How long does the pipe need to be to cool 60m3 of water from 60oC to 25oC?

In summary: If the power output of the radiator is 100 watts when the water is at 60 degrees, then the power output will be less when the water is at 35 degrees.
Hello there,

Firstly I would like to welcome myself to the forum, my name is Ben and I'm 21 and I'm from the Isle of Mull off the west coast of Scotland.

I am studying mechanical engineering and this question has been giving me a massive headache.
If I have 60m3 of water at 60oC in a tank which passes through a copper pipe in a room that is at 20 degrees C, how long would my pipe need to be to bring the water temperature down to 25oC.

This question comes from my current internship. We have a rig that we need to cool with water and we are looking at heating the building with the energy and we are considering passing it through a long snaking pipe at the side of the shop floor.

Help is much appreciated.

Welcome to PF!

First, we have an ME section, so I'll move it there.

Second, being a real world question instead of a homework question means it doesn't have an answer or perhaps has any answer. Literally: as worded, the pipe can be whatever length you want.

The key issue in real life problems is that you aren't provided with the constraints and assumptions required to calculate an answer, you have to generate them yourself. So, for a start, what restrictions, if any, do you know of regarding flow rate, time and number of passes through the room?

Worse, I rather suspect you are asking a wrong/irrelevant question. The key issue is probably how much heat you have available.

Yes this is the first time I've really applied knowledge such as thermodynamics I've been 'taught' at university to a real life situation.
Really what we want to know is if we can cool the water heat by heating the building. I'm looking for an answer as to whether or not its possible, which I assume could be achieved by passing it through a copper pipe snaking up the side of the factory floor.

To calculate the amount of potential heat energy in the water;

I did

m=p*V
= (983.2 * 60)
= 58992 kg

c = 4.18(kJ/(kg K))

deltaT = 60 - 25 = 35C

So the amount of potential heat energy is
Q=mc(delta)T
=(58992*4.18*35)
=8.631 MWAs for the flow rate, well this ##60m^3## (of water going from 25C to 60C) is expected to be passed through the test rig per day.
So at ##60m^3## per 24 hours,
That equals a flow rate of ##2.5m^3## per hour or ##0.0006944 m^3 /s##

As for real life constraints, well I've found some thin copper tube for sale that is 35mm in diameter, and the wall of the factory floor is absolutely effing huge so I just want to find out the amount of pipe I need first.

I was hoping if you could point me in the right direction for how to answer this question.
Thanks for any help :)

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The amount of heat you must remove is 8.631 MJ, not MW. (Always check the units in your calculations!)

Assuming this amount of energy is removed at a steady rate during the day, then the power would be:

8.631 × 106 J / (24 * 3600) = 99.89 J / s = 99.89 watts

Ok so I've got 8.631 MJ of heat to remove if I want ot go from 60 degrees to 25 degree.
And if I passed it through a long copper pipe then on average it would be producing 99.89 watts.

If a halogen light bulb at 100 watts produces like 3 watts of light and 97 watts of heat, then isn't 99.89 watts of heat from 60##m^3## of hot water really terrible?

How can I work out the length of copper pipe that I would need in order to transfer this heat from the 60 degree hot water to the ambient 25 degree hot air though.
So I imagine if I passed the water through 10 meters of pipe, the temperature might drop from 60 to 55 degrees, but how do I work out how much I would need to take it from 60 to 25 degrees, in open air at 25 degrees C.

How to dissipate the heat is simultaneously the hardest and easiest part. Hardest because if you want to calculate heat transfer from scratch, that is very hard. Easiest because commercial products exist to do it that have pre-calculated/tested capabilities. You can just look them up: fin-tube radiators.

100 W is a tiny amount of power for a radiator: one small section would do the job.

Yeah I was thinking I just pass it through a load of radiators.

