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Author Topic: How to calculate water cooling systems?  (Read 13105 times)
N2WQ
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Posts: 99




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« on: July 26, 2017, 10:19:23 AM »

Hi folks,

Working on a solid state amp with a pair of MRFX1K80 LDMOS chips with 65V drain voltage. This particular voltage makes it so much easier to design the output transformer and also place the chips on the same RF deck rather than run two separate decks. For now we are using the ceramic chips, but plan on switching to the plastic ones when available in decent quantities.

So my question is where to find good, practical details on how to size the water cooling system. Right now the RF deck with the pair of LDMOS fits on a 160 x 160mm board (roughly 6x6") and we plan on keeping it cool with 4 40x160mm water cooling blocks. So the question we are running into is how much water do we need and at what rate do we need to circulate it.

Our design objective is to build a 2500W out amp that can do forever key-down (RTTY really). So we want all components to have plenty of margin. Thus, we assume that the water cooling system must be able to absorb and dissipate 2500W worth of heat. Similarly, we plan on using a 5000W power supply.

Also, we were wondering about water vs oil based cooling system.

Full disclosure- neither I nor my friend have any experience with the thermal design aspects. My friend is a brilliant and established RF design guy, while I am thinking about sourcing, amp controller, etc. So we need some really good advice and hand holding.

73, Rudy N2WQ
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N3QE
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« Reply #1 on: July 26, 2017, 11:55:29 AM »

Eimac has some datasheets on tube cooling at the 5kW+ level. (Some of these are WAY MORE than 5kW).

http://www.relltubes.com/filebase/en/src/Technical_Notes/EimacAB16-Water_Purity_Requirements.pdf

http://www.cpii.com/docs/related/40/eikliquidcooling.pdf

https://frank.pocnet.net/sheets/084/y/YU191A.pdf
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W9IQ
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Posts: 1707




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« Reply #2 on: July 26, 2017, 12:10:34 PM »

Since you are specifying that you need to remove 2500 watts of heat on a continuous basis, the liquid in the cooling system is simply acting as a transportation medium for the heat. You will need to have a cooling system on the other end of the water loop that can cool the water on a continuous basis. This means you will need a chiller capable of about 8,300 BTU/hr (the equivalent cooling of 2500 watts of heating). You will find articles on PC web sites showing how you can re-purpose a regular window air conditioner for this type of application, for example.

The pumping rate of the liquid is determined by the allowable temperature rise of the cooling medium, the area of the liquid in contact with the heated surfaces of the LDMOS or its spreader (the heat exchanger), the turbidity within the heat exchanger,  the heat capacity of the liquid, and the similar effectiveness of the heat exchanger on the chiller end.

If you relax the 'continuous basis' requirement, you can consider the capacity of a liquid reservoir to transport and store the heat allowing the liquid cool naturally or with convection through a radiator and fan combination, for example during non-operating times.

The math starts with the heat capacity of the cooling liquid. It takes ~4.2 Joules of heat to raise one gram of fresh water 1 degree C. You can find a handy table of the heat capacity of other common liquids at http://www.engineeringtoolbox.com/specific-heat-fluids-d_151.html.

Armed with the specific heat capacity, you can then calculate how many hours of operation X gallons of water would afford before a cooling period must commence.

For example, let's assume we have 5 gallons (18.9 kg) of water. If the water starts at 20° C and we allow it to heat to 90° C (to prevent it from boiling) we have an allowable temperature rise of 70° C or K. 2500 watts is equal to 2.5 kiloJoules / second. We can then do a first order calculation of how long we can operate before cooling of the water is required:

Q (kiloJoules) = 4.2 j kJ/Kg °K * 18.9 kg * 70° K
Q (kiloJoules) = 5556 kJ

Operating Time = 5556 kJ / 2.5 kJ/s
Operating Time = 2222 seconds or 37 minutes

By using a liquid with a lower starting temperature, a higher boiling point, a higher heat capacity, or a higher density, the operating time before cooling could be extended.

Mineral oil has a lower heat capacity (1.67) than that of water. It can be taken up to a bulk temperature of 315° C but your maximum LDMOS case temperature is likely much less than this - say 120° C.  Five gallons of mineral oil has a mass of 26.9 kg. If you re-run the above formulas, you arrive at a first order operating time of ~30 minutes.

I hope that gets you started with the necessary math.

