soil heat transfer/temperature change air drain

[xiloergos]

Hello everyone. I'm new here. Really cool website. I'm just a carpenter, don't ask too much of me! = )

I'm designing a geothermal system for a possible greenhouse I'll build, but I don't have much money, so I need to make calculations very precisely.

I find myself in need of help, to figure out what the temperature change in air flowing through a corrugated pipe underground would be.

I have this information:

Air: Air will be entering the pipe at 10 degrees Fahrenheit.
Pipe: High density Polyethylene corrugated (solid - without any piercing or knifing) with a thermal conductivity coefficient of 0.5 W/m^K (temperature of the pipe is negligible - meaning it's not insulated or heated)
Surface area of the pipe: The pipe is 100' long with a diameter of 4" (so I believe the surface area is 2.777778 square feet - contact with the ground surrounding it)
Flow: The air is moved by a 270 CFM fan (so I believe traveling at 270 cubic feet per minute)
Surface: The surface surrounding the pipe is soil (I'm going to assume the soil will have a thermal conductivity coefficient of 1.5 W/m^K) and the temperature of the soil is permanently 45 degrees Fahrenheit.

I have attached a very simple diagram with most of this information.

I need to know at what temperature will the air come out?

I really tried to use my head for this, and even tried the chatgpt thingy online, but I think I need the help of real engineers!

Thank you in advance!

Xilo

 

[1503-44]

ChatGPT is not an engineering genius. It has difficulty doing simple addition.

This is what you need to know.

https://myengineeringtools.com/Thermodynamics/Heat_Conduction_Pipe.html

Let us know if you have trouble with that.


--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[xiloergos]

Hello again! Thanks you so much for replying so fast! And for the handy link! I need more formulas I think.


So I've been doing some math. This is what I've produced so far:




Φ = Q/A = = (Ti-To)/((Do/2λ)*ln(Do/Di))

Once I inserted my numbers:

Φ = Q/A = = (-10°c - 7°c) / ( ( 0.1016m / 2 * 0.50 W/m/K ) * ln ( 0.1016m / 0.0976m) )

Φ = -4165.778539739 W/m2.°c

That’s the heat flux in W/m2.


Then:

Q = Φ*A = Φ * (π.Do.L)

Q = 73.30 W/m2.°c * ( π * 0.1016m * 30.48m )

Q = 713 kW

That’s the heat transferred in watts.


Then:

U = 1/R = 1/((Do/2λ)*ln(Do/Di))
U = 1 / R = 1 / ( ( Do/2λ ) * ln( Do/Di ) )
U = 1 / R = 1 / ( ( 0.1016m / 2 * 0.5 W/m/K ) * ln( 0.1016m / 0.0976m ) )
U = 245.04579645524 W/m2.°c
That’s the overall heat transfer coefficient in W/m2.°c

Then:
Φ = Q/A = U.ΔT
ΔT = Φ / U
ΔT = -4165.778539739 W/m2.°c / 245.04579645524 W/m2.°c
ΔT = - 16.97145394053 °c (close enough, I’m probably missing decimals)
Which corroborates the previous numbers are correct, I think = )

But now I seem to be missing formulas or a way to proceed. Because, I have the fan propelling the air through the pipe, the fan being a 270 CFM capacity fan, I will assume 100% reliability, meaning it will propel 270 square feet per minute. That figure converted to the meter unit of measurement is 7.645548689991626 CMM (cubic meter minute).

Pipe length in m = 30.48m
Pipe radius in m = 0.0498m

The volume of the pipe is:
V = h * π * r²
V = 30.48m * π * (0.0498m)²
V = 23.747807555167 m3




So how do I actually figure out what the temperature of the air will be when coming out of the pipe? What formula should I apply? There should be a formula that would involve the application of the heat transferred over the time of interaction and volume? Or something like this?

Thank you so much!

Xilo

 

[1503-44]

OK. You seem to be running off the rails a bit.
It's 12pm here so I'll get bake with you tomorrow.


--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[xiloergos]

Thank you so much mate! I really appreciate your help! Are you in the UK? I'm Eastern Time in the US, but until last year was living in Norway = )

Thank you! What's your name?

Alejandro

 

[1503-44]

I lived in London in 9/2013-12/2015.
I'm somewhere in the Atlantic in the GMT zone.

I'm doing a spreadsheet on it.
Do you have MS Excel? Or maybe I can convert it to Google spreadsheet

--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[xiloergos]

Excellent thanks mate, I have Excel, that will work. Much appreciated!

 

[1503-44]

I've been all over Norway. Even up to Böda, Hammerfest and Kirkenes.
Beautiful, but COLD! 0 in July.

--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[xiloergos]

Huh yes, I was up there in Tromso for two years. US is home now.

 

[1503-44]

Still working on it. I forgot how difficult these are to solve when the heat flux is not constant. It drops as the air temperature in the pipe rises. I should have just loaded it up in the simulator, but no, I chose XL. Anyway its a good review heat transfer for me. Hope you are not in too much of a hurry.

