Heat gain in cool water pipe

[barters81]

I currently have a problem I'm trying to overcome which involves cool water heating up when being pumped through a pipe within an underground tunnel.

Knowns are:

Temperature of water into the pipe
Pipe dimensions
Ambient heat
Material properties of pipe
Flow rate

To gain a clearer picture, I will be pumping water into a pipe at 24.2 degrees and through 2km of piping. I need to get the water to come out at around 25 degrees. I'm trying to justify insulation, and maybe even a heat exchanger through these calculations.

Any help would be much appreciated. Thanks

 

[IRstuff]

And yet, you've only listed 2 of your knowns? How do you expect to get a cogent answer?

TTFN

 

[barters81]

Thanks for the helpful input....if you read above there are a fair few more than 2 knowns I've given. But anyway....

I don't expect someone to give me the complete answer as I won't learn for next time. I'd rather a push in the right direction, so quantitive values really aren't needed, but here they are anyway.

"knowns" are:

- flowrate of water through pipe (25 m3/hr)
- dimension of pipe (OD = 114.3mm, ID = 110.3mm L = 2.2km)
- material of pipe (galvanised pipe)
- ambient air temperature (assume constant 29 degrees celsius)
- temperature of water entering pipe (24.2 degrees celsius)

We can obviously assume full turbulent flow through the pipe, and work a Nusselt number accordingly. I'd like to work out the final temperature of the water given a number of different ambient air temperatures. I'm assuming convection within the pipe and conduction on the outside...although I've been wrong before and is why I'm asking the guys in the know. I'm sure I'm over complicating things.

At this stage a gut feeling is telling me I'm going to need another heat exchanger to get my water to the destination at the required temp. But we all know what projects are like....$$$ and without written justification I won't see any.

Thanks in advance for any "Helpful" advice.

 

[IRstuff]

OK, thanks for the additional info. A back of the envelope calc says that the water stays exposed for about 0.8 hrs, during which it accumulates about 40 MJ of energy, using 24.6ºC as the average temperature and a 4 W/m^2-K heat transfer coefficient. The net result is enough energy to raise the temperature about 0.46ºC, so it's close. You'll need to run a more accurate analysis.

The constant temperature situation is a little unclear. Is it forcibly made constant, i.e., is there a high velocity air flow? If so, the heat transfer coefficient may be a lot higher, in which case, the water temperature will climb more.



TTFN

 

[barters81]

Thanks for that, how did you work it out? I'd like to be able to write this stuff down, and maybe prove that insulating the pipe is all we need.

This is a pipeline going in a tunnel 70m underground. The ambient temperature is supposed to be between 25-28 degrees. I've just said 29 as a worst case scenario. We do have some air flow underground which would be roughly 1-2m/s. The machinery underground is being cooled by this water pipe so the tunnel shouldn't heat up. We have chillers down there which cool this water further before going through the TBM (tunneling machine).

I'd like to be able to make a spreadsheet showing the exit temperatures for a range of ambient temps and air rates. And for my own curiosity know how to work such problems. I dove into some text books that I haven't read for a few years, and the further I went the more complicated everything got.

Once again thanks for the help its much appreciated.

 

[IRstuff]

The standard equation of htc*area*delta_T as applied to the water, which raises its temperature. I cheated and made a bunch of simplifying assumptions to get a quick answer. It's an iterative process to get a temperature rise that you plug back into the heat transfer equation.

I don't know if what I did works well for heating, so lots of caveats here. The air flow would be enough to kick up the heat transfer coefficient up to something like 10 W/m^2-K, which changes things quite a bit

TTFN

 

[25362]

I may be wrong, but it seems to me that even with an OHTC=10 W/m²K the outlet temperature would be beteen 25 and 25.5^oC. Nevertheless, as IRstuff says, if the outlet temperature is critical a thorough analysis is advisable.

