مقالات تکنولوژی و تجهیزات ابکاری

finishing equipment & plant engineering

IMMERSION HEATER DESIGN

BY TOM RICHARDS

PROCESS TECHNOLOGY, MENTOR, OHIO; www.process-technology.com

The immersion heater represents a sound, economical method of heating processsolutions in the finishing industry.Classical heater installations consisted of hanging a steam coil on one tankwall, sized to heat up water to a “rule-of-thumb” temperature in two hours.While this method has proved adequate in providing heat and covering a multitudeof oversights, it has also proved unsatisfactory with regard to energy

costs and control. As the cost of energy rose, the finisher increased heat-uptimes in an attempt to conserve energy. Soon, heat losses prevented achievingdesired temperature levels so tank insulation, covers, and other methods of lossconservation were added. Again, adequate solutions to most of the challengeswere found, but the hanging steam coil remained unchanged. Today, we havethe knowledge that allows us to adequately plan, design, install, and operateeconomical, efficient heating systems.Molecular activity, chemical solubility, and surface activity are enhancedthrough temperature elevation. The reduced solution surface tension, low vaporpressure of some organic addition agents, and heat-sensitive decomposition orcrystallization of other additives are major considerations that modify the benefits

gained as solution temperature rises. To achieve a proper balance of all thesefactors, while providing economical installation and operation, it is necessary toanalyze the individual heating requirements of each process. Your best sourceof process information is your process chemical supplier, which can tell you:

1. Recommended materials of construction.

2. Maximum (minimum) solution temperature.

3. Maximum heater surface temperature.

4. Specific heat of the process solution.

5. Specific gravity of the process solution.

6. Recommended heel (sludge) allowance.To size the heater, first determine the tank size: space required for the part,parts rack or barrel, space required for busing (anodes), in-tank pumps andfiltration, sumps, overflow dams, level controls, air or solution agitation pipes,and any other accessories. From this data, a tank size and configuration can bedetermined.Calculate the weight of solution to be heated. For rectangular tanks:

Weight = L × W × D × S.G. × 62.4 lb/ft3

where L, W, and D are length, width, and depth in feet (substitute 0.036 lb/in.3 for dimensions in inches). S.G. is the specific gravity of the solution (wateris 1.0). For cylindrical tanks:

Weight = R2D × S.G. × 62.4 lb/ft3

where R is the radius of the tank.Calculate the temperature rise required by subtracting the average (or lowest) ambient temperature from the desired operating temperature (if the shoptemperature is kept very cool during winter months, it might be wise to use thistemperature as the average ambient temperature).

Temperature rise = T operating minus T ambient [To - Ta = T rise]

Determine an adequate heat-up time to suit your production requirements.The traditional 2-hour heatup may prove costly and unnecessary since using thisvalue usually provides a heater more than twice the size necessary for heat maintenance.A 4- to 6-hour heatup more closely approximates the heat maintenancevalue but may impose production constraints deemed impractical. Long heat-uptimes can be overcome through the use of 24-hr timers; however, unattendedheat-starts carry the responsibility of tank liquid level monitoring and approved

• Initial BTUH(Q) = Weight × Trise × where s.h. is specific heat. This should be the actual value from the processsupplier (water is 1.0).Calculate the approximate heat loss from the tank surface and tank walls.(Use the data shown in Tables I and II.)The losses from the tank surface can represent the most significant loss affectingheater sizing. The addition of even a partial or loose-fitting cover will reduce these losses.The tank surface area is simply the width in feet times the length in feet. You canuse inches instead of feet, but then must divide the results by 144 to obtain square feet.If you install partial covers, such as removable covers extending from the tankedge to the anode busing, use the remaining “open” dimensions. The covered area-uses the reduced loss values shown in Table III. The use of partial covers reducesexhaust 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 solutionto accelerate evaporative losses. Air agitation spargers sized at one cfm per footof length affect a 6 in. ( ft) wide path along their length. Thus, a three foot byfour foot tank surface with two lanes of air agitation running on the four footdimension has:

