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topic 51573

Hard anodizing chiller sizing

Disambiguation / See also --

Topic 36775, "Best way to cool anodizing tanks?"

Topic 58129, "Cooling of home/D-I-Y anodising baths"


Current question and answers:

April 7, 2021

Q. Hi All,

We have to maintain -5 °C temperature for Hard Anodizing bath, For that how much temperature we have to maintain in chiller. And how to calculate heat absorption in Titanium coil, that we are using in anodizing bath.

Thanks & Regards,

Shankar Rathod
- Bangalore, India
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April 2021

A. Hi Shankar. The heat exchange formula is
Q = U * A * ΔT
where Q is heat input, U is heat transfer coefficient, A is surface area of the coil, and ΔT is the difference in temperature between the solution in the tank (the outside surface of the coil) and the cooling fluid (the inside surface of the cooling coil). A complication there is that the cooling fluid has two temperatures, the temperature at which it leaves the chiller / enters the coil and the temperature at which it leaves the coil / enters the chiller. You can use the average/median/half-way temperature of the cooling water in your calculation.

But be careful with your units! If you look something up, make sure you stick with either Watts, BTUs or Calories, and square inches, square feet, or square meters, and degrees Fahrenheit or Centigrade throughout your calculation; do a units check. I always worked in BTUs, square feet, and degrees F.

So after you have determined how many BTUs the rectifier is adding to the tank (Q), and you get the U factor for the titanium coil from the manufacturer (titanmf.com implies that you can use 150 for BTU-square feet-degrees F calculations, although that strikes me as very high), you transpose to get
A * ΔT = Q / U
So you divide Q by U and now you have the product of A * ΔT

So if the area of the coil is already fixed, you can calculate the ΔT you must create. Or if the chiller manufacturer gives you the suggested operating temperatures, you can calculate the required coil area. Otherwise you can do the math to pick a reasonable pair of values. Remember that the flow of cooling water must be sufficient to remove Q as well. One final caution: chillers must be "de-rated" for such low temperatures; if you calculate that you need a chiller with "X BTU/hr" capacity, the chiller manufacturer may tell you that the chiller must actually be rated "2X BTU/hr" or so (see Chris Jurey's explanation below)

Some suppliers have on-line worksheets to get you started, for example,
https://www.titanmf.com/immersion-coil-surface-area-calculator-for-anodizing-applications/

Luck & Regards,

pic of Ted Mooney
Ted Mooney, P.E. RET
Aloha -- an idea worth spreading
finishing.com - Pine Beach, New Jersey
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Previous closely related Q&A's starting in:

2002

Q. Sir,

I am working in a Pvt Ltd company and wanted to know about the process of chilling plant, selection of pump, and spares required for Preparing New Chilling Plant and what books to be referred. FUNCTIONS OF CHILLER USED AND PARTS, PHE SELECTION AND PUMP, ETC., REQUIRED.

ADISESHAN
- Mumbai, India
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2002

A. Hi ADISESHAN. You start by sizing the chiller to remove the heat load impressed on the plating or anodizing tank by the rectifier. Multiply the Amps by the Volts times the percentage of time the tank is occupied and divide by 1000 to find the kiloWatt (KW) load. Then just apply the appropriate conversion factor to convert the KW to BTU/hour, or Calories/hour or Tons of Refrigeration as desired. Specify titanium and plastic construction. You can order a "packaged chiller" of a given Tonnage, which would include the required holding tank, cooling tower, heat exchanger, compressor, and pumps -- all properly sized together. Good luck.

Ted Mooney, finishing.com Teds signature
Ted Mooney, P.E.
Striving to live Aloha
finishing.com - Pine Beach, New Jersey
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March 13, 2009

Q. I am sizing new hard anodizing chillers and am having a hard time reconciling what the math tells me I need and what experience tells me we need. First of all the math... I size the chillers according to engineering custom: rectifier running volts and amps and time converted to kWh converted to btuh converted to tons of refrigeration. I reviewed the literature again as well as past Ted Mooney posts and verified that I was indeed doing it correctly. Now for the experience... I logged power consumption on existing appropriately sized cycling chillers and found that they are using only about half of the power that would be expected by the calculation. I can only assume that much of the heat is removed by the air agitation of the bath rather than being transferred into the solution to be removed by the chillers. Does anyone have experience similar to this that has allowed them to effectively down-size their chiller equipment?

Jon Barrows
Jon Barrows, MSF, EHSSC
Independence, Missouri

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March 15, 2009

A. Hi, Jon. You probably already realize this, but for the benefit of readers who may not: cooling towers function by evaporating a portion of the recirculated water to cool the rest, and the process is governed by a principle called the "wet bulb temperature". If you take a thermometer, cover the bulb with a wet cloth, and put it in a cloth sling to swing it in circles to maximize the evaporation, the thermometer will show a lower temperature than a dry thermometer. The difference will be something like 10 °F, depending on temperature and humidity. Average wet bulb temperature data for various geographical locations and seasons is available. So, to my knowledge, no agitation system could, even at best, cool a tank to below the wet bulb temperature inside the shop -- which is probably significantly higher than the operating temperature of the hard anodizing bath. And I think, except in the case of a dry air-conditioned shop, this contribution towards cooling will be unreliable and fairly minor.

