Make a power balance: the input power of the sun - losses via radiation - losses via convection to the air = power going via conduction to the fluid which is equal to the power via convection from the internal wall to the fluid.

The result of the balance equation will be the outer wall temperature.

The wall temperature will be function of :

- coefficient of absorption of the outer surface

- coefficient of convection to the environment as well as air temperature

- conduction of pipe wall and its diameter and wall thickness

- convection at the internal wall

- fluid temperature

If it is not clear enough ask for further explanations.

Dear Danielle Nel :

It is possible, if you assume the Heat flow to have just an unidirectional component (in the radial direction), and if you assume the whole system (pipe + inner fluid) having reached a steady stable state.

1. Is there something circulating inside the pipe ? Water? Air ? from what means is this fluid inside the pipe circulating ? work due to a mechanical pump, natural air convenction, thermosiphon, etc., etc.

2. Can you know the Temperature of the fluid inside the pipe? at the center ...

3. Also, you will have to compute some of the physical properties of the fluid inside the pipe and of the air surrounding the pipe: i.e. Density, viscosity, Pr.

One more thing:

4. You would have to know the global average solar Irradiance on the site.

Do you have some instrument to measure the solar total global Irradiance on site.

Or, is there a meteorological station nearby the site? Which you can have acces to the data from?

Try to look for these kind of data in a Solar Energy Engineering textbook too, so you can make yourself a better idea of these kind of variables and magnitudes.

Then, what you have to do is to take a Heat Transfer textbook and look at how to write a thermal equivalent circuit, between the center of the inner fluid to the ambient (far from the surface of the Pipe). You can do this considering the flow rate just in the radial direction (from the center of the inside of the pipe, radially to the ambient air outside the pipe). You will have to compute the Heat Transfer Coefficients and to evaluate the properties of the fluid considering a Taverage between the center of the pipe and the inner wall.

This is n't hard, you just have to get a good Heat Transfer textbook, I'll recommend :

HEAT TRANSFER; A Practical Approach. YUNUS A. CENGEL

If you cannot get it and you need it, send me a inbox msg.

For the part of the Radiation Heat Transfer I recommend you this:

Solar Engineering of Thermal Processes. John A. Duffie & William A. Beckman. WILEY

Best Regards !

Franklin Uriel Parás Hernández and Nick Hamburger thank you dearly for your assistance. I am only doing heat transfer this year, in my third year, so your help is much appreciated.

Dear Danielle Nel no prob. at all,

Continuing with the procedure to do this :

it's simple. What you have to do is to write a thermal equivalent circuit between the Temperature at the center of the pipe and the temperature of the atmosphere:

Draw a equivalent circuit of a series of thermal resistances in series (most of them in series, except maybe for the last two)

All the analysis has to be done on cylindrical coordinates:

1. Start with a Thermal Resistance by Heat Convection, placing the value of the heat convection resistance between the temperature at the center of the fluid and the other end corresponding to the Temperature of the inner wall of the pipe.

The resistance's expression is Rth,conv = 1/(4*pi()*rin2)hconv

2. Next, add a Thermal Heat Conduction Resistance to the first one. Place this second resistance in series with the first one. The difference with the first Rth is that the second Resistance corresponds to the Heat Transfer by Conduction through the pipe's wall. Hence, you need to assign the temperature of the right end of the Rth to the outer wall of the pipe. Because the second resistance correspong to the Heat Conduction through the pipe walls the value of Rth has to be (rout-rin)/(4*pi()rinroutk)

(Remeber you have to work the problem in cylindrical coordinates)

3. The last part has to be considered a little different, and the way of drawing the last part of the circuit will depend on how much simplification you are willing to have on your problem.