--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[xiloergos]

Oh sir I'm in no rush at all, I like looking ahead of my projects, run all calculations, budget, re budget and budget again! Measure twice cut once. So take your time! It's a pleasure having your assistance! Once you are done, I might have a few questions on your calculations, because I like understanding things. I'm really grateful for your efforts!

 

[CarlB]

The attached figure may be of assistance.
The approach is designed for temperature change of water, but seems the equations should be valid for air flow, just use properties for air and correct units...

 

[xiloergos]

Thank you very much CarlB! All help is welcomed! I'll have a look, cheers!

 

[1503-44]

Finally found the easy way to do it buried deep in one of my thermo texts.
Simplified a bit by assuming that the pipe inner wall is what is being maintained at 45F.

I will try to check that solution (pulling the heat txf value from the chart) against some other specific formulas. In any case, your pipe length seems to be longer than you need to reach the 45° soil temperature and will probably have some safety factor in actually getting there. I solved for lengths of pipe needed to get to various temperatures up to 44.9°. You will see that you can never actually get there, just very, very close to it.

Hopefully someone will check the spreadsheet. I included pictures of the Example problem I followed, to make that as easy as possible. All feedback welcome.

Link to the XLSX file in Dropbox....

https://www.dropbox.com/scl/fi/lmum...ipe.xlsx?rlkey=n6t2s209w681yhvlp8nu501va&dl=0

--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[xiloergos]

Hello again sir, I followed the link and it's the same excel sheet as before, I still don't see how to calculate the end temperature using that sheet. What am I missing? Is it the same excel sheet? Thanks!

 

[xiloergos]

OK now I've downloaded it! Thanks I'm having a look at it! Wow! Impressive work! Let me have a look and I'll get back!

THANK YOU KINDLY!

 

[xiloergos]

Ok first of all, THANK YOU and more THANK YOU for your help and hard work! This is impressive to me!

I have a few questions (which will seem very dumb to you) but if it's not too much trouble, because I love understanding things:

1.

The formula for calculating 'Pipe flow area - Ax' in cell F17 shows =PI()*F4^2/4/144.
I understand the "=PI()*F4^2" to get the area, but "/4/144". What does this mean?

2.

The formula for Velocity in seconds in cell F22 shows =ROUND(F21/60,3)
Why the ',3' is that an Excel thing?

2.

"h" I'm assuming is Heat transfer coefficient of air? Your note says "h" is obtained from Figure1, but when I expand the pictures attached I can't quite see what's written on them.

3.

Cell F30 has the following formula =$F23*PI()*F4/12*F29 to calculate 'h*p*d*(Ts-Tm)' what is the PI()*F4/12 part of the formula? Area of the pipe/cross section?

4.

Where does the formula for pressure drop in cell E41, Dp = 4 * U h L /d /cp come from? One of the books? And what does the "U" stand for? Or is it a "V"? Because in the formula in the cell next to it, the reference for "U" appears to be to cell F22, Velocity (in seconds).

5.

The formula in cell F41, has a /(1/12) , why is that the value of "d"?
Same formula in the same cell divides by 3600, I assume that is to make it "per second"?
Same formula in the same cell also divides by 32.174, but in it's values reference in the cell next to it E41, there's no mention to what this is.

6.

In the main calculations table (Column F) with an assumed 'To' of 44.9°F, the calculated 'L' ends up having a value of 51.16. But in the final calculation in Column 'P', the assumed 'To' has a value of 44°F (really close to 44.9°F) and the 'L' value is 872.16 linear feet. How is this possible?


Again I'm sorry if this questions are too simple or don't even make sense to you!

My understanding of the final results (by the way cool and useful graph!) is that I would need to bury the pipe around 900 feet in length to achieve exit air close to 45˚F. Which is challenging but not impossible. But I could run them only 150 feet, and still achieve 40˚F which is a lot easier, cheaper and more efficient right? It seems like the efficiency of the system sort of finds a plateau at around 150 feet anyway correct?


Again thanks, and I'm enjoying this a lot!

 

[1503-44]

Ok first of all, THANK YOU and more THANK YOU for your help and hard work! This is impressive to me!

I have a few questions (which will seem very dumb to you) but if it's not too much trouble, because I love understanding things:

1.

The formula for calculating 'Pipe flow area - Ax' in cell F17 shows =PI()*F4^2/4/144.
I understand the "=PI()*F4^2" to get the area, but "/4/144". What does this mean?

[..... Yes, just the area of a circle A= [π]/4 * d^2 the /144 converts in2 to ft2. Excel orders it's calculations by squaring all terms right to left first, multiplies and divides from left to right and lastly adds or subtracts from left to right, so first it does the d^2, then multiplies that by [π] then divides by 4. Last divides by 144 in2/ft2 to get the answer in ft2
....]


2.

The formula for Velocity in seconds in cell F22 shows =ROUND(F21/60,3)
Why the ',3' is that an Excel thing?
[. Round(the value in cell, to 3 places)
Just trims the long string of 12 something digits to fit neatly in the box.]
2.

