Problem 1 Solution:
Given
Tube: Polished copper;
Diameter, D: 0.15 m
Length, L: 1 m
Steam at (Tsat): 100°C= 373K
Tube surface temperature (Tw): 90°C= 363K
a) The steam properties at Tsat :100°C [from Table in any heat transfer book,
like Book of Incropera]
ρl = 957.9 kg/m3
h fg = 2257 ×103 J/kg
ρ v = 0.60 kg/m 3
µl = 279 ×10 −6 kg ⋅ m/s
σ = 0.0589 N/m
C pl = 4217 J/kg ⋅ °C
Prl = 1.76
Properties of saturated
(373K+363K)/2=368K:
ρ l = 961.7 kg/m 3
water
at
reference
temperature
(Tf):
k l = 0.678 W/m ⋅ K
µ l = 296 × 10 −6 kg ⋅ m/s
C pl = 4212 J/kg ⋅ °C
The modified latent heat of vaporization,
ℎ = ℎ + 0.68
−
= 2257 × 10 + 0.68 × 4212 100 − 90 = 2285 × 10
/!
For film condensation (FC) in horizontal tube, we have the heat transfer
coefficient (Nu) from
/.01
)* * − *+ ℎ´ &
ℎ$% &
$$$$
"#% =
= 0.729 (
.
'
-'
−
Thus,
' )* * − *+ ℎ´ &
$
ℎ% = 0.729 (
.
& -'
−
/.01
/.01
0.678 9.81 × 961.7 961.7 − 0.6 2285 × 10 × 0.15
ℎ$% = 0.729
(
.
0.15
296 × 1023 × 0.678 100 − 90
= 4556 W/m2K
a) For boiling in the system, the tube surface temperature (Ts) is 110°C= 383K.
Since the excess temperature (Ts -Tsat) is 10°C, it can be assumed as nucleate
boiling (NB) and heat flux can be determined from Rohsenow relation:
3
C (T − Tsat ) 
 p ,l s

q& nucleate
 C sf h fg Prln 


For water-copper(polished), we have
= 0.0128 and n= 1. Employing
properties at Tsat in above Rohsenow correlation
 g (ρ l − ρ v ) 
= µ l h fg 

