tubes per pass, L; and (b) Compute and plot the effectiveness, ε, fluid
outlet temperatures, Th,o and Tc,o, and water-side convection coefficient, hc , as a function of the water
c ≤ 15,000 kg / h for the tube length found in part (a) with all other conditions
flow rate for 5000 ≤ m
remaining the same.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible heat loss to surroundings, (2) Negligible kinetic and potential energy
changes, (3) Constant properties, (4) Fully-developed flow in tubes.
PROPERTIES: Table A-1, Brass (400 K): k = 137 W/m⋅K; Table A-5, Water (323 K): ρ = 998.1
kg/m3, k = 0.643 W/m⋅K, cp = 4182 J/kg⋅K, µ = 548 × 10-6 N⋅s/m2, Pr = 3.56.
ANALYSIS: (a) From an energy balance on the water, the heat rate required is
(
)
$
c
q=m
c c Tc,o − Tc,i = 10, 000 3600 kg s × 4182 J kg ⋅ K (84 − 16 ) C = 789, 933 W .
(1)
The required tube length may be obtained from Eqs. 11.14 and 11.15,
q = U o A o F∆T"m,CF
(2)
∆T"m,CF = (160 − 84 ) C − (94 − 16 ) C  / "n (160 − 84 94 − 16 ) = 77.0$ C .

$
$

From Fig. 11.10, F = 0.86 using P = (84 - 16)/(160 - 16) = 0.47 and R = (160 - 94)/(84 - 16) = 0.97. From
Eq. 11.5,
 1 ro ro ro 1 
Uo = 
+ "n +

ri ri h i 
 ho k
−1
where hi must be estimated from the appropriate correlation. With N = 11, the number of tubes,
N
4 × (10, 000 3600 ) kg s (11)
4m
Re D =
=
= 25, 621 .
π Dµ π × 22.9 × 10−3 m × 548 × 10−6 N ⋅ s m 2
For fully developed turbulent flow, the Dittus-Boelter correlation with n = 0.4 yields
0.4
Nu D = hi D k = 0.023 Re0.8
= 0.023 ( 25, 621)
D Pr
0.8
h i = Nu D ( k D ) = 128.6 × 0.643 W m ⋅ K
(3.56 )0.4 = 128.6
(22.9 ×10−3 m ) = 3610 W m2 ⋅ K .
Continued...
PROBLEM 11.44 (Cont.)


1
25.4 × 10−3 m
25.4 25.4
1
"n
+
+
×
Uo = 

 400 W m 2 ⋅ K 2 × 137 W m ⋅ K 22.9 22.9 3610 W m 2 ⋅ K 
−1
= 355 W m 2 ⋅ K .
Returning now to Eq. (2), find Ao, then the length,
A o = π Do L × No. of Passes × No.of Tubes = π × 25.4 × 10−3 m × 4 × 11L = 3.511 L
<
L = 789, 933 W 3.511 m × 355 W m 2 ⋅ K × 0.86 × 77.0$ C = 9.6 m
(b) Using the IHT Heat Exchanger Tool, Shell and Tube, One-shell pass and N tube passes, the
Correlation Tool, Forced Convection, Internal Flow for Turbulent, fully developed condition, and the
Properties Tool for Water, a model was developed using the effectiveness - NTU method to compute and
.
plot Tc,o , Th,o , ε, and hi as a function of m
c
Tco, Tho, eps*100, hi*10^-2
100
80
60
40
20
5000
10000
15000
Cold flow rate, mdotc (kg/h)
Tco (C)
Tho (C)
eps * 100
hci * 10^-2 (W/m^2.K)
In order to avoid a boiling condition in the cold fluid, the cold flow rate should not be less than 8000
kg/h. As expected, Tc,o and Th,o decrease and the internal convection coefficient increases nearly linearly
with increasing flow rate. The effectiveness increases with increasing flow rate since the overall
convection coefficient is increasing.
COMMENTS: (1) The thermal resistance of the brass tubes is negligible. Since L/Di = 400, fullydeveloped conditions are reasonable.
(2) In the analysis of part (b), you have to specify the capacity rate for the hot fluid in order to solve the
model. From the analysis of part (a) using the model, we found L = 9.56 m and Ch = 11,974 W/K.
PROBLEM 11.45
KNOWN: Properties and flow rate of computer coolant. Diameter and number of heat exchanger tubes.
Heat exchanger transfer rate and inlet temperature of computer coolant. Flow rate, specific heat, inlet
temperature, and average convection coefficient of water.
FIND: (a) Tube flow convection coefficient, h i , (b) Tube length/pass required to achieve prescribed
fluid outlet temperature, (c) Compute and plot the dielectric fluid outlet temperature, Tf,o , as a function of
f for the range 4 ≤ m
f ≤ 6 kg/s based upon the length/pass found in part (c), (d) the
its flow rate m
w , on Tf,o and (e) the effect of ±3°C change in inlet
effect of ±10% change in the water flow rate, m
water temperature, Tw ,i, on Tf,o. For parts (c, d, e), account for any changes in the overall convection
coefficient, while all other conditions remain the same.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible heat loss to surroundings, kinetic and potential energy changes, fouling
and tube wall resistance; (2) Constant properties; (3) Fully developed flow, (4) Convection coefficient
on shell side, h o , remains constant for all operating conditions.
PROPERTIES: Coolant (given): cp = 1040 J/kg⋅K, µ = 7.65 × 10-4 kg/s⋅m, k = 0.058 W/m⋅K, Pr = 14;
Water (given): cp = 4200 J/kg⋅K.
ANALYSIS: (a) For flow through a single tube,
Re D =
4m
f ,t
π Dµ
=
4 ( 4.81kg s ) / 72
π ( 0.01m ) 7.65 × 10−4 kg s ⋅ m
= 11,120
and using the Dittus-Boelter correlation, find
h i = ( k D ) 0.023 Re 4D/ 5 Pr 0.3 = 0.023
0.058 W m ⋅ K
0.01m
(11,120 )4 / 5 (14 )0.3 = 508 W
m2 ⋅ K .
<
(b) Find the capacity ratio
c
Cf = m
f p,f = 4.81kg s (1040 J kg ⋅ K ) = 5002 W K = Cmin
Cw = m
w cp,w = 2.5 kg s ( 4200 J kg ⋅ K ) = 10, 500 W K = C max
hence, Cr = Cmin /Cmax = 0.476 and
Cf Tf ,i − Tf ,o
( 25 − 15 )$ C
ε=
=
=
= 0.500 .
q max Cf Tf ,i − Tw,i
( 25 − 5 )$ C
q
(
(
)
)
Using Fig. 11.16 with NTU = (UA/Cmin ) = (UNπD2L/Cmin ) ≈ 0.85,
Continued...
PROBLEM 11.45 (Cont.)
(
U = h i−1 + h −o 1
)
−1

−1
= (508 )

(
−1  −1
( )
+ 104


W m 2 ⋅ K = 483 W m 2 ⋅ K
)
<
L = 0.85 (5002 W K ) 144π 483 W m 2 ⋅ K 0.01m = 1.95m .
(c) Using the IHT Heat Exchanger Tool, Shell and Tube, One-shell pass and N-tube passes, and the
Correlation Tool, Forced Convection, Internal Flow for Turbulent, fully developed conditions, a model
was developed using the effectiveness-NTU method employed above to compute and plot Tf,o as a
f .
function of m
Outlet temperature, Tfo (C)
16
15
14
4
5
6
Dielectric flow rate, mdotf (kg/s)
A change in the dielectric fluid flow rate of ±1 kg/s causes approximately ±0.5°C change in its outlet
temperature.
(d) Using the above IHT model with the base conditions for part (c), the effect of a ±10% change in the
w = 2.5 kg/s (2.25 ≤ m
w ≤ 2.75 kg/s) causes the dielectric fluid
water flow rate from its design value, m
outlet temperature to change as
Tf ,o = 15 ± 0.14$ C
<
(e) Using the IHT model of part (c) with the base case conditions for part (c), the effect of a ±3°C in the
water inlet temperature from its design value, Tc,i = 5°C ( 2 ≤ Tc,i ≤ 8°C) cause the dielectric fluid outlet
temperature to change as
Tf ,o = 15 ± 1.5$ C
<
COMMENTS: (1) For the analyses of part (a), Eq. 11.31 yields NTU = 0.85 and q = 50 kW and Tw,o =
9.76°C.
(2) The results of the analyses provide operating performance information on the effect of changes due to
f ), water fluid flow rate (≤ 10% of m
w ) and water inlet
dielectric fluid flow rate (±1 kg/s of m
temperature (±3°C of Tw,i) on the dielectric fluid outlet temperature, Tf,o, supplied to the computer. The
greatest effect on Tf,o, is that by the input water temperature.
PROBLEM 11.46
KNOWN: Shell and tube heat exchanger with 135 tubes (one shell, double pass) of inner diameter 12.5
mm and surface area 47.5 m2.
FIND: (a) Exchanger gas and water outlet temperatures, (b) Tube heat transfer coefficient, h i , assuming
fully developed flow, (c) Compute and plot the effectiveness and fluid outlet temperatures, Tc,o and Th,o
c ≤ 12 kg/s with all other conditions remaining the same, and (d) Hot
for the water flow rate range 6 ≤ m
gas inlet temperature, Th,i, required to supply 10 kg/s of hot water with an outlet temperature of 42°C
with all other conditions the same; determine also the effectiveness.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible heat lost to surroundings, (2) Negligible kinetic and potential energy
changes, (3) Fully-developed conditions for internal flow of water in tubes, (4) Exhaust gas properties are
those of air, and (5) The overall coefficient remains unchanged for the operating conditions examined.
PROPERTIES: Table A-6, Water ( Tc ≈ 300 K): ρ = 997 kg/m3, c = 4179 J/kg⋅K, k = 0.613
W/m⋅K, µ = 855 × 10-6 N⋅s/m2, Pr = 5.83; Table A-4, Air (1 atm, Th ≈ 400 K): ρ = 0.8711 kg/m3, c =
1014 J/kg⋅K.
,
ANALYSIS: (a) Using the ε-NTU method, first find the capacity rates, C = mc
Cc = 6.5 kg s × 4179 J kg ⋅ K = 27,164 W K
Ch = 5.0 kg s × 1014 J kg ⋅ K = 5, 070 W K .
Recognize that Ch = Cmin and determine
Cmin
Cmax
=
Ch
Cc
=
5, 070
27,164
= 0.19
NTU =
AU
Cmin
=
47.5m 2 × 200 W m 2 ⋅ K
5, 070 W K
= 1.87 .
From Fig. 11.16 for the shell and tube exchanger, find with NTU = 1.87 and Cmin/Cmax = 0.19 that ε ≈
0.78. From the definition of effectiveness,
Ch Th,i − Th,o
200 − Th,o
q
ε=
=
=
= 0.78 or Th,o = 55.7$ C .
q max Cmin Th,i − Tc,i
200 − 15
(
)
)
(
<
From energy balances on the two fluids, Ch (Th,i - Th,o) = Cc (Tc,o - Tc,i), find
(
)
$
Tc,o = Tc,i + ( Ch /Cc ) Th,i − Th,o = 15$ C + 0.19 ( 200 − 55.7 ) C = 42.4$ C .
<
(b) To estimate h i for the water, find first the Reynolds number. From Eq. 8.6,
Continued...
PROBLEM 11.46 (Cont.)
Re D =
i
4m
π Di µ
=
4m
c N
π Di µ
=
4 × 6.5 kg s 135
π 12.5 × 10
−3
m × 855 × 10−6 N s ⋅ m2
= 5736
While the flow is fully developed and turbulent, ReD = 10,000 such that Dittus-Boelter correlation is not
strictly applicable. However, its use allows a first estimate.
4 / 5 0.4
Nu Di = h Di k = 0.023 ReD
Pr
= 0.023 (5736 )
4/5
(5.83)0.4 = 47.3
hi = Nu D k Di = 47.3 × 0.613 W m 2⋅ K 12.5 × 10−3 m = 2320 W m 2 ⋅ K .
<
(c) Using the IHT Heat Exchanger Tool, Shell and Tube, One-shell pass and N-tube passes, and the
prescribed properties, a model was developed following the analysis of part (a) to compute and plot ε,
c.
Tc,o, and Th,o for a function of m
Tco, Tho (C), eps*100
100
80
60
40
20
6
8
10
12
Cold flow rate, mdotc (kg/s)
Cold outlet temperature, Tco (C)
Hot outlet temperature, Tho (C)
Effectiveness, eps * 100
The outlet temperatures decrease nearly linearly with increasing cold fluid flow rate; the decrease in the
cold outlet temperature is nearly twice that of the hot fluid. The change in the effectiveness with
increasing flow rate is only slightly increased.
c = 10 kg/s with Tc,o
(d) Using the above IHT model, the hot inlet temperature Th,i, required to provide m
= 42°C and the effectiveness for this operating condition are
Th,i = 74.4$ C
ε = 0.55
<
COMMENTS: (1) Check that assumptions for Th and Tc used in part (a) for evaluation of the fluid
properties are satisfactory as Th = 400.7 K and Tc = 301.5 K .
(2) From part (b), with h i = 2320 W/m2⋅K and U = 200 W/m2 ⋅K, the shell-side convection coefficient is
h o = 219 W/m2 ⋅K. As such, U is controlled by shell-side conditions. Assuming U as a constant in part
c is therefore reasonable. However, for part (d) with m
h doubling, we should
(c) with changes in m
expect U to increase.
PROBLEM 11.47
KNOWN: Power output and efficiency of an ocean energy conversion system. Temperatures and
overall heat transfer coefficient of shell-and-tube evaporator.
FIND: (a) Evaporator area, (b) Water flow rate.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible heat loss to surroundings, (2) Negligible kinetic and potential energy
changes, (3) Constant properties.
PROPERTIES: Table A-6, Water ( Tm = 296 K): cp = 4181 J/kg⋅K.
ANALYSIS: (a) The efficiency is
η=
& 2MW
W
=
= 0.03.
q
q
Hence the required heat transfer rate is
q=
2MW
= 66.7MW.
0.03
Also
∆ Tlm,CF =
( 300 − 290 ) − ( 292 − 290 ) °C
300 − 290
ln
292 − 290
= 5 °C
and, with P = 0 and R = ∞, from Fig. 11.10 it follows that F = 1. Hence
A=
q
6.67 × 107 W
=
U F ∆ Tl m,CF 1200 W / m2 ⋅ K ×1 × 5 °C
A = 11,100m 2 .
<
(b) The water flow rate through the evaporator is
&h =
m
(
q
c p,h Th,i − Th,o
)
=
6.67 × 107 W
4181J/kg ⋅ K ( 300 − 292 )
<
& h = 1994 kg/s.
m
7
COMMENTS: (1) From the ε-NTU method, (Cmin/Cmax) = 0, qmax = 8.34 × 10 W, ε = 0.80 and
2
from Fig. 11.16, NTU ≈ 1.65, giving A = 11,500 m . (2) The required heat exchanger size is enormous
due to the small temperature differences involved.
PROBLEM 11.48
KNOWN: Length and tube diameter for a shell-and-tube (one shell pass, multiple tube passes) heat
exchanger. Flow rate and temperature of saturated steam. Condensation convection coefficient.
Velocity and inlet and outlet temperatures of cooling water.
FIND: (a) Required number of tubes and if the heat exchanger length is not to exceed 1.5 m, the number
h as a
of tube passes; (b) Compute and plot water outlet temperature Tc,o, and condensation rate, m
function of the mean velocity for the range 0.5 ≤ um ≤ 3 m/s, for the heat transfer area found from part
(a), accounting for changes in the overall coefficient, but all other conditions remaining the same; and (c)
Repeat the analysis of part (b) for tube diameters of 15 and 25 mm.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible heat loss to surroundings and changes in kinetic and potential energy,
(2) Negligible tube wall conduction and fouling resistances, (3) Constant properties, and (4) The shellside coefficient h o remains unchanged for the operating conditions examined.
PROPERTIES: Table A.6, Sat. water (340 K): hfg = 2.342 × 106 J/kg; Sat. water ( Tc = 297 K): ρ = 998
kg/m3, cp = 4180 J/kg⋅K, µ = 917 × 10-6 kg/s⋅m, k = 0.609 W/m⋅K, Pr = 6.3.
ANALYSIS: (a) The required heat rate is
(
)
6
6
h
q=m
h fg = 2.73 kg s 2.342 × 10 J kg = 6.39 × 10 W .
Hence, from conservation of energy,
(
)
6
m
c = q  cp,c Tc,o − Tc,i  = 6.39 × 10 W 4180 J kg ⋅ K (12K ) = 127.5 kg s .
Hence the number of tubes is
N=m
c m
c,t = m
c
(π D2 4) ρ um = 4 ×127.5 kg s π (0.019m )2 998 kg m3 (1.5 m s ) = 300 .
<
To determine the heat transfer surface area A, use the ε - NTU method. Find first
Re D = ρ u m D µ = 998 kg m3 (1.5 m s )( 0.019m ) 917 × 10−6 kg s ⋅ m = 31, 017
and using the Dittus-Boelter equation,
h i = ( k D ) 0.023 Re D
4/5
Pr
0.4
= ( 0.609 W m ⋅ K 0.019m ) 0.023 (31, 017 )
−1
= [(1 6034 ) + (1 12, 500 )]
U = [1 h i + 1 h o ]
4/5
−1
(6.3 )0.4 = 6034 W
2
m ⋅K
W m 2 ⋅ K = 4070 W m 2 ⋅ K .
Continued...
PROBLEM 11.48 (Cont.)
c cp,c = 127.5 kg/s(5180 J/kg⋅K) = 5.33 × 105 W/K
With Cmin = m
ε = q q max = q Cmin (Th,i − Tc,i ) = 6.39 × 106 W 5.33 ×105 W K ( 49K ) = 0.245 .
Hence, with Cr = 0, Eq. 11.36b yields NTU = -ln(1 - ε) = -ln(1 - 0.245) = 0.281,
)
(
A = NTU (C min U ) = 0.281 5.33 × 105 W K 4070 W m 2 ⋅ K = 36.8m 2
The tube length, L, in terms of the number of tubes, N, and passes, P, is
L = A N ⋅ Pπ D
and if P = 2,
<
L = 36.8m 2 300 × 2 × π × 0.019m = 1.03m
which is less than the maximum length 1.5 m.
(b) Using the IHT Heat Exchanger Tool, All Exchangers, Cr = 0, the Properties Tool for Water, and the
Correlations Tool, Forced Convection, Internal Flow, for Turbulent fully developed conditions, a model
h as a function of um with A =
was developed following the foregoing analysis to compute Tc,o and m
2
36.8 m as determined from part (a). The plot is shown below with the results for part (c).
h as a function of um for tube diameters of
(c) The IHT model was used to compute and plot Tc,o and m
15, 19, and 25 mm.
4
310
3
Tco (C)
mdoth (kg/s)
320
300
290
2
1
0.5
1
1.5
2
Cold flow rate, mdotc (kg/s)
D = 15 mm
D = 19 mm
D = 25 mm
2.5
3
0.5
1
1.5
2
2.5
3
Cold flow rate, mdotc (kg/s)
D = 15 mm
D = 19 mm
D = 25 mm
The effect of tube diameter on Tc,o as a function of the water flow rate is significant. As D increases at
any flow rate, the outlet temperatures decreases. The effect of tube diameter on the condensation rate is
slight. However, the condensation rate increases markedly as the water flow rate increases.
PROBLEM 11.49
KNOWN: Shell(1)-and-tube (two passes, p = 2) heat exchanger for condensing saturated steam at 1 atm.
Inlet cooling water temperature and mean velocity. Thin-walled tube diameter and length prescribed, as
well as, convective heat transfer coefficient on outer tube surface, ho.
FIND: (a) Number of tubes/pass, N, required to condense 2.3 kg/s of steam, (b) Outlet water
temperature, Tc,o, (c) Maximum condensation rate possible for same water flowrate and inlet temperature,
h , for water mean velocity, um, in the range 1
and (d) Compute and plot Tc,o and the condensation rate, m
≤ um ≤ 5 ms/, using the heat transfer surface area found in part (a) assuming the shell-side convection
coefficient remains unchanged.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible heat loss to surroundings, (2) Negligible kinetic and potential energy
changes, (3) Negligible thermal resistance due to the tube walls.
PROPERTIES: Table A.6, Saturated steam (1 atm): Tsat = 100°C, hfg = 2257 kJ/kg; Water (assume Tc,o
≈ 25°C, Tm = (Th + Tc)/2 ≈ 295 K): ρ = 1/vf = 998 kg/m3, cc = cp,h = 4181 J/kg⋅K, µ = µf = 959 × 10-6
N⋅s/m2, k = kf = 0.606 W/m⋅K, Pr = Pr" = 6.62.
ANALYSIS: (a) The heat transfer rate for the heat exchanger is
3
6
h
q=m
h fg = 2.3 kg s × 2257 × 10 J kg = 5.191 × 10 W
(1)
Using the ε-NTU method, evaluate the following parameters:
Water-side heat transfer coefficient:
u D
3.5 m s × 0.014 m
Re D = m =
= 50, 993
µ ρ 959 × 10−6 N ⋅ s m 2 998 kg m3
hi =
(2)
k
k
0.606 W m⋅ K
2
0.8
1/ 3
Nu D = 0.023Re0.8
Pr1/ 3 =
× 0.023 (50, 993 ) ( 6.62 )
= 10, 906 W m ⋅ K (3)
D
D
D
0.014 m
using the Colburn equation for fully developed turbulent conditions.
Overall coefficient:
−1
U = (1 h i + 1 h o )
−1
= (1 10, 906 + 1 21,800 )
= 7269 W m 2⋅ K
(4)
c = ρ(πD /4)umN,
Effectiveness relations: With Cmin = Cc and m
2
(
q = ε q max = ε Cmin Th,i − Tc,i
c = 998 kg m
C min = m
c c
3
)
(π × 0.014
(5)
2
m
2
)
4 × 3.5 m s × N × 4181J kg ⋅ K = 2248 N
(6)
Continued...
PROBLEM 11.49 (Cont.)
5.191 × 106 W = ε × 2248 N (100 − 15 ) K
εN = 27.17
From Eq. 11.36a with Cr = 0, the effectiveness is
(7)
ε = 1 − exp ( − NTU ) = 1 − exp ( −0.142 ) = 0.132
(8)
where, using As = πDLNP, NTU is evaluated as,
NTU =
UAs
=
7269 W m 2⋅ K (π × 0.014 m × 0.5 m ) N × 2
Cmin
2248 N
Hence, using Eq. (7), the required number of tubes is
= 0.142
<
N = 27.17/ε = 205.8 ≈ 206
and the total surface area is
As = π DLNP = π × 0.014 m × 0.5 m × 206 × 2 = 9.06 m 2 .
(b) The water outlet temperature with Cmin = 2248 N = 463,090 W/K,
Tc,o = Tc,i + q Cmin = 15$ C + 5.191 × 106 W 463, 090 W K = 26.1$ C
<
(c) The maximum condensation rate will occur when q = qmax. Hence
463, 090 W K (100 − 15 ) K
q max Cmin Th,i − Tc,i
m
=
=
= 17.44 kg s .
h,max =
h fg
h fg
2257 × 103 J kg
(d) Using the IHT Heat Exchanger Tool, All Exchangers, Cr = 0, along with the Properties Tool for
using the heat transfer surface area As
Water, the foregoing analysis was performed to obtain Th,o and m
h
2
= 9.06 m (part a) as a function of um.
(
)
<
Tco (C), mdoth*10 (kg/s)
35
30
25
20
15
10
1
2
3
4
5
Water mean velocity, um (m/s)
Outlet temperature, Tco (C)
Condensation rate, mdoth*10 (kg/s)
Note that the condensation rate increases nearly linearly with the water mean velocity. The cold water
outlet temperature decreases nearly linearly with um. We should expect this behavior from energy
balance considerations. Since hh is nearly two times greater than hc, U is controlled by the water side
coefficient. Hence U will increase with increasing um.
COMMENTS: Note that the assumed value for Tm to evaluate water properties in part (a) was a good
choice.
PROBLEM 11.50
KNOWN: Feed water heater (single shell, two tube passes) with inlet temperature 20°C supplies
10,000 kg/h of water at 65°C by condensing steam at 1.30 bar. Overall heat transfer coefficient is
2
2000 W/m ⋅K.
FIND: (a) Required area using LMTD and NTU approaches, (b) Steam condensation rate.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible heat loss to surroundings, (2) Negligible kinetic and potential energy
changes, (3) Constant properties.
3
PROPERTIES: Table A-6, Steam (1.3 bar, saturated): Th = 380.3 K, hfg = 2238 × 10 J/kg⋅K;
Table A-6, Water ( Tc = 316 K): cp = 4179 J/kg⋅K.
ANALYSIS: (a) Using the LMTD approach, from Eqs. 11.14 and 11.18,
∆ Tl m,CF = [ ∆ T1 − ∆ T2 ] / ln ( ∆ T1 / ∆ T2 )
A = q / U F ∆ Tl m,CF
(1,2)
( 380.3 − 338 )
∆ Tlm,CF =  (380.3 − 338 ) − ( 380.3 − 293 )  K / ln
= 62.1K.
( 380.3 − 293)
Since Th is uniform throughout the HXer, F = 1. From an energy balance on the cold fluid,
(
)
& c c p,c Tc,o − Tc,i =
q=m
10,000kg
J
× 4179
( 338 − 293 ) K = 5.224 × 105 W.
3600 s
kg ⋅ K
Substituting numerical values into Eq. (1) find that
A = 5.224 ×10 5 W/2000W/m 2 ⋅ K ×1 × 62.1K = 4.21m 2.
<
Using the NTU approach, recognize that Cmin = Cc and Cmax = Ch → ∞ so that Cmin/Cmax = 0. The
effectiveness, defined by Eq. 11.20, is
ε≡
q
q max
=
(
Cc Tc,o − Tc,i
(
)
Cmin Th,i − Tc,i
)
=
( 338 − 293) K
= 0.515.
( 380.3 − 293) K
From Fig. 11.16, with ε = 0.52 and Cmin/Cmax = 0, find NTU ≈ 0.70. Hence,
A = Cmin NTU/U =
(10,000 ) / ( 3600 ) kg/s × 4179J/kg ⋅ K × 0.70
2000 W / m 2 ⋅ K
= 4.1m 2.
(b) The condensation rate of steam is
& h = q / h fg = 5.224 ×105 W / 2238 ×10 3 J / k g = 0.233 kg/s = 840 kg/h.
m
(
)
<
<
COMMENTS: Note both methods of solution given the same result. Eq. 11.31 could have been
used to obtain a more precise NTU value.
PROBLEM 11.51
KNOWN: Shell-and-tube HXer with one shell and one tube pass.
FIND: (a) Oil outlet temperature for prescribed conditions, (b) Effect of fouling and water flowrate on
oil outlet temperature.
SCHEMATIC:
ASSUMPTIONS: (1) Constant properties, (2) Negligible changes in kinetic and potential energies, (3)
Negligible fouling and losses to surroundings, (4) Uniform tube outer surface temperature.
PROPERTIES: Table A.5, Engine oil ( Th ≈ 350 K): ρh = 854 kg/m3, cp,h = 2118 J/kg⋅K, µh = 0.0356
N⋅s/m2, kh = 0.318 W/m⋅K, Prh = 546; ( Ts ≈ 330 K): µs = 0.0836 N⋅s/m2; Table A.6, Water ( Tc ≈ 320
K): cp,c = 4180 J/kg⋅K; Table A.1, Copper ( T ≈ 320 K): k = 399 W/m⋅K.
ANALYSIS: (a) To determine the outlet temperature of the oil, we will need to know the overall heat
transfer coefficient. From Eq. 11.5,
1
UA
=
1
h i Ai
+
ln ( Do Di )
2π kL t
+
R ′′f ,o
Ao
+
1
(1)
h o Ao
where ho = 500 W/m2⋅K (water-side) and hi (oil-side) must be estimated from an appropriate correlation.
Using properties evaluated at an estimated average mean temperature Th ≈ 350 K, find
Re D,h =
4m
h,1
π Di µ h
=
4 × (1kg s 100 )
π ( 0.006 m ) × 0.0356 N ⋅ s m 2
= 59.6 .
(2)
Since ReD < 2300, the flow is laminar. To assess flow conditions, evaluate
L Di
0.5 m 0.006 m
=
= 0.00256
Gz −1 =
Re D,h Prh
59.6 × 546
(3)
Since Gz −1 < 0.05, the flow is characterized by combined entry length conditions (Fig. 8.9), and
 Re Pr 1/ 3  µ
Nu D = 1.86  D  
 L D   µs

where [



0.14 



(4)
] ≥ 2. To evaluate µs, assume Ts ≈ 330 K. Hence,

−1/ 3  0.0356 
Nu D = 1.86 ( 0.00256 )



 0.0836 
0.14 
 = 1.86 × 6.48 = 12.1

Note that [ ] = 6.48 > 2 as required. Hence,
Continued...
PROBLEM 11.51 (Cont.)
k
hi = Nu D
D
= 12.1 × 0.138 W m⋅ K ( 0.006 m ) = 277 W m 2⋅ K .
With R ′′f ,o = 0 and Lt = NtL, Eq. 1 yields
1
UA
1
UA
=
=
ln ( Do Di )
 1
1 
+
+


π N t L  h i Di
2k
h o Do 
1
(
1
(5)
)
(
)
1 500 W m 2⋅ K × 0.008 m + ln (8 6 ) ( 2 × 399 W m⋅ K ) + 1 277 W m 2⋅ K × 0.006 m 

π × 100 × 0.5 m 
1
UA
= 6.366 × 10−3 [0.2500 + 0.0003 + 0.6017 ] = 5.424 × 10−3 K W
UA = 184 W/K
With knowledge of UA, we can now use the ε - NTU method to obtain the oil outlet temperature, Th,o.
p,
Find the capacity rates, C = mc
c
Cc = m
c p,c = 2 kg s × 4180 J kg ⋅ K = 8360 W K = Cmax
c
Ch = m
h p,h = 1kg s × 2118 J kg ⋅ K = 2118 W K = C min
Cr = Cmin Cmax = 2118 8360 = 0.253
From Eq. 11.25, find
NTU = UA Cmin = 184 W K ( 2118 W K ) = 0.0869 .
(6)
For this exchanger - one shell and one pass - there are no figures (11.14-19) or relations (Table 11.3) that
can be directly used to evaluate ε. However, the HXer approximates a CF concentric tube HXer; hence,
use Eq. 11.30a.
ε=
1 − exp [− NTU (1 − Cr )]
1 − Cr exp [− NTU (1 − Cr )]
=
1 − exp [−0.0869 (1 − 0.253 )]
1 − 0.253exp [−0.0869 (1 − 0.253 )]
= 0.0824
(7)
From the definition of effectiveness,
ε=
q
q max
=
(
)
)
Ch Th,i − Th,o
Cmin Th,i − Tc,i
(
(
)
$
Th,o = Th,i − ε Th,i − Tc,i = 140$ C − 0.0824 (140 − 15 ) C = 129.7$ C
<
The foregoing result indicates that Th ≈ 408 K, which is much larger than the assumed value of 350 K.
Since the properties of oil depend strongly on temperature, they should be re-evaluated and the foregoing
calculations repeated until convergence is achieved. Using the Correlations, Properties and Heat
Exchangers Toolpads of IHT, we obtain hi = 226 W/m2⋅K, UA = 159 W/K, ε = 0.064, and Th,o = 132°C.
Continued …..
PROBLEM 11.51 (Cont.)
(b) If the foregoing calculations are repeated with R ′′f ,o = 0.0003 m2⋅K/W, there is only a slight increase
in the oil outlet temperature to Th,o = 132.3°C. The effect is small because the fouling resistance is
approximately an order of magnitude smaller than the convection resistances. As shown below,
Oil outlet temperature, Tho(C)
132.5
132.3
132.1
131.9
1
1.2
1.4
1.6
1.8
2
Water flowrate, mdotc(kg/s)
R''fo = 0
R''fo = 0.0003 m^2.K/W
c = 1 kg/s, Tc,o is only approximately
the effect of the water flowrate is also small, because, even for m
c on ho has not been considered, it would also be small
4.5°C larger than Tc,i. Although the effect of m
since the water-side convection resistance is substantially larger than the oil side resistance.
COMMENTS: In Part (a), note that the Nusselt number for the oil entrance region flow is 12.1/3.66 ≈
3.3 times that for fully developed flow.
PROBLEM 11.52
KNOWN: Shell-and-tube heat exchanger with one shell pass and 20 tube passes.
FIND: Average convection coefficient for the outer tube surface.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible heat loss to surroundings, (2) Negligible changes in kinetic and
potential energies, (3) Constant properties, (4) Type of oil not specified, (5) Thermal resistance of tubes
negligible; no fouling.
PROPERTIES: Table A-6, Water, liquid ( Th = 330 K): cp = 4184 J/kg⋅K, k = 0.650 W/m⋅K, µ = 489
-6
2
× 10 N⋅s/m , Pr = 3.15.
ANALYSIS: To find the average coefficient for the outer tube surface, ho, we need to evaluate hi for
the internal tube flow and U, the overall coefficient. From Eq. 11.5,
1
1
1
1  1
1 
=
+
=
+

UA h i Ai h o Ao N tπ L  h iDi h oD o 
where Nt is the total number of tubes. Solving for ho,
−1
−1
h o = Do−1  ( UA ) N tπ L −1 / hi Di  .



(1)

Evaluate hi from an appropriate correlation; begin by calculating the Reynolds number.
Re D,i =
&h
4m
4 × 0.2 kg/s
=
= 26,038.
π Di µ π ( 0.020m ) 489 ×10−6 N ⋅ s / m 2
Hence, flow is turbulent and since L >> Di, the flow is likely to be fully developed. Use the Dittus0.3
Boelter correlation with n = 0.3 since Ts < Tm, NuD = 0.023 Re4D/ 5 Pr
k
0.650 W / m ⋅ K
× 0.023 ( 26,038 ) 4 / 5 ( 3.15 )0.3 = 3594 W / m 2 ⋅ K. (2)
h i = Nu D =
D
0.020m
To evaluate UA, we need to employ the rate equation, written as
UA = q / F ∆ Tln,CF
(3)
& h cp,h (Th,i – Th,o) = 0.2 kg/s × 4184 J/kg⋅K ) (87-27)°C = 50,208 W and ∆ Tln,CF =
where q = m
[∆ T1 - ∆ T2]/ l n (∆ T1/∆ T2) = [(87 – 37) – (27 – 7)]°C/ l n (87 – 37/27 – 7) = 32.7°C. Find F ≈ 0.5
using Fig. 11.10 with P = (27 – 87)/(7 – 87) = 0.75 and R = (7 – 37)/(27 – 87) = 0.50. Substituting
numerical values in Eqs. (3) and (1), find
UA = 50,208 W / 0 . 5 × 32.7° C = 3 0 7 1 W / K
(4)
−1 
h o = ( 0.024m )
( 3071 W / K )−1 × 20 × π × 3m − 1/3594 W / m 2 ⋅ K × 0.020m 

−1
2
= 878 W / m ⋅ K.
<
COMMENTS: Using the ε-NTU method: find Ch and Cc to obtain Cr = 0.5 and ε = 0.75. From Eq.
11.31b,c find NTU = 3.59 and UA = 3003 W/K.
PROBLEM 11.53
KNOWN: Engine oil cooled by air in a cross-flow heat exchanger with both fluids unmixed.
FIND: (a) Heat transfer coefficient on oil side of exchanger assuming fully-developed conditions and
constant wall heat flux, (b) Effectiveness, and (c) Outlet temperature of the oil.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible heat loss to surroundings, (2) Negligible KE and PE changes, (3)
Constant properties, (4) Oil flow and thermal conditions are fully developed, (5) Oil cooling process
approximates constant wall flux conditions.
PROPERTIES: Table A-5, Engine oil (assume Th,o ≈ 45°C, Th = (45 + 75)°C/2 = 333 K): ch =
-2
2
2047 J/kg⋅K, µ = 7.45 × 10 N⋅s/m , k = 0.140 W/m⋅K; Table A-4, Air (assume Tc,o ≈ 40°C, Tc = (30
+ 40)°C/2 = 308 K, 1 atm): cc = 1007 J/kg⋅K.
ANALYSIS: (a) For the oil side, using Eq. 8.6, find,
(
)
& / π D µ = 4 ( 0.026 kg/s ) / π ( 0.01m ) 7.45 ×10 −2 N ⋅s / m 2 = 44.4
Re D = 4 m
Since ReD < 2000 the flow is laminar. For the fully-developed conditions with constant wall flux,
hD
NuD = i = 4.36,
k
hi = 4.36
k
0.140 W / m ⋅ K
= 4.36
= 61.0 W / m 2 ⋅ K.
D
0.01m
<
(b) The effectiveness can be determined by the ε-NTU method.
& h ch = 0.026 kg/s× 2047 J / k g ⋅ K = 53.22 W / K
Ch = m
& c cc = 0.53 k g / s ×1007 J / kg ⋅ K = 533.7 W / K
Cc = m
Cmin = Ch
Cmin / Cmax = 0.10
NTU = U A / Cmin = 53 W / m 2 ⋅ K × 1m 2 /53.22 W / K = 1.00.
Using Fig. 11.18, with Cmin/Cmax = 0.1 and NTU = 1, find ε ≈ 0.64.
<
(c) From Eqs. 11.20 and 11.19,
ε=
q
q max
=
(
) = Th,i − Th,o .
Cmin (Th,i − Tc,i ) Th,i − Tc,i
Ch Th,i − Th,o
Solving for Th,o and substituting numerical values, find
(
)
Th,o = Th,i − ε Th,i − Tc,i = 75°C − 0.64 ( 75 − 30 ) ° C = 46.2°C.
COMMENTS: Note that the Th value at which the oil properties were evaluated is reasonable.
<
PROBLEM 11.54
KNOWN: Shell-tube heat exchanger with one shell and single tube pass; Tube side: exhaust gas with
specified flow rate and temperature change; Shell side: supply of saturated water at 11.7 bar; Tube
dimensions and thermal conductivity, and fouling resistance on gas side, R ′′f,h , specified.
FIND: Number of tubes and their length if the gas velocity is not to exceed um,i = 25 m/s.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible losses to the surroundings and kinetic
and potential energy changes, (3) Negligible water-side thermal resistance, (4) Exhaust gas properties
are those of atmospheric air, (5) Gas-side flow is fully developed, and (6) Constant properties.
%
PROPERTIES: Table A-4, Air Th = 581 K : ρ = 0.600 kg / m3 , c = 1047 J / kg ⋅ K,
ν = 4.991 × 10-5 m2 / s, k = 0.0457 W / m ⋅ K, Pr = 0.684. Table A-6, Water (11.7 bar,
saturated): Tc,i = 460 K = 187$ C.
ANALYSIS: We’ll employ the NTU-ε method to design the exchanger. Since Cr = 0, use Eq.
11.36b.
$
NTU = − n 1 − ε
where the effectiveness can be evaluated from Eqs. 11.19 and 11.20.
h c h = 2 kg / s × 1047 J / kg ⋅ K = 2094 W / K
C min = C h = m
!
& = 400 − 215$ C = 0.868
ε=
C min !Th,i − Tc,i & 400 − 187$ C
NTU = − n1 − 0.868$ = 2.029
$
C h Th,i − Th,o
$
From Eq. 11.25,
UA = C min ⋅ NTU = 2094 W / K × 2.029 = 4249 W / K
(1)
Considering the gas-side flow rate and velocity criteria, find the number of tubes required as
h = N ⋅ ρ h ⋅ A c ⋅ u m,i = N ⋅ ρ h π D2i / 4 u m,i
m
"
'
Continued …..
PROBLEM 11.54 (Cont.)
$
2 kg / s = N × 0.6009 kg / m3 × π 0.050 m 2 / 4 × 25 m / s
<
N = 67.8 tubes, specify 68
The overall coefficient, considering the convection process, fouling resistance and the tube thermal
resistance, is evaluated as
Ui = 1 / R ′′f,i + R cv,i
′′ + R ′′cd,t = 56.4 W / m2 ⋅ K
R ′′f,i = 0.0015 m2 ⋅ K / W
R cv,i
′′ = 1 / hi = 1 / 62 W / m2 ⋅ K = 0.0161 m2 ⋅ K / W
R ′′cd,t =
$
$
Di n Do / Di
0.050 m n 58 / 50
=
= 9.28 × 10−5 m2 ⋅ K / W
2k
2 × 40 W / m ⋅ K
where the gas-side convection coefficient estimate is explained in the Comments section. Substituting
numerical values, determine the required tube length
$
UA = U i ⋅ A i = Ui N π Di L
4249 W / K = 56.4 W / m2 ⋅ K × 68 × π × 0.050 m × L
L = 7.1 m
<
COMMENTS: (1) Is the assumption of negligible water-side thermal resistance reasonable?
Explain why.
(2) Knowing the tube gas-side velocity, the usual convection correlation calculation methodology is
followed. The flow is turbulent, Re Di = 2.5 × 104 , and assuming fully developed flow, use the
Dittius-Boelter correlation, Eq. 8.60, to find Nu Di = 67.8 and h i = 62.0 W / m2 ⋅ K.
PROBLEM 11.55
KNOWN: Hot and cold gas flow rates and inlet temperatures of a recuperator. Overall heat transfer
coefficient. Desired cold gas outlet temperature.
FIND: (a) Required surface area, (b) Effect of surface area on cold-gas outlet temperature.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible heat loss to surroundings, (2) Negligible kinetic and potential
energy and flow work changes, (3) Constant properties.
PROPERTIES: Given: cp,c = cp,h = 1040 W/m⋅K.
ANALYSIS: (a) With Cmin = Cc = 6.2 kg/s × 1040 J/kg⋅K = 6,448 W/K, Cmax = Ch = 6.5 kg/s ×
1040 J/kg⋅K = 6,760 W/K, Cr = Cmin/Cmax = 0.954, q = Cc (Tc,o – Tc,i) = 6,448 W/K (200 K) = 1.29 ×
6
6
10 W, qmax = Cmin (Th,i – Tc,i) = 6,448 W/K (400 K) = 2.58 × 10 W, and ε = q/qmax = 0.50, Fig.
11.18 yields NTU ≈ 1.10. Hence
A=
NTU × Cmin 1.10 × 6, 448 W / K
=
= 70.9 m 2
2
U
100 W / m ⋅ K
<
(b) Using the Heat Exchanger option of IHT, the following result was obtained
Air o u tle t te m p e ra tu re , K
600
550
500
450
400
350
300
0
50
100
150
200
H e a t e xch a n g e r a re a , m ^2
The air outlet temperature increases, of course, with increasing heat exchanger area, but the approach
to the maximum possible outlet temperature, Th,i, is slow and the heat exchanger size needed to
achieve a large outlet temperature may be prohibitively expensive.
PROBLEM 11.56
KNOWN: Inlet temperature and flow rates for a concentric tube heat exchanger. Hot fluid outlet
temperature.
FIND: (a) Maximum possible heat transfer rate and effectiveness, (b) Preferred mode of operation.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state operation, (2) Negligible KE and PE changes, (3) Negligible heat
loss to surroundings, (4) Fixed overall heat transfer coefficient.
PROPERTIES: Table A-5, Ethylene glycol ( Tm = 80°C): cp = 2650 J/kg⋅K; Table A-6, Water ( Tm
≈ 30°C): cp = 4178 J/kg⋅K.
ANALYSIS: (a) Using the ε-NTU method, find
& h cp,h = ( 0.5kg/s )( 2650 J / k g ⋅ K ) = 1325 W / K.
Cmin = C h = m
Hence from Eqs. 11.19 and 11.6,
(
)
q max = Cmin Th,i − Tc,i = (1325W/K )(100 −15 ) °C = 1.13 ×105 W.
(
)
& h c p,h Th,i − Th,o = 0.5 kg/s ( 2650J/kg ⋅ K )(100 − 60 ) ° C = 0.53 × 105 W.
q=m
<
Hence from Eq. 11.20,
ε = q / q max = 0.53 ×105 /1.13 ×10 5 = 0.47.
<
(b) From Eq. 11.7,
q
0.53× 105
= 15 °C +
= 40.4 °C.
Tc,o = Tc,i +
& c c p,c
m
0.5kg/s × 4178J/kg ⋅ K
&h
Since Tc,o < Th,o, a parallel flow mode of operation is possible. However, with (Cmin/Cmax) = ( m
& c cp,c) = 0.63,
cp,h/ m
Fig. 11.14 → (NTU)PF ≈ 0.95
Fig. 11.15 → (NTU)CF ≈ 0.75.
Hence from Eq. 11.15
( ACF / APF ) = ( NTU )CF / ( NTU)PF ≈ ( 0.75/0.95) = 0.79.
Because of the reduced size requirement, and hence capital investment, the counterflow mode of
operation is preferred.
PROBLEM 11.57
KNOWN: Single-pass, cross-flow heat exchanger with both fluids (water) unmixed; hot water enters
at 90°C and at 10,000 kg/h while cold water enters at 10°C and at 20,000 kg/h; effectiveness is 60%.
FIND: Cold water exit temperature, Tc,o.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible heat loss to surroundings, (2) Negligible kinetic and potential energy
changes, (3) Constant properties.
PROPERTIES: Table A-6, Water ( Tc ≈ (10 + 40)°C/2 ≈ 300 K): cc = 4179 J/kg⋅K; Table A-6,
Water ( Th ≈ (90 + 60)°C/2 ≈ 350 K): ch = 4195 J/kg⋅K.
ANALYSIS: From an energy balance on the cold fluid, Eq. 11.7, the outlet temperature can be
expressed as
& c C c.
Tc,o = Tc,i + q / m
The heat rate can be written in terms of the effectiveness and qmax. Using Eqs. 11.20 and 11.19,
(
)
q = ε qmax = ε Cmin Th,i − Tc,i .
By inspection, it can be noted that the hot fluid is the minimum capacity fluid. Substituting numerical
values,
(
& h c h ) Th,i − Tc,i
q = ε (m
)
q = 0.60 (10,000kg/h/3600s/h ) 4195J/kg ⋅ K (90 − 10 ) ° C = 559.3 ×103 W.
The exit temperature of the cold water is then
Tc,o = 10 °C + 559.3 ×103 W /
20,000
k g / s × 4179J/kg ⋅ K = 34.1°C.
3600
<
COMMENTS: (1) The properties of the cold fluid should be evaluated at T = (Tc,o + Tc,i)/2 = (34.1
+ 10)°C/2 = 295 K. Note the analysis assumed Tc ≈ 300 K, hence little error is incurred. For best
precision, one should check Th and Ch.
(2) From Fig. 11.18, the value of NTU could be determined. First evaluate the term
& h Ch / m
& c Cc =
Cmin / Cmax = m
and with ε = 0.60, find NTU ≈ 1.2.
10,000 × 4195
= 0.50
20,000 × 4179
PROBLEM 11.58
KNOWN: Hxer consisting of 32 tubes in 0.6m square duct. Hot water enters tubes at 150°C with
3
mean velocity 0.5 m/s. Atmospheric air at 10°C enters exchanger with volumetric flow rate of 1 m /s.
2
Heat transfer coefficient on tube outer surfaces is 400 W/m ⋅K.
FIND: Outlet temperatures of the fluids, Tc,o and Th,o.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible heat loss to surroundings, (2) Negligible potential and kinetic energy
changes, (3) Constant properties, (4) Hxer is a single-pass, cross-flow type with one fluid mixed (air)
and the other unmixed (water), (5) Tube water flow is fully developed, (6) Negligible thermal
resistance due to tube wall.
3
PROPERTIES: Table A-4, Air (Tc,i = 10°C = 283 K, 1 atm): ρ = 1.2407 kg/m ; Table A-4, Air
(assume Tc,o ≈ 40°C, Tc = (10 + 40)°C/2 = 298 K, 1 atm): cp = 1007 J/kg⋅K; Table A-6, Water
-3
3
(assume Th,o ≈ 140°C, Th = (140 + 150)°C/2 = 418 K): ρ = 1/vf = 1/1.0850 × 10 m /kg, cp = 4297
-6
2
J/kg⋅K, µf = 188 × 10 N⋅s/m , kf = 0.688 W/m⋅K, Prf = 1.18.
ANALYSIS: Using the ε-NTU method, first find the capacity rates.
& h c p,h = ( ρ Ac u m ) N ⋅ c p,h
Ch = m
h
Ch =
)
(
2
π
m
J
W
10.2 ×10 −3 m × 0.5 × 32 × 4297
= 5178
3
3
−
s
kg ⋅ K
K
1.0850 ×10 m /kg 4
1
×
kg
W
& c c p,c = ( ρ V ) c p,c = 1.2407
× lm 3 / s ×1007J/kg ⋅ K = 1249 .
Cc = m
c
K
m3
(1,2)
Note that the cold fluid is the minimum fluid, Cc = Cmin. The overall heat transfer coefficient follows
from Eq. 11.5,
−1
 1
1 
UoAo = 
+
(3)

 h i Ai h o A o 
where hi must be estimated from an appropriate internal flow correlation. The Reynolds number for
water flow is
1/1.0850 ×10 −3 m 3kg × 0.5m/s × 10.2 × 10−3 m
ρ u m Di
Re D =
=
= 25,002. (4)
µ
188 ×10−6 N ⋅ s / m2
(
)
(
)
Continued …..
PROBLEM 11.58 (Cont.)
-3
The flow is turbulent and since L/Di = 0.6m/10.2 × 10 m = 59, fully developed conditions may be
assumed. The Dittus-Boelter correlation with n = 0.3 is appropriate.
hD
0.8 Pr 0.3 = 0.023 25,002 0.8 1.18 0.3 = 79.7
Nu D = i i = 0.023Re D
(
) ( )
k
hi =
k
0.688W/m ⋅ K
Nu D =
× 79.7 = 5376W/m 2 ⋅ K.
−
3
Di
10.2 ×10 m
Substituting numerical values into Eq. (3), find
 12.5mm 

1
1
Uo = 
+


10.2mm  5376W/m2 ⋅ K 4 0 0 W / m2 ⋅ K 


−1
= 366.6W/m2 ⋅ K.
It follows from Eq. 11.25, with Ao = N(πDo L), that
NTU =
)
(
Uo Ao
W
W
= 366.6
× 32 ×π ×12.5 ×10−3 m × 0.6m /1249
= 0.22.
Cmin
K
m2 ⋅ K
From Fig. 11.19, noting that Cmin = Cc is the mixed fluid (solid curves),
Cmixed
C
C
1249W/K
= min = c =
= 0.24
Cunmixed Cmax Ch 5178W/K
and with NTU = 0.22 find ε ≈ 0.19. From the definition of effectiveness, Eq. 11.20,
ε=
q
q max
=
(
Cc Tc,o − Tc,i
(
)
Cmin Th,i − Tc,i
(
)
)
Tc,o = Tc,i + ε Th,i − Tc,i = 10 °C + 0.19 (150 −10 ) °C = 36.6°C.
<
Equating the energy balances on both fluids,
(
)
(
Cc Tc,o − Tc,i = Ch Th,i − Th,o
)
or
(
C
Th,o = Th,i − c Tc,o − Tc,i
Ch
Th,o = 150°C −
)
1249W/K
( 36.6 −10 ) °C = 143.5°C.
5178W/K
COMMENTS: (1) Note that the assumptions of Th,o and Tc,o used in evaluating properties are
reasonable.
& c from V, the density at 10°C is more appropriate than at Tc .
(2) Note that to calculate m
<
PROBLEM 11.59
KNOWN: Flow rates and inlet temperatures of exhaust gases and combustion air used in a crossflow (one fluid mixed) heat exchanger. Overall heat transfer coefficient. Desired air outlet
temperature.
FIND: Required heat exchanger surface area.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible heat loss to surroundings, (3) Negligible
kinetic and potential energy changes, (4) Constant properties, (5) Gas properties are those of air.
PROPERTIES: Table A-4, Air ( Tm ≈ 700 K, 1 atm): cp = 1075 J/kg⋅K.
ANALYSIS: From Eqs. 11.6 and 11.7,
Th,o = Th,i −
& c cp,c
m
& h c p,h
m
kg/s
( 850 − 300 ) K = 733K.
( Tc,o − Tc,i ) = 1100K − 10
15 kg/s
From Eqs. 11.15, 11.17 and 11.18,
∆ Tlm = F
( Th,i − Tc,o ) − ( Th,o − Tc,i ) = F 250 − 433 = F × 333K.
ln ( 250/433 )
ln  ( Th,i −Tc,o ) / ( Th,o − Tc,i ) 


From Fig. 11.13, with R = (300 – 850)/(733 – 1100) = 1.50 and P = (733 – 1100)/(300 – 1100) = 0.46,
F ≈ 0.73. With
(
)
& h c p,h Th,i − Th,o = 15kg/s ×1075J/kg ⋅ K ( 367K ) = 5.92 ×10 6 W
q=m
it follows from Eq. 11.14 that
A=
5.92 × 106 W
1 0 0 W / m 2 ⋅K × 0.73 ( 333K )
= 243m 2 .
COMMENTS: Using the effectiveness-NTU method, from Eq. 11.22
Tc,o − Tc,i (850 − 300 ) K
=
= 0.688.
ε=
Th,i − Tc,i (1100 − 300) K
Hence, with Cmixed/Cunmixed = Cc/Ch = 0.67, Fig. 11.19 gives NTU ≈ 2.3. From Eq. 11.25,
C
10 k g / s ×1075 J / k g ⋅ K
A = NTU min ≈ 2.3
≈ 247m 2 .
2
U
100 W / m ⋅ K
<
PROBLEM 11.60
KNOWN: Dimensions, configuration and material of a single-pass, cross-flow heat exchanger. Inlet
conditions of inner and outer flow. Fouling factor of inner surface.
FIND: (a) Percent fuel savings for prescribed conditions, (b) Effect of UA on air outlet temperature and
fuel savings.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible heat loss to surroundings and potential and kinetic energy changes, (2)
Air properties are those of atmospheric air at 300 K, (3) Gas properties are those of atmospheric air at
1400 K, (4) Tube wall temperature may be approximated as 800 K for treating variable property effects.
PROPERTIES: Table A.4, Air (1 atm, T = 300 K): ν = 15.89 × 10-6 m2/s, cp = 1007 J/kg⋅K, k = 0.0263
W/m⋅K, Pr = 0.707; (T = 1400 K): µ = 530 × 10-7 kg/s⋅m, cp = 1207 J/kg⋅K, k = 0.091 W/m⋅K, Pr =
0.703; (T = 800 K): µ = 370 × 10-7 kg/s⋅m, Pr = 0.709.
c cp,c = 1 kg/s × 1007 J/kg⋅K = 1007 W/K = Cmin and Ch
ANALYSIS: (a) With capacity rates of Cc = m
h c p,h = 1.05 kg/s × 1207 J/kg⋅K = 1267 W/K = Cmax, Cmin/Cmax = 0.795. The overall coefficient is
= m
1
UA
=
1
h i Ai
+
R ′′f ,i
Ai
ln ( Do Di )
1
.
+
( 2π kL ) N h o Ao
+
For flow through a single tube,
Re D =
4m
h
Nπ Di µ
=
4 × 1.05 kg s
80π ( 0.055 m ) 530 × 10 −7 kg s ⋅ m
= 5733 .
Assuming fully developed turbulent flow throughout and using the Sieder-Tate correlation,
0.14
4 / 5 1/ 3
Nu D = 0.027 ReD
Pr ( µ µs )
4/5
= 0.027 (5733 )
(0.703)1/ 3 (530 370 )0.14 = 25.6
h i = Nu D k Di = 25.6 ( 0.091W m⋅ K ) 0.055 m = 42.4 W m 2⋅ K .
For flow over the tube bank,
Vmax = [ST (ST − Do )] V = [0.12 m (0.12 − 0.08 ) m ]1m s = 3m s
3 m s ( 0.08 m )
V
D
Re D,max = max o =
= 15,100
ν
15.89 × 10−6 m 2 s
From the Zhukauskas correlation for a tube bank,
0.63
(0.707 )0.36 (0.707 0.709 )1/ 4 = 102.3
ho = Nu D ( k Do ) = 102.3 ( 0.0263 W m⋅ K ) 0.08 m = 33.6 W m 2⋅ K .
Nu D = 0.27 (15,100 )
Hence, based on the inner surface, the overall coefficient is
Continued...
PROBLEM 11.60 (Cont.)
1
Ui
1
Ui
=
1
hi
+ R ′′f ,i +
Di ln ( Do Di )
2k
+
Di
Do h o

0.055 ln ( 0.08 0.055 )

40
=  0.0236 + 0.0002 +
+
 2
m ⋅K W
0.08 × 33.6 
0.055
Ui = ( 0.0236 + 0.0002 + 0.0005 + 0.0246 ) m 2 ⋅ K W 

−1

= 22.3 W m 2⋅ K .
Hence, ( UA )i = Ui Nπ Di L = 22.3 W m 2⋅ K × 80π ( 0.055 m )1.4 m = 432 W K . The number of transfer
units is then NTU = UA/Cmin = 432 W/K/1007 W/K = 0.429, and with Cmixed/Cunmixed = Cc/Ch = Cmin/Cmax =
0.795, Fig. 11.19 yields ε ≈ 0.3 or, from Eq. 11.35 a,
(
)
ε = 1 − exp −C−r 1 {1 − exp [−Cr ⋅ NTU ]} = 0.305 .
Hence, with
(
)
q max = Cmin Th,i − Tc,i = 1007 W K (1100 K ) = 1.11 × 106 W
q = ε q max = 0.305 × 1.11 × 106 W = 337,800 W
Tc,o = Tc,i + q C min = 300 K + (337, 800 W 1007 W K ) = 635 K .
Hence,
750
45
700
40
Fuel Savings (%)
Air outlet temperature, Tco(K)
% fuel savings ≡ FS = ( ∆Tc 10 K ) × 1% = (335 K 10 K ) × 1% = 33.5%
(b) Using the Heat Exchangers Toolpad of IHT to perform the parametric calculations, the following
results are obtained.
650
600
<
35
30
550
25
300
350
400
450
500
550
600
300
Convection parameter, UA(W/K)
350
400
450
500
550
600
Convection parameter, UA(W/K)
Significant benefits are derived by increasing UA, with values of Tc,o = 716 K and FS = 41.6% obtained
for UA = 600 W/K. The major contributions to the total resistance are made by the inner and outer
convection resistances. These contributions could be reduced by using extended surfaces on both the
inner and outer surfaces.
COMMENTS: For part (a), properties of the flue gas should be evaluated at (Th,i + Th,o)/2 and the
calculations repeated. The Colburn equation yields
4 / 5 1/ 3
= 20.8
Nu D = 0.023 ReD
Pr
which is 19% less than the result of the Sieder-Tate correlation.
PROBLEM 11.61
KNOWN: Rate of thermal energy production in combustor and transfer to load in furnace. Cold air and
flue gas flowrates and specific heats in recuperator. Recuperator cold air inlet temperature.
FIND: Recuperator hot gas inlet and outlet temperatures and air outlet temperature for a recuperator
effectiveness of ε = 0.3. Value of ε needed to achieve a recuperator outlet temperature of 800 K.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible flowwork and potential and kinetic energy changes, (2) Constant
properties, (3) Negligible effect of fuel addition on flowrate.
PROPERTIES: Air and gas: cp,c = cp,h = 1200 J/kg⋅K.
ANALYSIS: With Cc = Ch = Cmin, the effectiveness of the recuperator, ε = q/qmax, may be expressed as
ε=
(
Cc Tc,o − Tc,i
(
)
Cmin Th,i − Tc,i
)
=
Tc,o − 300 K
Th,i − 300 K
= 0.3
The unknown temperatures, Tc,o and Th,i, are also related through an energy balance performed on the air
entering the combustor and leaving the furnace. Specifically,
(
)
C Th,i − Tc,o = q comb − q load = 0.6 × 106 W
where C = 1 kg/s × 1200 J/kg⋅K = 1200 W/K. Solving the foregoing equations, we obtain
Th,i = 1014 K
Tc,o = 514 K
<
Expressing the effectiveness as
ε=
(
)
)
Ch Th,i − Th,o
1014 K − Th,o
=
714 K
Cmin Th,i − Tc,i
we also obtain
(
Th,o = 800 K.
<
For a combustor air inlet temperature of Tc,o = 800 K and Th,i = 1014 K, the required effectiveness is
ε=
Tc,o − Tc,i
Th,i − Tc,i
=
(800 − 300 ) K
= 0.70
(1014 − 300 ) K
<
COMMENTS: The effectiveness of the recuperator may be increased by increasing NTU and hence
UA, as, for example, by increasing the number of tubes.
PROBLEM 11.62
2
KNOWN: Single-shell, two-tube pass heat exchanger with surface area 0.5 m and overall heat
2
transfer coefficient of 2000 W/m ⋅K; saturated steam at 100°C condenses on one side while water at a
flow rate of 0.5 kg/s enters at 15°C.
& h.
FIND: (a) Outlet temperature of the water, Tc,o, (b) Rate of condensation of steam, m
SCHEMATIC:
ASSUMPTIONS: (1) Negligible heat loss to surroundings, (2) Negligible, kinetic and potential energy
changes, (3) Constant properties.
PROPERTIES: Table A-6, Steam (100°C, 1 atm): hfg = 2257 kJ/kg; Table A-6, Water ( Tc ≈ (15 +
35)°C/2 ≈ 300 K): cc = 4179 J/kg⋅K.
ANALYSIS: (a) Using the ε-NTU method of analysis, recognize that the minimum capacity fluid is
the cold fluid since for the hot fluid, Ch → ∞. See Fig. 11.9a. That is,
& c cc = 0.5 kg/s × 4179 J / k g ⋅ K = 2090 W / K.
Cmin = m
It follows also that,
NTU = A U / Cmin = 0.5m 2 × 2000 W / m 2 ⋅ K / 2 0 9 0 W / K = 0.48.
& c is the minimum
Using NTU = 0.48 and Cmin/Cmax = 0, find from Fig. 11.16 that ε = 0.39. Since m
fluid, from Eq. 11.22
(
)(
)
Tc,o = Tc,i + ε ( Th,i − Tc,i ) = 15 °C + 0.39 (100 −15) °C = 48.2 °C.
ε = Tc,o − Tc,i / Th,i − Tc,i
<
(b) The rate of steam condensation can be expressed as
& h = q / h fg.
m
From Eqs. 11.19 and 11.20
(
q = ε q max = ε Cmin Th,i − Tc,i
)
q = 0.39 × 2090W/K (100 − 15 ) K = 69,284W.
Hence, the condensation rate is
& h = 69,284W/2257 ×103 J / k g = 0.031kg/s.
m
<
& h.
& c >> m
COMMENTS: (1) Be sure to recognize why Ch → ∞. Note also that m
(2) Note that Tc = (Tc,i + Tc,o)/2 = (15 + 48.2)°C/2 ≈ 305 K. This compares favorably with the value
of 300 K at which properties of the cold fluid were evaluated.
PROBLEM 11.63
KNOWN: Concentric tube heat exchanger with prescribed conditions.
FIND: (a) Maximum possible heat transfer, (b) Effectiveness, (c) Whether heat exchanger should be
run in PF or CF to minimize size or weight; determine ratio of required areas for the two flow
conditions.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible heat loss to surroundings, (2) Negligible kinetic and potential energy
changes, (3) Constant properties, (4) Overall heat transfer coefficient remains unchanged for PF or CF
conditions.
PROPERTIES: Hot fluid (given): c = 2100 J/kg⋅K; Cold fluid (given): c = 4200 J/kg⋅K.
ANALYSIS: (a) The maximum possible heat transfer rate is given by Eq. 11.19.
(
)
q max = Cmin Th,i − Tc,o .
& h ch, giving
The minimum capacity fluid is the hot fluid with Cmin = m
(
)
& h c h Th,i − Tc,o = 0.125
q max = m
kg
J
× 2100
( 210 − 40 ) K = 44,625W.
s
kg ⋅ K
<
(b) The effectiveness is defined by Eq. 11.20 and the heat rate, q, can be determined from an energy
balance on the cold fluid.
(
)
& c c c Tc,o − Tc,i / q max
ε = q / q max = m
ε = 0.125 kg/s× 4200 J / k g ⋅ K (95 − 40 ) K/44,625W = 0.65.
<
(c) Operating the heat exchanger under CF conditions will require a smaller heat transfer area than for
PF conditions. The ratio of the areas is
ACF q / U ∆ Tl m,CF ∆ Tl m,PF
.
=
=
A PF q / U ∆ Tl m,PF ∆ Tlm,CF
To calculate the LMTD, first find Th,o from overall energy balances on the two fluids.
Th,o = Th,i −
& c cc
m
0.125 × 4200
Tc,o − Tc,i = 210°C −
( 95 − 40 ) ° C = 100°C.
& h ch
m
0.125 × 2100
(
)
Using Eq. 11.15 with ∆T1 and ∆T2 as shown below, find ∆ Tlm = (∆T1 - ∆T2)/ l n (∆T1/∆T2).
Substituting values, find
ACF  ( 210 − 40 ) − (100 − 95 ) / ln (170/5) 46.8°C
=
=
= 0.55.
A PF ( 210 − 95 ) − (100 − 40)  ln (115/60 ) 84.5°C
<
COMMENTS: In solving part (c), it is also possible to use Figs. 11.15 and 11.16 to evaluate NTU
values for corresponding ε and Cmin/Cmax values. With knowledge of NTU it is then possible to find
ACF/APF.
PROBLEM 11.64
KNOWN: Concentric tube HXer with prescribed inlet fluid temperatures, fluid flow rates and overall
coefficient.
FIND: (a) Maximum heat transfer rate, qmax; (b) Outlet fluid temperatures when area is 0.33 m2 with CF
operation; (c) Compute and plot the effectiveness, ε, and fluid outlet temperatures, Tc,o and Th,o, as a
function of UA for the range 50 ≤ UA ≤ 1000 W/K for CF operation with all other conditions remaining
the same; as UA becomes very large, find asymptotic value for Th,o; (d) Largest heat transfer rate which
could be achieved if HXer is very long with PF operation; effectiveness for this arrangement; and (e)
Compute and plot ε, Tc,o and Th,o as a function of UA for the range 50 ≤ UA ≤ 1000 W/K for PF
operation with all other conditions remaining the same; as UA becomes very large, find asymptotic value
for Tc,o and Th,o.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible heat loss to surroundings, (2) Negligible kinetic and potential energy
changes, (3) Constant properties.
PROPERTIES: Table A-6, Water (Assume Tc,o ≈ 85°C, Tc ≈ 335 K): cc = 4186 J/kg⋅K, (Assume Th,o ≈
100°C, Th ≈ 100°C, Th ≈ 420 K): ch = 4302 J/kg⋅K.
ANALYSIS: (a) With Cmin = Ch, the maximum heat transfer rate from Eq. 11.19 is
(
)
(
)
q max = C min Th,i − Tc,i = C h Th,i − Tc,i =
42 kg
3600 s
× 4302 ×
J
kg ⋅ K
( 200 − 35 ) K = 8281 W .
<
(b) Using the ε - NTU method, find ε from values of Cmin, Cmin/Cmax, and NTU.
C min = 42 3600 kg s × 4302 J kg ⋅ K = 50.19 W K ,
C min C max =
42 kg h × 4302 J kg ⋅ K
84 kg h × 4186 J kg ⋅ K
= 0.514
NTU = UA Cmin = 180 W m 2 ⋅ K × 0.33m 2 50.19 W K = 1.184 .
Using Eq. 11.30 for counter flow operation, with Cr = Cmin/Cmax, find that
1 − exp [− NTU (1 − Cr )]
1 − exp [−1.18 (1 − 0.514 )]
=
= 0.616 .
ε=
1 − Cr exp [− NTU (1 − Cr )] 1 − 0.514 exp [−1.18 (1 − 0.514 )]
From the definition of effectiveness, ε = Ch (Th,i - Th,o)/Cmin (Th,i - Tc,i), it follows that
(
)
$
Th,o = Th,i − ε Th,i − Tc,i = 200$ C − 0.62 ( 200 − 35 ) C = 98.4$ C .
<
Equating the energy balances on both fluids, Ch (Th,i - Th,o ) = Cc (Tc,o - Tc,i ), find
(
)
$
Tc,o = ( Ch Ce ) Th,i − Th,o + Tc,i = 0.514 ( 200 − 98.4 ) C + 35$ C = 87.2$ C .
<
Continued...
PROBLEM 11.64 (Cont.)
(c) Using the IHT Heat Exchanger Tool, Concentric Tube, counter flow operation and the Properties
Tool for Water, a model was developed using the effectiveness NTU method employed in the previous
analysis to compute ε, Tc,o and Th,o as a function of UA for CF operation. The results are plotted and
discussed below.
(d) For PF with same prescribed inlet conditions, the temperature distributions appear as shown above
when A→ ∞ . At the outlet, Tc,o = Th,o, and from the sketch δTh,max + δTc,max = (200 - 35)°C = 165°C.
From the energy balance, find
Chδ Th,max = Ccδ Tc,max
and solving simultaneously, find
δ Th,max = 109.0$ C
Th,o = Th,i − δ Th,max = 200 − 109.0 = 91.0$ C .
The heat rate and effectiveness are
<
<
q = Ch ⋅ δ Th,max = 50.19 W K × 109.0K = 5471W
ε = q q max = 5471W 8, 281W = 0.661 .
Counterflow operation
Tco, Tho (C), eps*100
Tco , Tho (C), eps*100
(e) Using the IHT model from part (c), but for PF operation, the effectiveness, Tc,o and Th,o were
computed and plotted as a function of UA.
130
110
90
70
50
30
0
200
400
600
800
1000
Parallel flow operation
130
110
90
70
50
30
100
200
300
400
UA (W/K)
UA (W/K)
Tho (C), minumum fluid
Tco (C), maximum fluid
Effectiveness, eps*100
Tho (C), minumum fluid
Tco (C), maximum fluid
Effectiveness, eps*100
500
COMMENTS: (1) From the plot for CF operation as UA increases, the minimum (hot) fluid outlet
temperature, Th,o, decreases to the cold fluid temperature, Tc,i. That is when UA → ∞ , Th,o →Tc,i. As UA
→ ∞ , the effectiveness approaches unity as expected since a very large CF heat exchanger has a heat rate
qmax and ε = 1.
(2) From the plot for PF operation, as UA increases, Th,o and Tc,o approach an asymptotic value, 91.0°C.
Also, as UA → ∞ , the effectiveness increases, approaching 0.661, rather than unity as would be the case
for CF operation.
PROBLEM 11.65
KNOWN: Flow rates and inlet temperatures of water and glycol in counterflow heat exchanger.
Desired glycol outlet temperature. Heat exchanger diameter and overall heat transfer coefficient
without and with spherical inserts.
FIND: (a) Required length without spheres, (b) Required length with spheres, (c) Explanation for
reduction in fouling and pump power associated with using spheres.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible kinetic energy, potential energy and flow work changes, (2)
Negligible heat loss to surroundings, (3) Constant properties, (4) Negligible tube wall thickness.
(
)
PROPERTIES: Table A-5, Ethylene glycol Th = 70°C : cp,h = 2606 J/kg⋅K; Table A-6, Water
(Tc ≈ 35°C ) : cp,c = 4178 J/kg⋅K.
ANALYSIS: (a) With Ch = Cmin = 1303 W/K and Cc = Cmax = 2089 W/K, Cr = 0.624. With actual
and maximum possible heat rates of
(
)
q = Ch Th,i − Th,o = 1303 W / K (100 − 40 ) °C = 78,180 W
(
)
q max = Cmin Th,i − Tc,i = 1303 W / K (100 − 15 ) °C = 110, 755 W
the effectiveness is ε = q/qmax = 0.706. From Eq. 11.30b,
NTU =
 ε −1 
1
 0.294 
ln 
 = −2.66 ln 
 = 1.71
Cr − 1  ε Cr − 1 
 0.559 
Hence, with A = πDL and NTU = UA/Cmin,
C
NTU
1303 W / K × 1.71
L = min
=
= 9.46m
π Di U
π (0.075m )1000 W / m 2 ⋅ K
<
c, m
h, Th,i, Th,o and Tc,i are unchanged, Cr, ε and NTU are unchanged. Hence, with U
(b) Since m
2
= 2000 W/m ⋅K,
<
L = 4.73m
(c) Because the spheres induce mixing of the flows, the potential for contaminant build-up on the
surfaces, and hence fouling, is reduced. Although the obstruction to flow imposed by the spheres acts
to increase the pressure drop, the reduction in the heat exchanger length reduces the pressure drop.
The second effect may exceed that of the first, thereby reducing pump power requirements.
COMMENTS: The water outlet temperature is Tc,o = Tc,i + q/Cc = 15°C + 78,180 W/2089 W/K =
(
)
52.4°C. The mean temperature Tc = 33.7°C is close to that used to evaluate the specific heat of
water.
PROBLEM 11.66
KNOWN: Concentric tube, counter-flow heat exchanger.
FIND: Total heat transfer rate and outlet temperatures of both fluids.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible heat loss to surroundings, (2) Negligible kinetic and potential energy
changes, (3) Constant properties.
PROPERTIES: Table A-6, Water ( Th ≈ 68°C ≈ 340 K): ch = 4188 J/kg⋅K; Table A-6, Water ( Tc ≈
37°C = 310 K): cc = 4178 J/kg⋅K.
ANALYSIS: Using the ε-NTU method, begin by evaluating the capacity rates.
& h ch = 2.5kg/s × 4188 J / k g ⋅ K = 10,470 W / K
Ch = m
& ccc = 5.0kg/s × 4178 J / k g ⋅ K = 20,890 W / K
Cc = m
Hence, Cmin = Ch and Cmin/Cmax = 0.50
From the definition, Eq. 11.25,
NTU = U A / Cmin = 1000W/m 2 ⋅ K × 23m 2 / (10,470W/K ) = 2.20.
Using values of NTU and Cmin/Cmax, find from Fig. 11.15, that
ε ≈ 0.80.
From the definition of ε, Eq. 11.20, it follows that
(
)
q = ε qmax = ε Cmin Th,i − Tc,i = 0.80 ×10,470W/K (100 − 20 ) K = 670kW.
<
Performing energy balances on both fluids, find
Tc,o = Tc,i + q / Cc = 20 °C + 670kW/20,890W/K = 52.1°C
<
Th,o = Th,i − q / C h = 100°C − 670kW/10,470W/K = 36.0°C.
<
COMMENTS: (1) Note that Tc = (20 + 52.1)°C/2 ≈ 310 K and Th = (100 + 36)°C/2 = 341 K and
that these values agree well with those used to evaluate the properties.
(2) Eq. 11.30 could be used to evaluate ε; the result gives ε = 0.800.
PROBLEM 11.67
KNOWN: Shell and tube heat exchanger for cooling exhaust gases with water.
FIND: Required surface area using ε-NTU method.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible heat loss to surroundings, (2) Negligible changes in kinetic and
potential energies, (3) Constant properties, (4) Gases have properties of air.
PROPERTIES: Table A-6, Water, liquid ( Tc = (85 + 35)°C/2 = 333 K): cp = 4185 J/kg⋅K.
ANALYSIS: Using the ε-NTU method, the area can be expressed as
A = NTU ⋅ Cmin / U
(1)
where NTU must be found from knowledge of ε and Cmin/Cmax = Cr. The capacity rates are:
& c cp,c = 2.5kg/s × 4185 J / k g ⋅ K = 10,463W/K
Cc = m
Equating the energy balance relation for each fluid,
(
)(
)
Ch = Cc Tc,o − Tc,i / Th,i − Th,o = 10,463W/K ( 85 − 35 ) / ( 200 − 93) = 4889W/K.
Hence,
Cr = Cmin / Cmax = Ch / Cc = 4889/10,463 = 0.467.
The effectiveness of the exchanger, with qmax = Cmin (Th,i – Tc,i) and Cmin = Ch, is
(
)
(
)
ε = q / q max = C h Th,i − Th,o / Ch Th,i − Tc,i = ( 200 − 93 ) / ( 200 − 35 ) = 0.648.
Considering the HXer to be a single shell with 2,4….tube passes, Eqs. 11.31b,c are appropriate to
evaluate NTU.
−1/2 E −1
2 / ε1 − (1 + Cr )
NTU = − 1 + C2r
ln
E=
.
1/2
E +1
2
1 + Cr
(
)
(
)
Substituting numerical values,
E=
2/0.648 − (1 + 0.467 )
(1 + 0.4672 )
1/2
(
= 1.467 NTU = − 1 + ( 0.467 )2
)
−1/2
ln
1.467 − 1
= 1.51.
1.467 + 1
Using the appropriate numerical values in Eq. (1), the required area is
A = 1.51 × 4889W/K/180W/m 2 ⋅ K = 40.9m 2.
COMMENTS: Figure 11.16 could also have been used with Cr and ε to find NTU.
<
PROBLEM 11.68
KNOWN: Dimensions, fluid flow rates, and fluid temperatures for a counterflow heat exchanger used to
heat blood.
FIND: (a) Outlet temperature of the blood, (b) Effect of water flowrate and inlet temperature on heat
rate and blood outlet temperature.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible heat loss to surroundings, (2) Negligible kinetic and potential energy
changes, (3) Constant properties.
PROPERTIES: Table A.6, Water ( Tm ≈ 55°C): cp = 4183 J/kg⋅K.
c
ANALYSIS: (a) Using the ε - NTU method, we first obtain Ch = ( m
h p,h ) = (0.10 kg/s × 4183 J/kg⋅K)
c
= 418.3 W/K and Cc = ( m
c p,c ) = (0.05 kg/s × 3500 J/kg⋅K) = 175 W/K = Cmin. Hence, (Cmin/Cmax) =
0.418 and
NTU =
UA
Cmin
500 W m 2⋅ K )π ( 0.055 m )( 0.5 m )
(
=
= 0.247 .
175 W K
From Eq. 11.30, ε = 0.21. Hence, from Eq. 11.23
(
)
$
q = ε Cmin Th,i − Tc,i = 0.21(175 W K )( 60 − 18 ) C = 1544 W .
From Eq. 11.7,
q
1544 W
Tc,o = Tc,i +
= 18$ C +
= 26.8$ C
Cc
175 W K
<
h does not have a significant effect on ε for the prescribed
(b) Because the variation of Cmin/Cmax with m
h.
NTU, Tc,o and q increase only slightly with increasing m
2000
29
1800
28
Heat rate, q(W)
Blood outlet temperature, Tco(C)
30
27
26
1600
1400
1200
25
24
1000
0.05
0.1
0.15
Water mass flowrate, mdoth(kg/s)
Thi = 70 C
Thi = 60 C
Thi = 50 C
0.2
0.05
0.1
0.15
0.2
Water mass flowrate, mdoth(kg/s)
Tci = 70 C
Tci = 60 C
Tci = 50 C
However, the water inlet temperature does have a significant effect, and accelerated heating is achieved
with Th,i = 70°C.
h = 0.2 kg/s and Th,i = 70°C, the outlet temperature of the blood is still below
COMMENTS: With m
the desired level of Tc,o ≈ 37°C. This value of Tc,o could be increased by increasing L or Th,i.
PROBLEM 11.69
KNOWN: Inlet temperatures and flow rates of water (c) and ethylene glycol (h) in a shell-and-tube heat
exchanger (one shell pass and two tube passes) of prescribed area and overall heat transfer coefficient.
FIND: (a) Heat transfer rate and fluid outlet temperatures and (b) Compute and plot the effectiveness, ε,
h , for the
and fluid outlet temperatures, Tc,o and Th,o as a function of the flow rate of ethylene glycol, m
h ≤ 5 kg/s.
range 0.5 ≤ m
SCHEMATIC:
ASSUMPTIONS: (1) Negligible heat loss to surroundings, (2) Negligible kinetic and potential energy
changes, (3) Constant properties, and (4) Overall coefficient remains unchanged.
PROPERTIES: Table A-5, Ethylene glycol ( Tm ≈ 40°C): cp = 2474 J/kg⋅K; Table A-6, Water ( Tm ≈
15°C): cp = 4186 J/kg⋅K.
ANALYSIS: (a) Using the ε-NTU method we first obtain
c
Ch = m
h p,h = ( 2kg s × 2474J kg ⋅ K ) = 4948 W K
(
)
Cc = ( mc cp,c ) = (5kg s × 4186J kg ⋅ K ) = 20, 930 W K .
Hence with Cmin = Ch = 4948 W/K and Cr = Cmin/Cmax = 0.236,
NTU =
UA
Cmin
800 W m 2 ⋅ K )15m 2
(
=
= 2.43 .
4948 W K
From Fig. 11.16, ε = 0.81 and from Eq. 11.23
(
)
<
q = ε Cmin Th,i − Tc,i = 0.81( 4948 W K )( 60 − 10 ) K = 2 × 105 W .
From Eqs. 11.6 and 11.7, energy balances on the fluids,
Tc,o = Tc,i +
Ch
q
Cc
$
= 60 C −
= 10$ C +
2 × 105 W
4948 W K
<
= 19.6$ C
2 × 105 W
20, 930 W K
<
= 19.6$ C .
(b) Using the IHT Heat Exchanger Tool, Shell and
Tube, and the Properties Tool for Water and
Ethylene Glycol, Tc,o, Th,o, and ε as a function of
h were computed and plotted.
m
h ) note that ε → 1 while
At very low Cmin, (low m
h increases, both fluid outlet
Th,o → Tc,i. As m
temperatures increase and the effectiveness
decreases.
Tco, Tho (C), eps*100
Th,o = Th,i −
q
100
80
60
40
20
0
0
1
2
3
4
Hot fluid flow rate, mdoth (kg/s)
Tco (C), max fluid
Tho (C), min fluid
Effectiveness, eps*100
5
PROBLEM 11.70
KNOWN: Flow rate, specific heat and inlet temperature of gas in cross-flow heat exchanger. Flow
rate and temperature of water which enters as saturated liquid and leaves as saturated vapor. Number
of tubes, tube diameter and overall heat transfer coefficient.
FIND: Required tube length.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible kinetic and potential energy changes, (2) Negligible heat loss to
surroundings, (3) Constant gas specific heat.
6
PROPERTIES: Table A-6, Saturated Water, (T = 450 K): hfg = 2.024 ×10 J/kg.
ANALYSIS: Use effectiveness-NTU method
ε=
q
q max
=
(
q
Cmin Th,i − Tc,i
)
=
(
q
& h c p,h Th,i − Tc,i
m
)
& c h fg = 3 k g / s × 2.024 ×10 6 J / k g = 6.072 × 106 W
q=m
ε=
6.072 × 106 W
= 0.571
10 k g / s ×1120J/kg ⋅ K (1400 − 450 ) K
Cmin / Cmax = 0.
From Fig. 11.19, find
NTU ≈ 0.8 ≈ Uo Nπ D o L / Cmin
L≈
0.8 × 10 kg/s ×1120J/kg ⋅ K
50 W / m 2 ⋅ K × 500π × 0.025m
<
= 4.56m.
COMMENTS: (1) The gas outlet temperature is
& h c p,h = 1400K − 6.072 × 106 W/10kg/s ×1120J/kg ⋅ K = 857.9 K.
Th,o = Th,i − q / m
(2) Using the LMTD method,
∆ Tlm,CF =  (1400 − 450 ) − ( 858 − 450 ) / ln  (1400 − 450 ) / (858 − 450 )  = 641K.
From Fig. 11.13, find F = 1, so the area and length are
Ao = q / Uo F ∆Tlm,CF = 6.072 ×10 6 W / 50 W / m 2 ⋅ K ×1 × 641K = 189m 2
(
L = A / Nπ Do = 189m 2 /500π ( 0.025m ) = 4.82m.
)
PROBLEM 11.71
KNOWN: Gas flow conditions upstream of a tube bank of prescribed geometry. Flow rate and inlet
temperature of water passing through the tubes.
FIND: (a) Overall heat transfer coefficient, (b) Water and gas outlet temperatures, (c) Effect of water
flow rate on heat recovery and outlet temperatures.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Negligible heat loss to the
surroundings and kinetic and potential energy changes, (4) Negligible tube fouling and wall thermal
resistance, (5) Fully developed water flow, (6) Gas properties are those of air.
PROPERTIES: Table A.6, Water (Assume Tm ≈ 340 K): cp = 4188 J/kg⋅K, µ = 420 × 10-6 N⋅s/m2, k =
0.660 W/m⋅K, Pr = 2.66; Table A.4, Air (Assume Tm ≈ 600 K): cp = 1051 J/kg⋅K, ν = 52.7 × 10-6 m2/s,
k = 0.047 W/m⋅K, Pr = 0.69.
c,1 =
ANALYSIS: (a) For the prescribed conditions, U = (1/hi + 1/ho)-1. For the internal flow, with m
0.025 kg/s,
4m
4 × 0.025 kg s
c,1
Re D =
=
= 3032 .
π Dµ π ( 0.025 m ) 420 × 10−6 N ⋅ s m 2
Hence, assuming turbulent flow,
Nu D = 0.023 Re 4D/ 5 Pr 0.4 = 0.023 (3032 )
4/5
hi =
k
D
Nu D =
0.660 W m⋅ K
0.025 m
For the external flow, Vmax =
( 2.66 )0.4 = 20.8
20.8 = 548 .
0.05 m
(0.05 − 0.025 ) m
5.0 m s = 10.0 m s . Hence
V
D
10 m s × 0.025
Re D,max = max =
= 4744
ν
52.7 × 10−6 m 2 s
0.36
From the Zhukauskas correlation and Tables 7.7 and 7.8, Nu D = ( 0.97 ) 0.27 Re0.63
( Pr Prs )1/ 4 .
D,max Pr
Neglecting the Prandtl number ratio,
0.63
Nu D = ( 0.97 ) 0.27 ( 4744 )
ho =
k
D
Nu D =
0.047 W m⋅ K
0.025 m
(0.69 )0.36 = 47.4
47.4 = 89.1W m 2⋅ K .
Continued...
PROBLEM 11.71 (Cont.)
<
Hence, U = (1/548 + 1/89.1)-1 = 76.7 W/m2⋅K.
c = 2.5 kg/s, Cc =
(b) The fluid outlet temperatures may be determined from the ε-NTU method. With m
c cp,c = 2.5 kg/s × 4188 J/kg⋅K = 10,470 W/K. With Ch = m
h c p,h = 2.25 kg/s × 1051 J/kg⋅K = 2365
m
W/K, Cmin/Cmax = Cmixed/Cunmixed = 2365/10,470 = 0.23. Hence, with A = NπDL = 100π × 0.025 m × 4 m =
31.4 m2,
(
2
NTU =
76.7 W m ⋅ K 31.4 m
UA
=
C min
2
2365 W K
) = 0.95
From Fig. 11.19, ε ≈ 0.61. From Eq. 11.19, qmax = Cmin(Th,i - Tc,i) = 2365 W/K(800 - 300)K = 1.18 × 106
W. Hence, q = εqmax = 0.72 × 106 W. From Eq. 11.6b,
(Th,i − Th,o ) = C
q
(
=
h
0.72 × 106 W
2365 W K
)
q
=
<
Th,o = 496 K
0.72 × 106 W
<
Tc,o = 369 K
= 69 K
Cc 10, 470 W K
(c) Using the appropriate Heat Exchangers, Correlations and Properties Toolpads of IHT, the following
results were obtained.
From Eq. 11.7b,
Tc,o − Tc,i =
= 304 K
540
800000
460
750000
Heat rate, q(W)
Outlet temperatures (K)
500
420
380
340
700000
300
2
4
6
8
10
12
14
16
Water flow rate, mdotc(kg/s)
18
20
650000
2
Gas outlet temperature, Tho(K)
Water outlet temperature, Tco(K)
4
6
8
10
12
14
16
18
20
Water flow rate, mdotc(kg/s)
c (and m
c,1 ), hi increases, thereby increasing U and q. However, because the total
With increasing m
c = 20 kg/s only yields U = 83.9 W/m2⋅K, despite
resistance is dominated by the gas-side condition, m
c is much
the fact that hi = 2180 W/m2⋅K. Because the extent to which q increases with increasing m
c . Hence, there is a trade-off
c itself, Tc,o decreases with increasing m
smaller than the increase in m
between the amount of hot water and the temperature at which it is delivered. If, for example, the
c cannot exceed 8 kg/s. To maintain an acceptable
temperature must exceed 50°C (Tc,o > 323 K), m
h (and V) should be increased, thereby increasing ho, and hence U
c, m
value of Tc,o, while increasing m
and q.
COMMENTS: If the air and water property functions of IHT are used to evaluate properties at
appropriate mean values of the inlet and outlet fluid temperatures, the following, more accurate, results
would be obtained for Parts (a) and (b): ε = 0.582, q = 0.697 × 106 W, Tc,o = 366.6 K, Th,o = 508.8 K, hi =
523 W/m2⋅K, ho = 86.5 W/m2⋅K and U = 74.2 W/m2⋅K.
PROBLEM 11.72
c and inlet
KNOWN: Tube arrangement in steam-to-air, cross-flow heat exchanger. Flow rate m
temperature of air. Condensing temperature of steam.
= 12 kg/s, (b) Effect of m
c on air outlet temperature, heat rate
FIND: (a) Air outlet temperature for m
c
and condensation rate.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible heat loss to surroundings, (2) Negligible kinetic and potential energy
changes, (3) Negligible steam side convection and tube wall conduction resistance, (4) Mean air
temperature is 350 K.
(
)
PROPERTIES: Table A.4, Air (Assume Tc ≡ Tc,i + Tc,o 2 ≈ 350 K, 1 atm): ρ = 0.995 kg/m3, cp =
1009 J/kg⋅K, ν = 20.92 × 10-6 m2/s, k = 0.030 W/m⋅K, Pr = 0.700; Ts = 400 K: Pr = 0.690.
ANALYSIS: (a) For a single-pass, cross-flow heat exchanger with one fluid mixed and the other
unmixed, Fig. 11.19 can be used to obtain ε, where Cmin/Cmax = Cmixed/Cunmixed = 0 and NTU = UA/Cmin =
c cp. From Eq. 11.5, U = h o , and the Zhukauskas correlation may be used to estimate h o .
U(πDL)N/ m
c = ρVA ≈ ρVNTLST. Hence,
The upstream velocity may be obtained from m
m
12 kg s
c
V=
=
= 1.44 m s .
ρ N T LST 0.995 kg m3 × 30 × 2 m × 0.14 m
For aligned tubes,
ST
0.14 m
Vmax =
V=
1.44 m s = 2.88 m s
ST − D
( 0.14 − 0.07 ) m
V
D 2.88 m s × 0.07 m
Re D,max = max =
= 9637 .
−6 2
ν
20.92 × 10 m s
From Table 7.7, select values of C = 0.27 and m = 0.63. Hence,
0.25
0.36
Nu D = 0.27 Re0.63
( Pr Prs )
D,max Pr
0.63
Nu D = 0.27 (9637 )
ho = Nu D
k
D
= 77.1
(0.70 )0.36 ( 0.70
0.030 W m⋅ K
0.07 m
0.25
0.69 )
= 77.1
= 33.0 W m 2⋅ K .
Hence,
NTU =
h oπ DLN
c
m
c p
=
33.0 W m 2⋅ K × π ( 0.07 m ) 2 m (1200 )
12 kg s × 1009 J kg ⋅ K
= 1.44 .
From Fig. 11.19, find ε ≈ 0.77 and then determine
Continued...
PROBLEM 11.72 (Cont.)
(
) = Tc,o − Tc,i
c (T − T )
q max
Ts − Tc,i
m
c p s
c,i
Tc,o = Tc,i + ε ( Ts − Tc,i ) = 300 K + 0.77 ( 400 − 300 ) K = 377 K = 104$ C
ε=
q
=
c
m
c p Tc,o − Tc,i
<
(b) With q = εqmax = εCc(Ts - Tc,i) and the condensation rate given by Eqs. 10.33 and 10.26,
q
q
m
≈
cd =
h ′fg h fg
the foregoing model may be used with the Heat Exchangers, Correlations and Properties Toolpads of IHT
c on Tc,o, q and m
to determine the effect of m
cd .
380
3E6
370
Heat rate, q(W)
Air outlet temperature, Tco(K)
2.5E6
360
2E6
1.5E6
1E6
350
500000
10
20
30
40
50
10
20
Air flow rate, mdotc(kg/s)
30
40
50
Air flow rate, mdotc(kg/s)
c , q must also increase. However, since the increase in q is
Since h o increases with increasing m
c , Tc,o decreases with increasing m
c.
proportionally less than the increase in m
Condensation rate, mdotcd(kg/s)
1.4
1.2
1
0.8
0.6
0.4
0.2
10
20
30
40
50
Air flow rate, mdotc(kg/s)
The condensation rate increases proportionally with the increase in q, and if the objective is to maximize
c should be maintained.
the condensation rate, the largest value of m
COMMENTS: If the objective is to heat the air, there is obviously a trade-off between maintaining
elevated values of the flowrate and outlet temperature.
PROBLEM 11.73
KNOWN: Heat exchanger operating in parallel-flow configuration.
FIND: Expression for Rlm/Rt which doesn’t involve temperatures. Plot result.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible heat loss to surroundings, (2) Negligible change in kinetic and
potential energy.
ANALYSIS: (a) For the exchanger, the rate equation is
q = UA∆Tlm
and we can define thermal resistances as
(
)
R t = Th,i − Tc,i / q
or
R lm = ( ∆Tlm ) / q = 1/UA.
Using the rate equation and the definition of effectiveness, find the thermal resistance based upon the
inlet temperatures of the hot and cold fluids as
(
)
R t = C min T h,i −Tc,i / C min ⋅q = 1/ ε C min.
The ratio of these resistances is
R lm
1/UA
ε
ε
=
=
=
Rt
1/ ε Cmin U A / Cmin NTU
and for the parallel flow, concentric tube configuration using Eq. 11.29a,
R lm 1 − exp  − NTU ( 1 + Cr )  1 − exp ( −B )
=
=
Rt
NTU (1 + Cr )
B
<
where B = NTU(1 + Cr). Evaluating the ratio for various values of B, find
B
0.1
0.5
1.0
3.0
5.0
10.0
<
Rlm/Rt
0.95
0.79
0.63
0.32
0.20
0.10
COMMENTS: (1) For Cmax → ∞, Cr → 0; hence B → NTU. (2) For Cmax ≈ Cmin, B → 2NTU or
1
. (3) For B << 1, Rlm/Rt → 1. (4) For B >> 1, Rlm/Rt → B . (5) We conclude that care
B ~ C−min
must be taken in representing heat exchangers with a thermal resistance, recognizing that the
resistance will depend on flow rates for wide ranges of conditions.
-1
PROBLEM 11.74
KNOWN: Heat exchanger condensing steam at 100°C with cooling water supplied at 15°C.
FIND: (a) Thermal resistance of the exchanger, (b) Change in thermal resistance if fouling is 0.0002
2
m ⋅K/W on each of the inner and outer tube surfaces, and (c) Plot the thermal resistance as a function
of tube water inlet rate assuming all other conditions remain unchanged; comment on whether UA will
remain constant if the flow rate changes.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible heat loss to surroundings, (2) Negligible kinetic and potential energy
changes.
PROPERTIES: Table A-6, Water ( Tm = 42°C=315 K): cp = 4179 J/kg⋅K.
ANALYSIS: (a) For an exchanger, using the rate equation,
(
)
q = UA ∆Tlm = Th,i − Tc,i / Rt ,
the thermal resistance of the exchanger can be expressed as
Rt =
Th,i − Tc,i
UA∆ Tlm
=
(
)
Cmin Th,i − Tc,i
1 q max
1
.
=
⋅
=
Cmin ⋅ q
C min
q
ε Cmin
For the present exchanger with Cr = 0, use Eq. 11.36a with
& c cp,c = 0.5kg/s × 4179 J / k g ⋅ K = 2090W/K
Cmin = m
NTU = U A / Cmin = 2000 W / m 2 ⋅ K × 0.5m 2 /2090 W / K = 0.478
ε = 1 − exp ( − NTU ) = 0.380.
Hence, the thermal resistance is
R t = 1/0.380 × 2090 W / K = 1.258 ×10−3 K/W.
<
(b) With fouling present, the overall heat transfer coefficient will decrease.
No fouling:
With fouling:
1
UoA
=
1
1
+
h h A h h c Ac
R′′
R′′
1
1
1
2R f′′
=
+ f,c + f,h =
+
U f A U oA A c
Ah
Uo A
A
Continued …..
PROBLEM 11.74 (Cont.)
1
1
2 × 0.0002 m 2 ⋅ K / W
=
+
Uf A 2000 W / m 2 ⋅K × 0.5m 2
0.5m 2
Uf A = 555.6 W / K .
It follows that NTU = UfA/Cmin = 0.266 and ε f = 0.233 giving
R t,f = 1/ ε f C min = 2.050 ×10
−3
K/W
and hence the increase in thermal resistance due to fouling is
( R t,f − R t ) / R t = ( 2.050 − 1.258) /1.258 = 63%.
<
(c) With no fouling, the thermal resistance, when all other conditions (UoA = 1000 W/K) remain
unchanged, depends on Cmin only as NTU = UoA/Cmin,
1
1
=
Rt =
ε Cmin Cmin

 UA  
1 − exp  −

 Cmin  

Cmin (W/K)
3
Rt (K/W × 10 )
200
4.967
400
2.723
−1
600
2.055

 1000 W / K  
=
1 − exp  −

Cmin 
Cmin  

1
1000
1.582
1500
1.370
2000
1.271
−1
3000
1.176
From the plot note that Rt is a weak function of
Cmin above Cmin > 1000 W/K, from which we
conclude that using a constant Rt would be
reasonable.
Concerning the variability of UA with changing
0.8
& c0.8.
Cmin: if most of the resistance is on the water side and the flow is turbulent, hc ≈ Re0.8
D ≈ u m ≈m
It follows that hc will depend significantly on changes in Cmin. However, if hc and hh are of similar
magnitude, the effect of Cmin on U may not be significant.
PROBLEM 11.75
KNOWN: Air conditioner modeled as a reversed Carnot heat engine, with refrigerant as the working
fluid, operating between indoor and outdoor temperatures of 23 and 43°C, respectively, removing 5 kW
from a building. Compressor and fan motor efficiency of 80%.
FIND: (a) Required motor power assuming negligible thermal resistances between the refrigerant in
the condenser and the outside air and between the refrigerant in the evaporator and the inside air, and
-3
(b) Required power if thermal resistances are each 3 × 10 K/W.
SCHEMATIC:
ASSUMPTIONS: (1) Ideal heat exchanger with no losses, (2) Air conditioner behaves as reversed
Carnot engine.
ANALYSIS: (a) With negligible thermal resistances, the Carnot cycle and reversed heat engine can
be represented as shown above. Hence,
w
& ideal = qH − q L = qL ( TH / TL ) − 1 = 5kW ( 316 K / 2 9 6 K) − 1 = 0.3378 kW.
Considering the fan power requirement and the efficiency of the motor,
& act = ( w
& ideal + w& fan ) / ηc = ( 0.3378 + 0.200 ) kW/0.8 = 0.672 kW.
w
<
-3
(b) Consider now thermal resistances of Rt = 3 × 10 K/W on the high temperature (condenser) and
low temperature (evaporator) sides.
Low side: in order to remove heat from the
room, TC < Ti. That is
(
)
Ti − TC = qR t = 5 kW 3×10−3 K / W = 15 K
TC = Ti − 1 5 K = 23°C − 15 K = 8°C.
High side: in order to reject heat from the
condenser to the outside air, TH > To,
TH − T o = q HR t = q c (T H / Tc ) R t
TH − ( 43 + 273) K = 5 kW  TH / ( 8+ 273) 3 ×10 −3 K / W
TH = 333.9 K = 61 °C.
The work required for this cycle is
w
& ideal = qH − q L = qL ( TH / TL ) − 1 = 5 kW ( 61+ 273) K /( 8 + 273) K − 1 = 0.943kW
& act = ( w
& ideal + w& fan ) / ηc = ( 0.943 + 0.2 ) kW/0.8 = 1.43 kW.
w
<
The effect of finite thermal resistances in the evaporator and condenser is to increase the power by a
factor of two.
PROBLEM 11.76
KNOWN: Flow rate and pressure of saturated vapor entering a condenser. Number and diameter of
condenser tubes. Water flow rate and inlet temperature. Tube outside convection coefficient.
FIND: (a) Water outlet temperature, (b) Total tube length, (c) Effect of fouling on mass condensation,
(d) Effect of water flow rate and inlet temperature on condenser performance.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible heat loss to surroundings and potential and kinetic energy changes, (2)
Constant properties, (3) Negligible wall conduction resistance and fouling (initially).
PROPERTIES: Water (given): cp = 4178 J/kg⋅K, µ = 700 × 10-6 kg/s⋅m, k = 0.628 W/m⋅K, Pr = 4.6;
Table A.6, Sat. steam (355 K): hfg = 2.304 × 106 J/kg; With fouling: R′′f = 0.0003 m2⋅K/W.
(
)
(
)
h
c
ANALYSIS: (a) From an energy balance, qh = m
h i h,i − i h,o = m
h fg = q c = m
c p,c Tc,o − Tc,i , or
c
m
h c,p
Tc,o = Tc,i +
c
m
c p,c
= 280 K +
1.5 kg s × 2.304 × 106 J kg
15 kg s × 4178 J kg ⋅ K
<
= 335.1K .
(b) Since Cr = 0, NTU = -ln(1 - ε), where
ε=
q
q max
=
(
) = (335.1 − 280 ) K = 0.735
c
m
(355 − 280 ) K
c p,c (Th,i − Tc,i )
c
m
c p,c Tc,o − Tc,i
Hence, NTU = -ln(1 - 0.735) = 1.327 = UA/Cmin. The overall heat transfer coefficient is given by 1/U =
1/ h i + 1/ h o . For the internal tube flow,
Re D =
4m
c,1
4 × 15 kg s 100
= 27, 284
π ( 0.01m ) 700 × 10−6 kg s ⋅ m
Hence, assuming fully developed flow with the Dittus-Boelter correlation,
π Dµ
=
Nu D = 0.023 Re4D/ 5 Pr n = 0.023 ( 27, 284 )
4/5
hi = ( k D ) Nu D =
0.628 W m⋅ K
0.01m
−1
and U = [(1 9408 ) + (1 5000 )]
( 4.6 )0.4 = 149.8
149.8 = 9408 W m 2⋅ K
W m 2⋅ K = 3265 W m 2⋅ K . Hence, the heat transfer area is
(
)
2
2
c
A=m
c p,c ( NTU U ) = 15 kg s ( 4178 J kg ⋅ K ) 1.327 3265 W m ⋅ K = 25.5 m
and the tube length is L = A/NπD = 25.5 m2/100π(0.01 m) = 8.11 m.
<
(c) With fouling, the overall heat transfer coefficient is 1/Uw = 1/Uwo + R′′f . Hence,
Continued...
PROBLEM 11.76 (Cont.)
(
)
1 U w = 3.063 × 10−4 + 3 × 10 −4 m 2 ⋅ K W
Uw = 1649 W/m2⋅K.
(
NTU = UA Cmin = 1649 W m 2⋅ K × 25.5 m 2
) (15 kg s × 4178 J kg ⋅ K ) = 0.671
From Eq. 11.36a, ε = 1 - exp(-NTU) = 1 - exp(-0.671) = 0.489. Hence, q = εqmax = 0.489 × 15 kg/s ×
4178 J/kg⋅K(355 - 280)K = 2.30 × 106 W. Without fouling the heat rate was
6
6
h
q=m
h fg = 1.5 kg s × 2.304 × 10 J kg = 3.46 × 10 W .
<
6
6
Hence, m
h,w m
h,wo = 2.30 × 10 3.46 × 10 = 0.666 .
The condensation rate with fouling is then m
h,w = 0.666 × 1.5 kg s = 0.998 kg s .
(d) The prescribed water inlet temperature of Tc,i = 280 K is already at the lower limit of available
sources, and it would not be feasible to consider smaller values. In addition, with h i already quite large,
c is not likely to provide a significant improvement in performance. Using the Heat
an increase in m
c ≤ 25 kg/s.
Exchanger and Correlations Tools from IHT, the following results were obtained for 15 ≤ m
Condensation rate, mdoth(kg/s)
1.2
1.1
1
0.9
15
17
19
21
23
25
Water mass flow rate, mdotc(kg/s)
c , there is approximately an 18% increase in the heat rate, and hence in the
Over the specified range of m
condensation rate. This increase is, in part, due to the increase in h i from 9408 to 14,160 W/m2⋅K,
which increases U from 1649 to 1752 W/m2⋅K, as well as to a reduction in Tc,o from 316.6 to 306.0 K,
which increases the mean driving potential for heat transfer.
COMMENTS: There is a significant reduction in performance due to fouling, which can not be restored
c . The desired performance could be achieved by oversizing the condenser, that is, by
by increasing m
increasing the number of tubes and/or the tube length.
PROBLEM 11.77
KNOWN: Rankine cycle with saturated steam leaving the boiler at 2 MPa and a condenser pressure
of 10 kPa. Net reversible work of 0.5 MW.
FIND: (a) Thermal efficiency of ideal Rankine cycle, (b) Required cooling water flow rate to
condenser at 15°C with allowable temperature rise of 10°C, and (c) Design of a shell and tube heat
exchanger (one shell and multiple tube passes) to satisfy condenser flow rate and temperature rise.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible loss from condenser to surroundings, (2) Negligible kinetic and
potential energy changes in heat exchanger, (3) Ideal Rankine cycle, and (4) Negligible thermal
resistance on condensate side of exchanger tubes.
PROPERTIES: Steam Tables, (Wark, 4th Edition): (1) p1 = p4 = 10 kPa = 0.10 bar, Tsat = 45.8°C =
-3 3
319 K, vf = 1.0102 × 10 m /kg, hf = 191.83 kJ/kg; (3) p2 = p3 = 2 Mpa = 20 bar, hg = 2799.5 kJ/kg, sg
= 6.3409 kJ/kg⋅K; (4) s4 = s3 = 6.3409 kJ/kg⋅K, p4 = 0.10 bar, sf = 0.6493 kJ/kg⋅K, sg = 8.1502 kJ/kg⋅K,
hf = 191.83 kJ/kg⋅K, hfg = 2392.8 kJ/kg; Table A-6, Water (Tsat = 293 K): cp,c = 4182 J/kg⋅K, µ = 1007
-6
2
5
2
5
× 10 N⋅s/m , k = 0.603 W/m⋅K, Pr = 7.0. Note: 1 bar = 10 N/m = 10 Pa.
ANALYSIS: (a) Referring to your thermodynamics text, find that
η=
wnet w t − wp ( h3 − h4 ) − v1 ( p2 − p1 )
=
=
QH
QH
h3 − h 2
where the net work is the turbine minus the pump work. Assuming the liquid in the pump is
incompressible,
(
)
wp = v1 ( p2 − p1 ) = 1.0102 ×10−3 m3 /kg 2× 106 − 10 × 103 N / m 2 = 2.01kJ/kg.
To find the enthalpies at states 2, 3, and 4, consider the individual processes. For the pump,
h 2 = h 1 + w p = (191.83 + 2.01) kJ/kg = 193.84 kJ/kg.
Since the exit state of the boiler is saturated at p3 = 2 Mpa,
h 3 = h g = 2799.5 kJ/kg.
QH = h3 − h2 = ( 2799.5 −193.84 ) kJ/kg = 2605.7 kJ/kg.
Since the process from 3 to 4 is isentropic, s4 – s3, hence
(
)
x 4 = ( s4 − sf ) / s g − s f = ( 6.3409 − 0.6493) / ( 8.1502 − 0.6493) = 0.759
h 4 = h f + xh fg = 191.83 + 0.759 ( 2392.8 ) kJ/kg = 2007.5kJ/kg.
Continued …..
PROBLEM 11.77 (Cont.)
w t = h3 − h4 = ( 2799.5 − 2007.5) kJ/kg = 792.0 kJ/kg.
Substituting appropriate values, the thermal efficiency is
η=
( 792.0 − 2.01) kJ/kg
2605.7 kJ/kg
<
= 0.303 = 30.3%.
(b) From an overall balance on the cycle, the heat rejected to the condenser is
Qc = QH − w net =  2605.7 − ( 792.0 − 2.01) kJ/kg = 1815.7 kJ/kg.
Since the net reversible power is 0.5 MW, the required steam rate (h) is
6
&
&h =W
m
net / w net = 0.5 ×10 W / ( 792.0 − 2.01) kJ/kg = 0.6329 kg/s.
Hence, the heat rate to be removed by the cold water passing through the condenser is
(
&h=m
& c c p,c Tc,out − Tc,in
q c = Q c ⋅m
)
& c × 4182 J / k g ⋅ K ( 25 − 15) K
1815.7 kJ/kg × 0.6329 k g / s = 1.149 × 106 W = m
<
& c = 27.47 k g / s
m
where cp,c = cp,f is evaluated at T2, Tc,in = 15°C and Tc,out – Tc,in = 10°C, the specified allowable rise.
(c) To design the heat exchanger we need to
evaluate UA. Considering the shell-tube
configuration and since Cr = Cmin/Cmax = 0,
ε = 1 − exp ( − NTU ) = 1 − exp − ( U A / Cmin ) 
ε=
ε=
q
q max
=
qc
(
& c cp,c Th − Tc,i
m
)
1.149 × 106 W
= 0.326
27.47kg/s × 4182 J / k g ⋅ K ( 45.7 − 15) K

UA

0.326 = 1 − exp  −

 27.47 k g / s × 4182 J / k g ⋅ K 
UAs = 45,372 W / K
& c c p,c. Our design process will involve the following steps: select tube diameter, D =
where Cmin = m
15 mm; set um = 2 m/s in each tube and find number of tubes; perform internal flow calculation to
estimate h c and then determine the length.
(
& c = ρ Ac Nu m = 1.010 ×10 −3 m 3 / k g
m
) (π (0.015m )2 / 4) 2 m / s × N = 27.47 k g / s
−1
N = 78.5 ≈ 79.
Continued …..
PROBLEM 11.77 (Cont.)
For flow in a single tube,
Re D =
4 ( 27.47 kg/s/79 )
&t
4m
=
= 29,310.
π Dµ π ( 0.015m ) 1007 × 10−6 N ⋅ s / m 2
Assuming the flow is fully developed and using the Dittus-Boelter correlation,
Nu =
hD
0.4 = 0.023 29,310 0.8 7.00 0.4 = 187.7
= 0.023Re0.8
(
) ( )
D Pr
k
h = 0.603W/m ⋅ K ×187.7/0.015m = 7544 W / m 2 ⋅ K.
Hence, the tube length is
UAs = h (π DL) N = 45,372 W / K
L = 45,372 W / K / 7 5 4 4 W / m 2 ⋅ K ×π ( 0.015m ) 79 = 1.6m
and our design has the following parameters:
N = 79 tubes
L =1.6m
D = 15 mm.
<
COMMENTS: (1) The selection of the tube diameter and water velocity values (15 mm, 2 m/s) was
based upon prior experience; they seemed reasonable. We could, however, establish other
requirements which would influence these choices such as allowable pressure drop and standard tube
sizes.
PROBLEM 11.78
KNOWN: Rankine cycle with saturated steam leaving the boiler at 2 Mpa and a condenser pressure
of 10 kPa. Heat rejected to the condenser of 2.3 MW. Condenser supplied with cooling water at rate
of 70 kg/s at 15°C.
FIND: (a) Size of the condenser as determined by the parameter, UA, and (b) Reduction in thermal
efficiency of the cycle if U decreases by 10% due to fouling assuming water flow rate and inlet
temperature and the condenser steam pressure remain fixed.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible loss from condenser to surroundings, (2) Negligible kinetic and
potential energy changes in exchanger, (3) Ideal Rankine cycle, (4) For fouled operating condition,
m& c, Tc,i and p4 remain the same.
PROPERTIES: Steam Tables (Wark, 4th Edition): See previous problem for calculations to obtain
cycle enthalpies; h1 = 191.83 kJ/kg, h4 = 2007.5 kJ/kg.
ANALYSIS: (a) For the condenser, recognize that Cmin = Cc, and Cr = Cmin/Cmax = 0,
ε=
q
q max
= 1 − exp ( − NTU ) = 1 − exp ( − U A / C min )
& c cp,c = 70 kg/s × 4182 J / k g ⋅ K = 292,740 W / K
Cmin = m
(
)
q max = Cmin Th − Tc,i = 292,740 W / K ( 45.7 −15 ) K = 8.987 ×106 W.
q = q h = 2.30 ×106 W
2.30 × 106 W


UA
= 0.256 = 1 − exp  −

 292,740 W / K 
8.987 ×106 W
<
UA = 86,538 W / K.
(b) In the fouled condition, U is reduced 10%, hence
Uf A = 0.9UA = 77,884 W / K
and
NTUf =
Uf A
77,884 W / K
=
= 0.266
Cmin 292,740 W / K
ε f = 1 − exp ( −NTU f ) = 1 − exp ( −0.266 ) = 0.234.
Continued …..
PROBLEM 11.78 (Cont.)
If we operate the cycle at the same back pressure p4 = 10 kPa so that Th = 45.7°C, the heat removal
rate must decrease,
q h = ε q max = 0.234 × 8.987 ×10 6 W = 2.103 × 10 6 W
since qmax = Cmin (Th – Tc,i) remains the same. From the previous problem, we found the heat
rejected as
h 4 − h1 = ( 2007.5 −191.83 ) kJ/kg = 1815.7 kJ/kg
and hence the cycle steam rate through the fouled condenser is
& h,f = qh / ( h4 − h1 ) = 2.103 ×106 W/1815.7 kJ/kg =1.158 kg/s.
m
For the unfouled condenser of part (a), the steam rate was
& h = 2.3MW/1815.7 kJ/kg = 1.267 kg/s.
m
Hence, we see that fouling reduces the steam rate by 8.5% when U is decreased 10%. Since p4
remains the same, the thermal efficiency remains unchanged,
<
η = 30.3%
as calculated in the previous problem. However, the net work of the cycle will decrease 8.5%.
COMMENTS: Fouling of the condenser heat exchanger has no effect on the thermal efficiency of
the cycle since the back pressure at the condenser is maintained constant. The effect is, however, to
reduce the heat rejection rate while maintaining exchanger flow rate and inlet temperature fixed.
Comparing the conditions:
Parameter
UA, W/K
ε
qh, MW
& net
w
Clean
86,538
0.256
2.300
--
Fouled
77,884
0.234
2.103
--
Change (%)
-10.0
-8.6
-8.6
-8.6
PROBLEM 11.79
KNOWN: Compact heat exchanger (see Example 11.6) after extended use has prescribed fouling
factors on water and gas sides.
FIND: Gas-side overall heat transfer coefficient.
SCHEMATIC:
ASSUMPTIONS: (1) Heat transfer coefficients on the inside and outside (cold- and hot-sides) are
the same as for the unfouled condition, (2) Temperature effectiveness of the finned hot side surface is
the same as for the unfouled condition.
ANALYSIS: The overall heat transfer coefficient follows from Eq. 11.1 as
R′′f,c
R′′f,h
1
1
1
=
+
+Rw +
+
Uh Ah (ηo hA )c (ηo A)c
(ηo A )h (ηoh A )h
where Rw and R ′′f are the wall resistance and fouling factors, respectively. Multiply both sides by Ah
and recognizing that ηo,c = 1, obtain
R′′f,c
R′′
1
1
1
=
+
+ Ah R w + f,h +
.
Uh hc ( Ac / Ah ) ( Ac / Ah )
ηo,h ηo hh
Substitute numerical values from Example 11.6 results (hh, ηo,h, Ah Rw, Ac/Ah) and those from the
problem statement R′′f,h , R′′f,c , hc to find,
(
)
1
1
=
Uh 1500 W / m2 ⋅ K ( 0.143)
+
0.0005m 2 ⋅ K / W
0.001m 2 ⋅ K / W
1
+ 3.51× 10−5 m 2 ⋅ K / W +
+
0.91
( 0.143)
0.91× 1 8 3 W / m 2 ⋅ K
)
(
1
= 4.662 ×10−3 + 6.993× 10−3 + 3.51 ×10−5 +5.495 ×10 −4 +6.005 ×10−3 m 2 ⋅ K / W
Uh
Uh = 65.4 W / m 2 ⋅ K.
<
2
COMMENTS: For the unfouled condition, we found Uh = 93.4 W/m ⋅K from Example 11.6. Note
that the thermal resistance of the tube-fin material is negligible and that fouling has a significant effect,
reducing Uh by 41%.
PROBLEM 11.80
KNOWN: Compact heat exchanger with prescribed core geometry and operating parameters.
FIND: Required heat exchanger volume; number of tubes in the longitudinal and transverse directions,
NL and NT ; required tube length.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible heat loss to surroundings, (2) Negligible KE and PE changes, (3)
Single pass operation, (4) Gas properties are those of air.
3
PROPERTIES: Table A-6, Water ( Tc = 325 K): ρ = 987.2 kg/m , cp = 4182 J/kg⋅K; Table A-4,
Air (Assume Th,o ≈ 400 K, Th ≈ 550 K, 1 atm): cp = 1040 J/kg⋅K.
ANALYSIS: To find the Hxer volume, first find Ah using the ε-NTU method. By definition,
V =A h /α
A h = NTU ⋅ Cmin / U h.
and
(1,2)
Find the capacity rates, q, qmax and ε:
& c c p,c = 2 k g / s × 4182 J / k g ⋅ K = 8364 W / K
Cc = m
& h c p,h = 1.25 kg/s ×1040 J / k g ⋅ K = 1300 W / K ← Cmin
Ch = m
Hence,
Cr =
Cmin
= 0.155.
Cmax
It follows that
ε=
q
q max
=
(
Cc Tc,o − Tc,i
(
)
Cmin Th,i − Tc,i
)
=
8364 W / K ( 350 − 300 ) K
= 0.804.
1300 W / K ( 700 − 300) K
With ε = 0.804 and Cr = 0.155, find NTU ≈ 1.7 from Fig. 11.18 for a single-pass, cross flow Hxer with
both fluids unmixed. Using Eqs. (2) and (1), find
A h = 1.7 ×1300 W / K / 9 3 . 4 W / m 2 ⋅ K = 23.7m2
V = 23.7m 2 /269m 2 / m 3 = 0.0880m 3.
Continued …..
PROBLEM 11.80 (Cont.)
To determine the number of tubes in the
longitudinal direction, consider the tubular
arrangement in the sketch. The Hxer
volume can be written as
V = A fr × l L
(3)
l L = ( NL −1) l + Df
(4)
where
and NL is the number of tubes in the longitudinal direction. Combining Eqs. (3) and (4) and substituting
numerical values, find
NL = ( V / Afr − Df ) / l + 1
(5)
where Df is the overall diameter of the finned tube, and
NL = 0.0880m3 /0.20 m 2 − 0.0285m /0.0343 + 1 = 13.0 ≈ 13.
(
)
<
&c
To determine the number of tubes in the transverse direction, compare the overall water flow rate m
& t. That is,
with that for a single tube, m
& t = ρcA t Vi
m
(
)
(6)
where At is the tube inner cross-sectional area π Di2 / 4 and Vi the internal velocity. Hence,
& c /m
& t = ( 2 k g / s ) /987.2 k g / m 3 ×
N=m
π
( 0.0138m) 2 × 0.100 m / s = 135.4 ≈ 135.
4
The total number of tubes required, N, is 135; the number in the transverse direction is
<
NT = N / NL = 135/13 = 10.4 ≈ 11.
To determine the water tube length, recognize that the total area (Ah), less that of the finned surfaces
(Af), will be that of the water tube surface area. That is,
A h − A f = π D ol T ⋅ N.
From specification of the core geometry, we know Af/Ah = 0.830; solve for l T to obtain
l T = Ah (1 −A f / A h ) / π Do ⋅ N
(7)
l T = 23.7m 2 (1 − 0.830 ) / π ( 0.0164m ) ×135 = 0.58m.
<
COMMENTS: In summary we find that
Total number of tubes, N (NT × NL)
Tubes in longitudinal direction, NL
Tubes in transverse direction, NT
143
13
11
2
with a total surface area of 27.3 m . The length of the exchanger is
Length in longitudinal direction, l L
Length in transverse direction, l T
0.44 m
0.58 m.
PROBLEM 11.81
KNOWN: Compact heat exchanger geometry, gas-side flow rate and inlet temperature, water-side
convection coefficient, water flow rate, and water inlet and outlet temperatures.
FIND: Gas-side overall heat transfer coefficient. Required heat exchanger volume.
SCHEMATIC:
ASSUMPTIONS: (1) Gas has properties of atmospheric air at an assumed mean temperature of 700
K, (2) Negligible fouling, (3) Negligible heat exchange with the surroundings and negligible kinetic
and potential energy and flow work changes.
PROPERTIES: Table A-1, aluminum (T ≈ 300 K): k = 237 W/m⋅K. Table A-4, air (p = 1 atm, T =
-7
2
700 K): cp = 1075 J/kg⋅K, µ = 338.8 × 10 N⋅s/m , Pr = 0.695. Table A-6, water ( T = 330 K): cp =
4184 J/kg⋅K.
ANALYSIS: For the prescribed heat exchanger core,
1
1
1
=
+ Ah R w +
ηo,h h h
U h h c ( Ac / A h )
where
A c Di  Af ,h
≈
1 −
A h Do 
Ah
 8.2
(1 − 0.913) = 0.070
=
10.2

