. Railway mechanical engineer . ed. The value will be based on a conductivity of322 per sq. ft. per hour per degree F. per inch thickness. From formula 5 the following formula, No. 6, is derivedwhen the heat transmission or thermal conductivity isdesired for a difference of one degree in temperature. H = T d 1 d2 <*) — + — + — + — + c ct c2 The thermal resistance R will therefore equal the recipro-cal of conduction, or 1R = - (7) HIn the car section under consideration, shown in Fig. 2,we will assume the outer plate to be % in. thick and theinner plate 1/16 in. thick, wi
. Railway mechanical engineer . ed. The value will be based on a conductivity of322 per sq. ft. per hour per degree F. per inch thickness. From formula 5 the following formula, No. 6, is derivedwhen the heat transmission or thermal conductivity isdesired for a difference of one degree in temperature. H = T d 1 d2 <*) — + — + — + — + c ct c2 The thermal resistance R will therefore equal the recipro-cal of conduction, or 1R = - (7) HIn the car section under consideration, shown in Fig. 2,we will assume the outer plate to be % in. thick and theinner plate 1/16 in. thick, with three-ply Salamander insula-tion representing a normal thickness of % in. The value ofinsulation efficiencies for various insulating materials such asare used in railway equipment cars is shown in Table figures are the results of tests made by the UnionPacific at Omaha, Neb., in 1914 for the Standard Car Com-mittee. In referring to this table it will be observed that the Waterpmot Insulating Fabric ; Lining-. *-*- - -/£ Shea thing Waterproofff ~lnsulatingFabric - -2 Insulation Fig. 3—Representative Section of Side Wall of Refrigerator Car value assigned to Yi in. Salamander is per per degree F. difference in temperature per 24 hours per hour. In view of the fact that the conduc-tivity varies inversely as its thickness the transmissionfor the unit thickness of 1 in. will be: .274: X :: % = .2055 the foregoing values to the car section shown inFig. 2 we have the following: 1/16 in. and % in. steel 322 ^4 in. Air space (Resistance)... 1 Ya in. Salamander 2055 Surface resistance 5 Substituting in formula 6 we then have: 1 H=- .5 + .125 .75 322 .2055 1 + .0625 322= .194 per hr. .5 + .000388 + + 1 + .000194 For this typical section this would mean a total of transmitted for 24 hours, which is well within thefigure of 8 , as specified in t
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