Page 199 - Chemical Process Equipment - Selection and Design

8.1. CONDUCTION OF HEAT

EXAMPLE Basing on the inside area,
8.11
Effest of Ignoring the Radius Correction of the Overall Heat
Transfer Cmfficient
The two film coefficients are 100 each, the two fouling coefficients U, = [1/100 + 1/2000 + [(30/0.01)(0.0448/0.04)]~1
are 2000 each, the tube olutside diameter is 0.1 ft, wall thickness is f 0.8/100 + 0.8/2000]-1 = 52.0898.
0.01 ft, and thermal conductivity of the metal is 30:
ri/ro = 0.04,10.05 = 0.8, Ignoring the corrections,
r, = (0.05 - 6).04)/ln1.25 8.0448,
rm/ro = 0.898, U = (2/100 + 2/2000 + 1/30/0.01)-' = 46.8750.
U,, = [1/100(0.8) + 1/2000(0.8) + 1/(30/0.01)(0.8963)
+ 1,I100 + 1/21)00]-' = 41.6721. The last value is very nearly the average of the other two.

where r, is the mean radius of the cylinder, given by Eq. (8.12).
Since wall thicknesses of heat exchangers are relatively small
and the accuracy of heat transfer coefficients may not be great, the
(8'20) ratio of radii in Eq. (8.21) often is ignored, so that the equation for
the overall coefficient becomes simply
where L, is the thickness of the metal. 111 1 1 1
If the wa11 is that of hollow cylinder with radii r, and r,,, the _=- +-+-- +-+- (8.22)
overall heat transfer coefficient based on the outside surface is u hi 4 kJL3 h4 h,
The results of the typical case of Example 8.2, however, indicate
,
1
1
I
- ___- 1 -+ 11 (8.21) that the correction may be significant. A case with two films and
=
(k3/L3)(rml~,3) h, h, two solid cylindrical walls is examined in Example 8.3.
EXAMPLE 8.3
A Case of ai Camposite W : Optimum Insulation Thickness
for a Steam Line
A 3 in. IPS Schecl40 steel line carries steam at 500°F. Ambient air is
at 70°F. Steam side coefficient is 1000 and air side is
3Btu/(hr)(sqft)(C'F). Conductivity of the metal is 30 and that of
insulation is 0.135 Btu/(hr)(sqft)("F/ft). h7alue of the steam is
$S.OO/MBfu. cost of the insulation is $1S/(yr)(cuft). Operation is
8760 hr/yr. The optimum diameter d of insulation thickness will be
found.
Pipe:
do = 0.2917 ft,
d, = 0.2557 ft,
h(do/di) = 01.1317.

Insulation:

68 IJ=l#Ul~DZ
Heat transfer coefficient based on inside area:
78 Q=43B*U
0.05 '$1
98 CZ=l.Sf~D~2-Dl~2)/
u.=d. -+- 1 0. I317 + ln(d/0.2917) -' 88 C1=.84358tQ
'[lOOOdi 30 ' 188 C=Cl+CZ ! (re-?&
urn >
llB PRINT USING 129 j
E:
Steam cost: 128 IMAGE .DDDcX.x .DUD.*
an. OOl x> EID. ODDD
C, = 5(10~6)(8760)Q/Ai 138 GOTO 48
= 0.0438Q/Ai, $ (yr)(sqft inside). (3) 148 END