The product of Ah and the wall conduction resistance is
Ah R w =
ln ( Do / Di )
2π kL / A h
=
Di ln ( Do / Di )
2 k ( Ac / A h )
=
0.0082m × ln (10.2 / 8.2 )
2 × 237 W / m ⋅ K ( 0.070 )
= 5.39 × 10−4 m 2 ⋅ K / W
2
2
h / σ A fr = 1.25 kg/s/0.534 × 0.20 m = 11.7 kg/s⋅m ,
With a gas-side mass velocity of G = m
Re =
G D h 11.7 kg / s ⋅ m 2 × 0.00363m
=
= 1254
µ
338.8 × 10−7 N ⋅ s / m 2
and Fig. 11.21 yields jH ≈ 0.0096. Hence,
hh ≈
0.0096 G c p
Pr 2 / 3
=
)
(
0.0096 11.7 kg / s ⋅ m 2 (1075 J / kg ⋅ K )
2/3
(0.695 )
= 154 W / m 2 ⋅ K
Continued …..
PROBLEM 11.81 (Cont.)
With r2c = r2 + t/2 = 15.8 mm + 0.330 mm/2 = 15.97 mm, r2c/r1 = 15.97/5.1 = 3.13, L = r2 – r1 = 10.7
1/2
-6 2
2
mm, Lc = L + t/2 = 10.87 mm = 0.0109m, Ap = Lct = 3.59 × 10 m , and L3/
c (hh/kAp) = 0.484,
Fig. 3.19 yields ηf ≈ 0.77. Hence,
A
ηo,h = 1 − f (1 − ηf ) = 1 − 0.913 (1 − 0.77 ) = 0.790
A
−1
(
2
U h = 1500 W / m ⋅ K × 0.07
)
−1
+ 5.39 × 10
−4
2
(
2
m ⋅ K / W + 0.79 × 154 W / m ⋅ K
)
−1
2
= 0.0183 m ⋅ K / W
U h = 54.7 W / m 2 ⋅ K
<
5
With q = Cc (Tc,o – Tc,i) = 4184 W/K × 80 K = 3.35 × 10 W, qmax = Cmin (Th,i – Tc,i) = 1344 W/K ×
5
535 K = 7.19 × 10 W, ε = 0.466 and Cr = 0.321. From Figure 11.18, we then obtain NTU ≈ 0.65.
The required gas-side surface area is then
Ah =
NTU × Cmin 0.65 × 1344 W / K
=
= 16.0 m 2
2
Uh
54.7 W / m ⋅ K
2
3
With α = 587 m /m , the required volume is
V=
Ah
16 m 2
=
= 0.0273 m3
2
3
α
587 m / m
<
COMMENTS: (1) Although Uh is small and Ah larger for the continuous fins than for the circular
fins of Example 11.6, the much larger value of α, renders the volume requirement smaller.
(2) The heat exchanger length is L = V/Afr = 0.137 m, and the number of tube rows is
NL ≈
L
+ 1 = 7.23 ≈ 7.
SL
(3) The hypothetical fin radius (r2 = 15.8 mm) was estimated to be the arithmetic mean of one-half the
center-to-center spacing between one tube and its six neighbors.
PROBLEM 11.82
KNOWN: Cooling coil geometry. Air flow rate and inlet and outlet temperatures. Freon pressure
and convection coefficient.
FIND: Required number of tube rows.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible fouling, (2) Constant properties, (3) Negligible heat loss to
surroundings.
-7
2
PROPERTIES: Table A-4, Air ( Th = 300 K, 1 atm): cp = 1007 J/kg⋅K, µ = 184.6 × 10 N⋅s/m , k =
0.0263 W/m⋅K, Pr = 0.707; Table A-5, Sat. R-12 (1 atm): Tsat = Tc = 243 K, hfg = 165 kJ/kg.
ANALYSIS: The required number of tube rows is
NL = ( L − Df ) /SL + 1
where
L = V / Afr
A h = NTU ( Cmin / U h )
V =A h / α
1 / Uh = 1 / hc ( Ac / Ah ) + Ah Rw + 1/ ηo,h h h .
-5
2
From Ex. 11.6, (Ac/Ah) = 0.143 and AhRw = 3.51 × 10 m ⋅K/W. With
G=
&h
m
1.50 k g / s
=
= 20.9 k g / s ⋅ m 2
2
σ Afr 0.449 × 0.16m
Re =
GD h 20.9 kg/s ⋅ m 2 × 6.68 ×10 −3 m
=
= 7563
µ
184.6 × 10−7 N ⋅ s / m 2
and Fig. 11.20 gives jH ≈ 0.0068. Hence,
h h = jh
Gcp
Pr
2/3
= 0.0068
20.9 k g / s ⋅ m2 × 1007 J / k g ⋅ K
( 0.707 )
2/3
= 180 W / m 2 ⋅ K.
(
)
1/2
-6 2
With Lc = 6.18 mm and Ap = 1.57 × 10 m from Ex. 11.6, L3c / 2 hh /kA p
= 0.338 and, from
Fig. 3.19, ηf ≈ 0.89 for r2c/r1 = 1.75. Hence, as in Ex. 11.6, ηo,h = 0.91 and
(
)
(
1 / Uh = 1/ 5000 W / m 2 ⋅ K 0.143 + 3.51×10− 5 m 2 ⋅ K / W + 1/ 0.91 ×180 W / m 2 ⋅ K
Uh = 133 W / m 2 ⋅ K.
Continued …..
)
PROBLEM 11.82 (Cont.)
& h cp,h = 1511 W/K,
With Cmin/Cmax = 0 and Cmin = m
ε=
q
qmax
=
(
) = 20 K = 0.30
Ch ( Th,i − Tc,i ) 67 K
Ch Th,i − Th,o
NTU = − ln (1 − ε ) = 0.355
and
C
1511W/K
A h = NTU min = 0.355
= 4.03m 2 .
2
Uh
133 W / m ⋅ K
Hence,
L=
Ah
4.03m2
=
= 0.0937m
2
3
2
α Afr
269 m / m 0.16m
(
)
and
NL =
L − Df
0.0652
+ 1=
+ 1 = 2.9.
SL
0.0343m
Hence, three or more rows must be used.
COMMENTS: For the prescribed operating conditions, the heat rate would be
(
)
q = Ch Th,i − Th,o = 1 5 1 1 W / K ( 20 K ) = 30,220 W.
If R-12 enters the tubes as saturated liquid, a flow rate of at least
&c =
m
q
30,220 W
=
= 0.183 kg/s
h fg 165,000 J / k g
would be needed to maintain saturated conditions in the tubes.
<
PROBLEM 11.83
KNOWN: Cooling coil geometry. Air flow rate and inlet temperature. Freon pressure and
convection coefficient.
FIND: Air outlet temperature.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible fouling, (2) Constant properties, (3) Negligible heat loss to
surroundings.
-7
2
PROPERTIES: Table A-4, Air ( Th ≈ 300 K, 1 atm): cp = 1007 J/kg⋅K, µ = 184.6 × 10 N⋅s/m , k =
0.0263 W/m⋅K, Pr = 0.707; Table A-5, Sat. R-12 (1 atm): Tsat = Tc = 243 K, hfg = 165 kJ/kg.
ANALYSIS: To obtain the air outlet temperature, we must first obtain the heat rate from the ε-NTU
method. To find Ah, first find the heat exchanger length,
L ≈ ( NL −1) SL + D f = 3 ( 0.0343m ) + 0.0285m = 0.131m.
Hence,
V = Afr L = 0.16m 2 ( 0.131m ) = 0.021m3
(
)
A h = α V = 269m 2 / m 3 0.021m 3 = 5.65m 2 .
The overall coefficient is
1
1
1
=
+ Ah R w +
Uh hc ( Ac / Ah )
ηo,h h h
-5
2
where Ex. 11.6 yields (Ac/Ah) = 0.143 and AhRw = 3.51 × 10 m ⋅K/W. With
G=
&h
m
1.50 k g / s
=
= 20.9 kg/s ⋅ m 2
σ Afr 0.449 × 0.16 m 2
Re =
GD h 20.9 kg/s ⋅ m 2 × 6.68 ×10 −3 m
=
= 7563.
µ
184.6 × 10−7 N ⋅ s / m 2
Fig. 11.20 gives jH ≈ 0.0068. Hence,
h h = jh
Gc p
Pr
2/3
= 0.0068
20.9 kg/s ⋅ m 2 ×1007 J / k g ⋅ K
( 0.707 ) 2 / 3
h h = 180 W / m 2 ⋅ K.
Continued …..
PROBLEM 11.83 (Cont.)
(
)
1/2
-6 2
With Lc = 6.18 mm and Ap = 1.57 × 10 m from Ex. 11.6, L3c / 2 hh /kA p
= 0.338 and, from
Fig. 3.19, ηf ≈ 0.89 for r2c/r1 = 1.75. Hence, as in Ex. 11.6, ηo,h = 0.91 and
1
1
1
=
+ 3.51 ×10−5 m 2 ⋅ K / W +
2
2
Uh
5000 W / m ⋅ K 0.143
0.91 180 W / m ⋅ K
(
)
(
)
Uh = 133 W / m 2 ⋅ K.
With
& h cp,h = 1.5 kg/s (1007 J / k g ⋅ K ) = 1 5 1 1 W / K
Cmin = C h = m
NTU =
Uh Ah 1 3 3 W / m 2 ⋅K × 5.65m 2
=
= 0.497.
Cmin
1511W/K
With Cmin/Cmax = 0, Eq. 11.36a yields
ε = 1 − exp ( − NTU ) = 1 − exp ( −0.497 ) = 0.392.
Hence,
(
)
q = ε q max = ε Cmin Th,i − Tc,i = 0.392 (1 5 1 1 W / K ) 67 K
q = 39,685 W.
The air outlet temperature is
Th,o = Th,i −
q
39,685 W
= 310 K −
= 283.7 K.
Ch
1511W/K
COMMENTS: If R-12 enters the tubes as saturated liquid, a flow rate of at least
&c =
m
q
39,685 W
=
= 0.241kg/s
h fg 165,000 J / k g
would be needed to maintain saturated conditions in the tubes.
<
PROBLEM 11.84
KNOWN: Cooling coil geometry. Gas flow rate and inlet temperature. Water pressure, flow rate
and convection coefficient.
FIND: Required number of tube rows.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible fouling, (2) Constant properties, (3) Negligible heat loss to
surroundings.
-7
PROPERTIES: Table A-4, Air ( Th ≈ 725 K, 1 atm): cp = 1081 J/kg⋅K, µ = 346.7 × 10
2
N⋅s/m , k = 0.0536 W/m⋅K, Pr = 0.698; Table A-6, Sat. water (2.455 bar): Tsat = Tc = 400 K, hfg
= 2183 kJ/kg.
ANALYSIS: The required number of tube rows is
NL =
L − Df
+1
SL
where
L=
V
Afr
V=
Ah
α
C
A h = NTU min
Uh
1
1
1
=
+ Ah R w +
.
Uh hc ( Ac / Ah )
ηo,h hh
From Ex. 11.6, (Ac/Ah) ≈ 0.143 and
A hR w =
Di ln ( D o / D i ) ( 0.0138m ) ln (16.4/13.8 )
=
= 5.55 ×10 −4 m 2 ⋅ K/W.
2k ( Ac / A h )
2 (15 W / m ⋅ K)( 0.143)
With
G=
&h
m
3.0 k g / s
=
= 18.6 k g / s ⋅ m 2
2
σ Afr 0.449 × 0.36m
Re =
GDh 18.6 k g / s ⋅ m2 × 6.68× 10−3 m
=
= 3576
µ
346.7 × 10−7 N ⋅ s / m 2
and Fig. 11.20 gives jh ≈ 0.009. Hence,
h h = jh
Gc p
Pr
2/3
= 0.009
18.6 kg/s ⋅ m 2 ×1081J/kg ⋅ K
( 0.698)
2/3
= 230 W / m 2 ⋅ K.
Continued …..
PROBLEM 11.84 (Cont.)
(
)
1/2
-6 2
With r2c/r1 = 1.75, Lc = 6.18 mm and Ap = 1.57 × 10 m from Ex. 11.6, L3c / 2 hh /kA p
= 1.52
and Fig. 3.19 gives ηf ≈ 0.40. Hence,
ηo,h = 1 −
Af
(1 − ηf ) = 1 − 0.83 (1 − 0.4) = 0.50.
A
Hence,
1
1
1
=
+ 5.55 ×10−4 m 2 ⋅ K / W +
4
2
2
Uh
10 W / m ⋅ K 0.143
0.50 230 W / m ⋅ K
(
)
(
)
Uh = 100.5 W / m 2 ⋅ K.
With
(
)
q=m
& c hfg = 0.5 kg/s 2.183 ×106 J / k g = 1.092 ×106 W
Cmin = Ch = 3.0 kg/s (1081J/kg ⋅ K ) = 3243 W / K
(
)
q max = Cmin Th,i − Tc,i = 3 2 4 3 W / K (500 K ) = 1.622 ×106 W
find
ε=
q
q max
=
1.092 × 106 W
1.622 × 106 W
= 0.674.
From Eq. 11.36b
NTU = − ln (1 − ε ) = − ln (1 − 0.674 ) = 1.121.
Hence,
C
3243 W / K
= 36.17m 2
A h = NTU min = 1.121
Uh
100.5 W / m 2 ⋅ K
L=
Ah
36.17m 2
=
= 0.373m
Afrα 0.36m 2 269m 2 / m 3
NL =
(
)
L − Df
373 − 28.5
+ 1=
+ 1 = 11.06 ≈ 11.
SL
34.3
COMMENTS: The gas outlet temperature is
Th,o = Th,i −
q
Cmin
= 900 K −
1.092 ×106 W
= 564 K.
3243 W / K
Hence Th = (900 K + 564 K)/2 = 732 K is in good agreement with the assumed value.
<
PROBLEM 11.85
KNOWN: Cooling coil geometry. Gas flow rate and inlet temperature. Water pressure and
convection coefficient.
FIND: Gas outlet temperature.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible fouling, (2) Constant properties, (3) Negligible heat loss to
surroundings.
-7
2
PROPERTIES: Table A-4, Air ( Th ≈ 725 K, 1 atm): cp = 1081 J/kg⋅K, µ = 346.7 × 10 N⋅s/m , k =
0.0536 W/m⋅K, Pr = 0.698; Table A-6, Sat. water (2.455 bar): Tsat = Tc = 400 K, hfg = 2183 kJ/kg.
ANALYSIS: To obtain Th,o, first obtain q from the ε-NTU method. To determine NTU, Ah must be
found from knowledge of L.
L ≈ ( NL −1) SL + D f = 10 ( 0.0343m ) + 0.0285m = 0.372m.
Hence,
V = Afr L = 0.36m 2 ( 0.372m ) = 0.134m3
(
)
A h = α V = 269m 2 / m 3 0.134m 3 = 36.05m 2 .
The overall coefficient is
1
1
1
.
=
+ Ah R w +
Uh hc ( Ac / Ah )
ηo,h hh
From Ex. 11.6, (Ac/Ah) ≈ 0.143 and
A hR w =
Di ln ( D o / D i ) ( 0.0138m ) ln (16.4/13.8 )
=
= 5.55 ×10 −4 m 2 ⋅ K/W.
2k ( Ac / A h )
2 (15 W / m ⋅ K)( 0.143)
With
G=
&h
m
3.0 k g / s
=
= 18.6 k g / s ⋅ m 2
2
σ Afr 0.449 × 0.36m
Re =
GDh 18.6 k g / s ⋅ m2 × 6.68× 10−3 m
=
= 3576
µ
346.7 × 10−7 N ⋅ s / m 2
and Fig. 11.20 gives jH ≈ 0.009. Hence,
Continued …..
PROBLEM 11.85 (Cont.)
h h = jh
Gcp
Pr 2 / 3
= 0.009
18.6 k g / s ⋅ m 2 × 1081J/kg ⋅ K
( 0.698 )2 / 3
h h = 230 W / m 2 ⋅ K.
(
)
1/2
-6 2
With r2c/r1 = 1.75, Lc = 6.18 mm and Ap = 1.57 × 10 m from Ex. 11.6, L3c / 2 hh /kA p
= 1.52
and Fig. 3.19 gives ηf ≈ 0.40. Hence,
ηo,h = 1 −
Af
(1 − ηf ) = 1 − 0.83 (1 − 0.4) = 0.50.
A
Hence,
1
1
1
=
+ 5.55 × 10−4 m 2 ⋅ K / W +
4
2
Uh
10 W / m ⋅ K 0.143
0.50 230 W / m 2 ⋅ K
(
)
(
)
Uh = 100.5 W / m 2 ⋅ K.
With
Cmin = Ch = 3 k g / s (1081J/kg ⋅ K ) = 3243 W / K
(
)
2
2
Uh Ah 100.5 W / m ⋅ K 36.05m
NTU =
=
= 1.117.
Cmin
3243 W / K
Since Cmin/Cmax = 0, Eq. 11.36a gives
ε = 1 − exp ( − NTU ) = 1 − exp ( −1.117 ) = 0.673.
Hence,
(
)
q = ε Cmin Th,i − Tc,i = 0.673 (3243 W / K )( 500 K ) = 1.091×10 6 W
and
Th,o = Th,i −
q
Cmin
= 900 K −
1.091× 106 W
= 564 K.
3243 W / K
COMMENTS: (1) The assumption of Th = 725 K is good.
(2) If water enters the tubes as saturated liquid, a flow rate of at least
&c =
m
q
1.091× 106 W
=
= 0.50 kg/s
h fg 2.183 ×10 6 J/kg
would be need to maintain saturated conditions in the tubes.
<
PROBLEM 12.1
KNOWN: Rate at which radiation is intercepted by each of three surfaces (see (Example 12.1).
2
FIND: Irradiation, G[W/m ], at each of the three surfaces.
SCHEMATIC:
ANALYSIS: The irradiation at a surface is the rate at which radiation is incident on a surface per unit
area of the surface. The irradiation at surface j due to emission from surface 1 is
Gj =
q1− j
Aj
.
-3
2
With A1 = A2 = A3 = A4 = 10 m and the incident radiation rates q1-j from the results of Example
12.1, find
12.1 ×10−3 W
= 1 2 . 1 W / m2
G2 =
−3 2
10
G3 =
G4 =
<
m
28.0 ×10−3 W
10−3 m2
19.8 ×10 −3 W
10 −3 m 2
= 28.0 W / m 2
<
= 19.8 W / m 2 .
<
COMMENTS: The irradiation could also be computed from Eq. 12.15, which, for the present
situation, takes the form
G j = I1 cosθ j ω1 − j
2
where I1 = I = 7000 W/m ⋅sr and ω1-j is the solid angle subtended by surface 1 with respect to j. For
example,
G 2 = I1 cosθ 2 ω1 −2
G 2 = 7000 W / m 2 ⋅ sr ×
cos 30°
10−3 m2 × cos60 °
( 0.5m )2
G 2 = 12.1W/m 2 .
Note that, since A1 is a diffuse radiator, the intensity I is independent of direction.
PROBLEM 12.2
KNOWN: A diffuse surface of area A1 = 10-4m2 emits diffusely with total emissive power E = 5 × 104
W/m2 .
FIND: (a) Rate this emission is intercepted by small surface of area A2 = 5 × 10-4 m2 at a prescribed
location and orientation, (b) Irradiation G2 on A2, and (c) Compute and plot G2 as a function of the
separation distance r2 for the range 0.25 ≤ r2 ≤ 1.0 m for zenith angles θ2 = 0, 30 and 60°.
SCHEMATIC:
ASSUMPTIONS: (1) Surface A1 emits diffusely, (2) A1 may be approximated as a differential surface
area and that A 2 r22 << 1.
ANALYSIS: (a) The rate at which emission from A1 is intercepted by A2 follows from Eq. 12.5 written
on a total rather than spectral basis.
q1→ 2 = Ie,1 (θ , φ ) A1 cos θ1dω 2−1 .
(1)
Since the surface A1 is diffuse, it follows from Eq. 12.13 that
Ie,1 (θ , φ ) = Ie,1 = E1 π .
(2)
The solid angle subtended by A2 with respect to A1 is
dω 2−1 ≈ A 2⋅ cos θ 2 r22 .
(3)
Substituting Eqs. (2) and (3) into Eq. (1) with numerical values gives
q1→ 2 =
E1
π
⋅ A1 cos θ1 ⋅
A 2 cos θ 2
2
=
r2
(
5 × 104 W m 2
π sr
)
(
× 10
−4
 5 × 10−4 m 2 × cos 30$ 
 sr (4)
m × cos 60 × 


(0.5m )2
2
$
)
q1→ 2 = 15, 915 W m 2sr × 5 × 10−5 m 2 × 1.732 × 10−3 sr = 1.378 × 10−3 W .
<
(b) From section 12, 2.3, the irradiation is the rate at which radiation is incident upon the surface per unit
surface area,
q
1.378 × 10−3 W
(5)
G 2 = 1→ 2 =
= 2.76 W m 2
−4 2
A2
5 × 10 m
(c) Using the IHT workspace with the foregoing equations, the G2 was computed as a function of the
separation distance for selected zenith angles. The results are plotted below.
<
Continued...
Irradiation, G2 (W/m^2)
PROBLEM 12.2 (Cont.)
10
5
0
0.2
0.4
0.6
0.8
1
Separation distance, r2 (m)
theta2 = 0 deg
theta2 = 30 deg
theta2 = 60 deg
For all zenith angles, G2 decreases with increasing separation distance r2 . From Eq. (3), note that dω2-1
and, hence G2, vary inversely as the square of the separation distance. For any fixed separation distance,
G2 is a maximum when θ2 = 0° and decreases with increasing θ2, proportional to cos θ2.
COMMENTS: (1) For a diffuse surface, the intensity, Ie, is independent of direction and related to the
emissive power as Ie = E/ π. Note that π has the units of [sr ] in this relation.
(2) Note that Eq. 12.5 is an important relation for determining the radiant power leaving a surface in a
prescribed manner. It has been used here on a total rather than spectral basis.
(3) Returning to part (b) and referring to Figure 12.10, the irradiation on A2 may be expressed as
A cos θ1
G 2 = Ii,2 cos θ 2 1
r22
2
Show that the result is G2 = 2.76 W/m . Explain how this expression follows from Eq. (12.15).
PROBLEM 12.003
KNOWN: Intensity and area of a diffuse emitter. Area and rotational frequency of a second surface,
as well as its distance from and orientation relative to the diffuse emitter.
FIND: Energy intercepted by the second surface during a complete rotation.
SCHEMATIC:
ASSUMPTIONS: (1) A1 and A2 may be approximated as differentially small surfaces, (2) A1 is a
diffuse emitter.
ANALYSIS: From Eq. 12.5, the rate at which radiation emitted by A1 is intercepted by A2 is
(
q1− 2 = Ie A1 cos θ1 ω 2 −1 = Ie A1 A 2 cos θ 2 / r 2
)
where θ1 = 0 and θ2 changes continuously with time. The amount of energy intercepted by both sides
of A2 during one rotation, ∆E, may be grouped into four equivalent parcels, each corresponding to
rotation over an angular domain of 0 ≤ θ2 < π/2. Hence, with dt = dθ2/ θ2 , the radiant energy
intercepted over the period T of one revolution is
T
0
∆E = ∫ qdt =
∆E =
π /2
4Ie A1  A 2  π / 2
4I A  A 
cos θ 2dθ 2 = e 1  2  sin θ 2


∫
0
θ2  r 2  0
θ2  r 2 
4 ×100 W / m 2 ⋅ sr ×10−4 m 2  10−4 m 2 
sr = 8 × 10−6 J
2
2 rad / s
 ( 0.50m ) 


COMMENTS: The maximum rate at which A2 intercepts radiation corresponds to θ2 = 0 and is qmax
2
-6
= Ie A1 A2/r = 4 × 10 W. The period of rotation is T = 2π/ θ2 = 3.14 s.
PROBLEM 12.004
KNOWN: Furnace with prescribed aperture and emissive power.
FIND: (a) Position of gauge such that irradiation is G = 1000 W/m2, (b) Irradiation when gauge is tilted
θd = 20o, and (c) Compute and plot the gage irradiation, G, as a function of the separation distance, L, for
the range 100 ≤ L ≤ 300 mm and tilt angles of θd = 0, 20, and 60o.
SCHEMATIC:
ASSUMPTIONS: (1) Furnace aperture emits diffusely, (2) Ad << L2.
ANALYSIS: (a) The irradiation on the detector area is defined as the power incident on the surface per
unit area of the surface. That is
(1,2)
q f → d = Ie Af cos θ f ω d − f
G = q f →d Ad
where q f →d is the radiant power which leaves Af and is intercepted by Ad. From Eqs. 12.2 and 12.5,
ω d − f is the solid angle subtended by surface Ad with respect to Af,
ω d − f = Ad cos θ d L2 .
(3)
Noting that since the aperture emits diffusely, Ie = E/π (see Eq. 12.14), and hence
(
G = ( E π ) Af cos θ f Ad cos θ d L2
) Ad
(4)
Solving for L2 and substituting for the condition θf = 0o and θd = 0o,
L2 = E cos θ f cos θ d A f π G .
(5)
1/ 2
π

(20 × 10−3 ) 2 m 2 π × 1000 W m 2 
= 193 mm .
4

(b) When θd = 20o, qf→d will be reduced by a factor of cos θd since ωd-f is reduced by a factor cos θd.
Hence,


L = 3.72 × 105 W m 2 ×
<
<
G = 1000 W/m2 × cos θd = 1000W/m2 × cos 20o = 940 W/m2 .
(c) Using the IHT workspace with Eq. (4), G is computed and plotted as a function of L for selected θd.
Note that G decreases inversely as L2. As expected, G decreases with increasing θd and in the limit,
approaches zero as θd approaches 90o.
Irradiation, G (W/m^2)
3000
2000
1000
0
100
200
Separation distance, L (mm)
thetad = 0 deg
thetad = 20 deg
thetad = 60 deg
300
PROBLEM 12.005
5
2
KNOWN: Radiation from a diffuse radiant source A1 with intensity I1 = 1.2 × 10 W/m ⋅sr is
incident on a mirror Am, which reflects radiation onto the radiation detector A2.
FIND: (a) Radiant power incident on Am due to emission from the source, A1, q1→m (mW), (b)
2
Intensity of radiant power leaving the perfectly reflecting, diffuse mirror Am, Im (W/m ⋅sr), and (c)
Radiant power incident on the detector A2 due to the reflected radiation leaving Am, qm→2 (µW), (d)
Plot the radiant power qm→2 as a function of the lateral separation distance yo for the range 0 ≤ yo ≤
0.2 m; explain features of the resulting curve.
SCHEMATIC:
ASSUMPTIONS: (1) Surface A1 emits diffusely, (2) Surface Am does not emit, but reflects
perfectly and diffusely, and (3) Surface areas are much smaller than the square of their separation
distances.
ANALYSIS: (a) The radiant power leaving A1 that is incident on Am is
q1→ m = I1 ⋅ A1 ⋅ cosθ 1 ⋅ ∆ω m-1
where ωm-1 is the solid angle Am subtends with respect to A1, Eq. 12.2,
∆ω m-1 ≡
dA n
r2
=
A m cos θ m
x2o + y 2o
=
2 × 10−4 m2 ⋅ cos 45°
0.12 + 01
. 2 m2
= 7.07 × 10−3 sr
with θ m = 90°−θ 1 and θ 1 = 45° ,
. × 105 W / m2 ⋅ sr × 1 × 10-4 m2 × cos 45°×7.07 × 10-3 sr = 60 mW
q1→ m = 12
(b) The intensity of radiation leaving Am, after perfect and diffuse reflection, is
I m = q1→ m / A m / π =
$
60 × 10−3 W
π × 2 × 10-4 m2
. W / m2 ⋅ sr
= 955
(c) The radiant power leaving Am due to reflected radiation leaving Am is
q m→ 2 = q 2 = I m ⋅ A m ⋅ cos θ m ⋅ ∆ω 2 − m
where ∆ω2-m is the solid angle that A2 subtends with respect to Am, Eq. 12.2,
Continued …..
<
PROBLEM 12.005 (Cont.)
∆ω 2 − m ≡
dA n
r2
=
1 × 10−4 m2 × cos 45°
A 2 cos θ 2
. × 10 −3 sr
=
= 354
2
2
2
2
2
Lo − x o + y o
0.1 + 0.1 m
$
with θ2 = 90° - θm
. W / m2 ⋅ sr × 2 × 104 m2 × cos 45°×3.54 × 10-3 sr = 47.8 µW
q m→ 2 = q 2 = 955
<
(d) Using the foregoing equations in the IHT workspace, q 2 is calculated and plotted as a function of
yo for the range 0 ≤ yo ≤ 0.2 m.
100
E m itte d p o w e r fro m A1 re fle cte d fro m Am o n to A2
q 2 (u W )
80
60
40
20
0
0
0 .0 5
0 .1
0 .1 5
0 .2
yo (m )
From the relations, note that q2 is dependent upon the geometric arrangement of the surfaces in the
following manner. For small values of yo, that is, when θ1 ≈ 0°, the cos θ1 term is at a maximum,
near unity. But, the solid angles ∆ωm-1 and ∆ω2-m are very small. As yo increases, the cos θ1 term
doesn’t diminish as much as the solid angles increase, causing q2 to increase. A maximum in the
power is reached as the cos θ1 term decreases and the solid angles increase. The maximum radiant
power occurs when yo = 0.058 m which corresponds to θ1 = 30°.
PROBLEM 12.6
KNOWN: Flux and intensity of direct and diffuse components, respectively, of solar irradiation.
FIND: Total irradiation.
SCHEMATIC:
ANALYSIS: Since the irradiation is based on the actual surface area, the contribution due to the
direct solar radiation is
Gdir = q ′′dir ⋅ cos θ .
From Eq. 12.19 the contribution due to the diffuse radiation is
G dif = π I dif .
Hence
′′ ⋅ cos θ + π Idif
G = Gdir + Gdif = qdir
or
G = 1000W/m 2 × 0.866 + π sr × 7 0 W / m 2 ⋅ sr
G = ( 866 + 220 ) W / m2
or
G = 1086 W / m 2.
<
COMMENTS: Although a diffuse approximation is often made for the non-direct component of solar
radiation, the actual directional distribution deviates from this condition, providing larger intensities at
angles close to the direct beam.
PROBLEM 12.007
7
2
KNOWN: Daytime solar radiation conditions with direct solar intensity Idir = 2.10 × 10 W/m ⋅sr
-5
within the solid angle subtended with respect to the earth, ∆ωS = 6.74 × 10 sr, and diffuse intensity
2
Idif = 70 W/m ⋅sr.
FIND: (a) Total solar irradiation at the earth’s surface when the direct radiation is incident at 30°,
and (b) Verify the prescribed value of ∆ωS recognizing that the diameter of the earth is DS = 1.39 ×
9
11
10 m, and the distance between the sun and the earth is re-S = 1.496 × 10 m (1 astronomical unity).
SCHEMATIC:
ANALYSIS: (a) The total solar irradiation is the sum of the diffuse and direct components,
1
6
GS = G dif + G dir = 220 + 1226 W / m2 = 1446 W / m2
<
From Eq. 12.19 the diffuse irradiation is
G dif = π I dif = π sr × 70 W / m2 ⋅ sr = 220 W / m2
The direct irradiation follows from Eq. 12.15, expressed in terms of the solid angle
G dir = I dir cosθ ∆ω S
G dir = 2.10 × 107 W / m2 ⋅ sr × cos 30°×6.74 × 10-5 sr = 1226 W / m2
(b) The solid angle the sun subtends with respect to the earth is calculated from Eq. 12.2,
9 m 2 /4
π
139
.
10
2
×
dA n π DS / 4
∆ω S =
=
=
= 6.74 × 10−5 sr
2
2
2
r
re-S
1496
.
× 1011 m
4
4
9
<
9
where dAn is the projected area of the sun and re-S, the distance between the earth and sun. We are
2 >> D 2 .
assuming that re-S
S
COMMENTS: Can you verify that the direct solar intensity, Idir, is a reasonable value, assuming that
4
the solar disk emits as a black body at 5800 K? I b,S = σTS4 / π = σ 5800 K / π
1
6
= 2.04 × 107 W / m2 ⋅ sr . Because of local cloud formations, it is possible to have an appreciable
9
diffuse component. But it is not likely to have such a high direct component as given in the problem
statement.
PROBLEM 12.8
KNOWN: Directional distribution of solar radiation intensity incident at earth’s surface on an overcast
day.
FIND: Solar irradiation at earth’s surface.
SCHEMATIC:
ASSUMPTIONS: (1) Intensity is independent of azimuthal angle θ.
ANALYSIS: Applying Eq. 12.17 to the total intensity
2π π / 2
Ii (θ ) cosθ sinθ dθ d φ
0 0
G=∫
∫
π /2
cos 2 θ sinθ d θ
0
G = 2 π In ∫
 1
 π /2
G = ( 2 π sr ) × 8 0 W / m 2 ⋅ sr  − cos3 θ 
 3
 0
π
G = −167.6W/m 2 ⋅ sr  cos3 − cos3 0 
2


G = 167.6W/m 2.
<
PROBLEM 12.9
KNOWN: Emissive power of a diffuse surface.
FIND: Fraction of emissive power that leaves surface in the directions π/4 ≤ θ ≤ π/2 and 0 ≤ φ ≤ π.
SCHEMATIC:
ASSUMPTIONS: (1) Diffuse emitting surface.
ANALYSIS: According to Eq. 12.12, the total, hemispherical emissive power is
∞ 2π π / 2
I λ ,e ( λ, θ , φ ) cos θ sin θ d θ d φ dλ.
0 0
0
E=∫
∫ ∫
For a diffuse surface Iλ,e (λ, θ, φ) is independent of direction, and as given by Eq. 12.14,
E = π I e.
The emissive power, which has directions prescribed by the limits on θ and φ, is
π
π /2
∞
I λ ,e ( λ ) dλ  dφ  
cos θ sin θ d θ 
0
0
/
4
π



∆E = ∫
∫
∫
π /2
2
π  sin θ 
1

∆E = I e [φ ]0 × 

= I e [π ]  1 − 0.707 2 
2

 2 π / 4
(
)
∆E = 0.25 π I e .
It follows that
∆E 0.25 π Ie
=
= 0.25.
E
π Ie
COMMENTS: The diffuse surface is an important concept in radiation heat transfer, and the
directional independence of the intensity should be noted.
<
PROBLEM 12.10
KNOWN: Spectral distribution of Eλ for a diffuse surface.
FIND: (a) Total emissive power E, (b) Total intensity associated with directions θ = 0o and θ = 30o,
and (c) Fraction of emissive power leaving the surface in directions π/4 ≤ θ ≤ π/2.
SCHEMATIC:
ASSUMPTIONS: (1) Diffuse emission.
ANALYSIS: (a) From Eq. 12.11 it follows that
∞
5
10
15
∞
20
E = ∫ E λ (λ ) dλ = ∫ (0) dλ + ∫ (100) dλ + ∫ (200) dλ + ∫ (100) dλ + ∫ (0) dλ
0
0
5
10
15
20
E = 100 W/m2 ⋅µm (10 − 5) µm + 200W/m2 ⋅µm (15 − 10) µm + 100 W/m2 ⋅µm (20−15) µm
<
E = 2000 W/m2
(b) For a diffuse emitter, Ie is independent of θ and Eq. 12.14 gives
Ie =
E
π
=
2000 W m 2
π sr
<
Ie = 637 W m 2⋅ sr
(c) Since the surface is diffuse, use Eqs. 12.10 and 12.14,
E(π 4 → π 2)
E
E(π 4 → π 2)
E
E(π 4 → π 2)
E
2π
π /2
∫ ∫π / 4
= 0
Ie cos θ sin θ dθ dφ
π Ie
π /2
2π
∫ cosθ sin θ dθ ∫0
= π /4
π
=
dφ
π /2
=
1  sin 2 θ 

π 
2
2π
φ 0

π / 4
1 1 2

(1 − 0.707 2 )(2π − 0)  = 0.50
π  2

COMMENTS: (1) Note how a spectral integration may be performed in parts.
(2) In performing the integration of part (c), recognize the significance of the diffuse emission
assumption for which the intensity is uniform in all directions.
<
PROBLEM 12.11
KNOWN: Diffuse surface ∆Ao, 5-mm square, with total emissive power Eo = 4000 W/m2.
FIND: (a) Rate at which radiant energy is emitted by ∆Ao, qemit; (b) Intensity Io,e of the radiation field
emitted from the surface ∆Ao; (c) Expression for qemit presuming knowledge of the intensity Io,e
beginning with Eq. 12.10; (d) Rate at which radiant energy is incident on the hemispherical surface, r =
R1 = 0.5 m, due to emission from ∆Ao; (e) Rate at which radiant energy leaving ∆Ao is intercepted by
the small area ∆A2 located in the direction (40o, φ) on the hemispherical surface using Eq. 12.5; also
determine the irradiation on ∆A2; (f) Repeat part (e), for the location (0o, φ); are the irradiations at the
two locations equal? and (g) Irradiation G1 on the hemispherical surface at r = R1 using Eq. 12.5.
SCHEMATIC:
ASSUMPTIONS: (1) Diffuse surface, ∆Ao, (2) Medium above ∆Ao is also non-participating, (3)
R12 >> ∆A o , ∆A2.
ANALYSIS: (a) The radiant power leaving ∆Ao by emission is
<
qemit = Eo⋅∆Ao = 4000 W/m2 (0.005 m × 0.005 m) = 0.10 W
(b) The emitted intensity is Io,e and is independent of direction since ∆Ao is a diffuser emitter,
<
Io,e = E o π = 1273 W m 2⋅ sr
The intensities at points P1 and P2 are also Io,e and the intensity in the directions shown in the schematic
above will remain constant no matter how far the point is from the surface ∆Ao since the space is nonparticipating.
(c) From knowledge of Io,e, the radiant power leaving ∆Ao from Eq. 12.10 is,
2π
π /2
q emit = ∫ Io,e ∆Ao cos θ sin θ dθ dφ = Io,e ∆A o ∫
cos θ sin θ dθ dφ = π Io,e ∆A o = 0.10 W
h
φ = 0 ∫θ = 0
<
(d) Defining control surfaces above ∆Ao and on A1, the radiant power leaving ∆Ao must pass through A1.
That is,
<
q1,inc = E o ∆A o = 0.10 W
Recognize that the average irradiation on the hemisphere, A1, where A1 = 2π R12 , based upon the
definition, Section 12.2.3,
G1 = q1,inc A1 = E o ∆ A o 2π R12 = 63.7 mW m 2
where q1,inc is the radiant power incident on surface A1.
Continued...
PROBLEM 12.11 (Cont.)
(e) The radiant power leaving ∆Ao intercepted by ∆A2, where ∆A2 = 4×10-6 m2, located at (θ = 45o, φ) as
per the schematic, follows from Eq. 12.5,
q ∆A →∆A = Io,e ∆Ao cos θ o ∆ω 2 − o
o
2
where θo = 45o and the solid angle ∆A2 subtends with respect to ∆Ao is
∆ω 2 − o = ∆A 2 cos θ 2 R12 = 4 × 10−6 m 2 ⋅ 1 (0.5m)2 = 1.60 × 10−5 sr
where θ2 = 0o, the direction normal to ∆A2,
q ∆A →∆A = 1273 W m 2 ⋅ sr × 25 × 10−6 m 2 cos 45o × 1.60 × 10−5 sr = 3.60 × 10−7 W
o
2
<
From the definition of irradiation, Section 12.2.3,
G 2 = q ∆A →∆A ∆A 2 = 90 mW m 2
o
2
(f) With ∆A2, located at (θ = 0o, φ), where cosθo = 1, cosθ2 = 1, find
∆ω 2 − o = 1.60 × 10−5 sr
q ∆A →∆A = 5.09 × 10−7 W
o
2
G 2 = 127 mW m 2
<
Note that the irradiation on ∆A2 when it is located at (0o, φ) is larger than when ∆A2 is located at (45o,
φ); that is, 127 mW/m2 > 90 W/m2. Is this intuitively satisfying?
(g) Using Eq. 12.15, based upon Figure 12.10, find
G1 = ∫ I1,i dA1 ⋅ dω 0 −1 A1 = π Io,e ∆Ao ∆A1 = 63.7 mW m 2
h
<
where the elemental area on the hemispherical surface A1 and the solid angle ∆Ao subtends with respect
to ∆A1 are, respectively,
dA1 = R12 sin θ dθ dφ
dω o −1 = ∆A o cos θ R12
From this calculation you found that the average irradiation on the hemisphere surface, r = R1, is
G1 = 63.7 mW m 2 . From parts (e) and (f), you found irradiations, G2 on ∆A2 at (0o, φ) and (45o, φ) as
127 mW/m2 and 90 mW/m2, respectively. Did you expect G1 to be less than either value for G2? How
do you explain this?
COMMENTS: (1) Note that from Parts (e) and (f) that the irradiation on A1 is not uniform. Parts (d)
and (g) give an average value.
(2) What conclusions would you reach regarding G1 if ∆Ao were a sphere?
PROBLEM 12.12
KNOWN: Hemispherical and spherical arrangements for radiant heat treatment of a thin-film material.
Heater emits diffusely with intensity Ie,h = 169,000 W/ m2⋅sr and has an area 0.0052 m2.
FIND: (a) Expressions for the irradiation on the film as a function of the zenith angle, θ, and (b) Identify
arrangement which provides the more uniform irradiation, and hence better quality control for the
treatment process.
SCHEMATIC:
ASSUMPTIONS: (1) Heater emits diffusely, (2) All radiation leaving the heater is absorbed by the thin
film.
ANALYSIS: (a) The irradiation on any differential area, dAs, due to emission from the heater, Ah ,
follows from its definition, Section 12.2.3,
q
G = h →s
dAs
(1)
Where qh→s is the radiant heat rate leaving Ah and intercepted by dAs. From Eq. 12.5,
q h →s = Ie,h ⋅ dA h cos θ h ⋅ ωs − h
(2)
where ωs-h is the solid angle dAs subtends with respect to any point on Ah. From the definition, Eq. 12.2,
ω=
dA n
(3)
r2
where dAn is normal to the viewing direction and r is the separation distance.
For the hemisphere: Referring to the schematic above, the solid angle is
ωs − h =
dAs
R2
and the irradiation distribution on the hemispheric surface as a function of θh is
G = Ie,h A h cosθ h R 2
(1)
<
For the sphere: From the schematic, the solid angle is
ωs,h =
dAs cos θs
R o2
=
dAs
4R 2 cos θ h
where Ro, from the geometry of sphere cord and radii with θs = θh, is
Continued...
PROBLEM 12.12 (Cont.)
R o = 2R cos θ h
and the irradiation distribution on the spherical surface as a function of θh is
G = Ie,h d A h 4R 2
(2)
<
(b) The spherical shape provides more uniform irradiation as can be seen by comparing Eqs. (1) and (2).
In fact, for the spherical shape, the irradiation on the thin film is uniform and therefore provides for
better quality control for the treatment process. Substituting numerical values, the irradiations are:
G hem = 169, 000 W m 2⋅ sr × 0.0052m 2 cosθ h
(2m )2 = 219.7 cosθ h W
m2
2
Gsph = 169, 000 W m 2⋅ sr × 0.0052 m 2 4 ( 2m ) = 54.9 W m 2
(3)
(4)
COMMENTS: (1) The radiant heat rate leaving the diffuse heater surface by emission is
q tot = π Ie,h A h = 276.1W
The average irradiation on the spherical surface, Asph = 4πR2,
2
Gsph = q tot Asph = 276.1W 4π ( 2m ) = 54.9 W m 2
while the average irradiation on the hemispherical surface, Ahem = 2πR2 is
2
G hem = 276.1W 2π ( 2m ) = 109.9 W m 2
(2) Note from the foregoing analyses for the sphere that the result for G sph is identical to that found as
Eq. (4). That follows since the irradiation is uniform.
(3) Note that G hem > G sph since the surface area of the hemisphere is half that of the sphere.
Recognize that for the hemisphere thin film arrangement, the distribution of the irradiation is quite
variable with a maximum at θ = 0° (top) and half the maximum value at θ = 30°.
PROBLEM 12.13
KNOWN: Hot part, ∆Ap, located a distance x1 from an origin directly beneath a motion sensor at a
distance Ld = 1 m.
FIND: (a) Location x1 at which sensor signal S1 will be 75% that corresponding to x = 0, directly
beneath the sensor, So, and (b) Compute and plot the signal ratio, S/So, as a function of the part position
x1 for the range 0.2 ≤ S/So ≤ 1 for Ld = 0.8, 1.0 and 1.2 m; compare the x-location for each value of Ld at
which S/So = 0.75.
SCHEMATIC:
ASSUMPTIONS: (1) Hot part is diffuse emitter, (2) L2d >> ∆Ap, ∆Ao.
ANALYSIS: (a) The sensor signal, S, is proportional to the radiant power leaving ∆Ap and intercepted
by ∆Ad,
S ~ q p →d = I p,e ∆A p cos θ p ∆ω d − p
(1)
L
cos θ p = cos θ d = d = Ld (L2d + x12 )1/ 2
R
(2)
when
∆ω d − p =
∆A d ⋅ cos θ d
R
2
= ∆Ad ⋅ Ld (L2d + x12 )3 / 2
(3)
Hence,
q p → d = I p,e ∆A p ∆A d
L2d
(4)
(L2d + x12 )2
It follows that, with So occurring when x= 0 and Ld = 1 m,
 L2 
d
=
=

2
2
2
2
2
2
So L (L + 0 )
 Ld + x1 
d
d
S
L2d (L2d + x12 ) 2
2
(5)
so that when S/So = 0.75, find,
<
x1 = 0.393 m
(b) Using Eq. (5) in the IHT workspace, the signal ratio, S/So, has been computed and plotted as a
function of the part position x for selected Ld values.
Continued...
PROBLEM 12.13 (Cont.)
1
Signal ratio, S/So
0.8
0.6
0.4
0.2
0
0
1
2
Part position, x (m)
Sensor position, Ld = 0.8 m
Ld = 1 m
Ld = 1.2 m
When the part is directly under the sensor, x = 0, S/So = 1 for all values of Ld. With increasing x, S/So
decreases most rapidly with the smallest Ld. From the IHT model we found the part position x
corresponding to S/So = 0.75 as follows.
S/So
0.75
0.75
0.75
Ld (m)
0.8
1.0
1.2
x1 (m)
0.315
0.393
0.472
If the sensor system is set so that when S/So reaches 0.75 a process is initiated, the technician can use the
above plot and table to determine at what position the part will begin to experience the treatment process.
PROBLEM 12.14
KNOWN: Diameter and temperature of burner. Temperature of ambient air. Burner efficiency.
FIND: (a) Radiation and convection heat rates, and wavelength corresponding to maximum spectral
emission. Rate of electric energy consumption. (b) Effect of burner temperature on convection and
radiation rates.
SCHEMATIC:
ASSUMPTIONS: (1) Burner emits as a blackbody, (2) Negligible irradiation of burner from
surrounding, (3) Ambient air is quiescent, (4) Constant properties.
-6
2
PROPERTIES: Table A-4, air (Tf = 408 K): k = 0.0344 W/m⋅K, ν = 27.4 × 10 m /s, α = 39.7 ×
-1
-6 2
10 m /s, Pr = 0.70, β = 0.00245 K .
ANALYSIS: (a) For emission from a black body
(
)
q rad = As E b = π D 2 / 4 σ T 4 = π ( 0.2m ) / 4  5.67 × 10−8 W / m 2 ⋅ K 4 (523 K ) = 133 W


2
4
3
2
<
-1
With L = As/P = D/4 = 0.05m and RaL = gβ (Ts - T∞) L /αν = 9.8 m/s × 0.00245 K (230 K)
3
-12 4 2
5
(0.05m) /(27.4 × 39.7 × 10 m /s ) = 6.35 × 10 , Eq. (9.30) yields
h=
)
(
k
k
4 =  0.0344 W / m ⋅ K  0.54 6.35 × 105 1/ 4 = 10.5 W / m 2 ⋅ K
Nu L =   0.54 Ra1/


L
L
0.05m
L


2
q conv = h As ( Ts − T∞ ) = 19.4 W / m 2 ⋅ K π ( 0.2m ) / 4  230 K = 75.7 W


<
The electric power requirement is then
q + qconv (133 + 75.7 ) W
Pelec = rad
=
= 232 W
η
0.9
<
The wavelength corresponding to peak emission is obtained from Wien’s law, Eq. (12.27)
<
λmax = 2898µ m ⋅ K / 523K = 5.54 µ m
(b) As shown below, and as expected, the radiation rate increases more rapidly with temperature than
(
)
the convection rate due to its stronger temperature dependence Ts4 vs. Ts5 / 4 .
Continued …..
PROBLEM 12.14(Cont.)
500
Heat rate (W)
400
300
200
100
0
100
150
200
250
300
350
Surface temperature (C)
qconv
qrad
Pelec
COMMENTS: If the surroundings are treated as a large enclosure with isothermal walls at Tsur = T∞
2
4
= 293 K, irradiation of the burner would be G = σ Tsur
= 418 W/m and the corresponding heat rate
would be As G = 13 W. This input is much smaller than the energy outflows due to convection and
radiation and is justifiably neglected.
PROBLEM 12.15
KNOWN: Evacuated, aluminum sphere (D = 2m) serving as a radiation test chamber.
FIND: Irradiation on a small test object when the inner surface is lined with carbon black and at
600K. What effect will surface coating have?
SCHEMATIC:
ASSUMPTIONS: (1) Sphere walls are isothermal, (2) Test surface area is small compared to the
enclosure surface.
ANALYSIS: It follows from the discussion of Section 13.3 that this isothermal sphere is an enclosure
behaving as a blackbody. For such a condition, see Fig. 12.12(c), the irradiation on a small surface
within the enclosure is equal to the blackbody emissive power at the temperature of the enclosure.
That is
G1 = Eb ( Ts ) = σ Ts4
G1 = 5.67 ×10 −8 W / m 2 ⋅K 4 ( 600K ) = 7348W/m 2 .
4
<
The irradiation is independent of the nature of the enclosure surface coating properties.
COMMENTS: (1) The irradiation depends only upon the enclosure surface temperature and is
independent of the enclosure surface properties.
(2) Note that the test surface area must be small compared to the enclosure surface area. This allows
for inter-reflections to occur such that the radiation field, within the enclosure will be uniform (diffuse)
or isotropic.
(3) The irradiation level would be the same if the enclosure were not evacuated since, in general, air
would be a non-participating medium.
PROBLEM 12.16
KNOWN: Isothermal enclosure of surface area, As, and small opening, Ao, through which 70W
emerges.
FIND: (a) Temperature of the interior enclosure wall if the surface is black, (b) Temperature of the
wall surface having ε = 0.15.
SCHEMATIC:
ASSUMPTIONS: (1) Enclosure is isothermal, (2) Ao << As.
ANALYSIS: A characteristic of an isothermal enclosure, according to Section 12.3, is that the radiant
power emerging through a small aperture will correspond to blackbody conditions. Hence
q rad = A o E b ( Ts ) = A o σ Ts4
where qrad is the radiant power leaving the enclosure opening. That is,
1/4
q

Ts =  rad 
 Ao σ 
1/4


70W
=

 0.02m 2 × 5.670 ×10 −8 W / m 2 ⋅ K 4 


= 498K.
<
Recognize that the radiated power will be independent of the emissivity of the wall surface. As long as
Ao << As and the enclosure is isothermal, then the radiant power will depend only upon the
temperature.
COMMENTS: It is important to recognize the unique characteristics of isothermal enclosures. See
Fig. 12.12 to identify them.
PROBLEM 12.17
KNOWN: Sun has equivalent blackbody temperature of 5800 K. Diameters of sun and earth as well
as separation distance are prescribed.
FIND: Temperature of the earth assuming the earth is black.
SCHEMATIC:
ASSUMPTIONS: (1) Sun and earth emit as blackbodies, (2) No attenuation of solar irradiation
enroute to earth, and (3) Earth atmosphere has no effect on earth energy balance.
ANALYSIS: Performing an energy balance on the earth,
E& in − E& out = 0
Ae,p ⋅G S = Ae,s ⋅ Eb (Te )
(π D2e / 4 )GS = π D2eσ Te4
1/4
Te = ( G S / 4σ )
where Ae,p and Ae,s are the projected area and total surface area of the earth, respectively. To
determine the irradiation GS at the earth’s
surface, equate the rate of emission from the
sun to the rate at which this radiation passes
through a spherical surface of radius RS,e – De/2.
E& in − E& out = 0
2
π DS2 ⋅σ TS4 = 4π  RS,e − De / 2  GS
(
π 1.39 × 109 m
) × 5.67 ×10−8 W / m2 ⋅ K4 ( 5800 K)4
2
2
= 4π 1.5× 1011 − 1.29 × 107 / 2 m2 × GS


GS = 1377.5 W / m 2 .
Substituting numerical values, find
(
Te = 1377.5 W / m 2 / 4 × 5.67 × 10−8 W / m 2 ⋅ K4
)
1/4
= 279 K.
<
COMMENTS: (1) The average earth’s temperature is greater than 279 K since the effect of the
atmosphere is to reduce the heat loss by radiation.
(2) Note carefully the different areas used in the earth energy balance. Emission occurs from the total
spherical area, while solar irradiation is absorbed by the projected spherical area.
PROBLEM 12.18
2
KNOWN: Solar flux at outer edge of earth’s atmosphere, 1353 W/m .
FIND: (a) Emissive power of sun, (b) Surface temperature of sun, (c) Wavelength of maximum solar
emission, (d) Earth equilibrium temperature.
SCHEMATIC:
ASSUMPTIONS: (1) Sun and earth emit as blackbodies, (2) No attenuation of solar radiation
enroute to earth, (3) Earth atmosphere has no effect on earth energy balance.
ANALYSIS: (a) Applying conservation of energy to the solar energy crossing two concentric
spheres, one having the radius of the sun and the other having the radial distance from the edge of the
earth’s atmosphere to the center of the sun
2
De 

2
E s π D s = 4π  R s− e −
 q ′′s.
( )

2 
Hence
Es =
) ×1 3 5 3 W / m2 = 6.302 ×107 W / m 2.
9m 2
1.39
×
10
(
)
(
4 1.5 ×1011 m − 0.65 ×107 m
2
<
(b) From Eq. 12.28, the temperature of the sun is
1/4
E 
Ts =  s 
σ 
1/4
 6.302 ×107 W / m2 

=
 5.67 ×10 −8 W / m 2 ⋅ K 4 


= 5774 K.
<
(c) From Wien’s law, Eq. 12.27, the wavelength of maximum emission is
C
2897.6 µ m ⋅ K
λmax = 3 =
= 0.50 µ m.
T
5774 K
(d) From an energy balance on the earth’s surface
E e π De2 = q S′′ π De2 / 4 .
(
)
(
)
Hence, from Eq. 12.28,
1/4 
 q′′S 
=
Te = 


 4σ 
1/4


−8 W / m 2 ⋅ K 4 
×
×
4
5.67
10


1353 W / m 2
= 278K.
<
COMMENTS: The average earth temperature is higher than 278 K due to the shielding effect of the
earth’s atmosphere (transparent to solar radiation but not to longer wavelength earth emission).
PROBLEM 12.19
KNOWN: Small flat plate positioned just beyond the earth’s atmosphere oriented such that its normal
passes through the center of the sun. Pertinent earth-sun dimensions from Problem 12.18.
FIND: (a) Solid angle subtended by the sun about a point on the surface of the plate, (b) Incident
intensity, Ii , on the plate using the known value of the solar irradiation about the earth’s atmosphere, GS
= 1353 W/m2, and (c) Sketch of the incident intensity as a function of the zenith angle θ, where θ is
measured from the normal to the plate.
SCHEMATIC:
ASSUMPTIONS: (1) Plate oriented normal to centerline between sun and earth, (2) Height of earth’s
atmosphere negligible compared to distance from the sun to the plate, (3) Dimensions of the plate are
very small compared to sun-earth dimensions.
ANALYSIS: (a) The pertinent sun-earth dimensions are shown in the schematic (a) above while the
position of the plate relative to the sun and the earth is shown in (b). The solid angle subtended by the
sun with respect to any point on the plate follows from Eq. 12.2,
ωS − p =
As cos θ p
L2S,p
(
)
π DS2 4 cos θ p
=
( RS,e + De 2 )
2
(
π 1.39 × 109 m
=
(1.5 ×10
11
2
)
4 ×1
7
m + 1.29 × 10 m 2
2
)
= 6.74 × 10−5 sr (1)
<
where AS is the projected area of the sun (the solar disk), θp is the zenith angle measured between the
plate normal and the centerline between the sun and earth, and LS,p is the separation distance between the
plate at the sun’s center.
(b) The plate is irradiated by solar flux in the normal direction only (not diffusely). Using Eq. (12.7),
the radiant power incident on the plate can be expressed as
GS∆A p = Ii ⋅ ∆A p cos θ p ⋅ ωS− p
(2)
and the intensity Ii due to the solar irradiation GS with cos θp = 1,
Ii = GS ωS− p = 1353 W m 2 6.74 × 10
−5
sr = 2.01 × 107 W m 2 ⋅ sr
(c) As illustrated in the schematic to the right, the intensity Ii
will be constant for the zenith angle range 0 ≤ θp ≤ θp,o where
θ p,o =
DS 2
LS,p
=
1.39 × 109 m 2
(1.5 ×1011m + 1.29 ×107 m 2)
θ p,o = 4.633 × 10−3 rad ≈ 0.27$
For the range θp > θp,o, the intensity will be zero. Hence
the Ii as a function of θp will appear as shown to the
right.
<
PROBLEM 12.20
KNOWN: Various surface temperatures.
FIND: (a) Wavelength corresponding to maximum emission for each surface, (b) Fraction of solar
emission in UV, VIS and IR portions of the spectrum.
ASSUMPTIONS: (1) Spectral distribution of emission from each surface is approximately that of a
blackbody, (2) The sun emits as a blackbody at 5800 K.
ANALYSIS: (a) From Wien’s law, Eq. 12.27, the wavelength of maximum emission for blackbody
radiation is
λmax =
C3 2897.6 µ m ⋅ K
.
=
T
T
For the prescribed surfaces
Surface
Sun
(5800K)
Tungsten
(2500K)
Hot
metal
(1500K)
0.50
1.16
1.93
λmax(µm)
Cool
Skin metal
(305K) (60K)
9.50
48.3
<
(b) From Fig. 12.3, the spectral regions associated with each portion of the spectrum are
Spectrum
Wavelength limits, µ m
UV
VIS
IR
0.0 – 0.4
0.4 – 0.7
0.7 - 100
For T = 5800K and each of the wavelength limits, from Table 12.1 find:
λ(µm)
λT(µm⋅K)
F(0→λ)
-2
2
10
58
0.4
2320
0.7
4060
10
5
5.8 × 10
0
0.125
0.491
1
Hence, the fraction of the solar emission in each portion of the spectrum is:
FUV = 0.125 – 0 = 0.125
FVIS = 0.491 – 0.125 = 0.366
FIR = 1 – 0.491 = 0.509.
COMMENTS: (1) Spectral concentration of surface radiation depends strongly on surface
temperature.
(2) Much of the UV solar radiation is absorbed in the earth’s atmosphere.
<
<
<
PROBLEM 12.21
KNOWN: Visible spectral region 0.47 µm (blue) to 0.65 µm (red). Daylight and incandescent lighting
corresponding to blackbody spectral distributions from the solar disk at 5800 K and a lamp bulb at 2900
K, respectively.
FIND: (a) Band emission fractions for the visible region for these two lighting sources, and (b)
wavelengths corresponding to the maximum spectral intensity for each of the light sources. Comment
on the results of your calculations considering the rendering of true colors under these lighting
conditions.
ASSUMPTIONS: Spectral distributions of radiation from the sources approximates those of
blackbodies at their respective temperatures.
ANALYSIS: (a) From Eqs. 12.30 and 12.31, the band-emission fraction in the spectral range λ1 to λ2
at a blackbody temperature T is
F( λ1− λ 2, T ) = F ( 0→ λ 2, T) − F( 0→ λ1, T)
where the F( 0→λ T ) values can be read from Table 12.1 (or, more accurately calculated using the
IHT Radiation | Band Emission tool)
Daylight source (T = 5800 K)
F( λ1− λ 2, T ) = 0.4374 − 0.2113 = 0.2261
<
where at λ2⋅T = 0.65 µm × 5800 K = 3770 µm⋅K, find F(0 - λT) = 0.4374, and at λ1⋅T = 0.47 µm × 5800
K = 2726 µm⋅K, find F(0 - λ T) = 0.2113.
Incandescent source (T = 2900 K)
F( λ1− λ 2, T) = 0.05098 − 0.00674 = 0.0442
<
(b) The wavelengths corresponding to the peak spectral intensity of these blackbody sources can be
found using Wien’s law, Eq. 12.27.
λ max = C3 / T = 2898 µm ⋅ K
For the daylight (d) and incandescent (i) sources, find
λ max, d = 2898 µ m ⋅ K / 5800 K = 0.50 µ m
λ max, i = 2898 µm ⋅ K / 2800 K = 1.0 µm
<
<
COMMENTS: (1) From the band-emission fraction calculation, part (a), note the substantial
difference between the fractions for the daylight and incandescent sources. The fractions are a
measure of the relative amount of total radiant power that is useful for lighting (visual illumination).
(2) For the daylight source, the peak of the spectral distribution is at 0.5 µm within the visible spectral
region. In contrast, the peak for the incandescent source at 1 µm is considerably outside the visible
region. For the daylight source, the spectral distribution is “flatter” (around the peak) than that for the
incandescent source for which the spectral distribution is decreasing markedly with decreasing
wavelength (on the short-wavelength side of the blackbody curve). The eye has a bell-shaped relative
spectral response within the visible, and will therefore interpret colors differently under illumination by
the two sources. In daylight lighting, the colors will be more “true,” whereas with incandescent
lighting, the shorter wavelength colors (blue) will appear less bright than the longer wavelength colors
(red)
PROBLEM 12.22
KNOWN: Lamp with prescribed filament area and temperature radiates like a blackbody at 2900 K
when consuming 100 W.
FIND: (a) Efficiency of the lamp for providing visible radiation, and (b) Efficiency as a function of
filament temperature for the range 1300 to 3300 K.
SCHEMATIC:
ASSUMPTIONS: (1) Filament behaves as a blackbody, (2) Glass envelope transmits all visible
radiation incident upon it.
ANALYSIS: (a) We define the efficiency of the lamp as the ratio of the radiant power within the visible
spectrum (0.4 - 0.7 µm) to the electrical power required to operate the lamp at the prescribed
temperature.
η = q vis q elec .
The radiant power for a blackbody within the visible spectrum is given as
q vis = F(0.4 µ m → 0.7 µ m)Asσ Ts4 =  F(0 → 0.7 µ m) − F(0 →0.4 µ m)  Asσ Ts4
using Eq. 12.31 to relate the band emission factors. From Table 12.1, find
F(0→0.7 µ m) = 0.0719
λ2 Ts = 0.7 µm × 2900 K = 2030 µm⋅K,
λ1 Ts = 0.4 µm × 2900 K = 1600 µm⋅K,
F(0→0.4 µ m) = 0.0018
The efficiency is then
η = [0.0719 − 0.0018] × 2(2 × 10−3 m × 5 × 10−3 m)5.67 × 10−8 W m 2 ⋅ K 4 (2900 K) 4 100 W
<
η = 5.62 W 100 W = 5.6%
(b) Using the IHT Radiation Exchange Tool, Blackbody Emission Factor, and Eqs. (1) and (2) above, a
model was developed to compute and plot η as a function of Ts.
100
60
Efficiency, eta (%)
20
8
4
1
0.6
0.2
0.08
0.04
0.01
1000
1500
2000
2500
Filament temperature, Ts (K)
3000
3500
Continued...
PROBLEM 12.22 (Cont.)
Note that the efficiency decreases markedly with reduced filament temperature. At 2900 K, η = 5.6%
while at 2345 K, the efficiency decreases by more than a factor of five to η = 1%.
COMMENTS: (1) Based upon this analysis, less than 6% of the energy consumed by the lamp
operating at 2900 K is converted to visible light. The transmission of the glass envelope will be less than
unity, so the efficiency will be less than the calculated value.
(2) Most of the energy is absorbed by the glass envelope and then lost to the surroundings by convection
and radiation. Also, a significant amount of power is conducted to the lamp base and into the lamp base
socket.
(3) The IHT workspace used to generate the above plot is shown below.
// Radiation Exchange Tool - Blackbody Band Emission Factor:
/* The blackbody band emission factor, Figure 12.14 and Table 12.1, is */
FL1Ts = F_lambda_T(lambda1,Ts)
// Eq 12.30
// where units are lambda (micrometers, mum) and T (K)
/* The blackbody band emission factor, Figure 12.14 and Table 12.1, is */
FL2Ts = F_lambda_T(lambda2,Ts)
// Eq 12.30
// Efficiency and rate expressions:
eta = qvis / qelec
eta_pc = eta * 100
qelec = 100
qvis = (FL2Ts - FL1Ts) * As * sigma * Ts^4
sigma = 5.67e-8
//Assigned Variables:
Ts = 2900
As = 0.005 * 0.005
lambda1 = 0.4
lambda2 = 0.7
// Eq. (1)
// Efficiency, %
// Electrical power, W
// Eq (2)
// Stefan-Boltzmann constant, W/m^2.K
// Filament temperature, K
// Filament area, m^2
// Wavelength, mum; lower limit of visible spectrum
// Wavelength, mum; upper limit of visible spectrum
/*Data Browser Results - Part (a):
FL1Ts
FL2Ts
eta
eta_pc
lambda2
qelec
sigma
0.001771
0.07185 0.07026 7.026
0.7
100
5.67E-8 */
qvis
As
Ts
lambda1
7.026
2.5E-5
2900
0.4
PROBLEM 12.23
KNOWN: Solar disc behaves as a blackbody at 5800 K.
FIND: (a) Fraction of total radiation emitted by the sun that is in the visible spectral region, (b) Plot the
percentage of solar emission that is at wavelengths less than λ as a function of λ, and (c) Plot on the same
coordinates the percentage of emission from a blackbody at 300 K that is at wavelengths less than λ as a
function of λ; compare the plotted results with the upper abscissa scale of Figure 12.23.
ASSUMPTIONS: (1) Visible spectral region has limits λ1 = 0.40 µm and λ2 = 0.70 µm.
ANALYSIS: (a) Using the blackbody functions of Table 12.1, find F(0→λ), the fraction of radiant flux
leaving a black surface in the spectral interval 0→λ as a function of the product λT. From the tabulated
values for F(0→λ) with T = 5800 K,
λ2 = 0.70 µm
λ2T = 4060 µm⋅K
F( 0 → λ ) = 0.4914
λ1 = 0.40 µm
λ1T = 2320 µm⋅K
F( 0 → λ ) = 0.1245
1
2
Hence, for the visible spectral region, the fraction of total emitted solar flux is
F( λ → λ ) = F( 0→ λ ) − F( 0→ λ ) = 0.4914 − 0.1245 = 0.3669 or
1
2
2
1
37%
<
(b,c) Using the IHT Radiation Tool, Band Emission Factor, F(0-λT) are evaluated for the solar spectrum
(T = 5800 K) and that for a blackbody temperature (T = 300 K) as a function of wavelength and are
plotted below.
100
F(0 -> lambda) (%)
80
60
40
20
0
0.1
0.4
0.8
2
6
10
40
80
Wavelength, lambda (mum)
Solar spectrum, TbS = 5800 K (left)
Blackbody, Tbs = 300 K (right)
The left-hand curve in the plot represents the percentage of solar flux approximated as the 5800 Kblackbody spectrum in the spectral region less than λ. The right-hand curve represents the percentage of
300 K-blackbody flux in the spectral region less than λ. Referring to upper abscissa scale of Figure 12.23,
for the solar flux, 75% of the solar flux is at wavelengths shorter than 1 µm. For the blackbody flux (300
K), 75% of the blackbody flux is at wavelengths shorter than 20 µm. These values are in agreement with
points on the solar and 300K-blackbody curves, respectively, in the above plot.
PROBLEM 12.24
KNOWN: Thermal imagers operating in the spectral regions 3 to 5 µm and 8 to 14 µm.
FIND: (a) Band-emission factors for each of the spectral regions, 3 to 5 µm and 8 to 14 µm, for
temperatures of 300 and 900 K, (b) Calculate and plot the band-emission factors for each of the
spectral regions for the temperature range 300 to 1000 K; identify the maxima, and draw conclusions
concerning the choice of an imager for an application; and (c) Considering imagers operating at the
maximum-fraction temperatures found from the graph of part (b), determine the sensitivity (%)
required of the radiation detector to provide a noise-equivalent temperature (NET) of 5°C.
ASSUMPTIONS: The sensitivity of the imager’s radiation detector within the operating spectral
region is uniform.
ANALYSIS: (a) From Eqs. 12.30 and 12.31, the band-emission fraction F(λ1 → λ2, T) for blackbody
emission in the spectral range λ1 to λ2 for a temperature T is
F( λ1→λ 2, T ) = F( 0→λ 2, T) − F( 0→λ 1, T)
Using the IHT Radiation | Band Emission tool (or Table 12.1), evaluate F(0-λT) at appropriate λ⋅T
products:
3 to 5 µm region
F( λ1− λ 2, 300 K ) = 0.1375 − 0.00017 = 0.01359
<
F( λ1− λ 2, 900 K ) = 0.5640 − 0.2055 = 0.3585
<
8 to 14µm region
F( λ1− λ 2, 300 K ) = 0.5160 − 0.1403 = 0.3758
<
F( λ1− λ 2, 900 K ) = 0.9511 − 0.8192 = 0.1319
<
(b) Using the IHT Radiation | Band Emission tool, the band-emission fractions for each of the
spectral regions is calculated and plotted below as a function of temperature.
Band fractions for thermal imaging spectral regions
0.4
Band fraction for range
0.3
0.2
0.1
0
300
400
500
600
700
800
900
1000
Temperature, T (K)
3 to 5 um region
8 to 14 um region
Continued …..
PROBLEM 12.24 (Cont.)
For the 3 to 5 µm imager, the band-emission factor increases with increasing temperature. For low
temperature applications, not only is the radiant power σ T 4 , T ≈ 300 K low, but the band fraction
(
)
is small. However, for high temperature applications, the imager operating conditions are more
favorable with a large band-emission factor, as well as much higher radiant power
σ T4 , T → 900 K .
(
)
For the 8 to 14 µm imager, the band-emission factor decreases with increasing temperature. This is a
more favorable instrumentation feature, since the band-emission factor (proportionally more power)
becomes larger as the radiant power decreases. This imager would be preferred over the 3 to 5 µm
imager at lower temperatures since the band-emission factor is 8 to 10 times higher.
Recognizing that from Wien’s law, the peaks of the blackbody curves for 300 and 900 K are
approximately 10 and 3.3 µm, respectively, it follows that the imagers will receive the most radiant
power when the peak target spectral distributions are close to the operating spectral region. It is good
application practice to chose an imager having a spectral operating range close to the peak of the
blackbody curve (or shorter than, if possible) corresponding to the target temperature.
The maxima band fractions for the 3 to 5 µm and 8 to 14 µm spectral regions correspond to
temperatures of 960 and 355 K, respectively. Other application factors not considered (like smoke,
water vapor, etc), the former imager is better suited with higher temperature scenes, and the latter with
lower temperature scenes.
(c) Consider the 3 to 5 µm and 8 to 14 µm imagers operating at their band-emission peak temperatures,
355 and 960 K, respectively. The sensitivity S (% units) of the imager to resolve an NET of 5°C can
be expressed as
S (%) =
F( λ1−λ 2, T1) − F(λ1−λ 2, T2 )
F( λ1− λ 2, T1)
× 100
where T1 = 355 or 960 K and T2 = 360 or 965 K, respectively. Using this relation in the IHT
workspace, find
S3−5 = 0.035%
S8 −14 = 0.023%
<
That is, we require the radiation detector (with its signal-processing system) to resolve the output signal
with the foregoing precision in order to indicate a 5°C change in the scene temperature.
PROBLEM 12.25
KNOWN: Tube furnace maintained at Tf = 2000 K used to calibrate a heat flux gage of sensitive
2
area 5 mm mounted coaxial with the furnace centerline, and positioned 60 mm from the opening of
the furnace.
2
FIND: (a) Heat flux (kW/m ) on the gage, (b) Radiant flux in the spectral region 0.4 to 2.5 µm, the
sensitive spectral region of a solid-state (photoconductive type) heat-flux gage, and (c) Calculate and
plot the heat fluxes for each of the gages as a function of the furnace temperature for the range 2000 ≤
Tf ≤ 3000 K. Compare the values for the two types of gages; explain why the solid-state gage will
always indicate systematically low values; does the solid-state gage performance improve, or become
worse as the source temperature increases?
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Graphite tube furnace behaves as a blackbody, (3)
Areas of gage and furnace opening are small relative to separation distance squared, and (4) Extension
tube is cold relative to the furnace.
ANALYSIS: (a) The heat flux to the gage is equal to the irradiation, Gg, on the gage and can be
expressed as (see Section 12.2.3)
G g = I f ⋅ cos θ g ⋅ ∆ω f − g
where ∆ωf - g is the solid angle that the furnace opening subtends relative to the gage. From Eq. 12.2,
with θg = 0°
∆ω f − g ≡
dA n
r2
=
A f cos θ g
L2
=
π 0.0125 m 2 / 4 × 1
$
0.060 m$
2
= 3.409 × 10−2 sr
The intensity of the radiation from the furnace is
I f = E b,f Tf / π = σTf4 / π = 5.67 × 10−8 W / m2 ⋅ K4 2000 K 4 / π = 2.888 × 105 W / m2 ⋅ sr
$
$
Substituting numerical values,
G g = 2.888 × 105 W / m2 ⋅ sr × 1 × 3.409 × 10-2 sr = 9.84 kW / m2
<
(b) The solid-state detector gage, sensitive only in the spectral region λ1 = 0.4 µm to λ2 = 2.5 µm, will
receive the band irradiation.
G g, λ1−λ 2 = F(λ1→λ 2, Tf ) ⋅ G g,b =  F(0→λ 2, Tf ) − F( 0→λ1, Tf )  G g,b


Continued …..
PROBLEM 12.25 (Cont.)
where for λ1 Tf = 0.4 µm × 2000 K = 800 µm⋅K, F(0 - λ1) = 0.0000 and for λ2 ⋅ Tf = 2.5 µm × 2000 K
= 5000 µm⋅K, F(0 - λ2) = 0.6337. Hence,
G g,λ1− λ 2 = 0.6337 − 0.0000 × 9.84 kW / m2 = 6.24 kW / m2
<
(c) Using the foregoing equation in the IHT workspace, the heat fluxes for each of the gage types are
calculated and plotted as a function of the furnace temperature.
H e a t flu x , G g (k W /m ^ 2 )
60
H e a t h e a t flu x g a g e ca lib ra tio n s
40
20
0
2000
2200
2400
2600
2800
3000
Fu rn a c e te m p e ra tu re , Tf (K )
B la c k h e a t flu x g a g e
S o lid -s ta te g a g e , 0 .4 to 2 .5 u m
For the black gage, the irradiation received by the gage, Gg, increases as the fourth power of the
furnace temperature. For the solid-state gage, the irradiation increases slightly greater than the fourth
power of the furnace temperature since the band-emission factor for the spectral region, F(λ1 - λ2, Tf),
increases with increasing temperature. The solid-state gage will always indicate systematic low
readings since its band-emission factor never approaches unity. However, the error will decrease with
increasing temperature as a consequence of the corresponding increase in the band-emission factor.
COMMENTS: For this furnace-gage geometrical arrangement, evaluating the solid angle, ∆ωf - g,
and the areas on a differential basis leads to results that are systematically high by 1%. Using the
view factor concept introduced in Chapter 13 and Eq. 13.8, the results for the black and solid-state
2
gages are 9.74 and 6.17 kW/m , respectively.
PROBLEM 12.26
KNOWN: Geometry and temperature of a ring-shaped radiator. Area of irradiated part and distance
from radiator.
FIND: Rate at which radiant energy is incident on the part.
SCHEMATIC:
ASSUMPTIONS: (1) Heater emits as a blackbody.
ANALYSIS: Expressing Eq. 12.5 on the basis of the total radiation, dq = Ie dAh cosθ dω, the rate at
which radiation is incident on the part is
q h − p = ∫ dq = Ie ∫ ∫ cos θ dω p − h dA h ≈ Ie cos θ ⋅ ω p − h ⋅ A h
Since radiation leaving the heater in the direction of the part is oriented normal to the heater surface, θ =
0 and cos θ = 1. The solid angle subtended by the part with respect to the heater is ωp-h = Ap cos θ1/L2,
while the area of the heater is Ah ≈ 2πrhW = 2π(L sin θ1)W. Hence, with Ie = Eb/π = σ Th4 π ,
(
)
$
2
4
5.67 × 10−8 W m 2⋅ K 4 (3000 K ) 0.007 m cos 30
qh −p ≈
×
× 2π (1.5 m ) 0.03m
π
(3m )2
q h − p ≈ 278.4 W
<
COMMENTS: The foregoing representation for the double integral is an excellent approximation since
W << L and Ap << L2.
PROBLEM 12.27
KNOWN: Spectral distribution of the emissive power given by Planck’s law.
FIND: Approximations to the Planck distribution for the extreme cases when (a) C2/λT >> 1, Wien’s
law and (b) C2/λT << 1, Rayleigh-Jeans law.
ANALYSIS: Planck’s law provides the spectral, hemispherical emissive power of a blackbody as a
function of wavelength and temperature, Eq. 12.26,
E λ ,b ( λ, T ) = C1 / λ 5 exp ( C2 / λT ) −1 .
We now consider the extreme cases of C2/λT >> 1 and C2/λT << 1.
(a) When C2/λT >> 1 (or λT << C2), it follows exp(C2/λT) >> 1. Hence, the –1 term in the
denominator of the Planck law is insignificant, giving
E λ ,b ( λ, T ) ≈ C1 / λ 5 exp ( −C 2 / λT ) .
(
)
<
This approximate relation is known as Wien’s law. The ratio of the emissive power by Wien’s law to
that by the Planck law is,
Eλ ,b,Wien
Eλ ,b,Planck
=
1/exp ( C2 / λ T)
1/  exp ( C2 / λ T) − 1
.
For the condition λT = λmax T = 2898 µm⋅K, C2/λT =
14388 µm ⋅ K
2898 µ m ⋅ K
= 4.966 and
Eλ ,b Wien
1/exp ( 4.966 )
=
= 0.9930.
Eλ ,b Planck 1/ exp ( 4.966 ) − 1
<
That is, for λT ≤ 2898 µm⋅K, Wien’s law is a good approximation to the Planck distribution.
(b) If C2/λT << 1 (or λT >> C2), the exponential term may be expressed as a series that can be
approximated by the first two terms. That is,
x2 x3
x
e =1 +x +
+
+ ... ≈ 1 + x
2!
3!
when
x << 1.
The Rayleigh-Jeans approximation is then
E λ ,b ( λ, T ) ≈ C1 / λ 5 1 + ( C 2 / λT ) − 1 = C1T / C 2 λ 4.
For the condition λT = 100,000 µm⋅K, C2/λT = 0.1439
−1
C T / C 2λ 4
= 1
 exp ( C2 / λ T) − 1 = ( λ T / C2 )  exp ( C2 / λ T) − 1 = 1.0754. <
Eλ ,b,Planck
C1 / λ 5
Eλ ,b,R − J
That is, for λT ≥ 100,000 µm⋅K, the Rayleigh-Jeans law is a good approximation (better than 10%) to
the Planck distribution.
COMMENTS: The Wien law is used extensively in optical pyrometry for values of λ near 0.65 µm
and temperatures above 700 K. The Rayleigh-Jeans law is of limited use in heat transfer but of utility
for far infrared applications.
PROBLEM 12.28
KNOWN: Aperture of an isothermal furnace emits as a blackbody.
FIND: (a) An expression for the ratio of the fractional change in the spectral intensity to the fractional
change in temperature of the furnace aperture, (b) Allowable variation in temperature of a furnace
operating at 2000 K such that the spectral intensity at 0.65µm will not vary by more than 1/2%.
Allowable variation for 10µm.
SCHEMATIC:
ASSUMPTIONS: (1) Furnace is isothermal and aperture radiates as a blackbody.
ANALYSIS: (a) The Planck spectral distribution, Eq. 12.26, is
I λ ( λ, T ) = C1 / πλ5 exp (C 2 / λT ) −1 .
Taking natural logarithms of both sides, find l nIλ = l n C1 / πλ 5  − l n [ exp ( C2 / λT ) − 1] . Take the


total derivative of both sides, but consider the λ variable as a constant.
(
)
{exp ( C2 /λ T )} ( C2 / λ ) −1 / T 2 dT
d  exp ( C2 / λ T ) − 1
dIλ
=−
=−
 exp ( C2 / λ T ) − 1
 exp ( C2 / λ T ) − 1
Iλ
exp ( C2 /λ T)
d Iλ C2
dT
=
⋅
⋅
Iλ
λ T  exp ( C2 / λ T ) −1 T
or
d Iλ / Iλ C2
1
=
⋅
.
dT/T
λ T 1− exp ( −C2 / λT )
<
(b) If the furnace operates at 2000 K and the desirable fractional change of the spectral intensity is
0.5% at 0.65 µm, the allowable temperature variation is
dT d Iλ
=
T
Iλ
 C

1
/ 2

 λ T 1 − exp ( −C2 / λ T )  
 14,388µm ⋅ K 
 −14,388 µm ⋅ K  
dT
−4
= 0.005/ 
/  1− exp 
  = 4.517 ×10 .
T
 0.65 µ m × 2000K  
 0.65 µ m × 2000K 
That is, the allowable fractional variation in temperature is 0.045%; at 2000 K, the allowable
temperature variation is
∆T ≈ 4.517 ×10 −4 T = 4.517 ×10 −4 × 2000K = 0.90K.
<
Substituting with T = 2000 K and λ = 10 µm, find that
dT
= 3.565 × 10−3
and
<
T
∆T ≈ 3.565 ×10−3 T = 7.1K.
COMMENTS: Note that the power control requirements to satisfy the spectral intensity variation for
0.65 µm and 10 µm conditions are quite different. The peak of the blackbody curve for 2000 K is λmax
= 2898 µm⋅K/2000 K=1.45 µm.
PROBLEM 12.29
KNOWN: Spectral emissivity, dimensions and initial temperature of a tungsten filament.
FIND: (a) Total hemispherical emissivity, ε, when filament temperature is Ts = 2900 K; (b) Initial rate
of cooling, dTs/dt, assuming the surroundings are at Tsur = 300 K when the current is switched off;
(c) Compute and plot ε as a function of Ts for the range 1300 ≤ Ts ≤ 2900 K; and (d) Time required for
the filament to cool from 2900 to 1300 K.
SCHEMATIC:
ASSUMPTIONS: (1) Filament temperature is uniform at any time (lumped capacitance), (2) Negligible
heat loss by conduction through the support posts, (3) Surroundings large compared to the filament,
(4) Spectral emissivity, density and specific heat constant over the temperature range, (5) Negligible
convection.
PROPERTIES: Table A-1, Tungsten (2900 K); ρ = 19, 300 kg m3 , cp ≈ 185 J kg ⋅ K .
ANALYSIS: (a) The total emissivity at Ts = 2900 K follows from Eq. 12.38 using Table 12.1 for the
band emission factors,
∞
ε = ∫ ε λ E λ ,b (Ts )dλ = ε1F(0 → 2 µ m) + ε 2 (1 − F0 → 2 µ m )
0
(1)
<
ε = 0.45 × 0.72 + 0.1 (1 − 0.72) = 0.352
where F(0→ 2 µ m) = 0.72 at λT = 2µm × 2900 K = 5800 µm⋅K.
(b) Perform an energy balance on the filament at the instant of time at which the current is switched off,
dT
E in − E out = Mc p s
dt
As (α G sur − E ) = As α σ Ts4 − ε σ Ts4 = Mc p dTs dt
)
(
and find the change in temperature with time where As = πDL, M = ρ∀, and ∀ = (πD2/4)L,
dTs
dt
dTs
=−
4
π DLσ (ε Ts4 − α Tsur
)
(
2
)
ρ π D 4 Lc p
4
ε Ts4 − α Tsur
)
(
ρ cp D
4σ
4 × 5.67 × 10−8 W m 2 ⋅ K 4 (0.352 × 29004 − 0.1 × 300 4 )K 4
<
= −1977 K s
19, 300 kg m 2 × 185 J kg ⋅ K × 0.0008m
(c) Using the IHT Tool, Radiation, Band Emission Factor, and Eq. (1), a model was developed to
calculate and plot ε as a function of Ts. See plot below.
Continued...
dt
=−
=−
PROBLEM 12.29 (Cont.)
(d) Using the IHT Lumped Capacitance Model along with the IHT workspace for part (c) to determine ε
as a function of Ts, a model was developed to predict Ts as a function of cooling time. The results are
shown below for the variable emissivity case (ε vs. Ts as per the plot below left) and the case where the
emissivity is fixed at ε(2900 K) = 0.352. For the variable and fixed emissivity cases, the times to reach
Ts = 1300 K are
tvar = 8.3 s
<
tfix = 5.1 s
0.4
Ts (K)
3000
eps
0.3
0.2
2000
1000
0
0.1
1000 1500 2000 2500 3000
Filament temperature, Ts (K)
2
4
6
8
10
Elapsed time, t (s)
Variable emissivity
Fixed emissivity, eps = 0.35
COMMENTS: (1) From the ε vs. Ts plot, note that ε increases as Ts increases. Could you have
surmised as much by looking at the spectral emissivity distribution, ελ vs. λ?
(2) How do you explain the result that tvar > tfix?
PROBLEM 12.30
KNOWN: Spectral distribution of emissivity for zirconia and tungsten filaments. Filament
temperature.
FIND: (a) Total emissivity of zirconia, (b) Total emissivity of tungsten and comparative power
requirement, (c) Efficiency of the two filaments.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible reflection of radiation from bulb back to filament, (2) Equivalent
surface areas for the two filaments, (3) Negligible radiation emission from bulb to filament.
ANALYSIS: (a) From Eq. (12.38), the emissivity of the zirconia is
∞
ε = ∫ ε λ ( E λ / E b ) dλ = ε1 F(0→0.4 µ m ) + ε 2 F(0.4→0.7 µ m ) + ε 3 F(0.7 µ m →∞ )
o
) (
(
ε = ε1 F(0→0.4µ m ) + ε 2 F(0→0.7 µ m ) − F(0→0.4µ m ) + ε 3 1 − F(0→0.7 µ m )
)
From Table 12.1, with T = 3000 K
λ T = 0.4µ m × 3000 ≡ 1200µ m ⋅ K :
F(0→0.4 µ m ) = 0.0021
λ T = 0.7 µ m × 3000 K = 2100µ m ⋅ K : F(0→0.7 µ m ) = 0.0838
ε = 0.2 × 0.0021 + 0.8 (0.0838 − 0.0021) + 0.2 × (1 − 0.0838 ) = 0.249
<
(b) For the tungsten filament,
(
ε = ε1 F(0→ 2µ m ) + ε 2 1 − F(0→ 2µ m )
)
With λT = 6000µm⋅K, F(0 → 2µm) = 0.738
ε = 0.45 × 0.738 + 0.1(1 − 0.738) = 0.358
<
Assuming, no reflection of radiation from the bulb back to the filament and with no losses due to
′′ = εσ T 4 .
natural convection, the power consumption per unit surface area of filament is Pelec
Continued …..
PROBLEM 12.30 (Cont.)
Zirconia:
′′ = 0.249 × 5.67 ×10−8 W / m 2 ⋅ K 4 (3000 K ) = 1.14 × 106 W / m 2
Pelec
Tungsten:
′′ = 0.358 × 5.67 × 10−8 W / m 2 ⋅ K 4 (3000 K ) = 1.64 × 106 W / m 2
Pelec
4
4
Hence, for an equivalent surface area and temperature, the tungsten filament has the largest power
<
consumption.
(c) Efficiency with respect to the production of visible radiation may be defined as
∫ ε λ Eλ ,b dλ = ∫0.4 ε λ (Eλ ,b / E b ) = ε vis F
= 0.4
0.7
ηvis
0.7
E
ε
ε
(0.4→0.7 µ m )
With F(0.4 → 0.7 µm) = 0.0817 for T = 3000 K,
Zirconia:
ηvis = (0.8 / 0.249 ) 0.0817 = 0.263
Tungsten:
ηvis = (0.45 / 0.358 ) 0.0817 = 0.103
Hence, the zirconia filament is the more efficient.
COMMENTS: The production of visible radiation per unit filament surface area is Evis = ηvis
′′ . Hence,
Pelec
Zirconia:
E vis = 0.263 × 1.14 × 106 W / m 2 = 3.00 ×105 W / m 2
Tungsten:
E vis = 0.103 × 1.64 × 106 W / m 2 = 1.69 × 105 W / m 2
Hence, not only is the zirconia filament more efficient, but it also produces more visible radiation
with less power consumption. This problem illustrates the benefits associated with carefully
considering spectral surface characteristics in radiative applications.
<
PROBLEM 12.31
KNOWN: Variation of spectral, hemispherical emissivity with wavelength for two materials.
FIND: Nature of the variation with temperature of the total, hemispherical emissivity.
SCHEMATIC:
ASSUMPTIONS: (1) ε λ is independent of temperature.
ANALYSIS: The total, hemispherical emissivity may be obtained from knowledge of the spectral,
hemispherical emissivity by using Eq. 12.38
∞
∫
ε (T ) = 0
ε λ ( λ ) Eλ ,b ( λ , T ) dλ
E b (T )
E λ ,b ( λ, T )
∞
ελ ( λ )
d λ.
0
E b (T )
=∫
We also know that the spectral emissive power of a blackbody becomes more concentrated at lower
wavelengths with increasing temperature (Fig. 12.13). That is, the weighting factor, Eλ,b (λ,T)/Eb (T)
increases at lower wavelengths and decreases at longer wavelengths with increasing T. Accordingly,
Material A:
ε(T) increases with increasing T
Material B:
ε(T) decreases with increasing T.
<
<
PROBLEM 12.32
KNOWN: Metallic surface with prescribed spectral, directional emissivity at 2000 K and 1 µm (see
Example 12.6) and additional measurements of the spectral, hemispherical emissivity.
FIND: (a) Total hemispherical emissivity, ε, and the emissive power, E, at 2000 K, (b) Effect of
temperature on the emissivity.
SCHEMATIC:
ANALYSIS: (a) The total, hemispherical emissivity, ε, may be determined from knowledge of the
spectral, hemispherical emissivity, ε λ , using Eq. 12.38.
2 µ m E λ ,b (λ , T)dλ
4 µ m E λ ,b (λ , T)dλ
∞
ε λ (λ )E λ ,b (λ , T) dλ E b (T) = ε1
+ ε2
0
0
2µ m
E b (T)
E b (T)
∫
ε (T) = ∫
∫
or from Eqs. 12.28 and 12.30,
ε (T) = ε1F(0→ λ ) + ε 2  F(0→ λ ) − F(0→ λ ) 
1
2
1
From Table 12.1,
λ1 = 2 µ m, T = 2000 K : λ1T = 4000 µ m ⋅ K,
λ2 = 4 µ m, T = 2000 K : λ2T = 8000 µ m ⋅ K,
F(0 →λ ) = 0.481
1
F(0→λ ) = 0.856
2
Hence,
<
ε (T) = 0.36 × 0.481 + 0.20(0.856 − 0.481) = 0.25
From Eqs. 12.28 and 12.37, the total emissive power at 2000 K is
E(2000 K) = ε (2000 K) ⋅ Eb (2000 K)
E(2000 K) = 0.25 × 5.67 × 10 −8 W m 2⋅ K 4 × (2000 K) 4 = 2.27 × 105 W m 2 .
<
(b) Using the Radiation Toolpad of IHT, the following result was generated.
0.4
Emissivity, eps
0.3
0.2
0.1
0
500
1000
1500
2000
2500
3000
Surface temperature, T(K)
Continued...
PROBLEM 12.32 (Cont.)
At T ≈ 500 K, most of the radiation is emitted in the far infrared region (λ > 4 µm), in which case ε ≈ 0.
With increasing T, emission is shifted to lower wavelengths, causing ε to increase. As T → ∞, ε →
0.36.
COMMENTS: Note that the value of ε λ for 0 < λ ≤ 2 µm cannot be read directly from the ε λ
distribution provided in the problem statement. This value is calculated from knowledge of ε λ ,θ (θ ) in
Example 12.6.
PROBLEM 12.33
KNOWN: Relationship for determining total, hemispherical emissivity, ε, by integration of the
spectral emissivity distribution, ελ (Eq. 12.38).
FIND: Evaluate ε from ελ for the following cases: (a) Ex. 12.5, use the result to benchmark your
code, (b) tungsten at 2800 K, and (c) aluminum oxide at 1400 K. Use the intrinsic function
INTEGRAL of IHT as your solution tool.
SCHEMATIC:
ASSUMPTIONS: (1) Surfaces are diffuse emitters.
ANALYSIS: (a) Using IHT as the solution tool, Eq. 12.38 is entered into the workspace, and a lookup table created to represent the spectral emissivity distribution. See Comment 1 for the IHT
annotated code. The result is ε = 0.558, in agreement with the analysis of Ex. 12.5 using the bandemission factors.
(b, c) Using the same code as for the benchmarking exercise in Part (a), but with new look-up table
files (*.lut) representing the spectral distributions tabulated below, the total hemispherical emissivities
for the tungsten at 2800 K and aluminum oxide at 1400 K are:
ε W = 0.31
<
ε Al2O3 = 0.38
These results compare favorably with values of 0.29 and 0.41, respectively, from Fig. 12.19. See
Comment 2.
Tungsten, 2800 K
Aluminum oxide, 1400 K
λ (µm)
ελ
λ (µm)
ελ
λ (µm)
ελ
λ (µm)
ελ
0.3
0.47
2.0
0.26
0.6
0.19
4.5
0.50
0.4
0.48
4.0
0.17
0.8
0.18
5
0.70
0.5
0.47
6.0
0.05
1.0
0.175
6
0.88
0.6
0.44
8.0
0.03
1.5
0.175
10
0.96
1.0
0.38
10
0.03
2
0.19
12.5
0.9
3
0.29
15
0.53
4
0.4
20
0.39
Continued …..
PROBLEM 12.33 (Cont.)
COMMENTS: (1) The IHT code to obtain ε from ελ for the case of Ex. 12.5 spectral distribution is
shown below.
// Benchmarking use of INTEGRAL and LOOKUPVAL functions
// Calculating total emissivity from spectral distribution
/* Results: integration from 0.05 to 15 by steps of 0.02, tabulated every 10
lLb
eps
eps_t
T
lambda LLB
198.2
0.001
0.5579 1600
14.85
0.1982 */
// Emissivity integral, Eq. 12.38
eps_t = pi * INTEGRAL(lL,lambda) / (sigma * T^4)
sigma = 5.67 e-8
// Blackbody Spectral intensity, Tools | Radiation
/* From Planck’s law, the blackbody spectral intensity is */
lL = eps *lLb
lLb = l_lambda_b(lambda, T, C1, C2) // Eq. 12.25
// where units are lLb(W/m^2.sr.mum), lambda (mum) and T (K) with
C1 = 3.7420e8
// First radiation constant, W⋅mum^4/m^2
C2 = 1.4388e4
// Second radiation constant, mum⋅K
// and (mum) represents (micrometers).
// Emissivity function
eps = LOOKUPVAL(eps_L, 1, lambda, 2)
/* The table file name is eps_L.lut, with 2 columns and 6 rows. See Help | Solver |
Lookup Tables | Lookupval
0.05
0.4
1.99
0.4
2
0.8
4.99
0.8
5
0.001
100
0.001
*/
// Input variable
T = 1600
(2) For tungsten at 2800 K, the spectral limits for 98% of the blackbody spectrum are 0.51 and
8.3 µm. For aluminum at 1400 K, the spectral limits for 98% of the blackbody spectrum are
1.0 and 16.7 µm. For both cases, the foregoing tabulated spectral emissivity distributions are
adequately represented for integration within the 98% limits.
PROBLEM 12.34
KNOWN: Spectral directional emissivity of a diffuse material at 2000K.
FIND: (a) Total, hemispherical emissivity, (b) Emissive power over the spectral range 0.8 to 2.5 µm
and for directions 0 ≤ θ ≤ π/6.
SCHEMATIC:
ASSUMPTIONS: (1) Surface is diffuse emitter.
ANALYSIS: (a) Since the surface is diffuse, ε λ,θ is independent of direction; from Eq. 12.36, ε λ,θ =
ε λ. Using Eq. 12.38,
∞
ε ( T) =
ε ( λ ) Eλ ,b ( λ ,T ) dλ / Eb ( T)
0 λ
1.5
∞
E (T ) =
ε1 E λ ,b ( λ,2000 ) dλ / E b +
ε E
( λ ,2000 ) dλ / E b.
0
1.5 2 λ ,b
Written now in terms of F(0 → λ), with F(0 → 1.5) = 0.2732 at λT = 1.5 × 2000 = 3000 µm⋅K, (Table
12.1) find,
∫
∫
∫
ε ( 2000K ) = ε1 × F( 0→1.5) + ε 2 1 − F( 0→1.5 )  = 0.2 × 0.2732 + 0.8[1 − 0.2732 ] = 0.636.


<
(b) For the prescribed spectral and geometric limits, from Eq. 12.12,
2.5 2π π / 6
∆E =
ε λ ,θ Iλ ,b ( λ , T ) cosθ sin θ d θ d φ d λ
0.8 0
0
where Iλ,e (λ, θ, φ) = ε λ,θ Iλ,b (λ,T). Since the surface is diffuse, ε λ,θ = ε λ, and noting Iλ,b is
∫
∫
∫
independent of direction and equal to Eλ,b/π, we can write
2.5
 1.5

 2π π / 6
 E b ( T )  0.8 ε1 E λ ,b ( λ, T ) dλ 1.5 ε2 E λ ,b ( λ, T ) dλ 
cosθ sin θ d θ dφ 
∆E = 
+


0
Eb ( T)
Eb ( T )
 0
 π 

∫
∫
∫
∫


or in terms F(0 → λ) values,
 2π sin2 θ π / 6  σ T4
∆E = φ
×
{ε1[ F0→1.5 − F0→0.8] + ε 2 [ F0→2.5 −F 0→1.5]}.

0
0
π
2


From Table 12.1:

∆E = 2π ×

λT = 0.8 × 2000 = 1600 µm⋅K
F(0 → 0.8)= 0.0197
λT = 2.5 × 2000 = 5000 µm⋅K
F(0 → 2.5) = 0.6337
−8
4
sin 2 π / 6 
 5.67 × 10 × 2000 W
2


(
π
m2
)
⋅ {0.2 [0.2732 − 0.0197 ] + 0.8 [ 0.6337 − 0.2732]}
∆E = 0.25 × 5.67× 10−8 × 2000 4 W / m 2 × 0.339 = 76.89 k W / m2 .
<
PROBLEM 12.35
KNOWN: Directional emissivity, ε θ, of a selective surface.
FIND: Ratio of the normal emissivity, ε n, to the hemispherical emissivity, ε.
SCHEMATIC:
ASSUMPTIONS: Surface is isotropic in φ direction.
ANALYSIS: From Eq. 12.36 written on a total, rather than spectral, basis, the hemispherical
emissivity is
π /2
εθ (θ ) cos θ sin θ d θ.
0
ε = 2∫
Recognizing that the integral can be expressed in two parts, find
π /2
 π /4

ε = 2 ∫
ε (θ ) cos θ sin θ dθ + ∫
ε (θ ) cos θ sin θ dθ 
0
π
/
4


π /4
π /2

cosθ sin θ dθ + 0.3 ∫
cos θ sin θ dθ 
ε = 2  0.8 ∫
0
/
4
π


 sin 2 θ π / 4
sin 2 θ π / 2 

ε = 2  0.8
+0.3
2 0
2 π /4 


1
1
ε = 2  0.8 ( 0.50 − 0) + 0.3 × (1 − 0.50)  = 0.550.
2
 2

The ratio of the normal emissivity (ε n) to the hemispherical emissivity is
εn
0.8
=
= 1.45.
ε
0.550
COMMENTS: Note that Eq. 12.36 assumes the directional emissivity is independent of the φ
coordinate. If this is not the case, then Eq. 12.35 is appropriate.
<
PROBLEM 12.35
KNOWN: Directional emissivity, ε θ, of a selective surface.
FIND: Ratio of the normal emissivity, ε n, to the hemispherical emissivity, ε.
SCHEMATIC:
ASSUMPTIONS: Surface is isotropic in φ direction.
ANALYSIS: From Eq. 12.36 written on a total, rather than spectral, basis, the hemispherical
emissivity is
π /2
εθ (θ ) cos θ sin θ d θ.
0
ε = 2∫
Recognizing that the integral can be expressed in two parts, find
π /2
 π /4

ε = 2 ∫
ε (θ ) cos θ sin θ dθ + ∫
ε (θ ) cos θ sin θ dθ 
0
π
/
4


π /4
π /2

cosθ sin θ dθ + 0.3 ∫
cos θ sin θ dθ 
ε = 2  0.8 ∫
0
/
4
π


 sin 2 θ π / 4
sin 2 θ π / 2 

ε = 2  0.8
+0.3
2 0
2 π /4 


1
1
ε = 2  0.8 ( 0.50 − 0) + 0.3 × (1 − 0.50)  = 0.550.
2
 2

The ratio of the normal emissivity (ε n) to the hemispherical emissivity is
εn
0.8
=
= 1.45.
ε
0.550
COMMENTS: Note that Eq. 12.36 assumes the directional emissivity is independent of the φ
coordinate. If this is not the case, then Eq. 12.35 is appropriate.
<
PROBLEM 12.36
KNOWN: The total directional emissivity of non-metallic materials may be approximated as
εθ = εn cos θ where εn is the total normal emissivity.
FIND: Show that for such materials, the total hemispherical emissivity, ε, is 2/3 the total normal
emissivity.
SCHEMATIC:
ANALYSIS: From Eq. 12.36, written on a total rather than spectral basis, the hemispherical
emissivity ε can be determined from the directional emissivity εθ as
ε=2
1 0π / 2 εθ cos θ sin θ dθ
With ε θ = ε n cos θ , find
ε = 2 εn
1 0π /2 cos2θ sin θ dθ
ε = −2 ε n cos3 θ / 3 |
"
π /2
'0
= 2 / 3 εn
<
COMMENTS: (1) Refer to Fig. 12.17 illustrating on cartesian coordinates representative directional
distributions of the total, directional emissivity for nonmetallic and metallic materials. In the
schematic above, we’ve shown ε θ vs. θ on a polar plot for both types of materials, in comparison
with a diffuse surface.
(2) See Section 12.4 for discussion on other characteristics of emissivity for materials.
PROBLEM 12.37
KNOWN: Incandescent sphere suspended in air within a darkened room exhibiting these
characteristics:
initially:
brighter around the rim
after some time: brighter in the center
FIND: Plausible explanation for these observations.
ASSUMPTIONS: (1) The sphere is at a uniform surface temperature, Ts.
ANALYSIS: Recognize that in observing the
sphere by eye, emission from the central region
is in a nearly normal direction. Emission from
the rim region, however, has a large angle from
the normal to the surface.
Note now the directional behavior, ε θ, for conductors and non-conductors as represented in Fig. 12.17.
Assume that the sphere is fabricated from a metallic material. Then, the rim would appear brighter
than the central region. This follows since ε θ is higher at higher angles of emission.
If the metallic sphere oxidizes with time, then the ε θ characteristics change. Then ε θ at small angles of
θ become larger than at higher angles. This would cause the sphere to appear brighter at the center
portion of the sphere.
COMMENTS: Since the emissivity of non-conductors is generally larger than for metallic materials,
you would also expect the oxidized sphere to appear brighter for the same surface temperature.
PROBLEM 12.38
KNOWN: Detector surface area. Area and temperature of heated surface. Radiant power
measured by the detector for two orientations relative to the heated surface.
FIND: (a) Normal emissivity of heated surface, (b) Whether surface is a diffuse emitter.
SCHEMATIC:
ASSUMPTIONS: (1) Detector intercepts negligible radiation from surroundings, (2) A1 and A2 are
differential surfaces.
ANALYSIS: The radiant power leaving the heated surface and intercepting the detector is
q12 (θ ) = I1 (θ ) A1 cos θ ω2 −1
I1 (θ ) = ε1 (θ ) Ib,1 = ε1 (θ ) σ T14 / π
ω 2−1 =
A2 cos θ
( L/cos θ )
2
.
Hence,
q12 (θ ) = ε1 (θ )
σ T14
π
A1 cos θ
q (θ ) π
L2
ε1 (θ ) = 12
σ T14 A1A 2 ( cos θ ) 4
A 2 cos θ
( L/cos θ ) 2
(a) For the normal condition, θ = 0, cosθ = 1, and ε 1 (θ) ≡ ε 1,n is
ε1,n =
1.155 ×10−6 W π
( 0.5m ) 2
5.67 ×10−8 W / m 2 ⋅ K 4 (1000K ) 4 5 × 10−6 m2 × 4 × 10−6 m 2
<
ε1,n = 0.80.
(b) For the orientation with θ = 60° and cosθ = 0.5, so that ε 1 (θ = 60°) is
ε1 ( 60° ) =
5.415 × 10−8 W π
( 0.5m )2
5.67 ×10−8 W / m 2 ⋅ K 4 (1000K ) 5 ×10−6 m2 × 4 ×10−6 m 2 ( 0.5)
4
4
ε1 ( 60° ) = 0.60.
Since ε 1,n ≠ ε 1(60°), the surface is not a diffuse emitter.
COMMENTS: Even if ε 1 (60°) were equal to ε 1,n, it could not be concluded there was diffuse
emission until results were obtained for a wider range of θ.
<
PROBLEM 12.39
KNOWN: Radiation thermometer responding to radiant power within a prescribed spectral interval
and calibrated to indicate the temperature of a blackbody.
FIND: (a) Whether radiation thermometer will indicate temperature greater than, less than, or equal to
Ts when surface has ε < 1, (b) Expression for Ts in terms of spectral radiance temperature and
spectral emissivity, (c) Indicated temperature for prescribed conditions of Ts and ε λ.
SCHEMATIC:
ASSUMPTIONS: (1) Surface is a diffuse emitter, (2) Thermometer responds to radiant flux over
interval dλ about λ.
ANALYSIS: (a) The radiant power which reaches the radiation thermometer is
q λ = ε λI λ ,b ( λ ,Ts ) ⋅ A t ⋅ ωt
(1)
where At is the area of the surface viewed by the thermometer (referred to as the target) and ωt the
solid angle through which At is viewed. The thermometer responds as if it were viewing a blackbody
at Tλ, the spectral radiance temperature,
q λ = I λ ,b ( λ , Tλ ) ⋅ At ⋅ ω t .
(2)
By equating the two relations, Eqs. (1) and (2), find
I λ ,b ( λ ,Tλ ) = ε λ I λ ,b ( λ , Ts ) .
(3)
Since ε λ < 1, it follows that Iλ,b(λ, Tλ) < Iλ,b(λ, Ts) or that Tλ < Ts. That is, the thermometer will
always indicate a temperature lower than the true or actual temperature for a surface with ε < 1.
(b) Using Wien’s law in Eq. (3), find
1
I λ ( λ, T ) = C1λ −5 exp ( −C2 / λT )
π
1
1
C1λ −5 exp ( −C2 / λ Tλ ) = ε λ ⋅ C1λ −5 exp ( −C2 / λ Ts ) .
π
π
-5
Canceling terms (C1λ /π), taking natural logs of both sides of the equation and rearranging, the desired
expression is
1
1
λ
=
+
lnε λ .
Ts Tλ C2
(4)
<
(c) For Ts = 1000K and ε = 0.9, from Eq. (4), the indicated temperature is
1
1
λ
1
0.65 µ m
ln ε λ =
=
−
−
ln ( 0.9 )
Tλ Ts C 2
1000K 14,388 µ m ⋅ K
That is, the thermometer indicates 5K less than the true temperature.
Tλ = 995.3K.
<
PROBLEM 12.40
KNOWN: Spectral distribution of emission from a blackbody. Uncertainty in measurement of
intensity.
FIND: Corresponding uncertainities in using the intensity measurement to determine (a) the surface
temperature or (b) the emissivity.
ASSUMPTIONS: Diffuse surface behavior.
ANALYSIS: From Eq. 12.25, the spectral intensity associated with emission may be expressed as
ε λ C1 / π
I λ ,e = ε λ I λ ,b =
λ 5 exp ( C2 / λ T ) −1
(a) To determine the effect of temperature on intensity, we evaluate the derivative,
−C 2 / λ T 2 )
(
=−
2
∂T
{λ 5 exp ( C2 / λT ) −1}
2
∂ I λ ,e (C2 / λ T ) exp ( C2 /λ T)
Iλ ,e
=
∂ I λ ,e
( ε λ C1 / π ) λ 5 exp ( C2 /λ T )
exp ( C2 / λT ) − 1
∂T
Hence,
dT 1 − exp ( −C2 / λ T) d Iλ ,e
=
T
Iλ ,e
( C2 / λ T )
(
)
With d Iλ ,e / Iλ ,e = 0.1, C2 = 1.439 ×104 µ m ⋅ K and λ = 10 µ m,
dT  1 − exp ( −1439K/T) 
=
 × 0.1
T 
1439K/T

T = 500 K:
dT/T = 0.033 → 3.3% uncertainty
T = 1000 K:
dT/T = 0.053 → 5.5% uncertainty
<
<
(b) To determine the effect of the emissivity on intensity, we evaluate
∂ I λ ,e
I
= I λ ,b = λ ,e
ελ
∂ ελ
d ε λ d I λ ,e
=
= 0.10 → 10% uncertainty
Hence,
Iλ ,e
ελ
<
COMMENTS: The uncertainty in the temperature is less than that of the intensity, but increases with
increasing temperature (and wavelength). In the limit as C2/λT → 0, exp (- C2/λT) → 1 – C2/λT and
dT/T → d Iλ,e/Iλ,e. The uncertainty in temperature then corresponds to that of the intensity
measurement. The same is true for the uncertainty in the emissivity, irrespective of the value of T or
λ.
PROBLEM 12.41
KNOWN: Temperature, thickness and spectral emissivity of steel strip emerging from a hot roller.
Temperature dependence of total, hemispherical emissivity.
FIND: (a) Initial total, hemispherical emissivity, (b) Initial cooling rate, (c) Time to cool to prescribed
final temperature.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible conduction (in longitudinal direction), convection and radiation from
surroundings, (2) Negligible transverse temperature gradients.
3
PROPERTIES: Steel (given): ρ = 7900 kg/m , c = 640 J/kg⋅K, ε = 1200ε i/T (K).
ANALYSIS: (a) The initial total hemispherical emissivity is
∞
εi =
ε E (1200 ) / E b (1200 )  dλ
0 λ  λb
and integrating by parts using values from Table 12.1, find
∫
λ T = 1200 µm ⋅ K → F(0 −1µ m) = 0.002; λ T = 7200 µm ⋅ K → F(0 − 6 µ m ) = 0.819
εi = 0.6 × 0.002 + 0.4 ( 0.819 − 0.002 ) + 0.25 (1 −0.819 ) = 0.373.
<
(b) From an energy balance on a unit surface area of strip (top and bottom),
− E& out = dEst /dt
− 2εσ T4 = d( ρδ cT ) /dt
dT 
2εiσ Ti4
dt i
ρδ c
 =−
=
−2 ( 0.373 ) 5.67× 10 − 8 W / m 2 ⋅ K 4 (1200 K )4
7900 k g / m3 ( 0.003 m) ( 640 J / k g ⋅ K )
= −5.78 K / s.
<
(c) From the energy balance,

2ε (1200/T ) σ T 4 Tf dT
dT
2400ε iσ t
1 
ρδ c  1
=− i
,∫
=−
dt,
t
=
−
Ti T3
dt
4800ε iσ  T 2 T2 
ρδ c
ρδ c ∫0
i 
 f
t=
3
( 0.003m ) 640 J / k g ⋅ K
 1
1  −2
−

 K = 311s.
4800 K × 0.373 × 5.67 ×10−8 W / m 2 ⋅ K4  6002 12002 
7900 k g / m
<
2
COMMENTS: Initially, from Eq. 1.9, hr = εiσ Ti3 = 36.6 W/m ⋅K. Assuming a plate width of W =
3
1m, the Rayleigh number may be evaluated from RaL = gβ(Ti - T∞) (W/2) /να. Assuming T∞ = 300
8
K and evaluating properties at Tf = 750 K, RaL = 2.4 × 10 . From Eq. 9.31, NuL = 93, giving h = 10
2
W/m ⋅K. Hence heat loss by radiation exceeds that associated with free convection. To check the
2
validity of neglecting transverse temperature gradients, compute Bi = h(δ/2)/k. With h = 36.6 W/m ⋅K
and k = 28 W/m⋅K, Bi = 0.002 << 1. Hence the assumption is valid.
PROBLEM 12.42
KNOWN: Large body of nonluminous gas at 1200 K has emission bands between 2.5 – 3.5 µm and
between 5 – 8 µm with effective emissivities of 0.8 and 0.6, respectively.
FIND: Emissive power of the gas.
SCHEMATIC:
ASSUMPTIONS: (1) Gas radiates only in specified bands, (2) Emitted radiation is diffuse.
ANALYSIS: The emissive power of the gas is
( )
∞
ε E
T dλ
0 λ λ ,b g
( )
E g = ε E b Tg = ∫
( )
( )
3.5
8
ε λ ,1E λ ,b Tg dλ + ε λ ,2Eλ ,b Tg dλ
2.5
5
Eg = ∫
∫
E g = ε1F( 2.5 −3.5 µm ) + ε 2 F(5− 8 µm)  σ Tg4.


