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308
CHAPTER 13
1958-Guide
Table 11. Summer Coefficients of Heat Transmission U of Flat Roofs Covered With Built-Up Roofing*
Btu per (hour) (square fool) (F deg difference between the air on the two sides)
Insulation on Top op Deck (Covered With Built- Up Roofing)
Type op Roof Deck Ceiling not shown
Thickness of Roof Deck
(Inches)
No Ceiling-- Underside of Roof
Exposed
Furred Ceiling with Air Space, Metal Lath
and Plaster
No In
sula* tion
Insulating Boardd Thickness, In.
i 1 u2
No In sula-
tion
Insulating Board1* Thickness, In.
i1
2
Flat Metal Roof Deck
4 Ply Felt Roof
Ditto + 4 in* Slag
0.73 0.35 0.23 0.17 0.13 0.40 0.25 0.18 0.14 0.12 0.54 0.30 0.20 0.16 0.13 0.34 0.22 0.16 0.13 0.11
Precast Cement Tile ------ -5.
= Concrete
----- --------------
idt&Sgwfif 1
--
4 Ply
Felt Roof
li
Ditto + 4 in* Slag 1
0.67 0.33 0.22 0.17 0.13 0.38, 0:24 0.18 0.14 0.12 0.50 0.28 0.20 0.15 0.12 0.32 0.21 0.17 0.13 0.11
4 Ply Felt Roof
2 4
6
Ditto
2
4
+ 4 in. Slag 6
0.65 0.33 0.22 0.16 0.13 0.37 0.24 0.18 0.14 0.12 0.59 0.31 0.21 0.16 0.13 0.36 0.23 0.17 0.13 0.12 0.54 0.30 0.20 0.16 0.13 0.33 0.22 0.17 0.13 0.11 0.49 0.28 0.20 0.15 0.12 0.31 0.21 0.16 0.13 0.11 0.46 0.87 0.19 0.15 0.12 0.30 0.21 0.16 0.13 0.11 0.42 0.26 0.19 0.14 0.12 0.29 0.20 0.16 0.13 0.10
Gypsum and Wood Fiberb 4 Ply
on 4' Gypsum Board
Felt
Roof
21 31
^
Ditto
21
+ i m. Slag 31
0.34 0.23 0.17 0.13 0.12 0.25 0.18 0.14 0.12 0.097 0.28 0.20 0.15 0.12 0.11 0.21 0.16 0.13 0.11 0.094 0.29 0.20 0.16 0.13 0.11 0.22 0.16 0.13 0.11 0.093 0.25 0.18 0.14 0.12 0.10 0.19 0.15 0.13 0.10 0.090
Wood0 ____V---
n / /____
4 Ply Felt Roof
Ditto
+
S'
I
1 0.43 0.26 0.19 0.15 0.12 0.29 0.20 0.15 0.13 0.11 H 0.33 0.22 0.17 0.13 0.11 0.24 0.18 0.14 0.12 0.097 2 0.29 0.20 0.16 0.13 0.11 0.22 0.16 0.13 0.11 0.094 3 0.22 0.16 0.13 0.11 0.09 0.17 0.13 0.12 0.10 0.085 1 0.35 0.23 0.17 0.14 0.11 0.25 0.18 0.14 0.12 0.10 u 0.29 0.20 0.15 0.12 0.1C 0.21 0.17 0.13 0.11 0.093 2 0.26 0.19 0.14 0.12 0.1C 0.20 0.15 0.13 0.1C 0.090 3 0.20 0.15 0.12 0.10 0.09 0.16 0.13 0.11 0.09 0.081
a The summer coefficients are considered temporary, and have been calculated with an outdoor wind velocity of 8 mph. For.summer an inside surface conductance of 1.2 has been used instead of the regular 1.65 value. In all of these roofs a 4 ply felt roof been assumed f in. thick, thermal conductivity = 1.33. Pitch and slag have been assumed as an additional thickness of $ in. which hag been assigned thermal con ductivity = 1.0. In both cases thermal conductivity refers to one inch thickness.
b 874 percent gypsum, 124 percent wood fiber. Thickness indicated includes 4 in. gypsum board. This is a poured roof.
c Nominal thickness of wood is specified, but actual thickness was used in calculations.
d If corkboard insulation is used, the coefficient U may be decreased 10 percent.
A method of determining heat flow rates, when structure is not given in Tables 9 or 10, is illustrated in Example 7.
Example 7: A 4 in. stone concrete roof covered with an average depth of 4 in. cin der concrete (A = 4.9) on which is placed a f in. thick felt roof with $ in. pitch and slag surface, is exposed to the sun. The location is the central part of the United States.
Cooling Load
309
Design temperatures arc: outdoor 95 F; daily range 20 deg; indoor temperature 80 F. Find the heat flow rate at 2:00 p.m. for a day in July.
Solution: For the purpose of selecting the equivalent temperature differential, this construction is assumed to be equal approximately to an uninsulated 6 in. concrete roof, for which the equivalent temperature is found to be 38 deg in the 2:00 p.m. column of Table 9. Calculate the overall heat transmission coefficient U (see Equa tion 3 of Chapter 9) of the roof as follows:
XJ =---------------:--------1------- :---------------= 0.33.
_L 4 4 -375 | 0 50 , 1
1.2 + 12 + 4.9 + 1.33 + 1.00 + 4.0
The heat flow rate is then 38 X 0.33 equals 12.5 Btu per (hr) (sq ft).
OUTDOOR
GLASS
INDOOR
SOLAR RADIATION
REFLECTED 1 SOLAR RADIATION
RADIATION EXCHANGE BETWEEN GLASS AND SURROUNDINGS
TCONVECTION EXCHANGE BETWEEN GLASS AND SURROUNDINGS
\
V-----
TRANSMITTED
SOLAR RADIATION
ABSORBED ENERGY DUE TO THE HEAT CAPACITY OF THE GLASS
CONVECTION AND RADIATION EXCHANGE BETWEEN CLASS AND INDOORS
--A.
Fig. 2. Instantaneous Heat-Balance for a Glass or Glass Block Section
TABLES FOR CALCULATING SOLAR HEAT GAIN THROUGH GLASS AREAS
Basic Principles
In order to set forth the principles involved in calculating heat flow through glass areas, the general instantaneous heat-balance relation will be presented. It will be shown schematically in Fig. 2. The net heat gam for the indoor space is the result of several contributing factors.
The following observations concerning the behavior of glass with respect to radiant energy will lead to a better understanding of the.heat-balance relation.
,, Glass transmits, in varying degrees, radiation having wave lengths between au and 4.75 microns. The percentage of each wave length transmitted is dependent
upon the chemical and physical characteristics of the glass, and upon the angle of cidence of the radiant energy. Of the energy not transmitted, part is absorbed
and part reflected.
2. Glass is opaque to radiant energy emitted from sources below 450 F.
two pf6?6 the abve-principles, it is convenient to group radiant energy into classifications, solar radiant energy and low-temperature radiant energy.
The complete heat-balance for a glass section can be expressed for a unit time interval as follows:
Total heat flow, 1 Transmitted, through glass sectionJ " solar radiation
]Heat flow by convective
and radiative exchanges at (2a) the indoor surface
mi
ft h! *2TM term
right side of Equation 2a can also be expressed by
neat balance equation as follows: