Heat transfer coefficient

[Lepetit]

Hello all,

I have a questions, can be possible a heat transfer coefficient of 6 (W/m^2/K) between the air inside an office an a pipe of Polyurethane? I assumed natural flow convection. I have calculated the heat flow in a radial direction (everything is known: temperatures, thermal conductivity...) and with a normal parameter of the heat transfer coefficient (15 W/m^2/K)the calculation doesn't have sense.

Thank you in advance.

Cheers

 

[ione]

Yes it could be reasonable, maybe even lower than 6 (W/m^2/K).

For a horizontal pipe and natural convection (you might want to consider radiative heat transfer too):

Nu={ 0.60 + (0.387Ra^(1/6))/[1+(0.559/Pr)^(9/16)]^(8/27) }^2

Where:
Nu = Nusselt number
Ra = Rayleigh number
Pr = Prandtl number

All the above numbers calculated using air properties evaluated at film temperature. Please note that the pipe external wall temperature could be (probably is) a bit different from that of the fluid flowing inside.

 

[Lepetit]

Thank you very much for your fast and helpful answer ione.

The problem to calculate an accurate value of it is that the pipe it's not completely horizontal or vertical, also in some points it's touching some platforms...really difficult to be accurate.

Thank you for the info again!

Cheers

 

[IRstuff]

Are you talking strictly about the convective coefficient on the outside only? What are you assuming for the inside of the pipe?

TTFN
faq731-376

 

[Lepetit]

@IRstuff I know everything about the coolant which is inside the pipe, so I have already calculate the inner heat transfer coefficient. The "problem" appeara when I calculate the heat transfer coefficient outside the pipe (air-pipe) because if I calculate the heat flow with a normal value of this coefficient (10-15 W/m^2/K) the intermediates temperatures have no sense. I have inverted the process and calculate the coefficient assuming as known all the intermediates temperatures (I have measured them); the value I obtain it's around 6 W/m^2/K.

I don't know if it's clear enough, sorry.

Thanks, cheers.

 

[IRstuff]

Why are you assuming that these "intermediate" temperatures are more correct? Are they measured?

TTFN
faq731-376

 

[ione]

Lepetit,

If you have to deal with natural convection and inclined cylinders, you can look on the web for "Raithby and Hollands". They do have developed a correlation for the scenario above mentioned which seems to work quite well.

 

[Lepetit]

@IRstuff, yes all the temperatures have been measured before, after I made the inverse way just to validate the value of the Heat transfer coefficient.

@ione, thank you very much! I will check right now.

Thank you both.

Cheers.

 

[Lepetit]

@ione, please can you write the correlation or put the link where I can find it? I am trying to look for it and I can't.
Thanks

 

[ione]

Nu={ 0.772 + [0.228/(1 + 0.676*(2*L*cot(alpha)/D)^1.23]}*[cos(alpha) + D*sin(alpha)/L]^0.25*Ra^0.25

Valid for L/D>5

Where:
L = length
D = diameter
alpha = angle between the axis of the cylinder and the horizontal
Nu = Nusselt number
Ra = Rayleigh number

 

[Lepetit]

Perfect!! thank you very much

 

[IRstuff]

OK, so the bottom line is that almost ALL of heat transfer correlations are essentially curve fits to measured data; which is why Ra and all the geometrical factors have differing numerical coefficients and exponents. The end result is that what you measure describes the system exactly, for the conditions you have. If the closest possible model of the situation doesn't match, then you most likely have an erroneous assumption about what you have or the conditions within your system.

TTFN
faq731-376

 

[Lepetit]

Yes, the problem was that I was chosen a bad correlation to make the theoretically situation with my system. Now it seems so much accurate with the Raithby and Hollands correlation. Finally I think that i will use this one.

Thank you

Cheers.