Using the blackbody function F(0-λT) from Table 12.1 with Tg = 1200 K,
λT(µm⋅K)
F(0-λT)
2.5 × 1200
3000
3.5 × 1200
4200
5 × 1200
6000
8 × 1200
9600
0.273
0.516
0.738
0.905
so that
F( 2.5 −3.5 µm ) = F( 0− 3.5µ m ) − F( 0− 2.5 µ m ) = 0.516 − 0.273 = 0.243
F( 5− 8µ m) = F( 0− 8µ m) − F( 0− 5µ m) = 0.905 − 0.738 = 0.167.
Hence the emissive power is
E g = [0.8 ×0.243 + 0.6 × 0.167 ] 5.67 ×10 −8 W / m 2 ⋅ K 4 (1200 K )
E g = 0.295 ×117,573 W / m 2 = 34,684 W / m 2 .
4
<
COMMENTS: Note that the effective emissivity for the gas is 0.295. This seems surprising since
emission occurs only at the discrete bands. Since λmax = 2.4 µm, all of the spectral emissive power is
at wavelengths beyond the peak of blackbody radiation at 1200 K.
PROBLEM 12.43
KNOWN: An opaque surface with prescribed spectral, hemispherical reflectivity distribution is
subjected to a prescribed spectral irradiation.
FIND: (a) The spectral, hemispherical absorptivity, (b) Total irradiation, (c) The absorbed radiant flux,
and (d) Total, hemispherical absorptivity.
SCHEMATIC:
ASSUMPTIONS: (1) Surface is opaque.
ANALYSIS: (a) The spectral, hemispherical absorptivity, α λ, for an opaque surface is given by Eq.
12.58,
<
αλ = 1− ρλ
which is shown as a dashed line on the ρ λ distribution axes.
(b) The total irradiation, G, follows from Eq. 12.16 which can be integrated by parts,
∞
5 µm
10 µm
20 µ m
G=
G λ dλ =
G λd λ+
G λd λ+
G λd λ
0
0
5µm
10 µm
∫
G=
∫
∫
∫
1
W
W
1
W
× 600
5 − 0) µm + 600
10 − 5 ) µ m + × 600
× ( 20 − 10 ) µ m
(
(
2
2
m2 ⋅ µ m
m2 ⋅ µ m
m2 ⋅ µ m
G = 7500 W / m 2 .
<
(c) The absorbed irradiation follows from Eqs. 12.45 and 12.46 with the form
∞
5 µm
10 µ m
20 µm
Gabs =
α λ G λ dλ = α1
G λ dλ + G λ,2
α λ dλ + α 3
G d λ.
0
0
5µ m
10 µ m λ
∫
∫
∫
∫
2
Noting that α 1 = 1.0 for λ = 0 → 5 µm, Gλ,2 = 600 W/m ⋅µm for λ = 5 → 10 µm and α 3 = 0 for λ > 10
µm, find that
(
)
Gabs = 1.0 0.5 × 600 W / m 2 ⋅ µ m (5 − 0 ) µ m + 600 W / m 2 ⋅ µ m ( 0.5 × 0.5 )(10 − 5) µm + 0
Gabs = 2250 W / m 2 .
(d) The total, hemispherical absorptivity is defined as the fraction of the total irradiation that is
absorbed. From Eq. 12.45,
G
2250 W / m 2
= 0.30.
α = abs =
G
7500 W / m 2
<
<
COMMENTS: Recognize that the total, hemispherical absorptivity, α = 0.3, is for the given spectral
irradiation. For a different Gλ, one would then expect a different value for α.
PROBLEM 12.44
KNOWN: Temperature and spectral emissivity of small object suspended in large furnace of prescribed
temperature and total emissivity.
FIND: (a) Total surface emissivity and absorptivity, (b) Reflected radiative flux and net radiative flux to
surface, (c) Spectral emissive power at λ = 2 µm, (d) Wavelength λ1/2 for which one-half of total
emissive power is in spectral region λ ≥ λ1/2.
SCHEMATIC:
ASSUMPTIONS: (1) Surface is opaque and diffuse, (2) Walls of furnace are much larger than object.
ANALYSIS: (a) The emissivity of the object may be obtained from Eq. 12.38, which is expressed as
∞
ε ( Ts )
∫
= o
ε λ ( λ ) E λ ,b ( λ , Ts ) dλ
= ε1  F( 0→3µ m ) − F(0→1µ m )  + ε 2 1 − F( 0→3µ m ) 

Eb (T )



where, with λ1Ts = 400 µm⋅K and λ2Ts = 1200 µm⋅K, F(0→1µm) = 0 and F( 0→3µ m ) = 0.002. Hence,
<
ε ( Ts ) = 0.7 ( 0.002 ) + 0.5 (0.998 ) = 0.500
The absorptivity of the surface is determined by Eq. 12.46,
∞
∞
∫ αλ (λ ) G λ (λ ) dλ = ∫o αλ (λ ) E λ ,b (λ , Tf ) dλ
α= o
∞
∫o
E b ( Tf )
G λ ( λ ) dλ
Hence, with λ1Tf = 2000 µm⋅K and λ2Tf = 6000 µm⋅K, F(0→1µm) = 0.067 and F( 0→3µ m ) = 0.738. It
follows that
α = α1  F(0 →3µ m ) − F( 0 →1µ m )  + α 2 1 − F(0 →3µ m )  = 0.7 × 0.671 + 0.5 × 0.262 = 0.601




<
(b) The reflected radiative flux is
G ref = ρ G = (1 − α ) E b ( Tf ) = 0.399 × 5.67 × 10−8 W m 2⋅ K 4 ( 2000 K ) = 3.620 × 105 W m 2
4
<
The net radiative flux to the surface is
q′′rad = G − ρ G − ε E b ( Ts ) = α E b ( Tf ) − ε E b ( Ts )
4
4
q′′rad = 5.67 × 10−8 W m 2⋅ K 4  0.601( 2000 K ) − 0.500 ( 400 K )  = 5.438 × 105 W m 2


<
(c) At λ = 2 µm, λTs = 800 K and, from Table 12.1, Iλ,b(λ,T)/σT5 = 0.991 × 10-7 (µm⋅K⋅sr)-1. Hence,
Continued...
PROBLEM 12.44 (Cont.)
Iλ ,b = 0.991 × 10−7 × 5.67 × 10−8
W m 2⋅ K 4
µ m ⋅ K ⋅ sr
× ( 400 K ) = 0.0575
5
W
m 2 ⋅ µ m ⋅ sr
Hence, with Eλ = ελEλ,b = ελπIλ,b,
E λ = 0.7 (π sr ) 0.0575 W m 2 ⋅ µ m ⋅ sr = 0.126 W m 2 ⋅ µ m
<
(d) From Table 12.1, F(0→λ) = 0.5 corresponds to λTs ≈ 4100 µm⋅K, in which case,
λ1/ 2 ≈ 4100 µ m ⋅ K 400 K ≈ 10.3 µ m
<
COMMENTS: Because of the significant difference between Tf and Ts, α ≠ ε. With increasing Ts →
Tf, ε would increase and approach a value of 0.601.
PROBLEM 12.45
KNOWN: Small flat plate maintained at 400 K coated with white paint having spectral absorptivity
distribution (Figure 12.23) approximated as a stairstep function. Enclosure surface maintained at 3000 K
with prescribed spectral emissivity distribution.
FIND: (a) Total emissivity of the enclosure surface, εes, and (b) Total emissivity, ε, and absorptivity, α,
of the surface.
SCHEMATIC:
ASSUMPTIONS: (1) Coated plate with white paint is diffuse and opaque, so that αλ = ελ, (2) Plate is
small compared to the enclosure surface, and (3) Enclosure surface is isothermal, diffuse and opaque.
ANALYSIS: (a) The total emissivity of the enclosure surface at Tes = 3000 K follows from Eq. 12.38
which can be expressed in terms of the bond emission factor, F(0-λT), Eq. 12.30,
<
ε e,s = ε1F(0 − λ T ) + ε 2 1 − F(0 − λ T )  = 0.2 × 0.738 + 0.9 [1 − 0.738] = 0.383
1 es
1 es 

where, from Table 12.1, with λ1Tes = 2 µm × 3000 K = 6000 µm⋅K, F(0-λT) = 0.738.
(b) The total emissivity of the coated plate at T = 400 K can be expressed as
ε = α1F(0 − λ T ) + α 2  F(0 − λ T ) − F(0 − λ T )  + α3 1 − F(0 − λ T ) 
1 s
2 s
1 s 
2 s 


<
ε = 0.75 × 0 + 0.15 [0.002134 − 0.000 ] + 0.96 [1 − 0.002134] = 0.958
where, from Table 12.1, the band emission factors are: for λ1Ts = 0.4 × 400 = 160 µm⋅K, find
F( 0 − λ T ) = 0.000; for λ2Tes = 3.0 × 400 = 1200 µm⋅K, find F( 0 − λ T ) = 0.002134. The total
1 s
2 s
absorptivity for the irradiation due to the enclosure surface at Tes = 3000 K is
α = α1F( 0 − λ T ) + α 2  F( 0 − λ T ) − F( 0 − λ T )  + α 3 1 − F(0 − λ T ) 
1 es
2 es
2 es 
2 es 


<
α = 0.75 × 0.002134 + 0.15 [0.8900 − 0.002134] + 0.96 [1 − 0.8900] = 0.240
where, from Table 12.1, the band emission factors are: for λ1Tes = 0.4 × 3000 = 1200 µm⋅K, find
F( 0 − λ T ) = 0.002134; for λ2Tes = 3.0 × 3000 = 9000 µm⋅K, find F( 0 − λ T ) = 0.8900.
1 es
2 es
COMMENTS: (1) In evaluating the total emissivity and absorptivity, remember that ε = ε ε λ ,Ts and α
$
= α(αλ, Gλ) where Ts is the temperature of the surface and Gλ is the spectral irradiation, which if the
surroundings are large and isothermal, Gλ = Eb,λ(Tsur). Hence, α = α(αλ ,Tsur ). For the opaque, diffuse
surface, αλ = ε λ.
(2) Note that the coated plate (white paint) has an absorptivity for the 3000 K-enclosure surface
irradiation of α = 0.240. You would expect it to be a low value since the coating appears visually
“white”.
(3) The emissivity of the coated plate is quite high, ε = 0.958. Would you have expected this of a
“white paint”? Most paints are oxide systems (high absorptivity at long wavelengths) with pigmentation
(controls the “color” and hence absorptivity in the visible and near infrared regions).
PROBLEM 12.46
KNOWN: Area, temperature, irradiation and spectral absorptivity of a surface.
FIND: Absorbed irradiation, emissive power, radiosity and net radiation transfer from the surface.
SCHEMATIC:
ASSUMPTIONS: (1) Opaque, diffuse surface behavior, (2) Spectral distribution of solar radiation
corresponds to emission from a blackbody at 5800 K.
ANALYSIS: The absorptivity to solar irradiation is
∞
∞
αλ Gλ d λ
αλ Eλ b ( 5800 K) dλ
αs = 0
= 0
= α1F(0.5 →1 µm ) + α 2 F(2→∞ ).
G
Eb
∫
∫
From Table 12.1,
λT = 2900 µm⋅K:
F(0 → 0.5 µm) = 0.250
λT = 5800 µm⋅K:
F(0 → 1 µm) = 0.720
λT = 11,600 µm⋅K:
F(0 → 2 µm) = 0.941
αs = 0.8 ( 0.720 − 0.250 ) + 0.9 ( 1− 0.941) = 0.429.
(
)
Gabs = αSGS = 0.429 1200 W / m 2 = 515 W / m 2.
Hence,
<
The emissivity is
∞
ε=
ε E ( 400 K ) dλ / E b = ε1F( 0.5→1 µm ) + ε 2F( 2→∞ ).
0 λ λb
∫
From Table 12.1,
λT = 200 µm⋅K:
F(0 → 0.5 µm) = 0
λT = 400 µm⋅K:
F(0 → 1 µm) = 0
λT = 800 µm⋅K
F(0 → 2 µm) = 0.
Hence, ε = ε 2 = 0.9,
4
E = εσ Ts4 = 0.9 × 5.67 ×10 −8 W / m 2 ⋅ K 4 ( 400 K ) = 1306 W / m 2 .
<
The radiosity is
J = E + ρSGS = E + (1 −α s ) G S = [1306 + 0.571 ×1200 ] W / m 2 = 1991W/m 2 .
<
The net radiation transfer from the surface is
q net = ( E − αSGS ) As = (1306 − 515 ) W / m 2 × 4 m 2 = 3164 W.
COMMENTS: Unless 3164 W are supplied to the surface by other means (for example, by
convection), the surface temperature will decrease with time.
<
PROBLEM 12.47
KNOWN: Temperature and spectral emissivity of a receiving surface. Direction and spectral
distribution of incident flux. Distance and aperture of surface radiation detector.
FIND: Radiant power received by the detector.
SCHEMATIC:
2
ASSUMPTIONS: (1) Target surface is diffuse, (2) Ad/L << 1.
ANALYSIS: The radiant power received by the detector depends on emission and reflection from the
target.
q d = I e+ rAs cosθ d − s∆ ωd −s
qd =
εσ Ts4 + ρG
A
As d
π
L2
∞
∫
ε= 0
ε λ Eλ b ( 700 K) dλ
Eb ( 700 K )
From Table 12.1,
= ε1F( 3→10 µ m) + ε 2 F(10 →∞ ) .
λT = 2100 µm⋅K:
F(0 → 3 µm) = 0.0838
λT = 7000 µm⋅K:
F(0 → 10 µm) = 0.8081.
The emissivity can be expected as
ε = 0.5 ( 0.8081 − 0.0838 ) + 0.9 (1 − 0.8081) = 0.535.
Also,
∞
∫
ρ= 0
ρ λ G λ dλ
∞
(1 − ε λ ) q λ′′ d λ
∫
= 0
= 1× F
G
q′′
ρ = 1× 0.4 + 0.5 × 0.6 = 0.70.
( 0→3 µm ) + 0.5 × F( 3→ 6µm)
Hence, with G = q ′′ cosθ i = 866 W / m2 ,
0.535 × 5.67 × 10−8 W / m 2 ⋅ K 4 ( 700 K ) + 0.7 × 866 W / m 2 −4 2 10−5 m 2
10 m
π
(1m )2
4
qd =
q d = 2.51×10 −6 W.
<
COMMENTS: A total radiation detector cannot discriminate between emitted and reflected radiation
from a surface.
PROBLEM 12.48
KNOWN: Small disk positioned at center of an isothermal, hemispherical enclosure with a small
aperture.
FIND: Radiant power [µW] leaving the aperture.
SCHEMATIC:
ASSUMPTIONS: (1) Disk is diffuse-gray, (2) Enclosure is isothermal and has area much larger than
disk, (3) Aperture area is very small compared to enclosure area, (4) Areas of disk and aperture are
small compared to radius squared of the enclosure.
ANALYSIS: The radiant power leaving the aperture is due to radiation leaving the disk and to
irradiation on the aperture from the enclosure. That is,
q ap = q1→2 + G 2 ⋅ A 2.
(1)
The radiation leaving the disk can be written in terms of the radiosity of the disk. For the diffuse disk,
q1→2 =
1
J1 ⋅ A1 cosθ1 ⋅ ω2−1
π
(2)
and with ε = α for the gray behavior, the radiosity is
J1 = ε1 E b ( T1 ) + ρ G1 = ε1 σ T14 + (1 − ε1 ) σ T34
(3)
where the irradiation G1 is the emissive power of the black enclosure, Eb (T3); G1 = G2 = Eb (T3).
The solid angle ω2 – 1 follows from Eq. 12.2,
ω 2 −1 = A2 / R2 .
(4)
Combining Eqs. (2), (3) and (4) into Eq. (1) with G 2 = σ T34 , the radiant power is
q ap =
1 
A
σ  ε1 T14 + (1 − ε1 ) T34  A1 cos θ1 ⋅ 2 + A 2σ T34

π 
R2
q ap =
π
1
W 
5.67 ×10 −8
0.7 ( 900K ) 4 + (1 − 0.7 )( 300K ) 4  ( 0.005m ) 2 cos45 °×
 4
π
m 2 ⋅ K 4 
π / 4 ( 0.002m )
( 0.100m ) 2
2
+
π
( 0.002m) 2 5.67 × 10−8 W / m2 ⋅ K 4 ( 300K) 4
4
q ap = ( 36.2 + 0.19 +1443 ) µ W = 1479 µ W.
<
COMMENTS: Note the relative magnitudes of the three radiation components. Also, recognize that
the emissivity of the enclosure ε 3 doesn’t enter into the analysis. Why?
PROBLEM 12.49
KNOWN: Spectral, hemispherical absorptivity of an opaque surface.
FIND: (a) Solar absorptivity, (b) Total, hemispherical emissivity for Ts = 340K.
SCHEMATIC:
ASSUMPTIONS: (1) Surface is opaque, (2) ε λ = α λ, (3) Solar spectrum has Gλ = Gλ,S proportional
to Eλ,b (λ, 5800K).
ANALYSIS: (a) The solar absorptivity follows from Eq. 12.47.
∞
∞
αS =
α λ ( λ ) E λ,b ( λ,5800K ) dλ /
E
( λ ,5800K ) dλ.
0
0 λ ,b
∫
∫
The integral can be written in three parts using F(0 → λ) terms.
αS = α1 F(0 →0.3 ) + α2 F(0 →1.5 ) − F(0 →0.3 )  + α3 1 − F(0 →1.5)  .




From Table 12.1,
λT = 0.3 × 5800 = 1740 µm⋅K
F(0 → 0.3 µm) = 0.0335
λT = 1.5 × 5800 = 8700 µm⋅K
F(0 → 1.5 µm) = 0.8805.
Hence,
αS = 0 × 0.0355 + 0.9 [ 0.8805 − 0.0335 ] + 0.11
[ − 0.8805] = 0.774.
<
(b) The total, hemispherical emissivity for the surface at 340K will be
∞
ε ( λ ) Eλ ,b ( λ ,340K ) dλ / Eb ( 340K ) .
0 λ
ε=∫
If ε λ = α λ, then using the α λ distribution above, the integral can be written in terms of F(0 → λ) values.
It is readily recognized that since
F( 0→1.5 µm,340K ) ≈ 0.000
at
λ T = 1.5 × 340 = 510 µ m ⋅ K
there is negligible spectral emissive power below 1.5 µm. It follows then that
ε = ε λ = α λ = 0.1
<
COMMENTS: The assumption ε λ = α λ can be satisfied if this surface were irradiated diffusely or if
the surface itself were diffuse. Note that for this surface under the specified conditions of solar
irradiation and surface temperature α S ≠ ε. Such a surface is referred to as a spectrally selective
surface.
PROBLEM 12.50
KNOWN: Spectral distribution of the absorptivity and irradiation of a surface at 1000 K.
FIND: (a) Total, hemispherical absorptivity, (b) Total, hemispherical emissivity, (c) Net radiant flux to
the surface.
SCHEMATIC:
ASSUMPTIONS: (1) α λ = ε λ.
ANALYSIS: (a) From Eq. 12.46,
∞
2µm
4 µm
6 µm
∫ αλ Gλ dλ = ∫0 αλ Gλ dλ + ∫2 αλ Gλ d λ + ∫ 4 αλ Gλ dλ
α= 0
∞
2µm
4µm
6 µm
∫0 Gλ dλ
∫0 Gλ dλ + ∫2 Gλ dλ + ∫ 4 Gλ dλ
α=
0 × 1 / 2 ( 2 − 0) 5000 + 0.6 ( 4 − 2) 5000 + 0.6 × 1 / 2 ( 6 − 4 ) 5000
1 / 2 ( 2 − 0 ) 5000 + ( 4 − 2 )( 5000 ) + 1 / 2 ( 6 − 4 ) 5000
α=
9000
= 0.45.
20,000
<
(b) From Eq. 12.38,
∞
2µm
∞
ε λ Eλ ,b dλ 0
Eλ ,b dλ 0.6
E
dλ
2 λ ,b
ε= 0
= 0
+
Eb
Eb
Eb
∫
∫
∫
ε = 0.6F( 2 µ m→∞) = 0.6 1 − F( 0→2 µm )  .


From Table 12.1, with λT = 2 µm × 1000K = 2000 µm⋅K, find F(0 → 2 µm) = 0.0667. Hence,
ε = 0.6 [1 − 0.0667 ] = 0.56.
<
(c) The net radiant heat flux to the surface is
q′′rad,net = α G − E = α G − ε σ T 4
(
)
q′′rad,net = 0.45 20,000W/m 2 − 0.56 × 5.67 ×10 −8 W / m 2 ⋅ K 4 × (1000K )
q′′rad,net = ( 9000 − 31,751) W / m 2 = −22,751W/m 2 .
4
<
PROBLEM 12.51
KNOWN: Spectral distribution of surface absorptivity and irradiation. Surface temperature.
FIND: (a) Total absorptivity, (b) Emissive power, (c) Nature of surface temperature change.
SCHEMATIC:
ASSUMPTIONS: (1) Opaque, diffuse surface behavior, (2) Convection effects are negligible.
ANALYSIS: (a) From Eqs. 12.45 and 12.46, the absorptivity is defined as
∞
∞
α ≡ G abs / G = ∫ α λ G λ d λ / ∫ G λ d λ.
0
0
The absorbed irradiation is,
)
(
G abs = 0.4 5000W/m 2 ⋅ µm ×2 µm / 2 + 0.8 × 5000W/m2 ⋅ µm ( 5 − 2 ) µm + 0 = 14,000 W / m 2 .
The irradiation is,
(
)
G = 2 µ m × 5000 W / m 2 ⋅ µ m / 2 + (10 − 2 ) µ m × 5000 W / m 2 ⋅ µm = 45,000 W / m 2 .
α = 14,000 W / m 2 /45,000 W / m 2 = 0.311.
Hence,
<
(b) From Eq. 12.38, the emissivity is
∞
2
5
ε λ Eλ ,b dλ / Eb = 0.4
Eλ ,b dλ / Eb + 0.8 Eλ ,b dλ / Eb
0
0
2
ε=∫
From Table 12.1,
∫
λT =2 µm × 1250K = 2500K,
λT = 5 µm × 1250K = 6250K,
∫
F(0 – 2) = 0.162
F(0 – 5) = 0.757.
Hence, ε = 0.4 × 0.162 + 0.8 ( 0.757 − 0.162 ) = 0.54.
E = ε E b = ε σ T 4 = 0.54 × 5.67 ×10 −8 W / m 2 ⋅ K 4 (1250K ) = 74,751W/m 2.
4
<
(c) From an energy balance on the surface, the net heat flux to the surface is
q′′net = α G − E = (14,000 − 74,751) W / m 2 = −60,751W/m 2 .
Hence the temperature of the surface is decreasing.
COMMENTS: Note that α ≠ ε. Hence the surface is not gray for the prescribed conditions.
<
PROBLEM 12.52
KNOWN: Power dissipation temperature and distribution of spectral emissivity for a tungsten
filament. Distribution of spectral absorptivity for glass bulb. Temperature of ambient air and
surroundings. Bulb diameter.
FIND: Bulb temperature.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, (2) Uniform glass temperature, Ts, and uniform irradiation of
inner surface, (3) Surface of glass is diffuse, (4) Negligible absorption of radiation by filament due to
emission from inner surface of bulb, (5) Net radiation transfer from outer surface of bulb is due to
exchange with large surroundings, (6) Bulb temperature is sufficiently low to provide negligible
emission at λ < 2µm, (7) Ambient air is quiescent.
-5
2
-5
2
PROPERTIES: Table A-4, air (assume Tf = 323 K): ν = 1.82 × 10 m /s, α = 2.59 × 10 m /s,k =
-1
0.028 W/m⋅K, β = 0.0031 K , Pr = 0.704.
ANALYSIS: From an energy balance on the glass bulb,
)
(
4 + h T −T
q′′rad,i = q′′rad,o + q′′conv = ε bσ Ts4 − Tsur
( s ∞)
(1)
where ε b = ε λ > 2 µ m = α λ > 2 µ m = 1 and h is obtained from Eq. (9.35)
Nu D = 2 +
4
0.589 Ra1/
D
1 + ( 0.469 / Pr )9 /16 


4/9
=
hD
k
(2)
with Ra D = gβ ( Ts − T∞ ) D3 / να . Radiation absorption at the inner surface of the bulb may be
expressed as
(
q′′rad,i = α G = α Pelec / π D2
)
(3)
where, from Eq. (12.46),
0.4
α = α1 ∫
0
2.0
∞
(G λ / G ) dλ + α 2 ∫0.4 (G λ / G ) dλ + α3 ∫2 (G λ / G ) dλ
Continued …..
PROBLEM 12.52 (Cont.)
The irradiation is due to emission from the filament, in which case (Gλ/G) ~ (Eλ/E)f = (εf,λEλ,b/εfEb).
Hence,
α = (α1 / ε f
0.4
) ∫0
(
)
ε f , λ E λ ,b / E b dλ + (α 2 / ε f
∞
) ∫0.4 ε f , λ ( E λ ,b / E b ) dλ + (α 3 / ε f ) ∫2
2.0
(
)
ε f , λ E λ ,b / E b dλ (4)
where, from the spectral distribution of Problem 12.25, εf,λ ≡ ε1 = 0.45 for λ < 2µm and εf,λ ≡ ε2 =
0.10 for λ > 2µm. From Eq. (12.38)
(
∞
ε f = ∫ ε f ,λ ( E λ ,b / E b ) dλ = ε1 F(0→ 2µ m ) + ε 2 1 − F(0→ 2µ m )
0
)
With λ Tf = 2 µ m × 3000 K = 6000 µ m ⋅ K, F( 0→ 2 µ m ) = 0.738 from Table 12.1. Hence,
ε f = 0.45 × 0.738 + 0.1(1 − 0.738 ) = 0.358
Equation (4) may now be expressed as
(
)
(
α = (α1 / ε f ) ε1 F( 0 → 0.4 µ m ) + (α 2 / ε f ) ε1 F( 0 → 2 µ m ) − F( 0 → 0.4 µ m ) + (α 3 / ε f ) ε 2 1 − F( 0 → 2 µ m )
)
where, with λT = 0.4µm × 3000 K = 1200 µm⋅K, F(0→0.4µm) = 0.0021. Hence,
α = (1 / 0.358 ) 0.45 × 0.0021 + ( 0.1 / 0.358 ) 0.45 × ( 0.738 − 0.0021) + (1 / 0.358 ) 0.1 (1 − 0.738 ) = 0.168
Substituting Eqs. (2) and (3) into Eq. (1), as well as values of εb = 1 and α = 0.168, an iterative
solution yields
Ts = 348.1 K
<
COMMENTS: For the prescribed conditions, q′′rad,i = 713 W / m 2 , q′′rad,o = 385.5 W / m 2 and
q′′conv = 327.5 W / m 2 .
PROBLEM 12.53
KNOWN: Spectral emissivity of an opaque, diffuse surface.
FIND: (a) Total, hemispherical emissivity of the surface when maintained at 1000 K, (b) Total,
hemispherical absorptivity when irradiated by large surroundings of emissivity 0.8 and temperature 1500
K, (c) Radiosity when maintained at 1000 K and irradiated as prescribed in part (b), (d) Net radiation
flux into surface for conditions of part (c), and (e) Compute and plot each of the parameters of parts (a)(c) as a function of the surface temperature Ts for the range 750 < Ts ≤ 2000 K.
SCHEMATIC:
ASSUMPTIONS: (1) Surface is opaque, diffuse, and (2) Surroundings are large compared to the
surface.
ANALYSIS: (a) When the surface is maintained at 1000 K, the total, hemispherical emissivity is
evaluated from Eq. 12.38 written as
λ1
∞
E λ ,b (T) dλ E b (T) + ε λ ,2
E (T) dλ E b (T)
0
λ1 λ ,b
∞
0
ε = ∫ ε λ E λ ,b (T) dλ E b (T) = ε λ ,1 ∫
∫
ε = ε λ ,1F(0 − λ T) + ε λ ,2 (1 − F(0 − λ T) )
1
1
where for λT = 6µm × 1000 K = 6000µm⋅K, from Table 12.1, find F0 − λ T = 0.738 . Hence,
<
ε = 0.8 × 0.738 + 0.3(1 − 0.738) = 0.669.
(b) When the surface is irradiated by large surroundings at Tsur = 1500 K, G = Eb(Tsur).
From Eq. 12.46,
∞
α = ∫ α λ G λ dλ
0
∞
∫0
∞
G λ dλ = ∫ ε λ E λ ,b (Tsur ) dλ E b (Tsur )
0
α = ε λ ,1F(0 − λ T ) + ε λ ,2 (1 − F(0 − λ T ) )
1 sur
1 sur
where for λ1Tsur = 6 µm × 1500 K = 9000 µm⋅K, from Table 12.1, find F(0 − λ T) = 0.890 . Hence,
<
α = 0.8 × 0.890 + 0.3 (1 − 0.890) = 0.745.
Note that α λ = ε λ for all conditions and the emissivity of the surroundings is irrelevant.
(c) The radiosity for the surface maintained at 1000 K and irradiated as in part (b) is
J = εEb (T) + ρG = εEb (T) + (1 − α)Eb (Tsur)
J = 0.669 × 5.67 × 10-8 W/m2 ⋅K4 (1000 K)4 + (1 − 0.745) 5.67 × 10-8 W/m2 ⋅K4 (1500 K)4
J = (37,932 + 73,196) W/m2 = 111,128 W/m2
<
Continued...
PROBLEM 12.53 (Cont.)
4
(d) The net radiation flux into the surface with G = σTsur
is
q″rad,in = αG − εE b (T) = G − J
q″rad,in = 5.67 × 10-8 W/m2 ⋅K (1500 K)4 − 111,128 W/m2
<
q″rad,in = 175,915 W/m2.
(e) The foregoing equations were entered into the IHT workspace along with the IHT Radiaton Tool,
Band Emission Factor, to evaluate F( 0−λT ) values and the respective parameters for parts (a)-(d) were
computed and are plotted below.
1
eps or alpha
0.9
0.8
0.7
0.6
0.5
500
1000
1500
2000
Surface temperature, Ts (K)
Emissivity, eps
Absorptivity, alpha; Tsur = 1500K
Note that the absorptivity, α = α (α λ , Tsur ) , remains constant as Ts changes since it is a function of
α λ (or ε λ ) and Tsur only. The emissivity ε = ε (ε λ , Ts ) is a function of Ts and increases as Ts
increases. Could you have surmised as much by looking at the spectral emissivity distribution? At what
condition is ε = α?
J or q''radin (W/m^2)
1E6
500000
0
-5E5
500
1000
1500
2000
Surface temperature, Ts (K)
Radiosity, J (W/m^2)
Net radiation flux in, q''radin (W/m^2)
The radiosity, J1 increases with increasing Ts since Eb(T) increases markedly with temperature; the
reflected irradiation, (1 - α)Eb(Tsur) decreases only slightly as Ts increases compared to Eb(T). Since G is
independent of Ts, it follows that the variation of q′′rad,in will be due to the radiosity change; note the
sign difference.
COMMENTS: We didn’t use the emissivity of the surroundings (ε = 0.8) to determine the irradiation
onto the surface. Why?
PROBLEM 12.54
KNOWN: Furnace wall temperature and aperture diameter. Distance of detector from aperture and
orientation of detector relative to aperture.
FIND: (a) Rate at which radiation from the furnace is intercepted by the detector, (b) Effect of
aperture window of prescribed spectral transmissivity on the radiation interception rate.
SCHEMATIC:
ASSUMPTIONS: (1) Radiation emerging from aperture has characteristics of emission from a
blackbody, (2) Cover material is diffuse, (3) Aperture and detector surface may be approximated as
infinitesimally small.
ANALYSIS: (a) From Eq. 12.5, the heat rate leaving the furnace aperture and intercepted by the
detector is
q = IeAa cos θ1 ωs −a .
From Eqs. 12.14 and 12.28
E (T ) σ Tf4 5.67 ×10 −8 (1500 ) 4
Ie = b f =
=
= 9.14 ×10 4 W / m 2 ⋅sr.
π
π
π
From Eq. 12.2,
ωs −a =
An
r
2
=
As ⋅ cos θ 2
r
2
=
10−5 m2 × cos45 °
2
(1m )
= 0.707 × 10−5 sr.
Hence
q = 9.14 ×104 W / m 2 ⋅ sr π ( 0.02m ) 2 / 4  cos30° × 0.707 ×10 −5 sr = 1.76 ×10 −4 W.


<
(b) With the window, the heat rate is
q = τ ( I e Aa cos θ1 ωs −a )
where τ is the transmissivity of the window to radiation emitted by the furnace wall. From Eq. 12.55,
∞
∞
τ λ Gλ dλ
τ λ Eλ ,b ( Tf ) dλ
2
0
0
τ=
=
= 0.8
E
/ E dλ = 0.8F( 0→ 2 µm) .
∞
∞
0 λ ,b b
G λ dλ
E λ ,b dλ
0
0
∫
∫
∫
∫
∫ (
)
With λT = 2 µm × 1500K = 3000 µm⋅K, Table 12.1 gives F(0 → 2 µm) = 0.273. Hence, with τ = 0.273 ×
0.8 = 0.218, find
q = 0.218 ×1.76 × 10−4 W = 0.384 × 10−4 W.
<
PROBLEM 12.55
KNOWN: Thermocouple is irradiated by a blackbody furnace at 1500 K with 25 mm2 aperture. Optical
fiber of prescribed spectral transmissivity in sight path.
FIND: (a) Distance L from the furnace detector should be positioned such that its irradiation is G = 50
W/m2 and, (b) Compute and plot irradiation, G, vs separation distance L for the range 100 ≤ L ≤ 400 mm
for blackbody furnace temperatures of Tf = 1000, 1500 and 2000 K.
SCHEMATIC:
ASSUMPTIONS: (1) Furnace aperture emits diffusely, (2) Ad << L2.
ANALYSIS: (a) The irradiation on the detector due to emission from the furnace which passes through
the filter is defined as
G d = q f → d A d = 50 W m 2
where the power leaving the furnace and intercepted at the detector is
σ T4
q f → d = [If ⋅ Af cos θ f ⋅ ω d − f ]τ filter = 
 π
(1)
A 
⋅ A f cos θ f ⋅ d  τ filter .
L2 
(2)
The transmittance of the filter is
τ filter = τ λ1F0 − λ T + τ λ 2 (1 − F0 − λ T ) = 0 × 0.4036 + 0.8(1 − 0.4036) = 0.477
(3)
where F0 − λ T = 0.4036 with λT = 2.4×1500 = 3600 µm⋅K from Table 12.1. Combining Eqs. (1) and (2)
and substituting numerical values,
G d = (1 π )5.67 × 10−8 W m 2 ⋅ K 4 × (1500K)4 (25 × 10−6 m 2 × 1)(Ad L2 ) × 0.477 Ad = 50 W m 2
find
<
L = 147 mm.
(b) Using the foregoing equations in the IHT workspace along with the IHT Radiation Tool, Band
Emission Factor, G was computed and plotted as a function of L for selected blackbody temperatures.
Irradiation, G (W/m^2)
250
200
150
100
50
0
100
200
300
400
Separation distance, L (mm)
Tf = 1000 K
Tf = 1500 K
Tf = 2000 K
Continued...
PROBLEM 12.55 (Cont.)
The irradiation decreases with increasing separation distance x as the inverse square of the distance. At
any fixed separation distance, the irradiation increases as Tf increases. In what manner will G depend
upon Tf? Is G ~ Tf4 ? Why not?
COMMENTS: The IHT workspace used to generate the above plot is shown below.
// Irradiation, Eq (2):
G = qfd / Ad
qfd = Ief * Af * omegadf * tauf
omegadf = Ad / L^2
Ief = Ebf / pi
Ebf = sigma * Tf^4
sigma = 5.67e-8
// Transmittance, Eq (3):
tauf = tau1 * FL1Tf + tau2 * ( 1 - FL1Tf )
/* The blackbody band emission factor, Figure 12.14 and Table 12.1, is */
FL1Tf = F_lambda_T(lambda1,Tf) // Eq 12.30
// where units are lambda (micrometers, mum) and T (K)
// Assigned Variables:
G = 50
// Irradiation on detector, W/m^2
Tf = 1000
// Furnace temperature, K
//Tf = 1500
//Tf = 2000
Af = 25 * 1e-6
// Furnace aperture, m^2
Ad = 0.003 * 0.007
// Detector area, m^2
tau1 = 0
// Spectral transmittance, <= lambda1
tau2 = 0.8
// Spectral transmittance, >= lambda2
lambda1 = 2.4
// Wavelength, mum
//L = 0.194
// Separation distance, m
L_mm = L * 1000
// Separation distance, mm
/* Data Browser Results - Part (a)
Ebf
FL1Tf
Ief
L
omegadf qfd
tauf
Ad
Tf
lambda1 sigma
tau1
tau2
2.87E5 0.4036 9.137E4 0.1476 0.0009634
0.00105 0.4771
50
1500
2.4
5.67E-8 0
0.8 */
Af
Df
G
2.1E-5
2.5E-5
0.025
PROBLEM 12.56
KNOWN: Spectral transmissivity of a plain and tinted glass.
FIND: (a) Solar energy transmitted by each glass, (b) Visible radiant energy transmitted by each with
solar irradiation.
SCHEMATIC:
ASSUMPTIONS: (1) Spectral distribution of solar irradiation is proportional to spectral emissive
power of a blackbody at 5800K.
ANALYSIS: To compare the energy transmitted by the glasses, it is sufficient to calculate the
transmissivity of each glass for the prescribed spectral range when the irradiation distribution is that of
the solar spectrum. From Eq. 12.55,
∞
∞
∞
τ S = τ λ ⋅ G λ ,Sdλ / G λ,Sdλ = τ λ ⋅ E λ ,b ( λ ,5800K ) dλ / E b ( 5800K ) .
0
0
0
Recognizing that τλ will be constant for the range λ1 →λ2, using Eq. 12.31, find
∫
∫
∫
τ S = τ λ ⋅ F( λ → λ ) = τ λ  F( 0→ λ ) − F( 0→ λ )  .
1
2
2
1 

(a) For the two glasses, the solar transmissivity, using Table 12.1 for F, is then
Plain glass:
λ2 = 2.5 µm
λ1 = 0.3 µm
λ2 T = 2.5 µm × 5800K = 14,500 µm⋅K
λ1 T = 0.3 µm × 5800K = 1,740 µm⋅K
F(0 → λ2) = 0.966
F( 0 → λ ) = 0.033
1
<
τS = 0.9 [0.966 – 0.033] = 0.839.
Tinted glass:
λ2 = 1.5 µm
λ2 T = 1.5 µm × 5800K = 8,700 µm⋅K
F( 0 → λ
= 0.881
λ1 = 0.5 µm
λ1 T = 0.5 µm × 5800K=2,900 µm⋅K
F( 0 → λ
= 0.033
2)
1)
<
τS = 0.9 [0.886 – 0.250] = 0.568.
(b) The limits of the visible spectrum are λ1 = 0.4 and λ2 = 0.7 µm. For the tinted glass, λ1 = 0.5 µm
rather than 0.4 µm. From Table 12.1,
λ2 = 0.7 µm
λ2 T = 0.7 µm × 5800K = 4,060 µm⋅K
F( 0 → λ ) = 0.491
2
λ1 = 0.5 µm
λ1 T = 0.5 µm × 5800K = 2,900 µm⋅K
F( 0 → λ
= 0.250
λ1 = 0.4 µm
λ1 T = 0.4 µm × 5800K=2,320 µm⋅K
F( 0 → λ
= 0.125
Plain glass:
τvis = 0.9 [0.491 – 0.125] = 0.329
Tinted glass:
τvis = 0.9 [0.491 – 0.250] = 0.217
1)
1)
<
<
COMMENTS: For solar energy, the transmissivities are 0.839 for the plain glass vs. 0.568 for the
plain and tinted glasses. Within the visible region, τvis is 0.329 vs. 0.217. Tinting reduces solar flux by
32% and visible solar flux by 34%.
PROBLEM 12.57
KNOWN: Spectral transmissivity and reflectivity of light bulb coating. Dimensions, temperature
and spectral emissivity of a tungsten filament.
FIND: (a) Advantages of the coating, (b) Filament electric power requirement for different coating
spectral reflectivities.
SCHEMATIC:
ASSUMPTIONS: (1) All of the radiation reflected from the inner surface of bulb is absorbed by the
filament.
ANALYSIS: (a) For λc = 0.7 µm, the coating has two important advantages: (i) It transmits all of
the visible radiation emitted by the filament, thereby maximizing the lighting efficiency. (ii) It returns
all of the infrared radiation to the filament, thereby reducing the electric power requirement and
conserving energy. (b) The power requirement is simply the amount of radiation transmitted by the
bulb, or
(
)λ
Pelec = A f E(0→λ ) = π DL + D2 / 2 ∫ c ε λ Eλ ,b dλ
c
0
From the spectral distribution of Problem 12.25, ελ = 0.45 for both values of λc. Hence,
{
}
λ
2
Pelec = π 0.0008 × 0.02 + (0.0008 ) / 2  m 2 0.45 E b ∫ c E λ ,b / E b dλ
0


(
)
Pelec = 5.13 × 10−5 m 2 × 0.45 × 5.67 × 10−8 W / m 2 ⋅ K 4 (3000 K ) F(0→λ )
c
4
Pelec = 106 W F(0→λ )
c
For λc = 0.7 µ m, λc T = 2100 µ m ⋅ K and from Table 12.1, F(0→λ ) = 0.0838. Hence,
c
λc = 0.7 µ m : Pelec = 106 W × 0.0838 = 8.88 W
<
For λc = 2 µ m, λc T = 6000 µ m ⋅ K and F(0→λ ) = 0.738. Hence,
c
λc = 2.0 µ m : Pelec = 106 W × 0.738 = 78.2 W
<
COMMENTS: Clearly, significant energy conservation could be realized with a reflective coating
and λc = 0.7 µm. Although a coating with the prescribed spectral characteristics is highly idealized
and does not exist, there are coatings that may be used to reflect a portion of the infrared radiation
from the filament and to thereby provide some energy savings.
PROBLEM 12.58
KNOWN: Spectral transmissivity of low iron glass (see Fig. 12.24).
FIND: Interpretation of the greenhouse effect.
SCHEMATIC:
ANALYSIS: The glass affects the net radiation transfer to the contents of the greenhouse. Since
most of the solar radiation is in the spectral region λ < 3 µm, the glass will transmit a large fraction of
this radiation. However, the contents of the greenhouse, being at a comparatively low temperature,
emit most of their radiation in the medium to far infrared. This radiation is not transmitted by the glass.
Hence the glass allows short wavelength solar radiation to enter the greenhouse, but does not permit
long wavelength radiation to leave.
PROBLEM 12.59
KNOWN: Spectrally selective, diffuse surface exposed to solar irradiation.
FIND: (a) Spectral transmissivity, τλ, (b) Transmissivity, τS, reflectivity, ρ S, and absorptivity, α S, for
solar irradiation, (c) Emissivity, ε, when surface is at Ts = 350K, (d) Net heat flux by radiation to the
surface.
SCHEMATIC:
ASSUMPTIONS: (1) Surface is diffuse, (2) Spectral distribution of solar irradiation is proportional to
Eλ,b (λ, 5800K).
ANALYSIS: (a) Conservation of radiant energy requires, according to Eq. 12.56, that ρ λ + α λ + τλ
=1 or τλ = 1 - ρ λ - α λ. Hence, the spectral transmissivity appears as shown above (dashed line). Note
that the surface is opaque for λ > 1.38 µm.
(b) The transmissivity to solar irradiation, GS, follows from Eq. 12.55,
∞
∞
τ S = τ λ G λ ,S dλ / G S = τ λ E λ ,b ( λ,5800K ) dλ / E b ( 5800K )
0
0
∫
∫
1.38
E λ ,b ( λ,5800K ) dλ / E b ( 5800K ) = τ λ ,1F(0 →λ ) = 0.7 × 8.56 = 0.599
1
0
τ S = τ λ ,b ∫
<
where λ1 TS = 1.38 × 5800 = 8000 µm⋅K and from Table 12.1, F( 0 → λ ) = 0.856. From Eqs. 12.52
1
and 12.57,
∞
ρS = ∫ ρλ Gλ ,S dλ / GS = ρ λ ,1F( 0 → λ ) = 0.1× 0.856 = 0.086
1
0
<
α S = 1 − ρ S − τ S = 1 − 0.086 − 0.599 = 0.315.
<
(c) For the surface at Ts = 350K, the emissivity can be determined from Eq. 12.38. Since the surface
is diffuse, according to Eq. 12.65, α λ = ε λ, the expression has the form
∞
0
∞
0
ε = ∫ ελ E λ ,b ( Ts ) d λ / E b ( Ts ) = ∫ α λ Eλ ,b ( 350K ) dλ / E b ( 350K )
ε = α λ ,1F( 0 −1.38 µ m ) + αλ ,2 1 − F( 0 −1.38 µ m )  = α λ ,2 = 1


<
where from Table 12.1 with λ1 TS = 1.38 × 350 = 483 µm⋅K, F( 0 − λ T ) ≈ 0.
(d) The net heat flux by radiation to the
surface is determined by a radiation balance
q′′rad = GS − ρSG S − τ SGS − E
q′′rad = αSGS − E
q′′rad = 0.315 × 750 W / m 2 − 1.0 × 5.67 ×10−8 W / m 2 ⋅ K 4 ( 350K ) = −615 W / m 2 .
4
<
PROBLEM 12.60
KNOWN: Large furnace with diffuse, opaque walls (Tf, ε f) and a small diffuse, spectrally selective
object (To, τλ, ρ λ).
FIND: For points on the furnace wall and the object, find ε, α, E, G and J.
SCHEMATIC:
ASSUMPTIONS: (1) Furnace walls are isothermal, diffuse, and gray, (2) Object is isothermal and
diffuse.
ANALYSIS: Consider first the furnace wall (A). Since the wall material is diffuse and gray, it
follows that
<
ε A = ε f = α A = 0.85.
The emissive power is
E A = εA E b ( Tf ) = εAσ Tf = 0.85 × 5.67 × 10 − 8 W / m 2 ⋅ K 4 ( 3000 K ) = 3.904 ×10 6 W / m 2 .
Since the furnace is an isothermal enclosure, blackbody conditions exist such that
<
G A = J A = E b ( Tf ) = σ Tf4 = 5.67 ×10− 8 W / m 2 ⋅ K 4 ( 3000K ) 4 = 4.593 ×10 6 W / m 2 .
<
4
Considering now the semitransparent, diffuse, spectrally selective object at To = 300 K. From the
radiation balance requirement, find
α λ = 1 − ρ λ −τ λ or α1 = 1 − 0.6 − 0.3 = 0.1 and α 2 = 1 − 0.7 − 0.0 = 0.3
∞
α B = ∫ α λ Gλ dλ / G = F0 − λ T ⋅α1 + (1 − F0 −λ T ) ⋅ α 2 = 0.970 × 0.1 + (1 − 0.970 ) × 0.3 = 0.106
0
<
where F0 - λT = 0.970 at λT = 5 µm × 3000 K = 15,000 µm⋅K since G = Eb(Tf). Since the object is
diffuse, ε λ = α λ, hence
εB =
∞
∫0
ε λ E λ ,b ( To ) dλ / E b,o = F0− λ Tα1 + (1 − F0− λ T ) ⋅ α 2 = 0.0138 × 0.1 + (1 − 0.0138 ) × 0.3 = 0.297
<
where F0-λT = 0.0138 at λT = 5 µm × 300 K = 1500 µmK. The emissive power is
E B = ε BE b,B (To ) = 0.297 × 5.67 ×10 −8 W / m 2 ⋅ K 4 ( 300 K ) = 136.5 W / m 2 .
4
<
The irradiation is that due to the large furnace for which blackbody conditions exist,
G B = G A = σ Tf4 = 4.593 ×106 W / m 2 .
<
The radiosity leaving point B is due to emission and reflected irradiation,
J B = EB + ρBG B = 136.5 W / m 2 + 0.3 × 4.593 ×10 6 W / m 2 = 1.378 ×10 6 W / m 2 . <
6
2
If we include transmitted irradiation, JB = EB + (ρ B + τB) GB = EB + (1 - α B) GB = 4.106 × 10 W/m .
In the first calculation, note how we set ρ B ≈ ρ λ (λ < 5 µm).
PROBLEM 12.61
KNOWN: Spectral characteristics of four diffuse surfaces exposed to solar radiation.
FIND: Surfaces which may be assumed to be gray.
SCHEMATIC:
ASSUMPTIONS: (1) Diffuse surface behavior.
ANALYSIS: A gray surface is one for which α λ and ε λ are constant over the spectral regions of the
irradiation and the surface emission.
For λ = 3 µm and T = 5800K, λT = 17,400 µm⋅K and from Table 12.1, find F(0 → λ) = 0.984. Hence,
98.4% of the solar radiation is in the spectral region below 3 µm.
For λ = 6 µm and T = 300K, λT = 1800 µm⋅K and from Table 12.1, find F(0 → λ) = 0.039. Hence,
96.1% of the surface emission is in the spectral region above 6 µm.
Hence:
Surface A is gray:
α S ≈ ε = 0.8
Surface B is not gray:
α S ≈ 0.8, ε ≈ 0.3
Surface C is not gray:
α S ≈ 0.3, ε ≈ 0.7
Surface D is gray:
α S ≈ ε = 0.3.
<
<
<
<
PROBLEM 12.62
KNOWN: A gray, but directionally selective, material with α (θ, φ) = 0.5(1 - cosφ).
FIND: (a) Hemispherical absorptivity when irradiated with collimated solar flux in the direction (θ =
45° and φ = 0°) and (b) Hemispherical emissivity of the material.
SCHEMATIC:
ASSUMPTIONS: (1) Gray surface behavior.
ANALYSIS: (a) The surface has the directional absorptivity given as
α (θ ,φ ) = α λ ,φ = 0.5[1 − cos φ ] .
When irradiated in the direction θ = 45° and φ = 0°, the directional absorptivity for this condition is
α ( 45°, 0° ) = 0.5 1 − cos ( 0° )  = 0.
<
That is, the surface is completely reflecting (or transmitting) for irradiation in this direction.
(b) From Kirchhoff’s law,
αθ ,φ = εθ ,φ
so that
εθ ,φ = αθ ,φ = 0.5 (1 − cos φ ) .
Using Eq. 12.35 find
2π π / 2
∫ ∫ εθ ,φ ,λ cos θ sin θ d θ d φ
ε= 0 0
2π π / 2
∫0 ∫0 cos θ sin θ dθ dφ
2π
2π
0.5 (1 − cos φ ) dφ 0.5 ( φ − sin φ )
ε= 0
=
= 0.5.
2π
π
2
dφ
0
0
∫
∫
<
PROBLEM 12.63
KNOWN: Area and temperature of an opaque surface. Rate of incident radiation, absorbed radiation
and heat transfer by convection.
FIND: Surface irradiation, emissive power, radiosity, absorptivity, reflectivity and emissivity.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, (2) Adiabatic sides and bottom.
ANALYSIS: The irradiation, emissive power and radiosity are
G = 1300 W 3m 2 = 433 W m 2
E = G abs − q′′conv = (1000 − 300 ) W 3m 2 = 233 W m 2
J = E + G ref = E + ( G − G abs ) =  233 + ( 433 − 333) W m 2 = 333 W m 2
<
<
<
The absorptivity, reflectivity and emissivity are
(
α = G abs G = 333 W m 2
) (433 W m2 ) = 0.769
<
ρ = 1 − α = 0.231
ε = E E b = E σ Ts4 = 233 W m 2 5.67 × 10−8 W m 2⋅ K 4 (500 K ) = 0.066
4
COMMENTS: The expression for E follows from a surface energy balance for which the absorbed
irradiation is balanced by emission and convection.
<
<
PROBLEM 12.64
KNOWN: Isothermal enclosure at a uniform temperature provides a known irradiation on two small
surfaces whose absorption rates have been measured.
FIND: (a) Net heat transfer rates and temperatures of the two surfaces, (b) Absorptivity of the
surfaces, (c) Emissive power of the surfaces, (d) Emissivity of the surfaces.
SCHEMATIC:
ASSUMPTIONS: (1) Enclosure is at a uniform temperature and large compared to surfaces A and
B, (2) Surfaces A and B have been in the enclosure a long time, (3) Irradiation to both surfaces is the
same.
ANALYSIS: (a) Since the surfaces A and B have been within the enclosure a long time, thermal
equilibrium conditions exist. That is,
q A,net = q B,net = 0.
Furthermore, the surface temperatures are the same as the enclosure, Ts,A = Ts,B = Tenc. Since the
enclosure is at a uniform temperature, it follows that blackbody radiation exists within the enclosure
(see Fig. 12.12) and
4
G = Eb ( Tenc ) = σ Tenc
(
Tenc = ( G / σ )1/4 = 6300W/m 2 /5.67 ×10 −8 W / m 2 ⋅ K 4
(b) From Eq. 12.45, the absorptivity is Gabs/G,
5600 W / m 2
αA =
= 0.89
2
6300 W / m
αB =
630 W / m 2
6300 W / m 2
)
1/4
= 577.4K.
<
<
= 0.10.
(c) Since the surfaces experience zero net heat transfer, the energy balance is Gabs = E. That is, the
absorbed irradiation is equal to the emissive power,
E A = 5600 W / m 2
E B = 630 W / m 2 .
<
(d) The emissive power, E(T), is written as
E = ε E b (T ) = ε σ T 4
or
ε = E / σ T4 .
Since the temperature of the surfaces and the emissive powers are known,
W
4
2 
8
ε A = 5600 W / m /  5.67 × 10−
577.4K )  = 0.89
(


m2 ⋅ K4
COMMENTS: Note for this equilibrium condition, ε = α.
ε B = 0.10.
<
PROBLEM 12.65
KNOWN: Opaque, horizontal plate, well insulated on backside, is subjected to a prescribed irradiation.
Also known are the reflected irradiation, emissive power, plate temperature and convection coefficient
for known air temperature.
FIND: (a) Emissivity, absorptivity and radiosity and (b) Net heat transfer per unit area of the plate.
SCHEMATIC:
ASSUMPTIONS: (1) Plate is insulated on backside, (2) Plate is opaque.
ANALYSIS: (a) The total, hemispherical emissivity of the plate according to Eq. 12.37 is
E
E
1200 W / m 2
ε=
E b ( Ts )
=
=
= 0.34.
<
4
4
σ Ts 5.67 ×10 −8 W / m 2 ⋅ K 4 × ( 227 + 273) K 4
The total, hemispherical absorptivity is related to the reflectivity by Eq. 12.57 for an opaque surface.
That is, α = 1 - ρ. By definition, the reflectivity is the fraction of irradiation reflected, Eq. 12.51, such
that
α = 1 − Gref / G = 1 − 500 W / m 2 / 2500 W / m 2 = 1 − 0.20 = 0.80.
(
)
<
The radiosity, J, is defined as the radiant flux leaving the surface by emission and reflection per unit
area of the surface (see Section 12.24).
J = ρ G + ε Eb = Gref + E = 500 W / m 2 +1200 W / m 2 = 1700 W / m 2.
(b) The net heat transfer is determined from
an energy balance,
<
q′′net = q′′in − q′′out = G − G ref − E − q′′conv
q′′net = ( 2500 − 500 −1200 ) W / m 2 − 15 W / m 2 ⋅ K ( 227 −127 ) K = −700 W / m 2 .
<
An alternate approach to the energy balance using the
radiosity,
q′′net = G − J − q′′conv
q′′net = ( 2500 −1700 − 1500 ) W / m 2
q′′net = −700W/m 2 .
COMMENTS: (1) Since the net heat rate per unit area is negative, energy must be added to the
plate in order to maintain it at Ts = 227°C. (2) Note that α ≠ ε. Hence, the plate is not a gray body.
(3) Note the use of radiosity in performing energy balances. That is, considering only the radiation
processes, q ′′net = G − J.
PROBLEM 12.66
KNOWN: Horizontal, opaque surface at steady-state temperature of 77°C is exposed to a convection
process; emissive power, irradiation and reflectivity are prescribed.
FIND: (a) Absorptivity of the surface, (b) Net radiation heat transfer rate for the surface; indicate
direction, (c) Total heat transfer rate for the surface; indicate direction.
SCHEMATIC:
ASSUMPTIONS: (1) Surface is opaque, (2) Effect of surroundings included in the specified
irradiation, (3) Steady-state conditions.
ANALYSIS: (a) From the definition of the thermal radiative properties and a radiation balance for an
opaque surface on a total wavelength basis, according to Eq. 12.59,
<
α = 1 − ρ = 1 − 0.4 = 0.6.
(b) The net radiation heat transfer rate to the surface
follows from a surface energy balance considering only
radiation processes. From the schematic,
(
)
′′ − E& ′′out
q′′net,rad = E& in
rad
q′′
net,rad = G− ρ G − E= (1380− 0.4×1380− 628 )W / m2 = 200W/m2 .
<
Since q ′′net,rad is positive, the net radiation heat transfer rate is to the surface.
(c) Performing a surface energy balance considering all
heat transfer processes, the local heat transfer rate is
q′′tot = ( E& in
′′ − E& ′′out )
′′
′′
q′′tot = qnet,rad
− qconv
q ′′tot = 200 W / m2 − 28 W / m2 ⋅ K ( 77 − 27 ) K = −1200 W / m2 .
The total heat flux is shown as a negative value indicating the heat flux is from the surface.
COMMENTS: (1) Note that the surface radiation
balance could also be expresses as
q′′net,rad = G − J
α G − E.
or
Note the use of radiosity to express the radiation flux leaving the surface.
(2) From knowledge of the surface emissive power and Ts, find the emissivity as
ε ≡ E / σ Ts4 = 628 W / m 2 / 5.67 × 10−8 W / m 2 ⋅ K 4 ( 77 + 273)4 K4 = 0.74.
(
Since ε ≠ α, we know the surface is not gray.
)
<
PROBLEM 12.67
KNOWN: Temperature and spectral characteristics of a diffuse surface at Ts = 500 K situated in a large
enclosure with uniform temperature, Tsur = 1500 K.
FIND: (a) Sketch of spectral distribution of E λ and E λ,b for the surface, (b) Net heat flux to the
surface, q″rad,in (c) Compute and plot q″rad,in as a function of Ts for the range 500 ≤ Ts ≤ 1000 K; also plot
the heat flux for a diffuse, gray surface with total emissivities of 0.4 and 0.8; and (d) Compute and plot
ε and α as a function of the surface temperature for the range 500 ≤ Ts ≤ 1000 K.
SCHEMATIC:
ASSUMPTIONS: (1) Surface is diffuse, (2) Convective effects are negligible, (3) Surface irradiation
corresponds to blackbody emission at 1500 K.
ANALYSIS: (a) From Wien’s law, Eq. 12.27, λmax T =
2897.6 µm⋅K. Hence, for blackbody emission from the
surface at Ts = 500 K,
2897.6 µ m ⋅ K
λmax =
500 K
<
= 5.80 µ m .
(b) From an energy balance on the surface, the net heat
flux to the surface is
q″rad,in = αG − E = αEb (1500 K) − εEb (500 K).
(1)
From Eq. 12.46,
4 E λ ,b (1500)
∞ E λ ,b (1500)
dλ + 0.8
dλ = 0.4F(0 − 4) − 0.8[1 − F(0 − 4) ].
0
4
Eb
Eb
α = 0.4 ∫
∫
From Table 12.1 with λT = 4µm × 1500 K = 6000 µm⋅K, F(0-4) = 0.738, find
α = 0.4 × 0.738 + 0.8 (1 − 0.738) = 0.505.
From Eq. 12.38
4 E λ ,b (500)
∞ E λ ,b (500)
dλ + 0.8
dλ = 0.4F(0 − 4) + 0.8[1 − F(0 − 4) ] .
0
4
Eb
Eb
ε = 0.4 ∫
∫
From Table 12.1 with λT = 4µm × 500 K = 2000 µm⋅K, F(0-4) = 0.0667, find
ε = 0.4 × 0.0667 + 0.8 (1 − 0.0667) = 0.773.
Hence, the net heat flux to the surface is
q′′rad,in = 5.67 × 10−8 W m 2 ⋅ K 4 [0.505 × (1500 K)4 − 0.773 × (500 K)4 ] = 1.422 × 105 W m 2 .
<
Continued...
PROBLEM 12.67 (Cont.)
(c) Using the foregoing equations in the IHT workspace along with the IHT Radiation Tool, Band
Emission Factor, q′′rad,in was computed and plotted as a function of Ts.
q''radin (W/m^2)
250000
200000
150000
100000
50000
500
600
700
800
900
1000
Surface temperature, Ts (K)
Grey surface, eps = 0.4
Selective surface, eps
Grey surface, eps = 0.8
The net radiation heat rate, q′′rad,in decreases with increasing surface temperature since E increases with
Ts and the absorbed irradiation remains constant according to Eq. (1). The heat flux is largest for the
gray surface with ε = 0.4 and the smallest for the gray surface with ε = 0.8. As expected, the heat flux for
the selective surface is between the limits of the two gray surfaces.
(d) Using the IHT model of part (c), the emissivity and absorptivity of the surface are computed and
plotted below.
eps or alpha
0.8
0.7
0.6
0.5
0.4
500
600
700
800
900
1000
Surface temperature, Ts (K)
Emissivity, eps
Absorptivity, alpha
The absorptivity, α = α (α λ , Tsur ) , remains constant as Ts changes since it is a function of α λ (or ε λ )
and Tsur only. The emissivity, ε = ε (ε λ , Ts ) is a function of Ts and decreases as Ts increases. Could
you have surmised as much by looking at the spectral emissivity distribution? Under what condition
would you expect α = ε?
PROBLEM 12.68
KNOWN: Opaque, diffuse surface with prescribed spectral reflectivity and at a temperature of 750K
is subjected to a prescribed spectral irradiation, Gλ.
FIND: (a) Total absorptivity, α, (b) Total emissivity, ε, (c) Net radiative heat flux to the surface.
SCHEMATIC:
ASSUMPTIONS: (1) Opaque and diffuse surface, (2) Backside insulated.
ANALYSIS: (a) The total absorptivity is determined from Eq. 12.46 and 12.56,
∞
αλ = 1− ρ λ
and
α=
α G dλ /G.
0 λ λ
Evaluating by separate integrals over various wavelength intervals.
∫
α=
(1 − ρ λ ,1 ) ∫13 Gλ dλ + (1 − ρ λ ,2 ) ∫36 G λ dλ + (1 − ρ λ ,2 ) ∫68 Gλ dλ
3
∫1
6
8
G λ d λ + ∫ G λ dλ + ∫ G λ dλ
3
6
(1,2)
=
Gabs
G
G abs = (1 − 0.6 ) 0.5 × 500W/m2 ⋅ µ m ( 3 − 1) µ m + (1 − 0.2 ) 500W/m2 ⋅ µ m ( 6 − 3 ) µ m




+ (1 − 0.2 )  0.5 × 500W/m2 ⋅ µ m ( 8 − 6) µ m


G = 0.5 × 5 0 0 W / m ⋅ µ m × ( 3 − 1) µ m + 5 0 0 W / m ⋅ µ m ( 6 − 3 ) µm + 0.5 × 5 0 0 W / m ⋅ µ m ( 8 − 6 ) µ m
2
α=
2
2
[ 200 + 1200 + 400]W / m 2 1800W/m2
=
[ 500 + 1500 + 500] W / m2 2 5 0 0 W / m2
= 0.720.
<
(b) The total emissivity of the surface is determined from Eq. 12.38 and 12.65,
ελ = αλ
ε λ = 1 − ρλ .
and, hence
(3,4)
The total emissivity can then be expressed as
∞
∞
ε E
( λ , Ts ) dλ / E b ( Ts ) = 0 (1 − ρ λ ) E λ ,b ( λ , Ts ) dλ / E b ( Ts )
0 λ λ ,b
3
∞
ε = 1 − ρ λ ,1
E
( λ , Ts ) dλ / Eb ( Ts ) + 1 − ρλ ,2 3 Eλ ,b ( λ ,Ts ) d λ / E b ( Ts )
0 λ ,b
ε =∫
∫
(
)∫
(
)∫
ε = (1 − ρ λ ,1 ) F( 0 →3 µ m ) + (1 − ρ λ ,2 ) 1 − F( 0 →3 µ m ) 


ε = (1 − 0.6 ) × 0.111 + (1 − 0.2 )[1 −0.111] = 0.756
<
where Table 12.1 is used to find F(0 - λ) = 0.111 for λ1 Ts = 3 × 750 = 2250 µm⋅K.
(c) The net radiative heat flux to the surface is
q′′rad = α G − ε Eb ( Ts ) = αG − ε σ Ts4
q′′rad = 0.720 × 2500W/m2
4
−0.756 × 5.67 ×10−8 W / m 2 ⋅ K 4 750K = −11,763W/m 2 .
(
)
<
PROBLEM 12.69
KNOWN: Opaque, gray surface at 27°C with prescribed irradiation, reflected flux and convection
process.
FIND: Net heat flux from the surface.
SCHEMATIC:
ASSUMPTIONS: (1) Surface is opaque and gray, (2) Surface is diffuse, (3) Effects of surroundings
are included in specified irradiation.
ANALYSIS: From an energy balance on the surface, the net heat flux from the surface is
q′′net = E& ′′out − E& ′′in
′′
q′′net = qconv
+ E + G ref − G = h ( Ts − T∞ ) + ε σ Ts4 + G ref − G.
(1)
To determine ε, from Eq. 12.59 and Kirchoff’s law for a diffuse-gray surface, Eq. 12.62,
ε = α = 1 − ρ = 1 − ( Gref / G ) = 1 − ( 800/1000 ) = 1 − 0.8 = 0.2
(2)
where from Eq. 12.51, ρ = Gref/G. The net heat flux from the surface, Eq. (1), is
q′′net = 1 5 W / m 2 ⋅ K ( 27 − 17 ) K + 0.2 × 5.67 ×10−8 W / m 2 ⋅ K 4 ( 27 + 273) K 4
4
+8 0 0 W / m 2 − 1000W/m2
q′′net = (150 + 91.9 + 800 − 1000 ) W / m 2 = 4 2 W / m 2 .
<
COMMENTS: (1) For this situation, the radiosity is
J = Gref + E = ( 800 + 91.9 ) W / m 2 = 8 9 2 W / m 2.
The energy balance can be written involving the radiosity (radiation leaving the surface) and the
irradiation (radiation to the surface).
q′′net = J −G + q′′conv = ( 892 −1000 + 150 ) W / m 2 = 4 2 W / m2 .
(2) Note the need to assume the surface is diffuse, gray and opaque in order that Eq. (2) is applicable.
PROBLEM 12.70
KNOWN: Diffuse glass at Tg = 750 K with prescribed spectral radiative properties being heated in a
large oven having walls with emissivity of 0.75 and 1800 K.
FIND: (a) Total transmissivity r, total reflectivity ρ, and total emissivity ε of the glass; Net radiative
heat flux to the glass, (b) q′′rad,in ; and (c) Compute and plot q′′rad,in as a function of glass temperatures
for the range 500 ≤ Tg ≤ 800 K for oven wall temperatures of Tw = 1500, 1800 and 2000 K.
SCHEMATIC:
ASSUMPTIONS: (1) Glass is of uniform temperature, (2) Glass is diffuse, (3) Furnace walls large
compared to the glass; εw plays no role, (4) Negligible convection.
ANALYSIS: (a) From knowledge of the spectral transmittance,τw, and spectral reflectivity, ρ λ , the
following radiation properties are evaluated:
Total transmissivity, r: For the irradiation from the furnace walls, Gλ = Eλ,b (λ, Tw ). Hence
∞
<
4
≈ τ λ1F(0 − λ T ) = 0.9 × 0.25 = 0.225 .
τ = ∫ τ λ E λ ,b ( λ , Tw ) dλ σ Tw
0
where λT = 1.6 µm × 1800 K = 2880 µm⋅K ≈ 2898 µm⋅K giving F(0-λT) ≈ 0.25.
Total reflectivity, ρ: With Gλ = Eλ,b (λ,Tw), Tw = 1800 K, and F0 − λT = 0.25,
(
)
<
ρ ≈ ρλ1F(0 − λ T ) + ρλ 2 1 − F( 0 − λ T ) = 0.05 × 0.25 + 0.5 (1 − 0.25 ) = 0.388
Total absorptivity, α : To perform the energy balance later, we’ll need α. Employ the conservation
expression,
α = 1 − ρ − τ = 1 − 0.388 − 0.225 = 0.387 .
Emissivity, ε: Based upon surface temperature Tg = 750 K, for
F0 − λ T ≈ 0.002 .
λ T = 1.6 µ m × 750K = 1200 µ m ⋅ K,
Hence for λ > 1.6 µm,
ε ≈ ελ ≈ 0.5.
<
(b) Performing an energy balance on the glass, the net radiative heat flux by radiation into the glass is,
Continued...
PROBLEM 12.70 (Cont.)
q′′net,in = E′′in − E′′out
(
( ))
q′′net,in = 2 α G − ε E b Tg
4
where G = σ Tw
4
4
q′′net,in = 2 0.387σ (1800K ) − 0.5σ ( 750K ) 


q′′net,in = 442.8 kW m 2 .
(b) Using the foregoing equations in the IHT Workspace along with the IHT Radiation Tool, Band
Emission Factor, the net radiative heat flux, q′′rad,in , was computed and plotted as a function of Tg for
selected wall temperatures Tw .
700
q''radin (kW/m^2)
600
500
400
300
200
500
600
700
800
Glass temperature, Tg (K)
Tw = 1500 K
Tw = 1800 K
Tw = 2000 K
As the glass temperature increases, the rate of emission increases so we’d expect the net radiative heat
rate into the glass to decrease. Note that the decrease is not very significant. The effect of increased wall
temperature is to increase the irradiation and, hence the absorbed irradiation to the surface and the net
radiative flux increase.
PROBLEM 12.71
KNOWN: Temperature, absorptivity, transmissivity, radiosity and convection conditions for a
semitransparent plate.
FIND: Plate irradiation and total hemispherical emissivity.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Uniform surface conditions.
ANALYSIS: From an energy balance on the plate
E& in = E& out
2G = 2q ′′conv + 2J.
Solving for the irradiation and substituting numerical values,
G = 40 W / m 2 ⋅ K ( 350 − 300 ) K + 5000 W / m 2 = 7000 W / m 2 .
<
From the definition of J,
J = E + ρG + τ G = E + (1 −α ) G.
Solving for the emissivity and substituting numerical values,
ε=
J − (1 − α ) G
σT
4
5000 W / m 2 ) − 0.6 ( 7000 W / m 2 )
(
=
= 0.94.
4
5.67 ×10 −8 W / m 2 ⋅ K 4 ( 350K)
<
Hence,
α ≠ε
and the surface is not gray for the prescribed conditions.
COMMENTS: The emissivity may also be determined by expressing the plate energy balance as
2α G = 2q ′′conv + 2E.
Hence
ε σ T 4 = αG − h ( T − T∞ )
ε=
(
)
0.4 7000 W / m 2 − 40 W / m 2 ⋅ K ( 50 K)
4
5.67 ×10 −8 W / m 2 ⋅ K 4 ( 350 K )
= 0.94.
PROBLEM 12.72
KNOWN: Material with prescribed radiative properties covering the peep hole of a furnace and
exposed to surroundings on the outer surface.
FIND: Steady-state temperature of the cover, Ts; heat loss from furnace.
SCHEMATIC:
ASSUMPTIONS: (1) Cover is isothermal, no gradient, (2) Surroundings of the outer surface are
large compared to cover, (3) Cover is insulated from its mount on furnace wall, (4) Negligible
convection on interior surface.
PROPERTIES: Cover material (given): For irradiation from the furnace interior: τf = 0.8, ρ f = 0;
For room temperature emission: τ = 0, ε = 0.8.
ANALYSIS: Perform an energy balance identifying the modes of heat transfer,
E& in − E& out = 0
Recognize that
αf Gf + αsur Gsur − 2 ε E b ( Ts ) − h ( Ts − T∞ ) = 0.
(1)
Gf = σ Tf4
(2,3)
4 .
Gsur = σ Tsur
α f = 1 − τ f − ρ f = 1 − 0.8 − 0.0 = 0.2.
From Eq. 12.57, it follows that
(4)
Since the irradiation Gsur will have nearly the same spectral distribution as the emissive power of the
cover, Eb (Ts), and since Gsur is diffuse irradiation,
αsur = ε = 0.8.
(5)
This reasoning follows from Eqs. 12.65 and 12.66. Substituting Eqs. (2-5) into Eq. (1) and using
numerical values,
0.2 × 5.67 ×10−8 ( 450 + 273 ) W / m 2 + 0.8 × 5.67 ×10−8 ×300 4 W / m 2
4
−2 × 0.8 × 5.67 ×10 −8 Ts4 W / m 2 − 50 W / m 2 ⋅ K ( Ts − 300 ) K = 0
9.072 × 10−8 Ts4 + 50Ts = 18,466
or
(2-5)
<
Ts = 344K.
The heat loss from the furnace (see energy balance schematic) is
q f,loss = As αf G f +τ fG f − ε E b ( Ts )  =
π D2
(α f + τf ) G f − ε E b ( Ts ) 
4 
q f,loss = π ( 0.050m )2 / 4 ( 0.8 + 0.2 )( 723K ) 4

4
−0.8 ( 344K )  5.67 × 10−8 W / m 2 ⋅ K4 = 29.2 W.

<
PROBLEM 12.73
KNOWN: Window with prescribed τλ and ρ λ mounted on cooled vacuum chamber passing radiation
from a solar simulator.
FIND: (a) Solar transmissivity of the window material, (b) State-state temperature reached by
window with simulator operating, (c) Net radiation heat transfer to chamber.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Diffuse behavior of window material, (3) Chamber
and room surroundings large compared to window, (4) Solar simulator flux has spectral distribution of
5800K blackbody, (5) Window insulated from its mount, (6) Window is isothermal at Tw.
ANALYSIS: (a) Using Eq. 12.55 and recognizing that Gλ,S ~ Eb,λ (λ, 5800K),
1.9
τ S = τ1
E
( λ,5800K ) dλ / E b ( 5800K ) = τ 1  F(0→1.9µm ) − F( 0→0.38µm )  .
0.38 λ ,b
∫
From Table 12.1 at λT = 1.9 × 5800 = 11,020 µm⋅K, F(0 → λ) = 0.932; at λT = 0.38 × 5800 µm⋅K =
2,204 µm⋅K, F(0 → λ) = 0.101; hence
τ S = 0.90 [ 0.932 − 0.101] = 0.748.
<
Recognizing that later we’ll need α S, use Eq. 12.52 to find ρ S
ρS = ρ1 F( 0→0.38µ m) + ρ 2 F( 0→1.9 µ m) − F( 0→0.38µ m)  + ρ 3 1 − F( 0→1.9µ m) 




ρS = 0.15 × 0.101 + 0.05[ 0.932 − 0.101] + 0.15[1 −0.932 ] = 0.067
αS = 1 − ρS − τS = 1 − 0.067 − 0.748 = 0.185.
(b) Perform an energy balance on the window.
′′ −sur − q conv
′′
αSGS − q ′′w − c − q w
=0
(
) (
)
4 − T 4 − εσ T4 − T4 −h T − T = 0.
αSGS − εσ Tw
( w ∞)
c
w
sur
Recognize that ρ λ (λ > 1.9) = 0.15 and that ε ≈ 1 – 0.15 = 0.85 since Tw will be near 300K.
Substituting numerical values, find by trial and error,
4 − 2984 − 774  k 4 − 28 W / m2 ⋅ K T − 298 K = 0
0.185 × 3000 W / m 2 − 0.85 × σ 2Tw
( w
)


Tw = 302.6K = 29.6°C.
<
(c) The net radiation transfer per unit area of the window to the vacuum chamber, excluding the
transmitted simulated solar flux is
(
)
4
q ′′w − c = εσ Tw
− Tc4 = 0.85 × 5.67 ×10−8 W / m2 ⋅ K4 302.64 − 77 4  K4 = 402 W / m2 .


<
PROBLEM 12.74
KNOWN: Reading and emissivity of a thermocouple (TC) located in a large duct to measure gas stream
temperature. Duct wall temperature and emissivity; convection coefficient.
FIND: (a) Gas temperature, T∞ , (b) Effect of convection coefficient on measurement error.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible heat loss from TC sensing junction to
support, (3) Duct wall much larger than TC, (4) TC surface is diffuse-gray.
ANALYSIS: (a) Performing an energy balance on the thermocouple, it follows that
q′′w − s − q′′conv = 0 .
where radiation exchange between the duct wall and the TC is given by Eq. 1.7. Hence,
4
ε sσ (Tw
− Ts4 ) − h (Ts − T∞ ) = 0 .
Solving for T∞ with Ts = 180oC,
εσ 4
T∞ = Ts − s (Tw
− Ts4 )
h
T∞ = (180 + 273)K −
0.6(5.67 × 10−8 W m 2⋅ K 4 )
2
125 W m ⋅ K
([450 + 273]4 − [180 + 273]4 ) K4
<
T∞ = 453 K − 62.9 K = 390 K = 117$ C .
(b) Using the IHT First Law model for an Isothermal Solid Sphere to solve the foregoing energy balance
for Ts, with T∞ = 125oC, the measurement error, defined as ∆T = Ts − T∞ , was determined and is plotted
as a function of h .
300
Measurement error, delta(C)
250
200
150
100
50
0
0
200
400
600
800
1000
Convection coefficient, hbar(W/m^2.K)
The measurement error is enormous (∆T ≈ 270oC) for h = 10 W/m2⋅K, but decreases with increasing h .
However, even for h = 1000 W/m2⋅K, the error (∆T ≈ 8°C) is not negligible. Such errors must always be
considered when measuring a gas temperature in surroundings whose temperature differs significantly
from that of the gas.
Continued...
PROBLEM 12.74 (Cont.)
COMMENTS: (1) Because the duct wall surface area is much larger than that of the thermocouple, its
emissivity is not a factor. (2) For such a situation, a shield about the thermocouple would reduce the
influence of the hot duct wall on the indicated TC temperature. A low emissivity thermocouple coating
would also help.
PROBLEM 12.75
KNOWN: Diameter and emissivity of a horizontal thermocouple (TC) sheath located in a large room.
Air and wall temperatures.
FIND: (a) Temperature indicated by the TC, (b) Effect of emissivity on measurement error.
SCHEMATIC:
ASSUMPTIONS: (1) Room walls approximate isothermal, large surroundings, (2) Room air is
quiescent, (3) TC approximates horizontal cylinder, (4) No conduction losses, (5) TC surface is
opaque, diffuse and gray.
PROPERTIES: Table A-4, Air (assume Ts = 25 oC, Tf = (Ts + T∞)/2 ≈ 296 K, 1 atm):
ν = 15.53 × 10 −6 m 2 s, k = 0.026 W m ⋅ K, α = 22.0 × 10−6 m 2 s, Pr = 0.708, β = 1 Tf .
ANALYSIS: (a) Perform an energy balance on the thermocouple considering convection and radiation
processes. On a unit area basis, with q′′conv = h(Ts − T∞ ),
E in − E out = 0
(1)
α G − ε E b (Ts ) − h(Ts − T∞ ) = 0 .
Since the surroundings are isothermal and large compared to the thermocouple, G = Eb(Tsur). For the
gray-diffuse surface, α = ε. Using the Stefan-Boltzman law, Eb = σT4, Eq. (1) becomes
4
εσ (Tsur
− Ts4 ) − h(Ts − T∞ ) = 0 .
Using the Churchill-Chu correlation for a horizontal cylinder, estimate h due to free convection.
(2)
2


6

hD 
0.387Ra1/
gβ∆TD3

D
= 0.60 +
=
.
Nu D =
,
Ra
D
8 / 27 
k
να
9
/16


1 + ( 0.559 Pr )





(3,4)
To evaluate RaD and Nu D , assume Ts = 25oC, giving
Ra D =
9.8 m s2 (1 296 K)(25 − 20)K(0.004m)3
15.53 × 10−6 m 2 s × 22.0 × 10−6 m 2 s
= 31.0
2


1/ 6

0.026 W m ⋅ K 
0.387(31.0)

2
h=
0.60 +
 = 8.89 W m ⋅ K .
8
/
27
0.004m


1 + ( 0.559 0.708 )9 /16 




(5)
With ε = 0.4, the energy balance, Eq. (2), becomes
0.4 × 5.67 × 10−8 W m 2 ⋅ K 4 [(30 + 273)4 − Ts4 ]K 4 − 8.89 W m 2 ⋅ K[Ts − (20 + 273)]K = 0
where all temperatures are in kelvin units. By trial-and-error, find
Ts ≈ 22.2oC
(6)
<
Continued...
PROBLEM 12.75 (Cont.)
(b) The thermocouple measurement error is defined as ∆T =Ts − T∞ and is a consequence of radiation
exchange with the surroundings. Using the IHT First Law Model for an Isothermal Solid Cylinder with
the appropriate Correlations and Properties Toolpads to solve the foregoing energy balance for Ts, the
measurement error was determined as a function of the emissivity.
Measurement error, delta(C)
5
4
3
2
1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Emissivity, eps
The measurement error decreases with decreasing ε, and hence a reduction in net radiation transfer from
the surroundings. However, even for ε = 0.1, the error (∆T ≈ 1oC) is not negligible.
PROBLEM 12.76
KNOWN: Temperature sensor imbedded in a diffuse, gray tube of emissivity 0.8 positioned within a
room with walls and ambient air at 30 and 20 oC, respectively. Convection coefficient is 5 W m 2⋅ K .
FIND: (a) Temperature of sensor for prescribed conditions, (b) Effect of surface emissivity and using a
fan to induce air flow over the tube.
SCHEMATIC:
ASSUMPTIONS: (1) Room walls (surroundings) much larger than tube, (2) Tube is diffuse, gray
surface, (3) No losses from tube by conduction, (4) Steady-state conditions, (5) Sensor measures
temperature of tube surface.
ANALYSIS: (a) Performing an energy balance on the tube, E in − E out = 0 . Hence, q′′rad − q′′conv = 0 ,
4
or ε tσ (Tw
− Tt4 ) − h(Tt − Tw ) = 0 . With h = 5 W m 2⋅ K and εt = 0.8, the energy balance becomes
0.8 × 5.67 × 10−8 W m 2 ⋅ K 4 (30 + 273) − Tt4  K 4 = 5 W m 2 ⋅ K [Tt − (20 + 273)] K


4
4.5360 × 10−8 3034 − Tt4  = 5 [Tt − 293]


<
which yields Tt = 298 K = 25oC.
(b) Using the IHT First Law Model, the following results were determined.
Sensor temperature, Tt(C)
30
28
26
24
22
20
0
5
10
15
20
25
Convection coefficient, h(W/m^2.K)
epst = 0.8
epst = 0.5
epst = 0.2
The sensor temperature exceeds the air temperature due to radiation absorption, which must be balanced
by convection heat transfer. Hence, the excess temperature Tt − T∞ , may be reduced by increasing h or
by decreasing αt, which equals εt for a diffuse-gray surface, and hence the absorbed radiation.
COMMENTS: A fan will increase the air velocity over the sensor and thereby increase the convection
heat transfer coefficient. Hence, the sensor will indicate a temperature closer to T∞
PROBLEM 12.77
KNOWN: Diffuse-gray sphere is placed in large oven with known wall temperature and experiences
convection process.
FIND: (a) Net heat transfer rate to the sphere when its temperature is 300 K, (b) Steady-state temperature
of the sphere, (c) Time required for the sphere, initially at 300 K, to come within 20 K of the steady-state
temperature, and (d) Elapsed time of part (c) as a function of the convection coefficient for 10 ≤ h ≤ 25
W/ m2⋅K for emissivities 0.2, 0.4 and 0.8.
SCHEMATIC:
ASSUMPTIONS: (1) Sphere surface is diffuse-gray, (2) Sphere area is much smaller than the oven wall
area, (3) Sphere surface is isothermal.
PROPERTIES: Sphere (Given) : α = 7.25 × 10-5 m2 /s, k = 185 W/m⋅K.
ANALYSIS: (a) From an energy balance on the sphere find
q net = qin − q out
q net = α GAs + q conv − EAs
q net = ασ To4 As + hAs ( T∞ − Ts ) − εσ Ts4 As .
(1)
Note that the irradiation to the sphere is the emissive power of a blackbody at the temperature of the oven
walls. This follows since the oven walls are isothermal and have a much larger area than the sphere area.
Substituting numerical values, noting that α = ε since the surface is diffuse-gray and that As = πD2 , find

q net = 0.8 × 5.67 × 10−8 W m 2 ⋅ K

4
(600K )4
+ 15 W m 2⋅ K × ( 400 − 300 ) K
4
− 0.8 × 5.67 × 10 −8 W m 2 ⋅ K 300 K
#
4
π 30 × 10 −3 m
q net = [16.6 + 4.2 − 1.0] W = 19.8 W .
%
2
(1)
<
(b) For steady-state conditions, qnet in the energy balance of Eq. (1) will be zero,
4
0 = ασ To4 As + hAs ( T∞ − Tss ) − εσ Tss
As
(2)
Substitute numerical values and find the steady-state temperature as
Tss = 538.2K
<
Continued...
PROBLEM 12.77 (Cont.)
(c) Using the IHT Lumped Capacitance Model considering convection and radiation processes, the
temperature- time history of the sphere, initially at Ts (0) = Ti = 300 K, can be determined. The elapsed
time required to reach
Ts ( t o ) = (538.2 − 20 ) K = 518.2K
was found as
<
t o = 855s = 14.3 min
(d) Using the IHT model of part (c), the elapsed time for the sphere to reach within 20 K of its steadystate temperature, tf , as a function of the convection coefficient for selected emissivities is plotted below.
Time-to-reach within 20 K of steady-state temperature
2500
Time, tf (s)
2000
1500
1000
500
10
15
20
25
Convection coefficient, h (W/m^2.K)
eps = 0.2
eps = 0.4
eps = 0.8
For a fixed convection coefficient, tf increases with decreasing ε since the radiant heat transfer into the
sphere decreases with decreasing emissivity. For a given emissivity, the tf decreases with increasing h
since the convection heat rate increases with increasing h. However, the effect is much more significant
with lower values of emissivity.
COMMENTS: (1) Why is tf more strongly dependent on h for a lower sphere emissivity? Hint:
Compare the relative heat rates by convection and radiation processes.
(2) The steady-state temperature, Tss , as a function of the convection coefficient for selected
emmissivities calculated using (2) is plotted below. Are these results consistent with the above plot of tf
vs h ?
Steady-state temperature, Tss (K)
600
550
500
450
400
10
15
20
Convection coefficient, h (W/m^2.K)
eps = 0.2
eps = 0.4
eps = 0.8
25
PROBLEM 12.78
2
KNOWN: Thermograph with spectral response in 9 to 12 µm region views a target of area 200mm
with solid angle 0.001 sr in a normal direction.
FIND: (a) For a black surface at 60°C, the emissive power in 9 – 12 µm spectral band, (b) Radiant
power (W), received by thermograph when viewing black target at 60°C, (c) Radiant power (W)
received by thermograph when viewing a gray, diffuse target having ε = 0.7 and considering the
surroundings at Tsur = 23°C.
SCHEMATIC:
ASSUMPTIONS: (1) Wall is diffuse, (2) Surroundings are black with Tsur = 23°C.
ANALYSIS: (a) Emissive power in spectral range 9 to 12 µm for a 60°C black surface is
E t ≡ E b (9 − 12 µ m ) = E b  F ( 0 → 12 µ m ) − F ( 0 − 9 µ m )
where E b ( Ts ) = σ Ts4 . From Table 12.1:
λ2 Ts = 12 × ( 60 + 273 ) ≈ 4000 µ m K,
F ( 0− 12 µ m ) = 0.491
λ1 Ts = 9 × ( 60 + 273) ≈ 3000 µ m K,
F ( 0 − 9 µ m ) = 0.273.
Hence
E t = 5.667 ×10−8 W / m 2 ⋅ K4 × ( 60 + 273 ) K 4 [ 0.491 − 0.273 ] = 144.9 W / m 2 .
4
<
(b) The radiant power, qb (w), received by the thermograph from a black target is determined as
E
q b = t ⋅ A s cosθ1 ⋅ω
π
Et = emissive power in 9 – 12 µm spectral region, part (a) result
2
-4 2
As = target area viewed by thermograph, 200mm (2 × 10 m )
ω = solid angle thermograph aperture subtends when viewed
from the target, 0.001 sr
θ = angle between target area normal and view direction, 0°.
where
Hence,
qb =
)
144.9 W / m2
× 2 ×10−4 m 2 × cos0 ° × 0.001sr = 9.23 µ W.
π sr
(
<
Continued …..
PROBLEM 12.78 (Cont.)
(c) When the target is a gray, diffuse emitter, ε = 0.7, the thermograph will receive emitted power from
the target and reflected irradiation resulting from the surroundings at Tsur = 23°C. Schematically:
The power is expressed as
q = q e + q r = ε q b + I r ⋅ As cosθ1 ⋅ω  F( 0→12 µ m) − F( 0→9 µm ) 


where
qb = radiant power from black surface, part (b) result
F(0 - λ) = band emission fraction for Tsur = 23°C; using Table 12.1
λ2 Tsur = 12 × (23 + 273) = 3552 µm⋅K, F( 0 − λ ) = 0.394
2
λ1 Tsur = 9 × (23 + 273) = 2664 µm⋅K, F( 0 − λ ) = 0.197
1
Ir = reflected intensity, which because of diffuse nature of surface
E (T )
G
I r = ρ = (1 − ε ) b sur .
π
π
Hence
q = 0.7 × 9.23 µ W + (1 − 0.7 )
(
5.667 ×10−8 W / m2 ⋅ K4 × ( 273 + 23) 4 K
)
π sr
× 2 × 10−4 m 2 × cos0° × 0.001sr [ 0.394 − 0.197]
q = 6.46 µ W + 1.64 µW = 8.10 µW.
<
COMMENTS: (1) Comparing the results of parts (a) and (b), note that the power to the
thermograph is slightly less for the gray surface with ε = 0.7. From part (b) see that the effect of the
irradiation is substantial; that is, 1.64/8.10 ≈ 20% of the power received by the thermograph is due to
reflected irradiation. Ignoring such effects leads to misinterpretation of temperature measurements
using thermography.
(2) Many thermography devices have a spectral response in the 3 to 5 µm wavelength region as well
as 9 – 12 µm.
PROBLEM 12.79
KNOWN: Radiation thermometer (RT) viewing a steel billet being heated in a furnace.
FIND: Temperature of the billet when the RT indicates 1160K.
SCHEMATIC:
ASSUMPTIONS: (1) Billet is diffuse-gray, (2) Billet is small object in large enclosure, (3) Furnace
behaves as isothermal, large enclosure, (4) RT is a radiometer sensitive to total (rather than a
prescribed spectral band) radiation and is calibrated to correctly indicate the temperature of a black
body, (5) RT receives radiant power originating from the target area on the billet.
ANALYSIS: The radiant power reaching the radiation thermometer (RT) is proportional to the
radiosity of the billet. For the diffuse-gray billet within the large enclosure (furnace), the radiosity is
J = ε E b ( T ) + ρ G = ε E b ( T ) + ( 1− ε ) E b ( Tw )
4
J = ε σ T 4 + (1 − ε ) σ Tw
(1)
4
where α = ε, G = Eb (Tw) and Eb = σ T . When viewing the billet, the RT indicates Ta = 1100K,
referred to as the apparent temperature of the billet. That is, the RT indicates the billet is a blackbody
at Ta for which the radiosity will be
E b ( Ta ) = J a = σ Ta4.
(2)
Recognizing that Ja = J, set Eqs. (1) and (2) equal to one another and solve for T, the billet true
temperature.
1/4
1−ε 4 
1
T =  Ta4 −
Tw 
ε
ε

.
Substituting numerical values, find
1/4
1 − 0.8
 1
4
T =  (1160K ) 4 −
1500K
(
)  = 999K.
0.8
 0.8

<
COMMENTS: (1) The effect of the reflected wall irradiation from the billet is to cause the RT to
indicate a temperature higher than the true temperature.
(2) What temperature would the RT indicate when viewing the furnace wall assuming the wall
emissivity were 0.85?
(3) What temperature would the RT indicate if the RT were sensitive to spectral radiation at 0.65 µm
instead of total radiation? Hint: in Eqs. (1) and (2) replace the emissive power terms with spectral
intensity. Answer: 1365K.
PROBLEM 12.80
KNOWN: Irradiation and temperature of a small surface.
FIND: Rate at which radiation is received by a detector due to emission and reflection from the
surface.
SCHEMATIC:
ASSUMPTIONS: (1) Opaque, diffuse-gray surface behavior, (2) As and Ad may be approximated
as differential areas.
ANALYSIS: Radiation intercepted by the detector is due to emission and reflection from the surface,
and from the definition of the intensity, it may be expressed as
qs −d = Ie + rA s cosθ ∆ω .
The solid angle intercepted by Ad with respect to a point on As is
∆ω =
Ad
r2
= 10− 6 sr.
Since the surface is diffuse it follows from Eq. 12.24 that
I e+ r =
J
π
where, since the surface is opaque and gray (ε = α = 1 - ρ),
J = E + ρ G = ε Eb + (1 − ε ) G.
Substituting for Eb from Eq. 12.28
J = ε σ Ts4 + (1 − ε ) G = 0.7 × 5.67 ×10 −8
W
m2 ⋅ K
500K )4 + 0.3 ×1500
(
4
W
m2
or
J = ( 2481 + 450 ) W / m 2 = 2931W/m 2.
Hence
I e+ r =
and
2 9 3 1 W / m2
= 933W/m 2 ⋅sr
π sr
(
)
q s− d = 933 W / m 2 ⋅ sr 10 −4 m2 × 0.866 10 −6 sr = 8.08 × 10 −8 W.
<
PROBLEM 12.81
KNOWN: Small, diffuse, gray block with ε = 0.92 at 35°C is located within a large oven whose walls
are at 175°C with ε = 0.85.
FIND: Radiant power reaching detector when viewing (a) a deep hole in the block and (b) an area on
the block’s surface.
SCHEMATIC:
ASSUMPTIONS: (1) Block is isothermal, diffuse, gray and small compared to the enclosure, (2)
Oven is isothermal enclosure.
ANALYSIS: (a) The small, deep hole in the isothermal block approximates a blackbody at Ts. The
radiant power to the detector can be determined from Eq. 12.54 written in the form:
σ Ts4
q = Ie ⋅At ⋅ωt =
⋅ At ⋅ ω t
π
1 
4 W
q=
5.67 ×10 −8 × (35 + 273 ) 
×
π sr 
 m 2
(
π 3 ×10−3
)
2
m2
4
× 0.001sr = 1.15 µ W
<
where A t = π D 2t /4. Note that the hole diameter must be greater than 3mm diameter.
(b) When the detector views an area on the surface of the block, the radiant power reaching the
detector will be due to emission and reflected irradiation originating from the enclosure walls. In terms
of the radiosity, Section 12.24, we can write using Eq. 12.24,
q = Ie+ r ⋅ At ⋅ ωt =
J
⋅ A t ⋅ ω t.
π
Since the surface is diffuse and gray, the radiosity can be expressed as
J = ε E b ( Ts ) + ρ G = ε E b ( Ts ) + (1 − ε ) E b ( Tsur )
recognizing that ρ = 1 - ε and G = Eb (Tsur). The radiant power is
q=
q=
1
ε E b ( Ts ) + (1 − ε ) E b ( Tsur )  ⋅ At ⋅ ωt
π
1 
4
4
0.92 × 5.67 ×10−8 ( 35 + 273 ) + (1 − 0.92 ) × 5.67 ×10 −8 (175 + 273)  W / m 2 ×


π sr 
(
π 3 ×10 −3
4
) m 2 × 0.001sr = 1.47 µ W.
2
<
COMMENTS: The effect of reflected irradiation when ε < 1 is important for objects in enclosures.
The practical application is one of measuring temperature by radiation from objects within furnaces.
PROBLEM 12.82
KNOWN: Diffuse, gray opaque disk (1) coaxial with a ring-shaped disk (2), both with prescribed
temperatures and emissivities. Cooled detector disk (3), also coaxially positioned at a prescribed
location.
FIND: Rate at which radiation is incident on the detector due to emission and reflection from A1.
SCHEMATIC:
2
ASSUMPTIONS: (1) A1 is diffuse-gray, (2) A2 is black, (3) A1 and A3 << R , the distance of
separation, (4) ∆r << ri, such that A2 ≈ 2 π ri ∆r, and (5) Backside of A2 is insulated.
ANALYSIS: The radiant power leaving A1 intercepted by A3 is of the form
q1→3 = ( J1 / π ) A1 cosθ1 ⋅ω 3−1
where for this configuration of A1 and A3,
ω3 −1 = A 3 cos θ3 / ( LA + LB )
θ1 = 0°
2
θ3 = 0 °.
Hence,
q1→3 = ( J1 / π ) A1 ⋅A 3 / ( LA + LB ) 2
J1 = ρG1 + ε E b ( T1 ) = ρ G1 + εσ T14.
The irradiation on A1 due to emission from A2, G1, is
G1 = q 2 →1 / A1 = ( I 2 ⋅ A 2 cos θ 2′ ⋅ ω1−2 ) / A1
where
ω1− 2 = A 1cos θ1′ / R 2
is constant over the surface A2. From geometry,
θ1′ = θ2′ = tan −1 (r i + ∆r / 2 ) / L A  = tan −1 ( 0.500 + 0.005 ) /1.000  = 26.8°
R = L A /cos θ1′ = 1m/cos26.8 ° = 1.12m.
Hence,
)
(
2
G1 = σ T24 / π A 2 cos26.8°⋅  A1 cos26.8° / (1.12m )  / A1 = 360.2 W / m 2


-2
2
using A2 = 2πri∆r = 3.142 × 10 m and
J1 = (1 − 0.3 ) × 360.2 W / m 2 + 0.3 × 5.67 ×10− 8 W / m 2 ⋅ K 4 ( 400 K ) = 687.7 W / m 2.
4
Hence the radiant power is
(
)
2
2
2
q1→3 = 687.7 W / m 2 / π π ( 0.010 m ) / 4  / (1 m + 1 m ) = 337.6× 10 −9 W.


<
PROBLEM 12.83
KNOWN: Area and emissivity of opaque sample in hemispherical enclosure. Area and position of
detector which views sample through an aperture. Sample and enclosure temperatures.
FIND: (a) Detector irradiation, (b) Spectral distribution and maximum intensities.
SCHEMATIC:
ASSUMPTIONS: (1) Diffuse-gray surfaces, (2) Hemispherical enclosure forms a blackbody cavity
about the sample, Ah >> As, (3) Detector field of view is limited to sample surface.
ANALYSIS: (a) The irradiation can be evaluated as Gd = qs-d/Ad and qs-d = Is(e+r) As ωd-s.
2
2
2
-5
Evaluating parameters: ωd-s ≈ Ad/L = 2 mm /(300 mm) = 2.22 × 10 sr, find
4
2⋅ 4
× −8
Es ε sσ Ts4 0.1 5.67 10 W / m K ( 400 K )
I s( e ) =
=
=
= 46.2 W / m 2 ⋅ sr
(
π
π
(1 − ε s ) σ Th
ρ G
=
I s( r ) = s s =
π
π
4
)
π sr
(
)
0.9 5.67 × 10−8 W / m 2 ⋅ K 4 ( 273 K )
4
π sr
(
= 90.2 W / m 2 ⋅ sr
)
qs −d = ( 46.2 + 90.2 ) W / m 2 ⋅ sr 5× 10−6 m 2 × 2.22 ×10 −5 sr = 1.51× 10−8 W
Gd = 1.51×10 −8 W / 2 ×10 −6 m 2 = 7.57 ×10 −3 W / m 2.
<
(b) Since λmax T = 2898 µm⋅K, it follows that λmax(e) = 2898 µm⋅K/400 K = 7.25 µm and λmax(r) = 2898
µm⋅K/273 K = 10.62 µm.
<
-4
-8
5
2
λ = 7.25 µm: Table 12.1 → Iλ,b(400 K) = 0.722 × 10 (5.67 × 10 )(400) = 41.9 W/m ⋅µm⋅sr
-4
-8
5
2
Iλ,b(273 K) = 0.48 × 10 (5.67 × 10 )(273) = 4.1 W/m ⋅µm⋅sr
2
Iλ = Iλ,e + Iλ,r = ε sIλ,b(400 K) + ρIλ,b(273K) = 0.1 × 41.9 + 0.9 × 4.1 = 7.90 W/m ⋅µm⋅sr
-4
-8
5
<
2
λ = 10.62 µm: Table 12.1 → Iλ,b(400 K) = 0.53 × 10 (5.67 × 10 )(400) = 30.9 W/m ⋅µm⋅sr
-4
-8
5
2
Iλ,b(273 K) = 0.722 × 10 (5.67 × 10 )(273) = 6.2 W/m ⋅µm⋅sr
2
Iλ = 0.1 × 30.9 + 0.9 × 6.2 = 8.68 W/m ⋅µm⋅sr.
<
COMMENTS: Although Th is substantially smaller than Ts, the high sample reflectivity renders the
reflected component of Js comparable to the emitted component.
PROBLEM 12.84
KNOWN: Sample at Ts = 700 K with ring-shaped cold shield viewed normally by a radiation detector.
FIND: (a) Shield temperature, Tsh, required so that its emitted radiation is 1% of the total radiant power
received by the detector, and (b) Compute and plot Tsh as a function of the sample emissivity for the
range 0.05 ≤ ε ≤ 0.35 subject to the parametric constraint that the radiation emitted from the cold shield
is 0.05, 1 or 1.5% of the total radiation received by the detector.
SCHEMATIC:
ASSUMPTIONS: (1) Sample is diffuse and gray, (2) Cold shield is black, and (3) A d , Ds2 , D 2t << L2t .
ANALYSIS: (a) The radiant power intercepted by the detector from within the target area is
q d = qs → d + qsh → d
The contribution from the sample is
qs → d = Is,e As cos θ s ∆ω d − s
Is,e = ε s E b π = ε s σ Ts4 π
A cos θ d Ad
∆ω d −s = d
=
L2t
L2t
θ s = 0$
θ d = 0$
qs → d = ε sσ Ts4 As Ad π L2t
(1)
The contribution from the ring-shaped cold shield is
qsh → d = Ish,e A sh cos θ sh ∆ω d − sh
4
π
Ish,e = E b π = σ Tsh
and, from the geometry of the shield -detector,
π
Ash =
D 2t − Ds2
4
(
)
12
cos θ sh = L t ( D 2 ) = L2t 


2
Continued...
PROBLEM 12.84 (Cont.)
where D = ( Ds + D t ) 2
∆ω d −sh =
Ad cos θ d
cos θ d = cos θ sh
R2
12
where
R =  L2t + D 2 


2
4
σ Tsh
qsh → d =
Ash
π




Lt
Ad


1
2
 ( (D + D ) 4 )2 + L2   ( (Ds + D t ) 4 )2 + L2t 


t
t 
  s
 
(2)
The requirement that the emitted radiation from the cold shield is 1% of the total radiation intercepted by
the detector is expressed as
qsh − d
q tot
=
qsh − d
qsh − d + qs − d
= 0.01
(3)
By evaluating Eq. (3) using Eqs. (1) and (3), find
<
Tsh = 134 K
(b) Using the foregoing equations in the IHT workspace, the required shield temperature for qsh - d/qtot =
0.5, 1 or 1.5% was computed and plotted as a function of the sample emissivity.
Shield temperature, Th (K)
250
200
150
100
50
0.05
0.15
0.25
0.35
Sample emissivity, epss
Shield /total radiant power = 0.5 %
1.0%
1.5 %
As the shield emission-to-total radiant power ratio decreases ( from 1.5 to 0.5% ) , the required shield
temperature decreases. The required shield temperature increases with increasing sample emissivity for a
fixed ratio.
PROBLEM 12.84
KNOWN: Sample at Ts = 700 K with ring-shaped cold shield viewed normally by a radiation detector.
FIND: (a) Shield temperature, Tsh, required so that its emitted radiation is 1% of the total radiant power
received by the detector, and (b) Compute and plot Tsh as a function of the sample emissivity for the
range 0.05 ≤ ε ≤ 0.35 subject to the parametric constraint that the radiation emitted from the cold shield
is 0.05, 1 or 1.5% of the total radiation received by the detector.
SCHEMATIC:
ASSUMPTIONS: (1) Sample is diffuse and gray, (2) Cold shield is black, and (3) A d , Ds2 , D 2t << L2t .
ANALYSIS: (a) The radiant power intercepted by the detector from within the target area is
q d = qs → d + qsh → d
The contribution from the sample is
qs → d = Is,e As cos θ s ∆ω d − s
Is,e = ε s E b π = ε s σ Ts4 π
A cos θ d Ad
∆ω d −s = d
=
L2t
L2t
θ s = 0$
θ d = 0$
qs → d = ε sσ Ts4 As Ad π L2t
(1)
The contribution from the ring-shaped cold shield is
qsh → d = Ish,e A sh cos θ sh ∆ω d − sh
4
π
Ish,e = E b π = σ Tsh
and, from the geometry of the shield -detector,
π
Ash =
D 2t − Ds2
4
(
)
12
cos θ sh = L t ( D 2 ) = L2t 


2
Continued...
PROBLEM 12.84 (Cont.)
where D = ( Ds + D t ) 2
∆ω d −sh =
Ad cos θ d
cos θ d = cos θ sh
R2
12
where
R =  L2t + D 2 


2
4
σ Tsh
qsh → d =
Ash
π




Lt
Ad


1
2
 ( (D + D ) 4 )2 + L2   ( (Ds + D t ) 4 )2 + L2t 


t
t 
  s
 
(2)
The requirement that the emitted radiation from the cold shield is 1% of the total radiation intercepted by
the detector is expressed as
qsh − d
q tot
=
qsh − d
qsh − d + qs − d
= 0.01
(3)
By evaluating Eq. (3) using Eqs. (1) and (3), find
<
Tsh = 134 K
(b) Using the foregoing equations in the IHT workspace, the required shield temperature for qsh - d/qtot =
0.5, 1 or 1.5% was computed and plotted as a function of the sample emissivity.
Shield temperature, Th (K)
250
200
150
100
50
0.05
0.15
0.25
0.35
Sample emissivity, epss
Shield /total radiant power = 0.5 %
1.0%
1.5 %
As the shield emission-to-total radiant power ratio decreases ( from 1.5 to 0.5% ) , the required shield
temperature decreases. The required shield temperature increases with increasing sample emissivity for a
fixed ratio.
PROBLEM 12.85
KNOWN: Infrared scanner (radiometer) with a 3- to 5-micrometer spectral bandpass views a metal plate
maintained at Ts = 327°C having four diffuse, gray coatings of different emissivities. Surroundings at
Tsur = 87°C.
FIND: (a) Expression for the scanner output signal, So, in terms of the responsivity, R (µV⋅m2/W), the
black coating (εo = 1) emissive power and appropriate band emission fractions; assuming R = 1
µV⋅m2/W, evaluate So(V); (b) Expression for the output signal, Sc, in terms of the responsivity R, the
blackbody emissive power of the coating, the blackbody emissive power of the surroundings, the coating
emissivity, εc, and appropriate band emission fractions; (c) Scanner signals, Sc (µV), when viewing with
emissivities of 0.8, 0.5 and 0.2 assuming R = 1 µV⋅m2/W; and (d) Apparent temperatures which the
scanner will indicate based upon the signals found in part (c) for each of the three coatings.
SCHEMATIC:
ASSUMPTIONS: (1) Plate has uniform temperature, (2) Surroundings are isothermal and large
compared to the plate, and (3) Coatings are diffuse and gray so that ε = α and ρ = 1 - ε.
ANALYSIS: (a) When viewing the black coating (εo = 1), the scanner output signal can be expressed as
So = RF( λ − λ ,T )E b ( Ts )
1 2 s
(1)
where R is the responsivity (µV⋅m2/W), Eb(Ts) is the blackbody emissive power at Ts and F( λ − λ ,T ) is
1 2 s
the fraction of the spectral band between λ1 and λ2 in the spectrum for a blackbody at Ts,
F( λ − λ ,T ) = F(0 − λ ,T ) − F(0 − λ ,T )
1 2 s
2 s
1 s
(2)
where the band fractions Eq. 12.38 are evaluated using Table 12.1 with λ1Ts = 3 µm (327 + 273)K =
1800 µm ⋅ K and λ2Ts = 5 µm (327 + 273) = 3000 µm ⋅ K . Substituting numerical values with R = 1
µV⋅m2/W, find
So = 1µ V⋅ m 2 W [0.2732 − 0.0393] 5.67 × 10−8 W m 2⋅ K
4
(600K )4
<
So = 1718 µ V
(b) When viewing one of the coatings (εc < εo = 1), the scanner output signal as illustrated in the
schematic above will be affected by the emission and reflected irradiation from the surroundings,
{
Sc = R F(λ − λ ,T )ε c E b ( Ts ) + F(λ − λ ,T ) ρc G c
1 2 s
1 2 sur
}
(3)
where the reflected irradiation parameters are
Continued...
PROBLEM 12.85 (Cont.)
4
G c = σ Tsur
ρc = 1 − ε c
(4,5)
and the related band fractions are
F( λ − λ ,T ) = F( 0 − λ ,T
− F( 0 − λ ,T )
1 sur
Combining Eqs. (2-6) above, the scanner output signal when viewing a coating is
1
2 sur )
2 sur
{
Sc = R  F(0 − λ T ) − F(0 − λ T )  ε cσ Ts4 +  F(0 − λ T ) − F(0 − λ T )  (1 − ε c )σ T 4
2 s
1 s 
2 sur
2 sur 


sur
(6)
}
(7)
(c) Substituting numerical values into Eq. (7), find
{
Sc = 1 µ V⋅ m 2 W [0.2732 − 0.0393]ε cσ ( 600K ) + [0.0393 − 0.0010](1 − ε c )σ (360K )
4
4
}
where for λ2Tsur = 5 µm × 360 K = 1800 µm⋅K, F( 0 − λ T ) = 0.0393 and λ1Tsur = 3 µm × 360 K = 1080
2 sur
µm⋅K, F( 0 − λ T
1 sur )
= 0.0010. For εc = 0.80, find
<
<
<
Sc (ε c = 0.8 ) = 1 µ V⋅ m 2 W {1375 + 7.295} W m 2 = 1382 µ V
Sc (ε c = 0.5 ) = 1 µ V⋅ m 2 W {859.4 + 18.238} W m 2 = 878 µ V
Sc (ε c = 0.2 ) = 1 µ V⋅ m 2 W {343.8 + 29.180} W m 2 = 373 µ V
(d) The scanner calibrated against a black surface (ε1 = 1) interprets the radiation reaching the detector by
emission and reflected radiation from a coating target (εc < εo ) as that from a blackbody at an apparent
temperature Ta. That is,
(8)
Sc = RF(λ − λ ,T )E b ( Ta ) = R  F( 0 − λ T ) − F( 0 − λ T )  σ Ta4
1 2 a
2 a
1 a 

For each of the coatings in part (c), solving Eq. (8) using the IHT workspace with the Radiation Tool,
Band Emission Factor, the following results were obtained,
εc
0.8
0.5
0.2
Sc (µV)
1382
878
373
Ta (K)
579.3
539.2
476.7
Ta - Ts (K)
-20.7
-60.8
-123.3
COMMENTS: (1) From part (c) results for Sc, note that the contribution of the reflected irradiation
becomes relatively more significant with lower values of εc.
(2) From part (d) results for the apparent temperature, note that the error, (T - Ta), becomes larger with
decreasing εc. By rewriting Eq. (8) to include the emissivity of the coating,
S′c = R  F( 0 − λ T ) − F( 0 − λ T )  ε cσ Ta4
2 a
1 a 

The apparent temperature Ta′ will be influenced only by the reflected irradiation. The results correcting
only for the emissivity, εc, are
εc
Ta′ K
0.8
600.5
0.5
602.2
0.2
608.5
Ta′ − Ts (K)
+0.5
+2.2
+8.5
$
PROBLEM 12.86
KNOWN: Billet at Tt = 500 K which is diffuse, gray with emissivity εt = 0.9 heated within a large
furnace having isothermal walls at Tf = 750 K with diffuse, gray surface of emissivity εf = 0.8.
Radiation detector with sensitive area Ad = 5.0 × 10-4 m2 positioned normal to and at a distance R = 0.5 m
from the billet. Detector receives radiation from a billet target area At = 3.0 × 10-6 m2.
FIND: (a) Symbolic expressions and numerical values for the following radiation parameters associated
with the target surface (t): irradiation on the target, Gt; intensity of the reflected irradiation leaving the
target, It,ref; emissive power of the target, Et; intensity of the emitted radiation leaving the target, It,emit; and
radiosity of the target Jt; and (b) Expression and numerical value for the radiation which leaves the target
in the spectral region λ ≥ 4 µm and is intercepted by the radiation detector, qt→d; write the expression in
terms of the target reflected and emitted intensities It,ref and It,emit, respectively, as well as other geometric
and radiation parameters.
SCHEMATIC:
ASSUMPTIONS: (1) Furnace wall is isothermal and large compared to the billet, (2) Billet surface is
diffuse gray, and (3) At, Ad << R2.
ANALYSIS: (a) Expressions and numerical values for radiation parameters associated with the target
are:
Irradiation, Gt: due to blackbody emission from the furnace walls which are isothermal and large
relative to the billet target,
G t = E b ( Tf ) = σ Tf4 = 5.67 × 10−8 W m 2⋅ K 4 ( 750 K ) = 17, 940 W m 2
4
<
Intensity of reflected irradiation, It,ref: since the billet is diffuse, I = Gi/π from Eq. 12.19, and diffusegray, ρt = 1 - εt,
I t,ref = ρ t G t π = (1 − ε t ) G t π
I t,ref = (1 − 0.9 ) × 19, 740 W m 2 π = 571W m 2 ⋅ sr
<
Emissive power, Et: from the Stefan-Boltzmann law, Eq. 12.28, and the definition of the total emissivity,
Eq. 12.37,
E t = ε t E b ( Tt ) = ε tσ Tt4
E t = 0.9 × 5.67 × 10−8 W m 2⋅ K 4 (500 K ) = 3189 W m 2
4
<
Continued...
PROBLEM 12.86 (Cont.)
Intensity of emitted radiation, It,emit: since the billet is diffuse, Ie = E/π from Eq. 12.14,
I t,emit = E t π = 3189 W m 2 π = 1015 W m 2 ⋅ sr
<
Radiosity, Jt: the radiosity accounts for the emitted radiation and reflected portion of the irradiation; for
the diffuse surface, from Eq. 12.24,
(
J t = π I t,ref + I t,emit
)
J t = π sr (571 + 1015 ) W m 2 ⋅ sr = 4983 W m 2
<
(b) The radiant power in the spectral region λ ≥ 4 µm leaving the target which is intercepted by the
detector follows from Eq. 12.5,
q t → d =  Fref I t,ref + Femit I t,emit  A t cos θ tω d − t
The F factors account for the fraction of the total spectral region for λ ≥ 4 µm,
Fref = 1 − F ( 0 − λ Tf ) = 1 − 0.2732 = 0.727
Femit = 1 − F (0 − λ Tt ) = 1 − 0.06673 = 0.933
where from Eq. 12.30 and Table 12.1, for λTf = 4 µm × 750 K = 3000 µm⋅K, F(0 - λTf) = 0.2732 and for
λTt = 4 µm × 500 K = 2000 µm⋅K, F(0 - λTt) = 0.06673. Since the radiation detector is normal to the
billet, cos θt = 1. The solid angle subtended by the detector area with respect to the target area is
ωd − t =
Ad cos θ d
R
2
=
5 × 10 −4 m 2 × 1
(0.5 m )
2
= 2.00 × 10−3 sr
Hence, the radiant power is
q t → d = ( 0.727 × 571 + 0.933 × 1015 ) W m 2 ⋅ sr × 3.0 × 10−6 m 2 × 1 × 2.00 × 10−3 sr
q t → d = ( 2.491 + 5.682 ) × 10−6 W = 8.17 µ W
<
COMMENTS: (1) Why doesn’t the emissivity of the furnace walls, εf, affect the target irradiation?
(2) Note the importance of the diffuse, gray assumption for the billet target surface. In what ways was
the assumption used in the analysis?
(3) From the calculation of the radiant power to the detector, qt→d, note that the contribution of the
reflected irradiation is nearly a third of the total.
PROBLEM 12.87
KNOWN: Painted plate located inside a large enclosure being heated by an infrared lamp bank.
FIND: (a) Lamp irradiation required to maintain plot at Ts = 140oC for the prescribed convection and
enclosure irradiation conditions, (b) Compute and plot the lamp irradiation, Glamp, required as a function
of the plate temperature, Ts, for the range 100 ≤ Ts ≤ 300 oC and for convection coefficients of h = 15, 20
and 30 W/m2⋅K, and (c) Compute and plot the air stream temperature, T∞ , required to maintain the plate
at 140oC as a function of the convection coefficient h for the range 10 ≤ h ≤ 30 W/m2⋅K with a lamp
irradiation Glamp = 3000 W/m2.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) No losses on backside of plate.
ANALYSIS: (a) Perform an energy balance on the plate, per unit area,
E in − E out = 0
(1)
α wall ⋅ G wall + α lamp G lamp − q′′conv − Es = 0
(2)
where the emissive power of the surface and convective fluxes are
E s = ε s E b (Ts ) = ε s ⋅ σ Ts4
q′′conv = h(Ts − T∞ )
(3,4)
Substituting values, find the lamp irradiation
0.7 × 450 W m 2 + 0.6 × G lamp − 20 W m 2 ⋅ K(413 − 300 K
−0.8 × 5.67 × 10−8 W m 2 ⋅ K 4 (413 K)4 = 0
(5)
<
Glamp = 5441 W/m2
(b) Using the foregoing equations in the IHT workspace, the irradiation, Glamp, required to maintain the
plate temperature in the range 100 ≤ Ts ≤ 300 oC for selected convection coefficients was computed. The
results are plotted below.
Irradiation, Glamp (W/m^2)
20000
16000
12000
8000
4000
0
100
200
300
Plate temperature, Ts (C)
h= 15 W/m^2.K
h = 20 W/m^2.K
h = 30 W/m^2.K
Continued...
PROBLEM 12.87 (Cont.)
As expected, to maintain the plate at higher temperatures, the lamp irradiation must be increased. At any
plate operating temperature condition, the lamp irradiation must be increased if the convection
coefficient increases. With forced convection (say, h ≥ 20 W/m2⋅K) of the airstream at 27oC, excessive
irradiation levels are required to maintain the plate above the cure temperature of 140oC.
(c) Using the IHT model developed for part (b), the airstream temperature, T∞ , required to maintain the
plate at Ts = 140oC as a function of the convection coefficient with Glamp = 3000 W/m2⋅K was computed
and the results are plotted below.
Air temperature, Tinf (C)
120
100
80
60
10
20
30
Convection coefficient, h (W/m^2.K)
As the convection coefficient increases, for example by increasing the airstream velocity over the plate,
the required air temperature must increase. Give a physical explanation for why this is so.
COMMENTS: (1) For a spectrally selective surface, we should expect the absorptivity to depend upon
the spectral distribution of the source and α ≠ ε.
(2) Note the new terms used in this problem; use your Glossary, Section 12.9 to reinforce their meaning.
PROBLEM 12.88
KNOWN: Small sample of reflectivity, ρ λ, and diameter, D, is irradiated with an isothermal enclosure
at Tf.
FIND: (a) Absorptivity, α, of the sample with prescribed ρ λ, (b) Emissivity, ε, of the sample, (c) Heat
removed by coolant to the sample, (d) Explanation of why system provides a measure of ρ λ.
SCHEMATIC:
ASSUMPTIONS: (1) Sample is diffuse and opaque, (2) Furnace is an isothermal enclosure with area
much larger than the sample, (3) Aperture of furnace is small.
ANALYSIS: (a) The absorptivity, α, follows from Eq. 12.42, where the irradiation on the sample is G
= Eb (Tf) and α λ = 1 - ρ λ.
∞
∞
α=
αλ G λ d λ / G =
(1 − ρλ )Eλ ,b ( λ,1000K ) dλ / E b (1000K)
0
0
∫
(
∫
)
(
)
α = 1 − ρλ ,1 F(0 →λ ) + 1 − ρλ ,2 1 − F(0 →λ )  .
1
1 

Using Table 12.1 for λ1 Tf = 4 × 1000 = 4000 µm⋅K, F(0-λ) = 0.491 giving
α = (1 − 0.2) × 0.491 + (1 − 0.8) × (1 − 0.491) = 0.49.
<
(b) The emissivity, ε, follows from Eq. 12.37 with ε λ = α λ = 1 - ρ λ since the sample is diffuse.
∞
ε = E ( Ts ) / E b ( Ts ) =
ε E
( λ,300K ) dλ / Eb ( 300K )
0 λ λ ,b
∫
(
)
(
)
ε = 1 − ρλ ,1 F(0 − λ ) + 1 − ρλ ,2 1 − F(0 →λ )  .
1
1 

Using Table 12.1 for λ1 Ts = 4 × 300 = 1200 µm⋅K, F(0-λ) = 0.002 giving
ε = (1 − 0.2) × 0.002 + (1 − 0.8 ) × (1 − 0.002 ) = 0.20.
(c) Performing an energy balance on the sample, the
heat removal rate by the cooling water is
q cool = A s α G + q′′conv − ε E b ( Ts ) 
where
G = Eb ( Tf ) = E b (1000K )
q′′conv = h ( Tf − Ts ) As = π D2 / 4
2
4
q cool = (π / 4 )( 0.03m ) 0.49 × 5.67 ×10 −8 W / m 2 ⋅ K 4 × (1000K )

+1 0 W / m2 ⋅ K (1000 − 300 ) K − 0.20 × 5.67 × 10 −8 W / m 2 ⋅ K 4 × ( 300K )4  = 24.4 W.

