Metal Finishing Guide Book

2011-2012 Surface Finishing Guidebook

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Cover Style Loose partials Insulated Floating balls Still Air Metal tank values, shown in Table II Insulated tank values, shown in Table II Table III: Cover Loss Values (BTU/hr/ft2 ) area uses the reduced loss values shown in Table III. The use of partial covers reduces exhaust volume requirements and associated energy demands as well. Air agitation can be said to primarily affect losses from the tank surface. Breaking bubbles increase the surface area and expose a thin film of solution to accelerate evaporative losses. Air agitation spargers sized at one cfm per foot of length affect a 6 in. ( ft) wide path along their length. Thus, a three foot by four foot tank surface with two lanes of air agitation running on the four foot dimen- sion has: 3 4 = 12ft2 = 4 ft2 surface plus 2 ½ 4 agitation increase, a total 16ft2 effective Multiply the effective area by the values shown in Table I. Be sure to deduct any cover area (if used) and use the reduced loss values shown in Table III. The tank wall area equals the tank length in feet, times the depth of solution in feet, times two plus the tank width in feet, times the depth of solution in feet, times two plus the tank length in feet, times the tank width in feet. L D 2 + W D 2 + L W = wall area. (You can use inches instead of feet but you must divide the result by 144 to convert into square feet.) Multiply the tank wall area times the values shown in Table II. Calculate the heat loss through parts being immersed. Racks per hour, times the weight of the loaded racks, times the specific heat of the parts (use 0.1 for most metals, 0.2 for aluminum), times the temperature rise (use the same value used in calculating the tank temperature rise). racks/hr weight/rack s.h. T rise A plastic or metal plating barrel must be included with the parts weight. A met- al barrel has a specific heat value close to the average parts (0.1), and can be includ- ed in the parts weight, but a plastic barrel has a specific heat of 0.46 and will require an independent calculation. Weight of barrel, times barrel loads per hour, times the specific heat of the barrel, times the temperature rise. barrels/hr weight/barrel 0.46 T rise Add to this the parts per barrel barrels/hr weight of parts/barrel s.h. T rise The heat loading and the actual heat-up time for immersed parts are distinct values. The heated solution can lose temperature to the immersed parts in a mat- ter of seconds. This heat loss is replaced by the heater. To determine the tem- perature drop of the process solution, divide the heat loss through parts (barrels) being immersed by the weight, times the specific heat of the solution. Heat loss (parts)/[Weight (solution) s.h. (solution)] = Temperature drop Calculate the heat loss through solution additions such as drag-in and make- 667 Ventilated (150 fpm) Twice that for still air Same as still air 0.25 times the value obtained from Table I Twice that for still air

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