3 × 4 = 12ft2 surface plus 2 × ½ × 4= 4

ft2 agitation increase, a total 16ft2 effectiveMultiply 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 solutionin 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 mustdivide 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, timesthe weight of the loaded racks, times the specific heat of the parts (use 0.1 formost metals, 0.2 for aluminum), times the temperature rise (use the same valueused in calculating the tank temperature rise).racks/hr × weight/rack × s.h. « T riseA plastic or metal plating barrel must be included with the parts weight. Ametal barrel has a specific heat value close to the average parts (0.1), and can beincluded in the parts weight, but a plastic barrel has a specific heat of 0.46 andwill require an independent calculation. Weight of barrel, times barrel loads perhour, times the specific heat of the barrel, times the temperature rise.barrels/hr × weight/barrel × 0.46 × T riseAdd to this the parts per barrelbarrels/hr × weight of parts/barrel × s.h. × T riseThe heat loading and the actual heat-up time for immersed parts are distinctvalues. The heated solution can lose temperature to the immersed parts in amatter of seconds. This heat loss is replaced by the heater. To determine thetemperature 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 dropCalculate the heat loss through solution additions such as drag-in and make-up water when working on small process tanks with high operating temperatures.In some operations, it is customary to replenish evaporative losses byrinsing parts over the tank. This practice increases the heat loading. Gallons ofwater each hour (drag-in or add), times 8.33 (lb/gal), times the temperature rise(water temperature to tank operating temperature).gallons per hour × 8.33 × T riseNow determine total heating requirement by comparing initial heat-uprequirements with the sum of the various losses. Assuming no additions oroperating losses during the initial heatup, we can equate our heater size based onthe initial heat-up requirement, plus the tank surface losses, plus the tank walllosses. This value must be compared with the operating requirements—tank surfacelosses, plus the tank wall losses, plus the rack (barrel) losses, plus the drag-in

(make-up) losses. The larger value becomes the design basis for heater sizing.Heater sizing can proceed based on the heating method employed. Electricimmersion heaters are sized based on 3.412 BTUH per watt-hour (3,412 BTUHper kilowatt-hour). Divide the design heating requirement by 3,412 to findkilowatts of electric heat required.design heating requirements(BTUH)/3,412The immersion heater sheath temperature will be higher than the solutiontemperature. Consult your immersion heater supplier for its recommendations

where solutions have high temperature limits. Electric heaters have the potentialof achieving sheath temperatures, particularly in air, and are capable of ignitingflammable materials; therefore, it is essential that liquid level switches and highsheath temperature cutoffs be employed. Look for (or ask about) UnderwritersLaboratory or other independent agency listing labels on electric heaters forassurance that the product meets a recognized standard. Verify and install thesheath ground to minimize personnel shock hazard and, as with all heaters, usea quality temperature controller for economical operation.Steam immersion heaters are sized based on steam pressure, overall transfercoefficients, area, and log mean temperature difference.The overall transfer coefficient is a value determined by several basic values:the ability of the heater material to conduct heat, the ability of the two fluidfilms that form on the inside and outside of the heater to conduct heat, and the

resistance to the flow of heat caused by fouling or buildup. You can significantlyalter the performance of immersion heaters by the choice of materials and thesupply or the lack of supply of tank agitation. By selecting proper materials thefouling caused by corrosion is either reduced or eliminated. Clean quality steamwill reduce internal fouling while properly placed agitation can enhance overallthermal performance. The precise calculation of the overall transfer coefficient isdetailed and will not be covered here, but is available from your heater supplier.The following rule-of-thumb values can be used for estimating steam heatersize. For metal coils, the range of values for the overall heat transfer coefficientis 100-200 BTU/hr/ft2/OF. For plastic coils, the overall heat transfer coefficientranges from 20-50. Use 150 for metal and 40 for Teflon.Now calculate the log mean temperature difference (LMTD) because thedriving force for the heat exchange is a varying quantity that is expressed asthis value.