But remember that rectifiers rarely operate at their full rated amperage and voltage, that the tank is empty some portion of the time, and that during hard anodizing the voltage is significantly lower than the terminal voltage for a percentage of the processing time. Also, some of the power (maybe 5-10 percent?) is lost to bus bar resistance and resistance at the joints, and never makes it to the tank. Try to find the ramp curve, and the percentage of time the tank is in use, deduct a little of the power for bus bar losses, and maybe things will reconcile.

In my career as a plating equipment engineer I sized a lot of chillers, and re-sized a number of cooling systems that proved too small ... and I just never saw evidence that this simple calculation of Actual amps x Actual volts x % duty cycle is overly conservative for cold tanks (it's conservative for hot tanks like chrome plating tanks which lose a lot of heat to the atmosphere). Good luck.

Regards,

Ted Mooney, finishing.com Teds signature
Ted Mooney, P.E.
Striving to live Aloha
finishing.com - Pine Beach, New Jersey
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March 17, 2009

A. Assuming typical hardcoat bath operating conditions of 32 °F, Ted is correct. Air agitation would be a heat load due to the higher ambient temperature and the heat of compression.

If you use a water cooled condenser with a cooling tower, the compressor discharge condensing temperature will be lower than outside air due to the evaporative cooling described by Ted. If you use colder well or city water as a heat sink, your compressor will be even more efficient.

The low suction pressures required to cool a 32 °F bath will significantly derate a refrigeration compressor. One half of "nameplate" capacity would not be unusual. This will result in lower power consumption under normal operating conditions (i.e. 32 °F bath. If you let the bath return to room temperature, your power consumption will go up. Your motor and other electric gear needs to be sized for these higher load, startup conditions.

Chris Jurey, Past-President IHAA
Luke Engineering & Mfg. Co. Inc.
supporting advertiser
Wadsworth, Ohio

luke banner
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March 18, 2009

thumbs up signThanks guys, I think I was hoping that I had overlooked something that would allow me to downsize the chillers and save some money. I had estimated an average load, but to really know I need to do as Ted suggested and measure volts and amps that I am putting in over time and see if it balances the power output of the chiller. I bet Ted is right and it will balance out.

Jon Barrows
Jon Barrows, MSF, EHSSC
Independence, Missouri

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thumbs up sign Thank you Chris! I was confident that I was right but I couldn't really reconcile Jon's numbers. Although I already knew that chillers must be derated to about 50% nameplate capacity for hardcoat cooling, I never made the simple connection that a derated chiller isn't going to draw its full nameplate electricity rating :-)

Regards,

Ted Mooney, finishing.com Teds signature
Ted Mooney, P.E.
Striving to live Aloha
finishing.com - Pine Beach, New Jersey
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Process chillers

April 20, 2015 -- this entry appended to this thread by editor in lieu of spawning a duplicative thread

Q. Hello everyone, can anybody please help me with the calculation for a water cooled chiller, used for hard anodising. required temperature is -2 °C and inlet temp is 42 °C. Flow rate is 6000 lph. Please explain the calculations for the same.

shubham ghosh
- noida, India
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April 2015

Hi Shubham.

You are trying to hold an anodizing temperature of -2 °C (28 °F), but the rectifier is putting heat into the tank whenever you are anodizing. The first step, as described above, is to multiply the actual Amps x actual Volts x an estimate of the duty cycle to determine the Watts which the rectifier is adding to the tank. Then you convert that to Calories per hour or BTUs/hour.

For example, if the actual current is 1250 Amps and the final voltage is 40 Volts, you are adding 50,000 Watts or 50 KW per hour at that point. But the tank is empty for a while after each load, and you don't reach the final voltage until the end of the cycle, so you might estimate, after looking at the ramping profile and cycle time, etc., that your duty cycle is 60%. So the actual heat input from the rectifier is 30 KW.

You convert this to Calories per hour by multiplying it by 860 or to BTU/hour by multiplying by 3412. So you need to remove 25,800 Calories/hour or 102,360 BTU/hour. If the flow rate of cooling water to the water cooled chiller is 6000 lph, which is essentially 6000 Kg/hour or 13220 lbs/hour, the temperature of the cooling water will rise by about 25800/6000 = 4.3 °C. or 102360/13220 = 7.75 °F.

A "ton of refrigeration" is defined as 12,000 BTU/hour, so your chiller consumes 8.53 tons. As Chris Jurey warns, the nameplate data on chillers applies to room temperature operations, not sub freezing temperatures, so the nameplate must be derated by about half for hard anodizing. So a nominal 15-ton chiller, certainly no less, would be about right. Good luck.

Regards,

pic of Ted Mooney
Ted Mooney, P.E. RET
Aloha -- an idea worth spreading
finishing.com - Pine Beach, New Jersey
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December 19, 2016

Q. Dear,
As per your calculation if flow rate of coolant is lower say 8000 liter per hour then we can achieve good chilling temperature, am right?
If yes then if we adjust the coolant flow rate then is there any adverse affect on chiller machine?

Am using glycol as cooling agent.

Aijazullah Tajir
- Abu dhabi UAE
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December 2016

A. Hi Aijazullah. Please spend a few paragraphs explaining your whole situation. Sorry, but there are many arrangements and I don't understand yours -- but even with glycol it is hard to get the chilled water cool enough to remove the heat from a hard anodizing tank with low flow rates.

Regards,

pic of Ted Mooney
Ted Mooney, P.E. RET
Aloha -- an idea worth spreading
finishing.com - Pine Beach, New Jersey
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