Pipe Heat Loss Calculations

[skippybangkok]

My head is spinning - anyone good at heat transfer calculations... ?

 

Home work is

 

tube / pipe 0.45m diameter, and length 1.5m, wall thickness 12mm

Hot air 170 Deg C blows through 8m/s

Stainless Steel

 

How many Watt ( What ) can transfer through the pipe 

 

[eisfeld]

Q = 2 * PI * (coefficient of material) * (pipe length) * (temp difference) / (ln(outer radius / inner radius))

 

The radii you can easily calculate from the diameter. Temp difference as well assuming room temp. Then just look up the coefficient for stainless steel and plug it into the calculator.

[GammaGlobulin]

 

 

 

You will find a very thorough discussion on this good site:  https://engineerexcel.com/heat-losses-from-pipes/

 

 

Edited August 24, 2023 by GammaGlobulin

[setbkk]

Inlet temp is 170. Do you know outlet temp? If so very easy: rate of heat transfer is mass flow rate of air X heat capacity of air X temp difference.

[GammaGlobulin]

24 minutes ago, setbkk said:

Inlet temp is 170. Do you know outlet temp? If so very easy: rate of heat transfer is mass flow rate of air X heat capacity of air X temp difference.

Making that simple calculation would NOT generalize to all conditions if for example you tested when the ambient temp was high, one day, your determination would NOT apply to heat loss when ambient temp was colder. 

 

Please refer to the good site which I linked in my previous comment. 

 

 

[eisfeld]

1 minute ago, GammaGlobulin said:

Making that simple calculation would NOT generalize to all conditions if for example you tested when the ambient temp was high, one day, your determination would NOT apply to heat loss when ambient temp was colder. 

 

Please refer to the good site which I linked in my previous comment. 

 

 

No, setbkk's approach (if OP had the outlet temp) would work regardless of the ambient temp. It's very simple: Power Inlet = Power Outlet + Power Radiated. If you know the inlet temp + flowrate you have inlet power. Similar for the outlet power and so you can easily calculate the radiated power.

 

If you measure on a cold day instead of a hot day you will simply see a different outlet temp.

[GammaGlobulin]

4 minutes ago, eisfeld said:

No, setbkk's approach (if OP had the outlet temp) would work regardless of the ambient temp. It's very simple: Power Inlet = Power Outlet + Power Radiated. If you know the inlet temp + flowrate you have inlet power. Similar for the outlet power and so you can easily calculate the radiated power.

 

If you measure on a cold day instead of a hot day you will simply see a different outlet temp.

Sorry, but wrong, I think.

 

Example:

 

Inlet Temp = 170 Degrees C

Ambient Temp (around pipe) = 170 Degrees C

Outlet Temp = approx 170 Degrees C   (maybe slightly lower due to radiated heat from pipe)

 

So: You say you could calculate heat loss from the steel pipe in all ambient-air conditions using the approach you approve of?

 

Simply WRONG.

Because...You canNOT.

 

Correct?

 

 

[eisfeld]

Just now, GammaGlobulin said:

Sorry, but wrong, I think.

 

Example:

 

Inlet Temp = 170 Degrees C

Ambient Temp (around pipe) = 170 Degrees C

Outlet Temp = approx 170 Degrees C   (maybe slightly lower due to radiated heat from pipe)

 

So: You say you could calculate heat loss from the steel pipe in all ambient-air conditions using the approach you approve of?

 

Simply WRONG.

Because...You canNOT.

 

Correct?

 

 

Simple. Heat loss would be nada in your example (maybe slightly not nada if you actually specified an outlet temp that is not equal to inlet temp).

[GammaGlobulin]

6 minutes ago, eisfeld said:

Simple. Heat loss would be nada in your example (maybe slightly not nada if you actually specified an outlet temp that is not equal to inlet temp).

This is my point.

Do you not understand it?

 

If you test in conditions where the ambient air around the pipe matches the temp inside the pipe, then there would be zero heat loss (except for the radiative heat loss from the steel pipe).

 

Therefore, such a calculation based on the "approach" suggested earlier would be MEANINGLESS for practical purposes of calculating convective and radiative heat loss when the ambient air is at different temperatures, say zero degrees, 50 degrees, etc.  (Do you not GET this simple concept?!!!)

 

IF the OP had said that the AMBIENT AIR TEMP will ALWAYS remain the same, then, the approach mentioned earlier could give an APPROXIMATION, but ONLY based on measurement of outlet and inlet temps, and ONLY based on the accuracy of the temperature sensor.

 

In addition, the radiative loss from the steel pipe would DEFINITELY depend upon the COLOR of the steel pipe, since we all know about BLACK BODY radiation, do we not?  (I.E. a black pipe will radiate MORE HEAT than a SILVER PIPE.  NOW...do you GOT IT????)

 

Use the formulae in the Engineering Website that I linked...

Simple!

 

 

 

[GammaGlobulin]

31 minutes ago, eisfeld said:

Simple. Heat loss would be nada in your example (maybe slightly not nada if you actually specified an outlet temp that is not equal to inlet temp).

Just use Physics.Stackexchange IF you don't believe me...   https://physics.stackexchange.com/questions/91792/realistic-calculation-of-heat-loss-for-pipe

 

This is NOT an engineering problem, per se.

This is a simple Physics problem.

 

 

But, STILL, I already linked, previously, a very helpful link specifically focused on heat loss from a pipe, which will generalize to all environmental conditions.  (INCLUDING the radiative heat loss due to infrared radiation from the pipe, and the color of the pipe, by taking into account blackbody radiation, which IS mentioned on this site, and which I DID mention before in one of my previous posts, too.)

 

So then...what's the problem?

 

 

Edited August 24, 2023 by GammaGlobulin

[setbkk]

Whether a physics or engineering issue, conservation of mass and energy still apply which is what my basis is. Also elimates need to try to estimate HT film coefficients and blackbody radiation.

[GammaGlobulin]

37 minutes ago, setbkk said:

Whether a physics or engineering issue, conservation of mass and energy still apply which is what my basis is. Also elimates need to try to estimate HT film coefficients and blackbody radiation.

Sure, and so then you are suggesting that the user create a table based on temperature measurements for all possible ambient temperatures which may be encountered in the environment in question?

 

If so, there are tables which are already prepared available on the net, I guess.

 

But I thought that the question was "how to calculate"....

 

 

[GammaGlobulin]

1 hour ago, setbkk said:

Whether a physics or engineering issue, conservation of mass and energy still apply which is what my basis is. Also elimates need to try to estimate HT film coefficients and blackbody radiation.

What you FAIL to COMPREHEND, so strangely, is that your approach FAILS miserably.... 

 

In the case where the system is in the DESIGN phase, which is crucial. 

 

Probably EVEN YOU will not be able to to take temperature measurements on a pipe that has not yet been built. 

 

Or.... Can you? 

 

 

[eisfeld]

 

3 hours ago, GammaGlobulin said:

This is my point.

Do you not understand it?

No. I really don't see your point. Your example was nonsensical.

 

3 hours ago, GammaGlobulin said:

If you test in conditions where the ambient air around the pipe matches the temp inside the pipe, then there would be zero heat loss (except for the radiative heat loss from the steel pipe).

If your ambient temp is the same as inside the pipe then of course there wont be any real heat loss. It's an absurd example.

 

3 hours ago, GammaGlobulin said:

Therefore, such a calculation based on the "approach" suggested earlier would be MEANINGLESS for practical purposes of calculating convective and radiative heat loss when the ambient air is at different temperatures, say zero degrees, 50 degrees, etc.  (Do you not GET this simple concept?!!!)

What you are missing is the simple fact that setbkk suggested a very simple way to calculate the heat loss by utilizing the conservation of energy law. If you know how X amount of energy goes in and Y (Y being smaller than X) comes out then the difference is your heat loss due to radiation. It's simply an easier way to get the answer IF one knows in and outlet temp, velocity and material going through (air). Are you honestly disputing this super simple physics fact?

 

3 hours ago, GammaGlobulin said:

IF the OP had said that the AMBIENT AIR TEMP will ALWAYS remain the same, then, the approach mentioned earlier could give an APPROXIMATION, but ONLY based on measurement of outlet and inlet temps, and ONLY based on the accuracy of the temperature sensor.

Are you really coming up with details like sensor accuracy? It's a homework question! I could as well question your approach by saying that your formula is not taking into account changes in material density and surface areas when temperature changes. What if there is wind blowing over the pipe? What if it gets so hot that it melts? See you can come up will all kinds of contrived nonsense details that make things more complicated then the situation at hand demands.