But this was where it got confusing, surely the 100 Watt output of the radiator is only when it is being run with regular 60 degree hot water from the house supply.
Surely, when the water temperature drops, the heat output of the radiator drops too?
Doesn't the manufacturers heat rating only apply when there is a constant temperature of water being passed through.Since I am cooling the water, the temperature of the water is slowly decreasing from 60 to 25 degrees C.
So does this mean I can't really use the manufacturers rating.
Is the level of thermodynamics 'from scratch' required to answer this question very hard as in a Bachelours degree in Thermodynamics/Mechanical Engineering, or very hard as in post PHD research into thermodynamics.

Thanks,
Ben

Did you take Heat Transfer? It was an optional but core elective for me, and one of the harder ones, but shouldn't be beyond you. The thing is, though, accuracy requires knowing the input assumptions and for free convection that is pretty much impossible without testing. Of course, all of that testing has already been done, so...

The perhaps dirty little secret about professional engineering is that because of the above, the really difficult stuff is done once and then repeated so others don't have to do it. Instead, you just take what is already out there and modify it for your purposes. So on a day to day basis, the work is a lot easier than what you do in college even though the problems are vastly more complex.

And that makes Google a critical engineering skill (yes, really). So Google "fun tube radiator performance". The first pdf (5th link) is a spec sheet for a Modine fin tube. It includes performance at various temperatures for several fluids and a walk-through of how to do the selection for your purposes.

If I'm interpreting your situation correctly and based on what has been calculated above as the average energy you could theoretically transfer if the entire tank was brought to 25 C and given that you said your shop is very large, this seems like a futile endeavor to recover that energy to heat the building. Just for back-of-the-envelope comparison, I have a portable AC unit in my living room/kitchen that is rated at 11,000 BTU which is about equal to 12 MJ (more than your available energy). On a hot day, it can take the rooms from 81 degrees down to 77 deg F.

In any case, even if there was more energy to recover from the water to heat the room with, russ is right that that is only half the problem. After that, you have to deal with HOW you are going to get that energy to transfer into the room. As a general rule of thumb for heat exchangers, the minimum ΔT that you need to drive good heat transfer is around 10 deg C. Taking this into account, you actually only can recover from your starting point of 60 C down to, say, 30-35 C. This only can make it less worthwhile to try to recover that energy. As you noted, you can probably get the same amount of energy from running a few incandescent light bulbs all day to heat the room.

Large bodies of water, much larger than your tank, are good for trapping massive quantities of heat. The problem with trying to use that heat to do useful work is that the temperature of the water is only a few degrees different from ambient conditions, hence efficiency will be extremely low for any practical process using that energy and converting it to work. For high efficiency, you must have a large temperature difference between the hottest and coldest points in your thermodynamic cycle.

Most of the time, this is not a problem, but large tropical storms manage to convert the heat trapped in warm water into driving the cyclonic winds forming these storms, and these winds can be quite destructive when the storm initially moves over land. Once this occurs, however, the storm is cut off from its source of heat, and the wind speeds drop very rapidly as the storm dissipates.

This is a straight up heat transfer problem. A pipe with water flowing through it and suspended within a fluid requires a convection correlation for the internal flow and the external free convection (the convection on the reservoir side).

I recently posted these, but here are the correlations:

If the Reynolds number in the pipe is less than about 3000, then

Nu = 4.36 for constant flux boundary condition, Nu = 3.66 for constant temperature boundary condition. I'd probably use the constant temperature boundary condition
on this one even though neither exactly works perfectly in this case, I'm thinking isothermal would be a better approximation. I think...

If the Reynolds number is greater than 3000, then Nu = 0.023*Re^(4/5)*Pr^(0.3) (in the case of the pipe cooling the fluid in the pipe)

So, then you need to know the convection correlation on the outside of the pipe is Nu = {0.60 + (0.387*Ra^(1/6)) / [ ( 1 + (0.559/Pr)^(9/16) )^(8/27) ] }^2 for Ra > 10^12

So, once you have both of those, the heat transfer at a given cross section of pipe is going to be U(x) = [ 1/(h_internal*pi*D_internal) + LN(r_external/r_internal)/(2*pi*k) + 1/(h_external*pi*D_external) ]^(-1) In this case, U is the heat transfer per unit length of pipe and overall temperature difference at that point (Watts/meter Kelvin)

