Find the convective heat transfer coefficient

  • Thread starter Carlo09
  • Start date
Ok it's my first time here and I was hoping to get some help on some questions I have been given. I am a first year chem eng and I'm finding the work pretty hard so any help at all will be useful, thanks.

I need to find the convective heat transfer coefficient, h for petrol using this equation:

hD/Lamdaf = 0.37 Re^0.6

so using information I am given:

D = 3mm = 0.003m
Lamdaf (thermal conductivity) = 0.145 w/m k
(M)=viscosity = 0.0006 Pa s
u=Velocity of petrol = 19.2 m/s
P=density of petrol = 737.22 kg/m^3

Ok so to calculate Re I am using: Dup/(M) = (0.003*19.2*737.22)/0.0006 = 70773.12

Is this correct so far?

Then I put this back into the equation and rearrange for h which I get to be =14533.475! w/m^2 k

Is this correct because it seems very big to me! if not please can someone point me into the right direction.... thank you very much!


Science Advisor
Gold Member
I'm assuming this is flow through a pipe? You seem to have run the numbers correctly, and that convective coefficient doesn't seem out of the realm of possibility to me.

Your Reynold's number does seem a bit low, I caluclated 126,500 but I might have used some fuzzy numbers in there.
It says the temperature of petrol is monitored by a thermocouple in the flow, so i'm guessing pipes?

How did you get your Re at that value? Have i used the wrong values to calculate it?

Thank you for your help


Science Advisor
Gold Member
I calculated the Reynold's number using the equation:



With these inputs the Reynold's number works out to 125,200.


Science Advisor
Gold Member
So since I was a third of the way there anyway, I went ahead and tried calculating the convective heat transfer coefficient. The equations I used are out of my heat transfer text book, "Introduction to Heat Transfer" by Incropera and DeWitt.

The answer I got was h= 30,950 W/m^2*K

I attached the MathCAD sheet I used to calculate it rather than trying to type it out in Latex.


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