- Glenn W9IQ

« Last Edit: July 26, 2017, 12:14:58 PM by W9IQ » Logged

- Glenn W9IQ

I never make a mistake. I thought I did once but I was wrong.
AC2RY
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Posts: 280




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« Reply #3 on: July 26, 2017, 12:15:08 PM »

Hi folks,

Working on a solid state amp with a pair of MRFX1K80 LDMOS chips with 65V drain voltage. This particular voltage makes it so much easier to design the output transformer and also place the chips on the same RF deck rather than run two separate decks. For now we are using the ceramic chips, but plan on switching to the plastic ones when available in decent quantities.

So my question is where to find good, practical details on how to size the water cooling system. Right now the RF deck with the pair of LDMOS fits on a 160 x 160mm board (roughly 6x6") and we plan on keeping it cool with 4 40x160mm water cooling blocks. So the question we are running into is how much water do we need and at what rate do we need to circulate it.

Our design objective is to build a 2500W out amp that can do forever key-down (RTTY really). So we want all components to have plenty of margin. Thus, we assume that the water cooling system must be able to absorb and dissipate 2500W worth of heat. Similarly, we plan on using a 5000W power supply.

Also, we were wondering about water vs oil based cooling system.

Full disclosure- neither I nor my friend have any experience with the thermal design aspects. My friend is a brilliant and established RF design guy, while I am thinking about sourcing, amp controller, etc. So we need some really good advice and hand holding.

73, Rudy N2WQ

At that power level you will need to put the whole board into oil and then arrange pumping to circulate oil through heat exchanger. Honestly to make things reliable, you will be forced to use modular system with combiner. I would design for 30% power reserve - like use four 1000W modules, each with separate cooling and power supply.
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N2WQ
Member

Posts: 99




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« Reply #4 on: July 26, 2017, 12:21:43 PM »

Since you are specifying that you need to remove 2500 watts of heat on a continuous basis, the liquid in the cooling system is simply acting as a transportation medium for the heat. You will need to have a cooling system on the other end of the water loop that can cool the water on a continuous basis. This means you will need a chiller capable of about 8,300 BTU/hr (the equivalent cooling of 2500 watts of heating). You will find articles on PC web sites showing how you can re-purpose a regular window air conditioner for this type of application, for example.

The pumping rate of the liquid is determined by the allowable temperature rise of the cooling medium, the area of the liquid in contact with the heated surfaces of the LDMOS or its spreader (the heat exchanger), the turbidity within the heat exchanger,  the heat capacity of the liquid, and the similar effectiveness of the heat exchanger on the chiller end.

If you relax the 'continuous basis' requirement, you can consider the capacity of a liquid reservoir to transport and store the heat allowing the liquid cool naturally or with convection through a radiator and fan combination, for example during non-operating times.

The math starts with the heat capacity of the cooling liquid. It takes ~4.2 Joules of heat to raise one gram of fresh water 1 degree C. You can find a handy table of the heat capacity of other common liquids at http://www.engineeringtoolbox.com/specific-heat-fluids-d_151.html.

Armed with the specific heat capacity, you can then calculate how many hours of operation X gallons of water would afford before a cooling period must commence.

For example, let's assume we have 5 gallons (18.9 kg) of water. If the water starts at 20° C and we allow it to heat to 90° C (to prevent it from boiling) we have an allowable temperature rise of 70° C or K. 2500 watts is equal to 2.5 kiloJoules / second. We can then do a first order calculation of how long we can operate before cooling of the water is required:

Q (kiloJoules) = 4.2 j kJ/Kg °K * 18.9 kg * 70° K
Q (kiloJoules) = 5556 kJ

Operating Time = 5556 kJ / 2.5 kJ/s
Operating Time = 2222 seconds or 37 minutes

By using a liquid with a lower starting temperature, a higher boiling point, a higher heat capacity, or a higher density, the operating time before cooling could be extended.

Mineral oil has a lower heat capacity (1.67) than that of water. It can be taken up to a bulk temperature of 315° C but your maximum LDMOS case temperature is likely much less than this - say 120° C.  Five gallons of mineral oil has a mass of 26.9 kg. If you re-run the above formulas, you arrive at a first order operating time of ~30 minutes.

I hope that gets you started with the necessary math.

- Glenn W9IQ



Hi Glenn,

Thank you so much for taking the time to fully answer my question and point us in the right direction.

