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418 CHAPTER 24 1965 Guide And Data Book . For a wall of a.single homogeneous material of conduc tivity k and thickness z with surface coefficients/< and/,:. Then by definition: *, .1. ./ ' ' U - 1/Rr ; \_- For a wall with air space construction and consisting of two homogeneous materials.of conductivitics.Jb^and fc* thick* nesses xt and ij, respectively, and separated by an air space of conductance a: Ar = -- + ---1------ b t" + -- fi kt a kt f. (3) and U -.1/fir. The temperature at any interface can be calculated the temperature drop through any component of the wall is proportional to.its resistance. Thus, the temperature drop Afi. through fij in-Equation 1 is: '*'*** Afi TM Ri/Rr(ti -- <) (4) where U and t> are the indoor and outdoor temperatures respectively. Hence, the temperature at the interface between fit and fitis: <i-* - U - .(5) For .types of building materials having non-uniform or ir regular. sections such as hollow clay tile or concrete blocks, it is necessary to use the conductance C of rite section unit as manufactured instead of a conductivity k. The resistance of the section 1/C is substituted for xjk in Equations 2 and & = It will be noted that, in order to compute the U value of a construction, it is first necessary to know the conductivity and thickness of homogeneous materials,. the conductance of non-homogenequs materials (such as.concrete blocks), the surface, conductances of both sides of the construction, and the conductances of any air spaces. If the conductivities of the materials in a wall are highly dependent on temperature, the mean temperature in use must be known in order to assign the. correct value. In such cases it is perhaps most convenient to use a.trial and error procedure for the calculation of the total resistance, fir* first, the mean operating temperature for each layer is estimated and conductivities^ or conductances C selected. The total resistance fir is then calculated as in Equation 3 and then the temperature at each interface is calculated from Equations 4 and 5. r-.. 1 ' * . .. The mean temperature of each component (arithmetic mean of its surface temperatures) can then be-used to obtain con ductivities k or conductances C. This procedure nan then be Table I.... Thermal Conductivity Ur) Values of Soils In Approximate Order of Decreasing Values*- ` ' ......... ...................... ` MaaoTeaiperafore--40 f * ' SoO Deslgaatioa Meebcntzai Anefym % by Weigh} - Gravel ' Send Sat Cloy >- Moisture Content--% ,j io | .'.`.20.;* Over . 05 to . 2.0 ant. . 2.00[nun o4os to Under 0.05 mm - 0.005 atm . Dry PenttIHb ptr at ft 100_ . H2 120 90 110 90 Fine Crushed Quarts-.-:'..V: 0.0 ` Crushed Quarts.-.. ...;............. 15.5 Graded Ottawa Sand............... : 0.0 Fairbanks Sand..... .................. 27.5 Lowell Sand...................... . ..*,0.0 " 100:0 79.0' 1.99.9 70.0 .100.0. Cfiena Hirer Gravel....................... Crashed Feldspar...:..':;..-. Crashed Granite.-.. ................. > Dakota Sandy Loam............. Crushed Trap Hock....................... 80.0 , 25.5 16.2 . 10.9 . 27.0 19.4 *70.3 ; 77.0. 57.9 = 63.0" - '0.0 '-;u 6 .5 .0 .1 2 .5 0.0 0.0 0.0 0 .6 4 .2 6 .8 21.2 10.0 10 .0 .12.0 ,i6:o 11.5`, *,i6.o: 10.0 14.0 8.5 10.5 . 816 11-0 . 9.0+ 6.0 7.5* 5.5 !7.'5 6.5 5.0 6.0 22.0 13.5 13:0 9.5 10.0 9.5 7.0 15.0 : 13.5 13+ RamseyiSandy Loam.: .J:.1:'.'1.:'. Northwav Fine Sand 6.4 - 0.0 Northway Sand___ ' 3.0 Healy Clay:...............:.V.. . 0.0 Fairbanks Silt;Loain...................... v- :< 0.0,. - S3.6' . 97.0 9710" 1.9 =' J- -7.6;,- Fairbanks Silty Clay Loam-......... ,0.0 . . .9.2 Northwav Silt Loam...................... 1.0 21.0 27.5 3.0 - 0.0 20.1 .80.9 63.8 64.4 18.5 4.5 .6.5 0.0. . 4.5 5.5 0.0 . 4.5 6.0 :78.0 4.0+ 11.5 27.0 . 13.6 10.0 . 8.5 7.5+ 5.5 9.0+ 8.0 5.0 9.0+ 7.5 5.0 . 9.0+ 7.5 4.0+ *7.0+ 6.0+ 100 10.0 10.0 9-5.. 7.0+* Design, Heat; Transmission Coefficients 419 rwated until the conductivities or conductances, have been w-irrectly selected for the resulting meau temperatures, Gcn- this can be done in two or three trial calculations:. Series arid Parallel Heat Flow Paths (fi.,)(fiu) -- the combined resistance of fi-- and fi-- -- 4.71 Oiktt - Then fir - 0.5 + 1 + 4.71 + 0.5 + 0.25 In many practical installations the components are arranged so that parallel heat flow paths of different conductances re sult. If there is no lateral heat flow between paths, each path may be considered to extend from inside to outside, and the transmittance of each path may be calculated using Equation 1 or 3. The average transmittance is then: ! . o(U.) + b(Ub) + - - -, + n{U.) ` (6) where a,b, * , n are respective fractions of a typical-basic area composed of several different paths whose transmittances are t/*, V*- ", !.............., If beat can flow laterally m any continuous layer so that tnmsverse isothermal planes result, .the total average, resist- arice Rn*r) will be the sum of the resistances of the layers be-, tween such planes, each layer being calculated by the appro priate Equation 1 or Equation 6. This is a series combination of layers, of which one (or more) provides parallel paths.. ,. The transmittance, assuming parallel heat flow, only, is usually considerably lower than that calculated with the assumption of combined series-parallel heat flow. The actual transmittance will be some value between the two calculated values. In the absence of test values for the combi nation, an intermediate value should be used.