I'll recomend you to do the following :

-So far you should have just two thermal Resistances; one corresponding to the Heat Transfer by Convection from the inner fluid (from its center) to the inner walls, and a second one corresponding to the Heat Conduction from the inner wall of the pipe to the outer walls...-

To the last Thermal Resistance you drew, add two more resistance to the end of the Resistance you already have for the Heat Conduction through the Pipe. This will represent a Y-connection (if you have seen something about DC Electric Circuits). So these two Resistances will come out from the same node (corresponding to the end of the Heat Conducción Rth, I mean, the point at Touter wall). One of these two Resistances connected in Y-connection is going to correspond to the Convection Heat Transfer from the pipe's outer walls to the ambient air ... so you will have to write Touter wall and Tamb for the nodes between this one. And finally, place the last Rth (the other one you need to draw a Y-connection) between the node of Touter walls and another Temperature assigned as Tsky.

This last Resistance (the one between Touter walls and Tsky will correspond to the Radiation part. Which is the Heat Transfer (transferred) by Radiation from the Sun to the outer pipe (thats why the Tsky)

The expression for Rth, rad involves a euivalent Heat Transfer Coefficient for Radiation

Look in a textbook how to compute hrad,

Now you just have to solve the circuit for Touter wall, and voila !

Notes that should be considered:

You will have to get at least some data for the mass flow rate and the velocity (meaning you have to know some value for its volume flow, at least) of the inner fluid in the pipe, and the temperature of the inner fluid, maybe its distribution along the radial component. So you can evaluate the physical properties of the flud, and be able to compute the Heat Transfer coefficient with the corresponding expression.

Also the same, for the air nearby the outer surface of the pipe (in this case is simpler to evaluate the physical properties of the ambient air (near the surface), you can measure the temperature and velocity of the air in a normal common day in the outside on site (at the place where you have this pipe) and is easier to extrapole (or interpolate) the density, viscosity, Prandtl number, etc..

Finally, you have to look for a model for Tsky, you can find these models in a Solar Energy Engineering textbook.

Regards ! :)

@ Franklin Uriel Parás Hernández

Would you, please, look again at suggested equations ?

Nick Hamburger of Course, no problem. Just, can you point to me exactly what equations are you meaning ?

Regards !

@ Franklin Uriel Parás Hernández : You should at least have a look at your points 1 and 2....

Dear Erling Næss :

The errors I have in points 1 and 2 are the Resistance's expressions for Cylindrical Coordinates, I should have notice them through the '4*pi()' in both of them, so thank you very much for point them out :)

Hence, the correct expression for the Convection's Thermal Resistance is Rth,conv = 1/(2*pi()*L*hconv) (this is in point 1), where L would be the pipe's Length.

And the correct expression for Rth,conduction is Rth,cond = (ln(rout/rin))/(2*pi()*L*k) , again, where L is the Length of the pipe.

Anyways, thank you for the comments :)

I will draw a diagram to better visually represent what I tried to mean with all my text.

Best Regards ! :)

Since you are at your "start" in heat transfer problems here is a more detailed explanation because I consider that if you only apply formula you do not profit from your work.

I hope it will help.

Wall temperature of a sun heated pipe

📷

1- Only part of the pipe surface is under the sun.

2- The normal energy flow is proportional to cos (α).

3- Radiation losses are proportional to the difference of the fourth temperature power

4- External convection depends on the air velocity at the outer pipe wall. If there is no wind then it is a so called "free convection" which follows a different law for its coefficient

5- The heat flow is proportional to the convective coefficient and the temperature difference between wall outer temperature and air temperature

6- Conduction heat flow is proportional to the pipe geometry, the conduction coefficient and the temperature coefficient between outer wall (which you want to compute it is thus your unknown) and the internal wall temperature

7- The heat flow in the pipe –at the wall – is proportional to a convective coefficient which is function of the fluid nature and its velocity

8- The fluid temperature will be function of its entry temperature, its physical parameters and its mass flow, the injected energy will change the fluid temperature over the pipe length so that it will be better to work with mean values.

If the pipe is long the variations can be important.

As you see it is not a simple problem you are confronted with and you should first understand the physics before simply apply the equations.

As a hint if the ratio De/Di is near to 1 (i.e. the pipe wall thickness is small) the wall can be considered as plane.

If the pipe is metallic and has conduction coefficients in the range of over 50 W/m°K the temperature drop in the wall can be neglected and you can consider as resistances only the convections. If it is glass or polymer then it must be considered due to low conduction.