Obviously our continuous operation requirement is a bit of a stretch. The goal is to design an amp that can comfortably keep up during a 48-hour contest period. I stated continuous operation because I didn't want us to get distracted with discussions on what percentage of the time the amp will be actually transmitting. Thus the idea to design to extreme specs and later scale down as needed.

Also, we certainly plan on using a heat exchange a 2 or 3 fans to cool the water. The funny thing is that we decided to use a 5 gallon bucket while building and testing the amp; looks like we were spot on without even knowing :-)

Again, really appreciate your time and thoughtful response.

Rudy N2WQ
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N2WQ
Member

Posts: 99




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« Reply #5 on: July 26, 2017, 12:27:36 PM »


At that power level you will need to put the whole board into oil and then arrange pumping to circulate oil through heat exchanger. Honestly to make things reliable, you will be forced to use modular system with combiner. I would design for 30% power reserve - like use four 1000W modules, each with separate cooling and power supply.


Do you mind elaborating on the recommendation to use multiple RF boards and a combiner?  At 65V the 1k80 will be running at about 70% of its full 1800W power rating so we thought there is plenty of safety margin there.

Rudy N2WQ
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AC2RY
Member

Posts: 280




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« Reply #6 on: July 26, 2017, 05:49:07 PM »


At that power level you will need to put the whole board into oil and then arrange pumping to circulate oil through heat exchanger. Honestly to make things reliable, you will be forced to use modular system with combiner. I would design for 30% power reserve - like use four 1000W modules, each with separate cooling and power supply.


Do you mind elaborating on the recommendation to use multiple RF boards and a combiner?  At 65V the 1k80 will be running at about 70% of its full 1800W power rating so we thought there is plenty of safety margin there.

Rudy N2WQ

Problem is not with theoretical rating, but with power density. It would be hard to arrange sufficiently low thermal resistance. Also power supply for 100 amps will require a lot of attention too. It is easier to follow industry practice and limit unit power to a manageable level. Then sum the outputs. That is how most professional amplifiers are built.

I think most of new high power rated devices are actually designed for pulse use with very low duty cycle. You may be able to get 1800W from single amplifier, but for only few milliseconds.
 
« Last Edit: July 26, 2017, 06:10:32 PM by AC2RY » Logged
KC4ZGP
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Posts: 1637




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« Reply #7 on: July 26, 2017, 06:31:40 PM »


Now why didn't I think of that? I could submerge my EB-104 heat sink in a shallow pan of oil.
Or water and add ice cubes just before I begin to operate. No noisy fan.

You're a genius Kraus

Kraus
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K6AER
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Posts: 4666




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« Reply #8 on: July 26, 2017, 06:51:43 PM »

I built a 1 KW solid state amplifier and when in the RTTY mode it took about 1 gallon a minuet of water to cool the board. The radiator was a 12 by 18 inch unit from a small window air conditioner. Water bump was from a fish store. The copper spreader had about 18 inches of copper tubing soldered to the amplifier heat spreader in a figure S patter.  Water temp never went above 120 degrees "F".

No calculations were used but if the water got to hot I was just going to get a bigger radiator. My first choice worked out fine. You can buy rater cooling fixtures at hobby computer stores. Gamers use them on fast CPU's.

Don't forget a small fan for the amplifier board for the components.
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W8JX
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« Reply #9 on: July 26, 2017, 07:29:29 PM »

If you used a closed loop system with a large reservoir that held say 52 gallon (like a old hot water tank) and circulated it at 1 gallon a minute it would heat the 52 gallons up 20 degrees in one hour at full load. But given it is not a 100% duty cycle and if tank was not insulated either it would heat much slower and as tank got warmer it would shed more heat and slow increase until it reached a balance. This would be a simple reliable method.
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N2WQ
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Posts: 99




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« Reply #10 on: July 26, 2017, 08:12:03 PM »

I built a 1 KW solid state amplifier and when in the RTTY mode it took about 1 gallon a minuet of water to cool the board. The radiator was a 12 by 18 inch unit from a small window air conditioner. Water bump was from a fish store. The copper spreader had about 18 inches of copper tubing soldered to the amplifier heat spreader in a figure S patter.  Water temp never went above 120 degrees "F".

No calculations were used but if the water got to hot I was just going to get a bigger radiator. My first choice worked out fine. You can buy rater cooling fixtures at hobby computer stores. Gamers use them on fast CPU's.