σ


1/ 2 
/.1
9.81 957.3
789 = 279 × 1023 × 2257 × 10 (
.
0.0589
789 = 143444 W/m2
ℎ$=> =
?@9
AB 2ABCD
=
EF FFF
E/
×;
4217 × 10
<
0.0128 × 2257 × 10 × 1.76
= GHIHH. H W/m2K
Justification: we have much larger heat transfer coefficient (ℎ$=> ) from the
nucleate boiling than that of the film condensation (ℎ$% ) obtained in part (a). This
is due to the fact that during nucleate boiling the saturated liquid having
comparatively larger thermal conductivity (than vapor for condensation) come
in contract with the heater surface besides undergoing vigorous mixing through
bubble nucleation and flow (upward), generate local turbulence flow resulting
very high heat transfer coefficient. Whereas in film condensation on a horizontal
tube (in this case) the heat transfer is low due to the vapor with low thermal
conductivity cannot come in direct contact with the (cold) surface because of
the condensate (liquid) film and also due to slow condensation and flow of the
condensate over the surface leading to low heat transfer coefficient.
c) For vertical orientation of the tube and the film thickness is not known, it is
convenient to determine the dimensionless parameter, P in order to determine
the heat transfer coefficient from:
J=
Where, ℎ = 2285 × 10 /!; L =
m2/s, ' = 0.678 W/m⋅K
J=
'K
- ℎ´
-
−
L 0M
)
E/
M* = 296 × 1023 /961.7= 3.078× 102N
0.678 × 1 100 − 90
= 470.7
2N 0
3 3,078 × 10
23
E/
M
296 × 10 × 2285 × 10 [
9.81]
J = 470, which is within the condition of 15.8 < J < 2530 and have following
average Nusselt number correlation in terms of P:
/.T0
/.T0
1
1
$$$$
[0.68 × 470.7 + 0.89]
"#S = [0.68J + 0.89] =
470.7
J
= 0.2413
$$$$
"#S =
ℎ$S = 0.2413
'U
L2U M
)
1/3
L2
ℎ$S U M)
= 0.2413
1/3
'U
= 0.2413
0.678
[
3,078×10
−7 2
V
1/3
9.81]
= W64X. H W/m2K
Justification: Comparison of this value of ℎ$S of vertical tube’s with the ℎ$% of
horizontal tube obtained in part (a) shows that ℎ$% > ℎ$S . This is anticipated
especially for comparatively long tube (relative to its diameter) as in vertical
orientation the film decreases temperature gradient along the tube as well as
film thickness increases resulting in decreasing the heat transfer coefficient. In
addition, for horizontal tube the characteristic length (diameter) for the film is
smaller compared to the vertical one (height or length) under same conditions
resulting to larger ℎ$% .
[Note that this ℎ$S of vertical tube’s film condensation might also be determined
by other not-so accurate (large error) ways like using correlation with ℎ$% of
horizontal tube with a relation to its diameter and height ratio. Beside
inaccuracy, such correlations also have risk of giving completely wrong result in
case ℎ$% (horizontal or vice versa) is not correct]
Problem 2 Solution:
Given:
Cylindrical furnace (plain base), Diameter, D: 2.5 m and height, H: 5 m;
Air at atmospheric pressure (1 atm).
Gas temperature, Tg: 1000 ºC= 1273K
Wall temperature, Tw: 300 ºC= 573K
Surface emissivity of furnace: 1 (black)
Partial pressure of CO2 and H2O
PCO2 = 0.16 bar= 0.158 atm
PH2O= 1.5x0.16 bar= 0.24 bar=0.237 atm
Using Table for mean beam length (Lm) for radiation from a gas volume to a
surface on its boundary of cylindrical geometry for the condition of H/D= 2 and
plane base, we have:
Lm = 0.60D = 0.60x2.5= 1.5 m
[Lm = 0.73D can also be used for entire surface]
a) Now in order to determine the total emissivity (εg) and absorptivity (Z ) of
the gas mixture we need to employ Hottel’s charts and related parameters.
For total emissivity (εg):
At 1 atm pressure and considering correction factor, CCO2=CH2O=1 (Note: CCO2 and
CH2O can also be determined and expected to be close to unity in this case) and the total
emissivity of the gas mixture can then be obtained from,
εg = εH2O+εCO2 -∆ε
Lm PCO2 = 1.5x0.158 = 0.237 m.atm
Lm PH2O = 1.5x0.237 = 0.356 m.atm
Using above values of parameter and Tg= 1273K from Hottel’s charts for CO2
and H2O we have:
εCO2 =0.14
and
εH2O = 0.2
Now for the overlap parameter, ∆ε:
Lm (PCO2+ PH2O)=1.5x(0.158+0.237)=0.593 m.atm ;
and
PH2O /(PCO2+ PH2O) = 0.237/(0.158+0.237)=0.6
From the overlap parameter chart for Tg> 930 ºC, we have
∆ε=0.048
Thus, total emissivity εg = 0.2+0.14 -0.048=0.292
For total absorptivity (Z ) at 1 atm, we have:
α g = α H 2O + α CO2 − ∆α
Hottel’s chart can also be used to obtain the Z through emissivity of each
component at Tw (300 ºC =573K) using following corelation:
Z =[
Jb_0 Kd
J^0_ Kd
Ae
Af
Ae
Af
\
/.1
(]^0_ `
, J^0_ K
a + ]b_0
, Jb_0 K
. − ∆]
=0.158x1.5(573/1273) = 0.11 m.atm;
=0.237x1.5x(573/1273) = 0.16 m.atm
Using these above values in Hottel’s chart for Tw=573K, we have:
εCO2 = 0.1
εH2O = 0.22
For overlap ∆ε :
Kd Jgh0 + Ji0h0
Ae
Af
= 1.5 × 0.395 j
1N
E0N
k = 0.27 m.atm, and
PH2O /(PCO2+ PH2O) = 0.6
Using chart for Tg= 127 ºC (for Tw=300 ºC) for above values [note: charts Tg=
127 ºC and 538 ºC can also be used and average value to be used for ∆ε]
∆ε = 0.01
Thus, total absorptivity lm = (1273/573)0.5[0.22+0.1]-0.01 = 0.467
b) Radiant heat flux from the gas to the wall:
n"m→p = ] r
F
= 0.292 × 5.67 × 102T × 1273
F
= HI. H4 kW/m2
The total/net rate of heat transfer to the wall:
y
s = tru]
F
−Z
F
v
y
Where t = w&x + 2 j k &0 = w × 2.5 × 5 + 2 j k 2.50 = 49.1 m2
F
F
z = 49.1 × 5.67 × 102T [0.292 1273
F
− 0.467 573 F ] = G{{H. W kW