"h" I'm assuming is Heat transfer coefficient of air? Your note says "h" is obtained from Figure1, but when I expand the pictures attached I can't quite see what's written on them.
[..... yes pulls the heat transfer coefficient =11 from the chart.
I'll take a better picture..]

3.

Cell F30 has the following formula =$F23*PI()*F4/12*F29 to calculate 'h*p*d*(Ts-Tm)' what is the PI()*F4/12 part of the formula? Area of the pipe/cross section?
[..... [π] * Diameter = cylinder circumference. Its in inches, so /12 for feet. .....]
4.

Where does the formula for pressure drop in cell E41, Dp = 4 * U h L /d /cp come from? One of the books? And what does the "U" stand for? Or is it a "V"? Because in the formula in the cell next to it, the reference for "U" appears to be to cell F22, Velocity (in seconds).

[..... yes, a magical formula. They used U for Velocity. I prefer V. .....]
5.

The formula in cell F41, has a /(1/12) , why is that the value of "d"?
[..... dividing by the conversion factor d is the 4inch diameter. 1ft/12in to result in d being feet.....]
Same formula in the same cell divides by 3600, I assume that is to make it "per second"?
Same formula in the same cell also divides by 32.174, but in it's values reference in the cell next to it E41, there's no mention to what this is.
[..... Oh that's the acceleration of gravity. Needed to convert lbs, a force, to mass, or visa versa .....]

6.

In the main calculations table (Column F) with an assumed 'To' of 44.9°F, the calculated 'L' ends up having a value of 51.16. But in the final calculation in Column 'P', the assumed 'To' has a value of 44°F (really close to 44.9°F) and the 'L' value is 872.16 linear feet. How is this possible?
[..... I'll have a closer look at that one .....]

Again I'm sorry if this questions are too simple or don't even make sense to you!

My understanding of the final results (by the way cool and useful graph!) is that I would need to bury the pipe around 900 feet in length to achieve exit air close to 45˚F. Which is challenging but not impossible. But I could run them only 150 feet, and still achieve 40˚F which is a lot easier, cheaper and more efficient right? It seems like the efficiency of the system sort of finds a plateau at around 150 feet anyway correct?

[..... CORRECT! That was a test. GOOD THINKING! By plotting the graph, I was trying to give you exactly the idea that you speak of. Almost all of the heat transfer occurs in the first half of the pipe, so any additional heating comes with a lot of pipe cost and backbreaking work installing the remaining length of pipe, but at a very small gain in heat. You could go 10 miles and get to 44.999°, it never reaches 45, but why. You can reach 42, or 40° with a lot less effort and cost. Pick a temperature that you can live with and... save your back and pocketbook! ]

Again thanks, and I'm enjoying this a lot!

[. Welcome to the "Imperial system of units".
I have written my own XL AddOn to do conversions that adds many additional units to the built in XL conversion function. Even converting Imperial units into other Imperial units. XL does not know about J/s-m2, Btu/s = Horsepower etc, etc. As I have one foot in the US and another in EU, so its a great help writing these thermo spreadsheets. Its almost done, except I keep adding new conversions to it all the time. I just upgraded my computer for playing with AI programs and have not copied it to my XL installation here just yet. I can see that I need to do that today!

If you click on the blue 1503-44 just at the top of my responses , it will take you to my profile where you will find my e-mail address in the Google drive link. Copy it and paste it into your browser search bar. We can talk there about other things. ]

--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[1503-44]

Better picture? Just download it from this image or from the attached file.
Don't know why it's so dark. But it's readable.

Select your pipe diameter 4
Go up to hit the curve for your U = flow velocity 52
Now Left to get the h[sub]c[/sub] value. 11

Some of the simplifying assumptions
270 is a flow volume at the avg temp of Tm and atmospheric pressure, It may be at 70F, in which case actual flow velocity at low temperatures will be somewhat less. If 270 has a given reference temperature, T_ref, multiply 270 By (459+T_m)/(459 + T_ref) to correct it to the avg flow rate at T_m.

It is the pipe inside diameter surface that I assumed was kept at 45F.
The pipe insulation effect will reduce that a bit. I'll try to have a look at that.
I don't think it will be much of a reduction, but don't know for sure yet.

It might look tempting to use smaller pipe and get a high hc value, but your working pressure will increase. No free lunch. Need to keep as close to atmospheric pressure as possible, otherwise a lot of air property values, volumes and velocities will change..

--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[xiloergos]

Thank you very much for all your efforts!

That picture is much better. Thank you for replying to my questions, everything but point 6 is super clear now = )

It's so wonderful understanding how things work, and you do a great job at it, did you consider teaching when you retired?

I tried to find your email address in the shared profile, but I believe it's not there. If you can double check.

Can I ask you, if I had to have certain books of physics, that I could find tables like the above, and perhaps practical exercises, that are immediately applicable to real world scenarios like this one, which books do you think I should have?

Again, many thanks!