Then, the mean temperature of the fluid at a given location along the length of the pipe is given by the following expression:

(T_reservoir - T_fluid(x) )/(T_reservoir - T_fluid(x=0,inlet) ) = exp(-pi*U(x)*x / mass flow rate*specific heat)

So, you just have to create a spreadsheet or program that calculates the value of U for different locations along the pipe and see at what length it cools to 25 C. No small task, and I've only given you a quick overview that might not be easy to understand. Feel free to ask any questions and I'll try to explain in better detail.

jlefevre76 said:
This is a straight up heat transfer problem. A pipe with water flowing through it and suspended within a fluid requires a convection correlation for the internal flow and the external free convection (the convection on the reservoir side).

I recently posted these, but here are the correlations:

If the Reynolds number in the pipe is less than about 3000, then

Nu = 4.36 for constant flux boundary condition, Nu = 3.66 for constant temperature boundary condition. I'd probably use the constant temperature boundary condition
on this one even though neither exactly works perfectly in this case, I'm thinking isothermal would be a better approximation. I think...

If the Reynolds number is greater than 3000, then Nu = 0.023*Re^(4/5)*Pr^(0.3) (in the case of the pipe cooling the fluid in the pipe)

So, then you need to know the convection correlation on the outside of the pipe is Nu = {0.60 + (0.387*Ra^(1/6)) / [ ( 1 + (0.559/Pr)^(9/16) )^(8/27) ] }^2 for Ra > 10^12

So, once you have both of those, the heat transfer at a given cross section of pipe is going to be U(x) = [ 1/(h_internal*pi*D_internal) + LN(r_external/r_internal)/(2*pi*k) + 1/(h_external*pi*D_external) ]^(-1) In this case, U is the heat transfer per unit length of pipe and overall temperature difference at that point (Watts/meter Kelvin)

Then, the mean temperature of the fluid at a given location along the length of the pipe is given by the following expression:

(T_reservoir - T_fluid(x) )/(T_reservoir - T_fluid(x=0,inlet) ) = exp(-pi*U(x)*x / mass flow rate*specific heat)

So, you just have to create a spreadsheet or program that calculates the value of U for different locations along the pipe and see at what length it cools to 25 C. No small task, and I've only given you a quick overview that might not be easy to understand. Feel free to ask any questions and I'll try to explain in better detail.
Hey Jlefevre76.
Yes this is exactly what I was looking for.

I was going down this same route myself. I was thinking I could work out the rate of heat transfer per unit of pipe, and do this over and over again. This is where I got so confused though. My dirty secret is that I've never really learned how to do spreadsheets or really used Excel.

Looking at this problem I thought there would be some easy way to integrate the equation, and find a purely mathematical solution to the problem of how much pipe do you need to cool the water from 60C to 25C in air that is 20C.
How do I make a spreadsheet to solve this problem?

I know this will probably be a lot of pipe, but if it is my next step finding a heat exchanger made up of lots of coils of pipe that I can just blow air through to cool it down even more.
Thanks,
Ben

Using the selection guide I provided, it looks to me like a 1m long section of fin-tube would be plenty for this.

I'm not sure why one would try to design from scratch an inferior version of such an ancient, highly developed, off-the-shelf product as a fin tube radiator. Is this really what your boss is after?

Hey Jlefevre76.
Yes this is exactly what I was looking for.

I was going down this same route myself. I was thinking I could work out the rate of heat transfer per unit of pipe, and do this over and over again. This is where I got so confused though. My dirty secret is that I've never really learned how to do spreadsheets or really used Excel.

Yeah, it will be iterative, so whether you use Excel or make a simple program in MATLab or whatever, it will require an iterative solution to solve for the temperature of the pipe at a given location and use that to recalculate the free convection on the outside which is dependent on the temperature differences between the surface and the medium. So, it's still not a trivial task. If the convection coefficient on the inside is high and the pipe is conductive, you can make some simplifying assumptions, but I would just write a program in MATLab or whatever you have available to find the length of pipe needed. Or, just buy a heat exchanger that will cool it close to the temperature of the reservoir if your boss want's it done ASAP.