 

3 hours ago, GammaGlobulin said:

In addition, the radiative loss from the steel pipe would DEFINITELY depend upon the COLOR of the steel pipe, since we all know about BLACK BODY radiation, do we not?  (I.E. a black pipe will radiate MORE HEAT than a SILVER PIPE.  NOW...do you GOT IT????)

Where is color in your equation? Your rude attitude is totally misplaced.

 

1 hour ago, GammaGlobulin said:

In the case where the system is in the DESIGN phase, which is crucial. 

 

Probably EVEN YOU will not be able to to take temperature measurements on a pipe that has not yet been built. 

 

Or.... Can you? 

What does this have to do with anything in this thread? The OP is asking about a simple homework assignment of a very simple standard physics or engineering task. And here you are argueing about blackbody radiation and pissing on peoples legs. Lots of pseudo intellectual blabla with a rude tone that misses the point.

[GammaGlobulin]

4 minutes ago, eisfeld said:

then the difference is your heat loss due to radiation.

What about heat loss due to convection? 

 

Or, don't you worry about discriminateing between the two? 

 

 

[eisfeld]

1 minute ago, GammaGlobulin said:

What about heat loss due to convection? 

 

Or, don't you worry about discriminateing between the two? 

Are you trolling? Heat loss is heat loss. OPs homework question wants to know how many Watt. Power is power. Energy is energy. The method by which it is taken away does not change the amount. Radiation. convection... maybe it's being cooled through lasers? Whatever it is, it does not matter. That's the beauty of setbkk's approach. It lets you ignore all factors like material of the pipe, thinkness, color or if a fluid flows over it because it just looks how much goes in, how much comes out and the rest is your uncle, sorry is the heat loss as per question.

[GammaGlobulin]

10 minutes ago, eisfeld said:

 

No. I really don't see your point. Your example was nonsensical.

 

If your ambient temp is the same as inside the pipe then of course there wont be any real heat loss. It's an absurd example.

 

What you are missing is the simple fact that setbkk suggested a very simple way to calculate the heat loss by utilizing the conservation of energy law. If you know how X amount of energy goes in and Y (Y being smaller than X) comes out then the difference is your heat loss due to radiation. It's simply an easier way to get the answer IF one knows in and outlet temp, velocity and material going through (air). Are you honestly disputing this super simple physics fact?

 

Are you really coming up with details like sensor accuracy? It's a homework question! I could as well question your approach by saying that your formula is not taking into account changes in material density and surface areas when temperature changes. What if there is wind blowing over the pipe? What if it gets so hot that it melts? See you can come up will all kinds of contrived nonsense details that make things more complicated then the situation at hand demands.

 

Where is color in your equation? Your rude attitude is totally misplaced.

 

What does this have to do with anything in this thread? The OP is asking about a simple homework assignment of a very simple standard physics or engineering task. And here you are argueing about blackbody radiation and pissing on peoples legs. Lots of pseudo intellectual blabla with a rude tone that misses the point.

The POINT is answering the OP based on Physics. 

 

And, your approach of taking temperature measurements from a pipe that is still in the design phase, not yet built, is IMPOSSIBLE. 

 

But, you don't get it. 

 

And, don't think your mention of elementary basics such as conservation of mass, energy, and momentum will help you to sound more credible in this case. 

 

 

[eisfeld]

3 minutes ago, GammaGlobulin said:

The POINT is answering the OP based on Physics. 

 

And, your approach of taking temperature measurements from a pipe that is still in the design phase, not yet built, is IMPOSSIBLE. 

 

But, you don't get it. 

 

And, don't think your mention of elementary basics such as conservation of mass, energy, and momentum will help you to sound more credible in this case. 

What are you on about? Who is talking about design phase of a pipe? Man it's a school physics question. You are way overcomplicating things.

 

You claim people don't get things but you bring zero arguments of why the simple approach via conservation of energy (not mass or momentum) cannot be used. Please, just ONE argument. It's not about sounding credible. It's about basic physics. If you claim the approach is wrong then please state why.

[GammaGlobulin]

OK...

LET'S BE   CRYSTAL CLEAR HERE...

 

If this is a homework assignment...

 

There is NO MENTION of knowing the outlet temperature.

