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, U
o
, with air at T
o
=15
°
C and velocity V
o
=
20 m/s in cross-flow; compare thermal resistances due to convection, tube wall conduction and fouling;
(b) Overall coefficient, U
o
, with water (rather than air) at T
o
= 15
°
C and velocity V
o
= 1 m/s in cross-
flow; compare thermal resistances due to convection, tube wall conduction and fouling; (c) For the
water-air conditions of part (a), compute and plot U
o
as a function of the air cross-flow velocity for 5
≤
V
o
≤
30 m/s for water mean velocities of u
m,i
= 0.2, 0.5 and 1.0 m/s; and (d) For the water-water
conditions of part (b), compute and plot U
o
as a function of the water mean velocity for 0.5
≤
u
m,i
≤
2.5
m/s for air cross-flow velocities of V
o
= 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):
k
w
= 15.1 W/m
⋅
K;
Table A.6
, Water
(
m,i
T
= 348 K):
ρ
i
= 974.8 kg/m
3
,
µ
i
= 3.746
×
10
-4
N
⋅
s/m
2
, k
i
= 0.668 W/m
⋅
K, Pr
i
= 2.354;
Table A.4,
Air (assume
f,o
T
= 315K, 1 atm): k
o
= 0.02737 W/m
⋅
K,
ν
o
= 17.35
×
10
-6
m
2
/s, Pr
o
= 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):
o
o
tot
cv,i
f ,i
w
f ,o
cv,o
1 U
A
R
R
R
R
R
R
=
=
+
+
+
+
cv,i
i
i
cv,o
o
o
R
1 h A
R
1 h
A
=
=
f ,i
f ,i
i
f ,o
f,o
o
R
R
A
R
R
A
′′
′′
=
=
and from Eq. 3.28,
(
) (
)
w
o
i
w
R
ln D
D
2 Lk
π
=
The convection coefficients can be estimated from appropriate correlations.
Continued...