insulated pipe heat loss 3

[bencpv]

hi,

I wonder if anyone can help me.....

I have been tasked to find out the heat loss of buried insulated pipework. I have spent ages looking online for help in this and managed to find an equation (attached) that takes into consideration the service pipe, insulation, and jacket pipe.

In my efforts to find such an equation, I found other (more complex) equations that contain things like Reynolds number.... I think this will be too complex for what I need. I just would like something relatively simple, relatively accurate. Does anybody know of such an equation? Ideally one that considers flow and return pipes at x distance apart.

Now, going back to the attached example, I worked through this and expected to reach a result of 33.7 Btu/h lf. Instead I came up with the answer of 41.6 Btu/h lf! I would be grateful to hear if anyone can get 33.7.....and how! Where am I going wrong?!!!

Thank you in advance,

Ben

 

[zekeman]

Post your computation and we can find your error.

 

[IRstuff]

I get 34.21 from the numbers and equation supplied, which is a bit too much error, but it's certainly possible that there's a typo.

TTFN
faq731-376

 

[ione]

Ben,

I've got the same result as IRstuff's got. Just enter that formula in a spreadsheet, being sure you're using the right conversion from Btu inch/(h ft^2 °F) to Btu/(h ft °F).

 

[zekeman]

Sorry,
I got 33.54
which is very close to the author's.
I'll bet the last term was the problem
It is ln(d/r-[d/r-1}^-2)/2pik
which is very close to
ln(d/r)/(2piK)

 

[IRstuff]

Still can't get that answer. The two expressions evaluate to 2.058 and 2.061, and the difference is way smaller than the difference in the answers. Here's my calc (note, I use Rankines because M11 doesn't have a native degree Fahrenheit unit) :

TTFN
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[bencpv]

Thank you so much for the prompt assistance!

From reading the replies I think that I wasn't converting from Btu inch/(h ft^2 °F) to Btu/(h ft °F - to be honest that thought never occurred to me! I was merely plugging in the numbers to the formula! Can anybody tell me the conversion?

In IRstuff's latest response: from looking at the calculation (bottom left) of ln(d/r[sub]4[/sub]))/k[sub]4[/sub] = 2.0580 (Btu/(hr.ft.R)[sup]-1[/sup]

I dont get this at all, I get 0.01338 - obviously I've made a an error somewhere, but where?

I'm using the units Btu, in, ft etc because they were in the example. The data that I will be inputting is metric with the final result in W/m. The conductivity is in W/m°C. the diameters in m. I thought that once I get my head around the formula I would just be able to plug in the metric values and end up with the correct units. Is this the case? Or is this not the correct way to go about it?

I attach my (brief) calcs that I did in Excel.

I apologise for the questions - its been quite a few years since I have done these kind of equations!

Thanks
B

 

[ione]

1 Btu inch/(h ft^2 °F) = 0.08333 Btu/(h ft °F).

 

[zekeman]

My bad .
Made a few errors in calc and now get 34.27 essentially in concurrence with
Irstuff and Ione.

 

[zekeman]

"
In IRstuff's latest response: from looking at the calculation (bottom left) of ln(d/r4))/k4 = 2.0580 (Btu/(hr.ft.R)-1

I dont get this at all, I get 0.01338 - obviously I've "

Dissecting
ln(d/r4)=2.050 this is true also in metric units
k=12 btu-inch/hr-ft^2-degF
but inch/ft= 1/12
so K becomes
12/12=1 BTU/hr ft deg F and therefore
ln(d/r4)/K=2.050/1.0=2.050 (Btu/hr-ft deg)^-1

Hope this nor more confusing to you

 

[IRstuff]

not to beat a dead horse, but this is where a program like Mathcad, Studyworks, or SMath Studio shines. S

Sorry about the inverted units; I was too lazy to reformat the equations into the normal format, and I was too obtuse to tack the inversion on the LHS:

TTFN
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[bencpv]

Thank you once again,

Well, I'm embarrassed to say that I'm even more confused now!

I don't think I should have used the imperial units, BTU etc to start with. Having been taught everything in metric, the imperial world baffles me!

I find myself spending 'company' time in trying to get to grips with the example when I'll be using metric data anyway....

If I have the following:
Insulation: Thermal conductivity, k[sub]i[/sub]= 0.024 W/m°C (PUR foam)
Casing pipe: Thermal conductivity, k[sub]c[/sub]= 0.52 W/m°C (HDPE)
Carrier pipe: Thermal conductivity, k[sub]pp[/sub]= 0.1 W/m°C (polypropylene)
Soil: Thermal conductivity, k[sub]s[/sub]= 0.2 W/m°C

Carrier ID = 90mm
Carrier OD = 110mm
Casing OD = 200mm
Casing thickness = 40.1mm

Max temp = 140°C
Min temp = 100°C

Buried depth = 1m

From using the the simplified formula from IRstuff, I get a heat loss value of 17.22 W/m.

Can anyone spare a couple of minutes and see if they reach the same conclusion?

Thanks

 

[IRstuff]

Unclear whether your temperatures were the equivalent of the others, so I ran it two ways:

TTFN
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[bencpv]

i have just run he calculation again and am now faced with another result! I now get 6.080 W/m

I have broken down the calculation spreadsheet and attach it. I hope it easy to follow. Can you see where i've gone wrong?

Please tell me what you meant by: 'Unclear whether your temperatures were the equivalent of the others, so I ran it two ways:'

can you tell me why you have Q[sub]L[/sub](T[sub]1[/sub].22K)? where does 22K come from?

Help would be much appreciated.

 

[zekeman]

I have problems verifying irstuff results.
One thing , the OP sizes are diameters and the last term would be incorrect.
Beyond this, I estimated the bracketed term about 37
which gives me about
Q=6.6 w/m
Could we differ by that much?

 

[ione]

Ben,

You've entered a "2" in your cell E8 wich shouldn't be there (but this could be done to the buried depth's definition). This time I'm with zekeman as I've got 6.63 W/m

 

[IRstuff]

I used 22ºC in addition to the 100ºC because I wasn't positive that the 100ºC given was the "soil" temperature, since that's the boiling point of water. The answer for the 100ºC case was 7.28 W/m, since I had left it in English units.

The order in which the OP gave the new parameters is not consistent with with the textbook form... I added the 40.1mm to the casing radius, but I see now that it's the OD, so the 40.1 should be subtracted, which results in 9.6 W/m for T2=100ºC

TTFN
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[bencpv]

sorry for not being in touch sooner.

once again I'd like to express my gratitude for everyone's work. I've just been mailed an example calc which is used by our suppliers of steel pipe. When I work through the first example (p.1-3 which uses PUR insulation, PEX carrier and PE casing) I get the same results, so I am happy (i think). Example attached.

does this match your calculations? I'm not sure how exact the results we will ultimately use have to be, but i'm pretty sure a ball-park figure will do (+/-0.5 W/m).

Is this method good enough?

On sheets 4 and 5, another example is given but this time the thermal conductivity of the carrier pipe and casing has not been considered. I'm not clear why this is, i've contacted the source but have yet to hear a response (and don't think i will). I thought that perhaps the pipe was using a steel carrier - but the part number tells me otherwise.

the last 2 sheets (in polish) - i have no idea why these have been given!

any help would be gratefully received.

B

 

[IRstuff]

note that in the first example, the Rex term is less than 1% of the insulation resistance.

TTFN
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[bencpv]

thanks IRstuff,
note that in the first example, the Rex term is less than 1% of the insulation resistance??????