(d) Assume that reflection makes the dominant contribution to the radiosity of the sample. When
viewing in the direction A, the spectral radiant power is proportional to ρ λ Gλ. In direction B, the
spectral radiant power is proportional to Eλ,b (Tf). Noting that Gλ = Eλ,b (Tf), the ratio gives ρ λ.
<
PROBLEM 12.89
KNOWN: Small, opaque surface initially at 1200 K with prescribed α λ distribution placed in a large
enclosure at 2400 K.
FIND: (a) Total, hemispherical absorptivity of the sample surface, (b) Total, hemispherical emissivity,
(c) α and ε after long time has elapsed, (d) Variation of sample temperature with time.
SCHEMATIC:
ASSUMPTIONS: (1) Surface is diffusely radiated, (2) Enclosure is much larger than surface and at a
uniform temperature.
PROPERTIES: Table A.1, Tungsten (T ≈ 1800 K): ρ = 19,300 kg/m3, cp = 163 J/kg⋅K, k ≈ 102
W/m⋅K.
ANALYSIS: (a) The total, hemispherical absorptivity follows from Eq. 12.46, where
G λ = E λ ,b ( Tsur ) . That is, the irradiation corresponds to the spectral emissive power of a blackbody at
the enclosure temperature and is independent of the enclosure emissivity.
∞
α = ∫ α λ G λ dλ
0
∞
∫0
∞
G λ dλ = ∫ α λ E λ ,b ( λ , Tsur ) dλ E b ( Tsur )
0
2µ m
∞
4
4
E λ ,b ( λ , Tsur ) dλ σ Tsur
E
+ α2
(λ , Tsur ) dλ σ Tsur
0
2 µ m λ ,b
α = α1 ∫
∫
α = α1F(0→ 2 µ m) + α 2 1 − F(0 → 2 µ m)  = 0.1 × 0.6076 + 0.8[1 − 0.6076] = 0.375
<
where at λ T = 2 × 2400 = 4800 µ m ⋅ K, F(0→ 2 µ m) = 0.6076 from Table 12.1.
(b) The total, hemispherical emissivity follows from Eq. 12.38,
∞
ε = ∫ ε λ E λ ,b (λ , Ts ) dλ
0
∞
∫0
E λ ,b (λ , Ts )dλ .
Since the surface is diffuse, ε λ = α λ and the integral can be expressed as
2µ m
ε = α1 ∫
0
∞
E λ ,b (λ , Ts ) dλ σ Ts4 + α 2 ∫
E λ ,b (λ , Ts ) dλ σ Ts4
2µm
ε = α1F(0 → 2 µ m) + α 2 1 − F(0 → 2 µ m)  = 0.1 × 0.1403 + 0.8[1 − 0.1403] = 0.702
<
where at λT = 2 × 1200 = 2400 µm ⋅ K, find F(0→ 2 µ m) = 0.1403 from Table 12.1.
(c) After a long period of time, the surface will be at the temperature of the enclosure. This condition of
thermal equilibrium is described by Kirchoff’s law, for which
ε = α = 0.375.
<
Continued...
PROBLEM 12.89 (Cont.)
(d) Using the IHT Lumped Capacitance Model, the energy balance relation is of the form
ρ cp ∀
dT
dt
= As [α G − ε (T)E b (T)]
4
where T = Ts, ∀ = π D3 6 , As = π D 2 and G = σ Tsur
. Integrating over time in increments of ∆t =
0.5s and using the Radiation Toolpad to determine ε(t),
the following results are obtained.
2400
1
0.8
Radiative properties
Temperature, T(K)
2200
2000
1800
1600
1400
0.6
0.4
0.2
0
0
1200
10
20
30
40
50
60
Time, t(s)
0
10
20
30
Time, t(s)
40
50
60
Absorptivity, alpha
Emissivity, eps
The temperature of the specimen increases rapidly with time and achieves a value of 2399 K within t ≈
47s. The emissivity decreases with increasing time, approaching the absorptivity as T approaches Tsur.
COMMENTS: (1) Recognize that α always depends upon the spectral irradiation distribution, which, in
this case, corresponds to emission from a blackbody at the temperature of the enclosure.
2
3
(2) With h r = εσ (T + Tsur )(T 2 + Tsur
) = 0.375σ 4Tsur
= 1176 W m 2⋅ K , Bi = h r (ro /3) k
= (1176 W m 2⋅ K) .667 ×10−3 m 102 W m ⋅ K = 0.0192 << 1 , use of the lumped capacitance model is
justified.
PROBLEM 12.90
KNOWN: Vertical plate of height L = 2 m suspended in quiescent air. Exposed surface with diffuse
coating of prescribed spectral absorptivity distribution subjected to simulated solar irradiation, GS,λ.
Plate steady-state temperature Ts = 400 K.
FIND: (a) Plate emissivity, ε, plate absorptivity, α, plate irradiation, G, and using an appropriate
correlation, the free convection coefficient, h , and (b) Plate steady-state temperature if the irradiation
found in part (a) were doubled.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Ambient air is extensive, quiescent, (3) Spectral
distribution of the simulated solar irradiation, GS,λ , proportional to that of a blackbody at 5800 K, (4)
Coating is opaque, diffuse, and (5) Plate is perfectly insulated on the edges and the back side, and (6)
Plate is isothermal.
PROPERTIES: Table A.4, Air (Tf = 350 K, 1 atm): ν = 20.92 × 10-6 m2/s, k = 0.030 W/m⋅K, α = 29.90
× 10-6 m2/s, Pr = 0.700.
ANALYSIS: (a) Perform an energy balance on the plate
as shown in the schematic on a per unit plate width
basis,
E in − E out = 0
α G − εσ T 4 − h ( T − T ) L = 0
∞ 
s
s

(1)
where α and ε are determined from knowledge of αλ and h is estimated from an appropriate correlation.
Plate total emmissivity: From Eq. 12.38 written in terms of the band emission factor, F(0 - λT), Eq. 12.30,
ε = α1F(0 − λ T ) + α 2 1 − F( 0 − λ T ) 
1 s
1 s 

ε = 0.9 × 0 + 0.1[1 − 0] = 0.1
<
where, from Table 12.1, with λ,Ts = 1µm × 400 K = 400 µm⋅K, F(0-λT) = 0.000.
Plate absorptivity: With the spectral distribution of simulated solar irradiation proportional to Eb (Ts =
5800 K),
Continued...
PROBLEM 12.90 (Cont.)
α = α1F(0 − λ T ) + α 2 1 − F( 0 − λ T ) 
1 s
1 s 

<
α = 0.9 × 0.7202 + 0.1[1 − 0.7202] = 0.676
where, from Table 12.1, with λ1Ts = 5800 µm⋅K, F(0 -λT) = 0.7202.
Estimating the free convection coefficient, h : Using the Churchill-Chu correlation Eq. (9.26) with
properties evaluated at Tf = (Ts + T∞ )/2 = 350 K,
Ra L =
Ra L =
gβ ( Ts − T∞ ) L3
να
3
9.8 m s 2 (1 350 K ) × 100 K ( 2 m )
20.92 × 10
−6
m
2
s × 29.90 × 10
−6
2
= 3.581 × 1010
m s


6


0.387Ra1/
L
Nu L = 0.825 +
8 27 


1 + ( 0.492 Pr )9 16 




2
2


6


0.387Ra1/
L
Nu L = 0.825 +
= 377.6
8 27 
9
16


1 + ( 0.492 0.700 )





h L = Nu L k L = 377.6 × 0.030 W m ⋅ K 2 m = 5.66 W m 2⋅ K
<
Irradiation on the Plate: Substituting numerical values into Eq. (1), find G.
0.676G − 0.1 × 5.67 × 10−8 W m 2⋅ K
4
( 400 K )4 −5.66 W
m 2 ⋅ K ( 400 − 300 ) K = 0
<
G = 1052 W m 2
(b) If the irradiation were doubled, G = 2104 W/m2, the plate temperature Ts will increase, of course,
requiring re-evaluation of ε and h . Since α depends upon the irradiation distribution, and not the plate
temperature, α will remain the same. As a first approximation, assume ε = 0.1 and h = 5.66 W/m2⋅K and
with G = 2104 W/m2, use Eq. (1) to find
<
Ts ≈ 492 K
With Tf = (Ts + T∞ )/2 = (492 + 300)K/2 ≈ 400 K, use the correlation to reevaluate h . For Ts = 492 K, ε
= 0.1 is yet a good assumption. We used IHT with the foregoing equations in part (a) and found these
results.
Ts = 477K
Tf = 388.5
h = 6.38 W m 2 ⋅ K
ε = 0.1
<
PROBLEM 12.91
KNOWN: Diameter and initial temperature of copper rod. Wall and gas temperature.
FIND: (a) Expression for initial rate of change of rod temperature, (b) Initial rate for prescribed
conditions, (c) Transient response of rod temperature.
SCHEMATIC:
ASSUMPTIONS: (1) Applicability of lumped capacitance approximation, (2) Furnace approximates a
blackbody cavity, (3) Thin film is diffuse and has negligible thermal resistance, (4) Properties of nitrogen
approximate those of air (Part c).
PROPERTIES: Table A.1, copper (T = 300 K): cp = 385 J/kg⋅K, ρ = 8933 kg/m3, k = 401 W/m⋅K.
Table A.4, nitrogen (p = 1 atm, Tf = 900 K): ν = 100.3 × 10-6 m2/s, α = 139 × 10-6 m2/s, k = 0.0597
W/m⋅K, Pr = 0.721.
ANALYSIS: (a) Applying conservation of energy at an instant of time to a control surface about the
cylinder, E in − E out = E st , where energy inflow is due to natural convection and radiation from the
furnace wall and energy outflow is due to emission. Hence, for a unit cylinder length,
q conv + q rad,net =
where
dT
ρπ D 2
cp
4
dt
q conv = h (π D )( T∞ − T )
q rad,net = π D (α G − ε E b ) = π D [α E b ( Tw ) − ε E b ( T )]
Hence, at t = 0 (T = Ti),
(
)
dT dt )i = 4 ρ cp D [ h ( T∞ − Ti ) + α E b ( Tw ) − ε E b ( Ti )]
gβ ( T∞ − Ti ) D3 9.8 m s 2 (1 900 K )(1200 K )( 0.01m )
(b) With Ra D =
=
= 937 , Eq. (9.34) yields
αν
100.3 × 139 × 10−12 m 4 s2
3
2


6


0.387Ra1/
D
Nu D = 0.60 +
= 2.58
8 / 27 
9
/16


1 + ( 0.559 Pr )





h=k
Nu D
D
=
(0.0597 W
m⋅ K ) 2.58
0.01m
= 15.4 W m 2⋅ K
With T = Ti = 300 K, λT = 600 µm⋅K yields F(0→λ) = 0, in which case ε = ε1F( 0 → λ ) + ε 2 1 − F( 0 → λ )  = 0.4 .
With T = Tw = 1500 K, λT = 3000 K yields F(0→λ) = 0.273. Hence, with α λ = ε λ , α = ε1F(0→λ) + ε2[1 F(0→λ)] = 0.9(0.273) + 0.4(1 - 0.273) = 0.537. It follows that
Continued...
PROBLEM 12.91 (Cont.)
dT 
 =
dt i

W
15 2 (1500 − 300 ) K

 m ⋅K
8933
385
 0.01m
3 
⋅
kg
K

m
4
kg 
J
+0.537 × 5.67 × 10−4
W
m2 ⋅ K
(1500 K )4 − 0.4 × 5.67 ×10−8
4
dT dt )i = 1.163 × 10−4 m 2 ⋅ K J [18, 480 + 154,140 − 180] W m 2 = 20 K s
W
m2 ⋅ K

(300 K )4 
4

<
Defining a pseudo radiation coefficient as hr = (αG - εEb)/(Tw - Ti) = (153,960 W/m2)/1200 K = 128.3
W/m2⋅K, Bi = (h + hr)(D/4)/k = 143.7 W/m2⋅K (0.0025 m)/401 W/m⋅K = 0.0009. Hence, the lumped
capacitance approximation is appropriate.
(c) Using the IHT Lumped Capacitance Model with the Correlations, Radiation and Properties (copper
and air) Toolpads, the transient response of the rod was computed for 300 ≤ T < 1200 K, where the upper
limit was determined by the temperature range of the copper property table.
1200
1100
Temperature, T(K)
1000
900
800
700
600
500
400
300
0
10
20
30
40
50
60
Time, t(s)
The rate of change of the rod temperature, dT/dt, decreases with increasing temperature, in accordance
with a reduction in the convective and net radiative heating rates. Note, however, that even at T ≈ 1200
K, αG, which is fixed, is large relative to q′′conv and εEb and dT/dt is still significant.
PROBLEM 12.92
KNOWN: Temperatures of furnace wall and top and bottom surfaces of a planar sample.
Dimensions and emissivity of sample.
FIND: (a) Sample thermal conductivity, (b) Validity of assuming uniform bottom surface temperature.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in sample, (3)
Constant k, (4) Diffuse-gray surface, (5) Irradiation equal to blackbody emission at 1400K.
PROPERTIES: Table A-6, Water coolant (300K): cp,c = 4179 J/kg⋅K
ANALYSIS: (a) From energy balance at top surface,
′′
α G − E = q cond
= k s ( Ts − T c ) / L
where E = εsσ Ts4 , G = σ Tw4 , α = ε s giving
4
εsσ Tw
− εs σ Ts4 = ks ( Ts − Tc ) /L.
Solving for the thermal conductivity and substituting numerical values, find
(
)
ks =
εs L σ
4 − T4
Tw
s
Ts − Tc
ks =
0.85 × 0.015m × 5.67 × 10−8 W / m 2 ⋅ K 4 
(1400K ) 4 − (1000K ) 4 


(1000 − 300) K
k s = 2.93 W / m ⋅ K.
<
(b) Non-uniformity of bottom surface temperature depends
on coolant temperature rise. From the energy balance
& c c p,c∆Tc = (α G − E ) W 2
q=m
∆Tc = 0.85 × 5.67 ×10−8 W / m2 ⋅ K 4 14004

2
4
4
−1000  K ( 0.10m ) /0.1kg/s × 4179 J / k g ⋅ K

∆Tc = 3.3K.
The variation in Tc (~ 3K) is small compared to (Ts – Tc) ≈ 700K. Hence it is not large enough to
introduce significant error in the k determination.
<
PROBLEM 12.93
KNOWN: Thicknesses and thermal conductivities of a ceramic/metal composite. Emissivity of ceramic
surface. Temperatures of vacuum chamber wall and substrate lower surface. Receiving area of radiation
detector, distance of detector from sample, and sample surface area viewed by detector.
FIND: (a) Ceramic top surface temperature and heat flux, (b) Rate at which radiation emitted by the
ceramic is intercepted by detector, (c) Effect of an interfacial (ceramic/substrate) contact resistance on
sample top and bottom surface temperatures.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional, steady-state conduction in sample, (2) Constant properties, (3)
Chamber forms a blackbody enclosure at Tw, (4) Ceramic surface is diffuse/gray, (5) Negligible interface
contact resistance for part (a).
PROPERTIES: Ceramic: kc = 60 W/m⋅K, εc = 0.8. Substrate: ks = 25 W/m⋅K.
ANALYSIS: (a) From an energy balance at the exposed ceramic surface, q′′cond = q′′rad , or
( Ls
T1 − T2
k s ) + ( Lc k c )
(
4
= ε cσ T24 − Tw
)
1500 K − T2
= 0.8 × 5.67 × 10−8 W m 2⋅ K 4 T24 − 90 4 K 4
0.008 m
0.005 m
+
25 W m⋅ K 60 W m⋅ K
(
)
3.72 × 106 − 2479T2 = 4.54 × 10−8 T24 − 2.98
4.54 × 10 −8 T24 + 2479T2 = 3.72 × 106
Solving, we obtain
<
<
T2 = 1425 K
q′′h =
T1 − T2
( Ls
k s ) + ( Lc k c )
(1500 − 1425 ) K
=
4.033 × 10
−4
= 1.87 × 105 W m2
2
m ⋅K W
(b) Since the ceramic surface is diffuse, the total intensity of radiation emitted in all directions is Ie =
εcEb(Ts)/π. Hence, the rate at which emitted radiation is intercepted by the detector is
(
q c( em )− d = Ie ∆Ac A d L2s − d
)
4
q c( em )− d =
0.8 × 5.67 × 10−8 W m 2⋅ K 4 (1425 K )
π sr
× 10−4 m 2 × 10−5 sr = 5.95 × 10−5 W
Continued...
PROBLEM 12.93 (Cont.)
(c) With the development of an interfacial thermal contact resistance and fixed values of q′′h and Tw, (i)
(
)
4
T2 remains the same (its value is determined by the requirement that q′′h = ε cσ T24 − Tw
, while (ii) T1
increases (its value is determined by the requirement that q′′h = ( T1 − T2 ) R ′′tot , where R ′′tot = [(Ls/ks) +
R ′′t,c + (Lc/kc)]; if q′′h and T2 are fixed, T1 must increase with increasing R ′′tot ).
COMMENTS: The detector will also see radiation which is reflected from the ceramic. The
corresponding radiation rate is qc(reflection)-d = ρc G c ∆A c A d L2s − d = 0.2 σ(90 K)4 × 10-4 m2 × (10-5 sr) =
7.44 × 10-10 W. Hence, reflection is negligible.
PROBLEM 12.94
KNOWN: Wafer heated by ion beam source within large process-gas chamber with walls at uniform
temperature; radiometer views a 5 × 5 mm target on the wafer. Black panel mounted in place of wafer
in a pre-production test of the equipment.
FIND: (a) Radiant power (µW) received by the radiometer when the black panel temperature is Tbp
= 800 K and (b) Temperature of the wafer, Tw, when the ion beam source is adjusted so that the
radiant power received by the radiometer is the same as that of part (a)
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Chamber represents large, isothermal
surroundings, (3) Wafer is opaque, diffuse-gray, and (4) Target area << square of distance between
target and radiometer objective.
ANALYSIS: (a) The radiant power leaving the black-panel target and reaching the radiometer as
illustrated in the schematic below is
4 9
q bp − rad = E b,bp Tbp / π A t cos θ t ⋅ ∆ω rad − t
(1)
where θt = 0° and the solid angle the radiometer subtends with respect to the target follows from Eq.
12.2,
π D2o / 4 π 0.025 m 2 / 4
dA n
∆ω rad − t =
=
=
= 1.964 × 10−3 sr
2
2
2
r
r
4
9 1
6
10.500 m6
4 , find
With E b,bp = σTbp
!
1
6
q bp − rad = 5.67 × 10−8 W / m2 ⋅ K4 800 K 4 / π sr
1
"#
$
6
2
× 0.005 m × cos 30°×1.964 × 10-3 sr
<
q bp − rad = 314 µW
Continued …..
PROBLEM 12.94 (Cont.)
(b) With the wafer mounted, the ion beam source is adjusted until the radiometer receives the same
radiant power as with part (a) for the black panel. The power reaching the radiometer is expressed in
terms of the wafer radiosity,
q w − rad = J w / π A t cos θ t ⋅ ∆ω rad − t
(2)
Since q w − rad = q bp − rad (see Eq. (1)), recognize that
4 9
J w = E b,bp Tbp
(3)
where the radiosity is
1 6
1 6 1
6 1 6
J w = ε w E b,w Tw + ρ w G w = ε w E b,w Tw + 1 − ε w E b Tch
(4)
and Gw is equal to the blackbody emissive power at Tch. Using Eqs. (3) and (4) and substituting
numerical values, find
1
6
4 = ε σT4 + 1 − ε σT4
σTbp
w w
w
ch
1800 K64 = 0.7 Tw4 + 0.31400 K64
Tw = 871 K
<
COMMENTS: (1) Explain why Tw is higher than 800 K, the temperature of the black panel, when
the radiometer receives the same radiant power for both situations.
(2) If the chamber walls were cold relative to the wafer, say near liquid nitrogen temperature, Tch = 80
K, and the test repeated with the same indicated radiometer power, is the wafer temperature higher or
lower than 871 K?
(3) If the chamber walls were maintained at 800 K, and the test repeated with the same indicated
radiometer power, what is the wafer temperature?
PROBLEM 12.95
KNOWN: Spectrally selective workpiece placed in an oven with walls at Tf = 1000 K experiencing
convection with air at T∞ = 600 K.
FIND: (a) Steady-state temperature, Ts, by performing an energy balance on the workpiece; show
control surface identifying all relevant processes; (b) Compute and plot Ts as a function of the
convection coefficient h, for the range 10 ≤ h ≤ 120 W m 2⋅K ; on the same plot, show the behavior for
diffuse surfaces of emissivity 0.2 and 0.8.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Convection coefficient constant, independent of
temperature, (3) Workpiece is diffusely irradiated by oven wall which is large isothermal surroundings,
(4) Spectral emissivity independent of workpiece temperature.
ANALYSIS: (a) Performing an energy balance on the workpiece,
E in − E out = α G − ε E b − q ′′conv = 0
α E b (Tf ) − ε E b (Ts ) − h(Ts − T∞ ) = 0
and
G = Eb (Tf),
(1)
(2)
<
where the total absorptivity is, with α λ = ε λ ,
(
)
α = F(0 →5µ m,1000 K) ⋅ ε1 + 1 − F(0 →5µ m,1000 K) ε 2 = 0.6337 × 0.2 + (1 − 0.6337) × 0.8 = 0.419
using Table 12.1 with λT = 5 × 1000 = 5000 µm⋅K for which F(0 − λ T) = 0.6337. The total emissivity is,
(
)
(3)
ε = F(0 →5 µ m⋅T )ε1 + 1 − F(0 →5×T ) ⋅ ε 2
s
s
which requires knowing Ts. Hence, an iterative solution is required, beginning by assuming a value of Ts
to find ε using Eq. (3), and then using that value in Eq. (2) to find Ts. The result is,
<
Ts = 800 K
for which ε = 0.512.
(b) Using the IHT workspace with the foregoing equations and the Radiation Exchange Tool, Band
Emission Factor, a model was developed to calculate Ts as a function of the convection coefficient.
Additionally, Ts was plotted for the cases when the workpiece is diffuse, gray with ε = 0.2 and 0.8
Continued...
PROBLEM 12.95 (Cont.)
Temperature, Ts (K)
1000
900
800
700
600
0
20
40
60
80
100
120
Convection coefficient, h (W/m^2.K)
Gray surface, eps = 0.2
Selective surface
Gray surface, eps = 0.8
For the workpiece with the selective surface ( ε λ as shown in the schematic), the temperature decreases
with increasing convection coefficient. For the gray surface, ε = 0.2 or 0.8, the temperature is lower and
higher, respectively, than that of the workpiece. Recall from part (c), at h = 60 W m 2⋅K , ε = 0.512 and
α = 0.419, so that it is understandable why the curve for the workpiece is between that for the two gray
surfaces.
COMMENTS: (1) For the conditions in part (b), make a sketch of the workpiece emissivity and the
absorptivity as a function of its temperature.
(2) The IHT workspace model used to generate the plot is shown below. Note how we used this model to
also calculate Ts vs h for the gray surfaces by adjusting λ1.
// Energy Balance:
alpha * Gf - eps * Ebs - h * (Ts - Tinf) = 0
Gf = Ebf
// Irradiation from furnace, W/m^2
Ebf = sigma * Tf^4
// Blackbody emissive power, W/m^2; furnace wall
Ebs = sigma * Ts^4
// Blackbody emissive power, W/m^2; workpiece
sigma = 5.67e-8
// Stefan-Boltzmann constant, W/m^2.K^4
// Radiation Tool - Band Emission Factor, Total emissivity and absorptivity
eps = FL1Ts * eps1 + (1 - FL1Ts) * eps2
/* The blackbody band emission factor, Figure 12.14 and Table 12.1, is */
FL1Ts = F_lambda_T(lambda1,Ts)
// Eq 12.30
// where units are lambda (micrometers, mum) and T (K)
alpha = FL1Tf * eps1 + (1 - FL1Tf) * eps2
FL1Tf = F_lambda_T(lambda1,Tf)
// Eq 12.30
// Assigned Variables:
Ts > 0
Tf = 1000
Tinf = 600
h = 60
eps1 = 0.2
eps2 = 0.8
//lambda1 = 5
//lambda1 = 1e6
lambda1 = 0.5
// Workpiece temperature, K; assures positive value for Ts
// Furnace wall temperature, K
// Air temperature, K
// Convection coefficient, W/m^2.K
// Spectral emissivity, 0 <= epsL <= 5 micrometers
// Spectral emissivity, 5 <= epsL <= infinity
// Wavelength, micrometers; spectrally selective workpiece
// Wavelength; gray surface for which eps = 0.2
// Wavelength; gray surface for which eps = 0.8
PROBLEM 12.96
KNOWN: Laser-materials-processing apparatus. Spectrally selective sample heated to the operating
temperature Ts = 2000 K by laser irradiation ( 0.5 µm ), Glaser, experiences convection with an inert gas
and radiation exchange with the enclosure.
FIND: (a) Total emissivity of the sample, ε ; (b) Total absorptivity of the sample, α, for irradiation from
the enclosure; (c) Laser irradiation required to maintain the sample at Ts = 2000 K by performing an
energy balance on the sample; (d) Sketch of the sample emissivity during the cool-down process when
the laser and inert gas flow are deactivated; identify key features including the emissivity for the final
condition (t → ∞ ); and (e) Time-to-cool the sample from the operating condition at Ts (0) = 2000 K to a
safe-to-touch temperature of Ts (t) = 40°C; use the lumped capacitance method and include the effects of
convection with inert gas ( T∞ = 300 K , h = 50 W/ m2⋅K) as well as radiation exchange Tenc = T∞ .
SCHEMATIC:
ASSUMPTIONS: (1) Enclosure is isothermal and large compared to the sample, (2) Sample is opaque
and diffuse, but spectrally selective, so that ελ = αλ, (3) Sample is isothermal.
PROPERTIES: Sample (Given) ρ = 3900 kg/m3, cp = 760 J/kg , k = 45 W/m⋅K.
ANALYSIS: (a) The total emissivity of the sample, ε, at Ts = 2000 K follows from Eq. 12.38 which can
be expressed in terms of the band emission factor, F(0-λ,T) Eq. 12.30,
ε = ε1F(0 − λ T ) + ε 2 1 − F( 0 − λ T ) 
(1)
ε = 0.8 × 0.7378 + 0.2 [1 − 0.7378] = 0.643
<
1 s

1 s

where from Table 12.1, with λ1Ts = 3µm × 2000 K = 6000 µm⋅K, F(0-λT) = 0.7378.
(b) The total absorptivity of the sample, α, for irradiation from the enclosure at Tenc = 300 K, is
α = ε1F(0 − λ T ) + ε 2 1 − F( 0 − λ T ) 
1 enc
1 enc 

(2)
α = 0.8 × 0 + 0.2 [1 − 0] = 0.200
<
where, from Table 12.1, with λ1Tenc = 3 µm × 300 K = 900 µm⋅K, F(0-λT) =0.
Continued...
PROBLEM 12.96 (Cont.)
(c) The energy balance on the sample, on a per unit area
basis, as shown in the schematic at the right is
E in − E out = 0
+α las G laser + 2α G − 2ε E b ( Ts ) − q′′cv = 0
4
α las G laser + 2ασ Tenc
− 2ετ Ts4 − 2h ( Ts − T∞ ) = 0
(3)
Recognizing that αlas(0.5 µm) = 0.8, and substituting numerical values find,
0.8 × G laser + 2 × 0.200 × 5.67 × 10−8 W m 2⋅ K
−2 × 0.643 × 5.67 × 10 −8 W m 2⋅ K
4
4
(300 K )4
( 2000 K )4 −2 × 50 W
m 2⋅ K ( 2000 − 500 ) K = 0
0.8 × G laser =  −184.6 + 1.167 × 106 + 1.500 × 105  W m 2


<
G laser = 1646 kW m 2
(d) During the cool-down process, the total
emissivity ε will decrease as the temperature
decreases, Ts (t). In the limit, t → ∞, the sample
will reach that of the enclosure, Ts (∞) = Tenc for
which ε = α = 0.200.
(e) Using the IHT Lumped Capacitance Model
considering radiation exchange (Tenc = 300 K)
and convection ( T∞ = 300 K, h = 50 W/m2⋅K)
and evaluating the emissivity using Eq. (1) with
the Radiation Tool, Band Emission Factors, the
temperature-time history was determined and
the time-to-cool to T(t) = 40°C was found as
<
t = 119 s
0 .7
0 .7
0 .6
0 .6
E m is s ivity, e p s
E m is s ivity, e p s
COMMENTS: (1) From the IHT model used for part (e), the emissivity as a function of cooling time and sample
temperature were computed and are plotted below. Compare these results to your sketch of part (c).
0 .5
0 .4
0 .3
0 .5
0 .4
0 .3
0 .2
0 .2
0
20
40
C o o lin g tim e , t(s )
60
200
800
1400
2000
S a m p le te m p e ra tu re , T(K )
Continued...
PROBLEM 12.96 (Cont.)
(2) The IHT workspace model to perform the lumped capacitance analysis with variable emissivity is
shown below.
// Lumped Capacitance Model - convection and emission/irradiation radiation processes:
/* Conservation of energy requirement on the control volume, CV. */
Edotin - Edotout = Edotst
Edotin = As * ( + Gabs)
Edotout = As * ( + q''cv + E )
Edotst = rho * vol * cp * Der(T,t)
T_C = T - 273
// Absorbed irradiation from large surroundings on CS
Gabs = alpha * G
G = sigma * Tsur^4
sigma = 5.67e-8
// Stefan-Boltzmann constant, W/m^2·K^4
// Emissive power of CS
E = eps * Eb
Eb = sigma * T^4
//sigma = 5.67e-8
// Stefan-Boltzmann constant, W/m^2·K^4
//Convection heat flux for control surface CS
q''cv = h * ( T - Tinf )
/* The independent variables for this system and their assigned numerical values are */
As = 2 * 1
// surface area, m^2; unit area, top and bottom surfaces
vol = 1 * w
// vol, m^3
w = 0.001
// sample thickness, m
rho = 3900
// density, kg/m^3
cp = 760
// specific heat, J/kg·K
// Convection heat flux, CS
h = 50
// convection coefficient, W/m^2·K
Tinf = 300
// fluid temperature, K
// Emission, CS
//eps = 0.5
// emissivity; value used to test the model initially
// Irradiation from large surroundings, CS
alpha = 0.200
// absorptivity; from Part (b); remains constant during cool-down
Tsur = 300
// surroundings temperature, K
// Radiation Tool - Band emission factor:
eps = eps1 * FL1T + eps2 * ( 1 - FL1T )
/* The blackbody band emission factor, Figure 12.14 and Table 12.1, is */
FL1T = F_lambda_T(lambda1,T) // Eq 12.30
// where units are lambda (micrometers, mum) and T (K)
lambda1 = 3
// wavelength, mum
eps1 = 0.8
// spectral emissivity; for lambda < lambda1
eps2 = 0.2
// spectral emissivity; for lambda > lambda1
PROBLEM 12.97
KNOWN: Cross flow of air over a cylinder placed within a large furnace.
FIND: (a) Steady-state temperature of the cylinder when it is diffuse and gray with ε = 0.5, (b) Steadystate temperature when surface has spectral properties shown below, (c) Steady-state temperature of the
diffuse, gray cylinder if air flow is parallel to the cylindrical axis, (d) Effect of air velocity on cylinder
temperature for conditions of part (a).
SCHEMATIC:
ASSUMPTIONS: (1) Cylinder is isothermal, (2) Furnace walls are isothermal and very large in area
compared to the cylinder, (3) Steady-state conditions.
PROPERTIES: Table A.4, Air (Tf ≈ 600 K): ν = 52.69 × 10−6 m 2 s , k = 46.9 × 10-3 W/m⋅K, Pr =
0.685.
ANALYSIS: (a) When the cylinder surface is gray and diffuse with ε = 0.5, the energy balance is of the
form, q′′rad − q′′conv = 0 . Hence,
4
εσ (Tsur
− Ts4 ) − h(Ts − T∞ ) = 0 .
The heat transfer coefficient, h , can be estimated from the Churchill-Bernstein correlation,
5/8
  Re
 
D
1 +

Nu D = (h D k) = 0.3 +
1/ 4   282, 000 

1 + ( 0.4 Pr )2 / 3 



2 1/ 3
0.62 Re1/
D Pr
4/5
where Re D = V D ν = 3 m s × 30 × 10−3 m 52.69 × 10−6 m 2 s = 1710. Hence,
Nu D = 20.8
h = 20.8 × 46.9 × 10−3 W m ⋅ K 30 × 10−3 m = 32.5 W m 2⋅ K .
Using this value of h in the energy balance expression, we obtain
0.5 × 5.67 × 10−8 (10004 − Ts4 ) W m 2 − 32.5 W m 2⋅ K(Ts − 400) K = 0
<
which yields Ts ≈ 839 K.
(b) When the cylinder has the spectrally selective behavior, the energy balance is written as
α G − ε E b (Ts ) − q′′conv = 0
∞
0
where G = Eb (Tsur). With α = ∫ α λ G λ dλ G ,
(
)
α = 0.1 × F(0 →3) + 0.5 × 1 − F(0 →3) = 0.1 × 0.273 + 0.5(1 − 0.273) = 0.391
where, using Table 12.1 with λT = 3 ×1000 = 3000 µm⋅K, F( 0→3) = 0.273. Assuming Ts is such that
emission in the spectral region λ < 3 µm is negligible, the energy balance becomes
Continued...
PROBLEM 12.97 (Cont.)
0.391 × 5.67 × 10−8 × 10004 W m 2 − 0.5 × 5.67 × 10−8 × Ts4 W m 2 − 32.5 W m 2 ⋅ K(Ts − 400) K = 0
<
which yields Ts ≈ 770 K.
Note that, for λT = 3 × 770 = 2310 µm⋅K, F(0→λ ) ≈ 0.11; hence the assumption of ε = 0.5 is
acceptable. Note also that the value of h based upon Tf = 600 K is also acceptable.
(c) When the cylinder is diffuse-gray with air flow in the longitudinal direction, the characteristic length
for convection is different. Assume conditions can be modeled as flow over a flat plate of L = 150 mm.
With
Re L = V L ν = 3 m s × 150 × 10−3 m 52.69 × 10−6 m 2 s = 8540
2 1/ 3
= 0.664(8540)1/ 2 0.6851/ 3 = 54.1
Nu L = (h L k) = 0.664 Re1/
L Pr
h = 54.1 × 0.0469 W m ⋅ K 0.150 m = 16.9 W m 2⋅ K .
The energy balance now becomes
0.5 × 5.667 × 10−8 W m 2⋅ K 4 (10004 − Ts4 )K 4 − 16.9 W m 2⋅ K(Ts − 400)K = 0
<
which yields Ts ≈ 850 K.
(b) Using the IHT First Law Model with the Correlations and Properties Toolpads, the effect of velocity
may be determined and the results are as follows:
Cylinder temperature, Ts(K)
900
850
800
750
700
650
0
4
8
12
16
20
Air velocity, V(m/s)
Since the convection coefficient increases with increasing V (from 18.5 to 90.6 W/m2⋅K for 1 ≤ V ≤ 20
m/s), the cylinder temperature decreases, since a smaller value of (Ts − T∞ ) is needed to dissipate the
absorbed irradiation by convection.
COMMENTS: The cylinder temperature exceeds the air temperature due to absorption of the incident
radiation. The cylinder temperature would approach T∞ as h → ∞ and/or α → 0. If α → 0 and h has
a small to moderate value, would Ts be larger than, equal to, or less than T∞ ? Why?
PROBLEM 12.98
KNOWN: Instrumentation pod, initially at 87°C, on a conveyor system passes through a large vacuum
brazing furnace. Inner surface of pod surrounded by a mass of phase-change material (PCM). Outer
surface with special diffuse, opaque coating of ε λ. Electronics in pod dissipate 50 W.
FIND: How long before all the PCM changes to the liquid state?
SCHEMATIC:
ASSUMPTIONS: (1) Surface area of furnace walls much larger than that of pod, (2) No convection,
(3) No heat transfer to pod from conveyor, (4) Pod coating is diffuse, opaque, (5) Initially pod internal
temperature is uniform at Tpcm = 87°C and remains so during time interval ∆tm, (6) Surface area
provided is that exposed to walls.
PROPERTIES: Phase-change material, PCM (given): Fusion temperature, Tf = 87°C, hfg = 25
kJ/kg.
ANALYSIS: Perform an energy balance on the pod for an interval of time ∆tm which corresponds to
the time for which the PCM changes from solid to liquid state,
Ein − Eout +Egen = ∆E
(α G − ε E b ) A s + Pe  ∆t m = Mh fg
where P e is the electrical power dissipation
rate, M is the mass of PCM, and hfg is the
heat of fusion of PCM.
Irradiation:
-8
Emissive power:
Emissivity:
Absorptivity:
2
4
4
2
G = σTw = 5.67 × 10 W/m ⋅K (1200 K) = 117,573 W/m
4
Eb = σ Tm
= σ ( 87 +273 ) = 952 W / m 2
4
ε = ε 1F(0-λT) + ε 2(1-F(0-λT))
λT = 5 × 360 = 1800 µm⋅K
ε = 0.05 × 0.0393 + 0.9 (1 – 0.0393)
ε = 0.867
F0-λT = 0.0393
α = α 1 F(0-λT) + α 2(1 –F(0-λT))
λT = 5 × 1200 = 6000 µm⋅K
α = 0.005 × 0.7378 + 0.9 (1 – 0.7378)
α = 0.273
F0-λT = 0.7378
(Table 12.1)
(Table 12.1)
Substituting numerical values into the energy balance, find,
( 0.273 ×117,573 − 0.867 × 952 ) W / m 2 × 0.040 m 2 + 50 W  ∆t = 1.6kg × 25 × 103 J/kg

 m
∆t m = 30.7 s = 0.51min.
<
PROBLEM 12.99
KNOWN: Niobium sphere, levitated in surroundings at 300 K and initially at 300 K, is suddenly
2
irradiated with a laser (10 W/m ) and heated to its melting temperature.
FIND: (a) Time required to reach the melting temperature, (b) Power required from the RF heater
causing uniform volumetric generation to maintain the sphere at the melting temperature, and (c)
Whether the spacewise isothermal sphere assumption is realistic for these conditions.
SCHEMATIC:
ASSUMPTIONS: (1) Niobium sphere is spacewise isothermal and diffuse-gray, (2) Initially sphere is
at uniform temperature Ti, (3) Constant properties, (4) Sphere is small compared to the uniform
temperature surroundings.
PROPERTIES: Table A-1, Niobium (T = (300 + 2741)K/2 = 1520 K): Tmp = 2741 K, ρ = 8570
3
kg/m , cp = 324 J/kg⋅K, k = 72.1 W/m⋅K.
ANALYSIS: (a) Following the methodology of Section 5.3 for general lumped capacitance analysis,
the time required to reach the melting point Tmp may be determined from an energy balance on the
sphere,
4 = Mc dT/dt
E& in − E& out = E& st
q o′′ ⋅ A c − εσ A s T 4 − Tsur
)
p(
2
2
3
(
)
where Ac = πD /4, As = πD , and M = ρV = ρ(πD /6). Hence,
(
) ( )(
) (
)
4 = ρ π D3 / 6 c dT/dt .
q′′o π D2 / 4 − εσ π D 2 T 4 − Tsur
)
p(
Regrouping, setting the limits of integration, and integrating, find
t
T
dT
 q ′′o
4 
4 ρ DcdT
b dt = mp
 4εσ + Tsur  − T = 6εσ dt
0
Ti
a 4 − T4
∫
where a 4 =
b=
(
10 W / mm 2 103 m m / m
∫
(
)
)
2
q′′o
4
4 =
+ Tsur
+ ( 300 K )
4εσ
4 × 0.6 × 5.67 ×10 −8 W / m 2 ⋅ K
a = 2928K
6εσ
6 × 0.6 × 5.67 ×10−8 W / m2 ⋅ K 4
=
= 2.4504 ×10−11 K −1 ⋅ s −1
3
ρ Dcp 8570 k g / m × 0.003 m × 324 J / k g ⋅ K
which from Eq. 5.18, has the solution
t=
 a + Tmp

 Tmp 
a + Ti
−1  Ti
ln
− ln
+ 2  tan−1 

 − tan 
3
a − Ti
 a
4ba  a − Tmp

 a 
1
  
 
  
Continued …..
PROBLEM 12.99 (Cont.)
t=
1
(
−11
4 2.4504 ×10
K
−1 −1
⋅s

3

) (2928 K )
ln
2928 + 2741
2928 + 300
− ln
2928 − 2741
2928 − 300
 −1  2741 
−1  300  
+2  tan 
− tan 

 
 2928 
 2928  

t = 0.40604 ( 3.4117 − 0.2056 + 2 [ 0.7524 − 0.1021] ) = 1.83s.
<
(b) The power required of the RF heater to induce a uniform volumetric generation to sustain steadystate operation at the melting point follows from an energy balance on the sphere,
4 − T 4 = −E&
E& in − E& out +E& gen = 0
−εσ A s Tmp
sur
gen
(
(
)
) (
)
& = 0.6 ×5.67 ×10 −8 W / m 2 ⋅ K 4 π 0.0032 m2 27414 − 300 4 K 4 = 54.3W.
E& gen = qV
<
(c) The lumped capacitance method is appropriate when
h r (D / 6)
h L
< 0.1
Bi = r c =
k
k
where hr is the linearized radiation coefficient, which has the largest value when T = Tmp = 2741 K,
2
h r = εσ ( T + Tsur ) T2 + Tsur
(
)
(
)
h r = 0.6 × 5.67 ×10 −8 W / m 2 ⋅ K 4 ( 2741 + 300 ) 27412 + 300 2 K 3 = 787 W / m 2 ⋅ K.
Hence, since
Bi = 787 W / m 2 ⋅ K ( 0.003m/3 ) / 7 2 . 1 W / m ⋅ K = 1.09 ×10−2
we conclude that the transient analysis using the lumped capacitance method is satisfactory.
<
COMMENTS: (1) Note that at steady-state conditions with internal generation, the difference in
temperature between the center and surface, is
2
&
To = Ts =
3
q ( D / 2)
6k
and with V = πD /6, from the part (b) results,
q& = E& gen / V = 54.3 W / π × 0.0033 / 6 m 3 = 3.841×109 W / m 3 .
(
)
Find using an approximate value for the thermal conductivity in the liquid state,
2
9
3
∆T = To − Ts =
3.841×10 W / m ( 0.03m/2)
= 18K.
6 × 80 W / m ⋅ K
We conclude that the sphere is very nearly isothermal even under these conditions.
(2) The relation for ∆T in the previous comment follows from solving the heat diffusion equation
written for the one-dimensional (spherical) radial coordinate system. See the deviation in Section 3.4.2
for the cylindrical case (Eq. 3.53).
PROBLEM 12.100
KNOWN: Spherical niobium droplet levitated in a vacuum chamber with cool walls. Niobium surface
is diffuse with prescribed spectral emissivity distribution. Melting temperature, Tmp = 2741 K.
FIND: Requirements for maintaining the drop at its melting temperature by two methods of heating: (a)
Uniform internal generation rate, q (W/m3), using a radio frequency (RF) field, and (b) Irradiation, Glaser,
(W/mm2), using a laser beam operating at 0.632 µm; and (c) Time for the drop to cool to 400 K if the
heating method were terminated.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions during the heating processes, (2) Chamber is isothermal
and large relative to the drop, (3) Niobium surface is diffuse but spectral selective, (4) q is uniform, (5)
Laser bean diameter is larger than the droplet, (6) Drop is spacewise isothermal during the cool down.
PROPERTIES: Table A.1, Niobium ( T = (2741 + 400)K/2 ≈ 1600 K): ρ = 8570 kg/m3, cp = 336
J/kg⋅K, k = 75.6 W/m⋅K.
ANALYSIS: (a) For the RF field-method of heating,
perform an energy balance on the drop considering
volumetric generation, irradiation and emission,
E in − E out + E g = 0
[α G − ε E b (Ts )] As + q ∀ = 0
(1)
where As = πD2 and V = πD3/6. The irradiation and
blackbody emissive power are,
4
G = σ Tw
E b = σ Ts4
(2,3)
The absorptivity and emissivity are evaluated using Eqs. 12.46 and 12.48, respectively, with the band
emission fractions, Eq. 12.30, and
α = α (α λ , Tw ) = ε1F ( 0 − λ1Tw ) + ε 2 [1 − F (0 − λ1Tw )]
(4)
α = 0.4 × 0.000 + 0.2 (1 − 0.000 ) = 0.2
where, from Table 12.1, with λ1Tw = 1 µ m × 300 K = 300 µ m ⋅ K , F(0 - λT) = 0.000.
ε = ε (ε λ , Ts ) = ε1F ( 0 − λ1Ts ) + ε 2 [1 − F ( 0 − λ1Ts )]
(5)
ε = 0.4 × 0.2147 + 0.2 (1 − 0.2147 ) = 0.243
with λ1Ts = 1 µ m × 2741K = 2741 µ m ⋅ K , F(0 - λT) = 0.2147. Substituting numerical values with Ts =
Tmp = 2741 K and Tw = 300 K, find
Continued...
PROBLEM 12.100 (Cont.)
0.2 × 5.67 × 10−8 W m 2⋅ K 4 (300 K )4 − 0.243 × 5.67 × 10−8 W m 2⋅ K 4 ( 2741K )4 


π ( 0.003 m ) + q π (0.003 m ) 6 = 0
2
3
<
q = [−91.85 + 777, 724] W m 2 ( 6 0.003 m ) = 1.556 × 109 W m3
(b) For the laser-beam heating method, performing an
energy balance on the drop considering absorbed laser
irradiation, irradiation from the enclosure walls and
emission,
E in − E out = 0
[α G − ε E b (Ts )] As + αlasG laser A p = 0
(6)
where Ap represents the projected area of the droplet,
Ap = π D2 4
(7)
Laser irradiation at 10.6 µm. Recognize that for the laser irradiation, Glaser (10.6 µm), the spectral
absorptivity is
α las (10.6 µ m ) = 0.2
Substituting numerical values onto the energy balance, Eq. (6), find
0.2 × σ × (300 K )4 − 0.243 × σ × ( 2741 K )4  π (0.003 m )2


2
+0.2 × G laser × π ( 0.003 m ) / 4 = 0
G laser (10.6 µ m ) = 1.56 ×107 W / m2 = 15.6 W / mm2
<
Laser irradiation at 0.632 µm. For laser irradiation at 0.632 µm, the spectral absorptivity is
α laser (0.632 µ m ) = 0.4
Substituting numerical values into the energy balance, find
G laser ( 0.632 µ m ) = 7.76 × 106 W / m 2 = 7.8 W / mm 2
(c) With the method of heating terminated, the drop experiences only radiation exchange and begins
cooling. Using the IHT Lumped Capacitance Model with irradiation and emission processes and the
Radiation Tool, Band Emission Factor for estimating the emissivity as a function of drop temperature,
Eq. (5), the time-to-cool to 400 K from an initial temperature, Ts(0) = Tmp = 2741 K was found as
<
Ts ( t ) = 400 K
<
t = 772 s = 12.9 min
COMMENTS: (1) Why doesn’t the emissivity of the chamber wall, εw, affect the irradiation onto the
drop?
(2) The validity of the lumped capacitance method can be determined by evaluating the Biot number,
Continued …..
PROBLEM 12.100 (Cont.)
Bi =
hD 6
k
=
185 W m 2⋅ K × 0.003 m 6
75.6 W m⋅ K
= 0.007
where we estimated an average radiation coefficient as
h rad ≈ 4εσ T 3 = 4 × 0.2 × 5.67 × 10−8 W m 2⋅ K 4 (1600 K ) = 185 W m 2⋅ K
following the estimation method described in Problem 1.20. Since Bi << 0.1, the lumped capacitance
method was appropriate.
3
(3) In the IHT model of part (c), the emissivity was calculated as a function of Ts(t) varying from 0.243
atTs = Tmp to 0.200 at Ts = 300 K. If we had done an analysis assuming the drop were diffuse, gray with
α= ε = 0.2, the time-to-cool would be t = 773 s. How do you explain that this simpler approach predicts
a time-to-cool that is in good agreement with the result of part (c)?
(4) A copy of the IHT workspace with the model of part (c) is shown below.
// Lumped Capacitance Model: Irradiation and Emission
/* Conservation of energy requirement on the control volume, CV. */
Edotin - Edotout = Edotst
Edotin = As * ( + Gabs)
Edotout = As * ( + E )
Edotst = rho * vol * cp * Der(T,t)
// Absorbed irradiation from large surroundings on CS
Gabs = alpha * G
G = sigma * Tsur^4
sigma = 5.67e-8
// Stefan-Boltzmann constant, W/m^2·K^4
// Emissive power of CS
E = eps * Eb
Eb = sigma * T^4
//sigma = 5.67e-8 // Stefan-Boltzmann constant, W/m^2·K^4
/* The independent variables for this system and their assigned numerical values are */
As = pi * D^2
// surface area, m^2
vol = pi * D^3 / 6
// vol, m^3
D = 0.003
// sphere diameter, m
rho = 8570
// density, kg/m^3
cp = 336
// specific heat, J/kg·K
// Emission, CS
//eps = 0.4
// emissivity
// Irradiation from large surroundings, CS
alpha = 0.2
// absorptivity
Tsur = 300
// surroundings temperature, K
// Radiation Tool - Band Emission Fractions
eps = eps1 * FL1T + eps2 * ( 1 - FL1T )
/* The blackbody band emission factor, Figure 12.14 and Table 12.1, is */
FL1T = F_lambda_T(lambda1,T)
// Eq 12.30
// where units are lambda (micrometers, mum) and T (K)
lambda1 = 1
// wavelength, mum
eps1 = 0.4
// spectral emissivity, lambda < lambda1
eps2 = 0.2
// spectral emissivity, lambda > lambda1
PROBLEM 12.101
KNOWN: Temperatures of furnace and surroundings separated by ceramic plate. Maximum allowable
temperature and spectral absorptivity of plate.
FIND: (a) Minimum value of air-side convection coefficient, ho, (b) Effect of ho on plate temperature.
SCHEMATIC:
ASSUMPTIONS: (1) Diffuse surface, (2) Negligible temperature gradients in plate, (3) Negligible
inside convection, (4) Furnace and surroundings act as blackbodies.
ANALYSIS: (a) From a surface energy balance on the plate, α w G w + αsur G sur = 2E + q′′conv . Hence,
4
4
α w σ Tw
+ αsurσ Tsur
= 2εσ Ts4 + h o (Ts − T∞ ) .
4
α σ T 4 + α surσ Tsur
− 2εσ Ts4
ho = w w
(Ts − T∞ )
Evaluating the absorptivities and emissivity,
∞
0
∞
0
α w = ∫ α λ G λ dλ G = ∫ α λ Eλ b (Tw ) E b (Tw )dλ = 0.3F(0 −3µ m) + 0.8 1 − F(0 −3µ m) 
With λ Tw = 3 µ m × 2400 K = 7200 µ m ⋅ K , Table 12.1 → F(0 −3µ m) = 0.819 . Hence,
α w = 0.3 × 0.819 + 0.8(1 − 0.819) = 0.391
Since Tsur = 300 K, irradiation from the surroundings is at wavelengths well above 3 µm. Hence,
∞
αsur = ∫ α λ E λ b (Tsur ) E b (Tsur )dλ ≈ 0.800 .
0
∞
0
The emissivity is ε = ∫ ε λ E λ b (Ts ) E b (Ts )dλ = 0.3F(0 −3µ m) + 0.8 1 − F(0 −3µ m)  . With
λ Ts = 5400 µ m ⋅ K , Table 12.1 → F(0 −3µ m) = 0.680 . Hence, ε = 0.3 × 0.68 + 0.8(1 − 0.68) = 0.460.
For the maximum allowable value of Ts = 1800 K, it follows that
ho =
ho =
0.391 × 5.67 × 10−8 (2400) 4 + 0.8 × 5.67 × 10−8 (300)4 − 2 × 0.46 × 5.67 × 10−8 (1800)4
(1800 − 300)
7.335 × 105 + 3.674 × 10 2 − 5.476 × 105
1500
= 126 W m 2 ⋅ K .
<
(b) Using the IHT First Law Model with the Radiation Toolpad, parametric calculations were performed
to determine the effect of ho.
Continued...
PROBLEM 12.101 (Cont.)
Temperature, T(K)
1900
1800
1700
1600
50
100
150
200
250
Convection coefficient, ho(W/m^2.K)
With increasing ho, and hence enhanced convection heat transfer at the outer surface, the plate
temperature is reduced.
COMMENTS: (1) The surface is not gray. (2) The required value of h o ≥ 126 W m 2⋅ K is well within
the range of air cooling.
PROBLEM 12.102
KNOWN: Spectral radiative properties of thin coating applied to long circular copper rods of prescribed
diameter and initial temperature. Wall and atmosphere conditions of furnace in which rods are inserted.
FIND: (a) Emissivity and absorptivity of the coated rods when their temperature is Ts = 300 K, (b)
Initial rate of change of their temperature, dTs/dt, (c) Emissivity and absorptivity when they reach steadystate temperature, and (d) Time required for the rods, initially at Ts = 300 K, to reach 1000 K.
SCHEMATIC:
ASSUMPTIONS: (1) Rod temperature is uniform, (2) Nitrogen is quiescent, (3) Constant properties,
(4) Diffuse, opaque surface coating, (5) Furnace walls form a blackbody cavity about the cylinders, G =
Eb(Tf), (6) Negligible end effects.
PROPERTIES: Table A.1, Copper (300 K): ρ = 8933 kg/m3, cp = 385 J/kg⋅K, k = 401 W/m⋅K; Table
A.4, Nitrogen (Tf = 800 K, 1 atm): ν = 82.9 × 10-6 m2/s, k = 0.0548 W/m⋅K, α = 116 × 10-6 m2/s, Pr =
0.715, β = (Tf)-1 = 1.25 × 10-3 K-1.
ANALYSIS: (a) The total emissivity of the copper rod, ε, at Ts = 300 K follows from Eq. 12.38 which
can be expressed in terms of the band emission factor, F(0 - λT), Eq. 12.30,
ε = ε1F ( 0 − λ1Ts ) + ε 2 [1 − F ( 0 − λ1Ts )]
(1)
ε = 0.4 × 0.0021 + 0.8 [1 − 0.0021] = 0.799
<
where, from Table 12.1, with λ1Ts = 4 µm × 300 K = 1200 µm⋅K, F(0 - λT) = 0.0021. The total
absorptivity, α, for irradiation for the furnace walls at Tf = 1300 K, is
α = ε1F ( 0 − λ1Tf ) + ε 2 [1 − F ( 0 − λ1Tf )]
(2)
<
α = 0.4 × 0.6590 + 0.8 [1 − 0.6590] = 0.536
where, from Table 12.1, with λ1Tf = 4 µm × 1300 K = 5200 K, F(0 - λT) = 0.6590.
(b) From an energy balance on a control volume about the rod,
(
)
E st = ρ c p π D 2 4 L ( dT dt ) = E in − E out = π DL [α G + h ( T∞ − Ts ) − E ]
dTs dt = 4 α G + h ( T∞ − Ts ) − εσ Ts4  ρ c p D .


With
(
(3)
)
−3 −1
2
gβ ( T∞ − Ts ) D3 9.8 m s 1.25 × 10 K 1000 K ( 0.01m )
=
= 1274
Ra D =
να
82.9 × 10−6 m 2 s × 116 × 10−6 m 2 s
3
(4)
Eq. 9.34 gives
Continued...
PROBLEM 12.102 (Cont.)
2


1/ 6

0.387 (1274 )
0.0548 

2
h=
0.60 +
 = 15.1W m ⋅ K
8
/
27
0.01m 

1 + ( 0.559 0.715 )9 /16 




(5)
With values of ε and α from part (a), the rate of temperature change with time is
4 0.53 × 5.67 × 10

dTs dt =
−8
W m ⋅ K × (1300 K ) + 15.1 W m ⋅ K × 1000 K − 0.8 × 5.67 × 10
2
4
4
2
−8
2
4
W m ⋅ K × (300 K ) 

3
8933 kg m × 385 J kg ⋅ K × 0.01m
dTs dt = 1.16 × 10−4 [85,829 + 15,100 − 3767 ] K s = 11.7 K s .
<
(c) Under steady-state conditions, Ts = T∞ = Tf = 1300 K. For this situation, ε = α, hence
<
ε = α = 0.536
(d) The time required for the rods, initially at Ts(0) = 300 K, to reach 1000 K can be determined using the
lumped capacitance method. Using the IHT Lumped Capacitance Model, considering convection,
irradiation and emission processes; the Correlations Tool, Free Convection, Horizontal Cylinder;
Radiation Tool, Band Emission Fractions; and a user-generated Lookup Table Function for the nitrogen
thermophysical properties, find
<
Ts(to) = 1000 K
to = 81.8 s
COMMENTS: (1) To determine the validity of the lumped capacitance method to this heating process,
evaluate the approximate Biot number, Bi = hD k = 15 W/m2⋅K × 0.010 m/401 W/m⋅K = 0.0004. Since
Bi << 0.1, the method is appropriate.
(2) The IHT workspace with the model used for part (c) is shown below.
// Lumped Capacitance Model - irradiation, emission, convection
/* Conservation of energy requirement on the control volume, CV. */
Edotin - Edotout = Edotst
Edotin = As * ( + Gabs)
Edotout = As * ( + q''cv + E )
Edotst = rho * vol * cp * Der(Ts,t)
//Convection heat flux for control surface CS
q''cv = h * ( Ts - Tinf )
// Emissive power of CS
E = eps * Eb
Eb = sigma * Ts^4
sigma = 5.67e-8
// Stefan-Boltzmann constant, W/m^2·K^4
// Absorbed irradiation from large surroundings on CS
Gabs = alpha * G
G = sigma * Tf^4
/* The independent variables for this system and their assigned numerical values are */
As = pi * D * 1
// surface area, m^2
vol = pi * D^2 / 4 * 1
// vol, m^3
rho = 8933
// density, kg/m^3
cp = 433
// specific heat, J/kg·K; evaluated at 800 K
// Convection heat flux, CS
//h =
// convection coefficient, W/m^2·K
Tinf = 1300
// fluid temperature, K
// Emission, CS
//eps =
// emissivity
// Irradiation from large surroundings, CS
//alpha =
// absorptivity
Tf = 1300
// surroundings temperature, K
Continued...
PROBLEM 12.102 (Cont.)
// Radiative Properties Tool - Band Emission Fraction
eps = eps1 * FL1Ts + eps2 * (1 - FL1Ts)
/* The blackbody band emission factor, Figure 12.14 and Table 12.1, is */
FL1Ts = F_lambda_T(lambda1,Ts)
// Eq 12.30
// where units are lambda (micrometers, mum) and T (K)
alpha = eps1 * FL1Tf + eps2 * (1- FL1Tf)
/* The blackbody band emission factor, Figure 12.14 and Table 12.1, is */
FL1Tf = F_lambda_T(lambda1,Tf)
// Eq 12.30
// Assigned Variables:
D = 0.010
// Cylinder diameter, m
eps1 = 0.4
// Spectral emissivity for lambda < lambda1
eps2 = 0.8
// Spectral emissivity for lambda > lambda1
lambda1 = 4
// Wavelength, mum
// Correlations Tool - Free Convection, Cylinder, Horizontal:
NuDbar = NuD_bar_FC_HC(RaD,Pr)
// Eq 9.34
NuDbar = h * D / k
RaD = g * beta * deltaT * D^3 / (nu * alphan)
//Eq 9.25
deltaT = abs(Ts - Tinf)
g = 9.8 // gravitational constant, m/s^2
// Evaluate properties at the film temperature, Tf.
Tff =Tfluid_avg(Tinf,Ts)
// Properties Tool - Nitrogen: Lookup Table Function "nitrog"
nu = lookupval (nitrog, 1, Tff, 2)
k = lookupval (nitrog, 1, Tff, 3)
alphan = lookupval (nitrog, 1, Tff, 4)
Pr = lookupval (nitrog, 1, Tff, 5)
beta = 1 / Tff
/* Lookup table function, nitrog; from Table A.4 1 atm):
Columns: T(K), nu(m^2/s), k(W/m.K), alpha(m^2/s), Pr
300
1.586E-5
0.0259 2.21E-5 0.716
350
2.078E-5
0.0293 2.92E-5 0.711
400
2.616E-5
0.0327 3.71E-5 0.704
450
3.201E-5
0.0358 4.56E-5 0.703
500
3.824E-5
0.0389 5.47E-5 0.7
550
4.17E-5
0.0417 6.39E-5 0.702
600
5.179E-5
0.0446 7.39E-5 0.701
700
6.671E-5
0.0499 9.44E-5 0.706
800
8.29E-5
0.0548 0.000116 0.715
900
0.0001003
0.0597 0.000139 0.721
1000
0.0001187
0.0647 0.000165 0.721 */
PROBLEM 12.103
KNOWN: Large combination convection-radiation oven heating a cylindrical product of a prescribed
spectral emissivity.
at 300 K, (b) Steady-state
FIND: (a) Initial heat transfer rate to the product when first placed in oven
temperature of the product, (c) Time to achieve a temperature within 50oC of the steady-state
temperature.
SCHEMATIC:
ASSUMPTIONS: (1) Cylinder is opaque-diffuse, (2) Oven walls are very large compared to the
product, (3) Cylinder end effects are negligible, (4) ε λ is dependent of temperature.
PROPERTIES: Table A-4, Air (Tf = 525 K, 1 atm): ν = 42.2 × 10 −6 m 2 s, k = 0.0423 W m ⋅ K ,
Pr = 0.684; (Tf = 850 K (assumed), 1 atm): ν = 93.8 × 10−6 m 2 s , k = 0.0596 W/m⋅K, Pr = 0.716.
ANALYSIS: (a) The net heat rate to the product is q net = As (q′′conv + α G − ε E b ) , or
q net = π DL[h(T∞ − T) + α G − εσ T 4 ]
(1)
Evaluating properties at Tf = 525 K, Re D = VD ν = 5 m s × 0.025 m 42.2 × 10 −6 m 2 s = 2960 , and the
Churchill-Bernstein correlation yields
5/8
2 1/ 3 
0.62 Re1/
 ReD  
D Pr
1 + 
Nu D =
= 0.3 +
 
k

[1 + (0.4 Pr) 2 / 3 ]1/ 4   282, 000 
hD
4/5
= 27.5
Hence,
h=
0.0423 W m ⋅ K
× 27.5 = 46.5 W m 2 ⋅ K .
0.025m
The total, hemispherical emissivity of the diffuse, spectrally selective surface follows from Eq. 12.38,
∞
ε λ (λ , Ts )E λ ,b σ Ts4 = ε1F(0→ 4 µ m) + ε 2 (1 − F(0→ 4 µ m) ) , where λT = 4 µm × 300 K = 1200
0
ε =∫
µm⋅K and F(0 − λ T) = 0.002 (Table 12.1). Hence, ε = 0.8 × 0.002 + 0.2 (1 − 0.002) = 0.201.
The absorptivity is for irradiation from the oven walls which, because they are large and isothermal,
behave as a black surface at 1000 K. From Eq. 12.46, with G λ = E λ ,b (λ, 1000 K) and α λ = ε λ ,
α = ε1F(0→ 4 µ m) + ε 2 (1 − F(0→ 4 µ m) ) = 0.8 × 0.481 + 0.2(1 − 0.481) = 0.489
where, for λT = 4 × 1000 = 4000 µm⋅K from Table 12.1, F(0 − λ T) = 0.481. From Eq. (1) the net initial
heat rate is q net = π × 0.025m × 0.2 m[46.5 W m 2 ⋅ K(750 − 300) K + 0.489σ (1000) 4 K 4 − 0.201σ (300 K) 4 ]
Continued...
PROBLEM 12.103 (Cont.)
q = 763 W.
<
(b) For the steady-state condition, the net heat rate will be zero, and the energy balance yields,
0 = h (T∞ − T) + α G − εσ T 4
(2)
Evaluating properties at an assumed film temperature of Tf = 850 K, ReD = VD/ν = 5 m/s × 0.025
m/93.8 × 10-6 m2/s = 1333, and the Churchill-Bernstein correlation yields Nu D = 18.6 . Hence, h = 18.6
2
(0.0596 W/m ⋅ K)/0.025 m = 44.3 W/m ⋅ K. Since irradiation from the oven walls is fixed, the
absorptivity is unchanged, in which case α = 0.489. However, the emissivity depends on the product
temperature. Assuming T = 950 K, we obtain
ε = ε1F(0 → 4 µ m) + ε 2 (1 − F(0 → 4 µ m) ) = 0.8 × 0.443 + 0.2(1 − 0.443) = 0.466
where for λT = 4 × 950 = 3800 µm⋅K, F0 − λ T = 0.443 , from Table 12.1. Substituting values into Eq. (2)
with σ = 5.67 × 10-8 W/m2⋅K4,
0 = 44.3 (750 − T) + 0.489 σ (1000 K)4 − 0.466 σ T4.
A trial-and-error solution yields T ≈ 930 K.
<
(c) Using the IHT Lumped Capacitance Model with the Correlations, Properties (for copper and air) and
Radiation Toolpads, the transient response of the cylinder was computed and the time to reach T = 880 K
is
t ≈ 537 s.
COMMENTS: Note that h is relatively insensitive to T, while ε is not. At T = 930 K, ε = 0.456.
<
PROBLEM 12.104
KNOWN: Workpiece, initially at 25°C, to be annealed at a temperature above 725°C for a period of
5 minutes and then cooled; furnace wall temperature and convection conditions; cooling surroundings
and convection conditions.
FIND: (a) Emissivity and absorptivity of the workpiece at 25°C when it is placed in the furnace, (b)
Net heat rate per unit area into the workpiece for this initial condition; change in temperature with
time, dT/dt, for the workpiece; (c) Calculate and plot the emissivity of the workpiece as a function of
temperature for the range 25 to 750°C using the Radiation | Band Emission tool in IHT, (d) The time
required for the workpiece to reach 725°C assuming the applicability of the lumped-capacitance
method using the DER(T,t) function in IHT to represent the temperature-time derivative in your
energy balance; (e) Calculate the time for the workpiece to cool from 750°C to a safe-to-touch
temperature of 40°C if the cool surroundings and cooling air temperature are 25°C and the convection
2
coefficient is 100 W/m ⋅K; and (f) Assuming that the workpiece temperature increases from 725 to
750°C during the five-minute annealing period, sketch (don’t plot) the temperature history of the
workpiece from the start of heating to the end of cooling; identify key features of the process;
determine the total time requirement; and justify the lumped-capacitance method of analysis.
SCHEMATIC:
ASSUMPTIONS: (1) Workpiece is opaque and diffuse, (2) Spectral emissivity is independent of
temperature, and (3) Furnace and cooling environment are large isothermal surroundings.
ANALYSIS: (a) Using Eqs. 12.38 and 12.46, ε and α can be determined using band-emission
factors, Eq. 12.30 and 12.31.
Emissivity, workpiece at 25°C
4 1 69
ε = 0.3 × 1.6 × 10−5 + 0.8 × 41 − 1.6 × 10−5 9 = 0.8
ε = ε λ1 ⋅ F 0− λT + ε λ 2 1 − F 0− λT
1
6
<
where F(0-λT) is determined from Table 12.1 with λT = 2.5 µm × 298 K = 745 µm⋅K.
Absorptivity, furnace temperature Tf = 750°C
4 1
α = ε λ1 ⋅ F 0− λ , T + ε 2 ⋅ 1 − F 0 − λ , T
1
6
1
69
6
<
α = 0.3 × 0174
.
.
+ 0.8 × 1 − 0174
= 0.713
where F(0 - λT) is determined with λT = 2.5 µm × 1023 K = 2557.5 µm⋅K.
(b) For the initial condition, T(0) = Ti, the energy balance shown schematically below is written in
terms of the net heat rate in,
Continued …..
PROBLEM 12.104 (Cont.)
E ′′in − E out
′′ = E ′′st
q ′′net,in = E in
′′ − E out
′′
and
1 6
1 6
q ′′net,in = 2 q ′′cv − ε E b Ti + αE b Tf
where G = Eb (Tf). Substituting numerical values,
1
6
q ′′net,in = 2 h T∞ − Ti − εσTi4 + ασTf4
!
1
6
1
6
q ′′net,in = 2 100 W / m2 ⋅ K 750 − 25 K − 0.8 × 5.67 × 10-8 W / m2 ⋅ K4 298 K 4
6 "#$
1
+ 0.713 × 5.67 × 10−8 W / m2 ⋅ K4 1023 K 4
q ′′net,in = 2 × 116.4 kW / m2 = 233 kW / m2
<
Considering the energy storage term,
dT
E st
= q ′′net,in
′′ = ρcL
dt i
q ′′net,in
dT
233 kW / m2
=
=
= 9.75 K / s
ρcL
dt i
2700 kg / m3 × 885 J / kg ⋅ K × 0.010 m
<
(c) With the relation for ε of Part (a) in the IHT workspace, and using the Radiation | Band Emission
tool, ε as a function of workpiece temperature is calculated and plotted below.
E m iss iv ity
0 .8
W o rk p ie c e e m is s ivity a s a fu n c tio n o f its te m p e ra tu re
0 .7 5
0 .7
0
100
200
300
400
500
T e m p e ra tu re , T (C )
600
700
800
Continued …..
PROBLEM 12.104 (Cont.)
As expected, ε decreases with increasing T, and when T = Tf = 750°C, ε = α = 0.713. Why is that so?
(d) The energy balance of Part (b), using the lumped capacitance method with the IHT DER (T,t)
function, has the form,
1
6
1 6
2 h T∞ − T − εσT4 + ασTf4 = ρcL DER T, t
where ε = ε (T) from Part (c). From a plot of T vs. t (not shown) in the IHT workspace, find
1 6
T t a = 725$ C when t a = 186 s
<
(e) The time to cool the workpiece from 750°C to the safe-to-touch temperature of 40°C can be
determined using the IHT code from Part (d). The cooling conditions are T∞ = 25°C and h = 100
2
W/m ⋅K with Tsur = 25°C. The emissivity is still evaluated using the relation of Part (c), but the
absorptivity, which depends upon the surrounding temperature, is α = 0.80. From the results in the
IHT workspace, find
1 6
T t c = 40$ C
when
<
t c = 413 s
(f) Assuming the workpiece temperature increases from 725°C to 750°C during a five-minute
annealing period, the temperature history is as shown below.
The workpiece heats from 25°C to 725°C in ta = 186 s, anneals for a 5-minute period during which the
temperature reaches 750°C, followed by the cool-down process which takes 413 s. The total required
time is
1
6
<
t = t a + 5 × 60 s + t c = 186 + 300 + 413 s = 899 s = 15 min
Continued …..
PROBLEM 12.104 (Cont.)
The Biot number based upon convection only is
1 6
h L / 2 100 W / m2 ⋅ K × 0.005 m
Bi = cv
.
=
= 0.003 << 01
k
165 W / m ⋅ K
so the lumped-capacitance method of analysis is appropriate.
COMMENTS: The IHT code to obtain the heating time, including emissivity as a function of the
workpiece temperature, Part (b), is shown below, complete except for the input variables.
/* Analysis. The radiative properties and net heat flux in are calculated when the workpiece is
just inserted into the furnace. The workpiece experiences emission, absorbed irradiation and
convection processes. See Help | Solver | Intrinsic Functions for information on DER(T, t). */
/* Results - conditions at t = 186 s, Ts C - 725 C
FL1T
T_C
Tf
L
Tf_C
Tinf_C
lambda1 rho
t
T
0.1607 725.1
1023
0.01
750
750
2.5
2700
186
998.1
*/
eps1
eps2
h
k
0.3
0.8
100
165
// Energy Balance
2 * ( h * (Tinf - T) + alpha * G - eps * sigma * T^4) = rho * cp * L * DER(T,t)
sigma = 5.67e-8
G = sigma * Tf^4
// Emissivity and absorptivity
eps = FL1T * eps1 + (1 - FL1T) * eps2
FL1T = F_lambda_T(lambda1, T)
// Eq 12.30
alpha = 0.713
// Temperature conversions
T_C = T - 273
// For customary units, graphical output
Tf_C = Tf - 273
Tinf_C = Tinf - 273
PROBLEM 12.105
KNOWN: For the semiconductor silicon, the spectral distribution of absorptivity, αλ, at selected
temperatures. High-intensity, tungsten halogen lamps having spectral distribution approximating that
of a blackbody at 2800 K.
FIND: (a) 1%-limits of the spectral band that includes 98% of the blackbody radiation corresponding
to the spectral distribution of the lamps; spectral region for which you need to know the spectral
absorptivity; (b) Sketch the variation of the total absorptivity as a function of silicon temperature;
explain key features; (c) Calculate the total absorptivity at 400, 600 and 900°C for the lamp
irradiation; explain results and the temperature dependence; (d) Calculate the total emissivity of the
wafer at 600 and 900°C; explain results and the temperature dependence; and (e) Irradiation on the
upper surface required to maintain the wafer at 600°C in a vacuum chamber with walls at 20°C. Use
the Look-up Table and Integral Functions of IHT to perform the necessary integrations.
SCHEMATIC:
ASSUMPTIONS: (1) Silicon is a diffuse emitter, (2) Chamber is large, isothermal surroundings for
the wafer, (3) Wafer is isothermal.
ANALYSIS: (a) From Eqs. 12.30 and 12.31, using Table 12.1 for the band emission factors, F(0 - λT),
equal to 0.01 and 0.99 are:
F(0→λ1⋅T ) = 0.01 at λ1 ⋅ T = 1437 µ m ⋅ K
F(0→λ 2⋅T ) = 0.99 at λ2 ⋅ T = 23,324 µ m ⋅ K
So that we have λ1 and λ2 limits for several temperatures, the following values are tabulated.
T(°C)
T(K)
λ1(µm)
λ2(µm)
400
600
900
2800
673
873
1173
0.51
2.14
1.65
1.23
8.33
34.7
26.7
19.9
<
For the 2800 K blackbody lamp irradiation, we need to know the spectral absorptivity over the
spectral range 0.51 to 8.33 µm in order to include 98% of the radiation.
(b) The spectral absorptivity is calculated from Eq. 12.46 in which the spectral distribution of the
lamp irradiation Gλ is proportional to the blackbody spectral emissive power E λ ,b λ , T at the
temperature of lamps, T" 2800 K.
∞α E
∞ α G dλ
λ λ
λ λ ,b λ , 2800 K
0
0
α" =
=
∞
σT"4
0 G λ dλ
Continued …..
1 6
I
I
I
1
6
PROBLEM 12.105 (Cont.)
For 2800 K, the peak of the blackbody curve is at 1 µm; the limits of integration for 98% coverage are
0.5 to 8.3 µm according to part (a) results. Note that αλ increases at all wavelengths with
temperature, until around 900°C where the behavior is gray. Hence, we’d expect the total absorptivity
of the wafer for lamp irradiation to appear as shown in the graph below.
At 900°C, since the wafer is gray, we expect α " = αλ ≈ 0.68. Near room temperature, since αλ ≈ 0
beyond the band edge, α " is dependent upon αλ in the spectral region below and slightly beyond the
peak. From the blackbody tables, the band emission fraction to the short-wavelength side of the peak
is 0.25. Hence, estimate α " ≈ 0.68 × 0.25 = 0.17 at these low temperatures. The increase of α "
with temperature is at first moderate, since the longer wavelength region is less significant than is the
shorter region. As temperature increases, the αλ closer to the peak begin to change more noticeably,
explaining the greater dependence of α " on temperature.
(c) The integration of part (b) can be performed numerically using the IHT INTEGRAL function and
specifying the spectral absorptivity in a Lookup Table file (*.lut). The code is shown in the
Comments (1) and the results are:
Tw(°C)
400
600
900
α"
0.30
0.59
<
0.68
(d) The total emissivity can be calculated from Eq. 12.38, recognizing that ελ = αλ and that for silicon
temperatures of 600 and 900°C, the 1% limits for the spectral integration are 1.65 - 26.7 µm and 1.23
- 19.9 µm, respectively. The integration is performed in the same manner as described in part (c); see
Comments (2).
T(°C)
600
900
ε
0.66
<
0.68
(e) From an energy balance on the silicon wafer with irradiation on the upper surface as shown in the
schematic below, calculate the irradiation required to maintain the wafer at 600°C.
α "G " − 2 ε E b Tw − α sur E b Tsur = 0
E ′′in − E out
′′ = 0
1 6
1 6
Recognize that αsur corresponds to the spectral distribution of Eλ,b (Tsur); that is, upon αλ for long
wavelengths (λmax ≈ 10 µm). We assume αsur ≈ 0.1, and with Tsur = 20°C, find
1
!
6 "$#
0.59 G " − 2σ 0.66 600 + 273 4 K 4 − 01
. 20 + 273 4 K4 = 0
6
1
Continued …..
PROBLEM 12.105 (Cont.)
G " = 735
. kW / m2
<
where E b T = σT4 and σ = 5.67 × 10-8 W / m2 ⋅ K4 .
16
COMMENTS: (1) The IHT code to obtain the total absorptivity for the lamp irradiation, α " for a
wafer temperature of 400°C is shown below. Similar look-up tables were written for the spectral
absorptivity for 600 and 800°C.
/* Results; integration for total absorptivity of lamp irradiation
T = 400 C; find abs_t = 0.30
lLb
absL
abs_t
C1
C2
T
sigma
lambda
1773
0.45
0.3012 3.742E8 1.439E4 2800
5.67E-8 10
// Input variables
T = 2800
*/
// Lamp blackbody distribution
// Total absorptivity integral, Eq. 12.46
abs_t = pi * integral (lLsi, lambda) / (sigma * T^4)
sigma = 5.67e-8
// See Help | Solver
// Blackbody spectral intensity, Tools | Radiation
/* From Planck’s law, the blackbody spectral intensity is */
lLsi = absL * lLb
lLb = l_lambda_b(lambda, T, C1, C2) // Eq. 12.25
// where units are lLb(W/m^2.sr.mum), lambda (mum) and T (K) with
C1 = 3.7420e8
// First radiation constant, W⋅mum^4/m^2
C2 = 1.4388e4
// Second radiation constant, mum⋅K
// and (mum) represents (micrometers).
// Spectral absorptivity function
absL = LOOKUPVAL(abs_400, 1, lambda, 2)
//absL = LOOKUPVAL(abs_600, 1, lambda, 2)
//absL = LOOKUPVAL(abs_900, 1, lambda, 2)
// Silicon spectral data at 400 C
// Silicon spectral data at 600 C
// Silicon spectral data at 900 C
// Lookup table values for Si spectral data at 600 C
/* The table file name is abs_400.lut, with 2 columns and 10 rows
0.5
0.68
1.2
0.68
1.3
0.025
2
0.05
3
0.1
4
0.17
5
0.22
6
0.28
8
0.37
10
0.45 */
(2) The IHT code to obtain the total emissivity for a wafer temperature of 600°C has the same
organization as for obtaining the total absorptivity. We perform the integration, however, with the
blackbody spectral emissivity evaluated at the wafer temperature (rather than the lamp temperature).
The same look-up file for the spectral absorptivity created in the part (c) code can be used.
PROBLEM 12.106
KNOWN: Solar irradiation of 1100 W/m2 incident on a flat roof surface of prescribed solar absorptivity
and emissivity; air temperature and convection heat transfer coefficient.
FIND: (a) Roof surface temperature, (b) Effect of absorptivity, emissivity and convection coefficient on
temperature.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Back-side of plate is perfectly insulated, (3)
Negligible irradiation to plate by atmospheric (sky) emission.
ANALYSIS: (a) Performing a surface energy balance on the exposed side of the plate,
αSGS − q′′conv − ε E b (Ts ) = 0
αSGS − h (Ts − T∞ ) − εσ Ts4 = 0
Substituting numerical values and using absolute temperatures,
W
W
0.6 × 1100
(Ts − 300)K − 0.2(5.67 × 10−8 W m2⋅ K 4 )Ts4 = 0
− 25
2
2
m
m ⋅K
Regrouping , 8160 = 25Ts + 1.1340 × 10−8 Ts4 , and performing a trial-and-error solution,
Ts = 321.5 K = 48.5oC.
(b) Using the IHT First Law Model for a plane wall, the following results were obtained.
<
120
Plate temperature, T(C)
100
80
60
40
20
0
0
20
40
60
80
100
Convection coefficient, h(W/m^2.K)
eps = 0.2, alphaS = 0.6
eps = 0.8, alphaS = 0.6
eps = 0.8, alphaS = 0.2
Irrespective of the value of h , T decreases with increasing ε (due to increased emission) and decreasing
αS (due to reduced absorption of solar energy). For moderate to large αS and/or small ε (net radiation
transfer to the surface) T decreases with increasing h due to enhanced cooling by convection. However,
for small αS and large ε, emission exceeds absorption, dictating convection heat transfer to the surface
and hence T < T∞ . With increasing h , T → T∞ , irrespective of the values of αS and ε.
COMMENTS: To minimize the roof temperature, the value of ε/αS should be maximized.
PROBLEM 12.107
KNOWN: Cavity with window whose outer surface experiences convection and radiation.
FIND: Temperature of the window and power required to maintain cavity at prescribed temperature.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Cavity behaves as a blackbody, (3) Solar spectral
distribution is that of a blackbody at 5800K, (4) Window is isothermal, (5) Negligible convection on
lower surface of window.
PROPERTIES: Window material: 0.2 ≤ λ ≤ 4 µm, τλ = 0.9, ρ λ = 0, hence α λ = 1 - τλ = 0.1; 4 µm <
λ, τλ = 0, α = ε = 0.95, diffuse-gray, opaque
ANALYSIS: To determine the window temperature, perform an energy balance on the window,
E& in − E& out = 0
′′
[α surG sur + α SG S − ε E b − q conv
]upper + α cG c − ε E b ( T ) lower = 0.
(1)
Calculate the absorptivities for various irradiation conditions using Eq. 12.46,
∞
∞
α=
αλ G λ dλ / G λ dλ
0
0
where G(λ) is the spectral distribution of the irradiation.
∫
∫
(2)
4
Surroundings, α sur: Gsur = Eb (Tsur) = σ Tsur
αsur = 0.1  F(0 →4 µm ) − F(0 →0.2 µm )  + 0.95 1 − F(0 →4 µm ) 




where from Table 12.1, with T = Tsur = (25 + 273)K = 298K,
λ T = 0.2µ m× 298K = 59.6µm ⋅ K,
F( 0 − λT ) = 0.000
λ T = 4µ m ×298K = 1192µm ⋅ K,
F( 0 − λT ) = 0.002
αsur = 0.1[ 0.002 − 0.000 ] + 0.95 [1 − 0.002] = 0.948.
(3)
Solar, α S: GS ~ Eb (5800K)
αS = 0.1  F(0 →4 µm ) − F(0 →0.2 µm )  + 0.95 1 − F(0 →4 µm ) 




where from Table 12.1, with T = 5800K,
λ T = 0.2µ m × 5800K = 1160µ m ⋅ K, F( 0 − λT ) = 0.002
λ T = 4µ m × 5800K = 23,200µm ⋅ K, F( 0 − λT ) = 0.990
αS = 0.1[ 0.990 − 0.002 ] + 0.95 [1 − 0.990 ] = 0.108.
(4)
Continued …..
PROBLEM 12.107 (Cont.)
Cavity, α c: Gc = Eb(Tc) = σ Tc4
α c = 0.1  F( 0→ 4µ m) − F( 0→ 0.2µ m)  + 0.95 1 − F( 0→ 4µ m) 




where from Table 12.1 with Tc = 250°C = 523K,
λ T = 0.2µ m× 523K = 104.6µm ⋅ K,
λ T = 4µ m ×523K = 2092µm ⋅ K
F0→λ T = 0.000
F0→λ T = 0.082
α c = 0.1[ 0.082 − 0.000] + 0.95[1 − 0.082] = 0.880.
(5)
To determine the emissivity of the window, we need to know its temperature. However, we know that
T will be less than Tc and the long wavelength behavior will dominate. That is,
ε ≈ ελ ( λ > 4 µ m ) = 0.95.
(6)
With these radiative properties now known, the energy equation, Eq. (1) can now be evaluated using
q ′′conv = h(T - T∞) with all temperatures in kelvin units.
4
0.948 × σ ( 298K ) + 0.108 × 8 0 0 W / m 2 − 0.95 ×σ T 4 − 1 0 W / m 2 ⋅ K ( T − 298K )
4
+0.880σ ( 523K ) − 0.95 ×σ T4 = 0
1.077 × 10−7 T4 + 10T − 7223 = 0.
Using a trial-and-error approach, find the window temperature as
<
T = 413K = 139 °C.
To determine the power required to maintain the cavity
at Tc = 250°C, perform an energy balance on the cavity.
E& in − E& out = 0
q p + Ac [ ρ E b ( Tc ) + τS GS + ε E b ( T ) − E b ( Tc ) ] = 0.
For simplicity, we have assumed the window opaque to
irradiation from the surroundings. It follows that
τ S = 1 − ρS −α S = 1 − 0 − 0.108 = 0.892
ρ = 1 − α = 1 − ε = 1 − 0.95 = 0.05.
2
Hence, the power required to maintain the cavity, when Ac = (π/4)D , is
q p = Ac σ Tc4 − ρσ Tc4 −τ sG s − εσ T4 


qp =
π
( 0.050m) 2 σ ( 523K ) 4 − 0.05σ ( 523K )4 − 0.892 ×8 0 0 W / m2 − 0.95σ ( 412K ) 4 


4
q p = 3.47W.
<
COMMENTS: Note that the assumed value of ε = 0.95 is not fully satisfied. With T = 412K, we
would expect ε = 0.929. Hence, an iteration may be appropriate.
PROBLEM 12.108
KNOWN: Features of an evacuated tube solar collector.
FIND: Ideal surface spectral characteristics.
SCHEMATIC:
ANALYSIS: The outer tube should be transparent to the incident solar radiation, which is
concentrated in the spectral region λ ≤ 3µm, but it should be opaque and highly reflective to radiation
emitted by the outer surface of the inner tube, which is concentrated in the spectral region above 3µm.
Accordingly, ideal spectral characteristics for the outer tube are
Note that large ρ λ is desirable for the outer, as well as the inner, surface of the outer tube. If the
surface is diffuse, a large value of ρ λ yields a small value of ε λ = α λ = 1 - ρ λ. Hence losses due to
emission from the outer surface to the surroundings would be negligible.
The opaque outer surface of the inner tube should absorb all of the incident solar radiation (λ ≤ 3µm)
and emit little or no radiation, which would be in the spectral region λ > 3µm. Accordingly, assuming
diffuse surface behavior, ideal spectral characteristics are:
PROBLEM 12.109
KNOWN: Plate exposed to solar flux with prescribed solar absorptivity and emissivity; convection
and surrounding conditions also prescribed.
FIND: Steady-state temperature of the plate.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Plate is small compared to surroundings, (3)
Backside of plate is perfectly insulated, (4) Diffuse behavior.
ANALYSIS: Perform a surface energy balance on the top surface of the plate.
E& in − E& out = 0
αS GS + α G sur − q ′′conv − ε Eb ( Ts ) = 0
Note that the effect of the surroundings is to provide an irradiation, Gsur, on the plate; since the spectral
distribution of Gsur and Eλ,b (Ts) are nearly the same, accordiing to Kirchoff’s law, α = ε. Recognizing
4
that Gsur = σ Tsur
and using Newton’s law of cooling, the energy balance is
4
αS GS + εσ Tsur
− h ( Ts − T∞ ) − ε ⋅ σ Ts4 = 0.
Substituting numerical values,
0.9 × 900 W / m 2 + 0.1 × 5.67 ×10− 8 W / m 2 ⋅ K × (17 + 273) K 4
4
(
)
−2 0 W / m 2 ⋅ K ( Ts − 290 ) K − 0.1 5.67 × 10−8 W / m 2 ⋅ K4 Ts4 = 0
6650 W / m 2 = 2 0 Ts + 5.67 × 10−9 Ts4.
From a trial-and-error solution, find
<
Ts = 329.2 K.
COMMENTS: (1) When performing an analysis with both convection and radiation processes
present, all temperatures must be expressed in absolute units (K).
(2) Note also that the terms α Gsur - ε Eb (Ts) could be expressed as a radiation exchange term,
written as
(
)
4 − T4 .
q′′rad = q / A = εσ Tsur
s
The conditions for application of this relation were met and are namely: surroundings much larger than
surface, diffuse surface, and spectral distributions of irradiation and emission are similar (or the surface
is gray).
PROBLEM 12.110
KNOWN: Directional distribution of α θ for a horizontal, opaque, gray surface exposed to direct and
diffuse irradiation.
FIND: (a) Absorptivity to direct radiation at 45° and to diffuse radiation, and (b) Equilibrium
temperature for specified direct and diffuse irradiation components.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Opaque, gray surface behavior, (3) Negligible
convection at top surface and perfectly insulated back surface.
ANALYSIS: (a) From knowledge of α θ (θ) – see graph above – it is evident that the absorptivity of
the surface to the direct radiation (45°) is
α dir = αθ ( 45 ° ) = 0.8.
<
The absorptivity to the diffuse radiation is the hemispherical absorptivity given by Eq. 12.44. Dropping
the λ subscript,
π /2
α dir = 2
αθ (θ ) cos θ sin θ dθ
(1)
0
π /2

sin 2 θ π / 3
sin 2 θ


α dir = 2  0.8
+0.1

0
2
2
π / 3 

∫
<
α dir = 0.625.
(b) Performing a surface energy balance,
′′ = 0
E& ′′in − E& out
α dir G dir + αdif G dif − ε σ Ts4 = 0.
(2)
The total, hemispherical emissivity may be obtained from Eq. 12.36 where again the subscript may be
deleted. Since this equation is of precisely the same form as Eq. 12.44 – see Eq. (1) above – and since
α θ = ε θ, it follows that
ε = αdif = 0.625
and from Eq. (2), find
Ts4 =
( 0.8 × 600 + 0.625 ×100 ) W / m 2 = 1.53 ×1010 K 4,
0.625 × 5.67 ×10−8 W / m 2 ⋅ K 4
Ts = 352 K.
<
COMMENTS: In assuming gray surface behavior, spectral effects are not present, and total and
spectral properties are identical. However, the surface is not diffuse and hence hemispherical and
directional properties differ.
PROBLEM 12.111
KNOWN: Plate temperature and spectral and directional dependence of its absorptivity. Direction
and magnitude of solar flux.
FIND: (a) Expression for total absorptivity, (b) Expression for total emissivity, (c) Net radiant flux,
(d) Effect of cut-off wavelength associated with directional dependence of the absorptivity.
SCHEMATIC:
ASSUMPTIONS: (1) Diffuse component of solar flux is negligible, (2) Spectral distribution of solar
radiation may be approximated as that from a blackbody at 5800 K, (3) Properties are independent of
azimuthal angle φ.
ANALYSIS: (a) For λ < λc and θ = 45°, αλ = α1 cosθ = 0.707 α1. From Eq. (12.47) the total
absorptivity is then
 ∞

 λc

Eλ ,b (λ ,5800 K ) dλ 
λ
λ
E
,5800
K
d
(
)
∫

,b
λ
∫
λ


 c

αS = 0.707 α1  0
 + α2 

Eb
Eb








αS = 0.707 α1 F(0→λ ) + α 2 1 − F(0→λ ) 
c
c 

<
For the prescribed value of λc, λcT = 11,600 µm⋅K and, from Table 12.1, F(0→λc) = 0.941. Hence,
αS = 0.707 × 0.93 × 0.941 + 0.25 (1 − 0.941) = 0.619 + 0.015 = 0.634
<
(b) With ελ,θ = αλ,o, Eq (12.36) may be used to obtain ελ for λ < λc.
π /2
cos3 θ π / 2 2
ε λ (λ , T ) = 2α1
cos 2 θ sin θ dθ = −2α1
= α1
0
∫
3
3
0
From Eq. (12.38),
∞
λc
E
λ , Tp dλ
λ
λ
E
,
T
d
∫
p
λ ,b
∫
λc λ ,b
0
ε = 0.667 α1
+ α2
Eb
Eb
(
)
(
)
ε = 0.667 α1 F(0→λ ) + α 2 1 − F(0 −λ ) 
c
c 

<
For λc = 2 µm and Tp = 333 K, λcT = 666 µm⋅K and, from Table 12.1, F(0-λc) = 0. Hence,
<
ε = α 2 = 0.25
Continued …..
PROBLEM 12.111 (Cont.)
(c)
q′′net = αS qS′′ − εσ Tp4 = 634 W / m 2 − 0.25 × 5.67 × 10−8 W / m 2 ⋅ K 4 (333 K )
4
q′′net = 460 W / m 2
<
(d) Using the foregoing model with the Radiation/Band Emission Factor option of IHT, the following
results were obtained for αS and ε. The absorptivity increases with increasing λc, as more of the
incident solar radiation falls within the region of α1 > α2. Note, however, the limit at λ ≈ 3 µm,
beyond which there is little change in αS. The emissivity also increases with increasing λc, as more
of the emitted radiation is at wavelengths for which ε1 = α1 > ε2 = α2. However, the surface
temperature is low, and even for λc = 5 µm, there is little emission at λ < λc. Hence, ε only increases
from 0.25 to 0.26 as λc increases from 0.7 to 5.0 µm.
Surface properties
0.7
0.6
0.5
0.4
0.3
0.2
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
Cut-off wavelength (micrometer)
AlphaS
Epsilon
2
2
The net heat flux increases from 276 W/m at λc = 2 µm to a maximum of 477 W/m at λc = 4.2 µm
2
and then decreases to 474 W/m at λc = 5 µm. The existence of a maximum is due to the upper limit
on the value of αS and the increase in ε with λc.
COMMENTS: Spectrally and directionally selective coatings may be used to enhance the
performance of solar collectors.
PROBLEM 12.112
KNOWN: Spectral distribution of α λ for two roof coatings.
FIND: Preferred coating for summer and winter use. Ideal spectral distribution of α λ.
SCHEMATIC:
ASSUMPTIONS: (1) Opaque, diffuse surface behavior, (2) Negligible convection effects and heat
transfer from bottom of roof, negligible atmospheric irradiation, (3) Steady-state conditions.
ANALYSIS: From an energy balance on the roof surface
ε σ Ts4 = αS G S.
Hence
1/4
α G 
Ts =  S S 
 ε σ 
.
Solar irradiation is concentrated in the spectral region λ < 4µm, while surface emission is concentrated
in the region λ > 4µm. Hence, with α λ = ε λ
Coating A:
α S ≈ 0.8,
ε ≈ 0.8
Coating B:
α S ≈ 0.6,
ε ≈ 0.2.
Since (α S/ε)A = 1 < (α S/ε)B = 3, Coating A would result in the lower roof temperature and is
preferred for summer use. In contrast, Coating B is preferred for winter use. The ideal coating is one
which minimizes (α S/ε) in the summer and maximizes it in the winter.
PROBLEM 12.113
KNOWN: Shallow pan of water exposed to night desert air and sky conditions.
FIND: Whether water will freeze.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Bottom of pan is well insulated, (3) Water surface
4
is diffuse-gray, (4) Sky provides blackbody irradiation, Gsky = σ Tsky
.
PROPERTIES: Table A-11, Water (300 K): ε = 0.96.
ANALYSIS: To estimate the water surface temperature for these conditions, begin by performing an
energy balance on the pan of water considering convection and radiation processes.
′′ = 0
E& ′′in − E& out
α Gsky − ε E b − h ( Ts− T∞ ) = 0
(
)
4 − T 4 − h T − T = 0.
ε σ Tsky
( s ∞)
s
4
Note that, from Eq. 12.64, Gsky = σ Tsky
and from Assumption 3, α = ε. Substituting numerical
values, with all temperatures in kelvin units, the energy balance is
0.96 × 5.67 ×10 −8
W
 ( −40 + 273 )4 − T 4  K 4 − 5 W T − ( 20 + 273 )  K = 0
s
 s


m2 ⋅ K
m 2 ⋅ K4 
5.443 ×10 −8  233 4 − Ts4  −5 [ Ts − 293] = 0.


Using a trial-and-error approach, find the water surface temperature,
Ts = 268.5K.
<
Since Ts < 273 K, it follows that the water surface will freeze under the prescribed air and sky
conditions.
COMMENTS: If the heat transfer coefficient were to increase as a consequence of wind, freezing
might not occur. Verify that for the given T∞ and Tsky , that if h increases by more than 40%,
freezing cannot occur.
PROBLEM 12.114
KNOWN: Flat plate exposed to night sky and in ambient air at Tair = 15°C with a relative humidity
of 70%. Radiation from the atmosphere or sky estimated as a fraction of the blackbody radiation
corresponding to the near-ground air temperature, Gsky = εsky σ Tair, and for a clear night, εsky =
0.741 + 0.0062 Tdp where Tdp is the dew point temperature (°C). Convection coefficient estimated by
correlation, h W / m2 ⋅ K = 125
. ∆T1/3 where ∆T is the plate-to-air temperature difference (K).
4
9
FIND: Whether dew will form on the plate if the surface is (a) clean metal with εm = 0.23 and (b)
painted with εp = 0.85.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Surfaces are diffuse, gray, and (3) Backside of
plate is well insulated.
PROPERTIES: Psychrometric charts (Air), Tdp = 9.4°C for dry bulb temperature 15°C and relative
humidity 70%.
ANALYSIS: From the schematic above, the energy balance on the plate is
E ′′in − E out
′′ = 0
1 6
4/3
4 " + 125
W / m2 − εσTs4 W / m2 = 0
ε 0.741 + 0.0062 Tdp 4 C9 σ Tair
. 1Tair − Ts 6
#
!
$
α sky G sky + q cv
′′ − ε E b Ts = 0
$
where Gsky = εsky σ Tair, εsky = 0.741 + 0.062 Tdp (°C); Tdp has units (°C); and, other temperatures in
kelvins. Since the surface is diffuse-gray, αsky = ε.
(a) Clean metallic surface, εm = 0.23
!
4 9 1
6 "#$
0.23 0.741 + 0.0062 Tdp $ C σ 15 + 273 4 K4
3
.
289 − Ts,m
+125
4 W / m2 = 0
84 / 3 W / m2 − 0.23 σ Ts,m
Ts,m = 282.7 K = 9.7$ C
(b) Painted surface, εp = 0.85
<
Ts,p = 278.5 K = 5.5°C
<
COMMENTS: For the painted surface, εp = 0.85, find that Ts < Tdp, so we expect dew formation.
For the clean, metallic surface, Ts > Tdp, so we do not expect dew formation.
PROBLEM 12.115
KNOWN: Glass sheet, used on greenhouse roof, is subjected to solar flux, GS, atmospheric emission,
Gatm, and interior surface emission, Gi, as well as to convection processes.
FIND: (a) Appropriate energy balance for a unit area of the glass, (b) Temperature of the greenhouse
ambient air, T∞,i, for prescribed conditions.
SCHEMATIC:
ASSUMPTIONS: (1) Glass is at a uniform temperature, Tg, (2) Steady-state conditions.
PROPERTIES: Glass: τλ = 1 for λ ≤ 1µm; τλ = 0 and α λ = 1 for λ > 1 µm.
ANALYSIS: (a) Performing an energy balance on the glass sheet with E& in − E& out = 0 and
considering two convection processes, emission and three absorbed irradiation terms, find
(
)
(
)
αS GS + α atm G atm + ho T∞,o − Tg + αi Gi + h i T∞,i − Tg − 2 ε σ Tg4 = 0
(1)
α S = solar absorptivity for absorption of Gλ,S ~ Eλ,b (λ, 5800K)
α atm = α i = absorptivity of long wavelength irradiation (λ >> 1 µm) ≈ 1
where
ε = α λ for λ >> 1 µm, emissivity for long wavelength emission ≈ 1
(b) For the prescribed conditions, T∞,i can be evaluated from Eq. (1). As noted above, α atm = α i = 1
and ε = 1. The solar absorptivity of the glass follows from Eq. 12.47 where Gλ,S ~ Eλ,b (λ, 5800K),
∞
0
∞
0
αS = ∫ α λ G λ,S dλ / Gs = ∫ α λ Eλ ,b ( λ ,5800K ) dλ / E b ( 5800K )
αS = α1F(0 →1µm ) + α 2 1 − F(0 →1µm )  = 0 × 0.720+ 1.0 [1 − 0.720] = 0.28.


Note that from Table 12.1 for λT = 1 µm × 5800K = 5800 µm⋅K, F(0 - λ) = 0.720. Substituting
numerical values into Eq. (1),
0.28 × 1100W/m 2 + 1× 250W/m 2 + 5 5 W / m 2 ⋅ K ( 24 − 27 ) K +1 × 4 4 0 W / m2 +
(
)
1 0 W / m 2 ⋅ K T∞,i − 27 K −2 × 1 × 5.67 × 10−8 W / m 2 ⋅ K ( 27 + 273)4 K4 = 0
find that
T∞ ,i = 35.5°C.
<
PROBLEM 12.116
KNOWN: Plate temperature and spectral absorptivity of coating.
FIND: (a) Solar irradiation, (b) Effect of solar irradiation on plate temperature, total absorptivity, and
total emissivity.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, (2) Opaque, diffuse surface, (3) Isothermal plate, (4) Negligible
radiation from surroundings.
ANALYSIS: (a) Performing an energy balance on the plate, 2αSGS - 2E = 0 and
αSGS − εσ T 4 = 0
For λT = 4.5 µm × 2000 K = 9000 µm⋅K, Table 12.1 yields F(o→λ) = 0.890. Hence,
(
)
ε = ε1F(0 → λ ) + ε 2 1 − F( o → λ ) = 0.95 × 0.890 + 0.03 (1 − 0.890 ) = 0.849
For λT = 4.5 µm × 5800 K = 26,100, F(o→λ) = 0.993. Hence,
(
)
αS = α1F( 0→ λ ) + α 2 1 − F( 0→ λ ) = 0.95 × 0.993 + 0.03 × 0.007 = 0.944
Hence,
<
4
GS = (ε αS )σ T 4 = ( 0.849 0.944 ) 5.67 × 10−8 W m 2⋅ K 4 ( 2000 K ) = 8.16 × 105 W m 2
(b) Using the IHT First Law Model and the Radiation Toolpad, the following results were obtained.
4.5
1
Radiative property
Solar irradiation, GS(10^6W/m^2)
0.8
3
1.5
0.6
0.4
0.2
0
500
1000
0
1500
2000
2500
Temperature, T(K)
500
1000
1500
2000
Plate temperature, T(K)
2500
3000
alphaS
eps
The required solar irradiation increases with T to the fourth power. Since αS is determined by the
spectral distribution of solar radiation, its value is fixed. However, with increasing T, the spectral
distribution of emission is shifted to lower wavelengths, thereby increasing the value of ε.
3000
PROBLEM 12.117
KNOWN: Thermal conductivity, spectral absorptivity and inner and outer surface conditions for wall
of central solar receiver.
FIND: Minimum wall thickness needed to prevent thermal failure. Collector efficiency.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Outer surface is opaque and diffuse, (3) Spectral
distribution of solar radiation corresponds to blackbody emission at 5800 K.
ANALYSIS: From an energy balance at the outer surface, E& in = E& out ,
T
−T
) ( L /s,ok ) + (1∞/,ih )
i
(
4 +h T −T
αSqS′′ + αsur Gsur = εσ Ts,o
∞,o +
o s,o
Since radiation from the surroundings is in the far infrared, α sur = 0.2. From Table 12.1, λT = (3 µm ×
5800 K) = 17,400 µm⋅K, find F(0→3µm) = 0.979. Hence,
∞
αλ Eλ ,b ( 5800 K) dλ
= α1 F( 0 →3 µm) + α2 F( 3→∞) = 0.9 ( 0.979) + 0.2 ( 0.021) = 0.885.
αs = 0
Eb
From Table 12.1, λT = (3 µm × 1000 K) = 3000 µm⋅K, find F(0→3µm) = 0.273. Hence,
∞
ε λ Eλ ,b (1000 K) dλ
o
= ε1F(0 →3 ) + ε 2 F(3→∞ ) = 0.9 ( 0.273) + 0.2 ( 0.727 ) = 0.391.
εs =
Eb
Substituting numerical values in the energy balance, find
∫
∫
(
0.885 80,000 W / m
2
)
+ 0.2 ×5.67 × 10
−8
2
W/m ⋅K
4
( 300 K )4
= 0.391 × 5.67 × 10
−8
2
W/m ⋅K
4
(1000 K )4
(
)
+25 W / m 2 ⋅ K ( 700 K ) + ( 300 K ) /  ( L / 1 5 W / m ⋅ K ) + 1/1000 W / m2 ⋅ K 


<
L = 0.129 m.
The corresponding collector efficiency is
 T −T

q ′′
η = use =  s,o ∞,i  / q′′S
q′′S
 ( L / k ) + ( 1 / hi ) 

300 K
η=

( 0.129 m / 1 5 W / m ⋅ K ) + 0.001m2 ⋅ K / W

(
)

 /80,000 W / m 2 = 0.391or 39.1%.


