convective heat transfer coefficient
convective heat transfer coefficient
(OP)
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
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





RE: convective heat transfer coefficient
TTFN
FAQ731-376: Eng-Tips.com Forum Policies
RE: convective heat transfer coefficient
It's my first time writting to a forum .... I'm impressed !
any other comments ?
Thermodata
RE: convective heat transfer coefficient
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.
RE: convective heat transfer coefficient
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 !
RE: convective heat transfer coefficient
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
htt
RE: convective heat transfer coefficient
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 (ΔT/d)1/4
Turbulent flow: h = 1.24 (ΔT)1/3
d = diameter in m
ΔT in oC.
RE: convective heat transfer coefficient
RE: convective heat transfer coefficient
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
RE: convective heat transfer coefficient
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: Eng-Tips.com Forum Policies
RE: convective heat transfer coefficient
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)
RE: convective heat transfer coefficient
RE: convective heat transfer coefficient
Holman's definition of laminar for this case:
104<Gr×Pr<109
Turbulent: Gr×Pr > 109
RE: convective heat transfer coefficient
http://www.cheresources.com/convection.shtml
You can find this and more in many books of course.
RE: convective heat transfer coefficient
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
RE: convective heat transfer coefficient
TTFN
FAQ731-376: Eng-Tips.com Forum Policies
RE: convective heat transfer coefficient
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.
RE: convective heat transfer coefficient
RE: convective heat transfer coefficient
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. :-p