There is only mention of knowing the inlet temperature.

 

Therefore, since the HOMEWORK QUESTION does NOT supply the OUTLET Temp...

 

You guys CANNOT use this NONEXISTING DATA to CALCULATE the answer and Complete the Homework Assignment.

 

How many guys here have studied Physics at University?

 

My answer would be .... e) None of the above

 

This is NOT a troll post as some guy asked in a comment above.

 

This boggles the mind....

Sorry.

 

 

[eisfeld]

16 minutes ago, GammaGlobulin said:

OK...

LET'S BE   CRYSTAL CLEAR HERE...

 

If this is a homework assignment...

 

There is NO MENTION of knowing the outlet temperature.

There is only mention of knowing the inlet temperature.

 

Therefore, since the HOMEWORK QUESTION does NOT supply the OUTLET Temp...

 

You guys CANNOT use this NONEXISTING DATA to CALCULATE the answer and Complete the Homework Assignment.

 

How many guys here have studied Physics at University?

 

My answer would be .... e) None of the above

 

This is NOT a troll post as some guy asked in a comment above.

 

This boggles the mind....

Sorry.

 

 

Wait until you realize that it also doesn't state the ambient temp and so none of the formulas anyone - including you - posted can be actually applied to get the heat loss in Watt. And what about your formula for convective heat loss? What fluid are we talking here? Oh and what's the flow rate? Hm yea, we don't seem to know, do we. Let me put it like this: You guy CANNOT use this NONEXISTING DATA to CALCULATE the answer and Complete the Homework Assignment. This conversation has boggled the mind indeed.

Edited August 25, 2023 by eisfeld

[GammaGlobulin]

31 minutes ago, eisfeld said:

Wait until you realize that it also doesn't state the ambient temp and so none of the formulas anyone - including you - posted can be actually applied to get the heat loss in Watt. And what about your formula for convective heat loss? What fluid are we talking here? Oh and what's the flow rate? Hm yea, we don't seem to know, do we. Let me put it like this: You guy CANNOT use this NONEXISTING DATA to CALCULATE the answer and Complete the Homework Assignment. This conversation has boggled the mind indeed.

OK...

NOW I agree with you.

There isn't enough information in the "homework question" to be able to arrive at any answer.

 

However, It is still impossible to measure the outlet temp of a pipe that does not yet exist. 

And, the formulae of MOST value to calculate heat transfer (total including due to radiation and convection) are clearly explained in the website I linked initially, And further, this website should be of most interest to the student, or whomever is interested in these types of calculations.

 

Do we design first and then build?

Or, do we build first, using any old random materials and dimensions, and hope for the best?

 

Should we just build first, and then make revisions later, based on outlet temperature measurements taken after the fact?

Doing so surely would boggle the mind.

 

 

[GammaGlobulin]

1 hour ago, eisfeld said:

Wait until you realize that it also doesn't state the ambient temp and so none of the formulas anyone - including you - posted can be actually applied to get the heat loss in Watt. And what about your formula for convective heat loss? What fluid are we talking here? Oh and what's the flow rate? Hm yea, we don't seem to know, do we. Let me put it like this: You guy CANNOT use this NONEXISTING DATA to CALCULATE the answer and Complete the Homework Assignment. This conversation has boggled the mind indeed.

I actually think this could be a nice little homework problem if the ambient temp was known, and if the fluid type and pressure was known, etc...

 

Because, if the inlet temp was known, then one would need to calculate heat transfer through the wall of the pipe, as the fluid travelled along the pipe at 8 meters/sec. Which would necessitate calculating the instantaneous change in temperature of the fluid as it cooled while passing through the pipe, as this would also cause the instantaneous heat transfer through the wall of the pipe to change. So, sounds like a calculus problem to me....

 

 

[GammaGlobulin]

In order to do this homework question justice, then it would really be necessary to apply principles of Fluid Dynamics to determine variables such as turbulence of the fluid in the pipe, which would affect heat transfer through the wall of the pipe.