<
COMMENTS: The collector efficiency could be increased and the outer surface temperature
reduced by decreasing the value of L.
PROBLEM 12.118
KNOWN: Dimensions, spectral absorptivity, and temperature of solar receiver. Solar irradiation and
ambient temperature.
FIND: (a) Rate of energy collection q and collector efficiency η, (b) Effect of receiver temperature on q
and η.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, (2) Uniform irradiaton, (3) Opaque, diffuse surface.
PROPERTIES: Table A.4, air (Tf = 550 K): ν = 45.6 × 10-6 m2/s, k = 0.0439 W/m⋅K, α = 66.7 × 10-6
m2/s, Pr = 0.683.
ANALYSIS: (a) The rate of heat transfer to the receiver is q = As (αSGS − E − q′′conv ) , or
q = π DL αSGS − εσ Ts4 − h ( Ts − T∞ )


For λT = 3 µm × 5800 K = 17,400, F(0→λ) = 0.979. Hence,
(
)
αS = α1F( 0→ λ ) + α 2 1 − F( 0→ λ ) = 0.9 × 0.979 + 0.2 ( 0.021) = 0.885
For λT = 3 µm × 800 K = 2400 µm⋅K, F(0→λ) = 0.140. Hence,
(
)
ε = ε1F(0 → λ ) + ε 2 1 − F( 0 → λ ) = 0.9 × 0.140 + 0.2 ( 0.860 ) = 0.298 .
With RaL = gβ(Ts - T∞ )L3/αν = 9.8 m/s2(1/550 K)(500 K)(12 m)3/66.7 × 10-6 m2/s × 45.6 × 10-6 m2/s =
5.06 × 1012, Eq. 9.26 yields
2


1/ 6


0.387Ra L
Nu L = 0.825 +
= 1867
8 / 27 


1 + ( 0.492 Pr )9 /16 




h = Nu L
k
L
= 1867
0.0439 W m⋅ K
12 m
= 6.83 W m 2⋅ K
Hence,
q = π ( 7 m × 12 m )  0.885 × 80, 000 W m − 0.298 × 5.67 × 10
2

−8
2
W m ⋅K
4
q = 263.9 m 2 ( 70,800 − 6, 920 − 3415 ) W m2 = 1.60 × 107 W
(800 K )4 − 6.83 W
m ⋅ K (500 K )
2

<
The collector efficiency is η = q/AsGS. Hence
η=
1.60 × 107 W
(
263.9 m 2 80, 000 W m 2
)
= 0.758
<
Continued …..
PROBLEM 12.118 (Cont.)
(b) The IHT Correlations, Properties and Radiation Toolpads were used to obtain the following results.
0.9
1.8E7
1.7E7
0.8
Efficiency, eta
Heat rate, q(W)
1.6E7
1.5E7
1.4E7
1.3E7
0.7
0.6
1.2E7
0.5
1.1E7
600
700
800
900
Receiver temperature, Ts(K)
1000
600
700
800
900
1000
Receiver temperature, T(K)
Losses due to emission and convection increase with increasing Ts, thereby reducing q and η.
COMMENTS: The increase in radiation emission is due to the increase in Ts, as well as to the effect of
Ts on ε, which increases from 0.228 to 0.391 as Ts increases from 600 to 1000 K.
PROBLEM 12.119
KNOWN: Dimensions and construction of truck roof. Roof interior surface temperature. Truck speed,
ambient air temperature, and solar irradiation.
FIND: (a) Preferred roof coating, (b) Roof surface temperature, (c) Heat load through roof, (d) Effect
of velocity on surface temperature and heat load.
SCHEMATIC:
ASSUMPTIONS: (1) Turbulent boundary layer development over entire roof, (2) Constant properties,
(3) Negligible atmospheric (sky) irradiation, (4) Negligible contact resistance.
PROPERTIES: Table A.4, Air (Ts,o ≈ 300 K, 1 atm): ν = 15 × 10−6 m 2 s , k = 0.026 W m⋅ K ,
Pr = 0.71.
ANALYSIS: (a) To minimize heat transfer through the roof, minimize solar absorption relative to
<
surface emission. Hence, use zinc oxide white for which αS = 0.16 and ε = 0.93.
(b) Performing an energy balance on the outer surface of the roof, αSGS + q′′conv − E − q′′cond = 0 , it
follows that
4
αSGS + h(T∞ − Ts,o ) = εσ Ts,o
+ (k t)(Ts,o − Ts,i )
where it is assumed that convection is from the air to the roof. With
Re L =
VL
ν
=
30 m s(5 m)
15 × 10
−6
2
= 107
m s
Nu L = 0.037 Re 4L/ 5 Pr1/ 3 = 0.037(107 )4 / 5 (0.71)1/ 3 = 13,141
h = Nu L (k L) = 13,141(0.026 W m⋅ K/5 m = 68.3 W m 2⋅ K .
Substituting numerical values in the energy balance and solving by trial-and-error, we obtain
Ts,o = 295.2 K.
<
(c) The heat load through the roof is
q = (kAs t)(Ts,o − Ts,i ) = (0.05 W m ⋅ K × 10 m 2 0.025 m)35.2 K = 704 W .
<
(d) Using the IHT First Law Model with the Correlations and Properties Toolpads, the following results
are obtained.
Continued...
PROBLEM 12.119 (Cont.)
700
300
650
Heat load, q(W)
Temperature, Tso(K)
295
290
600
550
285
500
5
280
5
10
15
20
25
30
10
15
20
25
30
Velocity, V(m/s)
Velocity, V(m/s)
The surface temperature and heat load decrease with decreasing V due to a reduction in the convection
heat transfer coefficient and hence convection heat transfer from the air.
COMMENTS: The heat load would increase with increasing αS/ε.
PROBLEM 12.120
KNOWN: Sky, ground, and ambient air temperatures. Grape of prescribed diameter and properties.
FIND: (a) General expression for rate of change of grape temperature, (b) Whether grapes will
freeze in quiescent air, (c) Whether grapes will freeze for a prescribed air speed.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible temperature gradients in grape, (2) Uniform blackbody irradiation
over top and bottom hemispheres, (3) Properties of grape are those of water at 273 K, (4) Properties
of air are constant at values for T∞, (5) Negligible buoyancy for V = 1 m/s.
3
PROPERTIES: Table A-6, Water (273 K): cp = 4217 J/kg⋅K, ρ = 1000 kg/m ; Table A-4, Air (273
-6 2
-6 2
-3
K, 1 atm): ν = 13.49 × 10 m /s, k = 0.0241 W/m⋅K, α = 18.9 × 10 m /s, Pr = 0.714, β = 3.66 × 10
-1
K .
ANALYSIS: (a) Performing an energy balance for a control surface about the grape,
dTg
π D3
π D2
dEst
= ρg
= h π D 2 T∞ − Tg +
c p⋅g
G ea + Gsky − E πD 2.
dt
6
(
dt
)
2
(
)
Hence, the rate of temperature change with time is
dTg
dt
=
(
)
6
h T − T + σ
∞
g
ρg c p⋅gD 
(( Tea4 + Tsky4 ) / 2 − εgTg4 ) .
<
(b) The grape freezes if dTg/dt < 0 when Tg = Tfp = 268 K. With
3
2
−3 −1
gβ T∞ − Tg D3 9.8 m / s 3.66 ×10 K 5K ( 0.015 m )
=
= 2374
Ra D =
αν
18.9 × 10−6 ×13.49 ×10 −6 m 4 / s 2
(
(
)
)
using Eq. 9.35 find
Nu D = 2 +
0.589 ( 2374 )1/4
1 + ( 0.469/Pr )9/16 


4/9
= 5.17
h = ( k / D) Nu D = ( 0.0241W/m ⋅ K ) / ( 0.015m ) 5.17 = 8.31W/m 2 ⋅ K.
Hence, the rate of temperature change is
dTg
dt
=
(
)
6
8.31W/m 2 ⋅ K (5 K )

1000 k g / m 3 4217 J / k g ⋅ K ( 0.015 m ) 
(
)
+5.67 ×10 −8 W / m 2 ⋅ K 4  2734 + 235 4 / 2 − 268 4  K 4


Continued …..
PROBLEM 12.120 (Cont.)
dTg
dt
= 9.49 ×10−5 K ⋅ m2 / J [ 41.55 − 48.56] W / m 2 = −6.66 ×10−4 K / s
<
and since dTg/dt < 0, the grape will freeze.
(c) For V = 1 m/s,
Re D =
VD 1 m / s ( 0.015 m )
=
= 1112.
ν
13.49 × 10−6 m 2 / s
Hence with (µ/µs)
1/4
= 1,
(
)
2/3
0.4 = 21.8
Nu D = 2 + 0.4Re1/2
D + 0.06Re D Pr
h = Nu D
k
0.0241
= 21.8
= 35 W / m 2 ⋅ K.
D
0.015
Hence the rate of temperature change with time is
dTg
= 9.49 ×10−5 K ⋅ m2 / J  35 W / m 2 ⋅ K (5 K ) − 48.56 W / m 2  = 0.012 K / s


dt
and since dTg/dt > 0, the grape will not freeze.
<
COMMENTS: With GrD = RaD/Pr = 3325 and GrD /Re2D = 0.0027, the assumption of negligible
buoyancy for V = 1 m/s is reasonable.
PROBLEM 12.121
KNOWN: Metal disk exposed to environmental conditions and placed in good contact with the earth.
FIND: (a) Fraction of direct solar irradiation absorbed, (b) Emissivity of the disk, (c) Average free
convection coefficient of the disk upper surface, (d) Steady-state temperature of the disk (confirm the
value 340 K).
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Disk is diffuse, (3) Disk is isothermal, (4)
Negligible contact resistance between disk and earth, (5) Solar irradiance has spectral distribution of
Eλ,b (λ, 5800 K).
PROPERTIES: Table A-4, Air (1 atm, Tf = (Ts + T∞)/2 = (340 + 300) K/2 = 320 K): ν = 17.90 ×
-6 2
-6 2
10 m /s, k = 0.0278 W/m⋅K, α = 25.5 × 10 m /s, Pr = 0.704.
ANALYSIS: (a) The solar absorptivity follows from Eq. 12.49 with Gλ,S α Eλ,b (λ, 5800 K), and α λ =
ε λ, since the disk surface is diffuse.
∞
αS = ∫ α λ Eλ ,b ( λ , 5800K ) / Eb ( 5800 K )
0
(
)
αS = ε1F(0 →1µm ) + ε 2 1 − f (0 →1µm ) .
From Table 12.1 with
λ T = 1 µm × 5800 K = 5800 µ m ⋅ K
find
F( 0 →λT ) = 0.720
giving
αS = 0.9 × 0.720 + 0.2 (1 − 0.720 ) = 0.704.
<
Note this value is appropriate for diffuse or direct solar irradiation since the surface is diffuse.
(b) The emissivity of the disk depends upon the surface temperature Ts which we believe to be 340 K.
(See part (d)). From Eq. 12.38,
∞
0
ε = ∫ ε λ Eλ ,b ( λ, Ts ) / E b ( Ts )
(
ε = ε1F( 0→1µ m) + ε 2 1 − F( 0→1 µ m )
)
Continued …..
PROBLEM 12.121 (Cont.)
From Table 12.1 with
λ T = 1 µm × 340 K = 340 µ m ⋅ K
find
F( 0 →λT ) = 0.000
giving
ε = 0.9 × 0.000 + 0.2 (1 − 0.000 ) = 0.20.
<
(c) The disk is a hot surface facing upwards for which the free convection correlation of Eq. 9.30 is
appropriate. Evaluating properties at Tf = (Ts + T∞)/2 = 320 K,
Ra L = gβ∆TL3 / να
Ra L = 9.8 m / s
2
where L = A s / P = D / 4
(1/320 K )( 340 − 300 ) K ( 0.4 m / 4 )3 /17.90 × 10 −6 m 2 / s × 25.5 × 10−6 m 2 / s = 3.042 × 106
Nu L = h L / k = 0.54Ra1/4
L
104 ≤ Ra L ≤ 107
(
h = 0.0278 W / m ⋅ K / ( 0.4 m / 4 ) × 0.54 3.042 × 106
)
1/4
= 6.37 W / m2 ⋅ K.
<
(d) To determine the steady-state temperature, perform
an energy balance on the disk.
E& in − E& out = E& st
(αSGs,d +α Gsky − ε Eb − q ′′conv ) A s − q cond = 0.
Since Gsky is predominately long wavelength radiation, it
follows that α = ε. The conduction heat rate between
the disk and the earth is
q cond = kS ( Ts − Tea ) = k ( 2D )( Ts − Tea )
where S, the conduction shape factor, is that of an isothermal disk on a semi-infinite medium, Table 4.1.
2
Substituting numerical values, with As = πD /4,
0.704 × 745 W / m 2 + 0.20σ ( 280 K )4 − 0.20σ Ts4

−6.3 W / m2 ⋅ K ( Ts − 300 K )  π / 4 ( 0.4 m ) 2 − 0.52 W / m ⋅ K ( 2× 0.4 m )( Ts − 280 K ) = 0

65.908 W + 8.759 W − 1.425 ×10−9 Ts4 − 0.792 ( Ts − 300 ) − 0.416 ( Ts − 280 ) = 0.
By trial-and-error, find
Ts ≈ 339 K.
so indeed the assumed value of 340 K was proper.
COMMENTS: Note why it is not necessary for this situation to distinguish between direct and
diffuse irradiation. Why does α sky = ε?
<
PROBLEM 12.122
KNOWN: Shed roof of weathered galvanized sheet metal exposed to solar insolation on a cool, clear
spring day with ambient air at - 10°C and convection coefficient estimated by the empirical
2
correlation h = 1.0 ∆T1/3 (W/m ⋅K with temperature units of kelvins).
FIND: Temperature of the roof, Ts, (a) assuming the backside is well insulated, and (b) assuming the
backside is exposed to ambient air with the same convection coefficient relation and experiences
radiation exchange with the ground, also at the ambient air temperature. Comment on whether the
roof will be a comfortable place for the neighborhood cat to snooze for these conditions.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) The roof surface is diffuse, spectrally selective,
(3) Sheet metal is thin with negligible thermal resistance, and (3) Roof is a small object compared to
the large isothermal surroundings represented by the sky and the ground.
ANALYSIS: (a) For the backside-insulated condition, the energy balance, represented schematically
below, is
E ′′in − E out
′′ = 0
$
" '
α sky E b Tsky + α SG S − q ′′cv − ε E b Ts = 0
$
4/3
4 + α G − 10
α skyσTsky
− εσTs4 = 0
S S . Ts − T∞
With α sky = ε (see Comment 2) and σ = 5.67 × 10−8 W / m2 ⋅ K 4 , find Ts.
4/3
0.65 σ 233 K 4 W / m2 + 0.8 × 600 W / m2 − 10
. Ts − 283 K
W / m2 − 0.65 σ Ts4 = 0
$
$
Ts = 312.5 K = 39.5$ C
<
Continued …..
PROBLEM 12.122 (Cont.)
(b) With the backside exposed to convection with the ambient air and radiation exchange with the
ground, the energy balance, represented schematically above, is
" '
$
" '
α sky E b Tsky + α grd E b Tgrd + α SG S − 2q cv
′′ − 2ε E b Ts = 0
Substituting numerical values, recognizing that Tgrd = T∞, and αgrd = ε (see Comment 2), find Ts.
$
$
0.65 σ 233 K 4 W / m2 + 0.65 σ 283 K 4 W / m2 + 0.8 × 600 W / m2
. Ts − 283 K
−2 × 10
Ts = 299.5 K = 26.5$ C
$4 / 3 W / m2 − 2 × 0.65 σ Ts4 = 0
<
COMMENTS: (1) For the insulated-backside condition, the cat would find the roof quite warm
remembering that 43°C represents a safe-to-touch temperature. For the exposed-backside condition,
the cat would find the roof comfortable, certainly compared to an area not exposed to the solar
insolation (that is, exposed only to the ambient air through convection).
(2) For this spectrally selective surface, the absorptivity for the sky irradiation is equal to the
emissivity, αsky = ε, since the sky irradiation and surface emission have the same approximate
spectral regions. The same reasoning applies for the absorptivity of the ground irradiation, αgrd = ε.
PROBLEM 12.123
KNOWN: Amplifier operating and environmental conditions.
FIND: (a) Power generation when Ts = 58°C with diffuse coating ε = 0.5, (b) Diffuse coating from among three (A,
B, C) which will give greatest reduction in Ts , and (c) Surface temperature for the conditions with coating chosen
in part (b).
SCHEMATIC:
ASSUMPTIONS: (1) Environmental conditions remain the same with all surface coatings, (2)
Coatings A, B, C are opaque, diffuse.
ANALYSIS: (a) Performing an energy balance
on the amplifier’s exposed surface,
E& in − E& out = 0, find
′′
Pe + As αS GS + α sky G sky − ε E b − q conv
 = 0
4 
Pe = As εσ Ts4 + h ( Ts − T∞ ) − α SG S − α skyσ Tsky


Pe = 0.13 × 0.13 m2 0.5 × σ ( 331 ) 4 + 15 ( 331 − 300 ) − 0.5 × 800− 0.5× σ ( 253 ) 4  W / m2


Pe = 0.0169m2 [0.5 × 680.6 + 465 − 0.5 × 800 − 0.5 × 232.3 ]W / m2 = 4887 W.
<
(b) From above, recognize that we seek a coating with low αS and high ε to decrease Ts . Further, recognize that
αS is determined by values of αλ = ελ for λ < 3 µm and ε by values of ελ for λ > 3 µm. Find approximate values as
Coating
A
B
C
ε
0.5
0.3
0.6
αS
0.8
0.3
0.2
αS/ε
1.6
1
0.333
Note also that αsky ≈ ε. We conclude that coating C is likely to give the lowest Ts since its αS/ε is substantially
lower than for B and C. While αsky for C is twice that of B, because Gsky is nearly 25% that of GS, we expect
coating C to give the lowest Ts .
(c) With the values of α S, α sky and ε for coating C from part (b), rewrite the energy balance as
4
Pe / A s + α SGS + α skyσ Tsky
− εσ T s4 − h ( Ts − T∞ ) = 0
4.887 W / ( 0.13 m ) + 0.2 × 800 W / m2 + 0.6 × 232.3 W / m 2 − 0.6 × σ Ts4 −15 ( Ts − 300 ) = 0
2
Using trial-and-error, find Ts = 316.5 K = 43.5°C.
<
COMMENTS: (1) Using coatings A and B, find Ts = 71 and 54°C, respectively. (2) For more precise values of
αS, αsky and ε, use Ts = 43.5°C. For example, at λTs = 3 × (43.5 + 273) = 950 µm⋅K, F0-λT = 0.000 while at λTsolar
= 3 × 5800 = 17,400 µm⋅K, F0-λT ≈ 0.98; we conclude little effect will be seen.
PROBLEM 12.124
KNOWN: Opaque, spectrally-selective horizontal plate with electrical heater on backside is exposed
to convection, solar irradiation and sky irradiation.
FIND: Electrical power required to maintain plate at 60°C.
SCHEMATIC:
ASSUMPTIONS: (1) Plate is opaque, diffuse and uniform, (2) No heat lost out the backside of
heater.
ANALYSIS: From an energy balance on
the plate-heater system, per unit area basis,
′′ = 0
E& ′′in − E& out
q′′elec + α S GS + α G sky
′′
−ε Eb ( Ts ) − qconv
=0
4
′′
where Gsky = σ Tsky
, Eb = σ Ts4 , a n d qconv
= h ( Ts − T∞ ) . The solar absorptivity is
∞
∞
∞
∞
αS = α λ G λ,Sdλ / G λ,Sdλ = α λ E λ,b ( λ, 5800 K) dλ / E λ ,b ( λ, 5 8 0 0 K ) d λ
0
0
0
0
where Gλ,S ~ Eλ,b (λ, 5800 K). Noting that α λ = 1 - ρ λ,
∫
∫
∫
∫
(
αS = (1 − 0.2 ) F( 0 − 2µ m) + (1 − 0.7 ) 1− F( 0 − 2µ m)
)
where at λT = 2 µm × 5800 K = 11,600 µm⋅K, find from Table 12.1, F(0-λT) = 0.941,
αS = 0.80 × 0.941 + 0.3 (1 − 0.941) = 0.771.
The total, hemispherical emissivity is
(
)
ε = (1 − 0.2) F( 0 − 2 µm) + (1 − 0.7 ) 1 − F( 0 − 2µ m) .
At λT = 2 µm × 333 K = 666 K, find F(0-λT) ≈ 0.000; hence ε = 0.30. The total, hemispherical
absorptivity for sky irradiation is α = ε = 0.30 since the surface is gray for this emission and
irradiation process. Substituting numerical values,
4
q′′elec = εσ Ts4 + h ( Ts − T∞ ) − αS GS − ασ Tsky
q ′′elec = 0.30 × σ ( 333 K ) + 10 W / m2 ⋅ K ( 60 − 20 ) °C − 0.771 × 600 W / m2 − 0.30 × σ ( 233 K )
4
q′′elec = 209.2 W / m 2 + 400.0 W / m2 − 462.6 W / m 2 − 50.1W/m 2 = 96.5 W / m 2.
COMMENTS: (1) Note carefully why α sky = ε for the sky irradiation.
4
<
PROBLEM 12.125
KNOWN: Chord length and spectral emissivity of wing. Ambient air temperature, sky temperature and
solar irradiation for ground and in-flight conditions. Flight speed.
FIND: Temperature of top surface of wing for (a) ground and (b) in-flight conditions.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, (2) Negligible heat transfer from back of wing surface, (3) Diffuse
surface behavior, (4) Negligible solar radiation for λ > 3 µm (αS = αλ ≤ 3 µm = ελ ≤ 3 µm = 0.6), (5)
Negligible sky radiation and surface emission for λ ≤ 3 µm (αsky = αλ > 3 µm = ελ > 3 µm = 0.3 = ε), (6)
Quiescent air for ground condition, (7) Air foil may be approximated as a flat plate, (8) Negligible
viscous heating in boundary layer for in-flight condition, (9) The wing span W is much larger than the
5
chord length Lc, (10) In-flight transition Reynolds number is 5 × 10 .
-5
2
-5
2
PROPERTIES: Part (a). Table A-4, air (Tf ≈ 325 K): ν = 1.84 × 10 m /s, α = 2.62 × 10 m /s, k =
3
-5
2
0.0282 W/m⋅K, β = 0.00307. Part (b). Given: ρ = 0.470 kg/m , µ = 1.50 × 10 N⋅s/m , k = 0.021
W/m⋅K, Pr = 0.72.
ANALYSIS: For both ground and in-flight conditions, a surface energy balance yields
αsky Gsky + αS GS = εσ Ts4 + h (Ts − T∞ )
(1)
where αsky = ε = 0.3 and αS = 0.6.
(a) For the ground condition, h may be evaluated from Eq. 9.30 or 9.31, where L = As/P = Lc × W/2 (Lc
3
+ W) ≈ Lc/2 = 2m and RaL = gβ (Ts - T∞) L /να. Using the IHT software to solve Eq. (1) and accounting
for the effect of temperature-dependent properties, the surface temperature is
<
Ts = 350.6 K = 77.6°C
10
2
where RaL = 2.52 × 10 and h = 6.2 W/m ⋅K. Heat transfer from the surface by emission and
2
convection is 257.0 and 313.6 W/m , respectively.
3
-5
2
(b) For the in-flight condition, ReL = ρu∞Lc/µ = 0.470 kg/m × 200 m/s × 4m/1.50 × 10 N⋅s/m = 2.51
7
× 10 . For mixed, laminar/turbulent boundary layer conditions (Section 7.2.3 of text) and a transition
5
Reynolds number of Rex,c = 5 × 10 .
(
4/5
Nu L = 0.037 Re L
h=
k
L
Nu L =
)
1/ 3
− 871 Pr
= 26, 800
0.021 W / m ⋅ K × 26, 800
4m
2
= 141 W / m ⋅ K
Substituting into Eq. (1), a trial-and-error solution yields
<
Ts = 237.7 K = −35.3°C
2
Heat transfer from the surface by emission and convection is now 54.3 and 657.6 W/m , respectively.
COMMENTS: The temperature of the wing is strongly influenced by the convection heat transfer
coefficient, and the large coefficient associated with flight yields a surface temperature that is within 5°C
of the air temperature.
PROBLEM 12.126
KNOWN: Spectrally selective and gray surfaces in earth orbit are exposed to solar irradiation, GS, in
a direction 30° from the normal to the surfaces.
FIND: Equilibrium temperature of each plate.
SCHEMATIC:
ASSUMPTIONS: (1) Plates are at uniform temperature, (2) Surroundings are at 0K, (3) Steadystate conditions, (4) Solar irradiation has spectral distribution of Eλ,b(λ, 5800K), (5) Back side of plate
is insulated.
ANALYSIS: Noting that the solar irradiation is directional (at 30° from the normal), the radiation
balance has the form
αSGS cos θ − ε Eb ( Ts ) = 0.
(1)
Using Eb (Ts) = σ Ts4 and solving for Ts, find
1/4
Ts = (αS / ε ) ( GS cos θ / σ )  .
(2)
For the gray surface, α S = ε = α λ and the temperature is independent of the magnitude of the
absorptivity.
1/4
2
 0.95 1353 W / m × cos 30° 

Ts = 
×
 0.95 5.67 ×10−8 W / m 2 ⋅ K 4 


= 379 K.
<
For the selective surface, α S = 0.95 since nearly all the solar spectral power is in the region λ < 3µm.
The value of ε depends upon the surface temperature Ts and would be determined by the relation.
ε = 0.95 F(0 →λT ) + 0.05 1 − F( 0 →λT ) 
s
s 

(3)
where λ = 3µm and Ts is as yet unknown. To find Ts, a trial-and-error procedure as follows will be
used: (1) assume a value of Ts, (2) using Eq. (3), calculate ε with the aid of Table 12.1 evaluating
F(0→λT) at λTs = 3µm⋅Ts, (3) with this value of ε, calculate Ts from Eq. (2) and compare with
assumed value of Ts. The results of the iterations are:
Ts(K), assumed value
ε, from Eq. (3)
Ts(K), from Eq. (2)
633
700
666
650
655
0.098 0.125 0.110 0.104 0.106
656
629
650
659
656
Hence, for the coating, Ts ≈ 656K.
COMMENTS: Note the role of the ratio α s/ε in determining the equilibrium temperature of an
isolated plate exposed to solar irradiation in space. This is an important property of the surface in
spacecraft thermal design and analysis.
<
PROBLEM 12.127
KNOWN: Spectral, hemispherical emissivity distributions for two panels subjected to solar flux in the
deep space environment.
FIND: Steady-state temperatures of the panels.
SCHEMATIC:
ASSUMPTIONS: (1) Surfaces are opaque and diffuse, (2) Panels are oriented normal to solar flux
with backside insulated, (3) Steady-state conditions, (4) No convection.
ANALYSIS: An energy balance on the panel is
q′′in − q′′out = 0
αS GS − ε E b ( Ts ) = αS GS − εσ Ts4 = 0
1/4
Ts = (αS / ε )( GS / σ ) 
.
For each panel determine α S and ε. Recognizing that Gλ,S ~ Eλ,b (λ, 5800K), the solar absorptivity
from Eq. 12.47 is
∞
αS
∞
∫ αλ Gλ ,S dλ = ∫0
= 0
∞
∫0
Gλ ,S dλ
ελ Eλ ,b ( λ ,5800K )
E b ( 5800K )
.
Note that ε λ = α λ since the surface is diffuse. Using Eq. 12.65 and Table 12.1 find
Surface A:
Surface B:
}
α S = ε1 F( 0 → 2 µ m ) + ε 2 1 − F(0 → 2 µ m ) 
λT = 2 × 5800 = 11,600µm ⋅ K,
α S = 0.5 × 0.940 + 0.2 [1 − 0.940 ] = 0.482}
F( 0 → 2 µ m ) = 0.940
α S = 0.11 × 0.979 +0.02 [1 − 0.979 ] = 0.108}
λ T = 3 × 5800 = 17,400µm ⋅ K,


F( 0 →3 µm ) = 0.979.
To determine the total emissivity, we need to know Ts. If Ts ≤ 400K, then for λT = 3 µm × 400K =
1200K, F(0→λ) = 0.002. That is, there is negligible power for λ < 3 µm if Ts ≤ 400K, and hence
Surface B:
ε ≈ ε 2 = 0.02.
Surface A:
ε ≈ ε 2 = 0.2
Substituting the solar absorptivity and emissivity values, find
1/4
Surface A:
 0.482

1353 W / m2
Ts = 
×

 0.20 5.67 ×10− 8 W / m2 ⋅ K 


Surface B:
 0.108

1353 W / m2
Ts = 
×

 0.02 5.67 ×10− 8 W / m2 ⋅ K 


= 499K
<
= 599K.
<
1/4
COMMENTS: (1) Note the assumption that Ts ≤ 400K used for finding ε is not satisfied; for better
precision, it is necessary to perform an iterative solution. (2) Note the importance of the α S/ε ratio
which determines the surface temperature.
PROBLEM 12.128
KNOWN: Radiative properties and operating conditions of a space radiator.
FIND: Equilibrium temperature of the radiator.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible irradiation due to earth emission.
ANALYSIS: From a surface energy balance, E& ′′in − E& ′′out = 0.
q′′dis + αS GS − E = 0.
Hence
1/4
 q′′ + αS GS 
Ts =  dis

εσ


1/4
 1500W/m 2 + 0.5 ×1000W/m2 

Ts = 
 0.95 × 5.67 ×10−8 W / m 2 ⋅ K 4 


or
Ts = 439K.
<
COMMENTS: Passive thermal control of spacecraft is practiced by using surface coatings with
desirable values of α S and ε.
PROBLEM 12.129
2
KNOWN: Spherical satellite exposed to solar irradiation of 1353 m ; surface is to be coated with a
checker pattern of evaporated aluminum film, (fraction, F) and white zinc-oxide paint (1 - F).
FIND: The fraction F for the checker pattern required to maintain the satellite at 300 K.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Satellite is isothermal, and (3) No internal power
dissipation.
ANALYSIS: Perform an energy balance on the satellite, as illustrated in the schematic, identifying
absorbed solar irradiation on the projected area, Ap, and emission from the spherical area As.
E in − E out = o
"F ⋅ α S,f + 1 − F$ ⋅α S,p ' GS A p − "F ⋅ ε f + 1 − F$ ⋅ ε p ' E b Ts $ A s = 0
where A p = π D 2 / 4, A s = π D 2 , E b = σT4 and σ = 5.67 × 10-8 W / m2 ⋅ K 4 . Substituting
numerical values, find F.
$
$
F × 0.03 + 1 − F$ × 0.85%σ 300 K$4 × 1 = 0
%
F × 0.09 + 1 − F × 0.22 × 1353 W / m2 × 1/ 4
−
<
F = 0.95
COMMENTS: (1) If the thermal control engineer desired to maintain the spacecraft at 325 K, would
the fraction F (aluminum film) be increased or decreased? Verify your opinion with a calculation.
2
(2) If the internal power dissipation per unit surface area is 150 W/m , what fraction F will maintain
the satellite at 300 K?
PROBLEM 12.130
KNOWN: Inner and outer radii, spectral reflectivity, and thickness of an annular fin. Base temperature
and solar irradiation.
FIND: (a) Rate of heat dissipation if ηf = 1, (b) Differential equation governing radial temperature
distribution in fin if ηf < 1.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, (2) One-dimensional radial conduction, (3) Adiabatic tip and bottom
surface, (4) Opaque, diffuse surface ( α λ = 1 − ρλ , ε λ = α λ ).
ANALYSIS: (a) If ηf = 1, T(r) = Tb = 400 K across the entire fin and
q f = [ε E b ( Tb ) − αSGS ]π ro2
With λT = 2 µm × 5800 K = 11,600 µm⋅K, F(0→2µm) = 0.941. Hence αS = α1 F( 0→ 2 µ m ) +
α2 1 − F( 0→ 2 µ m )  = 0.2 × 0.941 + 0.9 × 0.059 = 0.241. With λT = 2 µm × 400 K = 800 µm⋅K,


F( 0→ 2 µ m ) = 0 and ε = 0.9. Hence, for GS = 0,
q f = 0.9 × 5.67 × 10−8 W m 2⋅ K 4 ( 400 K ) π ( 0.5 m ) = 1026 W
and for GS = 1000 W/m2,
4
(
)
2
q f = 1026 W − 0.241 1000 W m 2 π ( 0.5 m ) = (1026 − 189 ) W = 837 W
2
<
<
(b) Performing an energy balance on a differential element extending from r to r+dr, we obtain
q r + αSGS ( 2π rdr ) − q r + dr − E ( 2π rdr ) = 0
where
and
q r = − k ( dT dr ) 2π rt
q r + dr = q r + ( dq r dr ) dr .
Hence,
αSGS ( 2π rdr ) − d [− k ( dT dr ) 2π rt ] dr − E ( 2π rdr ) = 0
2π rtk
d 2T
dr
2
+ 2π tk
dT
dr
+ αSGS 2π r − E2π r = 0
 d 2T 1 dT 
+
 + α G − εσ T 4 = 0
 dr 2 r dr  S S


kt 
<
COMMENTS: The radiator should be constructed of a light weight, high thermal conductivity material
(aluminum).
PROBLEM 12.131
KNOWN: Rectangular plate, with prescribed geometry and thermal properties, for use as a radiator
in a spacecraft application. Radiator exposed to solar radiation on upper surface, and to deep space
on both surfaces.
FIND: Using a computer-based, finite-difference method with a space increment of 0.1 m, find the
tip temperature, TL, and rate of heat rejection, qf, when the base temperature is maintained at 80°C for
the cases: (a) when exposed to the sun, (b) on the dark side of the earth, not exposed to the sun; and
(c) when the thermal conductivity is extremely large. Compare the case (c) results with those
obtained from a hand calculation assuming the radiator is at a uniform temperature.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (b) Plate-radiator behaves as an extended surface with
one-dimensional conduction, and (c) Radiating tip condition.
ANALYSIS: The finite-difference network with 10 nodes and a space increment ∆x = 0.1 m is
shown in the schematic below. The finite-difference equations (FDEs) are derived for an interior
node (nodes 01 - 09) and the tip node (10). The energy balances are represented also in the schematic
below where qa and qb represent conduction heat rates, qS represents the absorbed solar radiation, and
qrad represents the radiation exchange with outer space.
Interior node 04
E in − E out = 0
q a + q b + q S + q rad = 0
kA c T03 − T04 / ∆x + kA c T05 − T04 / ∆x
1
6
1
1 6
6
4 − T4 = 0
+α S G S P / 2 ∆x + εP∆xσ Tsur
04
4
9
where P = 2W and Ac = W⋅t.
Tip node 10
q a + q S + q rad,1 + q rad,2 = 0
kA c T09 − T10 / ∆x + α SGS P / 2
1
6
1 6 1∆x / 26
4 − T4 + εP ∆x / 2 σ T4 − T4 = 0
+ε A cσ 4Tsur
1 6 4 sur 04 9
10 9
Continued …..
PROBLEM 12.131 (Cont.)
Heat rejection, qf. From an energy balance on the base node 00,
q f + q 01 + q S + q rad = 0
q f + kA c T01 − T00 / ∆x + α SGS P / 2
1
1 6 1∆x / 26
6
)
(
4 − T4 = 0
+ ε P ( ∆x/2 )σ Tsur
00
The foregoing nodal equations and the heat rate expression were entered into the IHT workspace to
obtain solutions for the three cases. See Comment 2 for the IHT code, and Comment 1 for code
validation remarks.
Case
k(W/m⋅K)
a
300
b
300
c
1 × 10
2
GS(W/m )
TL(°C)
qf(W)
1353
30.5
2766
0
-7.6
4660
0
80.0
9557
10
<
<
10
COMMENTS: (1) Case (c) using the IHT code with k = 1 × 10 W/m⋅K corresponds to the
condition of the plate at the uniform temperature of the base; that is T(x) = Tb. For this condition, the
heat rejection from the upper and lower surfaces and the tip area can be calculated as
4
q f,u = εσ Tb4 − Tsur
P ⋅ L + Ac
4
9
1
!
6
"#
$
q f,u = 0.65 σ 80 + 273 4 − 44 W / m2 12 + 6 × 0.012 m2
q f,u = 9565 W / m2
Note that the heat rejection rate for the uniform plate is in excellent agreement with the result of the
FDE analysis when the thermal conductivity is made extremely large. We have confidence that the
code is properly handling the conduction and radiation processes; but, we have not exercised the
portion of the code dealing with the absorbed irradiation. What analytical solution/model could you
use to validate this portion of the code?
(2) Selection portions are shown below of the IHT code with the 10-nodal FDEs for the temperature
distribution and the heat rejection rate.
// Finite-difference equations
// Interior nodes, 01 to 09
k * Ac * (T00 - T01) / deltax + k * Ac * (T02 - T01) / deltax + absS * GS * P/2 * deltax + eps * P *
deltax * sigma * (Tsur^4 - T01^4) = 0
…..
…..
k * Ac * (T03 - T04) / deltax + k * Ac * (T05 - T04) / deltax + absS * GS * P/2 * deltax + eps * P *
deltax * sigma * (Tsur^4 - T04^4) = 0
…..
…..
k * Ac * (T08 - T09) / deltax + k * Ac * (T10 - T09) / deltax + absS * GS * P/2 * deltax + eps * P *
deltax * sigma * (Tsur^4 - T09^4) = 0
// Tip node 10
k* Ac * (T09 - T10) / deltax + absS * GS * P/2 * (deltax / 2) + eps * P * (deltax / 2) * sigma *
(Tsur^4 - T10^4) - eps * Ac * sigma * (Tsur^4 - T00^4) = 0
// Rejection heat rate, energy balance on base node
qf + k * Ac * (T01 - T00) / deltax + absS * GS * (P/4) * (deltax /2) + eps * (P * deltax /2) *
sigma * (Tsur^4 - T00^4) = 0
Continued …..
PROBLEM 12.131 (Cont.)
(3) To determine the validity of the one-dimensional, extended surface analysis, calculate the Biot
number estimating the linearized radiation coefficient based upon the uniform plate condition, Tb =
80°C.
1 6
2 ≈ εσT3 = 2.25 W / m2 ⋅ K
h rad = εσ 1Tb + Tsur 6 4Tb2 + Tsur
9 b
Bi = h rad t / 2 / k
1
6
Bi = 2.25 W / m2 ⋅ K 0.012 m / 2 / 300 W / m ⋅ K = 4.5 × 10 −5
Since Bi << 0.1, the assumption of one-dimensional conduction is appropriate.
PROBLEM 12.132
KNOWN: Directional absorptivity of a plate exposed to solar radiation on one side.
FIND: (a) Ratio of normal absorptivity to hemispherical emissivity, (b) Equilibrium temperature of
plate at 0° and 75° orientation relative to sun’s rays.
SCHEMATIC:
ASSUMPTIONS: (1) Surface is gray, (2) Properties are independent of φ.
ANALYSIS: (a) From the prescribed α θ (θ), α n = 0.9. Since the surface is gray, ε θ = α θ. Hence
from Eq. 12.36, which applies for total as well as spectral properties.
π /2

2 π /3
π /2
sin 2 θ
 sin θ

ε=2
εθ cos θ sin θ dθ = 2  0.9
+0.1

0
2
2
0

π / 3 
∫
ε = 2  0.9 ( 0.375) + 0.1( 0.5 − 0.375 )  = 0.70.
Hence
α n 0.9
=
= 1.286.
ε
0.7
(b) Performing an energy balance on the plate,
αθ q ′′s cos θ − 2 ε σ Ts4 = 0
or
1/4
 α

Ts =  θ q s′′ cos θ 
2ε σ

.
Hence for θ = 0°, α θ = 0.9 and cosθ = 1,
1/4


0.9
× 1353
Ts = 
 2 × 0.7 × 5.67 ×10−8

= 352K.
<
For θ = 75°, α θ = 0.1 and cosθ = 0.259
1/4


0.1
Ts = 
× 1353 × 0.259
 2 × 0.7 × 5.67 ×10−8

= 145K.
<
COMMENTS: Since the surface is not diffuse, its absorptivity depends on the directional distribution
of the incident radiation.
PROBLEM 12.133
KNOWN: Transmissivity of cover plate and spectral absorptivity of absorber plate for a solar
collector.
FIND: Absorption rate for prescribed solar flux and preferred absorber plate coating.
SCHEMATIC:
ASSUMPTIONS: (1) Solar irradiation of absorber plate retains spectral distribution of blackbody at
5800K, (2) Coatings are diffuse.
ANALYSIS: At the absorber plate we wish to maximize solar radiation absorption and minimize
losses due to emission. The solar radiation is concentrated in the spectral region λ < 4µm, and for a
representative plate temperature of T ≤ 350K, emission from the plate is concentrated in the spectral
region λ > 4µm. Hence,
<
Coating A is vastly superior.
With Gλ,S ~ Eλ,b (5800K), it follows from Eq. 12.47
α A ≈ 0.85 F( 0 − 4 µm ) + 0.05 F( 4 µm −∞ ).
From Table 12.1, λT = 4µm × 5800K = 23,200µm⋅K,
F( 0 − 4µ m) ≈ 0.99.
Hence
α A = 0.85 ( 0.99 ) + 0.05 (1 − 0.99 ) ≈ 0.85.
2
With GS = 1000 W/m and τ = 0.76 (Ex. 12.9), the absorbed solar flux is
(
GS,abs = α A (τ G S ) = 0.85 0.76× 1000 W / m 2
GS,abs = 646 W / m 2 .
)
<
COMMENTS: Since the absorber plate emits in the infrared (λ > 4µm), its emissivity is ε A ≈ 0.05.
Hence (α/ε)A = 17. A large value of α/ε is desirable for solar absorbers.
PROBLEM 12.134
KNOWN: Spectral distribution of coating on satellite surface. Irradiation from earth and sun.
FIND: (a) Steady-state temperature of satellite on dark side of earth, (b) Steady-state temperature on
bright side.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Opaque, diffuse-gray surface behavior, (3)
Spectral distributions of earth and solar emission may be approximated as those of blackbodies at 280K
and 5800K, respectively, (4) Satellite temperature is less than 500K.
ANALYSIS: Performing an energy balance on the satellite,
E& in − E& out = 0
(
)
(
)
(
)
α E GE π D2 / 4 + αS GS π D2 / 4 − ε σ Ts4 π D2 = 0
1/4
 α G + αS GS 
Ts =  E E

4ε σ


.
From Table 12.1, with 98% of radiation below 3µm for λT = 17,400µm⋅K,
αS ≅ 0.6.
With 98% of radiation above 3µm for λT = 3µm × 500K = 1500µm⋅K,
ε ≈ 0.3
α E ≈ 0.3.
(a) On dark side,
1/4
α G 
Ts =  E E 
 4ε σ 
1/4
2


0.3 × 3 4 0 W / m


=
 4 × 0.3 × 5.67 × 10−8 W / m 2 ⋅ K 4 


<
Ts = 197 K.
(b) On bright side,
1/4
 α G + αS GS 
Ts =  E E

4ε σ


Ts = 340K.
1/4
 0.3 × 340 W / m 2 + 0.6 × 1353W/m2 

=
 4 × 0.3 × 5.67 × 10−8 W / m2 ⋅ K4 


<
PROBLEM 12.135
KNOWN: Space capsule fired from earth orbit platform in direction of sun.
FIND: (a) Differential equation predicting capsule temperature as a function of time, (b) Position of
capsule relative to sun when it reaches its destruction temperature.
SCHEMATIC:
ASSUMPTIONS: (1) Capsule behaves as lumped capacitance system, (2) Capsule surface is black,
(3) Temperature of surroundings approximates absolute zero, (4) Capsule velocity is constant.
ANALYSIS: (a) To find the temperature as a function of time, perform an energy balance on the
capsule considering absorbed solar irradiation and emission,
E& in − E& out = E& st
GS ⋅π R 2 − σ T 4 ⋅ 4 πR 2 = ρ c ( 4 / 3 ) π R 3 ( dT/dt ) .
(1)
2
2
Note the use of the projected capsule area (πR ) and the surface area (4πR ). The solar irradiation
will increase with decreasing radius (distance toward the sun) as
2
2
2
GS ( r ) = GS,e ( re / r ) = GS,e re / ( re − Vt ) = GS,e 1/ (1 − V t / re )
(2)
(
)
(
)
where re is the distance of earth orbit from the sun and r = re – Vt. Hence, Eq. (1) becomes


GS,e
dT
3 
4
=
−σ T  .
dt ρ cR  4 (1 − V t / r )2

e


The rate of temperature change is


2


dT
3
1353 W / m
=
− σ T4 

2
dt

4 ×106 J / m 3 ⋅ K ×1.5m  4 1 − 16 × 103 m / s × t/1.5 × 1011m


−2
dT
= 1.691 ×10−4 1 −1.067 ×10 −7 t
− 2.835 ×10−14 T 4
dt
(
) (
)
)
(
where T[K] and t(s). For the initial condition, t = 0, with T = 20°C = 293K,
dT
( 0 ) = −3.984 ×10−5 K/s.
dt
<
That is, the capsule will cool for a period of time and then begin to heat.
(b) The differential equation cannot be explicitly solved for temperature as a function of time. Using a
5
numerical method with a time increment of ∆t = 5 × 10 s, find
T ( t ) = 150 °C = 423 K
at
t ≈ 5.5 ×106 s.
<
3
6
Note that in this period of time the capsule traveled (re – r) = Vt = 16 × 10 m/s × 5.5 × 10 = 1.472 ×
10
11
10 m. That is, r = 1.353 × 10 m.
PROBLEM 12.136
KNOWN: Dimensions and spectral absorptivity of radiator used to dissipate heat to outer space.
Radiator temperature. Magnitude and direction of incident solar flux.
FIND: Power dissipation within radiator.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible heat loss through sides and bottom of
compartment, (3) Opaque, diffuse surface.
ANALYSIS: Applying conservation of energy to a control surface about the compartment yields
E& in + E& g = E& out
(
)
E& g = εσ Tp4 −α G s A.
The emissivity can be expressed as
∞
(
)
ε = ∫ ε λ Eλ ,b / Eb dλ = ε1F 0 → λ + ε 2F λ →∞ .
(
( 1 )
0
1)
From Table 12.1: λ1T = 1000 µm⋅K → F( 0 → λ ) = 0.000321
1
ε = 0.2 ( 0.000321) + 0.8 (1 − 0.00321 ) = 0.8.
The absorptivity can be expressed as
∞
∞
α = ∫ α λ ( G λ / G ) d λ = ∫ α λ E λ ,b (5800 K ) / E b ( 5800 K )  dλ.
0
0
From Table 12.1: λ1T = 11,600 µm ⋅ K → F( 0→λ ) = 0.941,
1
α = 0.2 ( 0.941) + 0.8 (0.059 ) = 0.235.
Hence,
(
)
E& g = 0.8 × 5.67 ×10 −8 W / m 2 ⋅ K 4 × ( 500 K ) 4 − 0.235 cos30° 1350 W / m 2  1 m 2


E& g = 2560 W.
<
COMMENTS: Solar irradiation and plate emission are concentrated at short and long wavelength
portions of the spectrum. Hence, α ≠ ε and the surface is not gray for the prescribed conditions.
PROBLEM 12.137
2
KNOWN: Solar panel mounted on a spacecraft of area 1 m having a solar-to-electrical power
conversion efficiency of 12% with specified radiative properties.
FIND: (a) Steady-state temperature of the solar panel and electrical power produced with solar
2
irradiation of 1500 W/m , (b) Steady-state temperature if the panel were a thin plate (no solar cells)
with the same radiative properties and for the same prescribed conditions, and (c) Temperature of the
solar panel 1500 s after the spacecraft is eclipsed by the earth; thermal capacity of the panel per unit
2
area is 9000 J/m ⋅K.
SCHEMATIC:
ASSUMPTIONS: (1) Solar panel and thin plate are isothermal, (2) Solar irradiation is normal to the
panel upper surface, and (3) Panel has unobstructed view of deep space at 0 K.
ANALYSIS: (a) The energy balance on the solar panel is represented in the schematic below and has
the form
E in − E out = 0
$ " '
α SG S ⋅ A p − ε a + ε b E b Tsp ⋅ A p − Pelec = 0
4
-8
2
(1)
4
where Eb (T) = σT , σ = 5.67 × 10 W/m ⋅K , and the electrical power produced is
Pelec = e ⋅ G S ⋅ A p
(2)
Pelec = 012
. × 1500 W / m2 × 1 m2 = 180 W
<
Substituting numerical values into Eq. (1), find
4 × 1 m2 − 180 W = 0
0.8 × 1500 W / m2 × 1 m2 − 0.8 + 0.7 σTsp
$
Tsp = 330.9 K = 57.9$ C
<
(b) The energy balance for the thin plate shown in the schematic above follows from Eq. (1) with
Pelec = 0 yielding
0.8 × 1500 W / m2 × /m2 − 0.8 + 0.7 σTp4 × 1 m2 = 0
(3)
$
Tp = 344.7 K = 71.7$ C
<
Continued …..
PROBLEM 12.137 (Cont.)
(c) Using the lumped capacitance method, the energy balance on the solar panel as illustrated in the
schematic below has the form
E in − E out = E st
4 ⋅ A = TC′′ ⋅ A
− (ε a + ε b )σ Tsp
p
p
dTsp
(4)
dt
where the thermal capacity per unit area is TC ′′ = Mc / A p = 9000 J / m2 ⋅ K.
"
'
Eq. 5.18 provides the solution to this differential equation in terms of t = t (Ti, Tsp). Alternatively,
use Eq. (4) in the IHT workspace (see Comment 4 below) to find
$
Tsp 1500 s = 242.6 K = −30.4$ C
<
COMMENTS: (1) For part (a), the energy balance could be written as
E in − E out + E g = 0
where the energy generation term represents the conversion process from thermal energy to electrical
energy. That is,
E g = − e ⋅ GS ⋅ A p
(2) The steady-state temperature for the thin plate, part (b), is higher than for the solar panel, part (a).
This is to be expected since, for the solar panel, some of the absorbed solar irradiation (thermal
energy) is converted to electrical power.
(3) To justify use of the lumped capacitance method for the transient analysis, we need to know the
effective thermal conductivity or internal thermal resistance of the solar panel.
(4) Selected portions of the IHT code using the Models Lumped | Capacitance tool to perform the
transient analysis based upon Eq. (4) are shown below.
// Energy balance, Model | Lumped Capacitance
/ * Conservation of energy requirement on the control volume, CV. * /
Edotin - Edotout = Edotst
Edotin = 0
Edotout = Ap * (+q”rad)
Edostat = rhovolcp * Ap * Der(T,t)
// rhovolcp = rho * vol * cp
// thermal capacitance per unit area, J/m^2⋅K
// Radiation exchange between Cs and large surroundings
q”rad = (eps_a + eps_b) * sigma * (T^4 - Tsur^4)
sigma = 5.67e-8 // Stefan-Boltzmann constant, W/m^2⋅K^4
// Initial condition
// Ti = 57.93 + 273 = 330.9
T_C = T - 273
// From part (a), steady-state condition
PROBLEM 12.138
KNOWN: Effective sky temperature and convection heat transfer coefficient associated with a thin
layer of water.
FIND: Lowest air temperature for which the water will not freeze (without and with evaporation).
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Bottom of water is adiabatic, (3) Heat and mass
transfer analogy is applicable, (4) Air is dry.
3
PROPERTIES: Table A-4, Air (273 K, 1 atm): ρ = 1.287 kg/m , cp = 1.01 kJ/kg⋅K, ν = 13.49 ×
-6 2
-3
3
10 m /s, Pr = 0.72; Table A-6, Saturated vapor (Ts = 273 K): ρ A = 4.8 × 10 kg/m , hfg = 2502
-4 2
kJ/kg; Table A-8, Vapor-air (298 K): DAB ≈ 0.36 × 10 m /s, Sc = ν/DAB = 0.52.
ANALYSIS: Without evaporation, the surface heat loss by radiation must be balanced by heat gain
due to convection. An energy balance gives
q′′conv = q′′rad
(
)
4 .
h ( T∞ − Ts ) = εs σ Ts4 − Tsky
or
At freezing, Ts = 273 K. Hence
εσ
5.67 ×10 −8 W / m2 ⋅ K 4 
4
T∞ = Ts + s Ts4 − Tsky
2744 − 2434  K4 = 4.69 °C.
= 273 K +
2


h
2 5 W / m ⋅K
(
)
<
With evaporation, the surface energy balance is now
(
)
4 .
q′′conv = q ′′evap + q′′rad or h ( T∞ − Ts ) = h m  ρ A,sat (Ts ) − ρ A,∞  h fg + ε sσ Ts4 − Tsky
)
(
h
εσ
4 .
T∞ = Ts + m ρ A,sat ( Ts ) h fg + s Ts4 − Tsky
h
h
Substituting from Eq. 6.92, with n ≈ 0.33,
(
h m / h = ρ cp Le
0.67
)
−1
=  ρ cp ( S c / P r )

0.67


−1
= 1.287 k g / m × 1010J/kg ⋅ K ( 0.52/0.72 )

3
0.67


−1
= 9.57 × 10
−4
3
m ⋅K/J,
T∞ = 273K + 9.57 ×10 −4 m 3 ⋅ K / J × 4.8 × 10−3 k g / m 3 × 2.5 × 106 J / k g + 4.69 K = 16.2°C.
<
COMMENTS: The existence of clear, cold skies and dry air will allow water to freeze for ambient
air temperatures well above 0°C (due to radiative and evaporative cooling effects, respectively). The
lowest air temperature for which the water will not freeze increases with decreasing φ ∞, decreasing
Tsky and decreasing h.
PROBLEM 12.139
KNOWN: Temperature and environmental conditions associated with a shallow layer of water.
FIND: Whether water temperature will increase or decrease with time.
SCHEMATIC:
ASSUMPTIONS: (1) Water layer is well mixed (uniform temperature), (2) All non-reflected
radiation is absorbed by water, (3) Bottom is adiabatic, (4) Heat and mass transfer analogy is
applicable, (5) Perfect gas behavior for water vapor.
3
PROPERTIES: Table A-4, Air (T = 300 K, 1 atm): ρ a = 1.161 kg/m , cp,a = 1007 J/kg⋅K, Pr =
3
0.707; Table A-6, Water (T = 300 K, 1 atm): ρ w = 997 kg/m , cp,w = 4179 J/kg⋅K; Vapor (T = 300 K,
3
6
1 atm): ρ A,sat = 0.0256 kg/m , hfg = 2.438 × 10 J/kg; Table A-8, Water vapor-air (T = 300 K, 1 atm):
-4 2
-6 2
DAB ≈ 0.26 × 10 m /s; with νa = 15.89 × 10 m /s from Table A-4, Sc = νa/DAB = 0.61.
ANALYSIS: Performing an energy balance on a control volume about the water,
(
)
E& st = G S,abs + GA,abs − E − q ′′evap A
(
d ρw cp,w LATw
) = (1 − ρ

dt
(
)
4

s ) GS + (1 − ρ A ) GA − εσ Tw − hm hfg ρ A,sat − ρ A,∞  A
or, with T∞ = Tw, ρ A,∞ = φ ∞ρ A,sat and
ρ w cp,w L
dTw
4 − h h 1−φ ρ
= (1 − ρs ) GS + (1 − ρ A ) GA − εσ Tw
∞ ) A,sat .
m fg (
dt
From Eq. 6.92, with a value of n = 1/3,
hm =
h
ρ a cp,a Le1 −n
=
h
ρ a cp,a ( Sc/Pr )1 −n
=
2 5 W / m2 ⋅ K ( 0.707 )2 / 3
1.161kg/m3 × 1007 J / k g ⋅ K ( 0.61)2 / 3
= 0.0236m/s.
Hence
ρ w cp,w L
dTw
= (1 − 0.3) 600 + (1 − 0) 300 − 0.97 × 5.67 × 10−8 ( 300) 4
dt
−0.0236 × 2.438 ×106 (1 − 0.5 ) 0.0256
ρ w cp,w L
dTw
= ( 420 + 300 − 445 − 736) W / m2 = −4 6 1 W / m 2 .
dt
Hence the water will cool.
<
COMMENTS: (1) Since Tw = T∞ for the prescribed conditions, there is no convection of sensible
energy. However, as the water cools, there will be convection heat transfer from the air. (2) If L =
-4
1m, (dTw/dt) = -461/(997 × 4179 × 1) = -1.11 × 10 K/s.
PROBLEM 12.140
KNOWN: Environmental conditions for a metal roof with and without a water film.
FIND: Roof surface temperature (a) without the film, (b) with the film.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Diffuse-gray surface behavior in the infrared (for
the metal, α sky = ε = 0.3; for the water, α sky = ε = 0.9), (3) Adiabatic roof bottom, (4) Perfect gas
behavior for vapor.
3
-6
PROPERTIES: Table A-4, Air (T ≈ 300 K): ρ = 1.16 kg/m , cp = 1007 J/kg⋅K, α = 22.5 × 10
2
3
3
m /s; Table A-6, Water vapor (T ≈ 303 K): νg = 32.4 m /kg or ρ A,sat = 0.031 kg/m ; Table A-8,
-4 2
Water vapor-air (T = 298 K): DAB = 0.26 × 10 m /s.
ANALYSIS: (a) From an energy balance on the metal roof
α SGS + α skyGsky = E + q ′′conv
)
(
0.5 700 W / m2 + 0.3 × 5.67 × 10 −8 W / m2 ⋅ K 4 ( 263 K ) 4
( )
= 0.3 × 5.67 × 10 −8 W / m2 ⋅ K4 Ts4 + 20 W / m2 ⋅ K ( Ts − 303 K )
4 3 1 W / m2 = 1.70 × 10 −8 Ts4 + 20 (Ts − 303 ).
<
From a trial-and-error solution, Ts = 316.1 K = 43.1°C.
(b) From an energy balance on the water film,
α SGS + α skyGsky = E + q ′′conv + q ′′evap
(
)
( )
0.8 700 W / m2 + 0.9 × 5.67 × 10 −8 W / m2 ⋅ K4 ( 263 K ) 4 = 0.9 × 5.67 × 10− 8 W / m2 ⋅ K4 Ts4
)
(
+20 W / m2 ⋅ K ( Ts − 303 ) + h m ρ A,sat ( Ts ) − 0.65 × 0.031kg/m3 h fg.
From Eq. 6.92, assuming n = 0.33,
h
hm =
=
ρ cp Le 0.67
2
h
ρ c p (α / D AB )
0.67
20 W / m ⋅ K
=
3
(
−4
1.16 k g / m × 1007 J / k g ⋅ K 0.225 × 10
/0.260 × 10
−4
)
0.67
= 0.019m/s.
804 W / m2 = 5.10 ×10−8 Ts4 + 20 ( Ts − 303) + 0.019  ρA,sat ( Ts ) − 0.020  hfg .
From a trial-and-error solution, obtaining ρ A,sat (Ts) and hfg from Table A-6 for each assumed value of
Ts, it follows that
Ts = 302.2 K = 29.2°C.
<
COMMENTS: (1) The film is an effective coolant, reducing Ts by 13.9°C. (2) With the film E ≈ 425
2
2
2
W/m , q ′′conv ≈ -16 W/m and q ′′evap ≈ 428 W/m .
PROBLEM 12.141
KNOWN: Solar, sky and ground irradiation of a wet towel. Towel dimensions, emissivity and solar
absorptivity. Temperature, relative humidity and convection heat transfer coefficient associated with
air flow over the towel.
FIND: Temperature of towel and evaporation rate.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, (2) Diffuse-gray surface behavior of towel in the infrared (αsky =
αg = ε = 0.96), (3) Perfect gas behavior for vapor.
3
PROPERTIES: Table A-4, Air (T ≈ 300 K): ρ = 1.16 kg/m , cp = 1007 J/kg⋅K, α = 0.225 × 10
2
-4
3
m /s; Table A-6, Water vapor (T∞ = 300 K): ρA,sat = 0.0256 kg/m ; Table A-8, Water vapor/air (T =
-4 2
298 K): DAB = 0.26 × 10 m /s.
ANALYSIS: From an energy balance on the towel, it follows that
αSGS + 2αsky G sky + 2α g G g = 2E + 2q′′evap + 2q′′conv
0.65 × 900W / m 2 + 2 × 0.96 × 200 W / m 2 × 2 × 0.96 × 250 W / m 2
= 2 × 0.96 σ Ts4 + 2n ′′A h fg + 2h (Ts − T∞ )
(1)
where n ′′A = h m  ρ A,sat ( Ts ) − φ∞ ρ A,sat ( T∞ )
From the heat and mass transfer analogy, Eq. 6.67, with an assumed exponent of n = 1/3,
h
20 W / m 2 ⋅ K
hm =
=
= 0.0189 m / s
2/3
2/3
ρ cp (α / DAB )
 0.225 
3
1.16 kg / m (1007 J / kg ⋅ K )


 0.260 
3
From a trial-and-error solution, we find that for Ts = 298 K, ρA,sat = 0.0226 kg/m , hfg = 2.442 × 10
-4
2
J/kg and n′′A = 1.380 × 10 kg/s⋅m . Substituting into Eq. (1),
4
2
2 4
−8
(585 + 384 + 480 ) W / m
= 2 × 0.96 × 5.67 ×10
W / m ⋅K
6
( 298 K )
+2 × 1.380 × 10−4 kg / s ⋅ m 2 × 2.442 × 106 J / kg
+2 × 20 W / m 2 ⋅ K ( −2 K )
1449 W / m 2 = (859 + 674 − 80 ) W / m 2 = 1453 W / m 2
The equality is satisfied to a good approximation, in which case
Ts ≈ 298 K = 25°C
and
(
)
n A = 2 As n ′′A = 2 (1.50 × 0.75 ) m 2 1.38 × 10−4 kg / s ⋅ m 2 = 3.11×10−4 kg / s
<
<
COMMENTS: Note that the temperature of the air exceeds that of the towel, in which case
convection heat transfer is to the towel. Reduction of the towel’s temperature below that of the air is
due to the evaporative cooling effect.
PROBLEM 12.142
KNOWN: Wet paper towel experiencing forced convection heat and mass transfer and irradiation from
radiant lamps. Prescribed convection parameters including wet and dry bulb temperature of the air
stream, Twb and T∞ , average heat and mass transfer coefficients, h and h m . Towel temperature Ts.
FIND: (a) Vapor densities, ρ A ,s and ρ A,∞ ; the evaporation rate nA (kg/s); and the net rate of radiation
transfer to the towel qrad (W); and (b) Emissive power E, the irradiation G, and the radiosity J, using the
results from part (a).
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible heat loss from the bottom side of the
towel, (3) Uniform irradiation on the towel, and (4) Water surface is diffuse, gray.
PROPERTIES: Table A.6, Water (Ts = 310 K): hfg = 2414 kJ/kg.
ANALYSIS: (a) Since Twb = T∞ , the free stream contains water vapor at its saturation condition. The
water vapor at the surface is saturated since it is in equilibrium with the liquid in the towel. From Table
A.6,
T (K)
vg (m3/kg)
ρg (kg/m3)
-2
T∞ = 290
69.7
ρ A, ∞ = 1.435 × 10
Ts = 310
ρ A,s = 4.361 × 10
22.93
-2
Using the mass transfer convection rate equation, the water evaporation rate from the towel is
(
)
n A = hm As ρ A,s − ρ A, ∞ = 0.027 m s ( 0.0925 m )
2
( 4.361 − 1.435 ) × 10−2 kg
3
m = 6.76 × 10
−6
kg s
<
To determine the net radiation heat rate q′′rad , perform an
energy balance on the water film,
E in − E out = 0
q rad − q cv − q evap = 0
q rad = q cv + q evap = hs As ( Ts − T∞ ) + n A h fg
and substituting numerical values find
q rad = 28.7 W m 2⋅ K ( 0.0925 m )
2
(310 − 290 ) K + 6.76 ×10−6 kg
s × 2414 × 103 J kg
q rad = ( 4.91 + 16.32 ) W = 21.2 W
(b) The radiation parameters for the towel surface are now evaluated. The emissive power is
<
E = ε E b ( Ts ) = εσ Ts4 = 0.96 × 5.67 × 10−8 W m 2⋅ K 4 (310 K ) = 502.7 W m 2
To determine the irradiation G, recognize that the net radiation heat rate can be expressed as,
<
4
q rad = (α G − E ) As
21.2 W = ( 0.96G − 502.7 ) W m 2 × ( 0.0925 m )
2
G = 3105 W/m2
<
where α = ε since the water surface is diffuse, gray. From the definition of the radiosity,
J = E + ρ G = [502.7 + (1 − 0.96 ) × 3105] W m 2 = 626.9 W m 2
where ρ = 1 - α = 1 - ε.
COMMENTS: An alternate method to evaluate J is to recognize that q′′rad = G - J.
<
PROBLEM 13.1
KNOWN: Various geometric shapes involving two areas A1 and A2.
FIND: Shape factors, F12 and F21, for each configuration.
ASSUMPTIONS: Surfaces are diffuse.
ANALYSIS: The analysis is not to make use of tables or charts. The approach involves use of the
reciprocity relation, Eq. 13.3, and summation rule, Eq. 13.4. Recognize that reciprocity applies to two
surfaces; summation applies to an enclosure. Certain shape factors will be identified by inspection.
Note L is the length normal to page.
(a) Long duct (L):
<
By inspection, F12 = 1.0
By reciprocity, F21 =
A1
A2
F12 =
2 RL
(3 / 4 ) ⋅ 2π RL
× 1.0 =
4
3π
= 0.424
<
(b) Small sphere, A1, under concentric hemisphere, A2, where A2 = 2A
Summation rule
F11 + F12 + F13 = 1
But F12 = F13 by symmetry, hence F12 = 0.50
By reciprocity,
F21 =
A1
A2
F12 =
A1
2A1
× 0.5 = 0.25.
<
<
(c) Long duct (L):
<
A1
Summation rule,
F22 = 1 − F21 = 1 − 0.64 = 0.363.
Summation rule,
F11 + F12 + F13 = 1
π RL
× 1.0 =
2
F21 =
A2
F12 =
2RL
By reciprocity,
π
= 0.637
<
<
By inspection,
F12 = 1.0
(d) Long inclined plates (L):
But F12 = F13 by symmetry, hence F12 = 0.50
By reciprocity,
(e) Sphere lying on infinite plane
Summation rule,
F21 =
A1
A2
F12 =
20L
10 ( 2 )
1/ 2
× 0.5 = 0.707.
L
F11 + F12 + F13 = 1
But F12 = F13 by symmetry, hence F12 = 0.5
By reciprocity,
<
<
F21 =
A1
A2
F12 → 0 since
A 2 → ∞.
Continued …..
<
<
PROBLEM 13.1 (Cont.)
(f) Hemisphere over a disc of diameter D/2; find also F22 and F23.
<
By inspection, F12 = 1.0
Summation rule for surface A3 is written as
F31 + F32 + F33 = 1. Hence, F32 = 1.0.
By reciprocity,
F23 =
A3
F32
A2
 2 π D / 2 2 
( )  / π D2 1.0 = 0.375.
 πD
F23 =  
−

4
 2 
  4


By reciprocity,
 π  D  2 π D2 
A1
F21 =
F12 =    /
 × 1.0 = 0.125.
A2
2 
 4  2 

<
F21 + F22 + F23 = 1 or
Summation rule for A2,
F22 = 1 − F21 − F23 = 1 − 0.125 − 0.375 = 0.5.
<
Note that by inspection you can deduce F22 = 0.5
(g) Long open channel (L):
Summation rule for A1
F11 + F12 + F13 = 0
<
but F12 = F13 by symmetry, hence F12 = 0.50.
By reciprocity,
F21 =
A1
A2
F12 =
2× L
( 2π 1) / 4 × L
=
4
π
× 0.50 = 0.637.
COMMENTS: (1) Note that the summation rule is applied to an enclosure. To complete the
enclosure, it was necessary in several cases to define a third surface which was shown by dashed
lines.
(2) Recognize that the solutions follow a systematic procedure; in many instances it is possible to
deduce a shape factor by inspection.
PROBLEM 13.2
KNOWN: Geometry of semi-circular, rectangular and V grooves.
FIND: (a) View factors of grooves with respect to surroundings, (b) View factor for sides of V
groove, (c) View factor for sides of rectangular groove.
SCHEMATIC:
ASSUMPTIONS: (1) Diffuse surfaces, (2) Negligible end effects, “long grooves”.
ANALYSIS: (a) Consider a unit length of each groove and represent the surroundings by a
hypothetical surface (dashed line).
Semi-Circular Groove:
F21 = 1;
F12 =
A2
W
F21 =
×1
A1
(π W / 2 )
<
F12 = 2 / π .
Rectangular Groove:
F4(1,2,3) = 1;
F(1,2,3)4 =
A4
W
F4(1,2,3) =
×1
A1 + A 2 + A3
H+W+H
F(1,2,3)4 = W / ( W + 2H ).
<
V Groove:
F3(1,2 ) = 1;
A3
W
F3(1,2 ) =
W/2 W/2
A1 + A 2
+
sin θ sin θ
F(1,2 )3 = sin θ .
F(1,2 )3 =
(b) From Eqs. 13.3 and 13.4,
F12 = 1 − F13 = 1 −
From Symmetry,
F31 = 1/ 2.
Hence, F12 = 1 −
W
1
×
( W / 2 ) / sin θ 2
A3
F31.
A1
or
F12 = 1 − sin θ .
<
(c) From Fig. 13.4, with X/L = H/W =2 and Y/L → ∞,
F12 ≈ 0.62.
<
COMMENTS: (1) Note that for the V groove, F13 = F23 = F(1,2)3 = sinθ, (2) In part (c), Fig. 13.4
could also be used with Y/L = 2 and X/L = ∞. However, obtaining the limit of Fij as X/L → ∞ from
the figure is somewhat uncertain.
PROBLEM 13.3
KNOWN: Two arrangements (a) circular disk and coaxial, ring shaped disk, and (b) circular disk
and coaxial, right-circular cone.
FIND: Derive expressions for the view factor F12 for the arrangements (a) and (b) in terms of the
areas A1 and A2, and any appropriate hypothetical surface area, as well as the view factor for coaxial
parallel disks (Table 13.2, Figure 13.5). For the disk-cone arrangement, sketch the variation of F12
with θ for 0 ≤ θ ≤ π/2, and explain the key features.
SCHEMATIC:
ASSUMPTIONS: Diffuse surfaces with uniform radiosities.
ANALYSIS: (a) Define the hypothetical surface A3, a co-planar disk inside the ring of A1. Using
the additive view factor relation, Eq. 13.5,
A 1,3 F 1,3 = A1 F12 + A 3 F32
1
F12 =
A 1,3 F 1,3 − A 3 F32
A1
$ $
$ $
<
where the parenthesis denote a composite surface. All the Fij on the right-hand side can be evaluated
using Fig. 13.5.
(b) Define the hypothetical surface A3, the disk at the bottom of the cone. The radiant power leaving
A2 that is intercepted by A1 can be expressed as
(1)
F21 = F23
That is, the same power also intercepts the disk at the bottom of the cone, A3. From reciprocity,
A1 F12 = A 2 F21
(2)
and using Eq. (1),
F12 =
A2
F23
A1
<
The variation of F12 as a function of θ is shown below for the disk-cone arrangement. In the limit
when θ → π/2, the cone approaches a disk of area A3. That is,
F12 θ → π / 2 = F13
$
When θ → 0, the cone area A
F12 θ → 0$ = 0
2
diminishes so that
PROBLEM 13.4
KNOWN: Right circular cone and right-circular cylinder of same diameter D and length L
positioned coaxially a distance Lo from the circular disk A1; hypothetical area corresponding to the
openings identified as A3.
FIND: (a) Show that F21 = (A1/A2) F13 and F22 = 1 - (A3/A2), where F13 is the view factor between
two, coaxial parallel disks (Table 13.2), for both arrangements, (b) Calculate F21 and F22 for L = Lo =
50 mm and D1 = D3 = 50 mm; compare magnitudes and explain similarities and differences, and (c)
Magnitudes of F21 and F22 as L increases and all other parameters remain the same; sketch and
explain key features of their variation with L.
SCHEMATIC:
ASSUMPTIONS: (1) Diffuse surfaces with uniform radiosities, and (2) Inner base and lateral
surfaces of the cylinder treated as a single surface, A2.
ANALYSIS: (a) For both configurations,
F13 = F12
(1)
since the radiant power leaving A1 that is intercepted by A3 is likewise intercepted by A2. Applying
reciprocity between A1 and A2,
A1 F12 = A 2 F21
(2)
Substituting from Eq. (1), into Eq. (2), solving for F21, find
<
F21 = A1 / A 2 F12 = A1 / A 2 F13
1
1
6
6
Treating the cone and cylinder as two-surface enclosures, the summation rule for A2 is
F22 + F23 = 1
(3)
Apply reciprocity between A2 and A3, solve Eq. (3) to find
F22 = 1 − F23 = 1 − A 3 / A 2 F32
1
6
and since F32 = 1, find
<
F22 = 1 − A 3 / A 2
Continued …..
PROBLEM 13.4 (Cont.)
(b) For the specified values of L, Lo, D1 and D2, the view factors are calculated and tabulated below.
Relations for the areas are:
Disk-cone:
A1 = π D12 / 4
A 2 = π D3 / 2 L2 + D 3 / 2
Disk-cylinder: A1 = π D12 / 4
1
62 1/ 2
A 2 = π D32 / 4 + π D 3L
A 3 = π D23 / 4
A 3 = π D23 / 4
The view factor F13 is evaluated from Table 13.2, coaxial parallel disks (Fig. 13.5); find F13 = 0.1716.
F21
0.0767
0.0343
Disk-cone
Disk-cylinder
F22
0.553
0.800
It follows that F21 is greater for the disk-cone (a) than for the cylinder-cone (b). That is, for (a),
surface A2 sees more of A1 and less of itself than for (b). Notice that F22 is greater for (b) than (a);
this is a consequence of A2,b > A2,a.
(c) Using the foregoing equations in the IHT workspace, the variation of the view factors F21 and F22
with L were calculated and are graphed below.
Right-circular cone and disk
1
1
0 .8
0.6
0 .6
Fij
Fij
0.8
R ig h t-c irc u la r c ylin d e r a n d d is k , L o = D = 5 0 m m
0.4
0 .4
0.2
0 .2
0
0
0
40
80
120
160
200
0
40
Cone height, L(mm)
F21
F22
80
120
160
C o n e h e ig h t, L (m m )
F2 1
F2 2
Note that for both configurations, when L = 0, find that F21 = F13 = 0.1716, the value obtained for
coaxial parallel disks. As L increases, find that F22 → 1; that is, the interior of both the cone and
cylinder see mostly each other. Notice that the changes in both F21 and F22 with increasing L are
greater for the disk-cylinder; F21 decreases while F22 increases.
COMMENTS: From the results of part (b), why isn’t the sum of F21 and F22 equal to unity?
200
PROBLEM 13.5
KNOWN: Two parallel, coaxial, ring-shaped disks.
FIND: Show that the view factor F12 can be expressed as
F12 =
1
A 1,3 F 1,3 2,4 − A 3 F3 2,4 − A 4 F4 1,3 − F43
A1
J 1 6 1 61 6
1 6
4 1 6
9L
where all the Fig on the right-hand side of the equation can be evaluated from Figure 13.5 (see Table
13.2) for coaxial parallel disks.
SCHEMATIC:
ASSUMPTIONS: Diffuse surfaces with uniform radiosities.
ANALYSIS: Using the additive rule, Eq. 13.5, where the parenthesis denote a composite surface,
1 6
F1 2,4 = F12 + F14
F12 = F1 2,4 − F14
1 6
(1)
Relation for F1(2,4): Using the additive rule
1 6 1 61 6
1 6
1 6
A 1,3 F 1,3 2,4 = A1 F1 2,4 + A 3 F3 2,4
(2)
where the check mark denotes a Fij that can be evaluated using Fig. 13.5 for coaxial parallel disks.
Relation for F14: Apply reciprocity
A1 F14 = A 4 F41
(3)
and using the additive rule involving F41,
A1 F14 = A 4 F4 1,3 − F43
1 6
(4)
Relation for F12: Substituting Eqs. (2) and (4) into Eq. (1),
F12 =
1
A 1,3 F 1,3 2,4 − A 3 F3 2,4 − A 4 F4 1,3 − F43
A1
J 1 6 1 61 6
1 6
4 1 6
9L
COMMENTS: (1) The Fij on the right-hand side can be evaluated using Fig. 13.5.
(2) To check the validity of the result, substitute numerical values and test the behavior at special
limits. For example, as A3, A4 → 0, the expression reduces to the identity F12 ≡ F12.
<
PROBLEM 13.6
KNOWN: Long concentric cylinders with diameters D1 and D2 and surface areas A1 and A2.
FIND: (a) The view factor F12 and (b) Expressions for the view factors F22 and F21 in terms of the
cylinder diameters.
SCHEMATIC:
ASSUMPTIONS: (1) Diffuse surfaces with uniform radiosities and (2) Cylinders are infinitely long
such that A1 and A2 form an enclosure.
ANALYSIS: (a) View factor F12. Since the infinitely long cylinders form an enclosure with surfaces
A1 and A2, from the summation rule on A1, Eq. 13.4,
F11 + F12 = 1
(1)
and since A1 doesn’t see itself, F11 = 0, giving
< (2)
F12 = 1
That is, the inner surface views only the outer surface.
(b) View factors F22 and F21. Applying reciprocity between A1 and A2, Eq. 13.3, and substituting
from Eq. (2),
A1 F12 = A 2 F21
F21 =
(3)
π D1L
A1
D
F12 =
×1= 1
π D2 L
A2
D2
< (4)
From the summation rule on A2, and substituting from Eq. (4),
F21 + F22 = 1
F22 = 1 − F21 = 1 −
D1
D2
<
PROBLEM 13.7
KNOWN: Right-circular cylinder of diameter D, length L and the areas A1, A2, and A3 representing
the base, inner lateral and top surfaces, respectively.
FIND: (a) Show that the view factor between the base of the cylinder and the inner lateral surface
has the form
4
!
F12 = 2 H 1 + H 2
1/ 2
9
−H
"#
$
where H = L/D, and (b) Show that the view factor for the inner lateral surface to itself has the form
F22 = 1 + H − 1 + H 2
4
1/ 2
9
SCHEMATIC:
ASSUMPTIONS: Diffuse surfaces with uniform radiosities.
ANALYSIS: (a) Relation for F12, base-to-inner lateral surface. Apply the summation rule to A1,
noting that F11 = 0
F11 + F12 + F13 = 1
F12 = 1 − F13
(1)
From Table 13.2, Fig. 13.5, with i = 1, j = 3,
F13 =
%K& K' !
1/ 2
1
S − S2 − 4 D3 / D1 2
2
S = 1+
1 + R 23
R12
=
1
1
R2
6 "#$
(K)
K*
(2)
+ 2 = 4 H2 + 2
(3)
where R1 = R3 = R = D/2L and H = L/D. Combining Eqs. (2) and (3) with Eq. (1), find after some
manipulation
Continued …..
PROBLEM 13.7 (Cont.)
%K
4
&K
!
'
1/ 2
"
F12 = 2 H 41 + H 2 9 − H #
!
$
1/ 2
2
1
2
2
F12 = 1 − 4 H + 2 − 4 H + 2 − 4
2
9 "#$
(K
)K
*
(4)
(b) Relation for F22, inner lateral surface. Apply summation rule on A2, recognizing that F23 = F21,
F21 + F22 + F23 = 1
F22 = 1 − 2 F21
(5)
Apply reciprocity between A1 and A2,
1
6
F21 = A1 / A 2 F12
(6)
and substituting into Eq. (5), and using area expressions
F22 = 1 − 2
1
A1
D
F12 = 1 − 2
F12 = 1 −
F12
2H
A2
4L
(7)
2
where A1 = πD /4 and A2 = πDL.
Substituting from Eq. (4) for F12, find
F22 = 1 −
4
!
"#
$
1/ 2
1/ 2
1
2 H 1 + H2
− H = 1+ H − 1 + H2
2H
9
4
9
<
PROBLEM 13.8
KNOWN: Arrangement of plane parallel rectangles.
FIND: Show that the view factor between A1 and A2 can be expressed as
F12 =
1
A 1,4 F 1,4 2,3 − A1 F13 − A 4 F42
2 A1
$ $ $
where all Fij on the right-hand side of the equation can be evaluated from Fig. 13.4 (see Table 13.2)
for aligned parallel rectangles.
SCHEMATIC:
ASSUMPTIONS: Diffuse surfaces with uniform radiosity.
ANALYSIS: Using the additive rule where the parenthesis denote a composite surface,
* + A F + A F + A F*
A(1,4 ) F(*1,4 )( 2,3) = A1 F13
1 12
4 43
4 42
(1)
where the asterisk (*) denotes that the Fij can be evaluated using the relation of Figure 13.4. Now,
find suitable relation for F43. By symmetry,
F43 = F21
(2)
and from reciprocity between A1 and A2,
F21 =
A1
F12
A2
(3)
Multiply Eq. (2) by A4 and substitute Eq. (3), with A4 = A2,
A 4 F43 = A 4 F21 = A 4
A1
F12 = A1 F12
A2
(4)
Substituting for A4 F43 from Eq. (4) into Eq. (1), and rearranging,
F12 =
1 
* − A F* 
A
F*
− A1 F13
4 42 
2 A1  (1,4 ) (1,4 )( 2,3)
<
PROBLEM 13.9
KNOWN: Two perpendicular rectangles not having a common edge.
FIND: (a) Shape factor, F12, and (b) Compute and plot F12 as a function of Zb for 0.05 ≤ Zb ≤ 0.4 m;
compare results with the view factor obtained from the two-dimensional relation for perpendicular
plates with a common edge, Table 13.1.
SCHEMATIC:
ASSUMPTIONS: (1) All surfaces are diffuse, (2) Plane formed by A1 + A3 is perpendicular to plane
of A2.
ANALYSIS: (a) Introducing the hypothetical surface A3, we can write
F2(3,1) = F23 + F21.
(1)
Using Fig. 13.6, applicable to perpendicular rectangles with a common edge, find
F23 = 0.19 :
with Y = 0.3, X = 0.5,
F2(3,1) = 0.25 :
Z = Za − Z b = 0.2, and
with Y = 0.3, X = 0.5, Za = 0.4, and
Y
X
=
0.3
0.5
Y
X
=
0.3
0.5
= 0.6,
= 0.6,
Z
=
0.2
X 0.5
Z 0.4
=
= 0.8
X 0.5
= 0.4
Hence from Eq. (1)
F21 = F2(3.1) − F23 = 0.25 − 0.19 = 0.06
By reciprocity,
F12 =
A2
0.5 × 0.3m 2
F21 =
× 0.06 = 0.09
2
A1
0.5 × 0.2 m
(2)
<
(b) Using the IHT Tool – View Factors for Perpendicular Rectangles with a Common Edge and Eqs.
(1,2) above, F12 was computed as a function of Zb. Also shown on the plot below is the view factor
F(3,1)2 for the limiting case Zb → Za.
PROBLEM 13.10
KNOWN: Arrangement of perpendicular surfaces without a common edge.
FIND: (a) A relation for the view factor F14 and (b) The value of F14 for prescribed dimensions.
SCHEMATIC:
ASSUMPTIONS: (1) Diffuse surfaces.
ANALYSIS: (a) To determine F14, it is convenient to define the hypothetical surfaces A2 and A3.
From Eq. 13.6,
( A1 + A2 ) F(1,2 )(3,4 ) = A1 F1(3,4 ) + A2 F2(3,4 )
where F(1,2)(3,4) and F2(3,4) may be obtained from Fig. 13.6. Substituting for A1 F1(3,4) from Eq. 13.5
and combining expressions, find
A1 F1(3,4 ) = A1 F13 + A1 F14
F14 =
1 
( A1 + A 2 ) F(1,2 )(3,4 ) − A1 F13 − A 2 F2(3,4 )  .
A1 
Substituting for A1 F13 from Eq. 13.6, which may be expressed as
( A1 + A2 ) F(1,2 )3 = A1 F13 + A2 F23 .
The desired relation is then
1 
F14 =
( A1 + A 2 ) F(1,2 )(3,4 ) + A 2 F23 − ( A1 + A 2 ) F(1,2 )3 − A 2 F2(3,4 )  .
A1 
(b) For the prescribed dimensions and using Fig. 13.6, find these view factors:
L +L
L +L
Surfaces (1,2)(3,4)
F(1,2 )(3,4 ) = 0.22
( Y / X ) = 1 2 = 1, ( Z / X ) = 3 4 = 1.45,
W
W
L
L
Surfaces 23
F23 = 0.28
( Y / X ) = 2 = 0.5, ( Z / X ) = 3 = 1,
W
W
L
L +L
Surfaces (1,2)3
F(1,2 )3 = 0.20
( Y / X ) = 1 2 = 1, ( Z / X ) = 3 = 1,
W
W
L +L
L
Surfaces 2(3,4)
F2(3,4 ) = 0.31
( Y / X ) = 2 = 0.5, ( Z / X ) = 3 4 = 1.5,
W
W
Using the relation above, find
1
F14 =
[( WL1 + WL2 ) 0.22 + ( WL2 ) 0.28 − ( WL1 + WL2 ) 0.20 − ( WL2 ) 0.31]
( WL1 )
<
F14 = [2 ( 0.22 ) + 1 ( 0.28 ) − 2 (0.20 ) − 1( 0.31)] = 0.01.
<
PROBLEM 13.11
KNOWN: Arrangements of rectangles.
FIND: The shape factors, F12.
SCHEMATIC:
ASSUMPTIONS: (1) Diffuse surface behavior.
ANALYSIS: (a) Define the hypothetical surfaces shown in the sketch as A3 and A4. From the
additive view factor rule, Eq. 13.6, we can write
√
√
√
A(1,3 ) F(1,3 )( 2,4 ) = A1F12 + A1 F14 + A3 F32 + A3F34
(1)
Note carefully which factors can be evaluated from Fig. 13.6 for perpendicular rectangles with a
common edge. (See √). It follows from symmetry that
A1F12 = A 4 F43 .
(2)
Using reciprocity,
(3)
A 4 F43 = A3F34,
then
A F =A F .
1 12
3 34
Solving Eq. (1) for F12 and substituting Eq. (3) for A3F34, find that
1
 A 1,3 F 1,3 2,4 − A1F14 − A3F32  .
F12 =

2A1  ( ) ( )( )
(4)
Evaluate the view factors from Fig. 13.6:
Y/X
Fij
(1,3) (2,4)
6
9
6
14
6
6
32
3
= 0.67
=1
=2
Z/X
6
9
6
6
6
3
Fij
= 0.67
0.23
=1
0.20
=2
0.14
Substituting numerical values into Eq. (4) yields
F12 =
1
( 6 × 9 ) m 2 × 0.23 − ( 6 × 6 ) m 2 × 0.20 − ( 6 × 3 ) m 2 × 0.14 

2 × (6 × 6 ) m 
2
<
F12 = 0.038.
Continued …..
PROBLEM 13.11 (Cont.)
(b) Define the hypothetical surface A3 and divide A2 into two sections, A2A and A2B. From the
additive view factor rule, Eq. 13.6, we can write
√
√
A1,3 F(1,3 )2 = A1F12 + A3F3( 2A ) + A3 F3( 2B ) .
(5)
Note that the view factors checked can be evaluated from Fig. 13.4 for aligned, parallel rectangles.
To evaluate F3(2A), we first recognize a relationship involving F(24)1 will eventually be required.
Using the additive rule again,
√
A 2A F( 2A )(1,3 ) = A 2A F( 2A )1 + A 2A F( 2A )3 .
(6)
Note that from symmetry considerations,
A 2A F( 2A )(1,3 ) = A1F12
(7)
and using reciprocity, Eq. 13.3, note that
A 2A F2A3 = A3F3( 2A ).
(8)
Substituting for A3F3(2A) from Eq. (8), Eq. (5) becomes
√
√
A(1,3 ) F(1,3 )2 = A1F12 + A 2A F( 2A )3 + A3 F3( 2B ) .
Substituting for A2A F(2A)3 from Eq. (6) using also Eq. (7) for A2A F(2A)(1,3) find that
√
√
√


A(1,3 ) F(1,3 )2 = A1F12 +  A1F12 − A 2A F( 2A )1  + A3 F3( 2B )



(9)

and solving for F12, noting that A1 = A2A and A(1,3) = A2
F12 =
1 
√
√
√

 A 2 F (1,3)2 + A 2A F ( 2A )1 − A3 F 3( 2B )  .
2A1 

(10)
Evaluate the view factors from Fig. 13.4:
Fij
(1,3) 2
X/L
Y/L
1
1.5
1
1
(2A)1
1
1
3(2B)
1
=1
=1
=1
Fij
= 1.5
0.25
= 0.5
1
1
=1
1
0.11
1
0.5
0.20
Substituting numerical values into Eq. (10) yields
F12 =
1
(1.5 × 1.0 ) m 2 × 0.25 + ( 0.5 × 1) m 2 × 0.11 − (1 × 1) m 2 × 0.20 

2 ( 0.5 × 1) m 
F12 = 0.23.
2
<
PROBLEM 13.12
KNOWN: Two geometrical arrangements: (a) parallel plates and (b) perpendicular plates with a
common edge.
FIND: View factors using “crossed-strings” method; compare with appropriate graphs and analytical
expressions.
SCHEMATIC:
(a) Parallel plates
(b) Perpendicular plates with common edge
ASSUMPTIONS: Plates infinite extent in direction normal to page.
ANALYSIS: The “crossed-strings” method is
applicable
to surfaces of infinite extent in one direction having an
obstructed view of one another.
F12 = (1/ 2w1 )[( ac + bd ) − (ad + bc )].
(a) Parallel plates: From the schematic, the edge and diagonal distances are
(
2
2
ac = bd = w1 + L
)
1/ 2
bc = ad = L.
With w1 as the width of the plate, find
F12 =

2
2w1 
1
(
2
2
w1 + L
)
1/ 2


1
− 2 (L) =
2×4

2
m
(
2
2
4 +1
)
1/ 2


m − 2 (1 m ) = 0.781.
<
Using Fig. 13.4 with X/L = 4/1 = 4 and Y/L = ∞, find F12 ≈ 0.80. Also, using the first relation of
Table 13.1,

Fij =   Wi + Wj

(
1/ 2
)2 + 4 
−  Wi − Wj
(
1/ 2 
)2 + 4 
 / 2 Wi

where wi = wj = w1 and W = w/L = 4/1 = 4, find
F12 =
1/ 2
1/ 2 

2
2
 ( 4 + 4 ) + 4  − ( 4 − 4 ) + 4   / 2 × 4 = 0.781.


(b) Perpendicular plates with a common edge: From the schematic, the edge and diagonal distances
are
ac = w1
(
bd = L
2
2
ad = w1 + L
)
bc = 0.
With w1 as the width of the horizontal plates, find




(
2
)
2
)
F12 = (1 / 2w1 ) 2 ( w1 + L ) −  w1 + L
F12 = (1 / 2 × 4 m )


( 4 + 1) m − 

2
(
2
1/ 2
1/ 2
4 +1


+ 0



m + 0  = 0.110.
From the third relation of Table 13.1, with wi = w1 = 4 m and wj = L = 1 m, find

Fij = 1 + w j / w i − 1 + w j / w i

(
)
(
1/ 2 
)2 
/ 2




2 1/ 2 
F12 = 1 + (1 / 4 ) − 1 + (1 / 4 ) 
/ 2 = 0.110.

 

<
PROBLEM 13.13
KNOWN: Parallel plates of infinite extent (1,2) having aligned opposite edges.
FIND: View factor F12 by using (a) appropriate view factor relations and results for opposing
parallel plates and (b) Hottel’s string method described in Problem 13.12
SCHEMATIC:
ASSUMPTIONS: (1) Parallel planes of infinite extent normal to page and (2) Diffuse surfaces with
uniform radiosity.
ANALYSIS: From symmetry consideration (F12 = F14) and Eq. 13.5, it follows that
F12 = (1/ 2 )  F1( 2,3,4 ) − F13 


where A3 and A4 have been defined for convenience in the analysis. Each of these view factors can
be evaluated by the first relation of Table 13.1 for parallel plates with midlines connected
perpendicularly.
W1 = w1 / L = 2
F13:
W2 = w 2 / L = 2
1/ 2
( W1 + W2 )2 + 4 


F13 =
F1(2,3,4):
1/ 2
− ( W2 − W1 ) + 4 
2


2W1
W1 = w1 / L = 2
Hence, find
1/ 2
− ( 2 − 2 ) + 4 

2× 2
2

= 0.618
W( 2,3,4 ) = 3w 2 / L = 6
1/ 2
F1( 2,3,4 ) =
1/ 2
( 2 + 2 )2 + 4 


=
( 2 + 6 )2 + 4 


1/ 2
− ( 6 − 2 ) + 4 

2× 2
2

= 0.944.
F12 = (1/ 2 ) [0.944 − 0.618] = 0.163.
<
(b) Using Hottel’s string method,
F12 = (1/ 2w1 )[( ac + bd ) − (ad + bc )]
(
ac = 1 + 42
)
1/ 2
= 4.123
bd = 1
(
ad = 12 + 22
)
1/ 2
= 2.236
bc = ad = 2.236
and substituting numerical values find
F12 = (1/ 2 × 2 ) [( 4.123 + 1) − ( 2.236 + 2.236 )] = 0.163.
COMMENTS: Remember that Hottel’s string method is applicable only to surfaces that are of
infinite extent in one direction and have unobstructed views of one another.
<
PROBLEM 13.14
KNOWN: Two small diffuse surfaces, A1 and A2, on the inside of a spherical enclosure of radius R.
FIND: Expression for the view factor F12 in terms of A2 and R by two methods: (a) Beginning with
the expression Fij = qij/Ai Ji and (b) Using the view factor integral, Eq. 13.1.
SCHEMATIC:
2
ASSUMPTIONS: (1) Surfaces A1 and A2 are diffuse and (2) A1 and A2 << R .
ANALYSIS: (a) The view factor is defined as the fraction of radiation leaving Ai which is
intercepted by surface j and, from Section 13.1.1, can be expressed as
Fij =
qij
(1)
Ai Ji
From Eq. 12.5, the radiation leaving intercepted by A1 and A2 on the spherical surface is
q1→ 2 = ( J1 / π ) ⋅ A1 cos θ1 ⋅ ω 2−1
(2)
where the solid angle A2 subtends with respect to A1 is
ω 2 −1 =
A 2,n
r2
=
A 2 cos θ 2
(3)
r2
From the schematic above,
cosθ1 = cosθ 2
r = 2R cos θ1
(4,5)
Hence, the view factor is
Fij
J1 / π ) A1 cos θ1 ⋅ A 2 cosθ 2 / 4R 2 cos θ1
(
=
=
A1J1
A2
4π R 2
<
(b) The view factor integral, Eq. 13.1, for the small areas A1 and A2 is
F12 =
1
cosθ1 cos θ 2
cosθ1 cosθ 2 A 2
dA1dA 2 =
∫
∫
A1 A1 A 2
π r2
π r2
and from Eqs. (4,5) above,
F12 =
A2
π R2
<
2
COMMENTS: Recognize the importance of the second assumption. We require that A1, A2, << R
so that the areas can be considered as of differential extent, A1 = dA1, and A2 = dA2.
PROBLEM 13.15
KNOWN: Disk A1, located coaxially, but tilted 30° of the normal, from the diffuse-gray, ring-shaped disk A2.
Surroundings at 400 K.
FIND: Irradiation on A1, G1, due to the radiation from A2.
SCHEMATIC:
ASSUMPTIONS: (1) A2 is diffuse-gray surface, (2) Uniform radiosity over A2, (3) The surroundings are large
with respect to A1 and A2.
ANALYSIS: The irradiation on A1 is
G1 = q 21 / A1 = ( F21 ⋅ J 2 A 2 ) / A1
(1)
where J2 is the radiosity from A2 evaluated as
4
J 2 = ε 2 E b,2 + ρ 2G 2 = ε 2σ T24 + (1 − ε 2 )σ Tsur
J 2 = 0.7 × 5.67 × 10−8 W / m 2 ⋅ K 4 (600 K ) + (1 − 0.7 ) 5.67 × 10−8 W / m 2 ⋅ K 4 ( 400 K )
4
4
J 2 = 5144 + 436 = 5580 W / m 2 .
(2)
Using the view factor relation of Eq. 13.8, evaluate view factors between A1′ , the normal projection of A1, and
A3 as
F1′ 3 =
Di2
Di2 + 4L2
=
(0.004 m )2
(0.004 m )2 + 4 (1 m )2
= 4.00 × 10−6
and between A1′ and (A2 + A3) as
Do2
( 0.012 )2
F1′( 23 ) =
=
= 3.60 × 10−5
2
2
2
2
Do + 4L
(0.012 ) + 4 (1 m )
giving
F1′ 2 = F1′( 23 ) − F1′ 3 = 3.60 × 10−5 − 4.00 × 10−6 = 3.20 × 10−5.
From the reciprocity relation it follows that
F21′ = A1′ F1′ 2 / A 2 = ( A1 cos θ1 / A 2 ) F1′ 2 = 3.20 × 10−5 cos θ1 ( A1 / A 2 ).
By inspection we note that all the radiation striking A1′ will also intercept A1; that is
F21 = F21′ .
(3)
(4)
Hence, substituting for Eqs. (3) and (4) for F21 into Eq. (1), find
(
)
G1 = 3.20 × 10−5 cos θ1 ( A1 / A 2 ) × J 2 × A 2 / A1 = 3.20 × 10−5 cos θ1 ⋅ J 2
(5)
G1 = 3.20 × 10−5 cos (30° ) × 5580 W / m 2 = 27.7 µ W / m 2 .
<
COMMENTS: (1) Note from Eq. (5) that G1 ~ cosθ1 such that G1is a maximum when A1 is normal to disk A2.
PROBLEM 13.16
KNOWN: Heat flux gauge positioned normal to a blackbody furnace. Cover of furnace is at 350 K
while surroundings are at 300 K.
FIND: (a) Irradiation on gage, Gg, considering only emission from the furnace aperture and (b)
Irradiation considering radiation from the cover and aperture.
SCHEMATIC:
ASSUMPTIONS: (1) Furnace aperture approximates blackbody, (2) Shield is opaque, diffuse and
2
2
gray with uniform temperature, (3) Shield has uniform radiosity, (4) Ag << R , so that ωg-f = Ag/R ,
(5) Surroundings are large, uniform at 300 K.
ANALYSIS: (a) The irradiation on the gauge due only to aperture emission is
(
)
G g = q f − g / A g = Ie,f ⋅ A f cos θ f ⋅ ω g − f / Ag =
Gg =
σ Tf4
5.67 × 10−8 W / m 2 ⋅ K 4 (1000 K )
4
Af =
Ag
σ Tf4
⋅ Af ⋅
/ Ag
π
R2
× (π / 4 )( 0.005 m ) = 354.4 mW / m 2 .
2
π (1 m )
(b) The irradiation on the gauge due to radiation from the aperture (a) and cover (c) is
Fc − g ⋅ J c A c
G g = G g,a +
Ag
πR
2
2
<
where Fc-g and the cover radiosity are
(
)
Fc − g = Fg − c A g / A c ≈
Dc2
⋅
Ag
4R 2 + Dc2 A c
J c = ε c E b ( Tc ) + ρcG c
4
= (170.2 + 387.4 ) W / m 2 .
but G c = E b ( Tsur ) and ρc = 1 − α c = 1 − ε c , J c = ε cσ Tc4 + (1 − ε c )σ Tsur
Hence, the irradiation is
G g = G g,a +
1 
Ag 
Dc2
4 
⋅
A

 ε cσ Tc4 + (1 − ε c )σ Tsur
2
2

 c


A g 4R + D A c


c


  0.2 × σ (350 )4 + (1 − 0.2 ) × σ (300 )4  W / m 2
 4 × 12 + 0.102  



G g = 354.4 mW / m 2 + 
0.102
G g = 354.4 mW / m 2 + 424.4 mW / m 2 + 916.2 mW / m 2 = 1, 695 mW / m 2 .
COMMENTS: (1) Note we have assumed Af << Ac so that effect of the aperture is negligible. (2) In
part (b), the irradiation due to radiosity from the shield can be written also as Gg,c = qc-g/Ag =
2
2
(Jc/π)⋅Ac⋅ωg-c/Ag where ωg-c = Ag/R . This is an excellent approximation since Ac << R .
PROBLEM 13.17
KNOWN: Temperature and diameters of a circular ice rink and a hemispherical dome.
FIND: Net rate of heat transfer to the ice due to radiation exchange with the dome.
SCHEMATIC:
ASSUMPTIONS: (1) Blackbody behavior for dome and ice.
ANALYSIS: From Eq. 13.13 the net rate of energy exchange between the two blackbodies is
(
q 21 = A 2F21 σ T24 − T14
)
(
)
From reciprocity, A2 F21 = A1 F12 = π D12 / 4 1
2
A 2F21 = (π / 4 )( 25 m ) 1 = 491 m 2 .
Hence
(
)
4
4
q 21 = 491 m 2 5.67 × 10−8 W / m 2 ⋅ K 4 ( 288 K ) − ( 273 K ) 


q 21 = 3.69 ×104 W.
<
COMMENTS: If the air temperature, T∞, exceeds T1, there will also be heat transfer by convection
to the ice. The radiation and convection transfer to the ice determine the heat load which must be
handled by the cooling system.
PROBLEM 13.18
KNOWN: Surface temperature of a semi-circular drying oven.
FIND: Drying rate per unit length of oven.
SCHEMATIC:
ASSUMPTIONS: (1) Blackbody behavior for furnace wall and water, (2) Convection effects are
negligible and bottom is insulated.
PROPERTIES: Table A-6, Water (325 K): h fg = 2.378 × 106 J / kg.
ANALYSIS: Applying a surface energy balance,
h fg
q12 = qevap = m
where it is assumed that the net radiation heat transfer to the water is balanced by the evaporative heat
loss. From Eq. 13.13
)
(
q12 = A1 F12 σ T14 − T24 .
From inspection and the reciprocity relation, Eq. 13.3,
F12 =
Hence
A2
D⋅L
× 1 = 0.637.
F21 =
A1
(π D / 2 ) ⋅ L
(
T14 − T24
πD
m
′= =
m
F12 σ
L
2
h fg
)
4
′=
m
π (1 m )
W (1200 K ) − (325 K )
× 0.637 × 5.67 ×10−8
2
m 2 ⋅ K 4 2.378 × 106 J / kg
or
′ = 0.0492 kg / s ⋅ m.
m
<
COMMENTS: Air flow through the oven is needed to remove the water vapor. The water surface
temperature, T2, is determined by a balance between radiation heat transfer to the water and the
convection of latent and sensible energy from the water.
PROBLEM 13.19
KNOWN: Arrangement of three black surfaces with prescribed geometries and surface temperatures.
FIND: (a) View factor F13, (b) Net radiation heat transfer from A1 to A3.
SCHEMATIC:
ASSUMPTIONS: (1) Interior surfaces behave as blackbodies, (2) A2 >> A1.
ANALYSIS: (a) Define the enclosure as the interior of the cylindrical form and identify A4.
Applying the view factor summation rule, Eq. 13.4,
F11 + F12 + F13 + F14 = 1.
(1)
Note that F11 = 0 and F14 = 0. From Eq. 13.8,
F12 =
D2
D2 + 4L2
=
(3m )2
= 0.36.
2
2
(3m ) + 4 ( 2m )
(2)
From Eqs. (1) and (2),
<
F13 = 1 − F12 = 1 − 0.36 = 0.64.
(b) The net heat transfer rate from A1 to A3 follows from Eq. 13.13,
(
q13 = A1 F13 σ T14 − T34
)
(
)
q13 = 0.05m 2 × 0.64 × 5.67 ×10−8 W / m 2 ⋅ K 4 10004 − 5004 K 4 = 1700 W.
COMMENTS: Note that the summation rule, Eq. 13.4, applies to an enclosure; that is, the total
region above the surface must be considered.
<
PROBLEM 13.20
KNOWN: Furnace diameter and temperature. Dimensions and temperature of suspended part.
FIND: Net rate of radiation transfer per unit length to the part.
SCHEMATIC:
ASSUMPTIONS: (1) All surfaces may be approximated as blackbodies.
ANALYSIS: From symmetry considerations, it is convenient to treat the system as a three-surface
enclosure consisting of the inner surfaces of the vee (1), the outer surfaces of the vee (2) and the
furnace wall (3). The net rate of radiation heat transfer to the part is then
(
)
(
4 − T 4 + A′ F σ T 4 − T 4
′ = A3′ F31 σ Tw
q1,2
p
3 32
w
p
)
From reciprocity,
A′3 F31 = A1′ F13 = 2 L × 0.5 = 1m
where surface 3 may be represented by the dashed line and, from symmetry, F13 = 0.5. Also,
A′3 F32 = A′2 F23 = 2 L ×1 = 2m
Hence,
(
)
′ = (1 + 2 ) m × 5.67 × 10−8 W / m 2 ⋅ K 4 10004 − 3004 K 4 = 1.69 × 105 W / m
q1,2
<
COMMENTS: With all surfaces approximated as blackbodies, the result is independent of the tube
diameter. Note that F11 = 0.5.
PROBLEM 13.21
KNOWN: Coaxial, parallel black plates with surroundings. Lower plate (A2) maintained at
prescribed temperature T2 while electrical power is supplied to upper plate (A1).
FIND: Temperature of the upper plate T1.
SCHEMATIC:
ASSUMPTIONS: (1) Plates are black surfaces of uniform temperature, and (2) Backside of heater
on A1 insulated.
ANALYSIS: The net radiation heat rate leaving Ai is
(
N
)
(
(
)
Pe = ∑ q ij = A1 F12σ T14 − T24 + A1F13σ T14 − T34
j=1
)
(
4 
Pe = A1σ  F12 T14 − T24 + F13 T14 − Tsur


From Fig. 13.5 for coaxial disks (see Table 13.2),
R1 = r1 / L = 0.10 m / 0.20 m = 0.5
S = 1+
F12 =
1 + R 22
R12
= 1+
1 + 12
( 0.5)2
)
(1)
R 2 = r2 / L = 0.20 m / 0.20 m = 1.0
= 9.0
1/ 2  1 
1/ 2 
1  2
2
2
2
S − S − 4 ( r2 / r1 )   = 9 − 9 − 4 ( 0.2 / 0.1)   = 0.469.
2
 2

From the summation rule for the enclosure A1, A2 and A3 where the last area represents the
surroundings with T3 = Tsur,
F12 + F13 = 1
F13 = 1 − F12 = 1 − 0.469 = 0.531.
Substituting numerical values into Eq. (1), with A1 = π D12 / 4 = 3.142 × 10−2 m 2 ,
(
)
17.5 W = 3.142 × 10−2 m 2 × 5.67 × 10−8 W / m 2 ⋅ K 4 0.469 T14 − 5004 K 4
(
)

+ 0.531 T14 − 3004 K 4 

(
)
(
9.823 × 109 = 0.469 T14 − 5004 + 0.531 T14 − 3004
find by trial-and-error that
)
T1 = 456 K.
COMMENTS: Note that if the upper plate were adiabatic, T1 = 427 K.
<
PROBLEM 13.22
KNOWN: Tubular heater radiates like blackbody at 1000 K.
FIND: (a) Radiant power from the heater surface, As, intercepted by a disc, A1, at a prescribed
location qs→1; irradiation on the disk, G1; and (b) Compute and plot qs→1 and G1 as a function of the
separation distance L1 for the range 0 ≤ L1 ≤ 200 mm for disk diameters D1 = 25, and 50 and 100 mm.
SCHEMATIC:
ASSUMPTIONS: (1) Heater surface behaves as blackbody with uniform temperature.
ANALYSIS: (a) The radiant power leaving the inner surface of the tubular heater that is intercepted
by the disk is
(1)
q 2→1 = ( A 2 E b2 ) F21
where the heater is surface 2 and the disk is surface 1. It follows from the reciprocity rule, Eq. 13.3, that
F21 =
A1
A2
F12 .
(2)
Define now the hypothetical disks, A3 and A4, located at the ends of the tubular heater. By
inspection, it follows that
F14 = F12 + F13
or
F12 = F14 − F13
(3)
where F14 and F13 may be determined from Fig. 13.5. Substituting numerical values, with D3 = D4 =
D2,
D /2
L L1 + L 2
200
ri
100 / 2
=
=
=8
= 3
=
= 0.25
F13 = 0.08
with
r⋅i
D1 / 2
50 / 2
L L1 + L 2
200
F14 = 0.20
L
with
ri
=
L1
D1 / 2
=
100
50 / 2
=4
rj
L
=
D4 / 2
L1
=
100 / 2
100
= 0.5
Substituting Eq. (3) into Eq. (2) and then into Eq. (1), the result is
q 2→1 = A1 ( F14 − F13 ) E b2

(
q 2→1 = π 50 × 10−3

)m
2
2

/ 4  ( 0.20 − 0.08 ) × 5.67 × 10−8 W / m 2 ⋅ K 4 (1000 K ) = 13.4W
4

where E b2 = σ Ts4 . The irradiation G1 originating from emission leaving the heater surface is
q
13.4 W
= 6825 W / m 2 .
(4)
G1 = s →1 =
2
A1
π ( 0.050 m ) / 4
Continued …..
<
<
PROBLEM 13.22 (Cont.)
(b) Using the foregoing equations in IHT along with the Radiation Tool-View Factors for Coaxial
Parallel Disks, G1 and qs→1 were computed as a function of L1 for selected values of D1. The results
are plotted below.
In the upper left-hand plot, G1 decreases with increasing separation distance. For a given separation
distance, the irradiation decreases with increasing diameter. With values of D1 = 25 and 50 mm, the
irradiation values are only slightly different, which diminishes as L1 increases. In the upper righthand plot, the radiant power from the heater surface reaching the disk, qs→2, decreases with
increasing L1 and decreasing D1. Note that while G1 is nearly the same for D1 = 25 and 50 mm, their
respective qs→2 values are quite different. Why is this so?
PROBLEM 13.23
KNOWN: Dimensions and temperatures of an enclosure and a circular disc at its base.
FIND: Net radiation heat transfer between the disc and portions of the enclosure.
SCHEMATIC:
ASSUMPTIONS: (1) Blackbody behavior for disc and enclosure surfaces, (2) Area of disc is much
less than that of the hypothetical surfaces, (A1/A2) << 1 and (A1/A3) << 1.
ANALYSIS: From Eq. 13.13 the net radiation exchange between the disc (1) and the hemispherical
dome (d) is
)
(
q1d = A1 F1d σ T14 − T 4 .
However, since all of the radiation intercepted by the dome must pass through the hypothetical area
A2, it follows from Eq. 13.8 of Example 13.1,
F1d = F12 ≈
D2
4 L2 + D 2
=
1
1
= 0.410.
( 2L / D )2 + 1 1.44 + 1
=
Hence
q1d =
π
(0.02 m )2 × 0.41× 5.67 ×10−8 W / m 2 ⋅ K (1000 K )4 − (300 K )4 
4
<
q1d = 7.24 W.
Similarly, the net radiation exchange between the disc (1) and the cylindrical ring (r) of length L/3 is
(
q1r = A1 F1r σ T14 − T 4
)
where
F1r = F13 − F12 =
D2
2
4 ( 2L / 3) + D 2
− 0.41 = 0.61 − 0.41 = 0.20.
Hence
q1r =
π
(0.02 m )2 × 0.2 × 5.67 ×10−8 W / m 2 ⋅ K 4 (1000 K )4 − (300 K )4 
4
q1r = 3.53 W.
<
PROBLEM 13.24
KNOWN: Circular plate (A1) maintained at 600 K positioned coaxially with a conical shape (A2)
whose backside is insulated. Plate and cone are black surfaces and located in large, insulated
enclosure at 300 K.
FIND: (a) Temperature of the conical surface T2 and (b) Electric power required to maintain plate at
600 K.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Plate and cone are black, (3) Cone behaves as
insulated, reradiating surface, (4) Surroundings are large compared to plate and cone.
ANALYSIS: (a) Recognizing that the plate, cone, and surroundings from a three-(black) surface
enclosure, perform a radiation balance on the cone.
(
)
(
q 2 = 0 = q 23 + q 21 = A 2 F23 σ T24 − T34 + A 2 F21 σ T24 − T14
)
where the view factor F21 can be determined from the coaxial parallel disks relation (Table 13.2 or
(
)
2
2
Fig. 13.5) with Ri = ri/L=250/500 = 0.5, Rj = 0.5, S = 1 + 1 + R 2j / R i2 = 1 + (1 + 0.5 )/0.5 = 6.00,
and noting F2′1 = F21,



(
F21 = 0.5 S − S2 − 4 rj / ri


2 1/ 2 

)


2 1/ 2 
2
 = 0.5 6 − 6 − 4 (0.5 / 0.5 )   = 0.172.



For the enclosure, the summation rule provides, F2′3 = 1 − F2′1 = 1 − 0.172 = 0.828. Hence,
(
)
(
0.828 T24 − 3004 = 0 + 0.172 T24 − 6004
)
<
T2 = 413 K.
(b) The power required to maintain the plate at T2 follows from a radiation balance,
(
)
(
q1 = q12 + q13 = A1F12σ T14 − T24 + A1F13σ T14 − T34
)
where F12 = A 2′ F2′1 / A1 = F21 = 0.172 and F13 = 1 − F12 = 0.828,
(
)
(
)
(
)
q1 = π 0.52 / 4 m 2σ  0.172 6004 − 4134 K 4 + 0.828 6004 − 3004 K 4 
q1 = 1312 W.


<
PROBLEM 13.25
KNOWN: Conical and cylindrical furnaces (A2) as illustrated and dimensioned in Problem 13.2 (S)
supplied with power of 50 W. Workpiece (A1) with insulated backside located in large room at 300
K.
FIND: Temperature of the workpiece, T1, and the temperature of the inner surfaces of the furnaces,
T2. Use expressions for the view factors F21 and F22 given in the statement for Problem 13.2 (S).
SCHEMATIC:
ASSUMPTIONS: (1) Diffuse, black surfaces with uniform radiosities, (2) Backside of workpiece is
perfectly insulated, (3) Inner base and lateral surfaces of the cylindrical furnace treated as single
surface, (4) Negligible convection heat transfer, (5) Room behaves as large, isothermal surroundings.
ANALYSIS: Considering the furnace surface (A2), the workpiece (A1) and the surroundings (As) as
an enclosure, the net radiation transfer from A1 and A2 follows from Eq. 13.14,
q1 = 0 = A1 F12 E b1 − E b2 + A1 F1s E b1 − E bs
Workpiece
(1)
1
Furnace
1
6
6
q 2 = 50 W = A 2 F21 E b2 − E b1 + A 2 F2s E b2 − E bs
1
4
-8
2
6
1
6
(2)
4
where Eb = σ T and σ = 5.67 × 10 W/m ⋅K . From summation rules on A1 and A2, the view
factors F1s and F2s can be evaluated. Using reciprocity, F12 can be evaluated.
F1s = 1 − F12
F2s = 1 − F21 − F22
F12 = A 2 / A1 F21
1
6
The expressions for F21 and F22 are provided in the schematic. With A1 = π D12 / 4 the A2 are:
1/ 2
Cone: A 2 = π D3 / 2 L2 + D3 / 2 2
Cylinder: A 2 = π D23 / 4 + π D 3L
1
6 Examine Eqs (1) and (2) and recognize that there are two unknowns, T1 and T2, and the equations
must be solved simultaneously. Using the foregoing equations in the IHT workspace, the results are
T1 = 544 K
T2 = 828 K
<
COMMENTS: (1) From the IHT analysis, the relevant view factors are: F12 = 0.1716; F1s = 0.8284;
Cone: F21 = 0.07673, F22 = 0.5528; Cylinder: F21 = 0.03431, F22 =0.80.
(2) That both furnace configurations provided identical results may not, at first, be intuitively obvious.
Since both furnaces (A2) are black, they can be represented by the hypothetical black area A3 (the
opening of the furnaces). As such, the analysis is for an enclosure with the workpiece (A1), the
furnace represented by the disk A3 (at T2), and the surroundings. As an exercise, perform this
analysis to confirm the above results.
PROBLEM 13.26
KNOWN: Furnace constructed in three sections: insulated circular (2) and cylindrical (3) sections,
as well as, an intermediate cylindrical section (1) with imbedded electrical resistance heaters.
Cylindrical sections (1,3) are of equal length.
FIND: (a) Electrical power required to maintain the heated section at T1 = 1000 K if all the surfaces
are black, (b) Temperatures of the insulated sections, T2 and T3, and (c) Compute and plot q1, T2 and
T3 as functions of the length-to-diameter ratio, with 1 ≤ L/D ≤ 5 and D = 100 mm.
SCHEMATIC:
ASSUMPTIONS: (1) All surfaces are black, (2) Areas (1, 2, 3) are isothermal.
ANALYSIS: (a) To complete the enclosure representing the furnace, define the hypothetical surface
A4 as the opening at 0 K with unity emissivity. For each of the enclosure surfaces 1, 2, and 3, the
energy balances following Eq. 13.13 are
q1 = A1F12 ( E b1 − E b2 ) + A1F13 ( E b1 − E b3 ) + A1F14 + ( E b1 − E b4 )
(1)
0 = A 2 F21 ( E b2 − E b1 ) + A 2 F23 ( E b2 − E b3 ) + A 2 F24 ( E b2 − E b4 )
(2)
0 = A3F31 ( E b3 − E b1 ) + A3F32 ( E b3 − E b2 ) + A3F34 ( E b3 − E b4 )
where the emissive powers are
E b1 = σ T14
E b2 = σ T24
E b3 = σ T34
E b4 = 0
(3)
(4 – 7)
2
For this four surface enclosure, there are N = 16 view factors and N (N – 1)/2 = 4 × 3/2 = 6 must be
directly determined (by inspection or formulas) and the remainder can be evaluated from the
summation rule and reciprocity relation. By inspection,
F22 = 0
F44 = 0
(8,9)
From the coaxial parallel disk relation, Table 13.2, find F24
S = 1+
1 + R 24
R 22
1 + ( 0.250 )
2
= 1+
= 18.00
(0.250 )
2
R 2 = r2 / L = 0.050 m / 0.200 m = 0.250
R 4 = r4 / L = 0.250

2
F24 = 0.5 S − S2 − 4 ( r4 / r2 ) 
1/ 2 

 
 

1/ 2 

2
F24 = 0.5 18.00 − 18.002 − 4 (1)   = 0.0557

 

(10)
Consider the three-surface enclosure 1 − 2 − 2′ and find F11 as beginning with the summation rule,
Continued …..
PROBLEM 13.26 (Cont.)
F11 = 1 − F12 − F12′
(11)
where, from symmetry, F12 = F12′ , and using reciprocity,
(
)
F12 = A 2 F21 / A1 = π D 2 / 4 F23 / (π DL / 2 ) = DF21 / 2L
(12)
and from the summation rule on A2
F21 = 1 − F22′ = 1 − 0.172 = 0.828 ,
(13)
Using the coaxial parallel disk relation, Table 13.2, to find F221,
S = 1+
1 + R 22′
R 22
= 1+
1 + 0.50 2
0.50 2
= 6.000
R 2 = r2 / L = 0.050 m / ( 0.200 / 2 m ) = 0.500

2
F22′ = 0.5 S − S2 − 4 ( r2′ / r2 ) 

1/ 2 



2
F22′ = 0.5 6 − 62 − 4 (1) 

R 2′ = 0.500


1/ 2 


 = 0.1716

Evaluating F12 from Eq. (12), find
F12 = 0.100 m × 0.828 / 2 × 0.200 m = 0.2071
and evaluating F11 from Eq. (11), find
F11 = 1 − 2 × F12 = 1 − 2 × 0.207 = 0.586
From symmetry, recognize that F33 = F11 and F43 = F21. To this point we have directly determined
six view factors (underlined in the matrix below) and the remaining Fij can be evaluated from the
summation rules and appropriate reciprocity relations. The view factors written in matrix form, [Fij]
are.
0.5858
0.8284
0.1781
0.1158
0.2071
0
0.02896
0.05573
0.1781
0.1158
0.5858
0.8284
0.02896
0.05573
0.2071
0
Knowing all the required view factors, the energy balances and the emissive powers, Eqs. (4-6), can
be solved simultaneously to obtain:
q1 = 255 W
E b2 = 5.02 × 104 W / m 2
T2 = 970 K
T3 = 837.5 K
E b3 = 2.79 × 104 W / m 2
<
<
Continued …..
PROBLEM 13.26 (Cont.)
(b) Using the energy balances, Eqs. (1-3), along with the IHT Radiation Tool, View Factors, Coaxial
parallel disks, a model was developed to calculate q1, T2, and T3 as a function of length L for fixed
diameter D = 100 m. The results are plotted below.
For fixed diameter, as the overall length increases, the power required to maintain the heated section
at T1 = 1000 K decreases. This follows since the furnace opening area is a smaller fraction of the
enclosure surface area as L increases. As L increases, the bottom surface temperature T2 increases as
L increases and, in the limit, will approach that of the heated section, T1 = 1000 K. As L increases,
the temperature of the insulated cylindrical section, T3, increases, but only slightly. The limiting
value occurs when Eb3 = 0.5 × Eb1 for which T3 → 840 K. Why is that so?
PROBLEM 13.27
KNOWN: Dimensions and temperature of a rectangular fin array radiating to deep space.
FIND: Expression for rate of radiation transfer per unit length from a unit section of the array.
SCHEMATIC:
ASSUMPTIONS: (1) Surfaces may be approximated as blackbodies, (2) Surfaces are isothermal, (3)
Length of array (normal to page) is much larger than W and L.
ANALYSIS: Deep space may be represented by the hypothetical surface A′3 , which acts as a
blackbody at absolute zero temperature. The net rate of radiation heat transfer to this surface is
therefore equivalent to the rate of heat rejection by a unit section of the array.
(
)
(
q′3 = A1′ F13 σ T14 − T34 + A′2 F23 σ T24 − T34
With
)
A′2 F23 = A3′ F32 = A1′ F12 , T1 = T2 = T and T3 = 0,
q′3 = A1′ ( F13 + F12 )σ T 4 = W σ T 4
<
Radiation from a unit section of the array corresponds to emission from the base. Hence, if blackbody
behavior can, indeed, be maintained, the fins do nothing to enhance heat rejection.
COMMENTS: (1) The foregoing result should come as no surprise since the surfaces of the unit
section form an isothermal blackbody cavity for which emission is proportional to the area of the
opening. (2) Because surfaces 1 and 2 have the same temperature, the problem could be treated as a
two-surface enclosure consisting of the combined (1, 2) and 3. It follows that q′3 = q′(1,2 )3 = A′(1,2 )
F(1,2 )3 σ T 4 = A′3 F3(1,2 ) σ T 4 = W σ T 4 , (3) If blackbody behavior cannot be achieved
(ε1, ε 2 < 1) , enhancement would be afforded by the fins.
PROBLEM 13.28
KNOWN: Dimensions and temperatures of side and bottom walls in a cylindrical cavity.
FIND: Emissive power of the cavity.
SCHEMATIC:
ASSUMPTIONS: (1) Blackbody behavior for surfaces 1 and 2
ANALYSIS: The desired emissive power is defined as
E = q3 / A3
where
q3 = A1 F13 E b1 + A 2 F23 E b2 .
From symmetry, F23 = F21, and from reciprocity, F21 = (A1/A2) F12. With F12 = 1 – F13, it follows
that
q3 = A1 F13 E b1 + A1 (1 − F13 ) E b2 = A1 E b2 + A1 F13 ( E b1 − E b2 ) .
Hence, with A1 = A3,
(
)
q
E = 3 = E b2 + F13 ( E b1 − E b2 ) = σ T24 + F13 σ T14 − T24 .
A3
From Fig. 13.15, with (L/ri) = 4 and (rj/L) = 0.25, F13 ≈ 0.05. Hence
(
)
E = 5.67 × 10−8 W / m 2 ⋅ K 4 + 0.05 × 5.67 × 10−8 W / m 2 ⋅ K 4 1000 4 − 7004 K 4
E = 1.36 ×104 W / m 2 + 0.22 × 104 W / m 2
E = 1.58 × 104 W / m 2 .
<
PROBLEM 13.29
KNOWN: Aligned, parallel discs with prescribed geometry and orientation.
FIND: Net radiative heat exchange between the discs.
SCHEMATIC:
ASSUMPTIONS: (1) Surfaces behave as blackbodies, (2) A1 << A2.
ANALYSIS: The net radiation exchange between the two black surfaces follows from Eq. 13.13
written as
(
)
q12 = A1 F12 σ T14 − T24 .
The view factor can be determined from Eq. 13.8 which is appropriate for a small disc, aligned and
parallel to a much larger disc.
Fij =
D 2j
D 2j + 4L2
where Dj is the diameter of the larger disk and L is the distance of separation. It follows that
F12 = F1o − F1i = 0.00990 − 0.00559 = 0.00431
where
(
)
(
)
F1i = Di / ( Di2 + 4L2 ) = 0.152 m 2 / (0.152 m 2 + 4 × 1 m 2 ) = 0.00559.
F1o = Do4 / Do2 + 4L2 = 0.22 m 2 / 0.22 m 2 + 4 × 1 m 2 = 0.00990
The net radiation exchange is then
π ( 0.03m )
2
q12 =
4
× 0.00431 × 5.67 × 10−8
W
2
m ⋅K
4
(5004 − 10004 ) K4 = −0.162 W.
COMMENTS: F12 can be approximated using solid angle concepts if Do << L. That is, the view
factor for A1 to Ao (whose diameter is Do) is
ω
A / L2 π Do2
D2
=
= o .
F1o ≈ o −1 = o
π
π
4π L2 4L2
Numerically, F1o = 0.0100 and it follows F1i ≈ Di2 / 4L2 = 0.00563. This gives F12 = 0.00437. An
analytical expression can be obtained from Ex. 13.1 by replacing the lower limit of integration by
Di/2, giving
(
) (
)
F12 = L2  −1/ Do2 / 4 + L2 + 1/ Di2 / 4 + L2  = 0.00431.


PROBLEM 13.30
KNOWN: Two black, plane discs, one being solid, the other ring-shaped.
FIND: Net radiative heat exchange between the two surfaces.
SCHEMATIC:
ASSUMPTIONS: (1) Discs are parallel and coaxial, (2) Discs are black, diffuse surfaces, (3)
Convection effects are not being considered.
ANALYSIS: The net radiative heat exchange between the solid disc, A1, and the ring-shaped disc,
A2, follows from Eq. 13.13.
(
q12 = A1F12 σ T14 − T24
)
The view factor F12 can be determined from Fig. 13.5 after some manipulation. Define these two
hypothetical surfaces;
A3 =
A4 =
π Do2
4
, located co-planar with A2, but a solid surface
π Di2
, located co-planar with A2, representing the missing center.
4
From view factor relations and Fig. 13.5, it follows that
F12 = F13 − F14 = 0.62 − 0.20 = 0.42
F14:
F13:
rj
L
rj
L
=
40 / 2
=
80 / 2
20
20
= 1,
L
= 2,
L
ri
ri
=
=
20
80 / 2
20
80 / 2
= 0.5,
F14 = 0.20
= 0.5,
F13 = 0.62.
Hence
(
)
(
)
q12 = π 0.802 / 4 m2 × 0.42 × 5.67 × 10−8 W / m 2 ⋅ K 4 3004 − 10004 K 4
q12 = −11.87 kW.
<
Assuming negligible radiation exchange with the surroundings, the negative sign implies that q1 = 11.87 kW and q2 = +11.87 kW.
PROBLEM 13.31
KNOWN: Radiometer viewing a small target area (1), A1, with a solid angle ω = 0.0008 sr. Target
2
has an area A1 = 0.004 m and is diffuse, gray with emissivity ε = 0.8. The target is heated by a ringshaped disc heater (2) which is black and operates at T2 = 1000 K.
FIND: (a) Expression for the radiant power leaving the target which is collected by the radiometer in
terms of the target radiosity, J1, and relevant geometric parameters; (b) Expression for the target
radiosity in terms of its irradiation, emissive power and appropriate radiative properties; (c)
Expression for the irradiation on the target, G1, due to emission from the heater in terms of the heater
emissive power, the heater area and an appropriate view factor; numerically evaluate G1; and (d)
Determine the radiant power collected by the radiometer using the foregoing expressions and results.
SCHEMATIC:
ASSUMPTIONS: (1) Target is diffuse, gray, (2) Target area is small compared to the square of the
separation distance between the sample and the radiometer, and (3) Negligible irradiation from the
surroundings onto the target area.
ANALYSIS: (a) From Eq. (12.5) with I1 = I1,e+r = J1/π, the radiant power leaving the target collected
by the radiometer is
J
(1)
q1→ rad = 1 A1 cos θ1ω rad −1
π
where θ1 = 0° and ωrad-1 is the solid angle the radiometer subtends with respect to the target area.
<
(b) From Eq. 13.16, the radiosity is the sum of the emissive power plus the reflected irradiation.
J1 = E1 + ρ G1 = ε E b,1 + (1 − ε ) G1
<
(2)
where E b1 = σ T14 and ρ = 1 - ε since the target is diffuse, gray.
(c) The irradiation onto G1 due to emission from the heater area A2 is
q
G1 = 2 →1
A1
where q2→1 is the radiant power leaving A2 which is intercepted by A1 and can be written as
q 2→1 = A 2 F21 E b2
(3)
where E b2 = σ T24 . F21 is the fraction of radiant power leaving A2 which is intercepted by A1. The
view factor F12 can be written as
Continued …..
PROBLEM 13.31 (Cont.)
F12 = F1− o
F1−i = 0.5 − 0.2 = 0.3
where from Eq. 13.8,
F1− o =
F1− i =
Do2
Do2 + 4L2
Di2
D12 + 4L2
0.52
=
=
0.52 + 4 ( 0.25 )
2
0.252
0.252 + 4 ( 0.25 )
2
= 0.5
(3)
= 0.2
and from the reciprocity rule,
F21 =
A1F12
A2
=
0.0004m 2 × 0.3
(
)
π / 4 0.52 − 0.252 m 2
= 0.000815
Substituting numerical values into Eq. (3), find
G1 =
(
)
π / 4 0.52 − 0.252 m 2 × 0.000815 × 5.67 × 10−8 W / m 2 ⋅ K 4 (1000 K )
4
0.0004 m 2
G1 = 17, 013 W / m 2
<
(d) Substituting numerical values into Eq. (1), the radiant power leaving the target collected by the
radiometer is
(
)
q1→ rad = 6238 W / m 2 / π sr × 0.0004 m 2 × 1 × 0.0008sr = 635 µ W
<
where the radiosity, J1, is evaluated using Eq. (2) and G1.
J1 = 0.8 × 5.67 × 10 −8 W / m 2 ⋅ K 4 × (500 K ) + (1 − 0.8 ) × 17, 013 W / m 2
4
J1 = ( 2835 + 3403) W / m 2 = 6238 W / m 2
<
COMMENTS: (1) Note that the emitted and reflected irradiation components of the radiosity, J 1, are
of the same magnitude.
(2) Suppose the surroundings were at room temperature, Tsur = 300 K. Would the reflected
irradiation due to the surroundings contribute significantly to the radiant power collected by the
radiometer? Justify your conclusion.
PROBLEM 13.32
KNOWN: Thin-walled, black conical cavity with opening D = 10 mm and depth of L = 12 mm that
is well insulated from its surroundings. Temperature of meter housing and surroundings is 25.0°C.
2
FIND: Optical (radiant) flux of laser beam, Go (W/m ), incident on the cavity when the fine-wire
thermocouple indicates a temperature rise of 10.1°C.
SCHEMATIC:
ASSUMPTIONS: (1) Cavity surface is black and perfectly insulated from its mounting material in
the meter, (2) Negligible convection heat transfer from the cavity surface, and (3) Surroundings are
large, isothermal.
ANALYSIS: Perform an energy balance on the walls of the cavity considering absorption of the
laser irradiation, absorption from the surroundings and emission.
E in − E out = 0
A o G o + A o G sur − A o E b Tc = 0
$
2
where Ao = π D /4 represents the opening of the cavity. All of the radiation entering or leaving the
cavity passes through this hypothetical surface. Hence, we can treat the cavity as a black disk at T c.
4
-8
2 4
Since Gsur = Eb (Tsur), and Eb = σ T with σ = 5.67 × 10 W/m ⋅K , the energy balance has the form
G o + σ 25.0 + 273 4 K4 − σ 25.0 + 101
. + 273 4 K 4 = 0
G o = 63.8 W / m2
$
$
<
PROBLEM 13.33
KNOWN: Electrically heated sample maintained at Ts = 500 K with diffuse, spectrally selective
coating. Sample is irradiated by a furnace located coaxial to the sample at a prescribed distance.
Furnace has isothermal walls at Tf = 3000 K with εf = 0.7 and an aperture of 25 mm diameter.
2
Sample experiences convection with ambient air at T∞ = 300 K and h = 20 W/m ⋅K. The
surroundings of the sample are large with a uniform temperature Tsur = 300 K. A radiation detector
sensitive to only power in the spectral region 3 to 5 µm is positioned at a prescribed location relative
to the sample.
FIND: (a) Electrical power, Pe, required to maintain the sample at Ts = 500 K, and (b) Radiant power
incident on the detector within the spectral region 3 to 5 µm considering both emission and reflected
irradiation from the sample.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state condition, (2) Furnace is large, isothermal enclosure with small
aperture and radiates as a blackbody, (3) Sample coating is diffuse, spectrally selective, (4) Sample
and detector areas are small compared to their separation distance squared, (5) Surroundings are large,
isothermal.
ANALYSIS: (a) Perform an energy balance on the sample mount, which experiences electrical
power dissipation, convection with ambient air, absorbed irradiation from the furnace, absorbed
irradiation from the surroundings and emission,
E′in − E′out = 0
Pe + [−h ( Ts − T∞ ) + α1G f + α sur G sur − ε E b ( Ts )] As = 0 (1)
where E b ( Ts ) = σ Ts4 and As = π Ds2 / 4.
Irradiations on the sample: The irradiation from the furnace aperture onto the sample can be written
as
A f Ffs E b,f A f Ffsσ Tf4
q
=
G f = f →s =
As
As
As
(2)
where A f = π Df2 / 4 and As = π Ds2 / 4. The view factor between the furnace aperture and sample
follows from the relation for coaxial parallel disks, Table 13.2,
R f = rf / Lsf = 0.0125 m / 0.750 m = 0.01667
S = 1+
1 + R s2
R f2
= 1+
1 + 0.013332
0.01667 2
R s = rs / Lsf = 0.0100 m / 0.750 m = 0.01333
= 3600.2
Continued …..
PROBLEM 13.33 (Cont.)