- An examination of the construction will usually reveal whether a value cldser to the higher or lower calculated value should be used. Gen erally, if the construction contains any highlyconducting layer in which lateral, conduction is very high compared to the transmittance through the wall, a value closer to the series- parallel calculation should be used. If, however, there is no layer of high lateral conductance, a value closer to the parallel heat flow calculation should be used. This is illustrated in Example 1. Example 1: Consider a construction consisting of:. 1. Inride surface having film coefficient/< -- 2. 2: A continuous layer of material of resistance fi( -- 1. 3. A parallel combination containing two beat flow,paths of pro- , portionste areas, a -- 0.1, and 6 = 0.9, with resistances .. fi<s 1 and fit* 8. 4. ' A continuous layer of material of resistance fit ' 6.5. ' . 5. Outride surfaceharing 61TM coefficient -- 4. - ' Solution: If parallel heat flow paths are assumed from air to air the total resistance through area a will be: * fi.r<- l/fi + fit +.fi< + fi* + 1/A:..,,:. - 0.5 +1 +.1 +.0.5 +0.25 - 3.25 and . -- U..~ 1/fi.r.- 1/3.25 The resistance and transmittance through area'5 will be: " fir = l/fi + fi + fi.+ fi* + l// . ' rU.,, . -- 6.5 + 1 + 8 + 0.5 + 0.25 -- 10.25 ( ,, Anij ...... UM - 1/6.96 = 0.144 If Ai and fi* are homogeneous materials, a value of about 0.125 might be selected, whereas, if they contain a highly.con ducting layer, then a value of 0.135 might be selected. .< When the construction contains one or more paths of small area having a very high conductance* compared to the con ductance of tire remaining area, the following method is sug gested: Heat Flow Through Panels Containing Metal The transmittance of a panel which includes metal or other highly conductive material extending wholly or partly through insulation should, if possible, be determined by test in the guarded hot box. When a calculation is required, a good ap-* . proximation can.be made by.a Zone Method. This involves two separate computations--one for a chosen limited portion, Zone A, containing the highly conductive element, and the., other for the remaining portion of simpler construction, called Zone B. The two computations are then combined, and the. average transmittance per unit of overall area is calculated. The basic laws of heat transfer are applied, by adding area conductances C* A of dementi in parallel, and adding area re sistances \/C- A of dements tn series. The surface shape of Zone A is determined by the'metal element. For a metal beam (Fig. 2) tire Zone A surface is a strip of width W, centered on the beam. For a rod perpendicu lar to panel surfaces it is a circle of diameter W. The value of W is calculated from Equation 7,.which is empirical., - W m + 2d (7) where m -- width or diameter of the metal. heat path - terminal; * inches!., : - . - ,/ d -- distance from panel surface to metal, inches. The value of d should not be taken- less than 0.5 in. (for still air). In general, the value of W should be calculated by Equa tion 7 for each end of the metal heat path,.and the larger value, within the limits of the baric area, should be used as illustrated in Example S'. Example t: Calculate the transmittance of the - roof deck shown in Figs. 2 and 3. Tee-bars on 24-in. centers support glass fiber form boards, gypsum concrete, and built-up roofing. The conductivities of components are: steel 312: gypsum concrete 1.66; gtiwi fiber 0.25. The conductance of built-up roofing is 3.0. Solution: The bade area is 2 sq ft (24 in. X 12 in.), with a tee- bar (12 in. long)! across the middle. This area'is divided into two sones, A ana B. Zone A is determined from Equation 7 as follows: . .- > -- BASIC AftCAJ 2 SQ FT -- Ui - 'l/fi*, - 1/10.25 Then the average calculated transmittances will be: .. VM - a(V.) + 6(W - L 0.118,;\ If, however, isothermal planes are assumed to occur at both sur faces of fi( and' at both surfaces of fit, the total calculated re-, nstance will be: ~ * ..... Rtl/fi + fit + (fi*)(fiM) ofitt + 5fir + fii` + Iff. ELEVATION .for eatargod action of Zona A, w* Fig. 3 . ,, ( Fig. 2,.... Gypsum.Roof Deck,on Bulb Tees . A