# Convective heat transfer coefficient

by Carlo09
Tags: coefficient, convective, heat, transfer
 P: 15 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!
 Sci Advisor PF Gold P: 2,234 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.
 P: 15 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
PF Gold
P: 2,234

## Convective heat transfer coefficient

I calculated the Reynold's number using the equation:

$$Re_{D}=\frac{\rho*u_{m}*D}{\mu}$$

where
$$\rho=719\frac{kg}{m^{3}}$$
$$u_{m}=19.2\frac{m}{s}$$
$$D=3mm$$
$$\mu=3.3*10^{-4}Pa*s$$

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