2
Fsf = 0.5 S − S2 − 4 ( rs / rf ) 
1/ 2 



1/ 2 

2
2
 = 0.5 3600 − 3600 − 4 ( 0.05 / 0.0625 )   = 0.000178



Hence the irradiation from the furnace is
π ( 0.025 m ) / 4 × 0.000178 × 5.67 × 10−8 W / m 2 ⋅ K 4 (3000 K )
= 1277 W / m 2
Gf =
π 0.0202 m 2 / 4
2
4
(
)
The irradiation from the surroundings which are large compared to the sample is
4
= 5.67 × 10−8 W / m 2 ⋅ K (300K ) = 459 W / m 2
G sur = σ Tsur
4
Emissivity of the Sample: The total hemispherical emissivity in terms of the spectral distribution can
be written following Eq. 12.38 and Eq. 12.30,
∞
ε E
(T ) dλ / σ T 4 = ε1F(0 − λ1Ts ) + ε 2 1 − F(0 − λ1Ts ) 
0 λ λ ,b s
ε =∫
ε = 0.8 × 0.066728 + 0.2 [1 − 0.066728] = 0.240
where, from Table 12.1, with λ1Ts = 4 µ m × 500 K = 2000 µ m ⋅ K, F( 0 − λ T ) = 0.066728.
Absorptivity of the Sample: The total hemispherical absorptivity due to irradiation from the furnace
follows from Eq. 12.46,
α f = ε1F( 0 − λ T ) + ε 2 1 − F( 0 − λ T )  = 0.8 × 0.945098 + 0.2 [1 − 0.945098] = 0.767
1 f
1 f 

where, from Table 12.1, with λ1Tf = 4 µ m × 3000 K = 12, 000 µ m ⋅ K, F(0 − λ T ) = 0.945098. The
total hemispherical absorptivity due to irradiation from the surroundings is
αsur = ε1F(0 − λ T ) + ε 2 1 − F(0 − λ T )  = 0.8 × 0.00234 + 0.2 [1 − 0.002134] = 0.201
1 sur
1 sur 

where, from Table 12.1, with λ1Tsur = 4 µ m × 300 K = 1200 µ m ⋅ K, F( 0 − λ T ) = 0.002134.
Evaluating the Energy Balance: Substituting numerical values into Eq. (1),
Pe =  +20 W / m 2 ⋅ K (500 − 300 ) K − 0.767 × 1277 W / m 2 


−0.201 × 459 W / m 2 + 0.240 × 5.67 × 10−8 W / m 2 ⋅ K 4 (500 K ) π ( 0.020 m ) / 4
4
2
<
Pe = 1.256 W − 0.308 W − 0.029 W + 0.267 W = 1.19 W
(b) The radiant power leaving the sample which is incident on the detector and within the spectral
region, ∆λ = 3 to 5µm, follows from Eq. 12.5 with Eq. 12.30,
qs − d, ∆λ =  Es, ∆λ + G f ,ref , ∆λ + Gsur,ref , ∆λ  (1/ π ) As cos θ s ⋅ A d cos θ d / L2sd
where θs = 45° and θd = 0°. The emitted component is
5µ m
ε λ ,b E λ ,b (Ts )
3
E s, ∆λ = ∫
{
}
E s, ∆λ = ε1  F( 0 − 4 µ m,T ) − F( 0 − 3µ m,T ) + ε 2  F(0 − 5µ m,T ) − F( 0 − 4 µ m,T )  σ Ts4
s
s
s
s 


Continued …..
PROBLEM 13.33 (Cont.)
E s, ∆λ = {0.8 [0.066728 − 0.013754] + 0.2 [0.16169 − 0.066728]}σ (500K ) = 217.5 W / m 2
4
where, from Table 12.1, F( 0 −3µ m,T ) = 0.013754 at λ T = 3 µ m × 500 K = 1500 µ m ⋅ K;
s
F( 0 − 4 µ m,T
s)
= 0.066728 at λ = 4 µ m × 500 K = 2000 µ m ⋅ K; and F( 0 −5µ m,T ) = 0.16169 at λT =
s
5µm × 500 K = 2500 µm⋅K.
The reflected irradiation from the furnace component is
5µ m
(1 − ε λ ) G f ,λ dλ
3
G f ,ref , ∆λ = ∫
where Gf,λ ≈ Eλ,b(Tf), using band emission factors,
{
G f ,ref , ∆λ = (1 − ε )  F( 0 − 4 µ m,T

f
}
) − F(0 − 3µ m,Tf )  + (1 − ε 2 )  F(0 − 5µ m,Tf ) − F(0 − 4µ m,Tf )  G f
G f ,ref , ∆λ = {0.2 [0.9451 − 0.8900] + 0.8 [0.9700 − 0.9451]}1277W / m 2 = 39.51W / m 2
where, from Table 12.1, F( 0 −3µ m,T ) = 0.8900 at λTf = 3 µm × 3000 K = 9000 µm⋅K;
f
F( 0 − 4 µ m,T
f
) = 0.9451 at λTf = 4 µm × 3000 K = 12,000 µm⋅K; and, F(0 −5µ m,Tf ) = 0.9700 at λTf =
5 µm × 3000 K = 15,000 µm⋅K.
The reflected irradiation from the surroundings component is
5µ m
(1 − ε λ )G ref ,λ dλ
3
G sur,ref , ∆λ = ∫
where Gref,λ ≈ Eλ (Tsur), using band emission factors,
{
G sur,ref , ∆λ = (1 − ε1 )  F( 0 − 4 µ m,T
sur )

− F( 0 −3µ m,T
+ (1 − ε 2 )  F(0 −5 µ m,T


sur ) 
sur )
 Gsur
− F(0 − 4 µ m,T
sur ) 
G sur,ref , ∆λ = {0.2 [0.002134 − 0.0001685] − 0.8 [0.013754 − 0.002134]} 459 W / m 2 = 4.44 W / m 2
where, from Table 12.1, F( 0 −3µ m,T
sur )
F( 0 − 4 µ m,T
sur )
= 0.0001685 at λTsur = 3 µm ×⋅300 K = 900 µm⋅K;
= 0.002134 at λTsur = 4 µm × 300 K = 1200 µm⋅K; and F( 0 − 5µ m,T ) = 0.013754 at
sur
λTsur = 5 µm ×300 K=1500 µm⋅K. Returning to Eq. (3), find
2
qsd, ∆λ = [217.5 + 39.51 + 4.44] W / m 2 (1/ π ) 8π ( 0.020 m ) / 4 


cos 45° × 8 × 10−5 m 2 × cos 0° / (1 m ) = 1.48 µ W
2
<
COMMENTS: (1) Note that Ffs is small, since Af, As << L2sf . As such, we could have evaluated
qf→s using Eq. 12.5 and found
Gf =
(
E b,f / π Af As / L2sf
) = 1276 W / m2
As
(2) Recognize in the analysis for part (b), Eq. (3), the role of the band emission factors in calculating
the fraction of total radiant power for the emitted and reflected irradiation components.
PROBLEM 13.34
KNOWN: Water-cooled heat flux gage exposed to radiant source, convection process and
surroundings.
FIND: (a) Net radiation exchange between heater and gage, (b) Net transfer of radiation to the gauge
per unit area of the gage, (c) Net heat transfer to the gage per unit area of gage, (d) Heat flux indicated
by gage described in Problem 3.98.
SCHEMATIC:
ASSUMPTIONS: (1) Heater and gauge are parallel, coaxial discs having blackbody behavior, (2) Ag
<< Ah, (3) Surroundings are large compared to Ah and Ag.
ANALYSIS: (a) The net radiation exchange between the heater and the gage, both with blackbody
behavior, is given by Eq. 13.13 having the form
)
(
)
(
q h − g = A h Fhg σ Th4 − Tg4 = A g Fgh σ Th4 − Tg .
Note the use of reciprocity, Eq. 13.3, for the view factors. From Eq. 13.8,
(
)
(
)
Fgh = D h2 / 4L2 + D 2h = ( 0.2m ) / 4 × 0.52 m 2 + 0.22 m 2 = 0.0385.
(
)
2
q h − g = π 0.012 m 2 / 4 × 0.0385 × 5.67 × 10−8 W / m 2 ⋅ K 4 8004 − 2904  K 4 = 69.0 mW.

<

(b) The net radiation to the gage per unit area will involve exchange with the heater and the
surroundings. Using Eq. 13.14,
q′′net,rad = −q g / A g = q h − g / A g + qsur − g / A g .
The net exchange with the surroundings is
(
)
(
(300
)
4
4
− Tg 4 = A g Fg −sur σ Tsur
− Tg4 .
qsur − g = Asur Fsur − g σ Tsur
q ′′net,rad =
69.0 × 10
−3
W
π ( 0.01m ) / 4
2
+ (1 − 0.0385 ) 5.67 × 10
−8
2
W/m ⋅K
4
4
− 290
4
)K
4
2
= 934.5 W / m .
<
(c) The net heat transfer rate to the gage per unit area of
the gage follows from the surface energy balance
q′′net,in = q′′) net,rad + q′′conv
q′′net,in = 934.5 W / m 2 + 15W / m 2 ⋅ K (300 − 290 ) K
q′′net,in = 1085 W / m 2 .
<
(d) The heat flux gage described in Problem 3.98 would experience a net heat flux to the surface of
2
2
1085 W/m . The irradiation to the gage from the heater is Gg = qh→g/Ag = Fgh σ Th4 = 894 W/m .
Since the gage responds to net heat flux, there would be a systematic error in sensing irradiation from
the heater.
PROBLEM 13.35
KNOWN: Long cylindrical heating element located a given distance above an insulated wall
exposed to cool surroundings.
FIND: Maximum temperature attained by the wall and temperature at location A.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Insulated wall, (3) Negligible conduction in wall,
(4) All surfaces are black.
ANALYSIS: Consider an elemental area at point
x = 0; this is the location that will attain the maximum
temperature. Since the wall is insulated and
conduction is negligible, the net radiation leaving
dAo is zero. From Eq. 13.13,
(
)
(
)
4
q′′o = q′′o,h + q′′o,sur = Fo,h σ To2 − Th4 + Fo,sur To5 − Tsur
=0
(1)
where Fo,sur = 1 – Fo,h and Fo,h can be found from the relation for a cylinder and parallel rectangle,
Table 13.1, with s1 = 2 mm, s2 = 0 mm, L = 40 mm, and r = R = 10 mm.
r  −1 s1
s 
10 mm  −1 2

− tan −1 2  =
− tan −1 0  = 0.25
Fo,h =
tan
tan
(2)


s1 − s 2 
L
L  2 mm − 0 
40

Rearranging Eq. (1) and substituting numerical values, find

To4 = Th4 +

(1 − Fo,h ) T 4
Fo,h
 
1 − Fo,h 
 
Fo,h 
sur  / 1 +

1 − 0.25 

4 1 − 0.25
To4 = ( 700 K ) +
(300 K )4  / 1 +
0.25
0.25 

 
(3)
To = 507 K.
<
For the point A located at x = 40 mm, use the same relation of Table 13.1 to find FA,h (for this point,
s1 = 41 mm, s2 = 39 mm, r = R = 10 mm, L = 40 mm),
10 mm
39 
 −1 41
− tan −1  = 0.125.
FA,h =
tan

40
40 
( 41 − 39 ) mm 
Substituting numerical values into Eq. (3), find
1 − 0.125 
4 1 − 0.125
4 
= ( 700 K ) +
TA
TA = 439 K.
(300 K )4  / 1 +
0.125
0.125 

 
COMMENTS: Note the importance of the assumptions that the wall is insulated and conduction is
negligible. In calculating Fo,h and FA,h we are finding the view factor for a small area or point.
Hence, we need only specify that s1 – s2 is very small compared to L.
<
PROBLEM 13.36
KNOWN: Diameter and pitch of in-line tubes occupying evacuated space between parallel plates of
of water through the tubes.
prescribed temperature. Temperature and flowrate m
= 0.20 kg/s, (b) Effect of m
on Ts.
FIND: (a) Tube surface temperature Ts for m
SCHEMATIC:
ASSUMPTIONS: (1) Surfaces behave as blackbodies, (2) Negligible tube wall conduction
resistance, (3) Fully-developed tube flow.
-6
2
PROPERTIES: Table A-6, water (Tm = 300 K): µ = 855 × 10 N⋅s/m , k = 0.613 W/m⋅K, Pr = 5.83.
ANALYSIS: (a) Performing an energy balance on a single tube, it follows that qps = qconv, or
(
)
A p Fpsσ Tp4 − Ts4 = hAs (Ts − Tm )
From Table 13.1 and D/S = 0.75, the view factor is
1/ 2
  D 2 
Fps = 1 − 1 −   
  S  
1/ 2
2
2
 D  −1  S − D 
+   tan 

 D2 
S


= 0.881
−6
/ π Dµ = 4 ( 0.20 kg / s ) / π ( 0.015 m ) 855 × 10
N ⋅ s / m 2 = 19,856, fully-developed
With Re D = 4m
turbulent flow may be assumed, in which case Eq. 8.60 yields
k
0.613 W / m ⋅ K
h=
0.023 Re 4D/ 5 Pr 0.4 =
(0.023)(19,856 )4 / 5 (5.83)0.4 = 5220 W / m2 ⋅ K
D
0.015 m
)
(
Hence, with (Ap/As) = 2S/πD = 0.849,
Ts − Tm =
Fpsσ A p
h
As
(T
)
0.881 × 5.67 × 10
4
4
p − Ts =
−8
2
W / m ⋅K
2
5220 W / m ⋅ K
4
(0.849 )
(T
4
4
p − Ts
)
With Tm = 300 K and Tp = 1000 K, a trial-and-error solution yields
<
Ts = 308 K
(b) Using the Correlations and Radiation Toolpads of IHT to evaluate the convection coefficient and
view factor, respectively, the following results were obtained.
is due to an increase in h and hence a reduction in the convection
The decrease in Ts with increasing m
resistance.
COMMENTS: Due to the large value of h, Ts << Tp.
PROBLEM 13.37
KNOWN: Insulated wall exposed to a row of regularly spaced cylindrical heating elements.
FIND: Required operating temperature of the heating elements for the prescribed conditions.
SCHEMATIC:
ASSUMPTIONS: (1) Upper and lower walls are isothermal and infinite, (2) Lower wall is insulated,
(3) All surfaces are black, (4) Steady-state conditions.
ANALYSIS: Perform an energy balance on the insulated wall considering convection and radiation.
E ′′in − E ′′out = −q1′′ − q′′conv = 0
where q1′′ is the net radiation leaving the insulated
wall per unit area. From Eq. 13.13,
)
(
(
′′ + q12
′′ = F1eσ T14 − Te4 + F12σ T14 − T24
q1′′ = q1e
)
where F12 = 1 – F1e. Using Newton’s law of cooling for q′′conv solve for Te,

(1 − F1e )

F1e
Te4 =  T14 +
(T
4
4
1 − T2
) + σh F1 (T − T
1e
1
∞ ).
The view factor between the insulated wall and the tube row follows from the relation for an infinite
plane and row of cylinders, Table 13.1,
1/ 2
  D 2 
F1e = 1 − 1 −   
  S  
1/ 2
  10 2 
F1e = 1 − 1 −   
  20  
1/ 2
 s2 − D2 
D
+   tan −1 

 D2 
S


1/ 2
 202 − 102 
 10 
+   tan −1 

 102 
 20 


= 0.658.
Substituting numerical values, find
)
200 W / m 2 ⋅ K
1


4 1 − 0.658
Te4 = (500 K ) +
5004 − 3004 K 4  +
×
(500 − 450 ) K
8
2
4
−
0.658

 5.67 × 10 W / m ⋅ K 0.658
(
Te = 774 K.
<
COMMENTS: Always express temperatures in kelvins when considering convection and radiation
terms in an energy balance. Why is F1e independent of the distance between the row of tubes and the
wall:
PROBLEM 13.38
KNOWN: Surface radiative properties, diameter and initial temperature of a copper rod placed in an evacuated
oven of prescribed surface temperature.
FIND: (a) Initial heating rate, (b) Time th required to heat rod to 1000 K, (c) Effect of convection on heating
time.
SCHEMATIC:
ASSUMPTIONS: (1) Copper may be treated as a lumped capacitance, (b) Radiation exchange between rod and
oven may be approximated as blackbody exchange.
3
PROPERTIES: Table A-1, Copper (300 K): ρ = 8933 kg/m , cp = 385 J/kg⋅K, k = 401 W/m⋅K.
ANALYSIS: (a) Performing an energy balance on a unit length of the rod, E
in = E st , or
q = Mc p
 π D 2  dT
× 1 cp
 4
 dt


dT
= ρ
dt
(
4
4
Neglecting convection, q = qrad = A2 F21 σ Tsur − T
F12 =1. It follows that
dT
dt
=
dT 
(
4
σπ D Tsur − T
(
4
)
2
ρ π D / 4 cp
) = 4σ (T
4
4
sur − T
1 12
1
)
(1)
ρ Dc p
4 (1650 K ) − (300 K )
4
4

 =
dt i
) = A F σ (Tsur4 − T4 ), where A = πD × 1 and
8933 kg / m
3

5.67 × 10
−8
2
W/m ⋅K
4
(0.01m ) 385 J / kg ⋅ K
<
= 48.8 K / s.
(b) Using the IHT Lumped Capacitance Model to numerically integrate Eq. (2), we obtain
t s = 15.0 s
(
4
(c) With convection, q = qrad + qconv = A1 F12 σ Tsur − T
dT
dt
=
(
4
4σ Tsur − T
ρ Dc p
4
) + 4h (T
4
) + hA (T
1
<
∞ - T), and the energy balance becomes
∞ − T)
ρ Dc p
Performing the numerical integration for the three values of h, we obtain
2
h (W/m ⋅K):
th (s):
10
14.6
100
12.0
(
4
COMMENTS: With an initial value of hrad,i = σ Tsur − T
500
6.8
4
) /(T
2
sur – T) = 311 W/m ⋅K, Bi = hrad (D/4)/k =
2
0.002 and the lumped capacitance assumption is justified for parts (a) and (b). With h = 500 W/m ⋅K and h + hr,i
= 811 W/m⋅K in part (c), Bi = 0.005 and the lumped capacitance approximation is also valid.
PROBLEM 13.39
KNOWN: Long, inclined black surfaces maintained at prescribed temperatures.
FIND: (a) Net radiation exchange between the two surfaces per unit length, (b) Net radiation transfer
to surface A2 with black, insulated surface positioned as shown below; determine temperature of this
surface.
SCHEMATIC:
ASSUMPTIONS: (1) Surfaces behave as blackbodies, (2) Surfaces are very long in direction normal
to page.
ANALYSIS: (a) The net radiation exchange between two black surfaces is given by Eq. 13.13,
(
q12 = A1F12σ T14 − T24
)
Noting that A1 = width×length ( " ) and that from symmetry, F12 = 0.5, find
(
)
q
′ = 12 = 0.1 m × 0.5 × 5.67 × 10−8 W / m 2 ⋅ K 4 10004 − 8004 K 4 = 1680 W / m.
q12
"
<
(b) With the insulated, black surface A3 positioned as
shown above, a three-surface enclosure is formed. From
an energy balance on the node representing A2, find
′ + q12
′
−q′2 = q32
−q 2 = A3F32 [E b3 − E b2 ] + A1F12 [E b1 − E b2 ].
To find Eb3, which at present is not known, perform an energy balance on the node representing A3.
Note that A3 is adiabatic and, hence q3 = 0, q13 = q32.
A1F13 [E b1 − E b3 ] = A3F32 [E b3 − E b2 ]
Since F13 = F23 = 0.5 and A1 = A3, it follows that
E b3 = (1/ 2 )[E b1 + E b2 ]
′
−q′2 = ( A3 / " ) F32 [( E b1 + E b2 ) / 2 − E b2 ] + q12
and
(
)
−q′2 = 0.1 m × 0.5 × 5.67 × 10−8 W / m 2 ⋅ K 4  1000 4 + 800 4 / 2 − 800 4  K 4

+1680 W / m = 2517 W / m

<
Noting that Eb3 = σ T34 = (1/2) [Eb1 +Eb2], it follows that
(
)
1/ 4
T3 =  T14 + T24 / 2 


(
)
1/ 4
=  10004 + 8004 / 2 


K = 916 K.
<
PROBLEM 13.40
KNOWN: Electrically heated tube suspended in a large vacuum chamber.
FIND: (a) Electrical power supplied to the heater, Pe, to maintain it at T1 = 127°C, and (b) Compute
and plot Pe as a function of tube length L for the range 25 ≤ L ≤ 250 mm for tube temperatures of T1
= 127, 177 and 227°C.
SCHEMATIC:
ASSUMPTIONS: (1) All surfaces are blackbodies, (2) Tube of area A1 is isothermal, (3) The
surroundings are very large compared to the tube.
ANALYSIS: (a) Recognize that the surroundings can be represented by the surfaces A2 and A3,
which are blackbodies at Tsur. This situation then permits calculation of necessary shape factors.
The net radiative heat rate from the heater, surface A1, follows from Eq. 13.14, with T2 = T3 = Tsur as
(
)
(
)
Pe = q1 = A1F12σ T14 − T24 + A1F13σ T14 − T34 .
(1)
Note that F12 = F13 from symmetry considerations. Write now the summation rule for surface A2
F21 + F22 + F23 = 1
or
F21 = 1 − F23
where F23 is determined from Fig. 13.5 using
rj
L
=
0.75 / 2
0.33
= 1.14
L
ri
=
0.33
0.75 / 2
= 0.88
giving F23 = 0.37. Hence F21 = 1 – 0.37 = 0.63. Using reciprocity, find now F12
(
)
π 752 / 4
π D2 / 4
× F21 =
× 0.63 = 0.36.
F12 =
F21 =
π DL
π ( 75 × 33)
A1
A2
Noting that F12 = F13 and that T2 = T3, the electrical power using Eq. (1) with numerical values can
be written as,
Continued …..
PROBLEM 13.40 (Cont.)
Pe = 2 π

× 0.75m × 0.33m × 0.36 × 5.67 × 10
−8
2
W / m ⋅K
4
(127 + 273 )4 − ( 27 + 273 )4  K 4 = 555W .


<
(b) Using the energy balance Eq. (1) of the foregoing analysis with the IHT Radiation Tool-View
Factors, Coaxial parallel disks, Pe was computed as a function of L for selected tube temperatures.
As the tube length L increases, the heater power Pe required to maintain the tube at T1 increases.
Note that for small values of L, say L < 100 mm, P is linear with L. For larger values of L, Pe is not
linear with L. Why is this so? What is the relationship between Pe and L for L >> 300 mm?
PROBLEM 13.41
KNOWN: Two horizontal, very large parallel plates with prescribed surface conditions and
temperatures.
FIND: (a) Irradiation to the top plate, G1, (b) Radiosity of the top plate, J1, (c) Radiosity of the lower
plate, J2, (d) Net radiative exchange between the plates per unit area of the plates.
SCHEMATIC:
ASSUMPTIONS: (1) Plates are sufficiently large to form a two surface enclosure and (2) Surfaces
are diffuse-gray.
ANALYSIS: (a) The irradiation to the upper plate is defined
as the radiant flux incident on that surface. The irradiation to
the upper plate G1 is comprised of flux emitted by surface 2
and reflected flux emitted by surface 1.
G1 = ε 2 E b2 + ρ 2 E b1 = ε 2σ T24 + (1 − ε 2 )σ T14
G1 = 0.8 × 5.67 × 10−8 W / m 2 ⋅ K 4 (1000 K ) + (1 − 0.8 ) × 5.67 × 10−8 W / m 2 ⋅ K 4 (500 K )
4
4
G1 = 2835 W / m 2 + 11, 340 W / m 2 = 14,175 W / m 2 .
<
(b) The radiosity is defined as the radiant flux leaving the surface by emission and reflection. For the
blackbody surface 1, it follows that
J1 = E b1 = σ T14 = 5.67 × 10−8 W / m 2 ⋅ K 4 (1000 K ) = 56, 700 W / m 2 .
4
<
(c) The radiosity of surface 2 is then,
J 2 = ε 2 E b2 + ρ 2G 2 .
Since the upper plate is a blackbody, it follows that G2 = Eb1 and
J 2 = ε 2 E b1 + ρ 2 E b1 = ε 2σ T24 + 1 (1 − ε 2 )σ T14 = 14,175 W / m 2 .
<
Note that J2 = G1. That is, the radiant flux leaving surface 2 (J2) is incident upon surface 1 (G1).
(d) The net radiation heat exchange per unit area can be found by three relations.
q1′′ = J1 − G1 = (56, 700 − 14,175 ) W / m 2 = 42, 525 W / m 2
′′ = J1 − J 2 = (56, 700 − 14,175 ) W / m 2 = 42, 525 W / m 2
q12
<
The exchange relation, Eq. 13.24, is also appropriate with ε1 = 1,
′′
q1′′ = −q′′2 = q12
(
)
(
)
q1′′ = ε 2σ T14 − T24 = 0.8 × 5.67 × 10−8 W / m 2 ⋅ K 4 10004 − 5004 K 4 = 42, 525 W / m 2 .
PROBLEM 13.42
KNOWN: Dimensions and temperature of a flat-bottomed hole.
FIND: (a) Radiant power leaving the opening, (b) Effective emissivity of the cavity, εe, (c) Limit of
εe as depth of hole increases.
SCHEMATIC:
ASSUMPTIONS: (1) Hypothetical surface A2 is a blackbody at 0 K, (2) Cavity surface is
isothermal, opaque and diffuse-gray.
ANALYSIS: Approximating A2 as a blackbody at 0 K implies that all of the radiation incident on A2
from the cavity results (directly or indirectly) from emission by the walls and escapes to the
surroundings. It follows that for A2, ε2 = 1 and J2 = Eb2 =0.
(a) From the thermal circuit, the rate of radiation loss through the hole (A2) is
1 − ε1
q1 = ( E b1 − E b2 ) / 
 ε1A1
+
1
A1F12
+
1 − ε2 
ε 2 A 2 
(1)
.
Noting that F21 = 1 and A1 F12 = A2 F21, also that
A1 = π D 2 / 4 + π DL = π D ( D / 4 + L ) = π ( 0.006 m )( 0.006 m / 4 + 0.024 m ) = 4.807 × 10 −4 m 2
A 2 = π D 2 / 4 = π ( 0.006 m ) / 4 = 2.827 × 10 −5 m 2 .
2
4
Substituting numerical values with Eb = σT , find
)
(

q1 = 5.67 × 10−8 W / m 2 ⋅ K 4 10004 − 0 K 4 / 
1 − 0.8
 0.8 × 4.807 × 10
−4
m
2
+

1
2.827 × 10
−5
m
2
+ 0

<
q1 = 1.580 W.
(b) The effective emissivity, εe, of the cavity is defined as the ratio of the radiant power leaving the
cavity to that from a blackbody having the same area of the cavity opening and at the temperature of
the inner surfaces of the cavity. For the cavity above,
εe =
q1
A 2σ T14
(
)
ε e = 1.580 W / 2.827 × 10−5 m 2 5.67 × 10−8 W / m2 ⋅ K 4 (1000 K ) = 0.986.
4
<
(c) As the depth of the hole increases, the term (1 - ε1)/ε1 A1 goes to zero such that the remaining term
in the denominator of Eq. (1) is 1/A1 F12 = 1/A2 F21. That is, as L increases, q1 → A2 F21 Eb1. This
implies that εe → 1 as L increases. For L/D = 10, one would expect εe = 0.999 or better.
PROBLEM 13.43
KNOWN: Long V-groove machined in an isothermal block.
FIND: Radiant flux leaving the groove to the surroundings and effective emissivity.
SCHEMATIC:
ASSUMPTIONS: (1) Groove surface is diffuse-gray, (2) Groove is infinitely long, (3) Block is
isothermal.
ANALYSIS: Define the hypothetical surface A2 with T2 = 0 K. The net radiation leaving A1, q1,
will pass to the surroundings. From the two surface enclosure analysis, Eq. 13.23,
(
σ T14 − T24
q1 = −q 2 =
)
1 − ε1
1
1− ε2
+
+
ε1A1 A1F12 ε 2 A 2
Recognize that ε2 = 1 and that from reciprocity, A1 F12 = A2 F21 where F21 = 1. Hence,
q1
A2
=
(
σ T14 − T24
)
1 − ε1 A 2
+1
ε1 A1
With A2/A1 = 2 tan20°/ ( 2" / cos 20° ) = sin20°, find
q1′′ =
(
)
5.67 × 10−8 W / m 2 ⋅ K 4 1000 4 − 0 K 4
(1 − 0.6 ) × sin 20° + 1
= 46.17 kW / m 2 .
<
0.6
The effective emissivity of the groove follows from the definition given in Problem 13.42 as the ratio
of the radiant power leaving the cavity to that from a blackbody having the area of the cavity opening
and at the same temperature as the cavity surface. For the present situation,
εe =
q1′′
E b ( T1 )
=
q1′′
σ T14
=
46.17 × 10+3 W / m 2
4
5.67 × 10−8 W / m 2 ⋅ K 4 (1000 K )
= 0.814.
<
COMMENTS: Note the use of the hypothetical surface defined as black at 0 K. This surface does
not emit and absorbs all radiation on it; hence, is the radiant power to the surroundings.
PROBLEM 13.44
KNOWN: Conical cavity formed in an isothermal, opaque, diffuse-gray material of emissivity ε and
temperature T.
FIND: Radiant power leaving the opening of the cavity in terms of T, ε, ro, and L.
SCHEMATIC:
ASSUMPTIONS: (1) Material is opaque, diffuse-gray, and isothermal, (2) Cavity opening is
hypothetical black surface at 0 K.
ANALYSIS: Define Ao, the opening of the cavity, as a black surface at To = 0 K. Considering Ao
and Ac as a two surface, diffuse-gray enclosure, the radiant power leaving the cavity opening is
1 − ε
1 − εo 
1
q cavity = −q o =  E b ( T ) − E b ( To ) / 
+
+

 ε Ac Ac Fco ε o Ao 
Recognizing that Eb(To) = 0 and εo = 1 and also, using reciprocity,
Ac Fco = Ao Foc
and from the enclosure, note Foc = 1. Hence,
E b (T )
Aoσ T 4
σ T4
q cavity =
.
=
=
1− ε
1
1− ε
1
1 − ε Ao
+
+
⋅
+1
ε Ac Ac Fco ε Ac Ao Foc
ε Ac
(
Noting that A o = π ro2 and A c = π ro L2 + ro2
q cavity =
π ro2 ⋅ σ T 4
1
)
(1)
1/ 2
1− ε
⋅
+1
1/ 2
ε 
2 
L / ro ) + 1
(

, find
.
<
COMMENTS: When L increases or Ao/Ac << 1, the radiant power approaches that of a blackbody
according to Eq. (1).
PROBLEM 13.45
KNOWN: Cavities formed by a cone, cylinder, and sphere having the same opening size (d) and
major dimension (L) with prescribed wall emissivity.
FIND: (a) View factor between the inner surface of each cavity and the opening of the cavity; (b)
Effective emissivity of each cavity as defined in Problem 13.42, if the walls are diffuse-gray with εw;
and (c) Compute and plot εe as a function of the major dimension-to-opening size ratio, L/d, over the
range from 1 to 10 for wall emissivities of εw = 0.5, 0.7, and 0.9.
SCHEMATIC:
ASSUMPTIONS: (1) Diffuse-gray surfaces, (2) Uniform radiosity over the surfaces.
ANALYSIS: (a) Using the summation rule and reciprocity, determine the view factor F12 for each of
the cavities considered as a two-surface enclosure.
Cone:
F21 + F22 = F21 + 0 = 1
F21 = 1
)
(
1/ 2
2
F12 = A 2 F21 / A1 = π d 2 / 4 / (π d / 2 )  L2 + ( d / 2 ) 
F21 = 1
Cylinder:

(
)
(
)

= (1/ 2 ) ( L / d ) + 1/ 4 
2

−1/ 2

−1
F12 = A 2 F21 / A1 = A 2 / A1 = π d 2 / 4 / π dL + π d 2 / 4  = (1 + 4L / d )
<
<

Sphere: F21 = 1
(

−1
)
F12 = A 2 F21 / A1 = A 2 / A1 = π d 2 / 4 / π D 2 − π d 2 / 4  = 4D 2 / d 2 − 1
<
.
(b) The effective emissivity of the cavity is defined as
ε eff = q12 / q c
where q c = A 2σ T14 which presumes the opening is a black surface at T1 and for the two-surface
enclosure,
q12 =
(
σ T14 − T24
)
A1σ T14
=
(1 − ε1 ) / ε1A1 + 1/ A1F12 + (1 − ε 2 ) / ε 2 A2 (1 − ε1 ) / ε1 + 1/ F12
since T2 = 0K and ε2 = 1. Hence, since A2/A1 = F12 for all the cavities, with ε1 = εw
1/ F12
1
ε eff =
=
(1 − ε w ) / ε w + 1/ F12 F12 (1 − ε w ) / ε w + 1

Cone: ε eff = 1 (1/ 2 ) ( L / d ) + 1/ 4 



2


−1/ 2

(1 − ε w ) / ε w + 1

(1)
<
Continued …..
PROBLEM 13.45 (Cont.)
{[1 + 4L / d]
−1
}
(1 − ε w ) / ε w + 1
Cylinder:
ε eff = 1
Sphere:
−1


2 2
ε eff = 1 /   4D / d − 1 (1 − ε w ) / ε w + 1



(2)
<
(3)
<
(c) Using the IHT Workspace with eqs. (1,2,3), the effective emissivity was computed as a function of
L/d (cone, cylinder and sphere) for selected wall emissivities. The results are plotted below.
Fig 2 C o n ica l ca vity
Fig. 1 C o ne , cylind e r, s p h e re ca vitie s , ep s = 0 .7
1
1
0 .9
e p s e ff
e ps e ff
0 .9 5
0 .9
0 .8
0 .8 5
0 .7
1
0 .8
1
2
3
4
5
6
7
8
9
2
3
4
5
10
6
7
8
9
10
L /d
L /d
e p s w = 0 .5
e p s w = 0 .7
e p s w = 0 .9
Cone
C ylin de r
Sp he re
In Fig. 1, εeff is shown as a function of L/d for εw = 0.7. For larger L/d, the sphere has the highest εeff
and the cone the lowest. Figures 2, 3 and 4 illustrate the εeff vs. L/d for each of the cavity types. As
expected, εeff increases with increasing wall emissivity.
Fig 4 Sp h e rica l ca vity
Fig 3 C ylind rica l ca vity
1
0 .9
0 .9
e ps e ff
e p s e ff
1
0 .8
0 .8
0 .7
0 .7
1
2
3
4
5
6
7
8
9
10
1
2
3
5
6
7
8
9
D /d
L /d
e p s w = 0 .5
e p s w = 0 .7
e p s w = 0 .9
4
e p s w = 0 .5
e p s w = 0 .7
e p s w = 0 .9
Note that for the spherical cavity, with L/d ≥ 5, εeff > 0.98 even with εw as low as 0.5. This feature
makes the use of spherical cavities for high performance radiometry applications attractive since εeff
is not very sensitive to εw.
COMMENTS: In Fig. 1, intercomparing εeff for the three cavity types, can you give a physical
explanation for the results?
10
PROBLEM 13.46
KNOWN: Very long diffuse, gray, thin-walled tube of 1-m radius contained inside a long black duct
of square cross-section, 3.2 m × 3.2 m. Top portion is open as shown schematically.
FIND: (a) Net radiant heat transfer rate per unit length of the tube from the opening, q1′ = q1 / L, and
the effective emissivity of the opening, εeff, for the condition when θ = 45° and (b) Compute and plot
q1′ and εeff as a function of θ for the range 0 ≤ θ ≤ 180°.
SCHEMATIC:
ASSUMPTIONS: (1) Tube is very long compared to its radius R and duct dimension, (2) Interior of
cylinder is diffuse, gray, (3) Interior of the duct is black.
ANALYSIS: (a) Consider the two-surface enclosure formed by the inner surface of the tube, A1, and
the hypothetical surface formed by the opening, A3. The surface A3 behaves as a blackbody (ε3 = 1)
at a temperature T3 = T2 = 300 K. The net heat rate leaving the opening follows from Eq. 13.23.
)
(1)
The view factor F13 can be determined from the reciprocity relation recognizing that F31 = 1.
F13 = A3F31 / A1 = 1.414L × 1/ 4.712L = 0.300
(2)
where the areas A1 and A3 are, with θ = 45°,
A1 = 2R (π − θ ) L = 2 × 1m [π − 45 × π /180] × L = 4.712L
(3)
q1 = −q1 =
(
σ T14 − T24
(1 − ε1 ) / ε1A1 + 1/ A1F13 + (1 − ε3 ) / ε3A3
A3 = 2R sin θ L = 2 × 1m × sin 45°× L = 1.414L
Substituting numerical values, find
q1 = −q3 =
(
(4)
)
5.67 × 10−8 W / m 2 ⋅ K 4 10004 − 3004 K 4
(1 − 0.1) / (0.1× 4.712L ) + 1/ ( 4.712L × 0.300 ) + 0
<
q1 / L = −q3 / L = 21, 487 W / m
Continued …..
PROBLEM 13.46 (Cont.)
The effective emissivity of the opening is the ratio of the radiant power leaving the opening to that of
a blackbody having the area of the opening (A3) and a temperature of the inner surface of the cavity
(T1).
ε eff =
ε eff =
q1
A3σ T14
=
q1 / L
(5)
( A3 / L )σ T14
21, 487 W / m
1.414 m × 5.67 × 10−8 W / m 2 ⋅ K 4 (1000 K )
4
= 0.268
<
(b) Using the foregoing equations, Eqs. (1-5), in the IHT workspace, q1′ = q1 / L and εeff as a function
of θ were computed and are plotted below.
Note that q1′ = 0 when θ = 0° since the tube is closed and no power leaves the tube. At θ = 180°, the
area of the tube has been reduced to zero and hence, q1′ = 0 . For small values of θ, εeff is highest and
decreases as θ increases, to the limit εeff = ε1 = 0.1.
PROBLEM 13.47
KNOWN: Temperature, emissivity and dimensions of a rectangular fin array radiating to deep space.
FIND: (a) Rate of radiation transfer per unit length from a unit section to space, (b) Effect of
emissivity on heat rejection.
SCHEMATIC:
ASSUMPTIONS: (1) Diffuse/gray surface behavior, (2) Length of array (normal to page) is much
larger than W and L, (3) Isothermal surfaces.
ANALYSIS: (a) Since the sides and base of the U-section have the same temperature and emissivity,
they can be treated as a single surface and the U-section becomes a two-surface enclosure. Deep
space may be represented by the hypothetical surface A′3 , which acts as a blackbody at absolute zero
temperature. From Eq. (13.23), with T1 = T2 = T and ε1 = ε2 = ε,
q′(1,2 )3 =
(
σ T 4 − T34
)
1− ε
1
1− ε
+
+
ε A(′1,2 ) A′(1,2 ) F(1,2 )3 ε A3′
where A′(1,2 ) = 2 L + W, A3′ = W, A(′ 1,2 ) F(1,2 )3 = A3′ F3(1,2 ) = W. Hence,
q′(1,2 )3 =
σ T4
1− ε
1 1− ε
+
+
ε (2 L + W ) W ε W
5, 67 × 10−8 W / m 2 ⋅ K 4 (325 K )
= 15.2 W / m
1 − 0.70
1
+
+0
0.70 ( 0.275m ) 0.025m
4
q′(1,2 )3 =
<
(b) For ε = 0.7 emission from the base of the U-section is q′b = ε A1′ σ T 4 = 0.7 × 0.025m
4
× 5.67 ×10−8 W / m 2 ⋅ K 4 (325 K ) = 11.1 W / m. The effect of ε on q′(1,2 )3 and q′b is shown as
follows.
Continued …..
PROBLEM 13.47 (Cont.)
Radiation heat rate (W/m)
16
14
12
10
8
6
4
2
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Emissivity, eps
Heat rejection with fins, q'(1,2)3 (W/m)
Emission from base, qb' (W/m)
The effect of the fins on heat transfer enhancement increases with decreasing emissivity.
COMMENTS: Note that, if the surfaces behaved as blackbodies (ε1 = ε2 = 1.0), the U-section
becomes a blackbody cavity for which heat rejection is simply A′3 Eb (T) = q′b . Hence, it is no
surprise that the q′b → q′(1,2 )3 as ε → 1 in the foregoing figure. For ε = 1, no enhancement is
provided by the fins.
PROBLEM 13.48
KNOWN: Power dissipation of electronic device and thermal resistance associated with attachment
to inner wall of a cubical container. Emissivity of outer surface of container and wall temperature of
service bay.
FIND: Temperatures of cubical container and device.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, (2) Device and container are isothermal, (3) Heat transfer from
the container is exclusively by radiation exchange with bay (small surface in a large enclosure), (4)
Container surface may be approximated as diffuse/gray.
ANALYSIS: From Eq. (13.27)
) (
(
4 − T4
Pe = q = σ 6W 2 ε c Ts,c
sur
)
1/ 4


q


4
Ts,c = 
+ Tsur 
2
 σ 6W ε c

( )
1/ 4


50 W
4
Ts,c = 
+ (150 K ) 
 5.67 × 10−8 W / m 2 ⋅ K 4 × 6 ( 0.12m )2 × 0.8



With
(
= 339.4 K = 66.4°C
<
)
q = Td − Ts,c / R t ,
Td = q R t + Ts,c = 50 W × 0.1K / W + 66.4°C = 71.4°C
<
COMMENTS: If the temperature of the device is too large to insure reliable operation, it may be
reduced by increasing ε c or W.
PROBLEM 13.49
KNOWN: Long, thin-walled horizontal tube with radiation shield having an air gap of 10 mm.
Emissivities and temperatures of surfaces are prescribed.
FIND: Radiant heat transfer from the tube per unit length.
SCHEMATIC:
ASSUMPTIONS: (1) Tube and shield are very long, (2) Surfaces at uniform temperatures, (3)
Surfaces are diffuse-gray.
ANALYSIS: The long tube and shield form a two surface enclosure, and since the surfaces are
diffuse-gray, the radiant heat transfer from the tube, according to Eq. 13.23, is
q12 =
(
σ T14 − T24
)
(1)
1 − ε1
1
1− ε2
+
+
ε1A1 A1F12 ε 2 A 2
By inspection, F12 = 1. Note that
A1 = π D1 "
A2 = π D2 "
and
where is the length of the tube and shield. Dividing Eq. (1) by , find the heat rate per unit length,
′ =
q12
q12
"
=
4
4
5.67 × 10−8 W / m 2 ⋅ K ( 273 + 120 ) − ( 273 + 35 )  K 4

(
1 − 0.8
0.8π 100 × 10−3 m
′ =
q12
+

1
1 − 0.1
+
π 100 × 10−3 m × 1 0.1π 120 × 10−3 m
) (
842.3 W / m 2
(0.7958 + 3.183 + 23.87 ) m−1
= 30.2 W / m.
)
(
)
<
COMMENTS: Recognize that convective heat transfer would be important in this annular air gap.
Suitable correlations to estimate the heat transfer coefficient are given in Chapter 9.
PROBLEM 13.50
KNOWN: Long electrical conductor with known heat dissipation is cooled by a concentric tube
arrangement.
FIND: Surface temperature of the conductor.
SCHEMATIC:
ASSUMPTIONS: (1) Surfaces are diffuse-gray, (2) Conductor and cooling tube are concentric and
very long, (3) Space between surfaces is evacuated.
ANALYSIS: The heat transfer by radiation exchange between the conductor and the concentric,
cooled cylinder is given by Eq. 13.25. For a unit length,
)
(
 1 1 − ε2
q
′ = 12 = σ ⋅ 2π r1 T14 − T24 /  +
q12
"
ε2
 ε1
 r1  
 
 r2  
(1)
where A1 = 2π r1 ⋅ ". Solving for T1 and substituting numerical values, find
1/ 4

 1 1 − ε 2  r1   
q′
T1 = T24 + 12  +
  
σ ⋅ 2π r1  ε1
ε 2  r2   

T1 =
{(27 + 273) K
4
4
+
1/ 4
 1 1 − 0.9  5   
+
  

0.9  25   
5.67 × 10−8 W / m 2 ⋅ K 4 × 2π ( 0.005m )  0.6

6 W/m
T1 =
{(300 K )
4
}
+ 3.368 × 109 K 4 [1.667 + 0.00222]
T1 = 342.3 K = 69°C.
1/ 4
(2)
<
COMMENTS: (1) Note that Eq. (1) implies that F12 = 1. From Eq. (2) by comparison of the second
term in the brackets involving ε2, note that the influence of ε2 is small. This follows since r1 << r2.
PROBLEM 13.51
KNOWN: Arrangement for direct thermophotovoitaic conversion of thermal energy to electrical
power.
FIND: (a) Radiant heat transfer between the inner and outer surface per unit area of the outer
surface, (b) Power generation per unit outer surface area if semiconductor has 10% conversion
efficiency for radiant power in the 0.6 to 2.0 µm range.
SCHEMATIC:
ASSUMPTIONS: (1) Surfaces are diffuse-gray, (2) Surfaces approximate long, concentric cylinder,
two-surface enclosure with negligible end effects.
ANALYSIS: (a) For this two-surface enclosure, the net radiation exchange per unit area of the outer
surface is,
qio
=
Ai
⋅
(
σ Ti4 − To4
)
(1)
1 1 − ε o  ri 
+
 
εi
ε o  ro 
and since Ai/Ao = 2πri " / 2π ro " = r1 / ro , the heat flux at surface Ao is
Ao
Ao
 0.0125 
=

A o  0.190 
qio
(5.67 ×10−8 W / m2 ⋅ K4 )(19484 − 2934 ) K4 = 45.62 kW / m2 .
1
+
1 − 0.5  0.125 

(2)
<

0.5  0.190 
(b) The power generation per unit area of surface Ao can be expressed as
0.9
Pe′′ = ηe ⋅ G abs ( 0.6 → 2.0 µ m )
(3)
where ηe is the semiconductor conversion efficiency and Gabs (0.6 → 2.0µm) represents the absorbed
irradiation on Ao in the prescribed wavelength interval. The total absorbed irradiation is Gabs,t =
qio/Ao and has the spectral distribution of a blackbody at Ti since To4 << Ti4 and Ai is gray. Hence,
we can write Eq. (3) as
Pe′′ = ηe ⋅ (q io / A o )  F( 0→ 2 µ m ) − F( 0→ 0.6 µ m )  .


(4)
From Table 12.2: λT = 2 × 1948 = 3896 µm⋅K, F(0-λT) = 0.461; λT = 0.6 × 1948 = 1169 µm⋅K, F(0λT)
= 0.0019. Hence
(
)
Pe′′ = 0.1 45.62 kW / m 2 [0.461 − 0.0019 ] = 2.09 kW / m 2 .
That is, the unit produces 2.09 kW per unit area of the outer surface.
<
PROBLEM 13.52
KNOWN: Temperatures and emissivities of spherical surfaces which form an enclosure.
FIND: Evaporation rate of oxygen stored in inner container.
SCHEMATIC:
5
PROPERTIES: Oxygen (given): hfg = 2.13 × 10 J/kg.
ASSUMPTIONS: (1) Opaque, diffuse-gray surfaces, (2) Evacuated space between surfaces, (3)
Negligible heat transfer along vent and support assembly.
ANALYSIS: From an energy balance on the inner container, the net radiation heat transfer to the
container may be equated to the evaporative heat loss
fg .
qoi = mh
Substituting from Eq. (13.26), where qoi = - qio and Fio = 1
( )(
)
−σ π Di2 Ti4 − To4
=
m
2

 ri  
1
ε
1
−
o
h fg  +
  
ε o  ro  
 εi


−5.67 × 10−8 W / m 2 ⋅ K 4 × π (0.8m )
2
=
m
(954 − 2804 ) K4
2
 1
0.95  0.4  
5
2.13 × 10 J / kg 
+

 
 0.05
0.05  0.6  

= 1.14 × 10−4 kg / s.
m
COMMENTS: This loss could be reduced by insulating the outer surface of the outer container
and/or by inserting a radiation shield between the containers.
<
PROBLEM 13.53
KNOWN: Emissivities, diameters and temperatures of concentric spheres.
FIND: (a) Radiation transfer rate for black surfaces. (b) Radiation transfer rate for diffuse-gray
surfaces, (c) Effects of increasing the diameter and assuming blackbody behavior for the outer sphere.
(d) Effect of emissivities on net radiation exchange.
SCHEMATIC:
ASSUMPTIONS: (1) Blackbody or diffuse-gray surface behavior.
ANALYSIS: (a) Assuming blackbody behavior, it follows from Eq. 13.13
(
)
q12 = A1F12σ T1 − T2 = π ( 0.8 m )
2
4
2
(1) 5.67 × 10−8 W / m 2 ⋅ K 4 ( 400 K )4 − (300 K )4  = 1995 W.
<
(b) For diffuse-gray surface behavior, it follows from Eq. 13.26
q12 =
(
σ A1 T14 − T24
)
1 1 − ε 2  r1 
+
 
ε1
ε 2  r2 
2
=
2
5.67 × 10−8 W / m 2 ⋅ K 4π ( 0.8 m )  4004 − 3004  K 4

1 1 − 0.05  0.4 
+


0.5
0.05  0.6 

2
= 191 W.
<
(c) With D2 = 20 m, it follows from Eq. 13.26
q12 =
2
4
4
5.67 × 10−8 W / m 2 ⋅ Kπ ( 0.8 m ) ( 400 K ) − (300 K ) 

1 1 − 0.05  0.4 
+


0.5
0.05  10 
 = 983 W.
2
<
With ε2 = 1, instead of 0.05, Eq. 13.26 reduces to Eq. 13.27 and
(
)
2
4
4
q12 = σ A1ε1 T14 − T24 = 5.67 × 10−8 W / m 2 ⋅ K 4π ( 0.8 m ) 0.5 ( 400 K ) − (300 K )  = 998 W.


Continued …..
<
PROBLEM 13.53 (Cont.)
(d) Using the IHT Radiation Tool Pad, the following results were obtained
Net radiation exchange increases with ε1 and ε2, and the trends are due to increases in emission from and
absorption by surfaces 1 and 2, respectively.
COMMENTS: From part (c) it is evident that the actual surface emissivity of a large enclosure has a
small effect on radiation exchange with small surfaces in the enclosure. Working with ε2 = 1.0 instead of
ε2 = 0.05, the value of q12 is increased by only (998 – 983)/983 = 1.5%. In contrast, from the results of
(d) it is evident that the surface emissivity ε2 of a small enclosure has a large effect on radiation
exchange with interior objects, which increases with increasing ε1.
PROBLEM 13.54
KNOWN: Two radiation shields positioned in the evacuated space between two infinite, parallel
planes.
FIND: Steady-state temperature of the shields.
SCHEMATIC:
ASSUMPTIONS: (1) All surfaces are diffuse-gray and (2) All surfaces are parallel and of infinite
extent.
ANALYSIS: The planes and shields can be represented by a thermal circuit from which it follows
that
q1′′ = −q′′2 =
(
σ T14 − T24
) = σ (T14 − Ts14 ) = σ (Ts14 − Ts24 ) = σ (Ts24 − T24 ) .
R1′′ + R ′′2 + R ′′3
R1′′
R ′′2
Since all the emissivities involved are equal, R1′′ =
4 T4
Ts1
= 1 −
(
R 3′′
A1
= 1 = R ′′2 = R ′′3 , so that
A1F12
)
(
R1′′
T14 − T24 = T14 − (1/ 3) T14 − T24
R1′′ + R ′′2 + R ′′3
)
(
4
4
Ts1
= ( 600 K ) − (1/ 3) 6004 − 3254 K 4
(
)
)
T1s = 548 K
(
R ′′3
4 T4
Ts2
T14 − T24 = T24 + (1/ 3) T14 − T24
= 2+
R ′′ + R ′′ + R ′′
1
2
4
3
(
)
4 = 325 K + 1/ 3 6004 − 3254 K 4
Ts2
(
) ( )
<
)
Ts2 = 474 K.
<
PROBLEM 13.55
KNOWN: Two large, infinite parallel plates that are diffuse-gray with temperatures and emissivities
of T1 and ε1 and T2 and ε2.
FIND: Show that the ratio of the radiation transfer rate with multiple shields, N, of emissivity εs to
that with no shields, N = 0, is
q12,N
q12,0
=
1 / ε1 + 1 / ε 2 − 1
1 / ε1 + 1 / ε 2 − 1 + N 2 / ε s − 1
where q12,N and q12,0 represent the radiation heat rate with N and N = 0 shields, respectively.
SCHEMATIC:
ASSUMPTIONS: (1) Plane infinite planes with diffuse-gray surfaces and uniform radiosities, and
(2) Shield has negligible thermal conduction resistance.
ANALYSIS: Representing the parallel plates by the resistance network shown above for the “noshield” condition, N = 0, with F12 = 1, the heat rate per unit area follows from Eq. 13.24 (see also Fig.
13.11) as
q12,0
′′ =
E b1 − E b2
1/ ε 1 + 1 / ε 2 − 1
(1)
With the addition of each shield as shown in the schematic above, three resistance elements are added
to the network: two surface resistances, (1 - εs)/εs, and one space resistance, 1/Fij = 1. Hence, for the
“N - shield” condition,
q12,N
=
′′
E b1 − E b2
1/ ε 1 + 1 / ε 2 − 1 + N 2 1 − ε s / ε s + 1
$
(2)
The ratio of the heat rates is obtained by dividing Eq. (2) by Eq. (1),
q12,N
′′
q12,0
=
1 / ε1 + 1 / ε 2 − 1
1 / ε1 + 1 / ε 2 − 1 + N 2 / ε s − 1
<
COMMENTS: Can you derive an expression to determine the temperature difference across pairs of
the N-shields?
PROBLEM 13.56
KNOWN: Emissivities of two large, parallel surfaces.
FIND: Heat shield emissivity needed to reduce radiation transfer by a factor of 10.
SCHEMATIC:
ASSUMPTIONS: (a) Diffuse-gray surface behavior, (b) Negligible conduction resistance for shield,
(c) Same emissivity on opposite sides of shield.
ANALYSIS: For this arrangement, F13 = F32 = 1.
Without (wo) the shield, it follows from Eq. 13.24,
(
A1σ T14 − T24
(q12 )wo =
1 1
+
−1
ε1 ε 2
).
With (w) the shield it follows from Eq. 13.28,
(q12 )w =
(
A1σ T14 − T24
).
1 1
2
+
+ −2
ε1 ε 2 ε 3
Hence, the heat rate ratio is
1 1
1
1
+
−1
+
−1
(q12 )w
ε1 ε 2
.
= 0.1 =
= 0.8 0.8
1 1
2
1
1
2
(q12 )wo
+
+ −2
+
+ −2
0.8 0.8 ε 3
ε1 ε 2 ε 3
Solving, find
ε 3 = 0.138.
COMMENTS: The foregoing result is independent of T1 and T2. It is only necessary that the
temperatures be maintained at fixed values, irrespective of whether or not the shield is in place.
<
PROBLEM 13.57
KNOWN: Surface emissivities of a radiation shield inserted between parallel plates of prescribed
temperatures and emissivities.
FIND: (a) Effect of shield orientation on radiation transfer, (b) Effect of shield orientation on shield
temperature.
SCHEMATIC:
ASSUMPTIONS: (1) Diffuse-gray surface behavior, (2) Shield is isothermal.
ANALYSIS: (a) On a unit area basis, the network representation of the system is
Hence the total radiation resistance,
R=
1 − ε s 1 − 2ε s
1 − ε1
1− ε2
+1+
+
+1+
2ε s
ε1
εs
ε2
is independent of orientation. Since q = (Eb1 – Eb2)/R, the heat transfer rate is independent of
orientation.
(b) Considering that portion of the circuit between Eb1 and Ebs, it follows that
q=
E b1 − E bs
1 − ε s 1 − 2ε s
, where f (ε s ) =
or
.
1 − ε1
2
ε
ε
s
s
+ 1 + f (ε s )
ε1
Hence,
1 − ε1

E bs = E b1 − 
+ 1 + f (ε s ) q.
 ε1

It follows that, since Ebs increases with decreasing f(εs) and (1 - 2εs)/2εs < (1 - εs)/εs, Ebs is larger
when the high emissivity (2εs) side faces plate 1. Hence Ts is larger for case (b).
<
PROBLEM 13.58
KNOWN: End of propellant tank with radiation shield is subjected to solar irradiation in space
environment.
)
(
FIND: (a) Temperature of the shield, Ts, and (b) Heat flux to the tank, q1′′ W / m 2 .
SCHEMATIC:
ASSUMPTIONS: (1) All surfaces are diffuse-gray, (2) View factor between shield and tank is unity,
Fst = 1, (3) Space surroundings are black at 0 K, (4) Resistance of shield for conduction is negligible.
ANALYSIS: (a) Perform a radiation balance on the shield. From the schematic,
αSGS − ε1E b (Ts ) − q′′st = 0
(1)
where q′′st is the net heat exchange between the shield and the tank. Considering these two surfaces as
large, parallel planes, from Eq. 13.24,
q′′st = σ Ts4 − Tt4 / [1/ ε 2 + 1/ ε1 − 1].
(2)
(
)
Substituting q′′st from Eq. (2) into Eq. (1), find
αs G s − ε1σ Ts4 − σ Ts4 − T14 / [1/ ε 2 + 1/ ε t − 1] = 0.
(
)
Solving for Ts, find
1/ 4
 α G + σ T 4 / [1/ ε + 1/ ε − 1] 
t
2
t

Ts =  S S
 σ (ε1 + 1/ [1/ ε 2 + 1/ ε t − 1]) 


.
Since the shield is diffuse-gray, αS = ε1 and then
1/ 4
 0.05 × 1250 W / m 2 + σ (100 )4 K 4 / [1/ 0.05 + 1/ 0.1 − 1] 

Ts = 


σ (0.05 + 1/ [1/ 0.05 + 1/ 0.1 − 1])


= 338 K.
<
(b) The heat flux to the tank can be determined from Eq. (2),
(
)
q′′st = 5.67 × 10−8 W / m 2 ⋅ K 4 3384 − 1004 K 4 / [1/ 0.05 + 1/ 0.1 − 1] = 25.3 W / m 2 .
<
PROBLEM 13.59
KNOWN: Black panel at 77 K in large vacuum chamber at 300 K with radiation shield having ε =
0.05.
FIND: Net heat transfer by radiation to the panel.
SCHEMATIC:
ASSUMPTIONS: (1) Chamber is large compared to shield, (2) Shape factor between shield and
plate is unity, (3) Shield is diffuse-gray, (4) Shield is thin, negligible thermal conduction resistance.
ANALYSIS: The arrangement lends itself to a network representation following Figs. 13.10 and
13.11.
Noting that F2s = Fs1 = 1, and that A2F2s = AsFs2, the heat rate is
(
)
 1
 1 − εs  1 
q1 = ( E b2 − E b1 ) / ΣR i = σ T24 − T1 4 / 
+ 2
.
+
 ε s As  As 
 As
Recognizing that As = A1 and multiplying numerator and denominator by A1 gives
(
)

 1 − εs 
q1 = A1σ T2 − T14  2 + 2 
 .
 ε s  

Substituting numerical values, find
q1 =
π 0.12 m 2

 1 − 0.05  
× 5.67 × 10−8 W / m 2 ⋅ K 4 3004 − 774 K 4 /  2 + 2 

4
 0.05  

q1 = 89.8 mW.
(
)
<
COMMENTS: In using the network representation, be sure to designate direction of the net heat
rate. In this situation, we have shown q1 as the net rate into the surface A1. The temperature of the
shield, Ts = 253 K, follows from the relation
1 − ε s
1 
q1 = ( E bs − E b1 ) / 
+
.
A
A
F
ε
1 s1 
 s s
PROBLEM 13.60
KNOWN: Dense cryogenic piping array located close to furnace wall.
FIND: Number of radiation shields, N, to be installed such that the temperature of the shield closest
to the array, Ts,N, is less than 30°C.
SCHEMATIC:
ASSUMPTIONS: (1) The ice-covered dense piping array approximates a plane surface, (2) Piping
array and furnace wall can be represented by infinite parallel plates, (3) Surfaces are diffuse-gray, and
(4) Convection effects are negligible.
ANALYSIS: Treating the piping array and furnace wall as infinite parallel plates, the net heat rate by
radiation exchange with N shields of identical emissivity, εs, on both sides follows from extending the
network of Fig. 13.12 to account for the resistances of N shields. (See Problem 13.14 (S).) For each
shield added, two surface resistances and one space resistance are added,
q fp =
σ Tf4 − Tp4 A f
"
'
1/ ε f + 1 / ε p − 1 + N 2 / ε s − 1
-8
2
(1)
4
where σ = 5.67 × 10 W/m ⋅K . The requirement that the N-th shield (next to the piping array) has a
temperature Ts,N ≤ 30°C will be satisfied when
q fp ≤
(
)
σ Ts,N 4 − Tp 4 Af
1/ε s + 1/ ε p − 1


(2)
Using the foregoing equations in the IHT workspace, find that Ts,N = 30°C when N = 8.60. So that
Ts,N is less than 30°C, the number of shields required is
N=9
<
COMMENTS: Note that when N = 0, Eq. (1) reduces to the case of two parallel plates. Show for
the case with one shield, N = 1, that Eq. (1) is identical to Eq. 13.28.
PROBLEM 13.61
KNOWN: Concentric tube arrangement with diffuse-gray surfaces.
FIND: (a) Heat gain by the cryogenic fluid per unit length of the inner tube (W/m), (b) Change in
heat gain if diffuse-gray shield with εs = 0.02 is inserted midway between inner and outer surfaces.
SCHEMATIC:
ASSUMPTIONS: (1) Surfaces are diffuse-gray, (2) Space between tubes is evacuated.
ANALYSIS: (a) For the no shield case, the thermal
circuit is shown at right. It follows that the net heat gain
per unit tube length is
 1 − εo
q
1
1 − εi 
−q1′ = oi = ( E bo − E bi ) / 
+
+

L
 επ Do π Di Fio ε iπ Di 
4
where A = πDL. Note that Fio = 1 and Eb = σT giving
− q1′ = 5.67 × 10
−8
2
W/m ⋅K
4
(
4
300 − 77
4
)
4
K /
1 − 0.05


 0.05π × 50 × 10 −3
1
+
π 20 × 10
−3
+
×1
1 − 0.02
0.02π × 20 × 10
−3
 −1
m

−q1′ = 457 W / m 2 / [121.0 + 15.9 + 779.8] m −1 = 0.501 W / m.
<
(b) For the with shield case, the thermal circuit will include three additional resistances.
From the network, it follows that −q i = ( E bo − E bi ) / ΣR t . With Fis = Fso = 1, find

−q′i = 457 W / m 2 / 121.0 +

1
+
2 (1 − 0.02 )
π 35 × 10−3 × 1 0.02π 35 × 10−3

+ 15.9 + 779.8  m −1

−q′i = 457 W / m 2 / [121.0 + 9.1 + 891.3 + 15.9 + 779.8] m −1 = 0.251W / m.
The change (percentage) in heat gain per unit length of the tube as a result of inserting the radiation
shield is
q′i,s − q′i,ns
(0.251 − 0.501) W / m 100 49%.
× 100 =
×
=−
q′i,ns
0.501W / m
<
PROBLEM 13.62
KNOWN: Heated tube with radiation shield whose exterior surface is exposed to convection and
radiation processes.
FIND: Operating temperature for the tube under the prescribed conditions.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) No convection in space between tube and shield,
(3) Surroundings are large compared to the shield and are isothermal, (4) Tube and shield are
infinitely long, (5) Surfaces are diffuse-gray, (6) Shield is isothermal.
ANALYSIS: Perform an energy balance on the shield.
E in − E out = 0
q12 − qconv − q rad = 0
where q12 is the net radiation exchange between the tube and inner surface of the shield, which from
Eq. 13.25 is,
A1σ T14 − T24
−q12 =
1 1 − ε 2,i D1
+
ε1
ε 2,i D2
)
(
Using appropriate rate equations for qconv and qrad, the energy balance is
A1σ T14 − T24
4
− hA 2 ( T2 − T∞ ) − ε 2,o A 2σ T24 − Tsur
=0
1 − ε 2,i D1
(
1+
)
)
(
ε 2,i D 2
where ε1 = 1. Substituting numerical values, with A1/A2 = D1/D2, and solving for T1,
( 20 / 60 ) × 5.67 ×10−8 W / m 2 ⋅ K 4 T14 − 3154 K 4
2
(
)
1 + (1 − 0.01/ 0.01)( 20 / 60 )
(
− 10 W / m ⋅ K (315 − 300 ) K
)
−0.1× 5.67 × 10−8 W / m 2 ⋅ K 4 3154 − 2904 K 4 = 0
T1 = 745 K = 472°C.
COMMENTS: Note that all temperatures are expressed in kelvins. This is a necessary practice
when dealing with radiation and convection modes.
<
PROBLEM 13.63
KNOWN: Cylindrical-shaped, three surface enclosure with lateral surface insulated.
FIND: Temperatures of the lower plate T1 and insulated side surface T3.
SCHEMATIC:
ASSUMPTIONS: (1) Surfaces have uniform radiosity or emissive power, (2) Upper and insulated
surfaces are diffuse-gray, (3) Negligible convection.
ANALYSIS: Find the temperature of the lower plate T1 from Eq. 13.30
q1 =
(
σ T14 − T24
)
−
−1  1
(1 − ε1 ) / ε1A1 + A1F12 + [(1/ A1F13 ) + (1/ A 2F23 )] 


From Table 13.2 for parallel coaxial disks,
R1 = r1 / L = 0.1/ 0.2 = 0.5
(
)
(
.
(1)
+ (1 − ε 2 ) / ε 2 A 2
R 2 = r2 / L = 0.1/ 0.2 = 0.5
)
S = 1 + 1 + R 22 / R12 = 1 + 1 + 0.52 / 0.52 = 6.0

2
F12 = 1/ 2 S − S2 − 4 ( r2 / r1 ) 
1/ 2 


1/ 2 

2
2
 = 1/ 2 6 − 6 − 4 ( 0.5 / 0.5 )   = 0.172.




Using the summation rule for the enclosure, F13 = 1 – F12 = 1 – 0.172 = 0.828, and from symmetry,
2
2
2
F23 = F13. With A1 = A2 = πD /4 = π(0.2 m) /4 = 0.03142 m and substituting numerical values into
Eq. (1), obtain
10, 000 W =
(
)
0.03142 m 2 × 5.67 × 10−8 W / m 2 ⋅ K 4 T14 − 4734 K 4
−1
0 +  0.172 + [(1/ 0.828 ) + (1/ 0.172 )] 

(
10, 000 = 4.540 × 109 T14 − 4734
)

−1
+ (1 − 0.8 ) / 0.8
<
T1 = 1225 K.
The temperature of the insulated side surface can be determined from the radiation balance, Eq. 13.31,
with A1 = A2,
J1 − J3 J3 − J 2
(2)
−
=0
1/ F13
1/ F23
where J1 = σ T14 and J2 can be evaluated from Eq. 13.19,
Continued …..
PROBLEM 13.63 (Cont.)
q2 =
E b2 − J 2
(1 − ε 2 ) / ε 2 A2
5.67 × 10−8 W / m 2 ⋅ K 4 ( 473 K ) − J 2
1 − 0.8 / 0.8 × 0.03142 m 2
4
−10, 000 W =
(
)
(
)
2
find J2 = 82,405 W/m . Substituting numerical values into Eq. (2),
5.67 × 10−8 W / m 2 ⋅ K 4 (1225 K ) − J 3 J3 − 82, 405 W / m 2
−
=0
1/ 0.172
1/ 0.172
4
2
find J3 = 105,043 W/m . Hence, for this insulated, re-radiating (adiabatic) surface,
E b3 = σ T34 = 105, 043 W / m 2
T3 = 1167 K.
<
PROBLEM 13.64
KNOWN: Furnace in the form of a truncated conical section, floor (1) maintained at T1 = 1000 K by
providing a heat flux q1,in
′′ = 2200 W / m2 ; lateral wall (3) perfectly insulated; radiative properties
of all surfaces specified.
FIND: (a) Temperature of the upper surface, T2, and of the lateral wall T3, and (b) T2 and T3 if all
the furnace surfaces are black instead of diffuse-gray, with all other conditions remain unchanged.
Explain effect of ε2 on your results.
SCHEMATIC:
ASSUMPTIONS: (1) Furnace is a three-surface, diffuse-gray enclosure, (2) Surfaces have uniform
radiosities, (3) Lateral surface is adiabatic, and (4) Negligible convection effects.
ANALYSIS: For the three-surface enclosure, write the radiation surface energy balances, Eq. 13.21,
to find the radiosities of the three surfaces.
E b,1 − J1
(1 − ε1 ) / ε1 A1
=
E b,2 − J 2
1 − ε 2 $ / ε 2 A2
E b,3 − J 3
1 − ε 3$ / ε 3 A3
J −J
J1 − J 2
+ 1 3
1/A1 F12 1/A1 F13
(1)
=
J 2 − J1
J −J
+ 2 3
1 / A 2 F21 1/ A 2 F23
(2)
=
J 3 − J1
J −J
+ 3 2
1/ A 3F31 1/ A 3 F32
(3)
4
-8
2
4
where the blackbody emissive powers are of the form Eb = σ T with σ = 5.67 × 10 W/m ⋅K . From
Eq. 13.19, the net radiation leaving A1 is
q1 =
E b,1 − J 1
(4)
1 − ε1$ / ε1 A1
2
q1 = q1,in
′′ ⋅ A1 = 2200 W / m2 × π 0.040 m / 4 = 2.76 W
$
Continued …..
PROBLEM 13.64 (Cont.)
Since the lateral surface is adiabatic,
q3 =
E b,3 − J 3
1 − ε 3$ / ε 3 A 3 = 0
(5)
from which we recognize Eb,3 = J3, but will find that as an outcome of the analysis. For the
2
enclosure, N = 3, there are N = 9 view factors, for which N (N - 1)/2 = 3 must be directly determined.
Calculations for the Fij are summarized in Comments.
With the foregoing five relations, we can determine the five unknowns: J1, J2, J3, Eb,2, and Eb,3. The
4
temperatures T2 and T3 will be evaluated from the relation Eb = σ T . Using this analysis approach
with the relations in the IHT workspace, the results for (a) the diffuse-gray surfaces and (b) black
surfaces are tabulated below.
2
J1 (kW/m )
2
J2 (kW/m )
2
J3 (kW/m )
T2 (K)
T3 (K)
(a) Diffuse-gray
55.76
45.30
53.48
896
986
(b) Black
56.70
46.24
54.42
950
990
COMMENTS: (1) From the tabulated results, it follows that the temperatures of the lateral and top
surfaces will be higher when the surfaces are black, rather than diffuse-gray as specified.
(2) From Eq. (5) for the net heat radiation leaving the lateral surface, A3, the rate is zero since the
wall is adiabatic. The consequences are that the blackbody emissive power and the radiosity are
equal, and that the emissivity of the surface has no effect in the analysis. That is, this surface emits
and absorbs at the same rate; the net is zero.
2
(3) For the enclosure, N = 3, there are N = 9 view factors, for which
N N −1 / 2 = 3× 2 / 2 = 3
$
must be directly determined. We used the IHT Tools | Radiation | View Factors Relations model that
sets up the summation rules and reciprocity relations for the N surfaces. The user is required to
specify the 3 Fij that must be determined directly; by inspection, F11 = F22 =0; and F12 can be
evaluated using the parallel coaxial disk relation, Table 13.2 (Fig. 13.5). This model is also provided
in IHT to simplify the calculation task. The results of the view factor analysis are:
F12 = 0.03348
F13 = 0.9665
F21 = 01339
.
F23 = 0.8661
(4) An alternative method of solution for part (a) is to treat the enclosure of part (a) as described in
Section 13.3.5. For part (b), the black enclosure analysis is described in Section 13.2. We chose to
use the net radiation method, Section 13.3.1, to develop a general 3-surface enclosure code in IHT that
can also handle black surfaces (caution: use ε = 0.999, not 1.000).
PROBLEM 13.65
KNOWN: Two aligned, parallel square plates with prescribed temperatures.
FIND: Net radiative transfer from surface 1 for these plate conditions: (a) black, surroundings at 0
K, (b) black with connecting, re-radiating walls, (c) diffuse-gray with radiation-free surroundings at 0
K, (d) diffuse-gray with re-radiating walls.
SCHEMATIC:
ASSUMPTIONS: (1) Plates are black or diffuse-gray, (2) Surroundings are at 0 K.
ANALYSIS: (a) The view factor for the aligned, parallel plates follows from Fig. 13.4, X/L = 0.4
m/0.8 m = 0.5, Y/L = 0.4 m/0.8 m = 0.5, F12 = F21 ≈ 0.075. When the plates are black with
surroundings at 0 K, from Eq. 13.13,
(
q1 = q12 + q1(sur ) = A1F12 ( E b1 − E b2 ) + A1F1(sur ) E b1 − E b(sur )
)
q1 = ( 0.4 × 0.4 ) m 2 [0.075 (3544 − 23, 224 ) + (1 − 0.075 )(3544 − 0 )] W / m 2 = 288 W.
<
(b) When the plates are black with connecting re-radiating walls, from Eq. 13.30 with F1R = R2R = 1
– F12 = 0.925,
A1 [E b1 − E b2 ]
q1 =
 F + (1/ F + 1/ F )−1 
1R
2R
 12

−1
=
(0.4 m )2 [3544 − 23, 224] W / m 2
0.075 + (1/ 0.925 + 1/ 0.925 )−1 


−1
= −1, 692 W.
<
(c) When the plates are diffuse-gray (ε1 = 0.6 and ε2 = 0.8) with the surroundings at 0 K, using Eq.
13.20 or Eq. 13.19, with Eb3 = J3 = 0,
q1 = A1F12 ( J1 − J 2 ) + A1F13 ( J1 − J3 ) = ( E b1 − J1 ) / [(1 − ε1 ) / ε1A1 ].
The radiosities must be determined from energy balances, Eq. 13.21, on each of the surfaces,
E b1 − J1
E b2 − J 2
= F12 ( J1 − J 2 ) + F13 ( J1 − J 3 )
= F (J − J ) + F (J − J )
(1 − ε1 ) / ε1
(1 − ε 2 ) / ε 2 21 2 1 23 2 3
3, 544 − J1
(1 − 0.6 ) / 0.6
2
23, 224 − J 2
= 0.075 ( J1 − J 2 ) + 0.925J1
(1 − 0.8 ) / 0.8
= 0.075 ( J 2 − J1 ) + 0.925J 2 .
2
Find J1 = 2682 W/m and J2 = 18,542 W/m . Combining these results,
q1 = ( 0.4 m )
2
(0.075 )( 2682 − 18, 542 ) W / m2 + ( 0.4 m )2 (0.925 )( 2682 − 0 ) W / m2 = 207 W.
<
(d) When the plates are diffuse-gray with connecting re-radiating walls, use Eq. 13.30,
A1 [E b1 − E b2 ]
q1 =
(1 − ε1 ) / ε1 + F12 + (1/ F1R + 1/ F2R )−1 
q1 =
−1
+ (1 − ε 2 ) / ε 2
(0.4 m )2 [35444 − 23, 244] W / m2
−1  −1
(1 − 0.6 ) / 0.6 + 0.075 + (1/ 0.925 + 1/ 0.925 ) 
= −1133 W.
+ (1 − 0.8 ) / 0.8
<
PROBLEM 13.66
KNOWN: Parallel, aligned discs located in a large room; one disk is insulated, the other is at a
prescribed temperature.
FIND: Temperature of the insulated disc.
SCHEMATIC:
ASSUMPTIONS: (1) Surfaces are diffuse-gray, (2) Surroundings are large, with uniform
temperature, behaving as a blackbody, (3) Negligible convection.
ANALYSIS: From an energy balance on surface A2,
q2 = 0 =
J −J
J 2 − J1
+ 2 3 .
1/ A 2 F21 1/ A 2 F23
(1)
Note that q2 = 0 since the surface is adiabatic. Since A3 is a blackbody, J3 = Eb3 = σ T34 ; since A2 is
adiabatic, J2 = Eb2 = σ T24 . From Fig. 13.5 and the summation rule for surface A1, find
F12 = 0.62 with
rj
L
=
0.2
L 0.1
= 2 and =
= 0.5,
0.1
ri 0.2
-8
4
F13 = 1 − F12 = 1 − 0.62 = 0.38.
2
Hence, Eq. (1) with J3 = 5.67 × 10 × 300 W/m becomes
J 2 − J1
J − 459.3 W / m 2
+ 2
=0
1/ A 2 × 0.62
1/ A 2 × 0.38
−0.62J1 + 1.00J 2 = 174.5
-8
4
(2,3)
2
The radiation balance on surface A1 with Eb3 = 5.67 × 10 × 500 W/m becomes
E b1 − J1
J −J
J −J
= 1 2 + 1 3
(1 − ε1 ) / ε1A1 1/ A1F12 1/ A1F13
(4)
3543.8 − J1
J1 − J 2
J − 459.3
=
+ 1
(1 − 0.6 ) / 0.6A1 1/ A1 × 0.62 1/ A1 × 0.38
2.50J1 − 0.62J 2 = 5490.2
(5,6)
2
Solve Eqs. (3) and (6) to find J2 = 1815 W/m and since Eb2 = J2,
1/ 4
E 
T2 =  b2 
 σ 
1/ 4


1815 W / m 2

=
 5.67 × 10−8 W / m 2 ⋅ K 4 


= 423 K.
<
COMMENTS: A network representation would help to visualize the exchange relations. However,
it is useful to approach the problem by recognizing there are two unknowns in the problem: J1 and J2;
hence two radiation balances must be written. Note also the significance of J2 = Eb2 and J3 = Eb3.
PROBLEM 13.67
KNOWN: Thermal conditions in oven used to cure strip coatings.
FIND: Electrical power requirement.
SCHEMATIC:
ASSUMPTIONS: (1) Diffuse-gray surfaces, (2) Furnace wall is reradiating, (3) Negligible end
effects.
ANALYSIS: The net radiant power leaving the heater surface per unit length is
E b1 − E b2
1 − ε1
1
1− ε2
+
+
ε1A1′ A′ F + (1/ A F ) + (1/ A′ F ) −1 ε 2A′2
1 12 
1 1R
2 2R 
where A1′ = π D = π ( 0.02 m ) = 0.0628 m and A′2 = 2 (s1 − s 2 ) = 0.08 m. The view factor
q1′ =
between the heater and one of the strips is
F21 =
D/2
s1 − s 2
 −1 s1
−1 s 2  0.01  −1 0.06
−1 0.02 
 tan L − tan L  = 0.04  tan 0.08 − tan 0.08  = 0.10




and using the view factor relations find
A1′ F12 = A′2 F21 = 0.08 m × 0.10 = 0.008 m
F12 = ( 0.080 / 0.0628 ) 0.10 = 0.127
F1R = 1 − F12 = 1 − 0.127 = 0.873
F2R = 1 − F21 = 1 − 0.10 = 0.90.
Hence, with E b = σ T 4 ,
q1′ =
q1′ =
4
4
5.67 × 10−8 (1700 ) − ( 600 ) 


1 − 0.9
1
1 − 0.4
+
+
1
−
0.9 × 0.0628 0.008 + [1/ (0.0628 × 0.873 ) + 1/ ( 0.08 × 0.90 )]
0.4 × 0.08
4.66 ×105
= 10,100 W / m.
1.77 + 25.56 + 18.75
<
COMMENTS: The radiosities for A1 and A2 follow from Eq. 13.19,
J1 = E b1 − (1 − ε1 ) q1′ / ε1A1′ = 4.56 × 105 W / m 2
J 2 = E b2 + (1 − ε 2 ) q1′ / ε 2 A′2 = 1.97 × 105 W / m 2 .
From Eq. 13.31, find JR and hence TR as
0.0628 × 0.873 ( J1 − J R ) − 0.08 × 0.90 ( J R − J 2 ) = 0
J R = 3.08 × 105 W / m 2 = σ TR4
TR = 1527 K.
PROBLEM 13.68
KNOWN: Surface temperature and emissivity of molten alloy and distance of surface from top of
container. Container diameter.
FIND: Net rate of radiation heat transfer from surface of melt.
SCHEMATIC:
ASSUMPTIONS: (1) Opaque, diffuse, gray behavior for surface of melt, (2) Large surroundings
may be represented by a hypothetical surface of temperature T = Tsur and ε = 1, (3) Negligible
convection at exposed side wall, (4) Adiatic side wall.
ANALYSIS: With negligible convection at an adiabatic side wall, the surface may be treated as
reradiating. Hence, from Eq. (13.30), with A1 = A2,
q1 =
With
A1 ( E b1 − E b2 )
1
1 − ε1
1− ε2
+
+
1
−
ε1
ε2
F12 + (1/ F1R ) + (1/ F2R )
)
(
R i = R j = ( D / 2 ) / L = 1.25 and S = 1 + 1 + R 2j / R i2  = 2.640, Table 13.2 yields


F12 =
1   2

2 1/ 2 
S − S − 4 ( r2 / r1 )   = 0.458
 
2  
Hence,
F1R = F2R = 1 − F12 = 0.542 and
q1 =
(
)
π (0.25m )2 × 5.67 × 10−8 W / m 2 ⋅ K 4 9004 − 3004 K 4
1 − 0.55
1
+
+0
−1
0.55
0.458 (3.69 )
= 3295 W
<
PROBLEM 13.69
KNOWN: Blackbody simulator design consisting of a heated circular plate with an opening over a
well insulated hemispherical cavity.
FIND: (a) Radiant power leaving the opening (aperture), Da = ro/2, (b) Effective emissivity of the
cavity, εe, defined as the ratio of the radiant power leaving the cavity to the rate at which the circular
plate would emit radiation if it were black, (c) Temperature of hemispherical surface, Thc, and (d)
Compute and plot εe and Thc as a function of the opening aperture in the circular plate, Da, for the
range ro/8 ≤ Da ≤ ro/2, for plate emissivities of εp = 0.5, 0.7 and 0.9.
SCHEMATIC:
ASSUMPTIONS: (1) Plate and hemispherical surface are diffuse-gray, (2) Uniform radiosity over
these same surfaces.
ANALYSIS: (a) The simulator can be treated as a three-surface enclosure with one re-radiating
surface (A2) and the opening (A3) as totally absorbing with no emission into the cavity (T3 = 300 K).
The radiation leaving the cavity is the net radiation leaving A1, q1 which is equal to –q3. Using Eq.
13.30,
q cav = q1 = −q3 =
(
σ T14 − T34
)
−1  −1
(1 − ε1 ) / ε1A1 +  A1F13 + [(1/ A1F12 ) + (1/ A3F32 )] 


(1)
+ (1 − ε 3 ) / ε 3A3
Using the summation rule and reciprocity, evaluate the required view factors:
F11 + F12 + F13 = 1
F13 = 0
F12 = 1
F31 + F32 + F33 = 1
F32 = 1.
(
Substituting numerical values with ε3 = 1, T3 = 300 K, A1 = π ro2 − ( ro / 4 )
2
) = 15π r
2
o /16 = 2.945 ×
10 m , A3 = π ra2 = π ( ro / 4 ) = 1.963 × 10 m and A1/A3 = 15, and multiplying numerator and
denominator by A1,
-2
2
-3
2
q cav = q1 =
(
2
A1σ T1 − T34
(1 − ε1 ) / ε1 +
{F
[
)
]
13 + (1/ F12 ) + ( A1 / A 3F32 )
}
−1 −1
(2)
+0
Continued …..
PROBLEM 13.69 (Cont.)
q cav = q1 =
(
)
2.945 × 10−2 m 2 × 5.67 × 10−8 W / m 2 ⋅ K 4 6004 − 300 4 K 4
(1 − 0.9 ) / 0.9 +
{0 + [1 + (15 /1)] }
−1 −1
<
= 12.6W
+0
(b) The effective emissivity is the ratio of the radiant power leaving the cavity to that from a
blackbody having the area of the opening and temperature of the inner surface of the cavity. That is,
εe =
q cav
A3σ T14
=
12.6 W
1.963 × 10
−3
2
m × 5.67 × 10
−8
W / m ⋅ K × ( 600 K )
2
4
4
= 0.873
(3)
<
(c) From a radiation balance on A1, find J1,
q1 = 12.6W =
E b1 − J1
=
(1 − ε1 ) / ε1A1
σ 6004 − J1
(1 − 09 ) / 0.9A1
J1 = 7301W / m 2
(4)
From a radiation balance on A2 with J3 = Eb3 = σ T34 = 459.9 W / m 2 and J 2 = σ T24 , find
J 2 − J1
+
J 2 − J3
(1/ A1F12 ) (1/ A3F32 )
J 2 = 6873 W / m 2
=
(
J 2 − 7301W / m 2
1/ 2.945 × 10−2 m 2
+
) (
J 2 − 459.9
1/1.963 × 10−3 m 2
)
=0
(5)
<
T2 = 590 K.
(d) Using the foregoing equations in the IHT workspace, εe and T2 were computed and plotted as a
function of the opening, Da, for selected plate emissivities, εp.
From the upper-left graph, εe decreases with increasing opening, Da, as expected. In the limit as Da
→ 0, ε3 → 1 since the cavity becomes a complete enclosure. From the upper-right graph, Thc, the
temperature of the re-radiating hemispherical surface decreases as Da increases. In the limit as Da →
0, T2 will approach the plate temperature, Tp = 600 K. The effect of decreasing the plate emissivity is
to decrease εe and decrease T2. Why is this so?
Continued …..
PROBLEM 13.69 (Cont.)
COMMENTS: The IHT Radiation, Tool, Radiation Tool, Radiation Exchange Analysis, ThreeSurface Enclosure with Re-radiating Surface, is especially convenient to perform the parametric
analysis of part (c). A copy of the IHT workspace that can generate the above graphs is shown below.
// Radiation Tool – Radiation Exchange Analyses, Reradiating Surface
/* For the three-surface enclosure A1, A3 and the reradiating surface A2, the net rate of radiation transfer
from the surface A1 to surface A3 is */
q1 = (Eb1 – Eb3) / ( (1 – eps1)/(eps1 * A1) + 1/(A1 * F13 + 1/(1/(A1 * F12) + 1/(A3 * F32))) + (1 –
eps3)/(eps3 * A3)) // Eq 13.30
/* The net rate of radiation transfer from surface A3 to surface A1 is */
q3 = q1
/* From a radiation energy balance on A2, */
(J2 – J1) / (1/(A2 * F21)) + (J2 – J3)/(1/(A2 * F23) ) = 0
// Eq 13.31
/* where the radiosities J1 and J3 are determined from the radiation rate equations expressed in terms of
the surface resistances, Eq 13.22 */
q1 = (Eb1 – J1) / ((1 – eps1) / (eps1 * A1))
q3 = (Eb3 – J3) / ((1 – eps3) / (eps3 * A3))
// The blackbody emissive powers for A1 and A3 are
Eb1 = sigma * T1^4
Eb3 = sigma * T3^4
// For the reradiating surface,
J2 = Eb2
Eb2 = sigma * T2^4
// Stefan-Boltzmann constant, W/m^2⋅K^4
sigma = 5.67E-8
// Effective emissivity:
epseff = q1 / (A3 * Eb1)
// Areas:
A1 = pi * ( ro^2 ⋅ ra^2)
A2 = 0.5 * pi * (2 * ro)^2
A3 = pi * ra^2
// Assigned Variables
T1 = 600
eps1 = 0.9
T3 = 300
eps3 = 0.9999
ro = 0.1
Da = 0.05
Da_mm = Da * 1000
Ra = Da / 2
// Eq (3)
// Hemisphere, As = 0.5 * pi * D^2
// Plate temperature, K
// Plate emissivity
// Opening temperature, K; Tsur
// Opening emissivity; not zero to avoid divide-by-zero error
// Hemisphere radius, m
// Opening diameter; range ro/8 to ro/2; 0.0125 to 0.050
// Scaling for plot
// Opening radius
PROBLEM 13.70
KNOWN: Long hemi-cylindrical shaped furnace comprised of three zones.
FIND: (a) Heat rate per unit length of the furnace which must be supplied by the gas burners and (b)
Temperature of the insulating brick.
SCHEMATIC:
ASSUMPTIONS: (1) Surfaces are opaque, diffuse-gray or black, (2) Surfaces have uniform
temperatures and radiosities, (3) Surface 3 is perfectly insulated, (4) Negligible convection, (5)
Steady-state conditions.
ANALYSIS: (a) From an energy balance on the ceramic plate, the power required by the burner is
q′burners = q1′ , the net radiation leaving A1; hence
q1′ = A1F12 ( J1 − J 2 ) + A1F13 ( J1 − J3 ) = 0 + A1F13 ( J1 − J3 )
(1)
since F12 = 0. Note that J2 = Eb2 = σ T24 and that J1 and J3 are unknown. Hence, we need to write
two radiation balances.
A1:
E b1 − J1
= q′ = 0 + A1FF13 ( J1 − J3 )
(1 − ε1 ) / ε1A1 1
A3:
0 = A3F31 ( J3 − J1 ) + A3F32 ( J3 − E b2 )
(2)
J + E b2
J3 = 1
2
(3)
since F31 = F23. Substituting Eq. (3) into (2), find
(371,589 − J1 ) / (1 − 0.85) / 0.85 = 1 J1 − ( J1 + 3,544 ) / 2 
J1 = 341, 748 W / m 2
J3 = 172, 646 W / m 2
using E b1 = σ T14 = 371, 589 W / m 2 and E b2 = σ T24 = 3544 W / m 2 . Substituting into Eq. (1), find
q1′ = 1 m × 1(341, 748 − 172, 646 ) W / m 2 = 169 kW / m.
<
(b) The temperature of the insulating brick, acting as a reradiating surface, is
J3 = E b3 = σ T34
1/ 4
T3 = ( J3 / σ )
(
= 172, 646 W / m 2 / 5.67 ×10−8 W / m 2 ⋅ K 4
)
1/ 4
= 1320 K.
<
PROBLEM 13.71
KNOWN: Steam producing still heated by radiation.
FIND: (a) Factor by which the vapor production could be increased if the cylindrical side of the
heater were insulated rather than open to the surroundings, and (b) Compute and plot the net heat rate
of radiation transfer to the still, as a function of the separation distance L for the range 15 ≤ L ≤ 100
mm for heater temperatures of 600, 800, 1000°C considering the cylindrical sides to be insulated.
SCHEMATIC:
ASSUMPTIONS: (1) Still and heater surfaces are black, (2) Surroundings are isothermal and large
compared to still heater surfaces, (3) Insulation is diffuse-gray, (4) Negligible convection.
ANALYSIS: (a) The vapor production will be proportional to the net radiation exchange to the still.
For the case when the sides are open (o) to the surroundings, the net radiation exchange leaving A2 is
from Eq. 13.13.
4
q 2,o = q 21 + q 2s = A 2 F21σ T24 − T14 + A 2 F2sσ T24 − Tsur
)
(
)
(
where F2s = 1 – F21 and F21 follows from Fig. 13.5 with L/ri = 100/100 = 1, rj/L = 100/100 = 1.
F21 = 0.38
With A 2 = π D 2 / 4, find
π ( 0.200m )
2
q 2,o =
4
× 5.67 × 10
−8
2
W / m ⋅K
q 2,o = −1752 W.
4
{0.38 (373
4
− 1273
4
)K
4
(
+ (1 − 0.38 ) 373 − 300
4
4
)K }
4
← w / oinsulation
With the cylindrical side insulated (i), a three-surface, re-radiating enclosure is formed. Eq. 13.30 can
be used to evaluate q2,i and with ε2 = ε1 = 1, the relation is
(
σ T24 − T1
q 2,i =
)
4
{
}σ (T24 − T14 )
−1
= A1 F12 + [1/ F1R + 1/ F2R ]
1
−1
A1F12 + [(1/ A1F1R ) + (1/ A 2 F2R )]
Recall F12 = 0.38 and F1R = 1 – F12 = 1 – 0.38 = 0.62, giving
2
q 2,i =
π ( 0.100 m )
4
q 2,i = −3204 W.
{0.38 + [1/ 0.62 + 1/ 0.62] }5.67 ×10
−1
−8
(
)
W / m 2 ⋅ K 3734 − 12734 K 4
← w insulation
Continued …..
PROBLEM 13.71 (Cont.)
Hence, the vapor production rate is increased by a factor
q 2 )insul
q 2 )open
=
3204 W
= 1.83
1752 W
That is, the vapor production is increased by 83%.
<
(b) The IHT Radiation Tool – Radiation Exchange Analysis for the Three-Surface Enclosure with a
reradiating surface can be used directly to compute the net heat rate to the still, q1 = q2, as a function
of the separation distance L for selected heater temperatures T1. The results are plotted below.
Note that the heat rate for all values of T1 decreases as expected with increasing separation distances,
but not markedly. For any separation distance, increasing the heater temperature greatly influences
the heat rate. For example, at L = 50 mm, increasing T1 from 600 to 800 K, causes a nearly 6 fold
increase in the heat rate. But increasing T1 from 800 to 1000 K causes only a 2 fold increase in the
heat rate.
COMMENTS: When assigning the emissivity variables (ε1, ε2, ε3) in the IHT model mentioned
above, set ε = 0.999, rather than 1.0, to avoid a “division by zero” error message. You could also call
up the Radiation Tool, View Factor Coaxial Parallel Disk to calculate F12.
PROBLEM 13.72
KNOWN: Furnace with cylindrical heater and re-radiating, insulated walls.
FIND: (a) Power required to maintain steady-state conditions, (b) Temperature of wall area.
SCHEMATIC:
ASSUMPTIONS: (1) Surfaces are diffuse-gray, (2) Furnace is of length where >> w, (3)
Convection is negligible.
ANALYSIS: (a) Consider the furnace as a three surface enclosure with the walls, AR, represented as
a re-radiating surface. The power that must be supplied to the heater is determined by Eq. 13.30.
q1 =
(
σ T14 − T24
)
(1 − ε1 ) / ε1A1 + A1F12 + [(1/ A1F1R ) + (1/ A 2F2R )]−1 

−1

+ (1 − ε 2 ) / ε 2 A 2
Note that A1 = πd and A2 = w . By inspection and the summation rule, find F12 = 60°/360° =
0.167, F1R = 1 – F12 = 1 – 0.167 = 0.833, and F2R ≈ 1. With q1′ = q1 / ",
q1′ =
5.67 × 10
(
−8
2
( (
)
(1500 − 500 ) K
m ) × 0.83 ) + (1 / 1m × 1)

W/m ⋅K

−3
−3
0 + π 10 × 10
m × 0.167 +  1 / π 10 × 10


4
4
4
4
−1 


−1
+ (1 − 0.6 ) / 0.6 × 1 m
<
q1′ = 8518 W / m.
(b) To determine the wall temperature, apply the radiation balance, Eq. 13.31,
J1 − J R
JR − J2
J1 − J R
J − J2
or
.
=
= R
(1/ A1F1R ) (1/ A 2 F2R )
(1/1m × 1)
1/ π 10 × 10−3 m × 0.833
)
(
J R = σ TR4 = ( J1 + 38.21J 2 ) / 39.21.
(1)
Since A1 is a blackbody, J1 = Eb1 = σ T14 . To determine J2, use Eq. 13.19. Noting that q1′ = −q′2 ,
find
q 2 = ( E b2 − J 2 ) / (1 − ε 2 ) / ε 2 A 2
or
J 2 = E b2 − q 2 (1 − ε 2 ) / ε 2 A 2
J 2 = 5.67 × 10−8 W / m 2 ⋅ K 4 (500 K ) −
4
( −8518 W / m )(1 − 0.6 )
= 9222 W / m 2 .
0.6 (1m )
Substituting this value for J2 into Eq. (1), the wall temperature can be calculated.
(
)
J R = 5.67 × 10 −8 W / m 2 ⋅ K 4 (1500K ) + 38.21 × 9222 W / m 2 / 39.21 = 16, 308 W / m 2
4
TR = ( J R / σ )
1/ 4
(
= 16, 308 W / m 2 / 5.67 × 10−8 W / m 2 ⋅ K 4
)
1/ 4
= 732 K.
<
COMMENTS: Considering the entire wall as a single re-radiating surface may be a poor assumption
since JR is not likely to be uniform over this large an area. It would be appropriate to consider several
isothermal zones for improved accuracy.
PROBLEM 13.73
KNOWN: Circular furnace with one end (A1) and the lateral surface (A2) black. Other end (A3) is
insulated.
FIND: (a) Net radiation heat transfer from each surface and (b) Temperature of A3, and (c) Compute
and plot T3 as a function of tube length L for the range 0.1 ≤ L ≤ 0.5 m with D = 0.3 m.
SCHEMATIC:
ASSUMPTIONS: (1) Surface A3 is diffuse-gray, (2) Uniform radiosity over A3.
ANALYSIS: (a) Since A3 is insulated, the net radiation from A3 is q3 = 0. Using Eq. 13.30, find
q1 = −q 2 =
(
4
4
σ T1 − T2
)
(1 − ε1 ) / ε1A1 +  A1F12 + [(1 / A1F13 ) + (1 / A 2 F23 )]−1 
−1
+ (1 − ε 2 ) / ε 2 A 2
)
}σ (T
−1
{
q1 = −q 2 = A1F12 + [(1/ A1F13 ) + (1/ A 2 F23 )]
and since ε1 = ε2 = 1,
4
4
1 − T2 .
Considering A1 and A3 as coaxial parallel disks, from Table 13.2 (Fig. 13.5) find F13,
R1 = r1 / L = 0.15 / 0.30 = 0.5
R 2 = r2 / L = 0.15 / 0.30 = 0.5
(
)
(
)
S = 1 + 1 + R 22 / R12 = 1 + 1 + 0.52 / 0.52 = 6

2
F13 = 0.5 S − S2 − 4 ( r2 / r1 ) 
1/ 2 



1/ 2 

2
 = 0.5 6 − 6 − 4 ( 0.5 / 0.5 )  = 0.172.