LMTD = (ΔT1 - ΔT2)/[ln(ΔT1/ΔT2)]

where ln = Naperian (natural) logarithms.

Steam pressure produces specific temperatures that will be used in the calculationof the LMTD. Typical values are given in Table IV.As an example, assume 10 psig steam is to be used to heat a solution from65OF (ambient shop temperature) to 140OF (solution operating temperature).Steam temperature (from Table IV): 240OF

ΔT1 = 240 - 65 = 175OF

ΔT2=240 - 140 = 100OF

LMTD = (175 - 200)/[ln(175/100)] = 75/0.55 = 134OF

The heater area required to steam heat a process solution equals the designheating requirement, divided by the overall heat transfer coefficient, times thelog mean temperature.Design heating requirement (BTUH)/Overall heating requirement LMTDAs with any immersion heater, the heater surface temperature will be higherthan the solution temperature. Obviously, it cannot exceed the steam temperature.

If the solution has a high temperature limit below available steamtemperatures, you may require a custom electric immersion heater or a hotwater (or thermal fluid) heater with a lower heating temperature.Although the heater temperature is limited to the steam temperature, damageto process tanks and accessories can result from overtemperature or lowliquid levels. It is wise to equip your process tank with overtemperature andlow liquid level cutoffs.Once a coil size is selected, piping size should be investigated. The quantityof steam used for a specific coil size varies with the steam pressure (see TableV) and the heat released is the heat of evaporation (latent heat) only. The valuesin the table are in BTUs per pound of steam. So the quantity of steam requiredequals the design heating requirement, divided by the heat of evaporation ofthe steam.Design heating requirement(BTUH)/Heat of evaporation (from Table IV)The result, in pounds of steam per hour, can be equated to pipe size asshown in Table V. The condensate generated (condensed steam) must be“trapped,” that is, equipped with a steam trap. Steam traps are sized based onpounds per hour times a safety factor. Since the amount of condensate varieswith the temperature of the solution, it is wise to use a safety factor of four orbetter. Trap capacity equals the steam required times four. The condensate piping is smaller than the steam pipe since the condensateis liquid. Some of the condensate will convert back to steam because of condensatetemperature and pressure. The use of piping smaller than in. nominal isnot recommended since scale and buildup inside the pipe is a factor in all steamlines. We recommend using in. nominal pipe for condensate lines. This size willhandle up to 1,920 lb/hr with a modest pressure drop.Steam coil valve sizing is usually smaller than the pipe size since a pressuredrop across the valve is required for proper operation. Some typical sizes fordiaphragm solenoid valves are shown in Table VI.Since the performance of the valve and trap can be affected by foreign matterin the steam, it is wise to place a 100-mesh strainer of the same pipe size as thesteam pipe ahead of the valve.

Metal steam heaters, when suspended in electrified tanks, may conductcurrent through the steam lines to ground so it is a good practice to installnonconductive couplings between the heater and the pipe lines. This can beaccomplished using a proprietary insulating coupling, dielectric union, or sectionof steam hose.Finally, because some steam heaters may be buoyant (tend to float) when inservice, it is necessary to secure these heaters through the use of ballasts or proprietaryhold-down fixtures.Hot water (thermal fluid) heating is similar to steam heating in the methodsused for sizing. The basic differences involve the usually lower heating solutiontemperatures and the lower performance, overall heat transfer coefficient of theheater. As in the case of steam heating, the overall transfer coefficient is subjectto varying performance and its precise computation is beyond the scope of thispresentation. The following rule-of-thumb values can be used for estimating hotwater heater sizes. For metal, the overall heat transfer coefficient is 70-100 BTU/hr/ft2/OF. For plastic, the range is 20-50. Use 95 for metal and 40 for Teflon.The calculation of the LMTD uses the same equation but now the heating fluidtemperature must change since it is yielding the fluid heat and not the evaporativeheat available in steam. It is wise to limit the heat drop of the heating fluid to 10OFsince greater drops may be impossible to achieve in a field-installed condition. Also, it is wise to design the exiting heating fluid temperature to be 15OF higherthan the final solution temperature to ensure field reproduction of design performance.Consult your heater supplier for assistance if you experience any difficultyin sizing a heater.As an example, heat a solution from 65OF (ambient shop temperature) to140OF (operating temperature) using 195OF hot water. Limit the hot watertemperature drop to 10OF or 185OF outlet. This temperature is more than 15OFabove the final bath temperature.

