Document 7RwDEOEqxaX3VYG1k6dq7wO8B

92 Chapter 4 1945 Guide Table 3. Conductivities (k) and Conductances '(C) Used~in~Calcul-ating----------Heat Loss Coefficients (U) in Tables 4 to 16 These constants are expressed in Blu per hour per square foot Per degree Fahrenheit temperature difference. Conductivities (ft) are per inch thickness and conductances (C) are for thickness or construction stated. not per inch thickness. _ MATERIAL DESCRIPTION Conductivity or Conductance (4) (C) ResrANCE Per Inch Thickness '(*> . For Thickness Listed (*) AIR SPACES atBounded bt ordinart materiai..,,. Vertical0, X orBounded bt aluminum foil Vertical0, in. more in width:.__________ in. mare in width............ --___ ' ------- 1.10 0.46 , -- 0.91 2.17 EXTERIOR FINISHES (Frame Walls) 4 in. thick (nominal)..... ..................... ...... ...... 12.50 2.27 1.28 1.28 0.08 0.44 0.78 0.78 INSULATING MATERIAL # Blankets:.__ _________ _________ ______ Made from mineral or vegetable fibers or Insulating Board.................. __________ 0.27 0.27 0.30 0.33 0.27 0.48 3.70 3.70 353 3.03 3.70 2.08 N ____ INTERIOR FINISHES _Gtpsum Board (5ds in.)_______________ Plain or decorated-- ___________ 0.50 3.30 Insulating Board Lath (^ in.) and Plaster................ ............ ........... ............ Plaster thickness assumed H io<-------- ----------- Insulating Board Lath (1 in.) and 04Plywood in.).... _________ ,,.____ Plain or decorated__________________________ MASONRY MATERIALS 1 Brick. .................. ...................... ........ Face____________----------------------- ------------------- 3 in. Clat tile (hollow)_________ -- ....... ...................... ....................................................... 4 in. Clat tile (hollow)___________ ............................... .................... ...................... 3.56 5.00 9.20 12.00 8 in. Clat tils (hollow)--_________ .......... ....... ........................ 10 in. Clat til* (hollow)......... ...... ..... . .................. ....... .................. 12 in. Clat tile (bollow)___________ .................................................................................... --... Concrete............... ..... .... light weight aggregate*..... .......... ...... .............. 250 12.00 8 in. Concrete blocks......... ........ ....... Hollow, cinder aggregate- ----------------- ..----- GtPSUM FIBER CONCRETE--____________ 87H per cent gypsum and 12H per cent wood chips_____________ .-- --------------- ------ 3 IN. GTPSUM TELE--____ --................... Hollow.................... _.............. .. ................ ....... -- 1.66 ........ - Tile and Terraezo.____________ ______ For flooring 1250 12.00 3.70 2.4 Q.66 0.60 0.31 4.40 2.12 250 ____ 1.28 1.00 0.64 0.60 0.58 0.40 051 158 1.00 1.00 050 0.60 053 050 0.47 0.61 0.46 2.00 050 0.28 0.20 0.11 0.08 ____ --__ __ _ ____ 0.40 0.08 ____ 0....6.0 0.08 0.08 057 0.42 152 1.67 3.18 053 0.47 0.40 ____ 0.78 1.00 157 1.67 1.72 250 353 0.78 1.00 1.00 155 1.66 158 2.00 2.13 ' 154 2.18 -- Conductance values for horizontal air spaces depend on whether the heat flow is upward or downward, but in most cases' it is sufficiently accurate to use the same values for horizontal as for vertical air spaces. ^Expanded slag, burned day or pumice. .. Heat Transmission Coefficients 93 Table 3. Conductivities (k) and Conductances (C) Used in Calculating ----------------Heat Loss-CoefficientsXWjnJTables 4_to 16--Concluded These constants are expressed in Btu per hour per square foot Per degree Fahrenheit temperature difference. Conductivities (k) are per inch thickness and conductances (C) are for thickness or construction stated, not per inch thickness. MATERIAL DESCRIPTION CONDUCnVTTT OR Conductance (*) (C) Resistance Per Inch For Thickness Thickness listed tt) (*) ROOFING MATERIALS Asphalt Shingles--.............................. Built-up roofing--_____________________ Heavy Roll roofing___ Slate_____________ Assumed thickness in._______ __ ______ 10.00 6.00 6.50 353 650 1.28 0.10 0.17 0.15 058 0.15 058 - SHEATHING Fib or Yellow Pine (1 m.)...... ...... .. 252 0.35 ... ::z v0.42 2.56 2.37 0.39 1.02 0.98 056 1.16 SURFACES Ordinary non-reflective materials, VerticalOrdinary non-reflective materials, vertical... 1.65 6.00 0.61 0.17 WOODS Fir sheathing (1 in.) building paper Yellow Pine or Fir.......... ..................... 1.15 0.80 050 057 155 2.00 building constructions. Lack of good judgment in the intelligent choice of an insulating material, or its improper installation, frequently repre sents the difference between good or unsatisfactory results. Computed Transmission Coefficients Computed heat transmission coefficients of many common types of building construction are given in Tables 4 to 16, inclusive, each con struction being identified by a serial number. For example, the coefficient of transmission (7) of an 8-in. brick wall and l/p, in. of plaster is 0.46, and the number assigned to a wall of this construction is 67-B, Tabled. Example 1. Calculate the coefficient of transmission (U) of an 8-in. brick wall with H in. of plaster applied directly to the interior surface, based on an outside wind exposure of 15 mph. It is assumed that the outside course is of hard (high density) brick having a conductivity of 9.20, and that the inside course is of common (low density) brick having a conductivity of 5.0, the thicknesses each being 4 in. The conductivity of the plaster is assumed to be 3.3, and the inside and outside surface coefficients are assumed to average 1.65 and 6.00, respectively, for still air and a 15 mph wind velocity. Solution, k (hard high density brick) = 9.20; x -- 4.0 in.; k (common low density brick) = 5.0; x = 4.0 in.; k (plaster) = 3.3; x = Hin.;/i = 1.65;/o ='6.0. Therefore. V =------------------------------------*___________ ___________ = ____________________ 1: J_ U)_ 41) 05 _1_ 6.0 + 9.20 + 5l) + 03 T65 0 167 + 0.435 + 0.80 + 0.152 + 0.606 = 0.46 Btu per hour per square foot per degree Fahrenheit difference in temperature between the air on the two sides. The coefficients in the tables were determined by calculations similar to those shown in Example 1, using fundamental Formulas 1, 2 and 3 and the values of k (or and a in Table 3. Actual thicknesses of lumber, are used in the computations rather than