F12 = 1 − F13 = 1 − 0.172 = 0.828.
From the summation rule and symmetry
From reciprocity, with F32 = F12
(
)
F23 = A3F32 / A 2 = π D 2 / 4 F12 / (π DL ) = DF12 / 4L = 0.3 m × 0.828 / 4 × 0.3 m = 0.207.
2
2
2
2
With A1 = πD /4 = π(0.3 m) /4 = 0.07069 m and A2 = πDL = π(0.3 m) (0.3 m) = 0.2827 m ,
{
(
q1 = −q 2 = 0.07069 m 2 × 0.828 +  1/ 0.07069 m 2 × 0.172

)
(
+ 1/ 0.2827 m 2 × 0.207 
−1 
 5.67 × 10

−8
(500
4
)
)
− 4004 K 4
<
q1 = −q 2 = 143 W
(b) From the radiation balance on A3, Eq. 13.31, find T3, with J1 = Eb1 and J2 = Eb2,
E b1 − J 3
−
J3 − E b2
(1/ A1F13 ) (1/ A2 F23 )
J3 − σ ( 400 K )
σ (500 K ) − J3
−
=0
(1/ 0.07069 m × 0.172 ) 1/ 0.2827 m2 × 0.207
4
=0
4
)
(
J 2 = E b3 = σ T34 = 1811 W / m 2
T3 = 423 K.
(c) Using the IHT Radiation Tools – Radiation Exchange Analysis, Three surface enclosure with a
reradiating surface and View Factors, Three-dimensional geometries, Coaxial parallel disks – and
<
Continued …..
PROBLEM 13.73 (Cont.)
appropriate view factor relations developed in part (a), T3 was computed as a function of L with the
diameter, D = 0.3 m, and is plotted below.
Note that T3 decreases with increasing tube length. In the limit as L → ∞, T3 will approach T2. In
the limit as L → 0, T3 will approach T1. Is this intuitively satisfying?
COMMENTS: The IHT workspace used for the part (c) analysis is copied below.
// Radiation Tool:
// Radiation Exchange Analysis, three-surface enclosure with a reradiating surface
/* For the three-surface enclosure A1, A2 and the reradiating surface A3, the net rate of radiation transfer
from the surface A1 to surface A2 is */
q1 = (Eb1 – Eb2) / ( (1 – eps1) / (eps1 * A1) + 1 / (A1 * F12 + 1 / (1 / (A1 * F13) + 1 / (A2 * F23))) + (1 –
eps2) / (eps2 * A2))
// Eq 13.30
/ * The net rate of radiation transfer from surface A2 to surface A1 is */
q2 = -q1
/ * From a radiation energy balance on A3, */
(J3 – J1) / (1 / (A3 * F31) ) + (J3 – J2) / (1 / (A3 * F32) ) = 0
// Eq 13.31
/* where the radiosities J1 and J2 are determined from the radiation rate equations expressed in terms of
the surface resistances, Eq 13.22 */
q1 = (Eb1 – J1) / ((1 – eps1) / (eps1 * A1))
q2 = (Eb2 – J2) / ((1 – eps2) / (eps2 * A2))
// The blackbody emissive powers for A1 and A2 are
Eb1 = sigma * T1^4
Eb2 = sigma * T2^4
// For the reradiating surface,
J3 = Eb3
Eb3 = sigma * T3^4
sigma = 5.67E-8
// Stefan-Boltzmann constant, W/m^2⋅K^4
// Radiation Tool – view factor, F13:
/* The view factor, F13, for coaxial parallel disks, is * /
F13 = 0.5 * (S – sqrt(S^2 – 4 * (r3 / r1)^2))
// where
R1 = r1 / L
R3 = r3 / L
r1 = D1 / 2
r3 = r1
S = 1 + (1 + R3^2) / R1^2
// See Table 13.2 for schematic of this three-dimensional geometry.
// View factor relations:
F12 = 1 – F13
// Summation rule, A1
F32 = F12
// Symmetry condition
F23 = A3 * F32 / A2
// Reciprocity relation
F31 = A1 * F13 / A3
// Reciprocity relation
// Areas:
A1 = pi * D1^2/4
A2 = pi * D2 * L
A3 = A1
// Assigned variables:
T1 = 500
// Temperature, K
D1 = 0.3
// Diameter, m
eps1 = 0.9999
// Emissivity; avoiding “divide-by-zero error”
D2 = D1
// Diameter, m
eps2 = 0.9999
// Emissivity; avoiding “divide-by-zero error”
T2 = 400
// Temperature, K
L = 0.3
// Length, m
PROBLEM 13.74
KNOWN: Very long, triangular duct with walls that are diffuse-gray.
FIND: (a) Net radiation transfer from surface A1 per unit length of duct, (b) The temperature of the
insulated surface, (c) Influence of ε3 on the results; comment on exactness of results.
SCHEMATIC:
ASSUMPTIONS: (1) Surfaces are diffuse-gray, (2) Duct is very long; end effects negligible.
ANALYSIS: (a) The duct approximates a three-surface enclosure for which the third surface (A3) is
re-radiating. Using Eq. 13.30 with A3 = AR, the net exchange is
E b1 − E b2
(1)
q1 = −q 2 =
1− ε2 )
(1 − ε1 ) +
(
1
+
−1
ε1A1
ε 2A 2
A F + (1/ A F + 1/ A F )
1 12
1 1T
2 2R
From symmetry, F12 = F1R = F2R = 0.5. With A1 = A2 = w⋅ , where is the length normal to the
page and w = 1 m,
q1′ = q1 / = ( q1 / A1 ) w
q1′ =
(56, 700 − 13, 614 ) W / m2 × 1m
= 9874 W / m.
1 − 0.5 )
1
(
(1 − 0.33) +
+
−1
0.33
0.5
0.5 + (1/ 0.5 + 1/ 0.5 )
(b) From a radiation balance on AR,
E −J
E −J
q R = q3 = 0 = b3 1 + b3 2
( A3F31 )−1 ( A3F32 )−1
or
<
J +J
E b3 = 1 2 .
2
(2)
To evaluate J1 and J2, use Eq. 13.19,
1 − 0.33

= 36, 653 W / m 2
J1 = 56, 700 − (9874 )

0.33
q (1 − ε i ) 
J i = E b,i − i

Ai ε i
1 − 0.5
J 13, 614
9874 )
= 23, 488 W / m 2
=
−
−
(
2

0.5
From Eq. (2), now find

2
1/ 4  (36, 653 + 23, 488 ) W / m
1/ 4
T3 = ( E b3 / σ )
= ([J1 + J 2 ] / 2σ )
=

 2 5.67 × 10−8 W / m 2 ⋅ K 4

(
)
1/ 4





= 853 K.
<
(c) Since A3 is adiabatic or re-radiating, J3 = Eb3. Therefore, the value of ε3 is of no influence on the
radiation exchange or on T3. In using Eq. (1), we require uniform radiosity over the surfaces. This
requirement is not met near the corners. For best results we should subdivide the areas such that they
represent regions of uniform radiosity. Of course, the analysis then becomes much more complicated.
PROBLEM 13.75
KNOWN: Dimensions for aligned rectangular heater and coated plate. Temperatures of heater, plate
and large surroundings.
FIND: (a) Electric power required to operate heater, (b) Heater power required if reradiating
sidewalls are added, (c) Effect of coating emissivity and electric power.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, (2) Blackbody behavior for surfaces and surroundings (Parts (a)
and (b)).
ANALYSIS: (a) For ε1 = ε2 = 1, the net radiation leaving A1 is
(
)
(
)
4
q elec = q1 = A1F12σ T14 − T24 + A1F1surσ T14 − Tsur
.
From Fig. 13.4, with Y/L = 1/0.5 = 2 and X/L = 2/0.5 = 4, the view factors are F12 ≈ 0.5 and Fsur ≈ 1 –
0.5 = 0.5. Hence,
( ) 0.5 × 5.67 × 10 W / m ⋅ K (700 K ) − (400 K ) 
4
+ ( 2m ) 0.5 × 5.67 × 10 W / m ⋅ K ( 700 K ) − (300 K )  = (12,162 + 13,154 ) W = 25, 316 W. <


q elec = 2m
−8
2
−8
2
2
4
2
4
4
4
4
(b) With the reradiating walls, the net radiation leaving A1 is qelec = q1 = q12. From Eq. 13.30 with ε1
= ε2 = 1 and A1 = A2,
(
q elec = A1σ T14 − T24
( )
){F12 + [(1/ F1R ) + (1/ F2R )]−1}
−1
4
4
q elec = 2 m 2 5.67 × 10−8 W / m 2 ⋅ K 4 ( 700 K ) − ( 400 K )  × 0.5 + [(1/ 0.5 ) + (1/ 0.5 )]


{
}
<
q elec = 18, 243 W.
(c) Separately using the IHT Radiation Tool Pad for a three-surface enclosure, with one surface
reradiating, and to perform a radiation exchange analysis for a three-surface enclosure, with one
surface corresponding to large surroundings, the following results were obtained.
Continued …..
PROBLEM 13.75 (Cont.)
In both cases, the required heater power decreases with decreasing ε2, and the trend is attributed to a
reduction in α2 = ε2 and hence to a reduction in the rate at which radiant energy must be absorbed by
the surface to maintain the prescribed temperature.
COMMENTS: With the reradiating walls in part (b), it follows from Eq. 13.31 that
J R = E bR = ( J1 + J 2 ) / 2 = ( E b1 + E b2 ) / 2.
Hence, TR = 604 K. The reduction in qelec resulting from use of the walls is due to the enhancement
of radiation to the heater, which, in turn, is due to the presence of the high temperature walls.
PROBLEM 13.76
KNOWN: Configuration and operating conditions of a furnace. Initial temperature and emissivity of
steel plate to be treated.
FIND: (a) Heater temperature, (b) Sidewall temperature.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Opaque, diffuse-gray surface behavior, (3)
Negligible convection, (4) Sidewalls are re-radiating.
ANALYSIS: (a) From Eq. 13.30
q1 =
1 − ε1
+
ε1A1
E b1 − E b2
1
−1
A1F12 + ( A1F1R )

−1  −1
+ ( A 2 F2R )
2
+

-8
2
1 − ε2
ε 2A2
= 1.5 × 105 W
4
4
2
Note that A1 = A2 = 4 m and Eb2 = σ T24 = 5.67 × 10 W/m ⋅K (300 K) = 459 W/m . From Fig.
13.4, with X/L = Y/L = 1, F12 = 0.2; hence F1R = 1 – F12 = 0.8, and F2R = F1R = 0.8. With (1-ε1)/ε1 =
0.25 and (1 - ε2)/ε2 = 1.5, find
1.5 × 105 W
4m 2
=
0.25 +
E b1 − 459 W / m 2
1
−1
0.2 + [1.25 + 1.25]
E b1 − 459 W / m 2
=
3.417
+ 1.5
E b1 = 1.28 × 105 W / m 2 + 459 W / m 2 = 1.29 × 105 W / m 2 = σ T14
(
T1 = 1.29 × 105 W / m 2 / 5.67 × 10−8 W / m 2 ⋅ K 4
)
1/ 4
= 1228 K.
(b) From Eq. 13.31,it follows that, with A1F1R = A2F2R,
J R = σ TR4 = ( J1 + J 2 ) / 2
From Eq. 13.19,
J1 = E b1 −
(1 − ε1 )
ε1A1
5
2
q1 = 1.29 × 10 W / m −
0.2 × 1.5 × 105 W
0.8 × 4m 2
J1 = 1.196 × 105 W / m 2 .
5
With q2 = q1 = - 1.5 × 10 W,
0.6
(1 − ε 2 )
J 2 = E b2 −
q 2 = 459 W / m 2 +
1.5 × 105 W = 5.67 × 104 W / m 2
ε 2A2
0.4 × 4m 2
1/ 4
 1.196 × 105 W / m 2 + 5.67 × 104 W / m 2 
TR = 

2
4
−8


2
5.67
10
W
/
m
K
×
×
⋅


= 1117 K.
<
COMMENTS: (1) The above results are approximate, since the process is actually transient. (2) T1
and TR will increase with time as T2 increases.
PROBLEM 13.76
KNOWN: Configuration and operating conditions of a furnace. Initial temperature and emissivity of
steel plate to be treated.
FIND: (a) Heater temperature, (b) Sidewall temperature.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Opaque, diffuse-gray surface behavior, (3)
Negligible convection, (4) Sidewalls are re-radiating.
ANALYSIS: (a) From Eq. 13.30
q1 =
1 − ε1
+
ε1A1
E b1 − E b2
1
−1
A1F12 + ( A1F1R )

−1  −1
+ ( A 2 F2R )
2
+

-8
2
1 − ε2
ε 2A2
= 1.5 × 105 W
4
4
2
Note that A1 = A2 = 4 m and Eb2 = σ T24 = 5.67 × 10 W/m ⋅K (300 K) = 459 W/m . From Fig.
13.4, with X/L = Y/L = 1, F12 = 0.2; hence F1R = 1 – F12 = 0.8, and F2R = F1R = 0.8. With (1-ε1)/ε1 =
0.25 and (1 - ε2)/ε2 = 1.5, find
1.5 × 105 W
4m 2
=
0.25 +
E b1 − 459 W / m 2
1
−1
0.2 + [1.25 + 1.25]
E b1 − 459 W / m 2
=
3.417
+ 1.5
E b1 = 1.28 × 105 W / m 2 + 459 W / m 2 = 1.29 × 105 W / m 2 = σ T14
(
T1 = 1.29 × 105 W / m 2 / 5.67 × 10−8 W / m 2 ⋅ K 4
)
1/ 4
= 1228 K.
(b) From Eq. 13.31,it follows that, with A1F1R = A2F2R,
J R = σ TR4 = ( J1 + J 2 ) / 2
From Eq. 13.19,
J1 = E b1 −
(1 − ε1 )
ε1A1
5
2
q1 = 1.29 × 10 W / m −
0.2 × 1.5 × 105 W
0.8 × 4m 2
J1 = 1.196 × 105 W / m 2 .
5
With q2 = q1 = - 1.5 × 10 W,
0.6
(1 − ε 2 )
J 2 = E b2 −
q 2 = 459 W / m 2 +
1.5 × 105 W = 5.67 × 104 W / m 2
ε 2A2
0.4 × 4m 2
1/ 4
 1.196 × 105 W / m 2 + 5.67 × 104 W / m 2 
TR = 

2
4
−8


2
5.67
10
W
/
m
K
×
×
⋅


= 1117 K.
<
COMMENTS: (1) The above results are approximate, since the process is actually transient. (2) T1
and TR will increase with time as T2 increases.
PROBLEM 13.77
KNOWN: Dimensions, surface radiative properties, and operating conditions of an electrical
furnace.
FIND: (a) Equivalent radiation circuit, (b) Furnace power requirement and temperature of a heated
plate.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Opaque, diffuse-gray surfaces, (3) Negligible plate
temperature gradients, (4) Back surfaces of heater are adiabatic, (5) Convection effects are negligible.
ANALYSIS: (a) Since there is symmetry about the plate, only one-half (top or bottom) of the system
need be considered. Moreover, the plate must be adiabatic, thereby playing the role of a re-radiating
surface.
<
2
2
(b) Note that A1 = A3 = 4 m and A2 = (0.5 m × 2 m)4 = 4 m . From Fig. 13.4, with X/L = Y/L = 4,
F13 = 0.62. Hence
F12 = 1 − F13 = 0.38,
and
F32 = F12 = 0.38.
It follows that
A1F12 = 4 (0.38 ) = 1.52 m 2
A1F13 = 4 (0.62 ) = 2.48 m 2 ,
A3F32 = 4 ( 0.38 ) = 1.52 m 2 ,
(1 − ε1 ) / ε1A1 = 0.1/ 3.6 m 2 = 0.0278 m−2
(1 − ε 2 ) / ε 2A 2 = 0.7 /1.2 m 2 = 0.583m −2 .
Also,
E b1 = σ T14 = 5.67 ×10−8 W / m 2 ⋅ K 4 (800 K ) = 23, 224 W / m 2 ,
4
E b2 = σ T24 = 5.67 ×10−8 W / m 2 ⋅ K 4 ( 400 K ) = 1452 W / m 2 .
4
The system forms a three-surface enclosure, with one surface re-radiating. Hence the net radiation
transfer from a single heater is, from Eq. 13.30,
q1 =
E b1 − E b2
1
1 − ε1
1− ε2
+
+
ε1A1 A F + [1/ A F + 1/ A F ]−1 ε 2A 2
1 12
1 13
3 32
Continued …..
PROBLEM 13.77 (Cont.)
q1 =
( 23, 224 − 1452 ) W / m 2
(0.0278 + 0.4061 + 0.583) m−2
= 21.4 kW.
The furnace power requirement is therefore qelec = 2q1 = 43.8 kW, with
q1 =
<
E b1 − J1
.
(1 − ε1 ) / ε1A1
where
J1 = E b1 − q1
1 − ε1
= 23, 224 W / m 2 − 21, 400 W × 0.0278 m −2
ε1A1
J1 = 22, 679 W / m 2 .
Also,
1− ε2
= 1, 452 W / m 2 − ( −21, 400 W ) × 0.583m −2
ε 2A 2
J 2 = E b2 − q 2
J 2 = 13,928 W / m 2 .
From Eq. 13.31,
J1 − J3
J −J
= 3 2
1/ A1F13 1/ A3F32
J1 − J3 A3F32 1.52
=
=
= 0.613
J 3 − J 2 A1F13 2.48
1.613J 3 = J1 + 0.613J 2 = 22, 629 + 8537 = 31,166 W / m 2
J3 = 19, 321W / m 2
Since J3 = Eb3,
1/ 4
T3 = ( E b3 / σ )
(
= 19, 321/ 5.67 ×10−8
)
1/ 4
= 764 K.
<
COMMENTS: (1) To reduce qelec, the sidewall temperature T2, should be increased by insulating it
from the surroundings. (2) The problem must be solved by simultaneously determining J1, J2 and J3
from the radiation balances of the form
E b1 − J1
= A F ( J − J ) + A1F13 ( J1 − J3 )
(1 − ε1 ) / ε1A1 1 12 1 2
E b2 − J 2
= A F ( J − J ) + A 2 F23 ( J 2 − J3 )
(1 − ε 2 ) / ε 2A 2 2 21 2 1
0 = A1F13 ( J3 − J1 ) + A 2 F23 ( J3 − J 2 ).
PROBLEM 13.78
KNOWN: Geometry and surface temperatures and emissivities of a solar collector.
FIND: Net rate of radiation transfer to cover plate due to exchange with the absorber plates.
SCHEMATIC:
ASSUMPTIONS: (1) Isothermal surfaces with uniform radiosity, (2) Absorber plates behave as
blackbodies, (3) Cover plate is diffuse-gray and opaque to thermal radiation exchange with absorber
plates, (4) Duct end effects are negligible.
ANALYSIS: Applying Eq. 13.21 to the cover plate, it follows that
E b1 − J1 =
1 − ε1 N
∑
ε1A1
J1 − J j
(
j=1 Ai Fij
−1
)
=
1 − ε1
 A1F12 ( J1 − J 2 ) + A1F13 ( J1 − J3 ) .
ε1A1 
From symmetry, F12 = F13 = 0.5. Also, J2 = Eb2 and J3 = Eb3. Hence
E b1 − J1 = 0.0556 ( 2J1 − E b2 − E b3 )
or with E b = σ T 4 ,
1.111J1 = E b1 + 0.0556 ( E b2 + E b3 )
(
)
4
4
4
1.111J1 = 5.67 × 10−8 ( 298 ) W / m 2 + 0.0556 5.67 × 10−8 (333) + (343)  W / m 2


J1 = 476.64 W / m 2
From Eq. 13.19 the net rate of radiation transfer from the cover plate is then
5.67 × 10−8 ( 298 ) − 476.64
E b1 − J1
q1 =
=
= ( −265.5 ) W.
(1 − 0.9 ) / 0.9 ( )
(1 − ε1 ) / ε1A1
4
The net rate of radiation transfer to the cover plate per unit length is then
q1′ = ( q1 / ) = 266 W / m.
<
COMMENTS: Solar radiation effects are not relevant to the foregoing problem. All such radiation
transmitted by the cover plate is completely absorbed by the absorber plate.
PROBLEM 13.79
KNOWN: Cylindrical peep-hole of diameter D through a furnace wall of thickness L. Temperatures
prescribed for the furnace interior and surroundings outside the furnace.
FIND: Heat loss by radiation through the peep-hole.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Furnace interior and exterior surroundings are
large, isothermal surroundings for the peep-hole openings, (3) Furnace refractory wall is adiabatic and
diffuse-gray with uniform radiosity.
ANALYSIS: The open-ends of the cylindrical peep-hole (A1 and A2) and the cylindrical lateral
surface of the refractory material (AR) form a diffuse-gray, three-surface enclosure. The hypothetical
areas A1 and A2 behave as black surfaces at the respective temperatures of the large surroundings to
which they are exposed. Since Ar is adiabatic, it behaves as a re-radiating surface, and its emissivity
has no effect on the analysis. From Eq. 13.30, the net radiation leaving A1 passes through the
enclosure into the outer surroundings.
q1 = − q 2 =
E b1 − E b2
1
1- ε 1
1− ε 2
+
+
−1 ε A
ε 1 A1 A F + 1 / A F + 1 / A F
2 2
1 1R
2 2R
1 12
1
4
6 1
6
-8
2
4
Since ε1 = ε2 = 1, and with Eb = σ T where σ = 5.67 × 10 W/m ⋅K ,
−1
q1 = A1 F12 + 1/ A1 F1R + 1 / A 2 F2R
σ T14 − T24
%&
'
1
6 1
2
6 ()* 4
9
where A1 = A2 = π D /4. The view factor F12 can be determined from Table 13.2 (Fig. 13.5) for the
coaxial parallel disks (R1 = R2 = 125/(2 × 250) = 0.25 and S = 17.063) as
F12 = 0.05573
From the summation rule on A1, with F11 = 0,
F11 + F12 + F1R = 1
F1R = 1 − F12 = 1 − 0.05573 = 0.9443
and from symmetry of the enclosure,
F2R = F1R = 0.9443
Substituting numerical values into the rate equation, find the heat loss by radiation through the peephole to the exterior surroundings as
q loss = q1 = 1046 W
<
COMMENTS: If you held your hand 50 mm from the exterior opening of the peep-hole, how would
that feel? It is standard, safe practice to use optical protection when viewing the interiors of high
temperature furnaces as used in petrochemical, metals processing and power generation operations.
PROBLEM 13.80
KNOWN: Composite wall comprised of two large plates separated by sheets of refractory insulation
of thermal conductivity k = 0.05 W/m⋅K; gaps between the sheets of width w = 10 mm, located at 1 m spacing, allow radiation transfer between the plates.
FIND: (a) Heat loss by radiation through the gap per unit length of the composite wall (normal to the
page), and (b) fraction of the total heat loss through the wall that is due to radiation transfer through
the gap.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Surfaces are diffuse-gray with uniform radiosities,
(3) Refractory insulation surface in the gap is adiabatic, and (4) Heat flow through the wall is onedimensional between the plates in the direction of the gap centerline.
ANALYSIS: (a) The gap of thickness w and infinite extent normal to the page can be represented by
a diffuse-gray, three-surface enclosure formed by the plates A1 and A2 and the refractory walls, AR.
Since AR is adiabatic, it behaves as a re-radiating surface, and its emissivity has no effect on the
analysis. From Eq. 13.30, the net radiation leaving the plate A1 passes through the gap into plate A2.
q1 = − q 2 =
E b1 − E b2
1
1− ε 2
1- ε 1
+
+
1
−
ε 1 A1 A F + 1 / A F + 1 / A F
ε 2 A2
1 12
1 1R
2 2R
4
-8
$ 2
$
4
where Eb = σ T with σ = 5.67 × 10 W/m ⋅K and A1 = A2 = w ⋅ " , but making " = 1 m to obtain
q1′ W / m .
$
The view factor F12 can be determined from Table 13.2 (Fig. 13.4) for aligned parallel rectangles
where X = X / L = ∞ since X → ∞ and Y = Y / L = W / L = 10 / 50 = 0.2 giving
F12 = 0.09902
From the summation rule on A1, with F11 = 0,
F11 + F12 + F1R = 1
F1R = 1 − F12 = 1 − 0.09902 = 0.901
and from symmetry of the enclosure,
F2R = F1R = 0.901 .
Continued …..
PROBLEM 13.80 (Cont.)
Substituting numerical values into the rate equation, find the heat loss through the gap due to radiation
as
<
q ′rad = q1′ = 37 W / m
(b) The conduction heat rate per unit length (normal to the page) for a 1 - m section is
400 − 35 K
T −T
q cond
= k 1 m 1 2 = 0.05 W / m ⋅ K × 1 m
′
L
0.050 mm
$
$
q ′cond = 365 W / m
The fraction of the total heat transfer through the 1 - m section due to radiation is
q ′rad
q ′rad
37
=
= 9.2%
=
q ′tot q cond
+ q rad 365 + 37
′
We conclude that if the installation process for the sheet insulation can be accomplished with a
smaller gap, there is an opportunity to reduce the cost of operating the furnace.
<
PROBLEM 13.81
KNOWN: Diameter, temperature and emissivity of a heated disk. Diameter and emissivity of a
hemispherical radiation shield. View factor of shield with respect to a coaxial disk of prescribed
diameter, emissivity and temperature.
FIND: (a) Equivalent circuit, (b) Net heat rate from the hot disk.
SCHEMATIC:
ASSUMPTIONS: (1) Surfaces may be approximated as diffuse/gray, (2) Surface 4 is reradiating, (3)
Negligible convection.
ANALYSIS: (a) The equivalent circuit is shown in the schematic. Since surface 4 is treated as
reradiating, the net transfer of radiation from surface 1 is equal to the net transfer of radiation to
surface 3 (q1 = -q3).
(b) From the thermal circuit, the desired heat rate may be expressed as
q1 =
E b1 − E b3


2
1
ε
−
(
)
1 − ε1
1
1
2
+
+
+  A 2 F23 +
1
1
ε1A1 A1F12
ε 2A 2

+

A 2 F24 A3F34
2
3
2





−1
+
1 − ε1
ε 3A 3
-3
2
where A1 = A3 = π D12 / 4 = π(0.05 m) /4 = 1.963 × 10 m , A2 = π D12 / 2 = 2A1 = 3.925 × 10 m ,
F12 = 1, and F24 = 1 – F23 = 0.7. With F34 = 1 – F32 = 1 – F23(A2/A3) = 1 – 0.3(2) = 0.4, it follows
that
Continued …..
PROBLEM 13.81 (Cont.)
(
A1σ T14 − T34
q1 =
2 (1 − ε 2 ) A1
1 − ε1 1
+
+
A2
ε1A1 F12
ε2
q1 =
(
A1σ T14 − T34


A

1

+  2 F23 +
A1
A1 
 A1
+

A 2 F24 A3F34 
)




1
0.667 + 1 + 49 +  0.6 +
1
1 


+
1.4 0.4 

(
)
)
=
−1
(
−1
1 − ε3
ε3
+
A1σ T14 − T34
)
0.667 + 1 + 49 + 1.098 + 1.5
+ 1.5
(
)
q1 = 0.0188 1.963 × 10−3 m 2 5.67 × 10−8 W / m 2 ⋅ K 4 9004 − 4004 K 4
q1 = 1.32 W
<
COMMENTS: Radiation transfer from 1 to 3 is impeded and enhanced, respectively, by the
radiation shield and the reradiating walls. However, the dominant contribution to the total radiative
resistance is made by the shield.
PROBLEM 13.82
KNOWN: Diameter, temperature and emissivity of a cylindrical heater. Dimensions, temperature,
and emissivity of a reflector. Temperature of large surroundings.
FIND: (a) Equivalent circuit and values of associated resistances and driving potentials, (b) Required
electric power per unit length of heater.
SCHEMATIC:
ASSUMPTIONS: (1) Surfaces are diffuse-gray, (2) Surroundings form a large enclosure which may
be represented by a hypothetical surface of temperature T3 = Tsur and emissivity ε3 = 1.
ANALYSIS: (a) The circuit is shown in the schematic, where the hypothetical surface is part of a
three-surface enclosure. On a unit length basis, the circuit resistances are
1 − ε1
0.2
=
= 2.65 m −1
ε1A1′ 0.8 (π )( 0.03 m )
<
1
1
=
= 16.98 m −1
A1′ F12 π ( 0.03 m ) 0.625
<
1
1
=
= 28.30 m −1
′
A1F13 π ( 0.03 m ) 0.375
<
1
A′2 F23
=
1
A′3F32
=
1
(0.15 m ) 0.764
= 8.73 m −1
<
1 − ε2
0.9
=
= 30 m −1
ε 2 A′2 0.1 ( 0.30 m )
<
The view factor F13 is obtained from the summation rule, where F13 = 1 – F12 = 1 – 0.625 = 0.375.
Similarly, F32 = 1 – F31, where F31 = F13 ( A1′ / A′3 ) = 0.375(πD/W) = 0.375π(0.03 m/0.15 m) =
0.236. Hence, F32 = 0.764. The potentials are
Continued …..
PROBLEM 13.82 (Cont.)
E b1 = σ T14 = 5.67 × 10−8 W / m 2 ⋅ K 4 (945 K ) = 45, 220 W / m 2
<
E b2 = σ T24 = 5.67 × 10−4 W / m 2 ⋅ K 4 (385 K ) = 1246 W / m 2
<
E b3 = σ T34 = 5.67 × 10 −8 W / m 2 ⋅ K 4 (300 K ) = 459 W / m 2
<
4
4
4
(b) The required heater power may be obtained by applying Eq. (13.19) to node 1. Hence,
q1′ =
E b1 − J1
(1 − ε1 ) / ε1A1′
The radiosity may be obtained by applying radiation balances to nodes J1 and J2:
E b1 − J1
(1 − ε1 ) / ε1A1′
E b2 − J 2
=
(1 − ε 2 ) / ε 2 A′2
J − E b3
J1 − J 2
+ 1
1/ A1′ F12 1/ A1′ F13
=
J 2 − J1 J 2 − E b3
+
1/ A1′ F12 1/ A′2 F23
2
2
Substituting the known potentials and solving, we obtain J1 = 37,600 W/m and J2 = 11,160 W/m .
Hence
q1′ =
( 45, 220 − 37, 600 ) W / m2
2.65 m −1
= 2875 W / m
<
COMMENTS: Additional power would have to be supplied to compensate for heat transfer by free
convection from the heater to the air.
PROBLEM 13.83
KNOWN: Cylindrical cavity with prescribed geometry, wall emissivity, and temperature.
FIND: Net radiation heat transfer from the cavity assuming the surroundings of the cavity are at 300 K.
SCHEMATIC:
A1 = A3 = π D 2 / 4 = 7.85 × 10−3 m 2
A 2 = π DL = 1.57 × 10 −2 m 2
F13 = F31 = 0.38
F21 = F23 = 0.310
F32 = F12 = 0.62
T1 = T2 = 1500 K
ε1 = ε 2 = 0.6
ASSUMPTIONS: (1) Cavity interior surfaces are diffuse-gray, (2) Surroundings are much larger than
the cavity opening A3; T3 = Tsur = 300 K and ε3 = 1.
ANALYSIS: The net radiation heat transfer from the cavity is –q3, which from Eq. 13.20 is,
q3 = A3F31 ( J3 − J1 ) + A3F32 ( J3 − J 2 ) .
(1)
While J3 = Eb3 since ε3 = 1, J1 and J2 are unknown and must be obtained from the radiation balances,
Eq. 13.21.
N
E bi − Ji
Ji − Ji
=∑
(1 − ε i ) / ε i Ai j=1 A F −1
i ij
(
(2)
)
From Fig. 13.5 with L/r1 = 0.050/0.050 = 1 and r3/L = 1, find F13 = 0.38. From summation rule and
reciprocity: F32 = F12 = 1 – F12 = 0.62 and F21 = F23 = (A3 F32)/A2 = 0.310. Note also, Eb1 = Eb2 =
4
2
2
σ T14 = σ(1500K) = 287,044 W/m and J3 = Eb3 = σ T34 = 459.3 W/m .
E b1 − J1
J −J
J −J
A1:
= 1 2 + 1 3
(1 − ε1 ) / ε1A1 ( A F )−1 ( A F )−1
1 12
287, 044 − J1
(1 − 0.6 ) / 0.6
A2:
=
E b2 − J 2
1 13
J1 − J 2
J − 459.3
+ 1
(0.62 )−1 (0.38 )−1
=
J 2 − J1
+
2.5J1 − 0.62J 2 = 430, 741
(3)
−0.31J1 + 2.140J 2 = 430, 708
(4)
J 2 − J3
(1 − ε 2 ) / ε 2 A 2 ( A 2 F21 )−1 ( A 2 F23 )−1
287, 044 − J 2
(1 − 0.6 ) / 0.6
=
J 2 − J1
J − 459.3
+ 2
(0.31)−1 (0.31)−1
2
2
Solving Eqs. (3) and (4) simultaneously, find J1 = 230,491 W/m and J2 = 234,654 W/m , and from Eq.
(1), find
q3 = 7.854 × 10−3 m2 [0.38 ( 459.3 − 230, 491) + 0.62 ( 459.3 − 234, 654 )] W / m 2 = −1827 W.
Hence, 1827 W are transferred from the cavity to the surroundings.
<
PROBLEM 13.84
KNOWN: Circular furnace with prescribed temperatures and emissivities of the lateral and end
surfaces.
FIND: Net radiative heat transfer from each surface.
SCHEMATIC:
ASSUMPTIONS: (1) Surfaces are isothermal and diffuse-gray.
ANALYSIS: To calculate the net radiation heat transfer from each surface, we need to determine its
radiosity. First, evaluate terms which will be required.
E b1 = σ T14 = 1452 W / m 2
A1 = A 2 = π D 2 / 4 = 0.07069 m 2
F12 = F21 = 0.17
E b2 = σ T24 = 3544 W / m 2
A3 = π DL = 0.2827 m 2
F23 = F13 = 0.83
E b3 = σ T34 = 23, 224 W / m 2
The view factor F12 results from Fig. 13.5 with L/ri = 2 and rj/L = 0.5. The radiation balances using
Eq. 13.21, omitting units for convenience, are:
1452 − J1
= 0.07069 × 0.17 ( J1 − J 2 ) + 0.07069 × 0.83 ( J1 − J3 )
A1 :
(1 − 0.4 )
0.4 × 0.07069
−2.500J1 + 0.2550J 2 + 1.2450J 3 = −1452
3544 − J 2
(1 − 0.5 )
A2 :
(1)
= 0.07069 × 0.17 ( J 2 − J1 ) + 0.07069 × 0.83 ( J 2 − J 3 )
0.5 × 0.07069
−0.1700J1 − 2.0000J 2 + 0.8300J 3 = −3544
(2)
23, 224 − J 3
= 0.07069 × 0.83 ( J 3 − J1 ) + 0.07069 × 0.83 ( J 3 − J 2 )
(1 − 0.8 )
A3 :
0.8 × 0.2827
0.05189J1 + 0.05189J 2 − 1.1037J 3 = −23, 224
Solving Eqs. (1) – (3) simultaneously, find
(3)
J1 = 12,877 W / m 2
J 2 = 12, 086 W / m 2
J 3 = 22, 216 W / m 2 .
Using Eq. 13.22, the net radiation heat transfer for each surface follows:
N
(
qi = ∑ Ai Fij Ji − J j
)
j=1
A1 : q1 = 0.07069 × 0.17 (12,877 − 12, 086 ) W + 0.07069 × 0.83 (12,877 − 22, 216 ) W = −538 W
A 2 : q 2 = 0.07069 × 0.17 (12, 086 − 12,877 ) W + 0.07069 × 0.83 (12, 086 − 22, 216 ) W = −603 W
A3 : q 3 = 0.07069 × 0.83 ( 22, 216 − 12,877 ) W + 0.07069 × 0.83 ( 22, 216 − 12, 086 ) W = 1141W
<
<
<
COMMENTS: Note that Σqi = 0. Also, note that J2 < J1 despite the fact that T2 > T1; note the role
emissivity plays in explaining this.
PROBLEM 13.85
KNOWN: Four surface enclosure with all sides of equal area; temperatures of three surfaces are
specified while the fourth is re-radiating.
FIND: Temperature of the re-radiating surface A4.
SCHEMATIC:
ASSUMPTIONS: (1) Surfaces are diffuse-gray, (2) Surfaces have uniform radiosities.
ANALYSIS: To determine the temperature of the re-radiating surface A4, it is necessary to recognize
that J4 = Eb4 = σ T44 and that the Ji (i = 1 to 4) values must be evaluated by simultaneously solving four
radiation balances of the form, Eq. 13.21,
Ji − j j
N
E bi − Ji
(1 − ε i ) / ε i Ai
=∑
(
j=1 Ai Fij
)−1
2
For simplicity, set A1 = A2 = A3 = A4 = 1 m and from symmetry, it follows that all view factors will be
Fij = 1/3. The necessary emissive powers are of the form Ebi = σ T14 .
4
2
Eb1 = σ(700 K) = 13,614 W/m ,
4
2
Eb2 = σ(500 K) = 3544 W/m ,
4
2
Eb3 = σ(300 K) = 459 W/m .
The radiation balances are:
A1:
A2:
A3:
13, 614 − J1
=
(1 − 0.7 ) / 0.7
3
3544 − J 2
(1 − 0.5 ) / 0.5
459 − J 3
(1 − 0.3 ) / 0.3
1
=
1
3
=
1
3
0=
A4:
1
1
3
3
( J1 − J 2 ) + ( J1 − J 3 ) + ( J1 − J 4 ) ; −1.42857J1 + 0.14826J 2
1
1
3
3
1
1
3
3
+ 0.14826J 3 + 0.14826J 4 = −13, 614
( J 2 − J1 ) + ( J 2 − J 3 ) + ( J 2 − J 4 ) 0.33333J1 − 2.00000J 2 + 0.33333J 3 + 0.33333J 4
( J 3 − J1 ) + ( J 3 − J 2 ) + ( J 3 − J 4 ) 0.77778J1 + 0.77778J 2 − 3.33333J 3 + 0.77778J 4
1
3
1
1
3
3
( J 4 − J1 ) + ( J 4 − J 2 ) + ( J 4 − J 3 )
= −3544
= −459
0.33333J1 + 0.33333J 2 + 0.33333J 3 − 1.00000J 4 = 0
Solving this system of equations simultaneously, find
2
J1 = 11,572 W/m ,
2
2
J2 = 6031 W/m ,
J3 = 6088 W/m ,
2
J4 = 7897 W/m .
Since the radiosity and emissive power of the re-radiating surface are equal,
T44 = J 4 / σ
(
T4 = 7897 W / m 2 / 5.67 × 10−8 W / m 2 ⋅ K 4
)
1/ 4
= 611 K.
COMMENTS: Note the values of the radiosities; are their relative values what you would have
expected? Is the value of T4 reasonable?
<
PROBLEM 13.86
KNOWN: A room with electrical heaters embedded in ceiling and floor; one wall is exposed to the
outdoor environment while the other three walls are to be considered as insulated.
FIND: Net radiation heat transfer from each surface.
SCHEMATIC:
ASSUMPTIONS: (1) Diffuse-gray surfaces, (2) Surfaces are isothermal and irradiated uniformly,
(3) Negligible convection effects, (4) A5 = A5A + A5B.
ANALYSIS: To determine the net radiation heat transfer from each surface, find the surface
radiosities using Eq. 13.20.
5
(
qi = ∑ Ai Fij Ji − J j
j=1
)
(1)
To determine the value of Ji, energy balances must be written for each of the five surfaces. For
surfaces 1, 2, and 3, the form is given by Eq. 13.21.
5 J −J
E bi − Ji
i
j
=∑
−
(1 − ε i ) / ε i Ai j=1 A F 1
i ij
(
i = 1, 2, and 3.
)
(2)
For the insulated or adiabatic surfaces, Eq. 13.22 is appropriate with qi = 0; that is
N
qi = ∑
Ji − J j
(
j=1 Ai Fij
−1
)
=0
i = 4 and 5.
(3)
In order to write the energy balances by Eq. (2) and (3), we will need to know view factors. Using
Fig. 13.4 (parallel rectangles) or Fig. 13.5 (perpendicular rectangles) find:
F12 = F21 = 0.39
X/L = 10/4 = 2.5,
Y/L = 6/4 = 1.5
F13 = F14 = 0.19
Z/X = 4/10 = 0.4,
Y/X = 6/10 = 0.6
F34 = F43 = 0.19
X/L = 10/6 = 1.66,
Y/L = 4/6 = 0.67
F24 = F13 = 0.19
Z/X = 4/10 = 0.4,
Y/X = 6/10 = 0.6
Note the use of symmetry in the above relations. Using reciprocity, find,
A
A
60
A
60
F32 = 2 F23 = 2 F13 =
F31 = 1 F13 =
× 0.19 = 0.285;
× 0.19 = 0.285
A3
A3
40
A3
40
F51 =
A1
F15 =
60
× 0.23 = 0.288;
F53 =
A3
F35 =
40
A5
48
A5
48
From the summation view factor relation,
F15 = 1 − F12 − F13 − F14 = 1 − 0.39 − 0.19 − 0.19 = 0.23
× 0.25 = 0.208.
F35 = 1 − F31 − F32 − F34 = 1 − 0.285 − 0.285 − 0.19 = 0.24
Continued …..
PROBLEM 13.86 (Cont.)
4
Using Eq. (2), now write the energy balances for surfaces 1, 2, and 3. (Note Eb = σT ).
J1 − J3
J1 − J 5
544.2 − J1
J1 − J 2
J1 − J 4
=
+
+
+
1 − 0.8 / 0.8 × 60 1/ 60 × 0.39 1/ 60 × 0.19 1/ 60 × 0.19 1/ 60 × 0.23
-1.2500J1 + 0.0975J2 + 0.0475J3 + 0.570J5 = − 544.2
617.2 − J 2
=
J 2 − J1
J 2 − J3
+
+
J2 − J4
J 2 − J5
+
1 − 0.9 / 0.9 × 60 1/ 60 × 0.39 1/ 60 × 0.19 1/ 60 × 0.19 1/ 60 × 0.23
+0.0433J1 – 1.111J2 + 0.02111J3 + 0.02111J4 + 0.02556J5 = − 617.2
390.1 − J3
=
J3 − J1
+
J3 − J 2
+
J3 − J 4
+
J 5 − J1
+
J5 − J 2
+
J5 − J3
+
(5)
J3 − J5
1 − 0.7 / 0.7 × 40 1/ 40 × 0.285 1/ 40 × 0.285 1/ 40 × 0.19 1/ 40 × 0.24
+0.1221J1 + 0.1221J2 – 1.4284J3 + 0.08143J4 + 0.1028J5 = − 390.1
Using Eq. (3), now write the energy balances for surfaces 4 and 5 noting q4 = q5 = 0.
J 4 − J3
J 4 − J5
J 4 − J1
J4 − J2
0=
+
+
+
1/ 40 × 0.285 1/ 40 × 0.285 1/ 40 × 0.19 1/ 40 × 0.24
0.285J1 + 0.285J2 + 0.19J3 – 1.0J4 + 0.24J5 = 0
0=
(4)
(6)
(7)
J5 − J 4
1/ 48 × 0.288 1/ 48 × 0.288 1/ 48 × 0.208 1/ 48 × 0.208
0.288J1 + 0.288J2 + 0.208J3 + 0.208J4 – 0.992J5 = 0
(8)
Note that Eqs. (4) – (8) represent a set of simultaneous equations which can be written in matrix
notation following treatment of Section 13.3.2. That is, [A] [J] = [C] with
 −1.250 0.0975 0.0475 0.0475 0.0575 
 −544.2 
545.1 
0.0433 − 1.111 0.02111 0.02111 0.02556 
 −617.2 
607.9 
A = 0.1221 0.1221 − 1.4284 0.08143 0.1028 
C =  −390.1  J =  441.5 
W / m2
0.285 0.285
 0 
542.3 
0.190 − 1.000 0.240 
0.288 0.288 0.208
 0 
5410 
0.208 − 0.992 
where the Ji were found using a computer routine. The net radiation heat transfer from each of the
surfaces can now be evaluated using Eq. (1).
q1 = A1F12(J1 – J2) + A1F13(J1 – J3) + A1F14(J1 – J4) + A1F15(J1 – J5)
2
q1 = 60 m [0.39(545.1 – 607.9)
2
+0.19(545.1 – 441.5) + 0.19(545.1 – 542.3) + 0.23(545.1 – 541.0)] W/m = - 200 W
<
2
q2 = 60 m [0.39(607.9 – 545.1)
2
+0.19(607.9 – 441.5) + 0.19(607.9 – 542.3) + 0.23(607.9 – 541.0) W/m = 5037 W
<
2
q3 = 40 m [0.285(441.5 – 545.1) + 0.285(441.5 – 607.9)
2
+0.19(441.5 – 542.3) + 0.24(441.5 – 541.0)] W/m = - 4,799 W
Since A4 and A5 are insulated (adiabatic), q4 = q5 = 0.
<
<
COMMENTS: (1) Note that the sum of q1 + q2 + q3 = + 38 W; this indicates a precision of less than
1% resulted from the solution of the equations. (2) The net radiation for the ceiling, A1, is into the
surface. Recognize that the embedded heaters function to offset heat losses to the room air by
convection.
PROBLEM 13.87
KNOWN: Cylindrical cavity closed at bottom with opening at top surface.
FIND: Rate at which radiation passes through cavity opening and effective emissivity for these
conditions: (a) All interior surfaces are black and at 600 K, (b) Bottom surface of the cavity ε = 0.6, T
= 600 K; other surfaces are re-radiating, (c) All surfaces are at 600 K with emissivity 0.6, (d) For the
cavity configurations of parts (b) and (c), compute and plot εe as a function of the interior surface
emissivity over the range 0.6 to 1.0 with all other conditions remaining the same.
SCHEMATIC:
ASSUMPTIONS: (1) Surfaces are opaque, diffuse and gray, (2) Surfaces as subsequently defined
have uniform radiosity, (3) Re-radiating surfaces are adiabatic, and (4) Surroundings are at 0 K so that
T1 = 0 K and ε1 = 1.0.
ANALYSIS: Define the hypothetical surface A1, the cavity opening, having T1 = 0 K for which Eb.1
= J1 = 0. The radiant power passing through the cavity opening A1 will be –q1 due to exchange
within the four-surface enclosure. The effective emissivity of the cavity is defined as the rate of the
radiant power leaving the cavity to that from a blackbody having the same area of the cavity opening
and at the temperature of the inner surface of the cavity. That is,
ε e = −q1 / A1σ T 4
(1)
where T is the cavity surface temperature. Recognizing that the analysis will require knowledge of
view factors, begin by evaluating them now.
For this four-surface enclosure (N=4), N(N-1)/2 = 6 view factors must be directly determined. The
2
remaining N – 6 = 10 can be determined by the summation rules and the reciprocity relations. By
inspection,
(1-4): F11 = 0
F14 = 0
F22 = 0
F44 = 0
Using the view factor equation for coaxial parallel disks, Table 13.2, or Fig. 13.5, evaluate F21,
(5):
F21 = 0.030
with rj/L = 7.5/40 = 0.188, L/ri = 40/15 = 2.67.
Considering the top and bottom surfaces, use the additive rule, Eq. 13.5,
(6):
F24 = F2(1.4) – F21
where F2(1.4) can be evaluated using the coaxial parallel disk relations again
F2(1.4) = 0.111
with r(1.4)/L = 15/40 = 0.375, L/r2 = 40/15 = 2.67
Substituting numerical values, find
F24 = 0.111 – 0.030 = 0.081
Continued …..
(2)
PROBLEM 13.87 (Cont.)
Using the summation rule for each surface, plus appropriate reciprocity relations, the remaining view
factors can be determined. Written as a matrix, the Fij are
0*
0.0300*
0.0413
0*
0.120
0.880
0*
0.167
0.108
0
0.889
0.667
0.892
0.0811*
0.125
0*
The Fij shown with an asterisk were independently determined.
(a) When all the internal surfaces of the cavity are black at 600 K, the cavity opening emits as a black
surface and the effective emissivity is unity. Using Eq. 13.14, the heat rate leaving A1 is
)
(3)
q1 = 5.67 × 10−8 W / m 2 ⋅ K 4 04 − 6004 K 4 × 1.767 × 10−4 m2 × [0.120 + 0.880 + 0] = −1.298 W
<
3
) (
(
q1 = ∑ A1Fijσ T14 − T14 = σ T14 − T 4 A1 [F12 + F13 + F14 ]
i =1
(
)
From Eq. (1), it follows that the effective emissivity must be unity.
<
ε1 = 1
(b) When the bottom surface of the cavity is T2 = 600 K with ε2 = 0.6 and all other surfaces are reradiating, an enclosure analysis to obtain q1 involves use of Eqs. 13.21 and 13.22. The former will be
used on A2 and the latter on the remaining areas.
A2:
E b2 − J 2
(1 − ε 2 ) / ε 2 A2
=
7384 − J 2
(1 − 0.6 ) / 0.6A 2
J 2 − J1
+
1/ A 2 F21
=
J2 − 0
1/ 0.03A 2
J 2 − J3
1/ A 2 F23
+
+
J2 − J4
1/ A 2 F24
J 2 − J3
1/ 0.889A 2
4
+
J2 − J4
(4)
1/ 0.811A 2
2
4
2
where Eb2 = σ T24 = σ(600K) = 7348 W/m and J1 = Eb1 = σ T14 = σ(0) = 0 W/m .
J −J
J −J
J −J
A3:
0= 3 1 + 3 2 + 3 4
1/ A3F31 1/ A3F32 1/ A3F34
0=
A4:
0=
J3 − 0
1/ 0.0413A3
J 4 − J1
1/ A 4 F41
0 = 0+
+
+
J3 − J 2
1/ 0.167A3
J4 − J2
1/ A 4 F42
+
+
J3 − J 4
(5)
1/ 0.125A3
J 4 − J3
1/ A 4 F43
J 4 − J3
J4 − J2
+
1/ 0.108A 4 1/ 0.892A 4
Solving Eqs. (4,5,6) simultaneously, find
J1
J2
0
6450
and the heat rate leaving surface A1 is
(6)
J3
5284
2
J4(W/m )
5378
Continued …..
PROBLEM 13.87 (Cont.)
q1 =
J1 − J 2
J −J
J −J
+ 1 3 + 1 4 = 0.9529 W
1/ A1F12 1/ A1F13 1/ A1F14
(7)
From Eq. (1), the cavity effective emissivity
εe =
−q1
A1σ T24
=
0.9529 W
1.767 × 10−4 m 2σ ( 600 K )
4
<
= 0.734
(c) When all the interior surfaces are at 600K (T2 = T3 = T4 = 600 K) and ε = 0.6 (ε2 = ε3 = ε4 = 0.6),
apply the radiation surfaces energy balances using Eq. 13.21 to A2, A3, and A4
A2:
E b2 − J 2
=
(1 − ε 2 ) / ε 2 A2
7384 − J 2
(1 − 0.6 ) / 0.6A 2
A3:
A4:
E b3 − J 3
(1 − ε 3 ) / ε 3A3
=
E b4 − J 4
(1 − ε 4 ) / ε 4 A 4
J 2 − J1
1/ A 2 F21
=
J2 − 0
1/ 0.03A 2
J 3 − J1
1 / A 3F31
=
+
+
J3 − J1
1/ A 4 F41
J 2 − J3
1/ A 2 F23
+
J2 − J4
1/ A 2 F24
J 2 − J3
1/ 0.889A 2
J3 − J 2
1 / A 3F32
+
+
+
+
J2 − J4
J3 − J 4
1 / A 3F34
J4 − J2
1/ A 4 F42
+
(8)
1/ 0.811A 2
=
J3 − 0
1 / 0.0413A 3
J 4 − J3
1/ A 4 F43
= 0+
+
J3 − J 2
1 / 0.167A 3
J4 − J2
1/ 0.108A 4
+
+
J3 − J 4
(9)
1 / 0.125A 3
J 4 − J3
1/ 0.892A 4
(10)
Solving Eqs. (8, 9, 10) simultaneously, find
J1
0
J2
7192
J3
7164
2
J4 (W/m )
7276
and the heat rate leaving surface A1 is
q1 =
J −J
J −J
+ 1 3 + 1 4 = −1.267 W
1/ A1F12 1/ A1F13 1/ A1F14
J1 − J 2
<
From Eq. (1), the cavity effectiveness is
εe =
−q1
A1σ T24
=
1.267W
1.767 × 10−4 m 2σ ( 600 K )
4
<
= 0.976
(d) For the cavity configurations of parts (b) and (c) and selected cavity depths, εe was computed as a
function of the interior surface emissivity ε = ε2 = ε4 using an IHT model. The model included the
following tools: Radiation-View Factors: Relations and Formulas (Coaxial parallel disks);
Radiation-Radiation Surface Energy Balance Relations. See comment 2 below.
Continued …..
PROBLEM 13.87 (Cont.)
For the cavity configurations of part (b) – re –radiating surfaces A3 and A4 – the εe vs. ε2 plot shows
that for all cavity depths, the effective emissivity increases as the emissivity of the bottom surface, ε2,
increases. Note that even when ε2 = 1, the cavity effective emissivity is always less than unity. Why
must that be so? The effect of increasing the cavity depth is to increase the effective emissivity.
For the cavity configuration of part (c) – all interior surfaces at 600 K and ε = 0.6 – the effective
emissivity increases with increasing interior surface emissivity. In the limit when ε → 1, εe → 1 as
expected, and the cavity performs as an isothermal enclosure. The effective emissivity increases with
increasing cavity depth.
COMMENTS: (1) This arrangement of a cavity, referred to as a cylindrical cavity with a lid (A4), is
widely used for radiometric applications to calibrate radiometers, radiation thermometers, and heat
flux gages. The effective emissivity can be improved by constructing the cavity with a conical bottom
surface (rather than a flat bottom). Why do you think this is so?
(2) The IHT model used to generate the part (b) graphical results is quite extensive. It is good practice
to build the code in pieces, beginning with evaluation of the view factors. To avoid divide-by-zero
-20
errors, use small values for variables which are zero, such F22 = 1e . Also, set unity emissivity
values as 0.9999 rather than 1.0. The set of equations is very stiff, especially because of the reradiating surface where T3 and T4 are unknowns. You should provide Initial Guess minimum values
for T3 and T4 (> 0, positive) and unknown radiosities (> 0, positive).
PROBLEM 13.88
KNOWN: Cylindrical furnace of diameter D = 90 mm and overall length L = 180 mm. Heating
elements maintain the refractory liming (ε = 0.8) of section (1), L1 = 135 mm, at T1 = 800°C. The
bottom (2) and upper (3) sections are refractory lined, but are insulated. Furnace operates in a
spacecraft environment.
FIND: Power required to maintain the furnace operating conditions with the surroundings at 23°C.
SCHEMATIC:
ASSUMPTIONS: (1) All surfaces are diffuse gray, (2) Uniform radiosity over the sections 1, 2, and
3, and (3) Negligible convection effects.
ANALYSIS: By defining the furnace opening as the hypothetical area A4, the furnace can be
represented as a four-surface enclosure as illustrated above. The power required to maintain A1 at T1
is q1, the net radiation leaving A1. To obtain q1 following the methodology of Section 13.2.2, we
must determine the radiosity at all surfaces by simultaneously solving the radiation energy balance
equations for each surface which will be of the form, Eqs. 13.20 or 13.21.
q1 =
N J −J
j
j
E bi − Ji
(1 − ε i ) / ε i Ai
=∑
j=1
(1,2)
1/ Ai Fij
Since ε4 = 1, J4 = Eb4, so we only need to perform three energy balances, for A1, A2, and A3,
respectively
E b1 − J1
J −J
J −J
J −J
A1:
= 1 2 + 1 3 + 1 4
(1 − ε1 ) / ε1A1 1/ A1F12 1/ A1F13 1/ A1F14
A2:
0=
A3:
0=
J 2 − J1
1/ A 2 F21
J3 − J1
1/ A3F31
+
+
J 2 − J3
1/ A 2 F23
J3 − J 2
1/ A3F32
+
+
J2 − J4
(3)
(4)
1/ A 2 F24
J3 − J 4
(5)
1/ A3F34
Note that q2 = q3 = 0 since the surfaces are insulated (adiabatic). Recognize that in the above
equation set, there are three equations and three unknowns: J1, J2, and J3. From knowledge of J1, q1
2
2
can be determined using Eq. (1). Next we need to evaluate the view factors. There are N = 4 = 16
view factors and N(N – 1)/2 = 6 must be independently evaluated, while the remaining can be
determined by the summation rule and appropriate reciprocity relations. The six independently
determined Fij are:
By inspection: (1) F22 = 0
(2) F44 = 0
Coaxial parallel disks: From Fig. 13.5 or Table 13.5,
Continued …..
PROBLEM 13.88 (Cont.)

2
F24 = 0.5 S − S2 − 4 ( r4 / r2 ) 
1/ 2 
(3)

 
 

1/
2


2
F24 = 0.5 18 − 182 − 4 (1)   = 0.05573

 

2
S =1+
1+ R4
2
= 1+
1 + 0.250
R2
0.250
2
= 18.00
2
R 2 = r2 / L = 45 / 180 = 0.250
R 4 = r4 / L = 0.250
Enclosure 1-2-2′: from the summation rule for A2,
(4)
F21 = 1 – F22’ = 1 – 0.09167 = 0.9083
where F22′ can be evaluated from the coaxial parallel disk relation, Table 13.5. For these surfaces, R2
= r2/L1 = 45/135 = 0.333, R2′ = r2/L1 = 0.333, and S = 11.00. From the summation rule for A1,
(5)
F11 = 1 – F12 – F12′ = 1 – 0.1514 – 0.1514 = 0.6972
and by symmetry F12 = F12′ and using reciprocity
F12 = A 2 F21 / A1 = [π ( 0.090m )( 2 / 4 )] × 0.9083 / π × 0.090m × 0.135m = 0.1514
Enclosure 2′ -3-4: from the summation rule for A4,
(6)
F43 = 1 – F42′ - F44 = 1 – 0.3820 – 0 = 0.6180
where F44 = 0 and using the coaxial parallel disk relation from Table 13.5, with R4 = r4/L2 = 45/45 =
1, R2′ = r2/L2 = 1, and S = 3.
The View Factors: Using summation rules and appropriate reciprocity relations, the remaining 10
view factors can be evaluated. Written in matrix form, the Fij are
0.6972*
0.1514
0.09704
0.05438
0.9083*
0*
0.03597
0.05573*
0.2911
0.01798
0.3819
0.3090
0.3262
0.05573
0.6180*
0*
The Fij shown with an asterisk were independently determined.
From knowledge of the relevant view factors, the energy balances, Eqs. (3, 4, 5), can be solved
simultaneously to obtain the radiosities,
J1 = 73, 084 W / m 2
J 2 = 67, 723 W / m 2
J 3 = 36, 609 W / m 2
The net heat rate leaving A1 can be evaluated using Eq. (1) written as
q1 =
E b1 − J1
(1 − ε1 ) / ε1A1
=
(75,159 − 73, 084 ) W / m2
(1 − 0.8 ) / 0.8 × 0.03817 m2
4
= 317 W
<
2
where Eb1 = σ T14 = σ(800 + 273K) = 75,159 W/m and A1 = πDL1 = π × 0.090m × 0.135m =
2
0.03817 m .
COMMENTS: (1) Recognize the importance of defining the furnace opening as the hypothetical
area A4 which completes the four-surface enclosure representing the furnace. The temperature of A4
is that of the surroundings and its emissivity is unity since it absorbs all radiation incident on it. (2)
To obtain the view factor matrix, we used the IHT Tool, Radiation, View Factor Relations, which
permits you to specify the independently determined Fij and the tool will calculate the remaining ones.
PROBLEM 13.89
KNOWN: Rapid thermal processing (RTP) tool consisting of a lamp bank to heat a silicon wafer
with irradiation onto its front side. The backside of the wafer (1) is the top of a cylindrical enclosure
whose lateral (2) and bottom (3) surfaces are water cooled. An aperture (4) on the bottom surface
provides for optical access to the wafer.
FIND: (a) Lamp irradiation, Glamp, required to maintain the wafer at 1300 K; heat removal rate by
the cooling coil, and (b) Compute and plot the fractional difference (Eb1 – J1)/Eb1 as a function of the
enclosure aspect ratio, L/D, for the range 0.5 ≤ L/D ≤ 2.5 with D = 300 mm fixed for wafer
emissivities of ε1 = 0.75, 0.8, and 0.85; how sensitive is this parameter to the enclosure surface
emissivity, ε2 = ε3.
SCHEMATIC:
ASSUMPTIONS: (1) Enclosure surfaces are diffuse, gray, (2) Uniform radiosity over the enclosure
surfaces, (3) No heat losses from the top side of the wafer.
ANALYSIS: (a) The wafer-cylinder system can be represented as a four-surface enclosure. The
aperture forms a hypothetical surface, A4, at T4 = T2 = T3 = 300 K with emissivity ε4 = 1 since it
absorbs all radiation incident on it. From an energy balance on the wafer, the absorbed lamp
irradiation on the front side of the wafer, αwGlamp, will be equal to the net radiation leaving the backside (enclosure-side) of the wafer, q1. To obtain q1, following the methodology of Section 13.2.2, we
must determine the radiosity of all the enclosure surfaces by simultaneously solving the radiation
energy balance equations for each surface, which will be of the form, Eqs. 13.20 or 13.21.
qi =
E bi − Ji
N J −J
i
j
(1 − εi ) / εi Ai
=∑
j= l
(1,2)
1/ Ai Fij
Since ε4 = 1, J4 = Eb4, we only need to perform three energy balances, for A1, A2 and A3,
respectively,
E b1 − J1
J −J
J −J
J −J
= 1 2 + 1 3 + 1 4
A1:
(1 − ε1 ) / A1 1/ A1F12 1/ A1F13 1/ A1F14
A2:
E b2 − J1
(1 − ε 2 ) / A2
=
J 2 − J1
1/ A 2 F21
+
J 2 − J3
1/ A 2 F23
+
J2 − J4
(3)
(4)
1/ A 2 F24
Continued …..
PROBLEM 13.89 (Cont.)
A3:
E b3 − J3
(1 − ε 3 ) / A3
=
J3 − J1
1/ A3F31
+
J3 − J 2
1/ A3F32
+
J3 − J 4
(5)
1/ A3F34
Recognize that in the above equation set, there are three equations and three unknowns: J1, J2, and J3.
From knowledge of the radiosities, the desired heat rate can be determined using Eq. (1). The
required lamp irradiation,
E b1 − J1
α w G lamp A1 = q1 =
(6)
(1 − ε1 ) / ε1A1
and the heat removal rate by the cooling coil, qcoil, on surfaces A2 and A3, is
q coil = − ( q 2 + q3 )
(7)
where the net radiation leaving A2 and A3 are, from Eq. (1),
E b2 − J 2
E b3 − J 3
q2 =
q1 =
(1 − ε 2 ) / ε 2 A 2
(1 − ε 3 ) / ε3A3
(8,9)
The surface areas are expressed as
A1 = π D12 / 4 = 0.07069 m 2
(
)
A3 = π D12 − D 42 = 0.06998 m 2
2
A 2 = π D1L = 0.2827
(10,11)
A 2 = π D 42 / 4 = 0.0007069 m 2
(12,13)
2
Next evaluate the view factors. There are N = 4 = 16 and N(N – 1)/2 = 6 must be independently
evaluated, and the remaining can be determined by summation rules and reciprocity relations. The six
independently determined Fij are:
By inspection: (1) F11 = 0
(2) F33 = 0
(3) F44 = 0
(4) F34 = 0
Coaxial parallel disks: from Fig. 13.5 or Table 13.5,

2
F14 = 0.5 S − S2 − 4 ( r4 / r1 ) 
1/ 2 
(5)

 
 

1/ 2 

2
F14 = 0.5 5.01 − 5.012 − 4 (15 /150 )   = 0.001997

 

S = 1+
1 + R 24
R12
= 1+
1 + 0.052
0.52
= 5.010
R1 = r1 / L = 150 / 300 = 0.5
R 4 = 15 / 300 = 0.05
Coaxial parallel disks: from the composite surface rule, Eq. 13.5,
(6)
F13 = F1(3,4) – F14 = 0.17157 – 0.01997 = 0.1696
where F1(3,4) can be evaluated from the coaxial parallel disk relation, Table 13.5. For these surfaces,
R1 = r1/L = 150/300 = 0.5, R(3,4) = r3/L = 150/300 = 0.5, and S = 6.000.
The view factors: Using summation rules and reciprocity relations, the remaining 10 view factors can
be evaluated. Written in matrix form, the Fij are
Continued …..
PROBLEM 13.89 (Cont.)
0*
0.2071
0.1713
0.1997
0.8284
0.5858
0.8287
0.8003
0.1696
0.2051
0*
0*
0.001997*
0.002001
0*
0*
The Fij shown with an asterisk were independently determined.
From knowledge of the relevant view factors, the energy balances, Eqs. (3, 4, 5) can be solved
simultaneously to obtain the radiosities,
2
J1
J2
J3
J4 (W/m )
5
5
5
1.514×10
1.097×10
1.087×10
576.8
From Eqs. (6) and (7), the required lamp irradiation and cooling-coil heat removal rate are
G lamp = 52, 650 W / m 2
<
q coil = 2.89 kW
(b) If the enclosure were perfectly reflecting, the radiosity of the wafer, J1, would be equal to its
5
2
blackbody emissive power. For the conditions of part (a), J1 = 1.514 × 10 W/m and Eb1 = 1.619 ×
5
2
10 W/m . As such, the radiosity would be independent of εw thereby minimizing effects due to
variation of that property from wafer-to-wafer. Using the foregoing analysis in the IHT workspace
(see Comment 1 below), the fractional difference, (Eb1 – J1)/Eb1, was computed and plotted as a
function of L/D, the aspect ratio of the enclosure.
Note that as the aspect ratio increases, the fractional difference between the wafer blackbody emissive
power and the radiosity increases. As the enclosure gets larger, (L/D increases), more power supplied
to the wafer is transferred to the water-cooled walls. For any L/D condition, the effect of increasing
the wafer emissivity is to reduce the fractional difference. That is, as εw increases, the radiosity
increases. The lowest curve on the above plot corresponds to the condition ε2 = ε3 = 0.03, rather than
0.07 as used in the εw parameter study. The effect of reducing ε2 is substantial, nearly halving the
fractional difference. We conclude that the “best” cavity is one with a low aspect ratio and low
emissivity (high reflectivity) enclosure walls.
COMMENTS: The IHT model developed to perform the foregoing analysis is shown below. Since
the model utilizes several IHT Tools, good practice suggests the code be built in stages. In the first
stage, the view factors were evaluated; the bottom portion of the code. Note that you must set the Fij
which
Continued …..
PROBLEM 13.89 (Cont.)
are zero to a value such as 1e-20 rather than 0. In the second stage, the enclosure exchange analysis
was added to the code to obtain the radiosities and required heat rate. Finally, the equations necessary
to obtain the fractional difference and perform the parameter analysis were added.
Continued …..
PROBLEM 13.89 (Cont.)
PROBLEM 13.90
KNOWN: Observation cabin located in a hot-strip mill directly over the line; cabin floor (f) exposed
to steel strip (ss) at Tss = 920°C and to mill surroundings at Tsur = 80°C.
FIND: Coolant system heat removal rate required to maintain the cabin floor at Tf = 50°C for the
following conditions: (a) when the floor is directly exposed to the steel strip and (b) when a radiation
shield (s) εs = 0.10 is installed between the floor and the strip.
SCHEMATIC:
ASSUMPTIONS: (1) Cabin floor (f) or shield (s), steel strip (ss), and mill surroundings (sur) form a
three-surface, diffuse-gray enclosure, (2) Surfaces with uniform radiosities, (3) Mill surroundings are
isothermal, black, (4) Floor-shield configuration treated as infinite parallel planes, and (5) Negligible
convection heat transfer to the cabin floor.
ANALYSIS: A gray-diffuse, three-surface enclosure is formed by the cabin floor (f) (or radiation
shield, s), steel strip (ss), and the mill surroundings (sur). The heat removal rate required to maintain
the cabin floor at Tf = 50°C is equal to - qf (or, -qs), where qf or qs is the net radiation leaving the
floor or shield. The schematic below represents the details of the surface energy balance on the floor
and shield for the conditions without the shield (floor exposed) and with the shield (floor shielded
from strip).
(a) Without the shield. Radiation surface energy balances, Eq. 13.21, are written for the floor (f) and
steel strip (ss) surfaces to determine their radiosities.
E b,f − J f
J f − E b,sur
J f − J ss
=
+
1 − ε f / ε f A f 1/ A f Ff − ss 1/ A f Ff − sur
(1)
$
E b,ss − J ss
1 − ε ss $ / ε ss A ss
=
J ss − E b,sur
J ss − J f
+
1/ A ss Fss− f 1/ A ss Fss-sur
(2)
Since the surroundings (sur) are black, Jsur = Eb,sur. The blackbody emissive powers are expressed as
4
-8
2 4
Eb = σ T where σ = 5.67 × 10 W/m ⋅K . The net radiation leaving the floor, Eq. 13.20, is
$
!
q f = A f Ff − ss J f − J ss + A f Ff − sur J f − E b,sur
&
(3)
Continued …..
PROBLEM 13.90 (Cont.)
The required view factors for the analysis are contained in the summation rule for the areas Af and
Ass,
Ff − ss + Ff − sur = 1
Fss− f + Fss− sur = 1
(4,5)
Ff-ss can be evaluated from Fig. 13.4 (Table 13.2) for the aligned parallel rectangles geometry. By
symmetry, Fss-f = Ff-ss, and with the summation rule, all the view factors are determined. Using the
foregoing relations in the IHT workspace, the following results were obtained:
.
Ff − ss = 01864
J f = 7959 W / m2
Ff − sur = 0.8136
J ss = 97.96 kW / m2
and the heat removal rate required of the coolant system (cs) is
<
q cs = − q f = 413
. kW
(b) With the shield. Radiation surface energy balances are written for the shield (s) and steel strip (ss)
to determine their radiosities.
E b,s − J s
1 − ε s $ / ε s As
=
E b,ss − J ss
1 − ε ss $ / ε ss Ass
J s − E b,sur
J s − J ss
+
1 / A s Fs− ss 1/ A s Fs− sur
=
(6)
J ss − E b,sur
J ss − J s
+
1/ A ss Fss− s 1/ A ss Fss− sur
(7)
The net radiation leaving the shield is
$
!
q s = A ss Fss− s J ss − J s + A ss Fss− sur J ss − E b,sur
&
(8)
Since the temperature of the shield is unknown, an additional relation is required. The heat transfer
from the shield (s) to the floor (f) - the coolant heat removal rate - is
−q s =
σ Ts4 − Tf4 A s
"
'
(9)
1 − 1/ ε s − 1/ ε f
where the floor-shield configuration is that of infinite parallel planes, Eq. 13.24. Using the foregoing
relations in the IHT workspace, with appropriate view factors from part (a), the following results were
obtained
J s = 18.13 kW / m2
J ss = 98.20 kW / m2
Ts = 377$ C
and the heat removal rate required of the coolant system is
q cs = − q s = 6.55 kW
<
COMMENTS: The effect of the shield is to reduce the coolant system heat rate by a factor of nearly
seven. Maintaining the integrity of the reflecting shield (εs = 0.10) operating at nearly 400°C in the
mill environment to prevent corrosion or oxidation may be necessary.
PROBLEM 13.91
KNOWN: Opaque, diffuse-gray plate with ε1 = 0.8 is at T1 = 400 K at a particular instant. The
bottom surface of the plate is subjected to radiative exchange with a furnace. The top surface is
subjected to ambient air and large surroundings.
FIND: (a) Net radiative heat transfer to the bottom surface of the plate for T1 = 400 K, (b) Change in
temperature of the plate with time, dT1/dt, and (c) Compute and plot dT1/dt as a function of T1 for the
range 350 ≤ T1 ≤ 900 K; determine the steady-state temperature of the plate.
SCHEMATIC:
ASSUMPTIONS: (1) Plate is opaque, diffuse-gray and isothermal, (2) Furnace bottom behaves as a
blackbody while sides are perfectly insulated, (3) Surroundings are large compared to the plate and
behave as a blackbody.
ANALYSIS: (a) Recognize that the plate (A1), furnace bottom (A2) and furnace side walls (AR)
form a three-surface enclosure with one surface being re-radiating. The net radiative heat transfer
leaving A1 follows from Eq. 13.30 written as
E b1 − E b2
1 − ε2
(1)
+
q1 =
1 − ε1
1
−1 ε 2 A 2
+
+ (1/ A1F1R + 1/ A 2 F2R )
ε1A1 A1F12
From Fig. 13.4 with X/L = 0.2/0.2 = 1 and Y/L = 0.2/0.2 = 1, it follows that F12 = 0.2 and F1R = 1 –
F12 = 1 – 0.2 = 0.8. Hence, with F1R = F2R (by symmetry) and ε2 = 1.
q1 =
(
)
5.67 × 10−8 W / m 2 ⋅ K 4 400 4 − 10004 K 4
1 − 0.8
0.8 × 0.4m 2
(
0.4m 2 × 0.20 + 2 / 0.04m 2 × 0.8
<
= −1153 W
1
+
)
−1
It follows the net radiative exchange to the plate is, qrad⋅f = 1153 W.
(b) Perform now an energy balance on the plate written as
E in − E out = E st
q rad.f − q conv − q rad,sur = Mc p
dT1
(
dt
)
4
= Mc p
q rad.f − hAs ( T1 − T∞ ) − ε1A1σ T14 − Tsur
dT1
. (2)
dt
Substituting numerical values and rearranging to obtain dT/dt, find
Continued …..
PROBLEM 13.91 (Cont.)
dT1
dt
=
1
 +1153W − 25 W / m 2 ⋅ K × 0.04 m 2 ( 400 − 300 ) K
2 kg × 900 J / kg ⋅ K 
(
)
−0.8 × 0.04m 2 × 5.67 × 10 −8 W / m 2 ⋅ K 4 4004 − 3004 K 4 

<
dT1
= 0.57 K / s.
dt
(c) With Eqs. (1) and (2) in the IHT workspace, dT1/dt was computed and plotted as a function of T1.
When T1 = 400 K, the condition of part (b), we found dT1/dt = 0.57 K/s which indicates the plate
temperature is increasing with time. For T1 = 900 K, dT1/dt is a negative value indicating the plate
temperature will decrease with time. The steady-state condition corresponds to dT1/dt = 0 for which
T1,ss = 715 K
<
COMMENTS: Using the IHT Radiation Tools – Radiation Exchange Analysis, Three Surface
Enclosure with Re-radiating Surface and View Factors, Aligned Parallel Rectangle – the above
analysis can be performed. A copy of the workspace follows:
// Energy Balance on the Plate, Equation 2:
M * cp * dTdt = - q1 – h * A1 * (T1 – Tinf) – eps1 * A1 * sigma * (T1^4 – Tsur^4)
/* Radiation Tool – Radiation Exchange Analysis,
Three-Surface Enclosure with Reradiating Surface: */
/* For the three-surface enclosure A1, A2 and the reradiating surface AR, the net rate of radiation transfer
from the surface A1 to surface A2 is */
q1 = (Eb1 – Eb2) / ( (1 – eps1) / (eps1 * A1) + 1 / (A1 * F12 + 1/(1/(A1 * F1R) + 1/(A2 * F2R))) + (1 –
eps2) / (eps2 * A2)) // Eq 13.30
/* The net rate of radiation transfer from surface A2 to surface A1 is */
q2 = -q1
/* From a radiation energy balance on AR, */
(JR – J1) / (1/(AR * FR1)) + (JR – J2) / (1/(AR *FR2)) = 0
// Eq 13.31
/* where the radiosities J1 and J2 are determined from the radiation rate equations expressed in terms of
the surface resistances, Eq 13.22 */
q1 = (Eb1 – J1) / ((1 – eps1) / (eps1 * A1))
q2 = (Eb2 – J2) / ((1-eps2) / (eps2 * A2))
// The blackbody emissive powers for A1 and A2 are
Eb1 = sigma * T1^4
Eb2 = sigma * T2^4
// For the reradiating surface,
JR = EbR
Continued …..
PROBLEM 13.91 (Cont.)
EbR = sigma *TR^4
sigma = 5.67E-8
// Stefan-Boltzmann constant, W/m^2⋅K^4
// Radiation Tool – View Factor:
/* The view factor, F12, for aligned parallel rectangles, is */
F12 = Fij_APR(Xbar, Ybar)
// where
Xbar = X/L
Ybar = Y/L
// See Table 13.2 for schematic of this three-dimensional geometry.
// View Factors Relations:
F1R = 1 – F12
FR1 = F1R * A1 / AR
FR2 = FR1
A1 = X * Y
A2 = X * Y
AR = 2 * (X * Z + Y * Z)
Z=L
F2R = F1R
// Assigned Variables:
T1 = 400
eps1 = 0.8
T2 = 1000
eps2 = 0.9999
X = 0.2
Y = 0.2
L = 0.2
M=2
cp = 900
h = 25
Tinf = 300
Tsur = 300
// Plate temperature, K
// Plate emissivity
// Bottom temperature, K
// Bottom surface emissivity
// Plate dimension, m
// Plate dimension, m
// Plate separation distance, m
// Mass, kg
// Specific heat, J/kg.K,
// Convection coefficient, W/m^2.K
// Ambient air temperature, K
// Surroundings temperature, K
PROBLEM 13.92
KNOWN: Tool for processing silicon wafer within a vacuum chamber with cooled walls. Thin wafer is
radiatively coupled on its back side to a chuck which is electrically heated. The top side is irradiated by
an ion beam flux and experiences convection with the process gas and radioactive exchange with the ionbeam grid control surface and the chamber walls.
FIND: (a) Show control surfaces and all relevant processes on a schematic of the wafer, and (b) Perform
an energy balance on the wafer and determine the chuck temperature Tc required to maintain the
prescribed conditions.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Wafer is diffuse, gray, (3) Separation distance
between the wafer and chuck is much smaller than the wafer and chuck diameters, (4) Negligible
convection in the gap between the wafer and chuck; convection occurs on the wafer top surface with the
process gas, (5) Surfaces forming the three-surface enclosure – wafer (εw = 0.8), grid (εg = 1), and
chamber walls (εc = 1) have uniform radiosity and are diffuse, gray, and (6) the chuck surface is black.
ANALYSIS: (a) The wafer is shown schematically above in relation to the key components of the tool:
the ion beam generator, the grid which is used to control the ion beam flux, q′′ib , the chuck which aids in
controlling the wafer temperature and the process gas flowing over the wafer top surface. The schematic
below shows the control surfaces on the top and back surfaces of the wafer along with the relevant
thermal processes: qcv, convection between the wafer and process gas; qa, applied heat source due to
absorption of the ion beam flux, q′′ib ; q1,top , net radiation leaving the top surface of the wafer (1) which
is part of the three-surface enclosure – grid (2) and chamber walls (3), and; q1,bac, net radiation leaving
the backside of the wafer (w) which is part of a two-surface enclosure formed with the chuck (c).
<
Continued …..
PROBLEM 13.92 (Cont.)
(b) Referring to the schematic and the identified thermal processes, the energy balance on the wafer has
the form,
E in − E out = 0
−q cv + q a − q1,bac − q1,top = 0
(1)
where each of the processes are evaluated as follows:
Convection with the process gas: with A w = π D42 = π ( 0.200m ) / 4 = 0.03142 m 2 ,
2
q cv = hA w ( Tw − T∞ ) = 10 W / m 2 × 0.03142m2 × ( 700 − 500 ) K = 62.84 W
(2)
Applied heat source – ion beam:
q a = q′′ib A w = 600 W / m 2 × 0.3142m 2 = 18.85 W
(3)
Net radiation heat rate, back side; enclosure (w,c): for the two-surface enclosure comprised of the back
side of the wafer (w) and the chuck, (c), Eq. 13.28, yields
q1,bac =
(
4
σ Tw
− Tc4
)
(1 − ε w ) / ε w A w + 1/ A w Fwc + (1 − ε c ) / ε c Ac
and since the wafer-chuck approximate large parallel plates, Fwc = 1, and since the chuck is black, εc = 1,
q1,bac =
q1,bac =
)
(
4
σ Tw
− Tc4 A w
(4)
(1 − ε w ) / ε w + 1
(
)
0.03142m 2 × σ 7004 − Tc4 K 4
(1 − 0.6 ) / 0.6 + 1
(
= 1.069 × 10−9 700 4 − Tc4
)
Net radiation heat rate, top surface; enclosure (1, 2, 3): from the surface energy balance on A1, Eq.
13.20.
E b1 − J1
(5)
q1,top =
(1 − ε1 ) / ε1A1
where ε1 = εw, A1 = Aw, Eb1 = σ T14 and the radiosity can be evaluated by an enclosure analysis
following the methodology of Section 13.2.2. From the energy balance, Eq. 13.21,
E b1 − J1
J −J
J −J
= 1 2 + 1 3
(6)
(1 − ε1 ) / ε1A1 1/ A1F12 1/ A1F13
4
since both surfaces are black (εg = εvc = 1). The view factor
where J2 = Eb2 = σ Tg4 and J3 = Eb3 = σ Tvc
F12 can be computed from the relation for coaxial parallel disks, Table 13.5.

2
F12 = 0.5 S − S2 − 4 ( r2 / r1 ) 

1/ 2 

S = 1+

1 + R 22
R12
= 1+
1 + 0.52
0.52
1/ 2 

2
2
 = 0.5 6.0 − 6.0 − 4 (1)   = 0.1716



= 6.00
Continued …..
PROBLEM 13.92 (Cont.)
R1 = r1 / L = 100 / 200 = 0.5
R 4 = r4 / L = 0.5
The view factor F13 follows from the summation rule applied to A1,
F13 = 1 − F12 = 1 − 0.1716 = 0.8284
Substituting numerical values into Eq. (6), with T1 = Tw = 700 K, T2 = Tg = 500 K, and T3 = Tvc =
300 K, find J1,
σ T14 − J1
(1 − ε1 ) / ε1A1
=
J1 − σ Tg4
1/ F12
+
4
J1 − σ Tvc
(7)
1/ F13
J1 = 8564 W / m 2
4
= 13, 614 W / m 2 and A1 = A w ,
Using Eq. (5), find q1,top with E b2 = σ Tw
q1,top =
(13, 614 − 8564 ) W / m2
(1 − 0.6 ) / 0.6 × 0.03142m 2
(
)
= 238 W
Evaluating Tc from the energy balance on the wafer, Eq. (1), and substituting appropriate expressions for
each of the processes, find
(
)
−62.84 W / m 2 + 18.85W − 1.069 × 10−9 700 4 − Tc4 − 238 W = 0
<
Tc = 842.5 K
From Eq. (4), with Tc = 815 K, the electrical power required to maintain the chuck is
(
)
Pc = −q1,bac = 1.069 × 10 −9 7004 − 842.5 = 282 W
COMMENTS: Recognize that the method of analysis is centered about an energy balance on the wafer.
Identifying the processes and representing them on the energy balance schematic is a vital step in
developing the strategy for a solution. This methodology introduced in Section 1.3.3 becomes important,
if not essential, in analyzing complicated physical systems.
PROBLEM 13.93
KNOWN: Ice rink with prescribed ice, rink air, wall, ceiling and outdoor air conditions.
FIND: (a) Temperature of the ceiling, Tc, having an emissivity of 0.05 (highly reflective panels) or
0.94 (painted panels); determine whether condensation will occur for either or both ceiling panel
types if the relative humidity of the rink air is 70%, and (b) Calculate and plot the ceiling temperature
as a function of ceiling insulation thickness for 0.1 ≤ t ≤ 1 m, identify conditions for which
condensation will occur on the ceiling.
SCHEMATIC:
ASSUMPTIONS: (1) Rink comprised of the ice, walls and ceiling approximates a three-surface,
diffuse-gray enclosure, (2) Surfaces have uniform radiosities, (3) Ice surface and walls are black, (4)
Panels are diffuse-gray, and (5) Thermal resistance for convection on the outdoor side of the ceiling is
negligible compared to the conduction thermal resistance of the ceiling insulation.
PROPERTIES: Psychometric chart (Atmospheric pressure; dry bulb temperature, Tdb = T∞,i =
15°C; relative humidity, RH = 70%): Dew point temperature, Tdp = 9.4°C.
ANALYSIS: The energy balance on the ceiling illustrated in the schematic below has the form
E in − E out = 0
− q o − q conv,c − q rad,c = 0
(1)
where the rate equations for each process are
(
)
q o = Tc − T∞,o / R cond
!
q conv,c = h A c Tc − T∞,i
R cond = t / kAc
(2,3)
&
(4)
$
$
$
q rad,c = ε E b Tc A c − α A w Fwc E b Tw − α A i Fic E b Ti
4
-8
2
(5)
4
The blackbody emissive powers are Eb = σ T where σ = 5.67 × 10 W/m ⋅K . Since the ceiling
panels are diffuse-gray, α = ε. The view factors required of Eq. (5): determine Fic (ice to ceiling)
from Table 13.2 (Fig. 13.5) for parallel, coaxial disks
Fic = 0.672
and Fwc (wall to ceiling) from the summation rule on the ice (i) and the reciprocity rule,
Fic + Fiw = 1
Fiw = Fcw (symmetry)
Fcw = 1 − Fic
Fwc = A c / A w Fcw = A c / A w 1 − Fic = 0.410
$
$
$
Continued …..
PROBLEM 13.93 (Cont.)
2
where Ac = π D /4 and Aw = π DL.
Using the foregoing energy balance, Eq. (1), and the rate equations, Eqs. (2-5), the ceiling temperature
is calculated using radiative properties for the two panel types,
ε
Tc (°C)
Reflective
0.05
14.0
Paint
0.94
8.6
Ceiling panel
Tc < Tdp
<
The dew point is 9.4°C corresponding to a relative humidity of 70% with (dry bulb) air temperature of
15°C. Condensation will occur on the painted panel since Tc < Tdp.
(b) The equations required of the analysis above were solved using IHT. The analysis is extended to
calculate the ceiling temperatures for a range of insulation thickness and the results plotted below.
Ceiling temperature, Tc (C)
15
10
5
0
0.2
0.4
0.6
0.8
1
Ceiling insulation thickness, t (m)
Painted ceiling, epsc = 0.94
Reflective panel, epsc = 0.05
For the reflective panel (ε = 0.05), the ceiling surface temperature is considerably above the dew
point. Therefore, condensation will not occur for the range of insulation thickness shown. For the
painted panel (ε = 0.94), the ceiling surface temperature is always below the dew point. We expect
condensation to occur for the range of insulation thickness shown.
COMMENTS: From the analysis, recognize that the radiative exchange between the ice and the
ceiling is the dominant process for influencing the ceiling temperature. With the reflective panel, the
rate is reduced nearly 20 times that with the painted panel. With the painted panel ceiling, for most of
the conditions likely to exist in the rink, condensation will occur.
PROBLEM 13.94
KNOWN: Diameter, temperature and emissivity of boiler tube. Thermal conductivity and emissivity of
ash deposit. Convection coefficient and temperature of gas flow over the tube. Temperature of
surroundings.
FIND: (a) Rate of heat transfer to tube without ash deposit, (b) Rate of heat transfer with an ash deposit
of diameter Dd = 0.06 m, (c) Effect of deposit diameter and convection coefficient on heat rate and
contributions due to convection and radiation.
SCHEMATIC:
ASSUMPTIONS: (1) Diffuse/gray surface behavior, (2) Surroundings form a large enclosure about the
tube and may be approximated as a blackbody, (3) One-dimensional conduction in ash, (4) Steady-state.
ANALYSIS: (a) Without an ash deposit, the heat rate per unit tube length may be calculated directly.
(
4
− Tt4
q′ = hπ D t ( T∞ − Tt ) + ε tσπ D t Tsur
)
q ′ = 100 W / m ⋅ K (π ) 0.05 m (1800 − 600 ) K + 0.8 × 5.67 × 10
2
−8
2
W/m ⋅K
4
(π )( 0.05 m )
(1500
4
− 600
4
q′ = (18,850 + 35,150 ) W / m = 54, 000 W / m
(b) Performing an energy balance for a control surface about the outer surface of the ash deposit,
q′conv + q′rad = q′cond , or
)
(
4
− Td4 =
hπ Dd (T∞ − Td ) + ε dσπ Dd Tsur
)K
<
2π k ( Td − Tt )
ln ( Dd / D t )
Hence, canceling π and considering an ash deposit for which Dd = 0.06 m,
(
)
100 W / m 2 ⋅ K ( 0.06 m )(1800 − Td ) K + 0.9 × 5.67 × 10−8 W / m 2 ⋅ K 4 ( 0.06 m ) 15004 − Td4 K 4
=
2 (1 W / m ⋅ K )( Td − 600 ) K
ln ( 0.06 / 0.05 )
A trial-and-error solution yields Td ≈ 1346 K, from which it follows that
(
4
− Td4
q′ = hπ Dd ( T∞ − Td ) + ε dσπ Dd Tsur
)
q ′ = 100 W / m ⋅ K (π ) 0.06 m (1800 − 1346 ) K + 0.9 × 5.67 × 10
2
−8
2
W /m ⋅K
4
(π ) 0.06 m
(1500
4
− 1346
Continued …..
4
)K
4
PROBLEM 13.94 (Cont.)
q′ = (8560 + 17,140 ) W / m = 25, 700 W / m
<
(c) The foregoing energy balance was entered into the IHT workspace and parametric calculations were
performed to explore the effects of h and Dd on the heat rates.
For Dd = 0.06 m and 10 ≤ h ≤ 1000 W / m 2 ⋅ K, the heat rate to the tube, q′cond , as well as the
contribution due to convection, q′conv , increase with increasing h. However, because the outer surface
temperature Td also increases with h, the contribution due to radiation decreases and becomes negative
(heat transfer from the surface) when Td exceeds 1500 K at h = 540 W / m 2 ⋅ K. Both the convection
and radiation heat rates, and hence the conduction heat rate, increase with decreasing Dd, as Td decreases
and approaches Tt = 600 K. However, even for Dd = 0.051 m (a deposit thickness of 0.5 mm), Td = 773
K and the ash provides a significant resistance to heat transfer.
COMMENTS: Boiler operation in an energy efficient manner dictates that ash deposits be minimized.
PROBLEM 13.95
KNOWN: Two parallel, large, diffuse-gray surfaces; top one maintained at T1 while lower one is
insulated and experiences convection.
FIND: (a) Temperature of lower surface, T2, when ε1 = ε2 = 0.5 and (b) Radiant flux leaving the
viewing port.
SCHEMATIC:
ASSUMPTIONS: (1) Surfaces are large, diffuse-gray, (2) Lower surface experiences convection and
radiation exchange, backside is perfectly insulated.
ANALYSIS: (a) Perform an energy balance on the lower surface, giving
q′′conv + q′′rad,1 = 0
(1)
′′ , the net radiant power per unit area exchanged between
where the latter term is equal to q1′′ or q12
surfaces 1 and 2. For this two surface enclosure,
)
(2)
h ( T∞ − T2 ) + σ T14 − T24 / [(1 − ε1 ) / ε1 + 1 + (1 − ε 2 ) / ε 2 ] = 0
(3)
q1′′ =
E b ( T1 ) − E b ( T2 )
(
σ T14 − T24
=
(1 − ε1 ) / ε1 + 1/ F12 + (1 − ε 2 ) / ε 2 (1 − ε1 ) / ε1 + 1 + (1 − ε 2 ) / ε 2
with F12 = 1. Combining Eqs. (1) and (2),
(
)
Substituting numerical values with ε1 = ε2 = 0.5,
(
)
50 W / m 2 ⋅ K (300 − T2 ) K + 5.67 × 10 −8 W / m 2 ⋅ K 4 4004 − T24 K 4 / [1 + 1 + 1] = 0
<
T2 ≈ 306 K.
(b) The radiant flux leaving the viewing port is q′′vp = G1. From an energy balance on the upper plate
q1′′ = E1 − α1G1
where q1′′ = q1′′− 2 , net exchange by radiation. But
(
q1′′ = (1/ 3)σ T14 − T24
)
E1 = ε E b1 = 0.5σ T14 .
Hence, the flux is
(
)
G1 = ( E1 − q1 ) / α = (1/ 0.5 ) 0.5σ T14 − (1/ 3)σ T14 − T24 

G1 = 2σ ( 0.5 − 0.333) T14 + 0.333T24  = 816 W / m 2 .


<
PROBLEM 13.96
KNOWN: Dimensions, emissivities and temperatures of heated and cured surfaces at opposite ends
of a cylindrical cavity. External conditions.
FIND: Required heater power and outside convection coefficient.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Opaque, diffuse-gray surfaces, (3) Negligible
convection within cavity, (4) Isothermal disk and heater surfaces, (5) One-dimensional conduction in
base, (6) Negligible contact resistance between heater and base, (7) Sidewall is reradiating.
ANALYSIS: The equivalent circuit is
From an energy balance on the heater surface, q1,elec = q1,cond + q1,rad,
(
2
q1,elec = k b π D / 4
2
)
T1 − Tb
Lb
+
(
σ T14 − T24
)
1 − ε 2,i
1 − ε1
1
+
+
ε1A1 A F + [(1/ A F ) + (1/ A F )]−1 ε 2,i A 2
1 12
1 1R
2 2R
2
2
where A1 = A2 = πD /4 = π(0.12 m) /4 = 0.0113 m and from Fig. 13.5, with Lc/r1 = 3.33 and r2/Lc =
0.3 find F12 = F21 = 0.077; hence, F1R = F2R = 0.923. The required heater power is
(800 − 300 ) K
q1,elec = 20 W / m ⋅ K × 0.0113 m 2
0.025 m
+
(
)
0.0113 m 2 × 5.67 × 10−8 W / m 2 ⋅ K 4 800 4 − 400 4 K 4
1 − 0.9
1
1 − 0.5
+
+
1
−
0.9
0.5
0.077 + [(1/ 0.923 ) + (1/ 0.923 )]
q1,elec = 4521 W + 82.9 W = 4604 W.
(
)
<
4
An energy balance for the disk yields, q rad,2 = q rad,1 = h o A 2 ( T2 − T∞ ) + ε 2,o A 2σ T24 − Tsur
,
ho =
(
)
82.9 W − 0.9 × 0.0113 m 2 × 5.67 × 10−8 W / m 2 ⋅ K 4 4004 − 3004 K 4
2
0.0113 m × 100 K
= 64 W / m 2 ⋅ K.
<
COMMENTS: Conduction through the ceramic base represents an enormous system loss. The base
should be insulated to greatly reduce this loss and hence the electric power input.
PROBLEM 13.97
KNOWN: Electrical conductors in the form of parallel plates having one edge mounted to a ceramic
insulated base. Plates exposed to large, isothermal surroundings, Tsur. Operating temperature is T1 =
500 K.
FIND: (a) Electrical power dissipated in a conductor plate per unit length, q1′ , considering only
radiative exchange with the surroundings; temperature of the ceramic insulated base T2; and, (b) q1′
and T2 when the surfaces experience convection with an airstream at T∞ = 300 K and a convection
2
coefficient of h = 24 W/m ⋅K.
SCHEMATIC:
ASSUMPTIONS: (1) Conductor surfaces are diffuse, gray, (2) Conductor and ceramic insulated
base surfaces have uniform temperatures and radiosities, (3) Surroundings are large, isothermal.
ANALYSIS: (a) Define the opening between the conductivities as the hypothetical area A3 at the
temperature of the surroundings, Tsur, with an emissivity ε3 = 1 since all the radiation incident on the
area will be absorbed. The conductor (1)-base (2)-opening (3) form a three surface enclosure with
one surface re-radiating (2). From Section 13.3.5 and Eq. 13.30, the net radiation leaving the
conductor surface A1 is
E b1 − E b2
(1)
q1 =
1 − ε3
1 − ε1
1
+
+
ε1A1 A F + [(1/ A F ) + (1/ A F )]−1 ε 3A3
1 13
1 12
3 32
where E b1 = σ T14 and E b1 = σ T34 . The view factors are evaluated as follows:
F32: use the relation for two aligned parallel rectangles, Table 13.2 or Fig. 13.4,
X = X / L = w / L = 10 / 40 = 0.25
Y = Y/L =∞
F32 = 0.1231
F13: applying reciprocity between A1 and A3, where A1 = 2L = 2 × 0.040 m = 0.080 and A3 =
w = 0.010 and is the length of the conductors normal to the page, >> L or w,
A F
F13 = 3 31 = 0.010" × 0.8769 / 0.080" = 0.1096
A1
where F31 can be obtained by using the summation rule on A3,
F31 = 1 − F32 = 1 − 0.1231 = 0.8769
F12: by symmetry
F12 = F13 = 0.1096
Continued …..
PROBLEM 13.97 (Cont.)
Substituting numerical values into Eq. (1), the net radiation leaving the conductor is
q1 =
(
)
5.67 × 10−8 W / m 2 ⋅ K 4 5004 − 3004 K 4
1 − 0.8
1
+
+0
0.8 × 0.080" 0.080" × 0.1096 + [(1/ 0.080" × 0.1096 ) + (1/ 0.010" × 0.123 )]−1
q1′ = q1 / " =
(3544 − 459.3) W
3.1250 + 101.557 + 0
<
= 29.5 W / m
(b) Consider now convection processes occurring at the conductor (1) and base (2) surfaces, and
perform energy balances as illustrated in the schematic below.
Surface 1: The heat rate from the conductor includes convection and the net radiation heat rates,
E b1 − J1
qin = q cv,1 + q1 = h A1 ( T1 − T∞ ) +
(2)
(1 − ε1 ) / ε1A1
and the radiosity J1 can be determined from the radiation energy balance, Eq. 13.21,
E b1 − J1
J −J
J −J
= 1 2 + 1 3
(1 − ε1 ) / ε1A1 1/ A1F12 1/ A1F13
(3)
where J 3 = E b3 = σ T34 since A3 is black.
Surface 2: Since the surface is insulated (adiabatic), the energy balance has the form
E b2 − J 2
0 = q cv,2 + q 2 = hA 2 ( T2 − T∞ ) +
1 − ε 2 / ε 2A2
(4)
and the radiosity J2 can be determined from the radiation energy balance, Eq. 13.21,
E b2 − J 2
J −J
J −J
= 2 1 + 2 3
(1 − ε 2 ) / ε 2 A 2 1/ A2 F21 1/ A 2 F23
(5)
There are 4 equations, Eqs. (2-5), with 4 unknowns: J2, J2, T2 and q1. Substituting numerical values,
the simultaneous solution to the set yields
J1 = 3417 W / m 2
J 2 = 1745 W / m 2
T2 = 352 K
′ = 441 W / m
q in
<
COMMENTS: (1) The effect of convection is substantial, increasing the heat removal rate from 29.5
W to 441 W for the combined modes.
(2) With the convection process, the current carrying capacity of the conductors can be increased.
Another advantage is that, with the presence of convection, the ceramic base operates at a cooler
temperature: 352 K vs. 483 K.
PROBLEM 13.98
KNOWN: Surface temperature and spectral radiative properties. Temperature of ambient air. Solar
irradiation or temperature of shield.
FIND: (a) Convection heat transfer coefficient when surface is exposed to solar radiation, (b)
Temperature of shield needed to maintain prescribed surface temperature.
SCHEMATIC:
ASSUMPTIONS: (1) Surface is diffuse (αλ = ελ), (2) Bottom
of surface is adiabatic, (3) Atmospheric irradiation is
negligible,
(4) With shield, convection coefficient is unchanged and
radiation losses at ends are negligible (two-surface enclosure).
ANALYSIS: (a) From a surface energy balance,
αSGS = ε sσ Ts4 + h ( Ts − T∞ ) .
Emission occurs mostly at long wavelengths, hence εs = α2 = 0.3. However,
∞
αS
∫ αλ Eλ ,b (λ , 5800 K ) dλ = α F
= 0
1 ( 0 −1µ m ) + α 2 F(1−∞ )
Eb
and from Table 12.1 at λT = 5800 µm⋅K, F(0-1µm) = 0.720 and hence, F(1 - ∞) = 0.280 giving
α = 0.9 × 0.72 + 0.3 × 0.280 = 0.732.
Hence
h=
αSGS − εσ Ts4
Ts − T∞
=
(
)
0.732 1200 W / m 2 − 0.3 × 5.67 × 10−8 W / m 2 ⋅ K 4 (320 K )
4
20 K
h = 35 W / m 2 ⋅ K.
<
(b) Since the plate emits mostly at long wavelengths, αs = εs = 0.3. Hence radiation exchange is
between two diffuse-gray surfaces.
q′′ps =
(
σ Tp4 − Ts4
)
1/ ε p + 2 / ε s − 1
= q′′conv = h ( Ts − T∞ )
(
)
Tp4 = ( h / σ )( Ts − T∞ ) 1/ ε p + 1/ ε s − 1 + Ts4
Tp4 =
35 W / m 2 ⋅ K ( 20 K )  1
1

4
+
− 1  + (320 K )

−8
2
4  0.8 0.3

5.67 × 10 W / m ⋅ K
Tp = 484 K.
<
COMMENTS: For Tp = 484 K and λ = 1 µm, λT = 484 µm⋅K and F(0-λ) = 0.000. Hence assumption
of αs = 0.3 is excellent.
PROBLEM 13.99
KNOWN: Long uniform rod with volumetric energy generation positioned coaxially within a larger circular tube
maintained at 500°C.
FIND: (a) Center T1(0) and surface T1s temperatures of the rod for evacuated space, (b) T1(0) and T1s for airspace,
(c) Effect of tube diameter and emissivity on T1(0) and T1s.
SCHEMATIC:
ASSUMPTIONS: (1) All surfaces are diffuse-gray.
PROPERTIES: Table A-4, Air ( T = 780 K): ν = 81.5 × 10
-6
2
m /s, k = 0.0563 W/m⋅K, α = 115.6 × 10
-6
2
m /s, β
-1
= 0.00128K , Pr = 0.706.
ANALYSIS: (a) The net heat exchange by radiation between the rod and the tube is
′ =
q12
(
4
4
σ T1 − T2
)
(1)
(1 − ε1 ) / ε1π D1 + 1 / π D1F12 + (1 − ε 2 ) / ε 2π D 2
′
and, from an energy balance on the rod, − E
out + E′gen = 0, or
(
)
2
′ = q π D1 / 4 .
q12
(2)
Combining Eqs. (1) and (2) and substituting numerical values, with F12 = 1, we obtain
(
)
4
4


σ T1 − T2


q=
D1  (1 − ε1 ) / ε1 + 1 + [(1 − ε 2 ) / ε 2 ]( D1 / D 2 ) 


4
4
3 W
=
20 × 10
3
0.050m
m
(
)
2
4
4
4
4
−8


5.67 × 10 W / m ⋅ K T1s − 773 K


 (1 − 0.2 ) / 0.2 + 1 + [(1 − 0.2 ) / 0.2 ]( 0.050 / 0.060 ) 


= 54.4 × 10
−8
(T
4
4
1s − 773
) W / m3
<
T1s = 792 K.
From Eq. 3.53, the rod center temperature is
T1 ( 0 ) =
T1 ( 0 ) ≈
q ( D1 / 2 )
2
4k
3
+ T1s
20 × 10 W / m
3
(0.050 m / 2 )2
4 × 15 W / m ⋅ K
<
+ 792 K = 0.21 K + 792 K = 792.2 K.
(b) The convection heat rate is given by Eqs. 9.58 to 9.60. However, assuming a maximum possible value of (Ts1 –
3
2
-1
3
-12
T2) = 19 K, RaL = gβ (Ts,1 – T2)L /αν = 9.8 m/s (0.00128 K )19 K (0.005 m) /115.6 × 81.5 × 10
4 3
-3/5
-3/5 5
4
3
-3/5
∗
3.16 and Ra c = {[ln(D2/D1)] /L [(D1)
+ (D2) ] } RaL = {[ln(1.2)] /(0.005 m) [(0.05 m)
4 2
m /s =
Continued …..
PROBLEM 13.99 (Cont.)
3/5 5
+(0.06 m)- ] } 3.16 = 0.14. It follows that buoyancy driven flow is negligible and heat transfer across
the airspace is by conduction. Hence, from Eq. 3.27, q′cond = 2 πk (T1s – T2)/ln(r2/r1).
q′cond =
2π k ( T1s − T2 ) 2π ( 0.0563 W / m ⋅ K )( T1s − 773 ) K
=
= 1.94 ( T1s − 773)
ln ( r2 / r1 )
ln (30 / 25 )
The energy balance then becomes
(
(
)
′ + q′cond , or
q π D12 / 4 = q12
)(q12′ + q′cond )
2 × 104 = 54.4 ×10 −8 (T1s4 − 7734 ) + 988 ( T1s − 773 )


q = 4 / π D12
T1s = 783 K
T1 ( 0 ) = 783.2 K
(c) Entering the foregoing model and the prescribed properties of air into the IHT workspace, the
<
parametric calculations were performed for D2 = 0.06 m and D2 = 0.10 m. For D2 = 1.0 m, Ra ∗c > 100
and heat transfer across the airspace is by free convection, instead of conduction. In this case, convection
was evaluated by entering Eqs. 9.58 – 9.60 into the workspace. The results are plotted as follows.
The first graph corresponds to the evacuated space, and the surface temperature decreases with
increasing ε1 = ε2, as well as with D2. The increased emissivities enhance the effectiveness of emission
at surface 1 and absorption at surface 2, both which have the effect of reducing T1s. Similarly, with
increasing D2, more of the radiation emitted from surface 1 is ultimately absorbed at 2 (less of the
radiation reflected by surface 2 is intercepted by 1). The second graph reveals the expected effect of a
reduction in T1s with inclusion of heat transfer across the air. For small emissivities (ε1 = ε2 < 0.2),
conduction across the air is significant relative to radiation, and the small conduction resistance
corresponding to D2 = 0.06 m yields the smallest value of T1s. However, with increasing ε,
conduction/convection effects diminish relative to radiation and the trend reverts to one of decreasing T1s
with increasing D2.
COMMENTS: For this situation, the temperature variation within the rod is small and independent of
surface conditions.
PROBLEM 13.100
KNOWN: Side wall and gas temperatures for adjoining semi-cylindrical ducts. Gas flow convection
coefficients.
FIND: (a) Temperature of intervening wall, (b) Verification of gas temperature on one side.
SCHEMATIC:
ASSUMPTIONS: (1) All duct surfaces may be approximated as blackbodies, (2) Fully developed
conditions, (3) Negligible temperature difference across intervening wall, (4) Gases are
nonparticipating media.
ANALYSIS: (a) Applying an energy balance to a control surface about the wall yields
E in = E out .
Assuming Tg,1 > Tw > Tg,2, it follows that
q rad (1→ w ) + q conv( g1→ w ) = q rad ( w → 2 ) + q conv( w → g2 )
(
)
(
(
)
)
(
4
4
A1F1wσ T14 − Tw
+ hA w Tg,1 − Tw = A w Fw2σ Tw
− T24 + hA w Tw − Tg,2
)
and with
A1F1w = A w Fw1 = A w Fw2 = A w
and substituting numerical values,
(
)
(
4
2σ Tw
+ 2hTw = σ T14 + T24 + h Tg,1 + Tg,2
)
4
11.34 × 10−8 Tw
+ 10Tw = 13, 900.
Trial-and-error solution yields
<
Tw ≈ 526 K.
(b) Applying an energy balance to a control surface about the hot gas (g,1) yields
E in = E out
(
)
(
hA1 T1 − Tg,1 = hA w Tg,1 − Tw
)
or
(
T1 − Tg,1 = [D / (π D / 2 )] Tg,1 − Tw
29°C = 29°C.
)
<
COMMENTS: Since there is no change in any of the temperatures in the axial direction, this scheme
simply provides for energy transfer from side wall 1 to side wall 2.
PROBLEM 13.101
KNOWN: Temperature, dimensions and arrangement of heating elements between two large parallel
plates, one insulated and the other of prescribed temperature. Convection coefficients associated with
elements and bottom surface.
FIND: (a) Temperature of gas enclosed by plates, (b) Element electric power requirement, (c) Rate
of heat transfer to 1 m × 1m section of panel.
SCHEMATIC:
ASSUMPTIONS: (1) Diffuse-gray surfaces, (2) Negligible end effects since the surfaces form an
enclosure, (3) Gas is nonparticipating, (4) Surface 3 is reradiating with negligible conduction and
convection.
ANALYSIS: (a) Performing an energy balance for a unit control surface about the gas space,
E in − E out = 0.
h1π D ( T1 − Tm ) − h 2s ( Tm − T2 ) = 0
Tm =
hπ DT1 + h 2sT2
h1π D + h 2s
=
10 W / m 2 ⋅ Kπ ( 0.025 m ) 600 K + 2 W / m 2 ⋅ K ( 0.05 m ) 400 K
10 W / m 2 ⋅ K π ( 0.025 m ) + 2 W / m 2 ⋅ K ( 0.05 m )
<
Tm = 577 K.
(b) The equivalent thermal circuit is
The energy balance on surface 1 is
′
′
′
= q1,conv
+ q1,rad
q1,elec
′
where q1,rad
can be evaluated by considering a unit cell of the form
A1′ = π D = π ( 0.025 m ) = 0.0785 m
A′2 = A′3 = s = 0.05 m
Continued …..
PROBLEM 13.101 (Cont.)
The view factors are:
1/ 2
2
F21 = 1 − 1 − ( D / s ) 


F21 = 1 − [1 − 0.25]
1/ 2
(
)
1/ 2
+ ( D / s ) tan −1  s 2 − D 2 / D2 

+ 0.5 tan −1 ( 4 − 1)
1/ 2

= 0.658 = F31
F23 = 1 − F21 = 0.342 = F32 .
For the unit cell,
A′2 F21 = sF21 = 0.05 m × 0.658 = 0.0329 m = A1′ F12 = A′3F31 = A1′ F13
A′2 F23 = sF23 = 0.05 m × 0.342 = 0.0171 m = A′3F32 .
Hence,
′
q1,rad
=
E b1 − E b2
R ′equiv + (1 − ε 2 ) / ε 2 A′2
−1
R ′equiv
= A1′ F12 +

1
1/ A1′ F13 + 1/ A′2 F23
=  0.0329 +



m
(0.0329 )−1 + (0.0171)−1 
1
R ′equiv = 22.6 m −1.
Hence
(
)
5.67 × 10−8 W / m 2 ⋅ K 4 6004 − 4004 K 4
′
q1,rad
=
[22.6 + (1 − 0.5) / 0.5 × 0.05] m−1
= 138.3 W / m
′
q1,conv
= h1π D ( T1 − Tm ) = 10 W / m 2 ⋅ Kπ ( 0.025 m )( 600 − 577 ) K = 17.8 W / m
′
= (138.3 + 17.8 ) W / m = 156 W / m.
q1,elec
<
(c) Since all energy added via the heating elements must be transferred to surface 2,
q′2 = q1′ .
Hence, since there are 20 elements in a 1 m wide strip,
′
q 2(1m×1m ) = 20 × q1,elec
= 3120 W.
<
COMMENTS: The bottom panel would have to be cooled (from below) by a heat sink which could
2
dissipate 3120 W/m .
PROBLEM 13.102
KNOWN: Flat plate solar collector configuration.
FIND: Relevant heat transfer processes.
SCHEMATIC:
The incident solar radiation will experience transmission, reflection and absorption at each of the
cover plates. However, it is desirable to have plates for which absorption and reflection are
minimized and transmission is maximized. Glass of low iron content is a suitable material. Solar
radiation incident on the absorber plate may be absorbed and reflected, but it is desirable to have a
coating which maximizes absorption at short wavelengths.
Energy losses from the absorber plate are associated with radiation, convection and conduction.
Thermal radiation exchange occurs between the absorber and the adjoining cover plate, between the
two cover plates, and between the top cover plate and the surroundings. To minimize this loss, it is
desirable that the emissivity of the absorber plate be small at long wavelengths. Energy is also
transferred by free convection from the absorber plate to the first cover plate and between cover
plates. It is transferred by free or forced convection to the atmosphere. Energy is also transferred by
conduction from the absorber through the insulation.
The foregoing processes provide for heat loss from the absorber, and it is desirable to minimize these
losses. The difference between the solar radiation absorbed by the absorber and the energy loss by
radiation, convection and conduction is the energy which is transferred to the working fluid. This
transfer occurs by conduction through the absorber and the tube wall and by forced convection from
the tube wall to the fluid.
PROBLEM 13.103
KNOWN: Operating conditions of a flat plate solar collector.
FIND: Expressions for determining the rate at which useful energy is collected per unit area.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Uniform surface heat fluxes and temperatures, (3)
Opaque, diffuse-gray surface behavior for long-wave thermal radiation, (4) Complete absorption of
solar radiation by absorber plate (αap,S = 1).
ANALYSIS: From an energy balance on the absorber plate, E ′′in = E ′′out ,
(
)
α ap,S τ cp,S GS = q′′u + q′′conv,i + q′′rad,ap − cp .
Hence with complete absorption of solar radiation by the absorber plate,
(
)
q′′u = τ cp,SGS − h i Tap − Tcp −
(
4
4
σ Tap
− Tcp
)
(1)
1/ ε ap + 1/ ε cp − 1
<
where Fap-cp ≈ 1 and Eq. 13.24 is used to obtain q′′rad,ap − cp . To determine q′′u from Eq. (1),
however, Tcp must be known. From an energy balance on the cover plate,
α cp,SGS + q′′conv,i + q′′rad,ap − cp = q′′conv,o + q ′′rad,cp −sky
or
(
)
α cp,SGS + h i Tap − Tcp +
(
4
4
σ Tap
− Tcp
)
1/ ε ap + 1/ ε c − 1
(
)
4
4
= h o Tcp − T∞ + ε cpσ Tcp
− Tsky
.
(
)
(2)
<
Eq. (2) may be used to obtain Tcp.
COMMENTS: With Tap presumed to be known, Tcp may be evaluated from Eq. (2) and q′′u from
Eq. (1).
PROBLEM 13.104
KNOWN: Ceiling temperature of furnace. Thickness, thermal conductivity, and/or emissivities of
alternative thermal insulation systems. Convection coefficient at outer surface and temperature of
surroundings.
FIND: (a) Mathematical model for each system, (b) Temperature of outer surface Ts,o and heat loss
q′′ for each system and prescribed conditions, (c) Effect of emissivity on Ts,o and q′′.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, (2) Diffuse/gray surfaces, (3) Surroundings form a large
enclosure about the furnace, (4) Radiation in air space corresponds to a two-surface enclosure of large
parallel plates.
-4
2
PROPERTIES: Table A-4, air (Tf = 730 K): k = 0.055 W/m⋅K, α = 1.09 × 10 m /s, ν = 7.62 ×
-5 2
-1
10 m /s, β = 0.001335 K , Pr = 0.702.
ANALYSIS: (a) To obtain Ts,o and q′′, an energy balance must be performed at the outer surface of
the shield.
q′′cond = q′′conv,o + q′′rad,o = q′′
Insulation:
k=
(Ts,i − Ts,o ) = h
(
)
L
′′
q conv,i + q′′rad,i = q′′conv,o + q′′rad,o = q′′
Air Space:
(
)
h i Ts,i − Ts,o +
(
4
4
o Ts,o − T∞ + ε oσ Ts,o − Tsur
(
4
4
σ Ts,i
− Ts,o
1
εi
+
1
εo
)=h
−1
(
)
(
)
4
4
o Ts,o − T∞ + ε oσ Ts,o − Tsur
)
where Eq. 13.24 has been used to evaluate q′′rad,i and hi is given by Eq. 9.49
hL
3 0.074
Nu L = i = 0.069Ra1/
L Pr
k
(b) For the prescribed conditions (εi = εo = 0.5), the following results were obtained.
The energy equation becomes
Insulation:
(
)
0.09 W / m ⋅ K 900 − Ts,o K
0.025 m
2
(
)
= 25 W / m ⋅ K Ts,o − 300 K + 0.5 × 5.67 × 10
−8
2
W/m ⋅K
4
(
4
Ts,o − 300
4
)
K
4
Continued …..
PROBLEM 13.104 (Cont.)
and a trial-and-error solution yields
q′′ = 1920 W / m 2
Ts,o = 366 K
<
The energy equation becomes
Air-Space:
(
)
5.67 × 10
−8
2
W / m ⋅K
h i 900 − Ts,o K +
4
(900
4
4
)
− Ts,o K
4
3
2
(
)
= 25 W / m ⋅ K Ts,o − 300 K + 0.5 × 5.67 × 10
−8
2
W /m ⋅K
4
(T
)
4
4
4
s,o − 300 K
where
hi =
0.055 W / m ⋅ K
0.025 m
3 0.074
0.069 Ra1/
L Pr
(1)
3
and RaL = gβ(Ts,i – Ts,o)L /αν. A trial-and-error solution, which includes reevaluation of the air
properties, yields
Ts,o = 598 K
q′′ = 10,849 W / m 2
<
The inner and outer heat fluxes are q′′conv,i = 867 W / m 2 , q′′rad,i = 9982 W / m 2 , q′′conv,o = 7452
2
W/m , and q′′rad,o = 3397 W / m 2 .
(c) Entering the foregoing models into the IHT workspace, the following results were generated.
Insulation:
Continued …..
PROBLEM 13.104 (Cont.)
As expected, the outer surface temperature decreases with increasing εo. However, the reduction in
Ts,o is not large since heat transfer from the outer surface is dominated by convection.
In this case Ts,o increases with increasing εo = εi and the effect is significant. The effect is due to an
increase in radiative transfer from the inner surface, with q′′rad,i = q′′conv,i = 1750 W / m 2 for εo = εi =
0.1 and q′′rad,i = 20,100 W / m 2 >> q′′conv,i = 523 W / m 2 for εo = εi = 0.9. With the increase in Ts,o,
(
)
the total heat flux increases, along with the relative contribution of radiation q′′rad,o to heat transfer
from the outer surface.
COMMENTS: (1) With no insulation or radiation shield and εi = 0.5, radiative and convective heat
2
fluxes from the ceiling are 18,370 and 15,000 W/m , respectively. Hence, a significant reduction in
the heat loss results from use of the insulation or the shield, although the insulation is clearly more
effective.
(2) Rayleigh numbers associated with free convection in the air space are well below the lower limit
of applicability of Eq. (1). Hence, the correlation was used outside its designated range, and the error
associated with evaluating hi may be large.
(3) The IHT solver had difficulty achieving convergence in the first calculation performed for the
radiation shield, since the energy balance involves two nonlinear terms due to radiation and one due
to convection. To obtain a solution, a fixed value of RaL was prescribed for Eq. (1), while a second
3
value of RaL,2 ≡ gβ(Ts,i – Ts,o)L /αν was computed from the solution. The prescribed value of RaL
was replaced by the value of RaL,2 and the calculations were repeated until RaL,2 = RaL.
PROBLEM 13.105
KNOWN: Dimensions of a composite insulation consisting of honeycomb core sandwiched between
solid slabs.
FIND: Total thermal resistance.
SCHEMATIC: Because of the repetitive nature of the honeycomb core, the cell sidewalls will be
adiabatic. That is, there is no lateral heat transfer from cell to cell, and it suffices to consider the heat
transfer across a single cell.
ASSUMPTIONS: (1) One-dimensional, steady-state conditions, (2) Equivalent conditions for each
cell, (3) Constant properties, (4) Diffuse, gray surface behavior.
PROPERTIES: Table A-3, Particle board (low density): k1 = 0.078 W/m⋅K; Particle board (high
density): k2 = 0.170 W/m⋅K; For both board materials, ε = 0.85; Table A-4, Air ( T ≈ 7.5°C, 1 atm):
-6 2
-6 2
-3 -1
ν = 14.15 × 10 m /s, k = 0.0247 W/m⋅K, α = 19.9 × 10 m /s, Pr = 0.71, β = 3.57 × 10 K .
ANALYSIS: The total resistance of the composite is determined by conduction, convection and
radiation processes occurring within the honeycomb and by conduction across the inner and outer
slabs. The corresponding thermal circuit is shown.
The total resistance of the composite and equivalent resistance for the honeycomb are
R = R cond,i + R eq + R cond,o
(
−1
−1
−1
−1
R eq
= R cond
+ R conv
+ R rad
)
.
hc
The component resistances may be evaluated as follows. The inner and outer slabs are plane walls,
for which the thermal resistance is given by Eq. 3.6. Hence, since L1 = L3 and the slabs are
constructed from low-density particle board.
R cond,i = R cond,o =
L1
k1W
2
=
0.0125 m
2
0.078 W / m ⋅ K ( 0.01 m )
= 1603 K / W.
Continued …..
PROBLEM 13.105 (Cont.)
Similarly, applying Eq. 3.6 to the side walls of the cell
R cond,hc =
L2
k 2 W − (W − t )

=
=
2
2
L2
(
k 2 2Wt − t

2
)
0.050 m
0.170 W / m ⋅ K  2 × 0.01 m × 0.002 m − ( 0.002 m )
2

= 8170 K / W.

