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Cool a circuit at near constant temperature

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RegiSTR

Mechanical
Mar 18, 2020
8
I have a closed circuit with polyester pipes, which circulates water at room temperature (~20deg C).
To circulate this water a pump is used that uses 5kW.
The flow rate of the pump is around 0.208m3/s.
The amount of water in the circuit is around 7m3.
The circuit has pipes with a diameter of 200mm.

To cool the circuit I want to use a concentric pipe around the circuit pipe, and run sea water through it (~10 deg C), to keep the water temperature in the circuit at a constant level.
The cooler sea water has a flow rate of around 0 to 0.0008m3/s.

Q=mcdT
c=4200J/kgK
Q=5000J
m1=0.208m3/s
m2=0.0008m3/s

Without getting too complicated: How would you calculate the diameter/length or flow rate I would need to keep the circuit at a near constant temperature (+/- 1deg C, if even possible).
 
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Have I seen this question before?

The numbers are nonsensical and need to be revised.

That water is flowing at 38 m/sec and the disparity between water flow and the dribble of seawater is bonkers.



Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
 
My bad, accidentally used m3/min instead of sec.

Q=mcdT=0.83*4200*10=34.8kW? But how is there no area/length involved.
 
So, the circuit does nothing except circulate?
And you just want to remove the pumping energy to keep a constant temperature?

What's the business case for this?
 
So this looks like it's 220m of 8" pipe?

What do you mean by "water temperature at a constant level?

20C at every part of the 220m?

What exactly is causing this to consume 5kW??
It's only now doing 6.5m/sec.

This looks very odd....

Also I've just worked out that if you do nothing, the end temperature of your flowing water will rise by all of 0.005 C That's going to be impossible to control to.

Far better to cool or control temperature i whatever tank you're using, but of course we have no diagram or sketch of this system....

The Q figure assumes a temperature rise of 10C. That's not going to happen as heat transfer requires a delta T. Also it removes too much heat. So to remove 5kW you only need a temp rise of approx 1.3 C in your seawater.

So what you need now is the overall heat transfer rate through your pipe expressed as W/m2/deg C, commonly called the U value. That depends on your material and wall thickness. Go look it up.

Your DT is fairly constant at an average of 9.5C so once you find U, you can find square area of pipe required.


Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
 
Any heat transfer at all involves a temperature change, so your stated goal is not possible.
 
There is around 30m of pipe and a 6m3 barrel in which the flow is circulated, the 5kW is the max output of the pump at the calculated resistance/flow.
There are all kinds of valves etc. in between. I was hoping to cool the system over a part of L=3m straight piping with D=200mm (HDPE: PE100 SDR17) against the pump flow direction, and was looking for a ball park figure to see if it's feasible or not to just cool around the outside.

Unfortunately I cannot find 'the' U value of this specific pipe. The pipe inner diameter is 200mm and is 7.7mm thick.



 
You need to calculate it.

see e.g.
Then adjust L and use say 9.4C as your average Delta T.

You need to know the heat transfer coefficient for your material - "polyester" is a bit vague. See if the vendor has a value.
You're suddenly switching to PE pipe? PE pipe is a constant OD with Id changing. 225mm SDR 17 is about 195mm ID. There is 180mm pipe, but this is OD, not ID.

My points are a bit more practical in that you will find it difficult to control the temperature and your heat transfer assumes 100% efficiency between seawater and your water. What level of temperature control are you looking for?

You're probably much better off with a plate HX.... Or a cooling coil in your "barrel"

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
 
You're essentially making a 1-tube shell-and-tube heat exchanger. There is a ton of information online to step you through the design process.

Here's an example that walks you through all of the terms, including dimensional analysis, U-value, etc. which should help you understand your problem a bit better and allow you to determine if you're on the right track for your project.
 
@LittleInch, but how do you know the delta T. And how would you know a cooling coil is better, as there is much more area around the pipe?
Sorry I meant Polyethelene but cannot change OP, I got no other info about the specs as what I stated, so I guess ill google them, probably does not deviate much.
 
Your inner pipe water is 20C

From one of the posts above I've calculated fairly roughly that the temp rise for the seawater flow assuming you start at 10C is only 1.3C when you add 5kW.

So an average delta T of 9.4 C.

Depends on how big your cooling coil is, plus you have more delta T to play with and with a larger volume your variance in heat and temperature should be more controllable if slower to react. Pipe in pipe / jacketed pipe is not easy to fabricate or test so beware of the pitfalls. Inserting a coil into a tank is much easier / you can buy them off the shelf.

PE is quite a good insulator so you really need to get your data correct. PE pipe does come in different sizes in different countries, but is all the same OD, but with different wall thicknesses depending on the SDR.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
 
Thanks for responding and your time, but this is where it gets difficult for me to understand. How did you calculate the temperature rise, as this seems the key to solving this so I can find the needed area.

As I understand, the 5kW pump adds heat to the circuit continuously, this is the exact heat that needs to be removed to keep the water at around 20 degrees. So each time the water passes the pump that amount needs to be cooled in the 'tube/shell' area, whatever its length/diameter/flow rate is.

c=4200J/kgK
Q=5000J
material: HDPE
thermal conductivity: ~0.5 W/(m K)
t=7.7mm
Di=0.2m

mo=0.0008 m3/s (max)
mi=0.208 m3/s

To1=10 deg C
To2=?
Ti1=?
Ti2=20 deg C

dTo=?
dti=?
Do=?
L=?


 
I took a simple bulk analysis to get me close.

The seawater at 0.8 kg/sec if heated with 5000w (J/sec) will rise in temperature by 5000/( 4200 *0.8) = 1.5 Dec C

Your issue I think is over cooling and how you control the cooling water when the flow rates are so different. Hence my idea of using a tank coil cooler

So

To1=10 deg C
To2=? 11.5
Ti1=? 20C
Ti2=20 deg C ( well so close to 20 it's not worth bothering about but would be 19.9 or something like that

dTo=? 10
dti=? 8.5
Do=? Dunno - say 400mm OD Or is Do the OD of your inner pipe? You know that surely?
L=? That's for you to calculate - I've given you the tools and the equations, but I'm not going to do this. My guess is about 3m.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
 
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