CHAPTER V HEAT TRANSFER (Part 3)

: 1. Type of construction 2. Character of refrigerated products 3. Temperatures to be maintained 4. Thickness of walls 5. Location of plant. The following table gives the most economic thickness of corkboard and other insulation of same unit transmission : —20° to — ur 8" —10° to — 0° 6" 0° to 15° 5" 15° to 35° 4" 35° to 45° 3" 45° and above , .2" ■ HEAT TRANSFER 119 Heat Transfer in Apparatus. — The heat transfer taking place in a refrigerating apparatus is similar to that occuring through insulation, in that the flow occurs from a region of high temperature to a region of low temperature. Whereas, a very slow rate of infiltration through insulation was desired just the reverse is true in the apparatus; the fastest possible transfer is wanted. Generally this transfer of heat is accomplished be- tween two fluids separated by a solid wall of good conductivity. ,2 FIG. 7. ,■3 ,4 ^ ^ n & .9 l.a -MEAN TEMPERATURE DIFFERENCE CURVE. Since the heat transfer may occur by means of conduction, radiation or convection, the fundamental law of heat transfer holds, the same as for insulation ; although the unit transmis- sion as determined by experiment combines all three methods, as well as the kind of material and thickness of separating wall. The formula then becomes — (2). 120 HOUSEHOLD REFRIGERATION B.t.u./hr. = "C" X sq. ft. of surface X temperature difference. (C is given in Table XLVIII.) The mean temperature difference for apparatus is some- what different than for insulation due to the fact that the tem- peratures on both sides of the insulation are comparatively constant, whereas in the apparatus they are changing con- stantly. Therefore in the first case an arithmetic degree mean difference can be used but a logarithmic mean temperature difference must be found in the latter case. Due to the character of the formula as given by Hausbrand and its attendent higher mathematics this logarithmic degree mean difference method has been put in a simple curve form, making it available to everyone. TABLE XLVII. THICKNESS OF INSULATION FOR COLD PIPES Thickness p Use With For Temperatures of Cork 154 in. Ice water, liquid ammonia, brine Above 25° F. and other cold lines. 2 in. to 3 in. Brine, ammonia and other cold 0° to 25" F. lines. 3 in. to 4 in. Brine, ammonia and other cold Below 0° F. lines. Ti = Inlet temperature of substance to be cooled T2 = Outlet temperature of substance to be cooled ti= Inlet temperature of cooling substance ta= Outlet temperature of cooling substance Ti — U and T2 — ti = Differences S = Smallest difference L = Largest difference S Factor — = Ordinate L Coefficient obtained from Curve = Abcissa Coefficient X largest difference = Mean temperature difference. Example: To cool milk from 120° to 80° with 72° water heated to 80° during the process. 2o^ SO' _c: do' S 8 L 40 FIG. 8. 72" HEAT TRANSFER 121 Running across on the .2 factor line to the curve and then projecting down at right angles the coefficient .5 is obtained. Then the mean temperature difference is .5 X 40 = 20. If an arithmetical degree mean difference had been used the result would have been the difference of average tempera- tures of (120 + 80) (80 + 72) = 24 ( 2 ) ( 2 ) which would have been a 20 per cent error. S It has been found that for all practical purposes if — is L greater than .5 an arithmetic degree 7nean difference can be used. Coefficients of Heat Transfer in Apparatus. — In table XLVIII is given overall unit heat transfer coefficients, as de- termined by experiments. They hold good for the general wall thicknesses found in refrigerating apparatus and when the surface is comparatively free from frost scale, oil and other foreign matter. The best heat transfer is obtained from liquid to liquid followed by liquid to gas and the worst transfer is from gas to gas. Copper has a considerably higher rate of transfer than steel while lead is very much worse than either. Some im- portant factors on which the rate of heat transfer depends are : 1. Velocity of fluids 2. Density and kinds of fluids 3. Temperatures at which they are handled 4. Thickness and material of separating wall 5. Smoothness and cleanliness of wall as regards foreign sub- stances, material, as well as gases. As an example to show the amount of steel pipe to be in- stalled in a small storage box having a load factor of 3000 B.t.u. hr. and held at a temperature of 40" with 15° average expansion through the coil, using the fundamental formula (2):- 3000 = 2.5 X sq. ft. X (40—15) sq. ft. = 48. 122 HOUSEHOLD REFRIGERATION w H M W a w Q a. 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From the foregoing it can readily be seen that many refrig- erating problems are really heat transfer problems, and can be solved by either formula — 1 or 2.