ΔT1=195 - 65=130OF

ΔT2=185 - 140=45OF

LMTD=(130 - 45)/ln(130/45)]=95/1.0607=80.56OF

The heater area required to heat a process solution equals the design heatingrequirement divided by the overall heat transfer coefficient times the LMTD.Design heating requirement/[Overall transfer coefficient LMTD]With hot water heaters, it is a wise precaution to install high liquid levelcutoffs that will shut off hot fluid flow in the event of a heater leak. If a hightemperature heating fluid is used, solution temperature sensitivity must beevaluated and high temperature, low liquid level cutoffs may be in order.Once the coil area has been selected, the hot water (thermal fluid) flow mustbe calculated. The flow is equal to the design heating requirement, divided bythe temperature drop of the heating fluid, times the specific heat of the heatingfluid, times the specific gravity of the heating fluid.Design heating requirement/[Temperature drop × s.h.× s.g.(all of the heating fluid)]This results in the pounds per hour of heating fluid. To convert this intogallons per minute, divide the pounds per hour by the weight of fluid per gallontimes 60 (water weighs 8.33 lb/gal). This value is used to evaluate pipe size(both inlet and outlet). Table VII gives a reasonable flow for water throughvarious pipe sizes.The control valve may be smaller than the pipe size. Some typical sizes fordiaphragm valves with a water pressure drop of 5 psig are given in Table VIII.As with steam heaters, it is a good practice to install a strainer to minimizeforeign particles that may affect valve performance. A 60-mesh strainer is usuallyfine enough for hot fluid systems.Metal heaters, when suspended in electrified tanks, may conduct currentthrough supply lines to ground so it is a good practice to install nonconductivecouplings between the heater and the pipe lines. A proprietary insulatingcoupling or dielectric union can be used.Plastic heaters and some empty metal heaters may be buoyant, so be sure toprovide adequate anchoring if floating is suspected.Thermal stratification is a fact of life in heated process tanks. To minimizethis effect good agitation (mixing) is required. Classic air agitation is sizedat one cfm per foot of length. When placed beneath a cathode (or anode) itprovides sufficient agitation to that surface to enhance deposition rates. Itdoes not, in this form, eliminate thermal stratification. Top-down mixing canbe provided through recirculation pumping. Pumps sized for 10 turnovers ormore per hour provide good mixing and uniform temperatures. Skimmingstyle pump inlets with sparger bottom discharges are best since higher temperaturesolutions are forced to the cooler areas.In tanks three feet deep and more, a vertical sump pump can be mounted on thetank flange with a length of discharge pipe anchored to the tank bottom. Thesecan often be coupled to in-tank filters for removal of particulates while providingmixing. Air agitation, when properly placed, can “average” temperature in their zoneof influence (usually 6-12 in.) and can be used to enhance response time for temperaturecontroller sensors. As the air agitation is increased, heat losses also increase,making air agitation a less desirable means of dealing with thermal stratification.Heat-sensitive solutions can be addressed by either electric or hot water(thermal fluid) heaters. Electric is the easiest to control since the heater surfacetemperature can be varied by varying the input voltage. A heater surfacetemperature controller can limit surface temperatures while still providingsufficient heat for the solution. Similarly, hot water systems can be sized formaximum hot water temperatures (and thus heater temperatures) but controland response are usually inferior to electric system