From Eq. 3.9 the convection resistance associated with the cellular airspace may be expressed as
R conv,hc = 1/ h ( W − t ) .
2
The cell forms an enclosure that may be classified as a horizontal cavity heated from below, and the
appropriate form of the Rayleigh number is Ra L = gβ ( T1 − T2 ) L32 / αν . To evaluate this parameter,
however, it is necessary to assume a value of the cell temperature difference. As a first
approximation, T1 − T2 = 15°C − ( −5°C ) = 20°C,
Ra L =
(
9.8 m / s 2 3.57 × 10−3 K −1
)(20 K )(0.05 m )3 = 3.11×105.
19.9 × 10−6 m 2 / s × 14.15 × 10−6 m 2 / s
Applying Eq. 9.49 as a first approximation, it follows that
h = ( k / L 2 )  0.069Ra L

1/ 3
Pr
0.074 

0.0247 W / m ⋅ K
=
0.05 m

 0.069

(
3.11 × 10
5
)
1/ 3

( 0.71)0.074  = 2.25

2
W / m ⋅ K.
The convection resistance is then
R conv,hc =
1
2.25 W / m ⋅ K ( 0.01 m − 0.002 m )
2
2
= 6944 K / W.
The resistance to heat transfer by radiation may be obtained by first noting that the cell forms a threesurface enclosure for which the sidewalls are reradiating. The net radiation heat transfer between the
2
end surfaces of the cell is then given by Eq. 13.30. With ε1 = ε2 = ε and A1 = A2 = (W – t) , the
equation reduces to
q rad =
( W − t )2 σ
(T14 − T24 )
2 (1/ ε − 1) +  F12 + [( F1R + F2R ) / F1R F2R 
−1
.
However, with F1R = F2R = (1 – F12), it follows that
q rad =
( W − t )2 σ
(T14 − T24 )
−
2 1
(1 − F12 ) 
1  
2  − 1  +  F12 +
2 (1 − F12 ) 
 ε  

=
( W − t )2 σ
(T14 − T24 ) .
2
 1 1
− +
 ε  1 + F12
2
The view factor F12 may be obtained from Fig. 13.4, where
X Y W − t 10 mm − 2 mm
= =
=
= 0.16.
L L
L2
50 mm
Hence, F12 ≈ 0.01. Defining the radiation resistance as
T −T
R rad,hc = 1 2
q rad
it follows that
Continued …..
PROBLEM 13.105 (Cont.)
R rad,hc =
(
2 (1/ ε − 1) + 2 / (1 + F12 )
( W − t )2 σ
) (
where T14 − T24 = T12 + T22
(T12 + T22 )(T1 + T2 )
)(T1 + T2 )(T1 − T2 ). Accordingly,
2 
  1

 2  0.85 − 1 + 1 + 0.01 
R rad,hc =
(0.01 m − 0.002 m )2 × 5.67 × 10−8 W / m2 ⋅ K 4 ( 288 K )2 + ( 268 K )2  ( 288 + 268 ) K
where, again, it is assumed that T1 = 15°C and T2 = -5°C. From the above expression, it follows that
0.353 + 1.980
= 7471 K / W.
R rad,hc =
−4
3.123 × 10
In summary the component resistances are
R cond,i = R cond,o = 1603 K / W
R cond,hc = 8170 K / W
R conv,hc = 6944 K / W
R rad,hc = 7471 K / W.
The equivalent resistance is then
1
1 
 1
+
+

 8170 6944 7471 
R eq = 
−1
= 2498 K / W
and the total resistance is
R = 1603 + 2498 + 1603 = 5704 K / W.
<
COMMENTS: (1) The problem is iterative, since values of T1 and T2 were assumed to calculate
Rconv,hc and Rrad,hc. To check the validity of the assumed values, we first obtain the heat transfer rate
q from the expression
Ts,1 − Ts,2 25°C − ( −10°C )
=
= 6.14 × 10−3 W.
q=
R
5704 K / W
Hence
T1 = Ts,i − qR cond,i = 25°C − 6.14 × 10−3 W × 1603 K / W = 15.2°C
T2 = Ts,o + qR cond,o = −10°C + 6.14 × 10−3 W × 1603 K / W = −0.2°C.
Using these values of T1 and T2, Rconv,hc and Rrad,hc should be recomputed and the process repeated
until satisfactory agreement is obtained between the initial and computed values of T1 and T2.
2
(2) The resistance of a section of low density particle board 75 mm thick (L1 + L2 + L3) of area W is
9615 K/W, which exceeds the total resistance of the composite by approximately 70%. Accordingly,
use of the honeycomb structure offers no advantages as an insulating material. Its effectiveness as an
insulator could be improved (Req increased) by reducing the wall thickness t to increase Rcond,
evacuating the cell to increase Rconv, and/or decreasing ε to increase Rrad. A significant increase in
Rrad,hc could be achieved by aluminizing the top and bottom surfaces of the cell.
PROBLEM 13.106
KNOWN: Dimensions and surface conditions of a cylindrical thermos bottle filled with hot coffee
and lying horizontally.
FIND: Heat loss.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible heat loss from ends (long infinite
cylinders), (3) Diffuse-gray surface behavior.
PROPERTIES: Table A-4, Air (Tf = (T1 + T2)/2 = 328 K, 1 atm): k = 0.0284 W/m⋅K, ν = 23.74 ×
-6 2
-6 2
-3 -1
10 m /s, α = 26.6 × 10 m /s, Pr = 0.0703, β = 3.05 × 10 K .
ANALYSIS: The heat transfer across the air space is
q = q rad + q conv .
From Eq. 13.25 for concentric cylinders
q rad =
(
σ (π D1L ) T14 − T24
1 1 − ε 2  r1 
+
 
ε1
ε 2  r2 
) = 5.67 ×10−8 W / m2 ⋅ K4π (0.07 × 0.3) m2 (3484 − 3084 ) K4
4 + 3 ( 0.035 / 0.04 )
q rad = 3.20 W.
From Eq. 9.25,
Ra L =
gβ ( T1 − T2 ) L3
αν
=
(
9.8 m / s 2 3.05 × 10−3 K −1
)(40 K )(0.005 m )3 = 236.7.
26.6 × 10−6 m 2 / s × 23.74 × 10−6 m2 / s
Hence from Eq. 9.60
[
]
[
]
4
4
ln ( D2 / D1 ) Ra L
ln ( 0.08 / 0.07 ) 236.7
∗
=
= 7.85.
Ra c =
5
5
3
−0.6
−0.6
−0.6
−0.6
−3
3
+ D2
L D1
m
(0.005 m ) 0.07 + 0.08
(
)
(
)
However, the implication of such a small value of Ra ∗c is that free convection effects are negligible.
Heat transfer across the airspace is therefore by conduction (keff = k). From Eq. 3.27
2π Lk ( T1 − T2 ) 2π × 0.3 m × 0.0284 W / m ⋅ K ( 75 − 35 ) K
=
= 16.04 W.
q cond =
ln ( r2 / r1 )
ln ( 0.04 / 0.035 )
Hence the total heat loss is
q = q rad + q cond = 19.24 W.
COMMENTS: (1) End effects could be considered in a more detailed analysis, (2) Conduction
losses could be eliminated by evacuating the annulus.
<
PROBLEM 13.107
KNOWN: Thickness and height of a vertical air space. Emissivity and temperature of adjoining
surfaces.
FIND: (a) Heat loss per unit area across the space, (b) Heat loss per unit area if space is filled with
urethane foam.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Diffuse-gray surface behavior, (3) Air space is a
vertical cavity, (4) Constant properties, (5) One-dimensional conduction across foam.
-6
2
PROPERTIES: Table A-4, Air (Tf = 4°C, 1 atm): ν = 13.84 × 10 m /s, k = 0.0245 W/m⋅K, α =
-6 2
-3 -1
19.5 × 10 m /s, Pr = 0.71, β = 3.61 × 10 K ; Table A-3, Urethane foam: k = 0.026 W/m⋅K.
ANALYSIS: (a) With the air space, heat loss is by radiation and free convection or conduction.
From Eq. 13.24,
(
σ T14 − T24
q′′rad =
)
1/ ε1 + 1/ ε 2 − 1
=
1.222
With
Ra L =
gβ ( T1 − T2 ) L3
)
(
5.67 × 10−8 W / m 2 ⋅ K 4 2914 − 2634 K 4
=
(
= 110.7 W / m 2 .
)
9.8 m 2 / s 3.61× 10−3 K −1 (18 + 10 ) K ( 0.1 m )
13.84 × 10−6 m 2 / s × 19.5 × 10−6 m 2 / s
and H/L = 30, Eq. 9.53 may be used as a first approximation to obtain
να
(
3
6
Nu L = 0.046Ra1/
L = 0.046 3.67 × 10
h=
k
Nu L =
0.0245 W / m ⋅ K
L
0.1 m
The convection heat flux is
)
1/ 3
3
= 3.67 × 106
= 7.10
7.10 = 1.74 W / m 2 ⋅ K.
q′′conv = h ( T1 − T2 ) = 1.74 W / m 2 ⋅ K (18 + 10 ) K = 48.7 W / m 2 .
The heat loss is then
q′′ = q′′rad + q′′conv = 110.7 + 48.7 = 159 W / m 2 .
(b) With the foam, heat loss is by conduction and
k
0.026 W / m ⋅ K
q′′ = q′′cond = ( T1 − T2 ) =
(18 + 10 ) K = 7.3 W / m2 .
L
0.1 m
<
COMMENTS: Use of the foam insulation reduces the heat loss considerably. Note the significant
effect of radiation.
PROBLEM 13.108
KNOWN: Temperatures and emissivity of window panes and critical Rayleigh number for onset of
convection in air space.
FIND: (a) The conduction heat flux across the air gap for the optimal spacing, (b) The total heat flux
for uncoated panes, (c) The total heat flux if one or both of the panes has a low-emissivity coating.
SCHEMATIC:
ASSUMPTIONS: (1) Critical Rayleigh number is RaL,c = 2000, (2) Constant properties, (3)
Radiation exchange between large (infinite), parallel, diffuse-gray surfaces.
-6
2
PROPERTIES: Table A-4, air [T = (T1 + T2)/2 = 1°C = 274 K]: ν = 13.6 × 10 m /s, k = 0.0242
-6 2
-1
W/m⋅K, α = 19.1 × 10 m /s, β = 0.00365 K .
ANALYSIS: (a) With Ra L,c = g β ( T1 − T2 ) L3op / αν
1/ 3
 αν Ra L,c 
Lop = 

 g β (T1 − T2 ) 
1/ 3

−12 m 4 / s 2 × 2000 
19.1×13.6 × 10

=

2
1
−
42°C 
 9.8 m / s 0.00365 K

)
(
= 0.0070m
The conduction heat flux is then
q′′cond = k ( T1 − T2 ) / Lop = 0.0242 W / m ⋅ K ( 42°C ) / 0.0070m = 145.2 W / m 2
<
(b) For conventional glass (εg = 0.90), Eq. (13.24) yields,
σ T14 − T24
5.67 × 10−8 W / m 2 ⋅ K 4 2954 − 2534 K 4
q′′rad =
=
= 161.3 W / m 2
2
)
(
εg
(
)
1.222
−1
and the total heat flux is
q′′tot = q′′cond + q′′rad = 306.5 W / m 2
<
(c) With only one surface coated,
5.67 × 10−8 W / m 2 ⋅ K 4 2954 − 2534
q′′rad =
= 19.5 W / m 2
1
1
(
0.90
+
0.10
)
−1
Continued …..
PROBLEM 13.108 (Cont.)
q′′tot = 164.7 W / m 2
<
With both surfaces coated,
5.67 × 10−8 W / m 2 ⋅ K 4 2954 − 2534
q′′rad =
= 10.4 W / m 2
1
1
(
0.10
q′′tot = 155.6 W / m 2
+
0.10
)
−1
<
COMMENTS: Without any coating, radiation makes a large contribution (53%) to the total heat
loss. With one coated pane, there is a significant reduction (46%) in the total heat loss. However, the
benefit of coating both panes is marginal, with only an additional 3% reduction in the total heat loss.
PROBLEM 13.109
KNOWN: Dimensions and emissivity of double pane window. Thickness of air gap. Temperatures
of room and ambient air and the related surroundings.
FIND: (a) Temperatures of glass panes and rate of heat transfer through window, (b) Heat rate if gap
is evacuated. Heat rate if special coating is applied to window.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, (2) Negligible glass pane thermal resistance, (3) Constant
properties, (4) Diffuse-gray surface behavior, (5) Radiation exchange between interior window
surfaces may be approximated as exchange between infinite parallel plates, (6) Interior and exterior
surroundings are very large.
PROPERTIES: Table A-4, Air (p = 1 atm). Obtained from using IHT to solve for conditions of Part
-6 2
-6 2
(a): Tf,i = 287.4 K: νi = 14.8 × 10 m /s, ki = 0.0253 W/m⋅K, αi = 20.8 × 10 m /s, Pri = 0.71, βi =
-6 2
-1
0.00348 K . T = (Ts,i + Ts,o)/2 = 273.7 K: ν = 13.6 × 10 m /s, k = 0.0242 W/m⋅K, α = 19.0 × 10
-6 2
6 2
-1
m /s, Pr = 0.71, β = 0.00365 K . Tf,o = 259.3 K: νo = 12.3 × 10 m /s, ko = 0.023 W/m⋅K, αo =
-6 2
-1
17.1 × 10 m /s, Pro = 0.72, βo = 0.00386 K .
ANALYSIS: (a) The heat flux through the window may be expressed as
)
(
(
4 − T4 + h T − T
q′′ = q′′rad,i + q′′conv,i = ε g σ Tsur,i
i ∞,i
s,i
s,i
q′′ = q′′rad,gap + q′′conv,gap =
(
4 − T4
σ Ts,i
s,o
1
1
+
−1
εg εg
(
)+h
)
)
(
gap Ts,i − Ts,o
(
4 − T4
q′′ = q′′rad,o + q′′conv,o = ε g σ Ts,o
sur,o + h o Ts,o − T∞,o
(1)
)
(2)
)
(3)
where radiation exchange between the window panes is determined from Eq. (13.24) and radiation
exchange with the surroundings is determined from Eq. (13.27). The inner and outer convection
coefficients, h i and h o , are determined from Eq. (9.26), and h gap is obtained from Eq. (9.52).
(
)
The foregoing equations may be solved for the three unknowns q′′, Ts,i , Ts,o . Using the IHT
software to effect the solution, we obtain
<
Ts,i = 281.8 K = 8.8°C
Continued …..
PROBLEM 13.109 (Cont.)
Ts,o = 265.6 K = −7.4°C
<
q = 91.3 W
<
(
)
(b) If the air space is evacuated h g = 0 , we obtain
Ts,i = 283.6 K = 10.6°C
<
Ts,o = 263.8 K = 9.2°C
<
q = 75.5 W
<
If the space is not evacuated but the coating is applied to inner surfaces of the window panes,
Ts,i = 285.9 K = 12.9°C
<
Ts,o = 261.3 K = −11.7°C
<
q = 55.9 W
<
If the space is evacuated and the coating is applied,
Ts,i = 291.7 K = 18.7°C
<
Ts,o = 254.7 K = −18.3°C
<
q = 9.0 W
<
COMMENTS: (1) For the conditions of part (a), the convection and radiation heat fluxes are
comparable at the inner and outer surfaces of the window, but because of the comparatively small
convection coefficient, the radiation flux is approximately twice the convection flux across the air
gap. (2) As the resistance across the air gap is progressively increased (evacuated, coated, evacuated
and coated), the temperatures of the inner and outer panes increase and decrease, respectively, and the
heat loss decreases. (3) Clearly, there are significant energy savings associated with evacuation of the
gap and application of the coating. (4) In all cases, solutions were obtained using the temperaturedependent properties of air provided by the software. The property values listed in the
PROPERTIES section of this solution pertain to the conditions of part (a).
PROBLEM 13.110
KNOWN: Absorber and cover plate temperatures and spectral absorptivities for a flat plate solar
collector. Collector orientation and solar flux.
FIND: (a) Rate of solar radiation absorption per unit area, (b) Heat loss per unit area.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Adiabatic sides and bottom, (3) Cover is
transparent to solar radiation, (4) Sun emits as a blackbody at 5800 K, (5) Cover and absorber plates
are diffuse-gray to long wave radiation, (6) Negligible end effects, (7) L << width and length.
-6
2
PROPERTIES: Table A-4, Air (T = Ta + Tc)/2 = 321.5 K, 1 atm): ν = 18.05 × 10 m /s, k =
-6 2
0.0279 W/m⋅K, α = 25.7 × 10 m /s.
ANALYSIS: (a) The absorbed solar irradiation is
GS,abs = αS,a GS
where
GS = q′′s cos 30° = 900 × 0.866 = 779.4 W / m 2
∞
αS,a
∞
∫ αλ ,a G λ ,Sdλ = ∫o αλ ,a Eλ ,b (5800 K ) dλ
= o
GS
E b (5800 K )
αS,a = α λ ,a,1F( 0→ 2 µ m ) + α λ ,a2 F( 2→∞ ).
For λT = 2 µm × 5800 K = 11,600 µm⋅K from Table 12.1, F(0→2λT) = 0.941, find
αS,a = 0.9 × 0.941 + 0.2 × (1 − 0.941) = 0.859.
Hence
GS,abs = 0.859 × 779.4 = 669 W / m 2 .
<
(b) The heat loss per unit area from the collector is
q′′loss = q′′conv + q′′rad .
The convection heat flux is
q′′conv = h ( Ta − Tc )
Continued …..
PROBLEM 13.110 (Cont.)
and with
gβ ( Ta − Tc ) L3
Ra L =
αν
−1
Ra L =
9.8 m / s 2 × (321.5 K )
(343 − 300 ) K ( 0.02 m )3
18.05 × 10−6 m 2 / s × 25.7 × 10−6 m 2 / s
= 22, 604
find from Eq. 9.54 with
H / L > 12, τ < τ ∗ , cos τ = 0.5, Ra L cos τ = 11, 302
Nu L
1708 

= 1 + 1.44 1 −

 11, 302 
h = Nu L
k
L
= 2.30 ×
 1708 (sin 108° )1.6   11, 302 1/ 3 
1 −
 + 
 − 1
11, 302

  5830 

0.0279 W / m ⋅ K
0.02 m
= 3.21 W / m 2 ⋅ K.
Hence, the convective heat flux is
q′′conv = 3.21 W / m 2 ⋅ K (343 − 300 ) K = 138.0 W / m 2 .
The radiative exchange can be determined from Eq. 13.24 treating the cover and absorber plates as a
two-surface enclosure,
q′′rad =
(
σ Ta4 − Tc4
)
1/ ε a + 1/ ε c − 1
=
4
4
5.67 × 10−8 W / m 2 ⋅ K 4 (343 K ) − (300 K ) 


1/ 0.2 + 1/ 0.75 − 1
q′′rad = 61.1 W / m 2 .
Hence, the total heat loss per unit area from the collector
q′′loss = (138.0 + 61.1) = 199 W / m 2 .
<
COMMENTS: (1) Non-solar components of radiation transfer are concentrated at long wavelength
for which αa = εa = 0.2 and αc = εc = 0.75.
(2) The collector efficiency is
η=
669.3 − 199.1
669.3
× 100 = 70%.
This value is uncharacteristically high due to specification of nearly optimum αa(λ) for absorber.
PROBLEM 13.111
KNOWN: Diameters and temperatures of a heated tube and a radiation shield.
FIND: (a) Total heat loss per unit length of tube, (b) Effect of shield diameter on heat rate.
SCHEMATIC:
ASSUMPTIONS: (1) Opaque, diffuse-gray surfaces, (2) Negligible end effects.
PROPERTIES: Table A-4, Air (Tf = 77.5°C ≈ 350 K): k = 0.030 W/m⋅K, Pr = 0.70, ν = 20.92 × 10
6 2
-6 2
-1
m /s, α = 29.9 × 10 m /s, β = 0.00286 K .
ANALYSIS: (a) Heat loss from the tube is by radiation and free convection
q′ = q′rad + q′conv
q′rad =
From Eq. (13.25)
(
σ (π Di ) Ti4 − To4
)
1 1 − ε o  ri 
+
 
εi
ε o  ro 
or
5.67 × 10−8
q′rad =
W
m ⋅ K4
(π × 0.1m )
(3934 − 3084 ) K 4
1 0.9  0.05 
+
0.8 0.1  0.06 
= 30.2
W
m
g β (Ti − To ) L3 9.8 m / s 2 × 0.00286 K −1 (85 K )( 0.01m )
=
= 3809
6
2
6
2
−
−
να
20.92 × 10 m / s 29.9 × 10 m / s
3
Ra L =
)(
(
)
Hence from Eq. (9.60)
4
4
 n ( Do / Di ) Ra L
 n (0.12 / 0.10 ) 3809
∗
=
= 171.6
Ra c =
5
5
L3 Di−3/ 5 + Do−3/ 5
(0.01m )3 (0.1m )−0.6 + (0.12m )−0.6 
(
)
and from Eq. (9.59)
1/ 4
Pr


k eff = 0.386 k 

 0.861 + Pr 
( )
Ra ∗c
1/ 4
Continued …..
-
PROBLEM 13.111 (Cont.)
1/ 4
W 
0.7


k eff = 0.386  0.03



m ⋅ K   0.861 + 0.7 

(171.6 )1/ 4 = 0.0343
W
m⋅K
Hence from Eq. (9.58)
W 

2π  0.0343
2π k eff
W
m ⋅ K 
q′conv =
(Ti − To ) = 
(120 − 35) K = 100.5
m
n ( Do / Di )
n (0.12 / 0.10 )
q′ = (30.2 + 100.5 )
W
W
= 130.7
m
m
(b) As shown below, both convection and radiation, and hence the total heat rate, increase with
increasing shield diameter. In the limit as Do → ∞, the radiation rate approaches that corresponding
to net transfer between a small surface and large surroundings at To. The rate is independent of ε.
Heat rates (W/m)
200
150
100
50
0
0.1
0.15
0.2
0.25
Shield diameter, Do(m)
Radiation heat rate (W/m)
Convection heat rate (W/m)
Total heat rate (W/m)
COMMENTS: Designation of a shield temperature is arbitrary. The temperature depends on the
nature of the environment external to the shield.
PROBLEM 13.112
KNOWN: Diameters of heated tube and radiation shield. Tube surface temperature and temperature
of ambient air and surroundings.
FIND: Temperature of radiation shield and heat loss per unit length of tube.
SCHEMATIC:
ASSUMPTIONS: (1) Opaque, diffuse-gray surfaces, (2) Negligible end effects, (3) Large
surroundings, (4) Quiescent air, (5) Steady-state.
PROPERTIES: Determined from use of IHT software for iterative solution. Air, (Ti + To)/2 =
-5 2
-5 2
-1
362.7 K: νi = 2.23 × 10 m /s, ki = 0.031 W/m⋅K, αi = 3.20 × 10 m /s, β i = 0.00276 K , Pri =
-5 2
-5 2
0.698. Air, Tf = 312.7 K: νo = 1.72 × 10 m /s, ko = 0.027 W/m⋅K, αo = 2.44 × 10 m /s, β o =
-1
0.0032 K , Pro = 0.705.
ANALYSIS: From an energy balance on the radiation shield, q′i = q′o or q′rad,i + q′conv,i
= q′rad,o + q′conv,o . Evaluating the inner and outer radiation rates from Eqs. (13.25) and (13.27),
respectively, and the convection heat rate in the air gap from Eq. (9.58),
(
σ π Di Ti4 − To4
) + 2π keff (Ti − To ) = σ π D
n ( Do / Di )
1 1 − ε o  Di 
+


εi
ε o  Do 
(
)
4
4
o ε o To − Tsur + π Do h o ( To − T∞ )
From Eqs. (9.59) and (9.60)
1/ 4


Pri
k eff = 0.386 k i 

 0.861 + Pri 
( )
Ra ∗c
1/ 4
4
 n ( Do / Di ) Ra L
∗
Ra c =
5
L3 Di−3/ 5 + Do−3/ 5
)
(
3
where RaL = g β i (Ti – To)L /νi αi and L = (Do – Di)/2. From Eq. (9.34), the convection coefficient
on the outer surface of the shield is
2


1/
6


0.387 Ra D
k 
h o = o 0.60 +
8 / 27 
Do 

1 + ( 0.559 / Pr )9 /16 




Continued …..
PROBLEM 13.112 (Cont.)
The solution to the energy balance is obtained using the IHT software, and the result is
To = 332.5 K = 59.5°C
<
The corresponding value of the heat loss is
q′i = 88.7 W / m
<
COMMENTS: (1) The radiation and convection heat rates are q ′rad,i = 23.7 W / m, q ′rad,o = 10.4 W / m,
q ′conv,i = 65.0 W / m, and q ′conv,o = 78.3 W / m. Convection is clearly the dominant mode of heat transfer.
(2)With a value of To = 59.5°C > 35°C, the heat loss is reduced (88.7 W/m compared to 130.7 W/m if the
shield is at 35°C).
PROBLEM 13.113
KNOWN: Dimensions and inclination angle of a flat-plate solar collector. Absorber and cover plate
temperatures and emissivities.
FIND: (a) Rate of heat transfer by free convection and radiation, (b) Effect of the absorber plate
temperature on the heat rates.
SCHEMATIC:
ASSUMPTIONS: (1) Diffuse-gray, opaque surface behavior.
PROPERTIES: Table A-4, air ( T = ( T1 + T2 ) / 2 = 323 K ) : ν = 18.2 × 10 m /s, k = 0.028
-6
-6
2
2
-1
W/m⋅K, α = 25.9 × 10 m /s, Pr = 704, β = 0.0031 K .
ANALYSIS: (a) The convection heat rate is
q conv = hA ( T1 − T2 )
2
where A = wH=4 m and, with H/L > 12 and τ < τ* = 70 deg, h is given by Eq. 9.54. With a
Rayleigh number of
Ra L =
gβ ( T1 − T2 ) L3
αν

Nu L = 1 + 1.44 1 −

(
9.8 m / s 2 0.0031 K −1
=
)(40°C)(0.03 m )3 = 69, 600
25.9 × 10−6 m 2 / s × 18.2 × 10−6 m 2 / s
  1708 ( 0.923)   0.5 × 69, 600 1/ 3 
 − 1
 1 −
 + 
0.5 ( 69, 600 )   0.5 ( 69, 600 )  
5830



1708
Nu L = 1 + 1.44 [0.951][0.955] + 0.814 = 3.12
h = ( k / L ) Nu L = ( 0.028 W / m ⋅ K / 0.03 m ) 3.12 = 2.91 W / m 2 ⋅ K
(
q conv = 2.91 W / m 2 ⋅ K 4 m 2
)(70 − 30)°C = 466 W
<
The net rate of radiation exchange is given by Eq. 13.24.
q=
(
Aσ T14 − T24
1
1
+
−1
ε1 ε 2
) = (4 m2 )5.67 ×10−8 W / m2 ⋅ K4 (3434 − 3034 ) K4 = 1088 W
1
0.96
+
1
0.92
<
−1
(b) The effect of the absorber plate temperature was determined by entering Eq. 9.54 into the IHT
workspace and using the Properties and Radiation Toolpads.
Continued …..
PROBLEM 13.113 (Cont.)
4
As expected, the convection and radiation losses increase with increasing Ti, with the T dependence
providing a more pronounced increase for the radiation.
COMMENTS: To minimize heat losses, it is obviously desirable to operate the absorber plate at the
lowest possible temperature. However, requirements for the outlet temperature of the working fluid
may dictate operation at a low flow rate and hence an elevated plate temperature.
PROBLEM 13.114
KNOWN: Disk heated by an electric furnace on its lower surface and exposed to an environment on
its upper surface.
FIND: (a) Net heat transfer to (or from) the disk qnet,d when Td = 400 K and (b) Compute and plot
qnet,d as a function of disk temperature for the range 300 ≤ Td ≤ 500 K; determine steady-state
temperature of the disk.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Disk is isothermal; negligible thermal resistance,
(3) Surroundings are isothermal and large compared to the disk, (4) Non-black surfaces are graydiffuse, (5) Furnace-disk forms a 3-surface enclosure, (6) Negligible convection in furnace, (7)
Ambient air is quiescent.
-6
2
PROPERTIES: Table A-4, Air (Tf = (Td + T∞)/2 = 350 K,1 atm): ν = 20.92 × 10 m /s, k = 0.30
-6 2
W/m⋅K, α = 29.9 × 10 m /s.
ANALYSIS: (a) Perform an energy balance on the disk identifying: qrad as the net radiation
exchange between the disk and surroundings; qconv as the convection heat transfer; and q3 as the net
radiation leaving the disk within the 3-surface enclosure.
(1)
q net,d = E in − E out = −q rad − q conv − q3
Radiation exchange with surroundings: The rate equation is of the form
(
4
q rad = ε d,2 Adσ Td4 − Tsur
)
(2)
(
)
q rad = 0.8 (π / 4 )( 0.400 m ) 5.67 × 10−8 W / m 2 ⋅ K 4 4004 − 3004 K 4 = 99.8 W.
2
Free convection: The rate equation is of the form
q conv = hAd ( Td − T∞ )
(3)
where h can be estimated by an appropriate convection correlation. Find first,
Ra L = gβ∆TL3 / να
(4)
Ra L = 9.8 m / s 2 (1/ 350 K )( 400 − 300 ) K ( 0.400 m / 4 ) / 20.92 × 10−6 m / s 2 × 29.9 × 10−6 m 2 / s
3
Ra L = 4.476 × 106
4
7
where L = Ac/P = D/4. For the upper surface of a heated plate for which 10 ≤ RaL ≤ 10 , Eq. 9.30 is
the appropriate correlation,
Continued …..
PROBLEM 13.114 (Cont.)
1/ 4
Nu L = hL / k = 0.54 Ra L
(5)
(
h = 0.030 W / m ⋅ K / ( 0.400 m / 4 ) × 0.54 4.476 × 106
)
1/ 4
= 7.45 W / m 2 ⋅ K
Hence, from Eq. (3),
q conv = 7.45 W / m 2 ⋅ K (π / 4 )( 0.400 m ) ( 400 − 300 ) K = 93.6 W.
Furnace-disk enclosure: From Eq. 13.20, the net radiation leaving the disk is
2
q3 =
J 3 − J1
+
J3 − J 2
( A3F31 )−1 ( A3F32 )−1
= A 3 [F31 ( J 3 − J1 ) + F32 ( J 3 − J 2 )].
(6)
The view factor F32 can be evaluated from the coaxial parallel disks relation of Table 13.1 or from
Fig. 13.5.
R i = ri / L = 200 mm / 200 mm = 1,
R j = rj / L = 1,
(
)
( )
S = 1 + 1 + R 2j / R 2j = 1 + 1 + 12 12 = 3



(
F31 = 1/ 2 S − S2 − 4 rj / ri


1/ 2 
)2 


2 1/ 2 
2
 = 1/ 2 3 − 3 − 4 (1)   = 0.382.



(7)
From summation rule, F32 = 1 – F33 – F31 = 0.618 with F33 = 0. Since surfaces A2 and A3 are black,
J 2 = E b2 = σ T24 = σ (500 K ) = 3544 W / m 2
4
J 3 = E b3 = σ T34 = σ ( 400 K ) = 1452 W / m 2 .
4
To determine J1, use Eq. 13.21, the radiation balance equation for A1, noting that F12 = F32 and F13 =
F31,
E b1 − J1
J −J
J −J
= 1 2 + 1 3
(1 − ε1 ) / ε1A1 ( A1F12 )−1 ( A1F13 )−1
3544 − J1
J − 3544 J1 − 1452
= 1
+
(1 − 0.6 ) / 0.6 (0.618 )−1 (0.382 )−1
J1 = 3226 W / m 2 .
(8)
Substituting numerical values in Eq. (6), find
q 3 = (π / 4 )( 0.400 m )
2
 0.382 (1452 − 3226 ) W / m 2 + 0.618 (1452 − 3544 ) W / m 2  = −248 W.


Returning to the overall energy balance, Eq. (1), the net heat transfer to the disk is
q net,d = −99.8 W − 93.6 W − ( −248 W ) = +54.6 W
<
That is, there is a net heat transfer rate into the disk.
(b) Using the energy balance, Eq. (1), and the rate equation, Eqs. (2) and (3) with the IHT Radiation
Tool, Radiation, Exchange Analysis, Radiation surface energy balances and the Correlation Tool,
Free Convection, Horizontal Plate (Hot surface up), the analysis was performed to obtain qnet,d as a
function of Td. The results are plotted below.
The steady-state condition occurs when qnet,d = 0 for which
<
Td = 413 K
Continued …..
PROBLEM 13.114 (Cont.)
COMMENTS: The IHT workspace for the foregoing analysis is shown below.
Continued …..
PROBLEM 13.114 (Cont.)
PROBLEM 13.115
KNOWN: Radiation shield facing hot wall at Tw = 400 K is backed by an insulating material of
known thermal conductivity and thickness which is exposed to ambient air and surroundings at 300 K.
FIND: (a) Heat loss per unit area from the hot wall, (b) Radiosity of the shield, and (c) Perform a
parameter sensitivity analysis on the insulation system considering effects of shield reflectivity ρs,
insulation thermal conductivity k, overall coefficient h, on the heat loss from the hot wall.
SCHEMATIC:
ASSUMPTIONS: (1) Wall is black surface of uniform temperature, (2) Shield and wall behave as
parallel infinite plates, (3) Negligible convection in region between shield and wall, (4) Shield is
2
diffuse-gray and very thin, (5) Prescribed coefficient h = 10 W/m ⋅K is for convection and radiation.
ANALYSIS: (a) Perform an energy balance on the shield to obtain
q′′w − s = q′′cond
But the insulating material and the convection process at the exposed surface can be represented by a
thermal circuit.
In equation form, using Eq.13.24 for the wall and shield,
q′′w − s =
(
(
4
σ Tw
− Ts4
)
1/ ε w + 1/ ε s − 1
σ 4004 − Ts4
)=
1 + 1/ 0.05 − 1
=
Ts − T∞
L / k + 1/ h
(1,2)
(Ts − 300 ) K
(0.025 / 0.016 + 1/10 ) m 2 ⋅ K / W
Ts = 350 K.
where εs = 1 - ρs. Hence,
q′′w − s =
(350 − 300 ) K
= 30 W / m 2 .
2
(0.025 / 0.016 + 1/10 ) m ⋅ K / W
<
(b) The radiosity of the shield follows form the definition,
( )
( )
4
J s = ρs G s + ε s E b (Ts ) = ρs σ Tw
+ (1 − ρs ) σ Ts4 .
(3)
J s = 0.95σ ( 400 K ) + (1 − 0.95 )σ (350 K ) = 1421 W / m 2 .
<
4
-8
2
4
4
with σ = 5.67 × 10 W/m ⋅K .
Continued …..
PROBLEM 13.115 (Cont.)
(c) Using the Eqs. (1) and (2) in the IHT workspace, q′′w − s can be computed and plotted for selected
ranges of the insulation system variables, ρs, k, and h. Intuitively we know that q′′w − s will decrease
with increasing ρs, decreasing k and decreasing h. We chose to generate the following family of
curves plotting q′′w − s vs. k for selected values of ρs and h.
Considering the base condition with variable k, reducing k by a factor of 3, the heat loss is reduced by
2
a factor of 2. The effect of changing h (4 to 24 W/m ⋅K) has little influence on the heat loss.
However, the effect of shield reflectivity change is very significant. With ρs = 0.98, probably the
upper limit of a practical reflector-type shield, the heat loss is reduced by a factor of two. To improve
the performance of the insulation system, it is most advantageous to increase ρs and decrease k.
PROBLEM 13.116
KNOWN: Diameter and surface temperature of a fire tube. Gas low rate and temperature.
Emissivity of tube and partition.
FIND: (a) Heat transfer per unit tube length, q′, without the partition, (b) Partition temperature, Tp,
and heat rate with the partition, (c) Effect of flow rate and emissivity on q′ and Tp. Effect of
emissivity on radiative and convective contributions to q′.
SCHEMATIC:
ASSUMPTIONS: (1) Fully-developed flow in duct, (2) Diffuse/gray surface behavior, (3)
Negligible gas radiation.
-7
2
PROPERTIES: Table A-4, air (Tm,g = 900 K): µ = 398 × 10 N⋅s/m , k = 0.062 W/m⋅K, Pr = 0.72;
-7
2
air (Ts = 385 K): µ = 224 × 10 N⋅s/m .
ANALYSIS: (a) Without the partition, heat transfer to the tube wall is only by convection. With m
−7
/ π Dµ = 4 ( 0.05 kg / s ) / π ( 0.07 m ) 398 × 10
= 0.05 kg/s and ReD = 4 m
N ⋅ s / m 2 = 22,850, the
g
flow is turbulent. From Eq. (8.61),
0.14
4 / 5 1/ 3
Nu D = 0.027 Re D
Pr ( µ / µs )
h=
k
D
Nu D =
0.062 W / m ⋅ K
0.07 m
(
4/5
= 0.027 ( 22,850 )
(0.72 )1/ 3 (398 / 224 )0.14 = 80.5
80.5 = 71.3 W / m 2 ⋅ K
)
q′ = hπ D Tm,g − Ts = 71.3 W / m 2 ⋅ K (π ) 0.07 m (900 − 385 ) = 8075 W / m
(b) The temperature of the partition is determined from an energy balance which equates net radiation
exchange with the tube wall to convection from the gas. Hence, q′′rad = q′′conv , where from Eq. 13.23,
q′′rad =
<
(
σ Tp4 − Ts4
1 − εp
εp
+
1
Fps
+
)
1 − εs Ap
ε s As
where F12 = 1 and Ap/As = D/(πD/2) = 2/π = 0.637. The flow is now in a noncircular duct for which
2
/ 2 = 0.025
Dh = 4Ac/P = 4(πD /8)/(πD/2+D) = πD/(π + 2) = 0.611 D = 0.0428 m and m
g
1/ 2 = m
2
2
kg/s. Hence, ReD = m
1/ 2 Dh/(πD /8)µ = 8(0.025 kg/s) (0.0428 m)/π(0.07 m) 398 ×
1/ 2 Dh/Acµ = m
-7
2
10 N⋅s/m = 13,970 and
4/5
Nu D = 0.027 (13, 970 )
h=
k
Dh
Nu D =
(0.72 )1/ 3 (398 / 224 )0.14 = 54.3
0.062 W / m ⋅ K
0.0428 m
54.3 = 78.7 W / m 2 ⋅ K
Continued …..
PROBLEM 13.116 (Cont.)
(
)
Hence, with εs = εp = 0.5 and q′′conv = h Tm,g − Tp ,
(
)
5.67 × 10−8 W / m 2 ⋅ K 4 Tp4 − 3854 K 4
1 + 1 + 0.637
(
)
= 78.7 W / m 2 ⋅ K 900 − Tp K
21.5 × 10−8 Tp4 + 78.7Tp − 71, 302 = 0
A trial-and-error solution yields
<
Tp = 796 K
The heat rate to one-half of the tube is then
(
Dσ Tp4 − Ts4
′ 2 = q′ps + q′conv =
q1/
1 − εp
εp
(
0.07 m 5.67 × 10
′ 2=
q1/
−8
2
W /m ⋅K
4
)
1 1 − εs Ap
+
+
ε s As
Fps
)(796.4
4
− 385
4
)K
2.637
(
+ h (π D / 2 ) Tm,g − Ts
)
4
2
+ 78.7 W / m ⋅ K ( 0.110 m )(900 − 385 ) K
′ 2 = 572 W / m + 4458 W / m = 5030 W / m
q1/
The heat rate for the entire tube is
<
′ 2 = 10, 060 W / m
q′ = 2q1/
(c) The foregoing model was entered into the IHT workspace, and parametric calculations were
performed to obtain the following results.
Radiation transfer from the partition increases with increasing εp = εs, thereby reducing Tp while
T and q ′ also increase with m.
increasing q′. Since h increases with increasing m,
p
Continued …..
PROBLEM 13.116 (Cont.)
Although the radiative contribution to the heat rate increases with increasing εp = εs, it still remains
small relative to convection.
COMMENTS: Contrasting the heat rate predicted for part (b) with that for part (a), it is clear that
use of the partition enhances heat transfer to the tube. However, the effect is due primarily to an
increase in h and secondarily to the addition of radiation.
PROBLEM 13.117
KNOWN: Height and width of a two-dimensional cavity filled with helium. Temperatures and
emissivities of opposing vertical plates.
FIND: (a) Heat rate per unit length, (b) Effect of L on heat rate.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Isothermal plates, (3) Diffuse-gray surfaces, (4)
Reradiating cavity sidewalls.
-6
2
PROPERTIES: Table A-4, Helium (T = 318 K, 1 atm): ν = 136 × 10 m /s, k = 0.158 W/m⋅K, α =
-6 2
-1
201 × 10 m /s, Pr = 0.679, β = 0.00314 K .
ANALYSIS: (a) The power generated by the electronics leaving the surface A1 is q′ = q′conv + q′rad ,
or
q′ = hH ( T1 − T2 ) +
(
Hσ T14 − T24
)
1 − ε1
1
1 − ε2
+
ε1 F + [(1/ F ) + (1/ F )]−1
ε2
12
1R
2R
The free convection coefficient can be estimated using Eq. 9.50 with
Ra L =
gβ ( T1 − T2 ) L3
αν
=
(
)
9.8 m / s 2 0.00314 K −1 60 K ( 0.02 m )
3
201 × 10−6 × 136 × 10−6 m 4 / s 2
= 540
However, since RaL < 1000, free convection effects can be neglected, in which case heat transfer is by
conduction and Nu L = 1. Hence,
h = Nu L ( k / L ) = 1.0 ( 0.158 W / m ⋅ K / 0.02m ) = 7.9 W / m 2 ⋅ K.
The view factor can be found from Fig. 13.4 with X/L = 8 and Y/L = ∞. Hence, F12 = 0.9 and F1R =
F2R = 0.1. It follows that
q′ = 7.9 W / m 2 ⋅ K ( 0.16 m )( 60°C ) +
)
(
0.16 m × 5.67 × 10−8 W / m 2 ⋅ K 3484 − 2884 K 4
0.25 +
1
0.9 + 10 + 10−1 

+ 0.25

<
q′ = 75.8 W / m + 45.5 W / m = 121 W / m.
Continued …..
PROBLEM 13.117 (Cont.)
(b) To assess the effect of plate spacing on convection heat transfer, q′conv = hH ( T1 − T2 ) was
(
)
computed by using both Eq. 9.50 and the conduction limit Nu L = 1 to determine h. These
expressions were entered into the IHT workspace, and the Radiation Toolpad was used to obtain the
appropriate radiation rate equation and view factor.
The cross-over at L = 28 mm marks the plate spacing below and above which, respectively, the
conduction limit and Eq. 9.50 are applicable. Although there is a slight increase in q′conv with
increasing L for L > 28 mm, the increase pales by comparison with that corresponding to a reduction
in L for L < 28 mm. As the two plates are brought closer to each other in the conduction limit, the
reduction in the corresponding thermal resistance significantly increases the heat rate. The total heat
rate and the conduction and radiative components are also plotted for 5 ≤ L ≤ 25 mm. There is an
increase in q′rad with decreasing L, due to an increase in F12 (F12 = 0.97 for L = 5 mm). However,
the increase is small, and conduction is the dominant heat transfer mode.
COMMENTS: Even for small values of L, the total heat rate is small and the scheme is poorly
suited for electronic cooling. Note that helium is preferred over air on the basis of its larger thermal
conductivity.
PROBLEM 13.118
KNOWN: Diameters, temperatures, and emissivities of concentric spheres.
FIND: Rate at which nitrogen is vented from the inner sphere. Effect of radiative properties on
evaporation rate.
SCHEMATIC:
ASSUMPTIONS: Diffuse-gray surfaces.
5
PROPERTIES: Liquid nitrogen (given): hfg = 2 × 10 J/kg; Table A-4, Helium ( T = (Ti + To)/2 =
-6
2
-6
2
180 K, 1 atm): ν = 51.3 × 10 m /s, k = 0.107 W/m⋅K, α = 76.2 × 10 m /s, Pr = 0.673, β =
-1
0.00556 K .
ANALYSIS: (a) Performing an energy balance for a control surface about the liquid nitrogen, it
follows that q = qconv + qrad = mh
fg . From the Raithby-Hollands expressions for free convection
between concentric spheres, qconv = keffπ(Di Do/L)(To – Ti), where
1/ 4


1/ 4

Pr
L

 
k eff = 0.74 k 

 
 0.861 + Pr  
4
−7 / 5
−7 / 5 5 
( D D ) Di + D o

 o i
Ra L =
gβ ( To − Ti ) L3
να
)
(
)(206 K )(0.05m )3 = 3.589 ×105.
(51.3 ×10−6 m2 / s )(76.2 ×10−6 m2 / s )
(
9.8 m / s 2 0.00556 K −1
=
Hence,
1/ 4 
0.673


k eff = 0.74 ( 0.107 W / m ⋅ K ) 

 0.861 + 0.673 
The heat rate by convection is
1/ 4
0.05 3.589 × 105 


 (1.1)4 (1 + 0.875 )5 
(
= 0.309 W / m ⋅ K.
)
q conv = ( 0.309 W / m ⋅ K )π 1.10 m 2 / 0.05 m 206 K = 4399 W.
From Table 13.3,
q rad = q oi =
(
σπ D12 To4 − Ti 4
)
1/ ε i + ((1 − ε o ) / ε o ) ( Di / Do )
2
2
5.67 × 10−8 W / m 2 ⋅ K 4 )π (1 m ) ( 2834 − 77 4 ) K 4
(
=
= 216 W.
1/ 0.3 + ( 0.7 / 0.3)(1/1.1)
2
Continued …..
PROBLEM 13.118 (Cont.)
Hence,
5
=q/h
m
fg = ( 4399 + 216 ) W / 2 × 10 J / kg = 0.023 kg / s.
<
With the cavity evacuated, IHT was used to compute the radiation heat rate as a function of εi = εo.
Clearly, significant advantage is associated with reducing the emissivities and qrad = 31.8 W for εi =
εo = 0.05.
COMMENTS: The convection heat rate is too large. It could be reduced by replacing He with a gas
of smaller k, a cryogenic insulator (Table A.3), or a vacuum. Radiation effects are second order for
small values of the emissivity.
PROBLEM 13.119
KNOWN: Dimensions, emissivity and upper temperature limit of coated panel. Arrangement and
power dissipation of a radiant heater. Temperature of surroundings.
FIND: (a) Minimum panel-heater separation, neglecting convection, (b) Minimum panel-heater
separation, including convection.
SCHEMATIC:
ASSUMPTIONS: (1) Top and bottom surfaces of heater and panel, respectively, are adiabatic, (2)
Bottom and top surfaces of heater and panel, respectively are diffuse-gray, (3) Surroundings form a
large enclosure about the heater-panel arrangement, (4) Steady-state conditions, (5) Heater power is
dissipated entirely as radiation (negligible convection), (6) Air is quiescent and convection from panel
may be approximated as free convection from a horizontal surface, (7) Air is at atmospheric pressure.
-6
2
PROPERTIES: Table A-4, Air (Tf = (400 + 298)/2 ≈ 350 K, 1 atm): ν = 20.9 × 10 m /s, k = 0.03
-6 2
-3 -1
W/m⋅K, Pr = 0.700, α = 29.9 × 10 m /s, β = 2.86 × 10 K .
ANALYSIS: (a) Neglecting convection effects, the panel constitutes a floating potential for which
the net radiative transfer must be zero. That is, the panel behaves as a re-radiating surface for which
Eb2 = J2. Hence
J − E b2 J1 − E b3
+
(1)
q1 = 1
1/ A1F12 1/ A1F13
and evaluating terms
E b2 = σ T24 = 5.67 × 10−8 W / m 2 ⋅ K 4 ( 400 K ) = 1452 W / m 2
4
E b3 = σ T34 = 5.67 × 10−8 W / m 2 ⋅ K 4 ( 298 K ) = 447 W / m 2
4
F13 = 1 − F12
A1 = 25 m 2
find that
75, 000 W
25 m
2
J − 1452
J − 447
= 1
+ 1
1/ F12
1/ (1 − F12 )
3000 W / m 2 = F12 ( J1 − 1452 ) + ( J1 − 447 ) − F12 ( J1 − 447 )
J1 = 3447 + 1005F12 .
(2)
Performing a radiation balance on the panel yields
J1 − E b2 E b2 − E b3
=
.
1/ A1F12
1/ A 2 F23
Continued …..
PROBLEM 13.119 (Cont.)
With A1 = A2 and F23 = 1 – F12
F1 ( J1 − 1452 ) = (1 − F12 )(1452 − 447 )
or
447F12 = F12 J1 − 1005.
(3)
Substituting for J1 from Eq. (2), find
447F12 = F12 (3447 + 1005F12 ) − 1005
2
1005F12
+ 3000F12 − 1005 = 0
F12 = 0.30.
Hence from Fig. 13.4, with X/L = Y/L and Fij = 0.3,
X / L ≈ 1.45
<
L ≈ 5 m /1.45 = 3.45 m.
(b) Accounting for convection from the panel, the net radiation transfer is no longer zero at this
surface and Eb2 ≠ J2. It then follows that
q1 =
J1 − J 2
J − E b3
+ 1
1/ A1F12 1/ A1F13
(4)
where, from an energy balance on the panel,
J 2 − E b2
(1 − ε 2 ) / ε 2 A 2
= q conv,2 = hA 2 ( T2 − T∞ ) .
(5)
2
With L ≡ As/P = 25 m /20 m = 1.25 m,
Ra L =
gβ ( Ts − T∞ ) L3
να
=
)
(
9.8 m / s 2 2.86 × 10−3 K −1 (102 K )(1.25 m )
( 20.9 × 29.9 )10
−12
3
4
m /s
2
= 8.94 × 109.
Hence
(
3
9
Nu L = 0.15Ra1/
L = 0.15 8.94 × 10
h = 311 k / L = 311
0.03 W / m ⋅ K
1.25 m
)
1/ 3
= 311
= 7.46 W / m 2 ⋅ K
q′′conv,2 = 7.46 W / m 2 ⋅ K (102 K ) = 761 W / m 2 .
From Eq. (5)
J 2 = E b2 +
1 − ε2
ε2
q′′conv,2 = 1452 +
0.7
0.3
761 = 3228 W / m 2 .
Continued …..
PROBLEM 13.119 (Cont.)
From Eq. (4),
75, 000
25
J − 3228
J − 447
= 1
+ 1
1/ F12
1/ (1 − F12 )
3000 = F12 ( J1 − 3228 ) + J1 − 447 − F12 ( J1 − 447 )
J1 = 3447 + 2781F12 .
(6)
From an energy balance on the panel,
E −J
J 2 − E b2
+ b3 2 =
= q conv,2
1/ A1F12 1/ A 2 F23 (1 − ε 2 ) / ε 2 A 2
J1 − J 2
F12 ( J1 − 3228 ) + (1 − F12 )( 447 − 3228 ) = 761
F12 J1 − 447F12 = 3542.
Substituting from Eq. (6),
F12 (3447 + 2781F12 ) − 447F12 = 3542
2
+ 3000F12 − 3542 = 0
2781F12
F12 = 0.71.
Hence from Fig. 13.4, with X/L = Y/L and Fij = 0.71,
X / L = 5.7
L ≈ 5 m / 5.7 = 0.88 m.
<
COMMENTS: (1) The results are independent of the heater surface radiative properties.
(2) Convection at the heater surface would reduce the heat rate q1 available for radiation exchange
and hence reduce the value of L.
PROBLEM 13.120
KNOWN: Diameter and emissivity of rod heater. Diameter and position of reflector. Width,
emissivity, temperature and position of coated panel. Temperature of air and large surroundings.
FIND: (a) Equivalent thermal circuit, (b) System of equations for determining heater and reflector
temperatures. Values of temperatures for prescribed conditions, (c) Electrical power needed to
operate heater.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Diffuse-gray surfaces, (3) Large surroundings act
as blackbody, (4) Surfaces are infinitely long (negligible end effects), (5) Air is quiescent, (6)
Negligible convection at reflector, (7) Reflector and panel are perfectly insulated.
-6
2
PROPERTIES: Table A-4, Air (Tf = 350 K, 1 atm): k = 0.03 W/m⋅K, ν = 20.9 × 10 m /s, α = 29.9
-6 2
-6 2
× 10 m /s, Pr = 0.70; (Tf = (1295 + 300)/2 = 800 K): k = 0.0573 W/m⋅K, ν = 84.9 × 10 m /s, α =
-6 2
120 × 10 m /s.
ANALYSIS: (a) We have assumed blackbody behavior for A1 and A4; hence, J = Eb. Also, A2 is
insulated and has negligible convection; hence q = 0 and J2 = Eb2. The equivalent thermal circuit is:
(b) Performing surface energy balances at 1, 2 and 3:
E − E b2 E b1 − J3 E b1 − E b4
+
+
q1 − q conv,1 = b1
1/ A1F12
1/ A1F13
1/ A1F14
E b1 − E b2
(2)
E b1 − J3
(3a)
= q conv,3 .
(3b)
0=
J3 − E b3
(1 − ε 3 ) / ε 3A3
(1)
=
J − E b2 E b4 − E b2
+ 3
+
1/ A 2 F21 1/ A 2 F23 1/ A 2 F24
E −J
E −J
+ b2 3 + b4 3
1/ A3F31 1/ A3F32 1/ A3F34
where
J3 − E b3
(1 − ε 3 ) / ε 3A3
Continued …..
PROBLEM 13.120 (Cont.)
Solution procedure with Eb3 and Eb4 known: Evaluate qconv,3 and use Eq. (3b) to obtain J3; Solve
Eqs. (2) and (3a) simultaneously for Eb1 and Eb2 and hence T1 and T2; Evaluate qconv,1 and use Eq.
(1) to obtain q1.
For free convection from a heated, horizontal plate:
Lc =
As
P
Ra L =
=
(W × L)
( 2L + 2W )
≈
gβ ( T3 − T∞ ) L3c
αν
W
2
= 0.5 m
−1
=
9.8 m / s 2 (350 K )
20.9 × 29.9 × 10
(
3
8
Nu L = 0.15Ra1/
L = 0.15 5.6 × 10
h3 =
k
Lc
Nu L =
(100 K )(0.5 m )3
)
1/ 3
0.03 W / m ⋅ K × 123.6
0.5 m
−12
4
m /s
2
= 5.6 × 108
= 123.6
= 7.42 W / m 2 ⋅ K.
q′′conv,3 = h3 ( T3 − T∞ ) = 742 W / m 2 .
Hence, with
E b3 = σ T34 = 5.67 × 10 −8 W / m 2 ⋅ K 4 ( 400 K ) = 1451 W / m 2
4
using Eq. (3b) find
J 3 = E b3 +
1 − ε3
ε 3A3
q conv,3 = (1451 + [0.3 / 0.7 ] 742 ) = 1769 W / m 2 .
-1
-1
View Factors: From symmetry, it follows that F12 = 0.5. With θ = tan (W/2)/H = tan (0.5) =
26.57°, it follows that
F13 = 2θ / 360 = 0.148.
From summation and reciprocity relations,
F14 = 1 − F12 − F13 = 0.352
F21 = ( A1 / A 2 ) F12 = ( 2D1 / D 2 ) F12 = 0.02 × 0.5 = 0.01
F23 = ( A3 / A 2 ) F32 = ( 2 / π )( F32′ − F31 ).
For X/L = 1, Y/L ≈ ∞, find from Fig. 13.4 that F32′ ≈ 0.42. Also find,
F31 = ( A1 / A3 ) F13 = (π × 0.01/1) 0.148 = 0.00465 ≈ 0.005
F23 = ( 2 / π )(0.42 − 0.005 ) = 0.264
F22 ≈ 1 − F22′ = 1 − ( A′2 / A 2 ) F2′2 = 1 − ( 2 / π ) = 0.363
F24 = 1 − F21 − F22 − F23 = 0.363
Continued …..
PROBLEM 13.120 (Cont.)
F31 = 0.005,
F32 = 0.415
F34 = 1 − F32′ = 1 − 0.42 = 0.58.
With E b4 = σ T44 = 5.67 × 10−8 W / m 2 ⋅ K 4 (300 K ) = 459 W / m 2 ,
Eq. (3a) → 0.005(Eb1 – 1769) + 0.415(Eb2 – 1769) + 0.58(459 – 1769) = 742
0.005Eb1 + 0.415Eb2 = 2245
4
Eq. (2) → 0.01(Eb1 – Eb2) + 0.264(1769 – Eb2) + 0.363(459 – Eb2) = 0
0.01Eb1 – 0.637Eb2 + 633.6 = 0.
(4)
(5)
Hence, manipulating Eqs. (4) and (5), find
E b2 = 0.0157E b1 + 994.7
0.005E b1 + ( 0.415 )( 0.0157E b1 + 994.7 ) = 2245.
T1 = ( E b1 / σ )
1/ 4
E b1 = 159, 322 W / m 2
E b2 = 0.0157 (159, 322 ) + 994.7 = 3496 W / m 2
<
= 1295 K
T2 = ( E b2 / σ )
1/ 4
= 498 K.
<
(c) With T1 = 1295 K, then Tf = (1295 + 300)/2 ≈ 800 K, and using
Ra D =
gβ ( T1 − T∞ ) D13
αν
9.8 m / s 2 (1/ 800 K )(1295 − 300 ) K ( 0.01 m )
3
=
120 × 84.9 × 10−12 m 4 / s 2
= 0.85 (1196 )
Nu D = 0.85Ra 0.188
D
0.188
= 1196
= 3.22
h1 = ( k / D1 ) Nu D = (0.0573 / 0.01) × 3.22 = 18.5 W / m 2 ⋅ K.
The convection heat flux is
q′′conv,1 = h1 ( T1 − T∞ ) = 18.5 (1295 − 300 ) = 18, 407 W / m 2 ,
Using Eq. (1), find
q1′′ = q′′conv,1 + F12 ( E b1 − E b2 ) + F13 ( E b1 − J3 ) + F14 ( E b1 − E b4 )
q1′′ = 18, 407 + 0.5 (159, 322 − 3496 )
+0.148 (159, 322 − 1769 ) + 0.352 (159, 322 − 459 )
q1′′ = 18, 407 + ( 77, 913 + 23, 314 + 55, 920 )
q1′′ = 18, 407 + 236, 381 = 254, 788 W / m 2
q1′ = π D1q1′′ = π ( 0.01) 254, 788 = 8000 W / m.
<
COMMENTS: Although convection represents less than 8% of the net radiant transfer from the
heater, it is equal to the net radiant transfer to the panel. Since the reflector is a re-radiating surface,
results are independent of its emissivity.
PROBLEM 13.121
KNOWN: Temperature, power dissipation and emissivity of a cylindrical heat source. Surface
emissivities of a parabolic reflector. Temperature of air and surroundings.
FIND: (a) Radiation circuit, (b) Net radiation transfer from the heater, (c) Net radiation transfer from
the heater to the surroundings, (d) Temperature of reflector.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Heater and reflector are in quiescent and infinite
air, (3) Surroundings are infinitely large, (4) Reflector is thin (isothermal), (5) Diffuse-gray surfaces.
-6
2
PROPERTIES: Table A-4, Air (Tf = 750 K, 1 atm): ν = 76.37 × 10 m /s, k = 0.0549 W/m⋅K, α =
-6 2
109 × 10 m /s, Pr = 0.702.
ANALYSIS: (a) The thermal circuit is
(b) Energy transfer from the heater is by radiation and free convection. Hence,
′
P1′ = q1′ + q1,conv
where
′
= hπ D1 ( T1 − T∞ )
q1,conv
and
Ra D =
gβ ( T1 − T∞ ) D3
−1
=
9.8 m / s 2 ( 750 K )
(900 K )(0.005 m )3
να
76.37 × 109 × 10−12 m 4 / s 2
Using the Churchill and Chu correlation, find
= 176.6.
2
Nu D
2

 

1/ 6
6




0.387
176.6
0.387Ra1/
(
)
 
D
= 0.6 +
= 0.6 +
= 1.85

8 / 27
8 / 27 
9
/16
9
/16

 

1 + ( 0.559 / Pr )

1 + ( 0.559 / 0.702 )






 

h = Nu D ( k / D ) = 1.85 (0.0549 W / m ⋅ K / 0.005 m ) = 20.3 W / m 2 ⋅ K.
Continued …..
PROBLEM 13.121 (Cont.)
Hence,
′
= 20.3 W / m 2 ⋅ Kπ ( 0.005 m )(1200 − 300 ) K = 287 W / m
q1,conv
<
q1′ = 1500 W / m − 287 W / m = 1213 W / m.
(c) The net radiative heat transfer from the heater to the surroundings is
q1′ (sur ) = A1′ F1sur ( J1 − Jsur ) .
The view factor is
F1sur = (135 / 360 ) = 0.375
and the radiosities are
4
J sur = σ Tsur
= 5.67 × 10−8 W / m 2 ⋅ K 4 (300 K ) = 459 W / m 2
4
J1 = E b1 − q1′ (1 − ε1 ) ε1A1′ = 5.67 × 10−8 W / m 2 ⋅ K 4 (1200 K )
4
−1213 W / m [0.2 / 0.8π ( 0.005 m )]
J1 = 98, 268 W / m 2 .
Hence
q1′ (sur ) = π ( 0.005 m ) 0.375 (98, 268 − 459 ) W / m 2 = 576 W / m.
<
(d) Perform an energy balance on the reflector,
q′2i = q′2o + q′2,conv
J 2i − E b2
(1 − ε 2i ) / ε 2i A′2
=
E b2 − Jsur
+ 2h 2 A′2 ( T2 − T∞ ) .
(1 − ε 2o ) / ε 2o A′2 + 1/ A′2 F2o(sur )
The radiosity of the reflector is
q1′ ( 2i )
(1213 − 576 ) W / m
J 2i = J1 −
= 98, 268 W / m 2 −
A1′ F1( 2i )
π ( 0.005 m )( 225 / 360 )
J 2i = 33, 384 W / m 2 .
Hence
( )=
( )
33, 384 − 5.67 × 10−8 T24
5.67 × 10−8 T24 − 459
(0.9 / 0.1 × 0.2 m )
(0.2 / 0.8 × 0.2 m ) + (1/ 0.2m ×1)
+ 2 × 0.4 ( T2 − 300 )
741.9 − 0.126 × 10−8 T24 = 0.907 × 10−8 T24 − 73.4 + 0.8T2 − 240
1.033 × 10−8 T24 + 0.8T2 = 1005
and from a trial and error solution, find
T2 = 502 K.
<
COMMENTS: Choice of small ε2i and large ε2o insures that most of the radiation from heater is
reflected to surroundings and reflector temperature remains low.
PROBLEM 13.122
KNOWN: Geometrical conditions associated with tube array. Tube wall temperature and pressure
of water flowing through tubes. Gas inlet velocity and temperature when heat is transferred from
products of combustion in cross-flow, or temperature of electrically heated plates when heat is
transferred by radiation from the plates.
FIND: (a) Steam production rate for gas flow without heated plates, (b) Steam production rate with
heated plates and no gas flow, (c) Effects of inserting unheated plates with gas flow.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible gas radiation, (3) Tube and plate
surfaces may be approximated as blackbodies, (4) Gas outlet temperature is 600 K.
3
PROPERTIES: Table A-4, Air ( T = 900 K, 1 atm): ρ = 0.387 kg/m , cp = 1121 J/kg⋅K, ν = 102.9 ×
-6 2
3
10 m /s, k = 0.062 W/m⋅K, Pr = 0.720; (T = 400 K): Pr = 0.686; (T = 1200 K): ρ = 0.29 kg/m ;
6
Table A-6, Sat. water (2.5 bars): hfg = 2.18 × 10 J/kg.
ANALYSIS: (a) With
Vmax = [ST / (ST − D )] V = 20 m / s
V
D 20 m / s (0.01 m )
Re D = max =
= 1944
ν
102.9 × 10−6 m 2 / s
and from the Zhukauskas correlation with C = 0.27 and m = 0.63,
Nu D = 0.27 (1944 )
0.63
(0.720 )0.36 (0.720 / 0.686 )1/ 4 = 28.7
h = 0.062 W / m ⋅ K × 28.7 / 0.01 m = 178 W / m 2 ⋅ K.
The outlet temperature may be evaluated from
 hA
Ts − Tm,o
= exp  −
 mc
Ts − Tm,i
 p


hNπ DL 
 = exp  −


 ρ VN TST Lc p 


178 W / m 2 ⋅ K × 100 × π × 0.01 m
= exp  −

 0.29 kg / m3 × 10 m / s × 5 × 0.02 m × 1121 J / kg ⋅ K 
400 − 1200


400 − Tm,o
Tm,o = 543 K.
Continued …..
PROBLEM 13.122 (Cont.)
With
∆T"m =
(Ts − Tm,i ) − (Ts − Tm,o ) = −800 − ( −143) = −382 K
ln (800 /143)
ln ( Ts − Tm,i ) / ( Ts − Tm,o )
find
q = hA∆T"m = 178 W / m 2 ⋅ K (100 )π ( 0.01 m )1 m ( −382 K )
q = −214 kW.
hfg,
If the water enters and leaves as saturated liquid and vapor, respectively, it follows that –q = m
hence
214, 000 W
=
= 0.098 kg / s.
m
2.18 × 106 J / kg
<
(b) The radiation exchange between the plates and tube walls is
(
)
q =  A p Fpsσ Tp4 − Ts4  ⋅ 2 ⋅ NT

where the factor of 2 is due to radiation transfer from two plates. The view factor and area are
1/ 2
2
Fps = 1 − 1 − ( D / S ) 


(
)
1/ 2
+ ( D / S ) tan −1  S2 − D 2 / D2 


Fps = 1 − 0.866 + 0.5 tan −1 1.732 = 1 − 0.866 + 0.524
Fps = 0.658
A p = N L ⋅ SL ⋅ 1 m = 20 × 0.02 m × 1 m = 0.40 m 2 .
Hence,
(
)
q = 5 ×  0.80 m 2 × 0.658 × 5.67 × 10−8 W / m 2 ⋅ K 4 12004 − 4004 K 4 


q = 305, 440 W
and the steam production rate is
=
m
305, 440 W
2.18 × 106 J / kg
= 0.140 kg / s.
<
(c) The plate temperature is determined by an energy balance for which convection to the plate from
the gas is equal to net radiation transfer from the plate to the tube. Conditions are complicated by the
fact that the gas transfers energy to both the plate and the tubes, and its decay is not governed by a
simple exponential. Insertion of the plates enhances heat transfer to the tubes and thereby increases
the steam generation rate. However, for the prescribed conditions, the effect would be small, since in
case (a), the heat transfer is already ≈ 80% of the maximum possible transfer.
PROBLEM 13.123
KNOWN: Gas-fired radiant tube located within a furnace having quiescent gas at 950 K. At a
particular axial location, inner wall and gas temperature measured by thermocouples.
FIND: Temperature of the outer tube wall at the axial location where the thermocouple
measurements are being made.
SCHEMATIC:
ASSUMPTIONS: (1) Silicon carbide tube walls have negligible thermal resistance and are diffusegray, (2) Tubes are positioned horizontally, (3) Gas is radiatively non-participating and quiescent, (4)
Furnace gas behaves as ideal gas, β = 1/T.
3
-6
2
PROPERTIES: Gas (given): ρ = 0.32 kg/m , ν = 130 × 10 m /s, k = 0.070 W/m⋅K, Pr = 0.72, α =
-4 2
ν/Pr = 1.806 × 10 m /s.
ANALYSIS: Consider a segment of the outer tube at
the
Prescribed axial location and perform an energy balance,
E ′in − E ′out = 0
q′rad,i + q′conv,i − q′rad,o − q′conv,o = 0
(1)
The heat rates by radiative transfer are:
Inside: For long concentric cylinders, Eq. 13.25,
q′rad,i =
q′rad,i =
(
4
4
σπ Di Ts,i
− Ts,o
)
1/ ε1 + (1 − ε 2 ) / ε 2 ( Di / Do )
(
)
4
5.67 × 10−8 W / m 2 ⋅ K 4π ( 0.10 m ) 12004 − Ts,o
K4
1/ 0.6 + (1 − 0.6 ) / 0.6 ( 0.10 / 0.20 )
(
)
4
q′rad,i = 8.906 × 10−9 12004 − Ts,o
.
Outside: For the outer tube surface to large surroundings,
(
)
(2)
(
)
4
4
4
− Tsur
= 0.6π ( 0.20 m ) 5.67 × 10−8 W / m 2 ⋅ K 4 Ts,o
− 9504 K 4
q′rad,o = επ Doσ Ts,o
(
)
4
− 9504 .
q′rad,o = 2.138 × 10−8 Ts,o
(3)
The heat rates by convection processes are:
Continued …..
PROBLEM 13.123 (Cont.)
Inside: The rate equation is
q′conv,i = h iπ Do Tm,g − Ts,o .
(
)
(4)
(
)
Find the Reynolds number with A c = π Do2 − Di2 / 4 and D h = 4 A c / P,
Dh =
(
/ ρ A = 0.13 kg / s /  0.32 kg / m × π / 4 0.2 − 0.1
um = m
c
Re D = u m D h / ν
(
4 (π / 4 ) D o − Di
4
2
π ( D o + Di )
) = π (0.2
2
2
− 0.1
)m
3

2
( 0.2 + 0.1) m
= 0.100 m
Re D =
2
2
) m  = 17.2 m / s
2
17.2 m / s × 0.100 m
130 × 10
−6
2
= 13, 231.
m /s
The flow is turbulent and assumed to be fully developed; from the Dittus-Boelter correlation,
0.3
Nu D = hDh / k = 0.023 Re0.8
D Pr
0.070 W / m ⋅ K
0.8
0.3
× 0.023 (13, 231) ( 0.720 ) = 28.9 W / m 2 ⋅ K
hi =
0.100 m
Substituting into Eq. (4),
(
)
(
)
q′conv,i = 28.9 W / m 2 ⋅ K × π ( 0.20 m ) 1050 − Ts,o K = 18.16 1050 − Ts,o .
(5)
Outside: The rate equation is
q′conv,o = h oπ Do Ts,o − T∞ .
(
)
Evaluate the Rayleigh number assuming Ts,o = 1025 K so that Tf = 987 K,
Ra D =
gβ∆TD3o
9.8 m 2 / s 2 (1/ 987 K )(1025 − 950 ) K ( 0.20 m )
3
=
−6
να
130 × 10 m / s × 1.806 × 10
For a horizontal tube, using Eq. 9.33 and Table 9.1,
2
(
n
= 0.48 2.537 × 105
Nu D = h o Do / k = CRa D
)
1/ 4
−4
2
= 2.537 × 105.
m /s
= 10.77
h o = ( 0.070 W / m ⋅ K ) / 0.20 m × 10.77 = 3.77 W / m 2 ⋅ K.
Substituting into Eq. (6)
(
)
(
)
q′conv,o = 3.77 W / m 2 ⋅ K × π ( 0.20 m ) Ts,o − 950 K = 2.369 Ts,o − 950 .
(7)
Returning to the energy balance relation on the outer tube, Eq. (1), substitute for the individual rates
from Eqs. (2, 5, 3, 7),
8.906 × 10
−9
(1200
4
4
)
(
)
− Ts,o + 18.16 1050 − Ts,o − 2.138 × 10
By trial-and-error, find
−8
(T
)
(
)
4
4
s,o − 950 − 2.369 Ts,o − 950 = 0
(8)
<
Ts,o = 1040 K.
COMMENTS: (1) Recall that in estimating ho we assumed Ts,o = 1025 K, such that ∆T = 75 K (vs.
92 K using Ts,o = 1042 K) for use in evaluating the Rayleigh number. For an improved estimate of
Ts,o, it would be advisable to recalculate ho.
(2) Note from Eq. (8) that the radiation processes dominate the heat transfer rate:
q′rad ( W / m )
Inside
Outside
7948
7839
q′conv ( W / m )
136
219
PROBLEM 13.124
KNOWN: Temperature and emissivity of ceramic plate which is separated from a glass plate of
equivalent height and width by an air space. Temperature of air and surroundings on opposite side of
glass. Spectral radiative properties of glass.
FIND: (a) Transmissivity of glass, (b) Glass temperature Tg and total heat rate qh, (c) Effect of
external forced convection on Tg and qh.
SCHEMATIC:
ASSUMPTIONS: (1) Spectral distribution of emission from ceramic approximates that of a
blackbody, (2) Glass surface is diffuse, (3) Atmospheric air in cavity and ambient, (4) Cavity may be
approximated as a two-surface enclosure with infinite parallel plates, (5) Glass is isothermal.
PROPERTIES: Table A-4, air (p = 1 atm): Evaluated at T = (Tc + Tg)/2 and Tf = (Tg +T∞)/2 using
IHT Properties Toolpad.
ANALYSIS: (a) The total transmissivity of the glass is
∞
τ λ E λ b dλ λ2 =1.6 µ m
∫
o
=
τ=
∫ (Eλ ,b / E b ) dλ = F(0→λ2 ) − F(0→λ1 )
E
b
λ1 = 0.4 µ m
With λ2T = 1600 µm⋅K and λ1T = 400 µm⋅K, respectively, Table 12.1 yields F( 0→ λ ) = 0.0197 and
2
F( 0→ λ
1)
= 0.0. Hence,
<
τ = 0.0197
With so little transmission of radiation from the ceramic, the glass plate may be assumed to be opaque
to a good approximation. Since more than 98% of the incident radiation is at wavelengths exceeding
1.6 µm, for which αλ = 0.9, αg ≈ 0.9. Also, since Tg < Tc, nearly 100% of emission from the glass is
at λ > 1.6 µm, for which ελ = αλ = 0.9, εg = 0.9 and the glass may be approximated as a gray surface.
(b) The glass temperature may be obtained from an energy balance of the form q′′conv,i + q′′rad,i =
q′′conv,o + q′′rad,o . Using Eqs. 13.24 and 13.27 to evaluate q′′rad,i and q′′rad,o , respectively, it follows
that
(
)
hi Tc − Tg +
(
σ Tc4 − Tg4
1
εc
+
1
εg
)=h
−1
(
)
(
4
4
o Tg − T∞ + ε gσ Tg − Tsur
)
Continued …..
PROBLEM 13.124 (Cont.)
where, assuming 104 ≤ Ra L ≤ 107 , hi and h o are given by Eqs. 9.52 and 9.26, respectively,
k
4 0.012
hi = i Ra1/
( H / L )−0.3
L Pri
L


1/ 6

ko 
0.387Ra

H
ho =
0.825 +
8 / 27 
H 

1 + ( 0.492 / Pr )9 /16 
o




3
2
3
with RaL = gβ i (Tc – Tg)L /νiαi and RaH = gβ o (Tg - T∞) H /νoαo. Entering the energy balance into
the IHT workspace and using the Correlations, Properties and Radiation Toolpads to evaluate the
convection and radiation terms, the following result is obtained.
Tg = 825 K
<
The corresponding value of qh is
q h = 108 kW
<
where qconv,i = 3216 W, qrad,i = 104.7 kW, qconvo,o = 15,190 W and qrad,o = 92.8 kW. The
convection coefficients are hi = 4.6 W / m 2 ⋅ K and ho = 7.2 W / m 2 ⋅ K.
(c) For the prescribed range of ho , IHT was used to generate the following results.
With increasing ho , the glass is cooled more effectively and Tg must decrease. With decreasing Tg,
qconv,i, qrad,i and hence qh must increase. Note that radiation makes the dominant contribution to heat
transfer across the airspace. Although qrad,o decreases with decreasing Tg, the increase in qconv,o
exceeds the reduction in qrad,o.
PROBLEM 13.125
KNOWN: Conditions associated with a spherical furnace cavity.
FIND: Cooling rate needed to maintain furnace wall at a prescribed temperature.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, (2) Blackbody behavior for furnace wall, (3) N2 is non-radiating.
ANALYSIS: From an energy balance on a unit surface area of the furnace wall, the cooling rate per
unit area must equal the absorbed irradiation from the gas (Eg) minus the portion of the wall’s
emissive power absorbed by the gas
q′′c = E g − α g E b ( Ts )
q′′c = ε gσ Tg4 − α gσ Ts4 .
Hence, for the entire furnace wall,
(
)
q c = Asσ ε g Tg4 − α g Ts4 .
The gas emissivity, εg, follows from Table 13.4 with
Le = 0.65D = 0.65 × 0.5 m = 0.325 m = 1.066 ft.
p c L e = 0.25 atm × 1.066 ft = 0.267 ft − atm
and from Fig. 13.18, find εg = εc = 0.09. From Eq. 13.42,
 Tg 
α g = α c = Cc  
 Ts 
0.45
(
)
× ε c Ts , p c L e Ts / Tg  .
With Cc = 1 from Fig. 13.19,
α g = 1 (1400 / 50 )
0.45
× ε c (500K, 0.095 ft − atm )
where, from Fig. 13.18,
ε c (500K, 0.095 ft − atm ) = 0.067.
Hence
α g = 1(1400 / 500 )
0.45
× 0.067 = 0.106
and the heat rate is
2
4
4
q c = π ( 0.5 m ) 5.67 × 10−8 W / m 2 ⋅ K 4 0.09 (1400 K ) − 0.106 (500 K ) 

q c = 15.1 kW.

<
PROBLEM 13.126
KNOWN: Diameter and gas pressure, temperature and composition associated with a gas turbine
combustion chamber.
FIND: Net radiative heat flux between the gas and the chamber surface.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Blackbody behavior for chamber surface, (3)
Remaining species are non-radiating, (4) Chamber may be approximated as an infinitely long tube.
ANALYSIS: From Eq. 13.40 the net rate of radiation transfer to the surface is
(
q net = Asσ ε g Tg4 − α g Ts4
)
or
(
q′net = π Dσ ε g Tg4 − α g Ts4
)
with As = πDL. From Table 13.4, Le = 0.95D = 0.95 × 0.4 m = 0.380 m = 1.25 ft. Hence, pwLe =
pcLe = 0.152 atm × 1.25 ft = 0.187 atm-ft.
(
)
Fig.13.18 ( Tg = 1273 K ) , → ε c ≈ 0.085.
Fig.13.20 ( p w / ( p c + p w ) = 0.5, Lc ( p w + pc ) = 0.375 ft − atm, Tg ≥ 930°C ) , → ∆ε ≥ 0.01.
Fig.13.16 Tg = 1273 K , → ε w ≈ 0.069.
From Eq. 13.38,
ε g = ε w + ε c − ∆ε = 0.069 + 0.085 − 0.01 ≈ 0.144.
From Eq. 13.41 for the water vapor,
(
α w = C w Tg / Ts
)0.45 × ε w (Ts , p w Lc Ts / Tg  )
where from Fig. 13.16 (773 K, 0.114 ft-atm), → εw ≈ 0.083,
0.45
α w = 1(1273 / 773)
× 0.083 = 0.104.
From Eq. 13.42, using Fig. 13.18 (773 K, 0.114 ft-atm), → εc ≈ 0.08,
0.45
α c = 1(1273 / 773)
× 0.08 = 0.100.
From Fig. 13.20, the correction factor for water vapor at carbon dioxide mixture,
(p w / ( pc + p w ) = 0.1, Le (p w + pc ) = 0.375, Tg ≈ 540°C ) , → ∆α ≈ 0.004
and using Eq. 13.43
α g = α w + α c − ∆α = 0.104 + 0.100 − 0.004 ≈ 0.200.
Hence, the heat rate is
4
4
q′net = π ( 0.4 m ) 5.67 × 10−8 W / m 2 ⋅ K 4  0.144 (1273) − 0.200 ( 773 )  = 21.9 kW / m.


<
PROBLEM 13.127
KNOWN: Pressure, temperature and composition of flue gas in a long duct of prescribed diameter.
FIND: Net radiative flux to the duct surface.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Duct surface behaves as a blackbody, (3) Other
gases are non-radiating, (4) Flue may be approximated as an infinitely long tube.
ANALYSIS: With As = πDL, it follows from Eq. 13.40 that
(
q′net = π Dσ ε g Tg4 − α g Ts4
)
From Table 13.4, Le = 0.95D = 0.95 × 1 m = 0.95 m = 3.12 ft. Hence
p w Le = 0.12 atm × 3.12 m = 0.312 atm − ft
p c L e = 0.05 atm × 3.12 m = 0.156 atm − ft.
With Tg = 1400 K, Fig. 13.16 → εw = 0.083; Fig. 13.18 → εc = 0.072. With pw/(pc + pw) = 0.67,
Le(pw +pc) = 0.468 atm-ft, Tg ≥ 930°C, Fig. 13.20 → ∆ε = 0.01. Hence from Eq. 13.38,
ε g = ε w + ε c − ∆ε = 0.083 + 0.072 − 0.01 = 0.145.
From Eq. 13.41,
(
α w = C w Tg / Ts
)0.45 × ε w (Ts , pw Le Ts / Tg  )
α w = 1(1400 / 400 )
0.45
× ε w Fig. 13.16 → ε w ( 400 K, 0.0891 atm − ft ) = 0.1
α w = 0.176.
From Eq. 13.42,
(
α c = Cc Tg / Ts
)0.45 × ε c (Ts , pcLeTs / Tg )
α c = 1(1400 / 400 )
0.45
× ε c Fig. 13.18 → ε c ( 400 K, 0.0891 atm − ft ) = 0.053
α c = 0.093.
With pw/(pc + pw) = 0.67, Le(pw + pc) = 0.468 atm-ft, Tg ≈ 125°C, Fig. 13.20 gives
∆α ≈ 0.003.
Hence from Eq. 13.43,
α g = α w + α c − ∆α = 0.176 + 0.093 − 0.003 = 0.266.
The heat rate per unit length is
4
4
q′net = π (1 m ) 5.67 × 10−8 W / m 2 ⋅ K 4 0.145 (1400 K ) − 0.266 ( 400 K ) 

q′net = 98 kW / m.

<
PROBLEM 13.128
KNOWN: Gas mixture of prescribed temperature, pressure and composition between large parallel
plates of prescribed separation.
FIND: Net radiation flux to the plates.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Furnace wall behaves as a blackbody, (3) O2 and
N2 are non-radiating, (4) Negligible end effects.
ANALYSIS: The net radiative flux to a plate is
(
)
′′ = Gs − Es = ε gσ Tg4 − 1 − τ g σ Ts4
qs,1
where G s = ε gσ Tg4 + τ g Es , E s = σ Ts4 and τ g = 1 − α g ( Ts ) . From Table 13.4, Le = 1.8L = 1.8 ×
0.75 m = 1.35 m = 4.43 ft. Hence pwLe = pcLe = 1.33 atm-ft. From Figs. 3.16 and 3.18 find εw ≈ 0.22
and εc ≈ 0.16 for p = 1 atm. With (pw + p)/2 = 1.15 atm, Fig. 13.17 yields Cw ≈ 1.40 and from Fig.
13.19, Cc ≈ 1.08. Hence, the gas emissivities are
ε w = C w ε w (1 atm ) ≈ 1.40 × 0.22 = 0.31
ε c = Ccε c (1 atm ) ≈ 1.08 × 0.16 = 0.17.
From Fig. 13.20 with pw/(pc + pw) = 0.5, Le(pc + pw) = 2.66 atm-ft and Tg > 930°C, ∆ε ≈ 0.047.
Hence, from Eq. 13.38,
ε g = ε w + ε c − ∆ε ≈ 0.31 + 0.17 − 0.047 ≈ 0.43.
To evaluate αg at Ts, use Eq. 13.43 with
(
α w = C w Tg / Ts
)0.45 ε w (Ts , pw L2Ts / Tg ) = Cw (1300 / 500 )0.45 ε w (500, 0.51)
0.45
α w ≈ 1.40 (1300 / 500 )
0.45
α c = Cc (1300 / 500 )
0.22 = 0.47
0.45
ε c (500, 0.51) ≈ 1.08 (1300 / 500 )
0.11 = 0.18.
From Fig. 13.20, with Tg ≈ 125°C and Le(pw + pc) = 2.66 atm-ft, ∆α = ∆ε ≈ 0.024. Hence
α g = α w + α c − ∆α ≈ 0.47 + 0.18 − 0.024 ≈ 0.63 and τ g = 1 − α g ≈ 0.37.
Hence, the heat flux from Eq. (1) is
q′′s,1 = 0.43 × 5.67 × 10−8 W / m 2 ⋅ K 4 (1300 K ) − 0.63 × 5.67 × 10−8 W / m 2 ⋅ K 4 (500 K )
4
4
q′′s,1 ≈ 67.4 kW / m 2 .
The net radiative flux to both plates is then q′′s,2 ≈ 134.8 kW / m 2 .
<
PROBLEM 13.129
KNOWN: Flow rate, temperature, pressure and composition of exhaust gas in pipe of prescribed
diameter. Velocity and temperature of external coolant.
FIND: Pipe wall temperature and heat flux.
SCHEMATIC:
ASSUMPTIONS: (1) L/D >> 1 (infinitely long pipe), (2) Negligible axial gradient for gas
temperature, (3) Gas is in fully developed flow, (4) Gas thermophysical properties are those of air, (5)
Negligible pipe wall thermal resistance, (6) Negligible pipe wall emission.
3
-7
PROPERTIES: Table A-4: Air (Tm = 2000 K, 1 atm): ρ = 0.174 kg/m , µ = 689 × 10 kg/m⋅s, k =
3
-6
0.137 W/m⋅K, Pr = 0.672; Table A-6: Water (T∞ = 300 K): ρ = 997 kg/m , µ = 855 × 10 kg/s⋅m, k
= 0.613 W/m⋅K, Pr = 5.83.
ANALYSIS: Performing an energy balance for a control surface about the pipe wall,
q′′r + q′′c,i = q′′c,o
ε gσ Tg4 + h i ( Tm − Ts ) = ho ( Ts − T∞ )
The gas emissivity is
ε g = ε w + ε c = ∆ε
where
Le = 0.95D = 0.238 m = 0.799 ft
p c L e = p w L e = 0.1 atm × 0.238 m = 0.0238 atm − m = 0.0779 atm − ft
and from Fig. 13.16 → εw ≈ 0.017; Fig. 13.18 → εc ≈ 0.031; Fig. 13.20 → ∆ε ≈ 0.001. Hence εg =
0.047. Estimating the internal flow convection coefficient, find
Re D =
4m
π Dµ
=
4 × 0.25 kg / s
π ( 0.25 m ) 689 × 10−7 kg / m ⋅ s
= 18, 480
and for turbulent flow,
4 / 5 0.3
Nu D = 0.023 ReD
Pr
= 0.023 (18, 480 )
4/5
h i = Nu D
k
D
= 52.9
0.137 W / m ⋅ K
0.25 m
(0.672 )0.3 = 52.9
= 29.0 W / m 2 ⋅ K.
Continued …..
PROBLEM 13.129 (Cont.)
Estimating the external convection coefficient, find
Re D =
ρ VD 997 kg / m3 × 0.3 m / s × 0.25 m
=
= 87, 456.
µ
855 × 10−6 kg / s ⋅ m
Hence
0.37
Nu D = 0.26 Re0.6
( Pr/ Prs )
D Pr
1/ 4
.
Assuming Pr/Prs ≈ 1,
Nu D = 0.26 (87, 456 )
0.6
(5.83)0.37 = 461
ho = Nu D ( k / D ) = 461 (0.613 W / m ⋅ K / 0.25 m ) = 1129 W / m 2 ⋅ K.
Substituting numerical values in the energy balance, find
0.047 × 5.67 × 10−8 W / m 2 ⋅ K 4 ( 2000 K ) + 29 W / m 2 ⋅ K ( 2000 − Ts ) K
4
= 1129 W / m 2 ⋅ K ( Ts − 300 ) K
Ts = 380 K.
<
The heat flux due to convection is
q′′c,i = h i ( Tm − Ts ) = 29 W / m 2 ⋅ K ( 2000 − 379.4 ) K = 46, 997 W / m2
and the total heat flux is
qs′′ = q′′r + q′′c,i = 42, 638 + 46, 997 = 89, 640 W / m 2 .
<
COMMENTS: Contributions of gas radiation and convection to the wall heat flux are approximately
the same. Small value of Ts justifies neglecting emission from the pipe wall to the gas. Prs = 1.62 for
Ts = 380 → (Pr/Prs)1/4 = 1.38. Hence the value of ho should be corrected. The value would ↑, and
Ts would ↓.
PROBLEM 13.130
KNOWN: Flowrate, composition and temperature of flue gas passing through inner tube of an
annular waste heat boiler. Boiler dimensions. Steam pressure.
.
FIND: Rate at which saturated liquid can be converted to saturated vapor, m
s
SCHEMATIC:
ASSUMPTIONS: (1) Inner wall is thin and steam side convection coefficient is very large; hence Ts
= Tsat(2.455 bar), (2) For calculation of gas radiation, inner tube is assumed infinitely long and gas is
approximated as isothermal at Tg.
-7
PROPERTIES: Flue gas (given): µ = 530 × 10 kg/s⋅m, k = 0.091 W/m⋅K, Pr = 0.70; Table A-6,
Saturated water (2.455 bar): Ts = 400 K, hfg = 2183 kJ/kg.
ANALYSIS: The steam generation rate is
=q/h
m
s
fg = ( q conv + q rad ) / h fg
where
(
q rad = Asσ ε g Tg4 − α g Ts4
)
with
ε g = ε w + ε c − ∆ε
α g = α w + α c − ∆α .
From Table 13.4, find Le = 0.95D = 0.95 m = 3.117 ft. Hence
p w Le = 0.2 atm × 3.117 ft = 0.623 ft − atm
p c L e = 0.1 atm × 3.117 ft = 0.312 ft − atm.
From Fig. 13.16, find εw ≈ 0.13 and Fig. 13.18 find εc ≈ 0.095. With pw/(pc + pw) = 0.67 and Le(pw +
pc) = 0.935 ft-atm, from Fig. 13.20 find ∆ε ≈ 0.036 ≈ ∆α. Hence εg ≈ 0.13 + 0.095 – 0.036 = 0.189.
Also, with pwLe(Ts/Tg) = 0.2 atm × 0.95 m(400/1400) = 0.178 ft-atm and Ts = 400 K, Fig. 13.16
yields εw ≈ 0.14. With pcLe(Ts/Tg) = 0.1 atm × 0.95 m(400/1400) = 0.089 ft-atm and Ts = 400 K, Fig.
13.18 yields εc ≈ 0.067. Hence
(
α w = Tg / Ts
)0.45 ε w (Ts , p w LeTs / Tg )
0.45
α w = (1400 / 400 )
0.14 = 0.246
and
(
α c = Tg / Ts
)0.65 ε c (Ts , pcLeTs / Tg )
Continued …..
PROBLEM 13.130 (Cont.)
α c = (1400 / 400 )
0.65
0.067 = 0.151
α g = 0.246 + 0.151 − 0.036 = 0.361.
Hence
4
4
q rad = π (1 m ) 7 m × 5.67 × 10−8 W / m 2 ⋅ K 4  0.189 (1400 K ) − 0.361 ( 400 K ) 


q rad = (905.3 − 11.5 ) kW = 893.8 kW.
For convection,
(
q conv = hπ DL Tg − Ts
)
with
Re D =
4m
π Dµ
=
4 × 2 kg / s
= 48, 047
π × 1 m × 530 × 10−7 kg / s ⋅ m
and assuming fully developed turbulent flow throughout the tube, the Dittus-Boelter correlation gives
4/5
Nu D = 0.023 Re 4D/ 5 Pr 0.3 = 0.023 ( 48, 047 )
(0.70 )0.3 = 115
h = ( k / D ) Nu D = ( 0.091 W / m ⋅ K /1 m )115 = 10.5 W / m 2 ⋅ K.
Hence
q conv = 10.5 W / m 2 ⋅ Kπ (1 m ) 7 m (1400 − 400 ) K = 230.1 kW
and the vapor production rate is
=
m
s
q
h fg
=
(893.8 + 230.1) kW
2183 kJ / kg
= 0.515 kg / s.
m
s
=
1123.9 kW
2183 kJ / kg
<
COMMENTS: (1) Heat transfer is dominated by radiation, which is typical of heat recovery devices
having a large gas volume.
(2) A more detailed analysis would account for radiation exchange involving the ends (upstream and
downstream) of the inner tube.
(3) Using a representative specific heat of cp = 1.2 kJ/kg⋅K, the temperature drop of the gas passing
through the tube would be ∆Tg = 1123.9 kW/(2 kg/s × 1.2 kJ/kg⋅K) = 468 K.
PROBLEM 13.131
KNOWN: Wet newsprint moving through a drying oven.
FIND: Required evaporation rate, air velocity and oven temperature.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible freestream turbulence, (3) Heat and
mass transfer analogy applicable, (4) Oven and newsprint surfaces are diffuse gray, (5) Oven end
effects negligible.
3
PROPERTIES: Table A-6, Water vapor (300 K, 1 atm): ρsat = 1/vg = 0.0256 kg/m , hfg = 2438
-6 2
kJ/kg; Table A-4, Air (300 K, 1 atm): η = 15.89 × 10 m /s; Table A-8, Water vapor-air (300 K, 1
-4 2
atm): DAB = 0.26 × 10 m /s, Sc = η/DAB = 0.611.
ANALYSIS: The evaporation rate required to completely dry the newsprint having a water content
of m′′A = 0.02 kg / m 2 as it enters the oven (x = L) follows from a species balance on the newsprint.
M
A,in − M A,out = M st
M
L − M 0 − M A,s = 0.
The rate at which moisture enters in the newsprint is
M
L = m′′A VW
hence,
2
−3
M
A,s = m′′A VW = 0.02 kg / m × 0.2 m / s × 1 m = 4 × 10 kg / s.
<
The required velocity of the airstream through the oven, u∞, can be determined from a convection
analysis. From the rate equation,
(
)
M
A,s = h m WL ρ A,s − ρ A,∞ = h m WLρ A,sat (1 − φ∞ )
hm = M
A,s / WLρ A,sat (1 − φ∞ )
h m = 4 × 10−3 kg / s /1 m × 20 m × 0.0256 kg / m3 (1 − 0.2 ) = 9.77 × 10−3 m / s.
Now determine what flow velocity is required to produce such a coefficient. Assume flow over a flat
plate with
Sh L = h m L / DAB = 9.77 × 10−3 m / s × 20 m / 0.26 × 10−4 m 2 / s = 7515
Continued …..
PROBLEM 13.131 (Cont.)
and
1/ 3  2
2
Re L = Sh L / 0.664Sc1/ 3  =  7515 / 0.664 ( 0.611)



8
 = 1.78 × 10 .
5
Since ReL > ReLc = 5 × 10 , the flow must be turbulent. Using the correlation for mixed laminar and
turbulent flow conditions, find
Re 4L/ 5 = Sh L / Sc1/ 3 + 871 / 0.037


Re 4L/ 5 =  7515 / (0.611)
1/ 3

+ 871 / 0.037

Re L = 5.95 × 106
noting ReL > ReLc. Recognize that u ∗∞ is the velocity relative to the newsprint,
u ∗∞ = Re L ν / L = 5.95 × 106 × 15.89 × 10 −6 m 2 / s / 20 m = 4.73 m / s.
The air velocity relative to the oven will be,
<
u ∞ = u ∗∞ − V = ( 4.73 − 0.2 ) m / s = 4.53 m / s.
The temperature required of the oven surface follows
from an energy balance on the newsprint. Find
E in − E out = 0
q rad − q evap = 0
where
−3
3
q evap = M
A,s h fg = 4.0 × 10 kg / s × 2438 × 10 J / kg = 9752 W
and the radiation exchange is that for a two surface enclosure, Eq. 13.23,
q rad =
(
σ T14 − T24
)
(1 − ε1 ) / ε1A1 + 1/ A1F12 + (1 − ε 2 ) / ε 2A 2
Evaluate,
A1 = π / 2 WL,
A 2 = WL,
.
F21 = 1, and A1F12 = A 2 F21 = WL
hence, with ε1 = 0.8,
(
)
q rad = σ WL T14 − T24 / [(1/ 2π ) + 1]
T14 = T24 + q rad [(1/ 2π ) + 1] / σ WL
4
T14 = (300 K ) + 9752 W [(1/ 2π + 1)] / 5.67 × 10−8 W / m 2 ⋅ K 4 × 1 m × 20 m
T1 = 367 K.
COMMENTS: Note that there is no convection heat transfer since T∞ = Ts = 300 K.
<
PROBLEM 13.132
KNOWN: Configuration of grain dryer. Emissivities of grain bed and heater surface. Temperature
of grain.
FIND: (a)Temperature of heater required for specified drying rate, (b) Convection mass transfer
coefficient required to sustain evaporation, (c) Validity of assuming negligible convection heat
transfer.
SCHEMATIC:
ASSUMPTIONS: (1) Diffuse/gray surfaces, (2) Oven wall is a reradiating surface, (3) Negligible
convection heat transfer, (4) Applicability of heat/mass transfer analogy, (5) Air is dry.
3
6
PROPERTIES: Table A-6, saturated water (T = 330 K): vg = 8.82 m /kg, hfg = 2.366 × 10 J/kg.
3
-6 2
Table A-4, air (assume T ≈ 300 K): ρ = 1.614 kg/m , cp = 1007 J/kg⋅K, α = 22.5 × 10 m /s. Table
-4 2
A-8, H2O(v) – air (T = 298 K): DAB = 0.26 × 10 m /s.
ANALYSIS: (a) Neglecting convection, the energy required for evaporation must be supplied by net
radiation transfer from the heater plate to the grain bed. Hence,
(
)
6
′
q′rad = m
evap h fg = ( 2.5 kg / h ⋅ m ) 2.366 × 10 J / kg / 3600 s / h = 1643 W / m
where q′rad is given by Eq. 13.30. With A′p = A′g ≡ A′,
q′rad =
(
A′ E bp − E bg
1 − εp
εp
)
+
1
+
(
) (
)
Fpg +  1/ FpR + 1/ FgR 
1 − εg
εg
−1
where A′ = R = 1 m, Fpg = 0 and FpR = FgR = 1. Hence,
q′rad =
(
) = 2.40 ×10−8 T4 − 3204 = 1643 W / m
(p )
0.25 + 2 + 0.111
σ Tp4 − 3204
2.40 × 10 −8 Tp4 − 2518 = 1643
<
Tp = 530 K
′evap , and ρA,∞ = 0,
(b) The evaporation rate is given by Eq. 6.12, and with A′s = 1 m, n ′A = m
Continued …..
PROBLEM 13.132 (Cont.)
hm =
n ′A vg 2.5 kg / h ⋅ m
n ′A
1
m3
=
=
×
× 8.82
= 6.13 × 10−3 m / s
′
′
ρ
As A,s
As
1m
3600 s
kg
(c) From the heat and mass transfer analogy, Eq. 6.92,
h = h m ρ c p Le 2 / 3
where Le = α/DAB = 22.5/26.0 = 0.865. Hence
(
)
h = 6.13 × 10 −3 m / s 1.161kg / m3 1007 J / kg ⋅ K ( 0.865 )
2/3
= 6.5 W / m 2 ⋅ K.
The corresponding convection heat transfer rate is
(
)
q′conv = hA′ Tg − T∞ = 6.5 W / m 2 ⋅ K (1 m )(330 − 300 ) K = 195 W / m
Since q′conv << q′rad , the assumption of negligible convection heat transfer is reasonable.
<
PROBLEM 13.133
KNOWN: Diameters of coaxial cylindrical drum and heater. Heater emissivity. Temperature and
emissivity of pellets covering bottom half of drum. Convection mass transfer coefficient associated
with flow of dry air over the pellets.
FIND: (a) Evaporation rate per unit length of drum, (b) Surface temperatures of heater and top half
of drum.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, (2) Negligible heat transfer from ends of drum, (3) Diffuse-gray
surface behavior, (4) Negligible heat loss from the drum to the surroundings, (5) Negligible
convection heat transfer from interior surfaces of the drum, (6) Pellet surface area corresponds to that
of bottom half of drum.
PROPERTIES Table A-6, sat. water (T = 325 K): ρ A,sat = ν g−1 = 0.0904 kg/m , hfg = 2378 kJ/kg.
3
ANALYSIS: (a) The evaporation rate is
( )
n ′A = h m (π Dd / 2 )  ρ A,sat Tp − ρ A,∞ 


n ′A = 0.024 m / s (π ×1m / 2 ) × 0.0904 kg / m3 = 0.00341 kg / s ⋅ m
<
(b) From an energy balance on the surface of the pellets,
q′p = q′evap = n′A h fg = 0.00341 kg / s ⋅ m × 2.378 × 106 J / kg = 8109 W / m
<
where q′p may be determined from analysis of radiation transfer in a three surface enclosure. Since
the top half of the enclosure may be treated as reradiating, net radiation transfer to the pellets may be
obtained from Eq. 13.30, which takes the form
q′p =
1− εh
+
ε h A′h
E bh − E bp
1
(
)
A′h Fhp + (1/ A′h Fhd ) + 1/ A′p Fpd 


−1
+
1− εp
ε p A′p
where Fhp = Fhd = 0.5, A′h = π D h and A′p = π Dd / 2.
The view factor Fpd may be obtained from the summation rule,
Fpd = 1 − Fph − Fpp
Continued …..
PROBLEM 13.133 (Cont.)
where Fph = A′h Fhp / A′p = (π D h × 0.5 ) / (π Dd / 2 ) = 0.10 and


2 1/ 2
Fpp = 1 − ( 2 / π )  1 − ( 0.1) 
+ 0.1 sin −1 (0.1) = 0.360


 

Hence, Fpd = 1 – 0.10 – 0.360 = 0.540, and the expression for the heat rate yields
4
8109 W / m =
0.25
+
π × 0.1m
E bh − σ (325 K )
1
0.053
1  π × 0.5m

−1
−1 − 
π 0.1m × 0.5 + (0.1m × 0.5 ) + (0.5m × 0.54 )  

 


+
E bh = σ Th4 = 35, 359 W / m 2
<
Th = 889 K
Applying Eq. (13.19) to surfaces h and p,
J h = E bh − q′h (1 − ε h ) / ε h A′h = 35, 359 W / m 2 − 6, 453 W / m 2 = 28, 906 W / m 2
(
)
J p = E bp + q′p 1 − ε p / ε p A′p = 633 W / m 2 + 272 W / m 2 = 905 W / m 2
Hence, from
Jd − J p
Jh − Jd
−
=0
−1
−1
′
A
F
( h hd )
A′p Fpd
(
28,906 W / m 2 − J d
(π × 0.1m × 0.5 )−1
)
−
J d − 905 W / m 2
(π × 0.5m × 0.54 )−1
=0
J d = σ Td4 = 24,530 W / m 2
Td = 811 K
<
COMMENTS: The required value of Th could be reduced by increasing Dh, although care must be
taken to prevent contact of the plastic with the heater.
PROBLEM 14.1
KNOWN: Mixture of O2 and N2 with partial pressures in the ratio 0.21 to 0.79.
FIND: Mass fraction of each species in the mixture.
SCHEMATIC:
pO2
p N2
MO
M
2
=
0.21
0.79
= 32 kg/kmol
N2 = 28 kg/kmol
ASSUMPTIONS: (1) Perfect gas behavior.
ANALYSIS: From the definition of the mass fraction,
ρ
ρ
mi = i = i
ρ Σ ρi
Hence, with
ρi =
pi
pi
M p
=
= i i.
R iT ( ℜ / M i ) T
ℜT
mi =
/ ℜT
Σ M i pi / ℜ T
Hence
M i pi
or, cancelling terms and dividing numerator and denominator by the total pressure p,
mi =
M i xi
ΣM i x i
.
With the mole fractions as
0.21
= 0.21
xO2 = pO2 / p =
0.21 + 0.79
x N2 = pN2 / p = 0.79,
find the mass fractions as
mO =
2
32 × 0.21
= 0.233
32 × 0.21 + 28 × 0.79
m N2 = 1 − m O2 = 0.767.
<
<
PROBLEM 14.2
KNOWN: Partial pressures and temperature for a mixture of CO2 and N2.
FIND: Molar concentration, mass density, mole fraction and mass fraction of each species.
SCHEMATIC:
A → CO 2 , M A = 44 kg/kmol
B → N2 , M B = 28 kg/kmol
ASSUMPTIONS: (1) Perfect gas behavior.
ANALYSIS: From the equation of state for an ideal gas,
Ci =
pi
.
ℜT
Hence, with pA = pB,
C A = CB =
1bar
8.314 ×10−2 m3 ⋅ bar/kmol ⋅ K × 2 9 8 K
CA = C B = 0.040 kmol/m 3.
<
With ρi = M i Ci , it follows that
ρ A = 44 kg/kmol × 0.04 kmol/m 3 = 1.78kg/m3
<
ρ B = 28 kg/kmol × 0.04 kmol/m 3 = 1.13 k g / m 3.
<
Also, with
xi = Ci / Σi Ci
find
x A = x B = 0.04/0.08 = 0.5
<
and with
m i = ρ i / Σρ i
find
m A = 1.78/ (1.78 +1.13 ) = 0.61
<
m B = 1.13/ (1.78 + 1.13 ) = 0.39.
<
PROBLEM 14.3
KNOWN: Mole fraction (or mass fraction) and molecular weight of each species in a mixture of n
species. Equal mole fractions (or mass fractions) of O2, N2 and CO2 in a mixture.
FIND:
SCHEMATIC:
xO2 = xN 2 = xCO2 = 0.333
or
m O = m N = m CO = 0.333
2
M CO
MO
2
2
2
2
= 44
= 32, M N = 28
2
ASSUMPTIONS: (1) Perfect gas behavior.
ANALYSIS: (a) With
ρ
mi = i =
ρ
ρi
p /RT
p M / ℜT
= i i
= i i
∑ ρi ∑ pi / Ri T ∑ pi M i / ℜT
i
i
i
and dividing numerator and denominator by the total pressure p,
mi =
M ixi
∑ M i xi
<
.
i
Similarly,
xi =
( ρi / M i ) ℜT
pi
ρR T
= i i
=
∑ pi ∑ ρi R i T ∑ ( ρi / M i ) ℜ T
i
i
i
or, dividing numerator and denominator by the total density ρ
xi =
mi / M i
.
∑ mi / M i
<
i
(b) With
MO
x
+ M N 2 x N2 + M CO2 x CO2 = 32 × 0.333 + 28 × 0.33 + 44 × 0.333 = 34.6
2 O2
m O2 = 0.31,
mN 2 = 0.27,
m CO2 = 0.42.
<
With
m O / M O + mN / M N + m CO / M
= 0.333/32 + 0.333/28 + 0.333/44
2
2
2
2
2
CO 2
m O2 = 2.987 ×10−2.
find
xO = 0.35,
2
xN = 0.40,
2
xCO = 0.25.
2
<
PROBLEM 14.4
KNOWN: Temperature of atmospheric air and water. Percentage by volume of oxygen in the air.
FIND: (a) Mole and mass fractions of water at the air and water sides of the interface, (b) Mole and
mass fractions of oxygen in the air and water.
SCHEMATIC:
ASSUMPTIONS: (1) Perfect gas behavior for air and water vapor, (2) Thermodynamic equilibrium
at liquid/vapor interface, (3) Dilute concentration of oxygen and other gases in water, (4) Molecular
weight of air is independent of vapor concentration.
PROPERTIES: Table A-6, Saturated water (T = 290 K): pvap = 0.01917 bars. Table A-9, O2/water,
H = 37,600 bars.
ANALYSIS: (a) Assuming ideal gas behavior, pw,vap = (Nw,vap/V) •T and p = (N/V) •T, in which
case
(
)
x w,vap = p w,vap / pair = ( 0.01917 /1.0133) = 0.0194
<
With mw,vap = (ρw,vap/ρair) = (Cw,vap Μ w/Cair Μ air) = xw,vap ( Μ w/ Μ air). Hence,
mw,vap = 0.0194 (18/29) = 0.0120
Assuming negligible gas phase concentrations in the liquid,
<
<
xw,liq = mw,liq = 1
(b) Since the partial volume of a gaseous species is proportional to the number of moles of the
species, its mole fraction is equivalent to its volume fraction. Hence on the air side of the interface
x O2,air = 0.205
(
)
mO2,air = x O2,air ΜO2 / Μair = 0.205 (32 / 29 ) = 0.226
<
<
The mole fraction of O2 in the water is
x O2,liq = pO2,air / H = 0.208 bars / 37, 600 bars = 5.53 × 10−6
where pO
water is
2,air
<
= x O2,air patm = 0.205 × 1.0133 bars = 0.208 bars. The mass fraction of O2in the
(
)
mO2,liq = x O2,liq Μ02 / Μw = 5.53 × 10−6 (32 /18 ) = 9.83 × 10−6
<
COMMENTS: There is a large discontinuity in the oxygen content between the air and water sides
of the interface. Despite the low concentration of oxygen in the water, it is sufficient to support the
life of aquatic organisms.
PROBLEM 14.5
KNOWN: Air is enclosed at uniform pressure in a vertical, cylindrical container whose top and
bottom surfaces are maintained at different temperatures.
FIND: (a) Conditions in air when bottom surface is colder than top surface, (b) Conditions when
bottom surface is hotter than top surface.
SCHEMATIC:
ASSUMPTIONS: (1) Uniform pressure, (2) Perfect gas behavior.
ANALYSIS: (a) If T1 > T2, the axial temperature gradient (dT/dx) will result in an axial density
gradient. However, since dρ/dx < 0 there will be no buoyancy driven, convective motion of the
mixture.
There will also be axial species density gradients, d ρO /dx and d ρN /dx. However, there is no
(
2
2
)
gradient associated with the mass fractions dmO /dx = 0, d mN /dx = 0 . Hence, from Fick’s
2
2
law, Eq. 14.1, there is no mass transfer by diffusion.
(b) If T1 < T2, dρ /dx > 0 and there will be a buoyancy driven, convective motion of the mixture.
However, dmO /dx = 0 and dmN /dx = 0, and there is still no mass transfer. Hence, although
2
2
there is motion of each species with the convective motion of the mixture, there is no relative motion
between species.
COMMENTS: The commonly used special case of Fick’s law,
jA = −D AB
dρ A
dx
would be inappropriate for this problem since ρ is not uniform. If applied, this special case indicates
that mass transfer would occur, thereby providing an incorrect result.
PROBLEM 14.6
KNOWN: Pressure and temperature of hydrogen stored in a spherical steel tank of prescribed
diameter and thickness.
FIND: (a) Initial rate of hydrogen mass loss from the tank, (b) Initial rate of pressure drop in the tank.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional species diffusion in a stationary medium, (2) Uniform total
molar concentration, C, (3) No chemical reactions.
ANALYSIS: (a) From Table 14.1
NA,r =
NA,r =
CA,o − CA,L
R m,dif
(
=
CA,o
(1 / 4π D AB ) (1 / ri − 1 / ro )
)
4π 0.3 ×10−12 m2 / s 1.5 kmol/m3
(1/0.05 m − 1/0.052 m )
= 7.35 ×10−12 kmol/s
or
−
−
n A,r = M A N A,r = 2 kg/kmol × 7.35 ×10 12 kmol/s = 14.7 ×10 12 kg/s.
<
(b) Applying a species balance to a control volume about the hydrogen,
&
&
M
A,st = −M A,out = −n A,r
3
3
3
& A,st = d ( ρAV ) = π D d ρA = π D dp A = π D M A dp A
M
dt
6
dt
6R A T dt
6ℜT
dt
Hence
(
)
6 0.08314 m 3 ⋅ bar/kmol ⋅ K ( 300 K )
dpA
6ℜ T
=−
×14.7 ×10 −12 kg/s
n A,r = −
3
dt
πD M A
π 0.1m 3 2 kg/kmol
(
)
dpA
= −3.50 × 10−7 bar/s.
dt
<
2
COMMENTS: If the spherical shell is appoximated as a plane wall, Na,x = DAB(CA,o) πD /L = 7.07
-12
× 10 kmol/s. This result is 4% lower than that associated with the spherical shell calculation.
PROBLEM 14.7
KNOWN: Molar concentrations of helium at the inner and outer surfaces of a plastic membrane.
Diffusion coefficient and membrane thickness.
FIND: Molar diffusion flux.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional diffusion in a plane wall, (3)
Stationary medium, (4) Uniform C = CA + CB.
ANALYSIS: The molar flux may be obtained from Eq. 14.50,
N′′A,x =
DAB
10 −9 m 2 / s
CA,1 − CA,2 =
( 0.02 − 0.005 ) kmol/m3
L
0.001m
(
N′′A,x = 1.5 ×10
)
−8
2
kmol/s ⋅ m .
COMMENTS: The mass flux is
′′ = 4 kg/kmol ×1.5 ×10 −8 kmol/s ⋅ m 2 = 6 ×10 −8 k g / s ⋅ m2 .
n ′′A,x = M A NA,x
<
PROBLEM 14.8
KNOWN: Mass diffusion coefficients of two binary mixtures at a given temperature, 298 K.
FIND: Mass diffusion coefficients at a different temperature, T = 350 K.
ASSUMPTIONS: (a) Ideal gas behavior, (b) Mixtures at 1 atm total pressure.
-4
2
PROPERTIES: Table A-8, Ammonia-air binary mixture (298 K), DAB = 0.28 × 10 m /s;
-4 2
Hydrogen-air binary mixture (298 K), DAB = 0.41 × 10 m /s.
ANALYSIS: According to treatment of Section 14.1.5, assuming ideal gas behavior,
D AB ~ T3 / 2
where T is in kelvin units. It follows then, that for
NH3 − Air:
DAB ( 350 K ) = 0.28 ×10 −4 m 2 / s ( 350 K / 2 9 8 K)
3/2
DAB ( 350 K ) = 0.36 ×10 −4 m 2 / s
H 2 − Air:
<
D AB (350 K ) = 0.41 ×10−4 m 2 / s ( 350/298 )
3/2
DAB ( 350 K ) = 0.52 ×10 −4 m 2 / s
COMMENTS: Since the H2 molecule is smaller than the NH3 molecule, it follows that
DH 2−Air > DNH3−Air
as indeed the numerical data indicate.
<
PROBLEM 14.9
KNOWN: The inner and outer surfaces of an iron cylinder of 100-mm length are exposed to a
carburizing gas (mixtures of CO and CO2). Observed experimental data on the variation of the carbon
composition (weight carbon, %) in the iron at 1000°C as a function of radius. Carbon flow rate under
steady-state conditions.
$% is a constant if the diffusion
FIND: (a) Beginning with Fick’s law, show that dρ c / d n r
coefficient, DC-Fe, is a constant; sketch of the carbon mass density, ρc(r), as function of ln(r) for such
a diffusion process; (b) Create a graph for the experimental data and determine whether DC-Fe for this
diffusion process is constant, increases or decreases with increasing mass density; and (c) Using the
experimental data, calculate and tabulate DC-Fe for selected carbon compositions over the range of the
experiment.
SCHEMATIC:
3
PROPERTIES: Iron (1000°C). ρ = 7730 kg/m .Experimental observations of carbon composition
r (mm)
Wt. C (%)
4.49
1.42
4.66
1.32
4.79
1.20
4.91
1.09
5.16
0.82
5.27
0.65
5.40
0.46
5.53
0.28
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional, radial diffusion in a stationary
medium, and (3) Uniform total concentration.
ANALYSIS: (a) For the one-dimensional, radial (cylindrical) coordinate system, Fick’s law is
jA = − D AB A r
dρ A
dr
(1)
where Ar = 2πrL. For steady-state conditions, jA is constant, and if DAB is constant, the product
r
dρ A
= C1
dr
(2)
must be a constant. Using the differential relation dr/r = d (ln r), it follows that
dρ A
= C1
d ln r
(3)
$
so that on a ln(r) plot, ρA is a straight line. See the graph below for this behavior.
Continued …..
PROBLEM 14.9 (Cont.)
(b) To determine whether DC-Fe is a constant for the experimental diffusion process, the data are
represented on a ln(r) coordinate.
Wt. carbon distribution - experimental observations
1.6
1.4
Exp data
Wt Carbon (%)
1.2
1
0.8
0.6
0.4
0.2
0
1.45
1.5
1.55
1.6
1.65
1.7
1.75
ln (r, mm)
Since the plot is not linear, DC-Fe is not a constant. From the treatment of part (a), if DAB is not a
constant, then
DAB
dρ A
= C2
d ln r
$
must be constant. We conclude that DC-Fe will be lower at the radial position where the gradient is
higher. Hence, we expect DC-Fe to increase with increasing carbon content.
(c) From a plot of Wt - %C vs. r (not shown), the mass fraction gradient is determined at three
locations and Fick’s law is used to calculate the diffusion coefficient,
jc = − ρ ⋅ A r ⋅ DC − Fe
∆ Wt − % C
∆r
$
where the mass flow rate is
jc = 3.6 × 10−3 kg / 100 h 3600 s / h = 1 × 10−8 kg / s
$
3
and ρ = 7730 kg/m , density of iron. The results of this analysis yield,
Wt - C (%)
1.32
0.955
0.37
r (mm)
4.66
5.04
5.47
∆ Wt-C/∆r (%/mm)
-0.679
-1.08
-1.385
DC-Fe × 10
11
6.51
3.79
2.72
2
(m /s)
PROBLEM 14.10
KNOWN: Three-dimensional diffusion of species A in a stationary medium with chemical reactions.
FIND: Derive molar form of diffusion equation.
SCHEMATIC:
ASSUMPTIONS: (1) Uniform total molar concentration, (2) Stationary medium.
ANALYSIS: The derivation parallels that of Section 14.2.2, except that Eq. 14.33 is applied on a
molar basis. That is,
& A,g − NA,x + dx − N A,y+ dy − N A,z+ dz = N
& A,st .
NA,x + N A,y + N A,z + N
With
NA,x + dx = NA,x +
∂NA,x
NA,x = −DAB ( dydz )
& A,g = N
& A ( dxdydz ) ,
N
∂x
dx,
∂C A
,
∂x
NA,y + dy = ....
NA,y = ....
& A,st = ∂CA dxdydz
N
∂t
It follows that
∂ 
∂CA  ∂ 
∂C A  ∂ 
∂C A  &
∂C A
.
 DAB
 +  D AB
 +  DAB
 + NA =
∂x 
∂x  ∂y 
∂y  ∂z 
∂z 
∂t
COMMENTS: If DAB is constant, the foregoing result reduces to Eq. 14.38b.
<
PROBLEM 14.11
KNOWN: Gas (A) diffuses through a cylindrical tube wall (B) and experiences chemical reactions at
& .
a volumetric rate, N
A
FIND: Differential equation which governs molar concentration of gas in plastic.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional radial diffusion, (2) Uniform total molar concentration, (3)
Stationary medium.
ANALYSIS: Dividing the species conservation requirement, Eq. 14.33, by the molecular weight, A,
and applying it to a differential control volume of unit length normal to the page,
& A,g − NA,r +dr = N
& A,st
NA,r + N
where
NA,r = ( 2πr ⋅1) N′′A,r = −2π rDAB
NA,r + dr = NA,r +
∂NA,r
∂r
∂C A
∂r
dr
& A,g = − N
& A ( 2π r ⋅ dr ⋅ 1)
N
∂ C 2π rdr ⋅1) 
& A,st =  A (
N
.
∂t
Hence
& A ( 2π rdr ) + 2π DAB
−N
∂  ∂C A 
∂C
r
dr = 2π rdr A


∂r  ∂r 
∂t
or
DAB ∂  ∂ CA  &
∂ CA
.
r
 − NA =
r ∂r  ∂r 
∂t
COMMENTS: (1) The minus sign in the generation term is necessitated by the fact that the
reactions deplete the concentration of species A.
& ( r , t ) , the foregoing equation could be solved for CA (r,t).
(2) From knowledge of N
A
(3) Note the agreement between the above result and the one-dimensional form of Eq. 14.39 for
uniform C.
<
PROBLEM 14.12
KNOWN: One-dimensional, radial diffusion of species A in a stationary, spherical medium with
chemical reactions.
FIND: Derive appropriate form of diffusion equation.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional, radial diffusion, (2) Uniform total molar concentration, (3)
Stationary medium.
ANALYSIS: Dividing the species conservation requirement, Eq. 14.33, by the molecular weight, A,
and applying it to the differential control volume, it follows that
& A,g − NA,r +dr = N
& A,st
NA,r + N
where
NA,r = −DAB 4π r 2
NA,r + dr = NA,r +
(
∂CA
∂r
∂NA,r
∂r
dr
(
)
∂ CA 4π r 2 dr 

& A,st = 
N
.
∂t
)
& A,g = N
& A 4π r 2 dr ,
N
Hence
(
)
& A 4π r 2 dr + 4π ∂  DABr 2 ∂CA  dr = 4π r 2 ∂CA dr
N
∂r 
∂r 
∂t
or
1
∂
 DAB r
r 2 ∂r 
2 ∂CA  + N&
∂r 
A =
∂C A
.
∂t
<
COMMENTS: Equation 14.40 reduces to the foregoing result if C is independent of r and variations
in φ and θ are negligible.
PROBLEM 14.13
KNOWN: Oxygen pressures on opposite sides of a rubber membrane.
FIND: (a) Molar diffusion flux of O2, (b) Molar concentrations of O2 outside the rubber.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional, steady-state conditions, (2) Stationary medium of uniform
total molar concentration, C = CA + CB, (3) Perfect gas behavior.
-9
2
PROPERTIES: Table A-8, Oxygen-rubber (298 K): DAB = 0.21 × 10 m /s; Table A-10, Oxygen-3
3
rubber (298 K): S = 3.12 × 10 kmol/m ⋅bar.
ANALYSIS: (a) For the assumed conditions
∗ = −D
′′ = J A,x
NA,x
AB
C ( 0 ) − CA ( L)
dCA
= DAB A
.
dx
L
From Eq. 14.33,
CA ( 0) = SpA,1 = 6.24 ×10−3 kmol/m3
CA ( L ) = Sp A,2 = 3.12 × 10−3 lmol/m 3 .
Hence
N′′A,x
= 0.21 ×10−9 m 2 / s
(6.24 ×10−3 − 3.12× 10−3 ) kmol/m3
0.0005 m
N′′A,x = 1.31 ×10−9 kmol/s ⋅ m2 .
<
(b) From the perfect gas law
CA,1 =
pA,1
ℜT
=
(
2 bar
3
)
0.08314 m ⋅ bar/kmol ⋅ K 298 K
CA,2 = 0.5CA,1 = 0.0404 kmol/m3 .
= 0.0807 kmol/m3
<
<
COMMENTS: Recognize that the molar concentrations outside the membrane differ from those
within the membrane; that is, CA,1 ≠ CA(0) and CA,2 ≠ CA(L).
PROBLEM 14.14
KNOWN: Water vapor is transferred through dry wall by diffusion.
FIND: The mass diffusion rate through a 0.01 × 3 × 5 m wall.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional species diffusion, (3)
Homogeneous medium, (4) Constant properties, (5) Uniform total molar concentration, (6) Stationary
medium with xA << 1, (7) Negligible condensation in the dry wall.
ANALYSIS: From Eq. 14.46,
′′ = −CDAB
NA,x
− CA,2
C
dx A
dC
= −DAB A = DAB A,1
.
dx
dx
L
From Eq. 14.33
CA,1 = Sp A,1 = 0.15 ×10 −3 kmol/m3
3
CA,2 = Sp A,2 = 0 kmol/m .
Hence
N′′A = 10−9 m 2 / s ×
0.15 ×10 −3 kmol/m3
= 0.15 ×10−10 kmol/s ⋅ m2 .
0.01m
Therefore
′′ ) = 18kg/kmol ×15 m 2 × 0.15 ×10 −10 kmol/s ⋅ m 2
n A = M A ( A ⋅ NA
or
n A = 4.05 ×10 −9 kg/s.
<
PROBLEM 14.15
KNOWN: Pressure and temperature of CO2 in a container of prescribed volume. Thickness and
surface area of rubber plug.
FIND: (a) Mass rate of CO2 loss from container, (b) Reduction in pressure over a 24 h period.
SCHEMATIC:
ASSUMPTIONS: (1) Loss of CO2 is only by diffusion through the rubber plug, (2) One-dimensional
diffusion through a stationary medium, (3) Diffusion rate is constant over the 24 h period, (4) Perfect
gas behavior, (5) Negligible CO2 pressure outside the plug.
-9
2
PROPERTIES: Table A-8, CO2-rubber (298 K, 1 atm): DAB = 0.11 × 10 m /s; Table A-10, CO2-3
3
rubber (298 K, 1 atm): S = 40.15 × 10 kmol/m ⋅bar.
ANALYSIS: (a) For diffusion through a stationary medium,
NA = ADAB
CA,1 − CA,2
L
CA,1 = SpA,1 = 40.15 ×10−3 kmol/m3 ⋅ bar × 5bar = 0.200 kmol/m3
where
CA,2 = Sp A,2 = 0.
Hence
(
NA = 3 × 10−4 m 2 0.11× 10−9 m 2 / s
)
( 0.200 − 0) kmol/m3 = 3.30 ×10 −13 kmol/s
0.02 m
and
n A = M A NA = 44 kg/kmol × 3.30 ×10 −13 kmol/s = 1.45 ×10 −11 kg/s.
<
(b) Applying conservation of mass to a control volume about the container
d ( ρA V )
dt
= −n A
d ( CA V )
or
dt
= − NA .
Hence, with CA = pA/ℜT,
dp A
dt
=−
N Aℜ T
V
=−
3.3 × 10
−13
kmol/s × 8.314 × 10
10
−2
−2
m ⋅ bar/kmol ⋅ K ( 298K )
3
3
= − 8.18 × 10
−10
bar/s.
m
Hence
dp
∆p A =  A  ∆t = −8.18 × 10−10 bar/s × 24h × 3600s/h = 7.06 ×10−5 bar.
 dt 
<
PROBLEM 14.16
KNOWN: Pressure and temperature of helium in a glass cylinder of 100 mm inside diameter and 5
mm thickness.
FIND: Mass rate of helium loss per unit length.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional radial diffusion through cylinder
wall, (3) Negligible end losses, (4) Stationary medium, (5) Uniform total molar concentration, (6)
Negligible helium concentration outside cylinder.
PROPERTIES: Table A-8, He-SiO2 (298 K): DAB ≈ 0.4 × 10
-3
3
K): S ≈ 0.45 × 10 kmol/m ⋅bar.
-13
2
m /s; Table A-10, He-SiO2 (298
ANALYSIS: From Table 14.1,
N′A,r =
CA,S1 − CA,S2
ln ( r2 / r1 ) / 2π D AB
where, from Eq. 14.44, CA,S = SpA. Hence
CA,S1 = SpA,1 = 0.45 × 10−3 kmol / m3 ⋅ bar × 4 bar = 1.8 × 10−3 kmol / m3
CA,S2 = SPA,2 = 0.
Hence
N′A,r =
1.8 × 10−3 kmol / m3
(
ln (0.055 / 0.050 ) / 2π 0.4 × 10−13 m 2 / s
)
N′A,r = 4.75 × 10−15 kmol / s ⋅ m.
The mass loss is then
n′A,r = ΜA N′A,r = 4 kg / kmol × 4.75 × 10−15 kmol / s ⋅ m
n ′A,r = 1.90 × 10−14 kg / s ⋅ m.
<
PROBLEM 14.17
KNOWN: Temperature and pressure of helium stored in a spherical pyrex container of prescribed
diameter and wall thickness.
FIND: Mass rate of helium loss.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Helium loss by one-dimensional diffusion in radial
direction through the pyrex, (3) C = CA + CB is independent of r, and xA << 1, (4) Stationary medium.
-13
PROPERTIES: Table A-8, He-SiO 2 (293 K): DAB = 0.4 × 10
-3
3
K): S = 0.45 × 10 kmol/m ⋅bar.
2
m /s; Table A-10, He-SiO 2 (293
ANALYSIS: From Table 14.1, the molar diffusion rate may be expressed as
NA,r =
CA,S1 − CA,S2
R m,dif
where
R m,dif =
1 1 
 1
1
1
1 
12
3
−
 − =

 = 1.81 ×10 s / m
−
13
2
4π DAB  r1 r2  4π 0.4 ×10 m / s  0.1m 0.11m 
(
)
with
CA,S1 = Sp A = 0.45 ×10 −3 kmol/m 3 ⋅ bar × 4 bar = 1.80 ×10 −3 kmol/m3
CA,S2 = 0
find
NA,r =
1.80 ×10−3 kmol/m3
1.81× 1012 s / m3
= 10−15 kmol/s.
Hence
n A,r = M A N A,r = 4 kg/mol ×10 −15 kmol/s = 4 ×10 −15 kg/s.
<
COMMENTS: Since r1 ≈ r2, the spherical shell could have been approximated as a plane wall with L
2
= 0.139 m2 . From Table 14.1,
= 0.01 m and A ≈ 4π rm
R m,dif =
L
DAB A
=
(
0.01m
)(
0.4 × 10−13 m 2 / s 0.137 m 2
)
= 1.8 ×1012 s / m 3
and
NA,x =
CA,S1 − CA,S2
R m,dif
=
1.80 × 10−3 kmol/m3
Hence the approximation is excellent.
1.8 ×1012 s / m3
= 10−15 kmol/s.
PROBLEM 14.18
KNOWN: Pressure and temperature of hydrogen inside and outside of a circular tube. Diffusivity
and solubility of hydrogen in tube wall of prescribed thickness and diameter.
FIND: Rate of hydrogen transfer through tube per unit length.
SCHEMATIC:
ASSUMPTIONS: (1) Steady diffusion in radial direction, (2) Uniform total molar concentration in
wall, (3) No chemical reactions.
ANALYSIS: The mass transfer rate per unit tube length is
′ =
NA,r
CA ( r1 ) − C A ( r2 )
ln ( r2 / r1 ) / 2π DAB
where from Eq. 14.44, CA,s = Spa,
CA ( r1 ) = Sp A,1 = 160 kmol/m3 ⋅ atm × 2 atm = 320 kmol/m3
CA ( r2 ) = Sp A,2 = 160 kmol/m 3 ⋅ atm × 0.1atm = 16 kmol/m 3 .
Hence,
N′A,r =
304 kmol/m3
( 320 − 16 ) kmol/m3
=
ln ( 20.5/20 ) / 2π × 1.8 × 10−11 m2 / s 2.18 × 108 s / m 2
′ = 1.39 ×10 −6 kmol/s ⋅ m.
NA,r
<
COMMENTS: If the wall were assumed to be plane,
R ′m,dif =
L
5 × 10−4 m
=
= 2.21 ×108 s / m2
11
2
−
DABπ D 1.8 ×10
m / sπ ( 0.04 m )
8
2
which is close to the value of 2.18 × 10 s/m for the cylindrical wall.
PROBLEM 14.19
KNOWN: Dimensions of nickel tube and pressure of hydrogen flow through the tube. Diffusion
coefficient.
FIND: Mass rate of hydrogen diffusion per unit tube length.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, (2) One-dimensional diffusion through tube wall, (3) Negligible
pressure of H2 in ambient air, (4) Tube wall is a stationary medium of uniform total molar
concentration, (5) Constant properties.
-3
3
PROPERTIES: Table A-10 (H2 – Ni): S = 9.01 × 10 kmol/m ⋅bar.
ANALYSIS: From Table 14.1, the resistance to diffusion per unit tube length is
Rm,dif = ln (Do/Di)/2π DAB, and the molar rate of hydrogen diffusion per unit length is
N A,r =
(
2π DAB CA,si − CA,so
ln ( Do / Di )
)
From Eq. (14.44), the tube wall molar concentrations are
CA,si = S pA,i = 9.01× 10−3 kmol / m3 ⋅ bar × 4 bar = 0.036 kmol / m3
CA,so = S p A,o = 0
N A,r =
With
2π × 10−12 m 2 / s × 0.036 kmol / m3
= 2.00 × 10−12 kmol / s ⋅ m
ln (0.028 / 0.025 )
M A = 2kg / kmol for H 2 ,
n A,r = M A N A,r = 2 kg / kmol × 2.00 × 10−12 kmol / s ⋅ m = 4.00 × 10−12 kg / s ⋅ m
COMMENTS: The hydrogen loss is miniscule.
<
PROBLEM 14.20
KNOWN: Conditions of the exhaust gas passing over a catalytic surface for the removal of NO.
FIND: (a) Mole fraction of NO at the catalytic surface, (b) NO removal rate.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional species diffusion through the film,
(3) Effects of bulk motion on NO transfer in the film are negligible, (4) No homogeneous reactions of
NO within the film, (5) Constant properties, including the total molar concentration, C, throughout the
film.
ANALYSIS: Subject to the above assumptions, the transfer of species A (NO) is governed by
diffusion in a stationary medium, and the desired results are obtained from Eqs. 14.60 and 14.61.
Hence
x A,s
x A,L
=
1
1 + ( Lk1′′ / D AB )
Also
N′′A,s = −
x A,s =
0.15
1 + 0.001m × 0.05 m / s 10−4 m2 / s
= 0.10.
<
k1′′Cx A,L
1 + ( Lk1′′ / D AB )
where, from the equation of state for a perfect gas,
C=
p
1.2 bar
=
= 0.0187 kmol/m 3.
2
3
−
ℜT 8.314 × 10 m ⋅ bar/kmol ⋅ K× 773 K
Hence
N′′A,s = −
0.05 m / s × 0.0187 kmol/m 3 × 0.15
(
1 + 0.001m × 0.05m/s 10
or
−4
2
m /s
)
(
= −9.35 ×10−5 kmol/s ⋅ m 2
)
n ′′A,S = M A N′′A,S = 30 kg/kmol −9.35 ×10−5 kmol/s ⋅ m 2 = −2.80 ×10−3 kg/s ⋅ m 2 .
The molar rate of NO removal for the entire surface is then
NA,s = N′′A,s A = −9.35 × 10−5 kmol/s ⋅ m 2 × 0.02 m 2 = −1.87 ×10 −6 kmol/s
or
−
n A,S = −5.61 ×10 5 kg/s.
<
COMMENTS: Because bulk motion is likely to contribute significantly to NO transfer within the film,
the above results should be viewed as a first approximation.
PROBLEM 14.