convective heat transfer coefficient

[thermodata]

Hi,
I've been reading through your forum a bit....maybe one of you can help me.

I am trying to calculate the heat loss of a hot pipe in still air through radiation and convection.

I don't have a problem with the radiation part, but I do have one with the convection part.

What will be the heat transfer coefficient H in the equation Q=H*S*(Th-Tc) ???

Temperature of the object: 140 C approx.
Temperature of the air: 20 C approx.

Thanks for your help !!

Thermodata

 

[IRstuff]

For natural convection, it's on the order of 3.2W/m^2-K

TTFN

FAQ731-376

 

[thermodata]

that is nice !... thanks to IRstuff for this fast response.
It's my first time writting to a forum .... I'm impressed !

any other comments ?

Thermodata

 

[ione]

It would be opportune to know the external diameter of the pipe.
For a 1.5” horizontal pipe I’ve got 7.8 W/m^2K, that is a value a bit less conservative than that reported by IRstuff. Anyway the order of magnitude is right.

 

[thermodata]

Thanks Ione,
I knew it was not one unique number, and simple .....
What I am trying to acheive is to get one number for averaging different conditions....

- vertical and honrizontal piping
- from 1 to 10 inches in diameter

I rather be conservative a little then too much !
What will be the % of error ..... 10 or 25%?

Thanks again !

 

[ione]

You have to use the following formula:

h = Nu*k/D

Where Nu is the Nusselt number, k the conductivity of the fluid and D the diameter of the cylinder.
Fluid properties have to be evaluated at film temperature Tfilm = (Tamb + Twall)/2
Take a look at the link below which reports a correlation from Incoprera and Dewitt to calculate Nu for your case

http://www.egr.msu.edu/~somerton/Nusselt/ii/ii_a.html

 

[25362]

J,P.Holman's Heat Transfer McGraw-Hill gives simplified equations for h, W/(m².^oC), for free convection to air from a horizontal cylinder as follows:

Laminar flow: h = 1.32 ([Δ]T/d)^1/4
Turbulent flow: h = 1.24 ([Δ]T)^1/3

d = diameter in m
[Δ]T in ^oC.

 

[ione]

I’ve heard about those correlations reported on Holman’s book and I have to admit they really are attractive (no need to calculate those boring non-dimensional numbers), but I’ve always had some difficulties to grasp them. Probably my limit, but I can’t easily deal with the concepts of laminar and turbulent flows when coupled to natural convection. When I think about laminar or turbulent flows I immediately associate those definitions to Reynolds number, as I was taught the Reynolds number is the parameter which characterizes the flow regime. In this scenario (natural convection across a cylinder) I do not know how to calculate Reynolds number.

 

[thermodata]

Thanks very much 25362 .....

I like this more simplier approach.
But I'm a little lost with the laminar and turbulent flow,here !
Do you mean I have to calculate both(laminar & turbulent) to obtain the final H coefficient I am looking for ?

Thermodata

 

[IRstuff]

As a general rule, you need to determine the Reynolds number to determine the flow regime, and most in "correlations" you choose the equation that fits your Reynolds number regime. That's assuming you have forced convection.

If there's no forced air flow, then the only determinant is the Rayleigh number, which is a measure of the buoyancy of the hot air rising from the object being cooled.

TTFN

FAQ731-376

 

[ione]

The approach could seem more convenient, but you have anyway to establish whether you’re in laminar or turbulent regime.The two formulas presented are for two different scenarios. You have to use the Rayleigh number (Ra) to characterize your regime. Ra is defined as the product of Grashof number (Gr) and Prandtl number Pr:

Ra = Gr *Pr

You have laminar natural convection in the range 10^4<Ra<10^9

You have the turbulent natural convection in the range 10^9<Ra<10^12.

With the data you’ve reported and for a 1.5” horizontal pipe you’re in laminar natural convection

h = 9.3 W/(m^2K)

 

[sspeare]

You might check out this pipe heat loss graph at engineeringtoolbox.com - see the link below. The graph is for steel pipe, but the site also has copper.

 

[25362]

Holman's definition of laminar for this case:
10⁴<Gr[&times;]Pr<10⁹
Turbulent: Gr[&times;]Pr > 10⁹

 

[Japo]

Some more notes, that I used a while ago:

http://www.cheresources.com/convection.shtml

You can find this and more in many books of course.

 

[thermodata]

Sorry for not answering for a while, vacations.....
Thanks a lot for your answers !
You really help me a lot with this !

Would it be right to say that a vertical cylinder will dissipate heat(convective) in the same manners(calculating) as a verticale plane surface ?

Thermodata

 

[IRstuff]

Similar, but not exactly the same. The heat transfer occurs in the boundary layer next to the surface, but the physical size of the surface affects how the air rises and leaves the surface.

TTFN

FAQ731-376

 

[dgsbsme]

Trouble is, this calls for an iterative solution.

Q = Delta T divide by Overall Thermal Resistance.

We have Tinside & Toutside for overall Delta T.

As for the Overall Thermal Resistance:

• We have enough to calculate Nusselt Number (inside) and get Convection Coefficient (inside). Straightforward enough.

• We know all the geometric parameters involved and can easily look up thermal conductivity numbers for the pipe and any insulation. Conduction - Duck soup.

• We need the Outside Convection Coefficient and to get it, we need the Nusselt Number for the outside fluid (air most commonly). To get that, we need the Rayleigh Number. To get that, we need (among other things) the outside surface temperature. OOPS! To get that, we need Q… to get that, we need to be finished with the whole problem.

Hmmm… not as simple as it seemed at first glance.

Never really noticed, all those years ago, in school… all those problems in the books always state the outside surface temperature as a given. Good thing too… imagine how discouraging it would have been had we realized what comes as “given” out here in the real world.

 

[dgsbsme]

I don't have my copy of Kern's "Process Heat Transfer" with me at the office. I wonder if there is a nice neat way to solve this in there.

 

[dgsbsme]

Don't seem to be able to edit a post once it's up...

Forgot to add the Radiant Heat Loss in there. That would be duck soup as well if only we had that pesky Outside Surface Temperature.


In case anyone was thinking of taking a shortcut and "intuiting" that the outside surface temp of the pipe (assuming uninsulated) is the same as the inside. If that were the case, pipe would make a perfect insulator:
     Tin=Tout is just another way of saying Tin-Tout=0
      and that is just another way of saying Q=0... no heat
      transfered.