SteamKing said:
8.631 × 106 J / (24 * 3600) = 99.89 J / s = 99.89 watts
Are you sure that shouldn't be 99,890 watts (about 100kW)?

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Anyway, even if you could maintain the tank at 60oC and pump 2.5m3/hr through your 35mm diameter tubing in your 20oC room, you'd need 5000m of tubing to cool the water to 25oC. So, yes, finned radiators (with fans) are the only possibility.

Questions: How is the water in this tank being cooled now? Is this water being thrown away? If it is reused, what temperature is required for reuse?

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insightful said:
Are you sure that shouldn't be 99,890 watts (about 100kW)?
Yep, yer right. I used the OP's calculation without checking his numbers for the heat rejection.

Is the 60m3 tank part of the rig or is it a supply tank for cooling? What is the heat output of the rig and what is the target temperature you need to keep it at?

BoB

Ok so to complete restart the conversation. I'm going to through out the pipe idea. 1500 metres of pipe sounds like too much. I've seen in some buildings they run central heating water through a pipe around the room and its heats the building like that.

So, I want to cool 60 metres cubic of hot (60 degrees C) water back down to a reusable temperature (25 degrees C) with in one day.
How do I calculate how many fan radiators I would need for this using the technical manuals of heaters. I'm confused because once the temperature drops to say 30 degrees, surely the rate of heat the radiators put out drops, so its harder to use the data tables.

Thanks for putting me in the right direction so far.

So, I want to cool 60 metres cubic of hot (60 degrees C) water back down to a reusable temperature (25 degrees C) with in one day.
How do I calculate how many fan radiators I would need for this using the technical manuals of heaters. I'm confused because once the temperature drops to say 30 degrees, surely the rate of heat the radiators put out drops, so its harder to use the data tables.
Did you read the one I pointed you to? It includes varying or non standard water temperature in the selection procedure.

This is what I had in mind:

http://www.houseneeds.com/hvac-heat...ter-heat-exchanger-heating-coil-htl2225-c2225

Using their data, I calculated this has about 16 m2 of surface area.
You could put four of these (with fans added) in series and it should work.
You need to check the pressure drop for your 2.5 m3/hr flow.
If pressure drop is too high, run them with the water circuit in parallel with a 10 m3/hr pump feeding all four.
Given the uncertainties, I'd go with the parallel setup to maximize the delta T between the air and the water.
I'd also leave room to double everything if needed.
Can a $15k to$30k (US) project be justified?

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Thanks insightful.

I'm completely blanking on how you worked out the surface area. I absolutely cannot figure it out from the website. I just don't understand because once you cool the volume of water to a lower delta T that the heat the radiator is taking out the water (the heat output) doesn't significantly change.

Does the answer lie in the pump?

I don't even understand the guide Russ posted with the fin radiators

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The data in the selection guide is simplified and probably based on testing. So the fact that the water temperature changes from one part of the HX to another is already accounted for for you. All you need for the input is the average temperature.

I can give more guidance later, but you should try to follow the selection procedure step by step, and we can help you as you go.

Ok so I followed the selection guide

This is where I get confused.
Lets say the water has cooled down from 33C (91.4F) to 25C and air temperature is 20C (68F)
Water velocity is 3ft/sec
Calculated heat loss required to cool from 33C to 25C = 1,512,466 BTU

Air and water correction = 1,512,466/0.06
=25207766BTU

From the table I pick my radiator and the maximum water temperature is 100F, so above my temperature of 91.4F, but whatever.I pick the best radiator possible for water at 100F and I get
18671.3 feet or 5,691metres!

Sweet jesus.

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I get
18671.3 feet or 5,691metres!
I think this confirms why you don't want to use natural air convection. A good heat transfer coefficient for natural air convection on a horizontal cylinder or vertical surface is 2 Btu/(hr.ft2.oF). For forced air convection, the coefficient is 20 (yes, 10 times the heat transfer). The water side coefficient is about 1000, so the air side coefficient is the controlling factor.