Don't forget a small fan for the amplifier board for the components.

So let's say I use one of these radiators and have a gallon of water:

http://www.ebay.com/itm/Quality-480-Radiator-Copper-65mm-Ultra-Thick-5-G1-4-Thread-For-PC-Liquid-Cooling-/191887476818?hash=item2cad628852

How do I calculate how this thing will perform a) without and b) with fans that together move N cfm? The PC sites don't really offer much in terms of calculations, just this and that guy's experience with this CPU and that graphics card. I am not afraid of crunching the numbers as long as I can find the right formulas.

I really appreciate everyone's genuine help!

Rudy N2WQ
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N2WQ
Member

Posts: 99




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« Reply #11 on: July 26, 2017, 08:13:40 PM »


At that power level you will need to put the whole board into oil and then arrange pumping to circulate oil through heat exchanger. Honestly to make things reliable, you will be forced to use modular system with combiner. I would design for 30% power reserve - like use four 1000W modules, each with separate cooling and power supply.


Do you mind elaborating on the recommendation to use multiple RF boards and a combiner?  At 65V the 1k80 will be running at about 70% of its full 1800W power rating so we thought there is plenty of safety margin there.

Rudy N2WQ

Problem is not with theoretical rating, but with power density. It would be hard to arrange sufficiently low thermal resistance. Also power supply for 100 amps will require a lot of attention too. It is easier to follow industry practice and limit unit power to a manageable level. Then sum the outputs. That is how most professional amplifiers are built.

I think most of new high power rated devices are actually designed for pulse use with very low duty cycle. You may be able to get 1800W from single amplifier, but for only few milliseconds.
 

This is the second or third time we come across "power density"; perhaps we should study this further and reconsider the single board design.

Rudy N2WQ
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W9IQ
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Posts: 1707




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« Reply #12 on: July 27, 2017, 04:13:36 AM »

Quote
So let's say I use one of these radiators and have a gallon of water:

Unfortunately, the website does not provide any technical usable information for that radiator that I could spot. You would have to conduct some experiments with it to determine its efficacy.

For a well specified radiator, you would find a BTU/hr specification at a given CFM. You can convert BTU/hr to watts by multiplying by 0.293. If you are still working on a continuous operation design then you would want 2500 watts / 0.293 or > 8,500 BTU/hr of radiative cooling (same number as with a chiller).

- Glenn W9IQ
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- Glenn W9IQ

I never make a mistake. I thought I did once but I was wrong.
WA7PRC
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WWW

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« Reply #13 on: July 27, 2017, 06:09:18 AM »

At the largest OEM of RF-driven sealed CO2 Lasers, we used water cooling on our RF sources, starting at the kilowatt level. I can't allude to knowing a lot about how to cool your FETs but, I can suggest a way to facilitate monitor the temperature...

We used an LM35 'Precision Centigrade Temperature Sensor' (datasheet) in a TO-220 package, affixed to the output water tube. It produces a very accurate and linear 10mV/°C DC output. Mounted next to the sensor was a TO-220 power resistor (Caddock MP800, etc) dissipating about 10W.

As long as there is ANY water flow, the sensor very closely assumes the water temperature. If/when water flow drops too low, the resistor heats the sensor, and an alarm is output.

Bryan WA7PRC
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N3QE
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« Reply #14 on: July 27, 2017, 07:17:01 AM »

This is the second or third time we come across "power density"; perhaps we should study this further and reconsider the single board design.

The 1800W devices are really capable of that dissipation as long as you can suck heat away fast enough to keep the junction at the required temperature. If you took that as a lesson, that thermal conduction interface right at the device is a super critical parameter, you would be right!

With water cooling you have some advantages over the air-cooled guys in being able to reach the theoretical dissipation. Especially when the air-cooled guys are trying to reduce blower noise. Most of the air-cooled designs I've on the interweb seem to be using puny PC-clone fans and massive amounts of aluminum/copper. You would go the other way, with a super-optimized small thermal interface at the device, efficiently coupling to the cooling water that goes to a much larger radiator (I'm thinking very small automotive radiator) with a large but slower turning fan.

See, now that you've got me thinking about water cooling, I'm starting to think that I should be playing with these 1800W devices and small radiators. I do a LOT of RTTY contesting.
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