 

[MortenA]

barter81

I have said this quite often: Maybe you should seek professional help? How will you document this work (allthough im sure IRstuff is correct in his calculations). If somebody questions your work will you then say: Look this helpfull guy on the inter gave me the answer?

I would not be impressed (no bad will intended towards IRstuff).

A good textbook on heat transfer is Kern: Process heat transfer.

For a buried pipe The heat resistance can often be estimated from:

Rs=Do/2Ks*ln((2Zb+sqrt(4Zb^2-Do^2))/Do)

where Rs=heat resistance (K/W)
Do= Outside diameter (remember to define to your reference point)
Zb=burial depth

But how well this fits a pipe in a buried duct im not sure.

The rest (resistance from inside film and thourgh pipewall and insulation is standard.

Try to look for textbooks on district heating since this is similar to what you describe.

Best regards

Morten

 

[magicme]

The problem you describe is simple conceptually but there feels like "death in the details".

A good text book would cover all these topics.

I can offer the following clips from my heat transfer course notes (please excuse the tutorial wording, I left them as-is):

<Exit temp>
The exit temperature of the fluid (assuming constant outside temperature) would be ...

T(m,0) = T(amb) - [T(amb) - T(m,i)] * exp [ - U * A / mdot / Cp ]
where
T(m,0) is the fluid exit temp
T(amb) is the external ambient temp
T(m,i) is the fluid inlet temp
mdot is the mass flow rate (kg / s)
A is the total internal pipe surface area ( 2 * pi * r(inner) * L )
U is the overall heat transfer coefficient from the flowing fluid, through the wall into the constant outside temperature on the other side. (This includes convection and conduction terms.)

The overall heat loss would be....

Q = mdot * Cp * [ T(m,0) - T(m,i) ]


<Overall HTC>

The equation for the overall heat transfer coefficient of the pipe (or cylinder) is slightly different in various textbooks, depending on the way the surface area is referenced. One favorite version for U is
U = 1 / [ ( 1/h1 ) + ( r1/k1 ) *ln ( r2/r1 ) + ( r1/k2 ) *ln ( r3/r2 ) + ( r1/k3 ) *ln ( r4/r3 ) + ( r1/r4 ) * ( 1/h2 ) ]

This example is for a 3-layer pipe. Whatever the number of layers, the last term must always have this form:
( rinner / router ) * ( 1 / houter )

It is important to know that in this setup, the reference area, A, to be used in the heat transfer calculation is the inside surface area of the pipe = 2 * pi * r1 * L.

Also, in practice, most engineers calculate q/L the heat transfer rate per unit of pipe length, and using the above equations, this becomes...
q/L = U * 2 * pi * r1 * (Tinside - Toutside)


<Internal HTC>

The equation used here for the average convection heat transfer coefficient along the inside wall of a pipe with fluid flowing inside:
H = .023 * Re^.8 * Pr^ n * k / D

<External HTC>
I use cylinder in cross-flow average HTC equation .... but I never wrote that up.... and refer people to the textbook (equation is a mess!).



regards,

magicme

 

[barters81]

Thanks guys, your help is much appreciated.

With regards to getting professional help, with my qualifications I should be able to figure this out. I only use what people say here to give me a push in the right direction. Take Magicme's response, which was great by the way, I will now go through my texts and confirm what he has shown me. Once I've done that, which I'm sure will be correct or close to as it definitely rings a bell from my uni years and matches how I first thought of modelling the problem, I will put together my own calculations and justifications.

This is my first thread on this site and I must say I think its great.

 

[25362]

I repeat this statement every time I refer to the estimation of U: it will be smaller than the smallest htc considered, a fact that can be seen from the equation by magicme:

1/U = 1/h_smallest + ....​

Therefore, in this particular case, by using the external htc, apparently the lowest of all, to be on the safe side one can quickly estimate (the accent on quickly) a "maximum fictitious" heat transfer using the external pipe surface and the maximum [&Delta;]T.