I back-calculated a fin-tube area of 175ft2 from Q=UAdeltaTlm (look up "log-mean deltaT").
For the unit I showed, using their data:
Q=167000 Btu/hr
U=20 Btu/(hr.ft2.oF)
water flow = 10 gpm
water in = 180oF
water out = 147oF
air flow = 2400 cfm
air in = 70oF
air out = 132oF
Since this is cross-flow, I used deltaT1 = 180-70=110 and deltaT2 = 147-132=15.
DeltaTlm = (110-15)/ln(110/15)=48oF

Hello, I have no background in Thermodynamics or even science, but I was wondering if someone could explain something that happens daily to me?
I heat water in a standard ceramic coffee/tea mug using the microwave. The water is not boiling but very hot. I then drop a room temperature metal tea ball partially filled with loose leaf tea into the water and it erupts into a roiling boil. Why?
I have searched many places and have been steered to the explanation that hot water freezes faster than cold water. This does not answer my question without some further clarification. Thank you for any insight and I apologize if I did not follow proper protocol. This is the first time I have ever participated in any type of blog.

Hypocritease

Hypocritease said:
Hello, I have no background in Thermodynamics or even science, but I was wondering if someone could explain something that happens daily to me?
I heat water in a standard ceramic coffee/tea mug using the microwave. The water is not boiling but very hot. I then drop a room temperature metal tea ball partially filled with loose leaf tea into the water and it erupts into a roiling boil. Why?
I have searched many places and have been steered to the explanation that hot water freezes faster than cold water. This does not answer my question without some further clarification. Thank you for any insight and I apologize if I did not follow proper protocol. This is the first time I have ever participated in any type of blog.

Hypocritease
Welcome to PF!

That phenomena is called "superheating":
https://en.m.wikipedia.org/wiki/Superheating

You probably could have started your own thread on it...it isn't very related to this one.

russ_watters said:
Welcome to PF!

That phenomena is called "superheating":
https://en.m.wikipedia.org/wiki/Superheating

You probably could have started your own thread on it...it isn't very related to this one.

Thank you. Sorry, I was not aware how to start a new thread, but I will investigate ,if I have a question in the future.

russ_watters
Yes, the answer is that the water in the microwave gets superheated: heated to a temperature beyond its boiling point, but it does not boil. Then, when you add your tea ball to the water, it provides the "activation energy" and starts the boiling process via nucleation sites. Also, FYI, the statement that hot water freezes faster than cold is false.

EDIT: I guess I stand corrected in that hotter water can, under certain circumstances, freeze faster than cooler water. But this still has nothing to do with your original question.

Thank you Kayan

1. What is thermodynamics?

Thermodynamics is the branch of physics that deals with the relationship between heat, energy, and work. It studies how energy is transferred and transformed in systems, and how these processes affect the properties of matter.

2. What are the laws of thermodynamics?

The laws of thermodynamics are fundamental principles that govern the behavior of energy in a system. The first law states that energy cannot be created or destroyed, only transferred or converted. The second law states that the total entropy of a closed system will always increase over time. The third law states that the entropy of a perfect crystal at absolute zero temperature is zero.

3. What is the difference between heat and temperature?

Heat is a form of energy that is transferred between objects or systems due to a temperature difference. Temperature, on the other hand, is a measure of the average kinetic energy of the particles in a substance. In other words, heat is the transfer of energy, while temperature is a measure of the amount of energy present.

4. What is an example of a thermodynamic process?

An example of a thermodynamic process is the expansion of a gas in a piston-cylinder system. As the gas expands, it does work on the piston, and its internal energy decreases. This process can be described by the first and second laws of thermodynamics.

5. How is thermodynamics related to other fields of science?

Thermodynamics is a fundamental concept that is applicable to many fields of science, including chemistry, physics, engineering, and biology. It is used to understand and analyze energy transfer and transformation in various systems, from microscopic particles to large-scale processes such as climate change.

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