[skippybangkok]

10 hours ago, GammaGlobulin said:

 

 

 

You will find a very thorough discussion on this good site:  https://engineerexcel.com/heat-losses-from-pipes/

 

 

Thanks  - been on that web site and a gazillion others. the trick is the Hconv constant

 

Hot air, 170 Deg C and 8 m/s

Edited August 25, 2023 by skippybangkok

[skippybangkok]

9 hours ago, eisfeld said:

No, setbkk's approach (if OP had the outlet temp) would work regardless of the ambient temp. It's very simple: Power Inlet = Power Outlet + Power Radiated. If you know the inlet temp + flowrate you have inlet power. Similar for the outlet power and so you can easily calculate the radiated power.

 

If you measure on a cold day instead of a hot day you will simply see a different outlet temp.

Thanks.... the question is not what's the exhaust temp of the air, rather how much heat loss would go through the pipe - realise they are both linked 

[skippybangkok]

7 hours ago, GammaGlobulin said:

Sure, and so then you are suggesting that the user create a table based on temperature measurements for all possible ambient temperatures which may be encountered in the environment in question?

 

If so, there are tables which are already prepared available on the net, I guess.

 

But I thought that the question was "how to calculate"....

 

 

Thanks... seen that one too.....  but this does not cover 17" pipe, stainless steel, nor speed of air nor thickness of pipe 

[skippybangkok]

4 hours ago, eisfeld said:

Wait until you realize that it also doesn't state the ambient temp and so none of the formulas anyone - including you - posted can be actually applied to get the heat loss in Watt. And what about your formula for convective heat loss? What fluid are we talking here? Oh and what's the flow rate? Hm yea, we don't seem to know, do we. Let me put it like this: You guy CANNOT use this NONEXISTING DATA to CALCULATE the answer and Complete the Homework Assignment. This conversation has boggled the mind indeed.

ok ...30 Deg C

[skippybangkok]

Probably best I can find is this

 

https://calcdevice.com/forced-convection-of-inner-pipe-surface-id130.html

 

 

[GammaGlobulin]

4 hours ago, skippybangkok said:

Thanks.... the question is not what's the exhaust temp of the air, rather how much heat loss would go through the pipe - realise they are both linked 

YES!

The guys on this thread just DO NOT GET IT!

 

If we are talking about some sort of homework problem for either an engineering school or for some Physics student, at an advanced level, and worthy of consideration, there is NO WAY in the WORLD that some REAL WORLD engineer is going to face the following scenario:

 

a. We are building a building with a steam conduit

b. We need to know the heat transfer of the steel pipe so that we do not too much heat up the room through which the pipe passes.

c. But, don't worry, because we know you know nothing about Physics.

d. So, we will build the building and steam conduit first.

e. After completion, we will send you up to the 28th floor so that you can take your itsy-bitsy thermometer up and measure the outlet temp from the steel conduit, and then use what little Arithmetic you might know to do your multiplications.

 

Outlet Temp is NOT part of this homework exercise if it's to be for either an Engineering school or Physics class.

 

Also, what's with all this emphasis on.... CONSERVATION OF MASS, CONSERVATION OF ENERGY, CONSERVATION OF MOMENTUM, etc? !!!

It's like we are back in 7th-grade science class, or something.

 

YES, of course, we are speaking of Classical Physics, and so OBVIOUSLY we know these things are conserved!  It's not as if we will be converting mass into energy in this steam conduit, or whatever!

 

And why do these guys insist upon turning this homework problem into one of taking measurements? They are not talking about calculating. They are talking about MEASURING!!!  What good is that, in this case?

 

Therefore, if someone is learning about how to do calculations of heat transfer through a wall, a cylindrical wall, etc., they should FIRST look at the site https://engineerexcel.com/heat-losses-from-pipes/  and learn the basics.

 

Then, they can post some intelligent questions on https://physics.stackexchange.com/

 

 

But, please, when posting your questions on the physics.stackeschange, do have your questions well conceived.  And, BE POLITE and RESPECTFUL... And do what I say, and not as I do...

 

 

Edited August 25, 2023 by GammaGlobulin

[GammaGlobulin]

4 hours ago, skippybangkok said:

Thanks  - been on that web site and a gazillion others. the trick is the Hconv constant

 

Hot air, 170 Deg C and 8 m/s

OK...then...

Would it not be good to carefully formulate you questions, and then post them on https://physics.stackexchange.com/

 

You will get better feedback and learn more by doing so; there is little doubt.

 

Homework is for LEARNING.

It's not for cheating yourself by finding shortcuts.

Learning is Good.

Shortcuts do not work if one is trying to learn.