Document 4vajKyZKj8odJybBxGvRN2ORx
54
CHAPTER 5
1960 Guide
Oats
1 2 3
4 5
Table 3 .... Radiation Factors or Emisuvities, *
Far determination of factor fg m Fgvofroa $
Surfaces
Fraction of Block-Bod)' Badkitiao
At 50-100F
At 1000F
A small hole in a large box, sphere, furnace, or enclosure...............
0.97 to 0.99
Black Qon-metallic surfaces such as asphalt, carbon, slate, paint, paper. 0.90 to 0.98
Red brick and tile, concrete and stone, rusty steel and iron, dark paints
(red, brown, green, etc.)............................................................................. 0.85 to 0.95
Yellow and buff brick and stone, firebrick, fire clay................................ 0.85 to 0.95
White or light-cream brick, tile, paint or paper, plaster, whitewash... 0.85 to 0.95
0.97 to 0.99 0.90 to 0.98
0.75 to 0.90 0.70 to 0.85 0.60 to 0.75
Absorptirity for Solar Radiation
0.97 to 0.99 0.85 to 0.98
0.65 to 0.80 0.50 to 0.70 0.30 to 0.50
8 Dull brass, copper, or aluminum; galvanized steel; polished iron......... 0.20 to 0.30 0.30 to 0.50 9 Polished brass, copper, mooei metal............................................................ 0.02 to 0.05 0.05 to 0.15 10 Highly polished aluminum, tin plate, nickel, chromium.......................... 0.02 to 0.04 0.05 to 0.10
SmtsxirHiea of otbw mtraii 1007 be found is Reference 4. * Reflect* about 8 percent.
0.40 to 0.65 0.30 to 0.50 0.10 to 0.40
The temperature drop, Af, through an individual resistance may then be calculated from the relation:
A* - Rq,,
(15)
where R is the resistance in question. The problem is now reduced to one of evaluating the in
dividual resistances of the system. This entails suitable manipulation of the rate Equations 1, 2, and 5 to produce expressions of the form:
dicate the magnitudes of the thermal conductivities, k, to be employed in the expressions of Table 6, after dividing k by 12.
The solution applicable to the problem depicted in Fig. 8, for the calculation of R\ and Rt, is Case 2 in Table 6. Thus for a 1-fl length of 2 in. nominal size pipe (I. O. = 2.067 in., 0. D. = 2.375 in.) insulated with 1 in. of material hav ing a conductivity of 0.025:
1.188
ft - 2rX
" 8-5 X 10- (hr) (P deg) per Btu.
where q is the heat transfer rate, and At is the potential drop or temperature difference through the resistance RTable 6 lists such solutions for six different conduction sys tems, Table 4 in Chapter 9 and Table 1 of this chapter in
, 2.188 R' - ixoixl - 39 (hr) (F de8) per Btu. The convection resistances to heat transfer from the pipe
Tutu p f Dog
0
Table 4 .... Heal Transmission by Radiation for Black-Body Conditions*
Expressed m Btu per (square foot) (boor)
-I -2 -5 -4 --5 -6 -7
-a --9
-30 59.3 -20 . 65.2 -10 71.4
0 78.0
58.7 64.7 70.8 77.4
58.2 64.1 70.1 76.7
57.7 63.5 69.5 76.0
57.2
62.9 68.9 75.4
56.7 62.3 683 74.7
56.2 61.7 67.7 74.0
55.7 *61.1
67.1 73.4
55.2 60.5 66.4 72.7
54.7 59.9 65.8 72.1
- 0 + 1 +2 +3 +4 +5 +6 +7 +8 +9
0
78.0
78.7
79.4
80.1
80)8
81.5
82.2
- 82.9
83.6
84.3
10
85.0
85.7
86.5
87.2
88.0
88.7
89.4
90.2
90.9
91.7
20
92.4
93.3
94.0
94.8
95.6
96.4
97.2
98.0
98.8
99.6
30 100 101 102 103 104 105 105 106 107 108
40 109 110 111 112 112 113 114 115 116 117
50 118 119 120 121 122 123 123 124 125 126
60 127 12S 129 130 131 132 133 134 135 136
70 137 138 139 140 142 143 144 145 146 147
80 148 149 150 151 152 153 154 155 156 157
90 159 160 161 162 163 164 166 167 168 169
100 170 171 173 174 175 176 178 179 180 182
110 183 184 185 187 188 189 191 192 193 195
120 196 197 199 200 201 203 204 206 207 209
130 211 212 214 215 217 218 220 221 222 224
* ErarnfU: Rxitutktp (ram will* of room at $2 F to surface at --25 F for cffectire cmiasiritjr of 0.05 -- (102 -- C2.2) 0.05 " 37.7 Btu per (aqusrc foot) (hour).
Heat Transfer
55
Table 5 ... Met Radiation Solutions
Sydoa
Solution
.. l| |e*
Two infinite parallel planes.
a-
~ T,')
o- + -* - 1
eJ cjj jj
l* e2 j
One radiation shield between two infinite parallel planes.
o *
Remo** Considering interrefiections. (Refer
ence 5)
Considering interrefiections. (Refer ence 5)
........ ... tp
n radiation shields between two infinite
parallel planes.
dnA "
(a). When!(l).
is the net radiation exchange without the shields.
Considering ence 5)
interrefiections.
(Refer
.--.
/ /"Vi 2 l^)
Two concentric spheres
or two infinitely long cylinders.
c At \j /
Considering interrefiections and diffuse surfaces. (Reference 5)
| >C\4A2 Two areas dAt and dA i
fr, -
- T'->
Surface diffuse, neglecting interrefiectiou. (Reference 5)
..--v | ) I2
y'
Tube of infinite length parallel to an infinite wall.
w " (i)- T')
where N is the length of cylinder from which q, is exchanged.
Neglecting interrefiections. (Reference 5)
,--m-. Surfaces are perfect radiators. fw __/ Surface element dA and rec1 tangle above and parallel to
it, with one corner of rectangle contained in normal to dA.
See Fig. 4
(Reference 4)
lb*. Surfaces are perfect radiators. l^w>T Adjacent rectangles in perpen-
dicutar planes.
See Fig. 5
(Reference 4)
1j 1j [|
Surfaces are perfect radiators. Opposed parallel rectangles and discs of equal size.
See Fig. 6
(Reference 4)
wall to the cold water, Rt, and from the air to the surface of the insulating material, Rc, are dependent on the flow conditions prevailing at these surfaces, and on the thermal properties of the fluids. These resistances are also directly dependent upon the temperature distribution, and for this reason it is necessary first to guess, on the basis of the prob lem statement, a temperature distribution upon which to base the initial calculations. Since the values of A* for heat transfer between water and pipe walla are relatively high in this temperature range, it is logical to assume only a small temperature difference between the temperature of the fluid body and the temperature of the pipe wall. For the purpose
of an initial guess, this temperature difference will be as sumed to be 2 deg. On the other hand, he for heat transfer
from air to a body is relatively small, and a higher tempera
ture difference would be expected between these masses.
The value initially assumed here will be 20 deg. In summary the temperature distribution in the system is assumed as follows:
Fluid temperature -- 34 F. Inner pipe wall temperature = 36 F. Outer insulation surface temperature ** 100 F. Ambient air temperature -- 120 F.
/'
With these assumptions and the problem statement, it is now possible to calculate values for the convective resist ances. If it is found in the ultimate solution of the problem that the temperature distribution is different from that as sumed, it will then be necessary to repeat the solution pro cedure.
If reference now be made to Table 2, it is found that Case 3 of this table is a system similar to that encountered in the