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how to find heat loss in hollow cylinder???
Posted Dec 24, 2011, 5:22 a.m. EST Heat Transfer & Phase Change Version 4.1 1 Reply
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Hi everyone,
Now i am using comsol 4.1. i have finished one Heat transfer problem (hollow cylinder) with dimensions ri=0.05m,r0=0.1m,and length is 1 m. i have applied temperature 573 k on inner radius and 323 k on outer surface.but,now i have some trouble in find the rate of heat loss from inner to outer...
can you help me??
Thanking you
parma
Now i am using comsol 4.1. i have finished one Heat transfer problem (hollow cylinder) with dimensions ri=0.05m,r0=0.1m,and length is 1 m. i have applied temperature 573 k on inner radius and 323 k on outer surface.but,now i have some trouble in find the rate of heat loss from inner to outer...
can you help me??
Thanking you
parma
1 Reply Last Post Dec 25, 2011, 4:09 a.m. EST
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Posted:
1 decade ago
Hi
you problem is rather symmetric, depending only on "r" so applying Fourier heat law
q[W/m^2] = -k[W/m/K]*Grad(T)[K/m]
and taking into account the cylindrical coordinate transform, you get the heat flux (W per area) as
q = -k*(573-323)/(0.1-0.05)/ln(0.1/0.05)
or a q/k of about 7213.5 [K/m]
and that is what you see if you plot the heat flux magnitude in a cut plane, no ?
as you say nothing about the heat conductance "k" I have normalised over "k" (or you can use k=1 in COMSOL, and as you are in steady state "rho" and "Cp" are irrelevant, you can also use "1" as values there.
for the ln(Rext/Rint) I leave it to you to search on the web, there are several references there, and probably in your heat transfer text books too ;)
By the way you need to play a little with the mesh density to get a resonnable value, you could also try to use boundary elements, on the inner and outer surface, an easy way to get a denser mesh for better boundary temperature gradient, hence heat flux estimations, particularly in time dependent solving (less critical in steady state, ... why ?)
--
Good luck
Ivar
you problem is rather symmetric, depending only on "r" so applying Fourier heat law
q[W/m^2] = -k[W/m/K]*Grad(T)[K/m]
and taking into account the cylindrical coordinate transform, you get the heat flux (W per area) as
q = -k*(573-323)/(0.1-0.05)/ln(0.1/0.05)
or a q/k of about 7213.5 [K/m]
and that is what you see if you plot the heat flux magnitude in a cut plane, no ?
as you say nothing about the heat conductance "k" I have normalised over "k" (or you can use k=1 in COMSOL, and as you are in steady state "rho" and "Cp" are irrelevant, you can also use "1" as values there.
for the ln(Rext/Rint) I leave it to you to search on the web, there are several references there, and probably in your heat transfer text books too ;)
By the way you need to play a little with the mesh density to get a resonnable value, you could also try to use boundary elements, on the inner and outer surface, an easy way to get a denser mesh for better boundary temperature gradient, hence heat flux estimations, particularly in time dependent solving (less critical in steady state, ... why ?)
--
Good luck
Ivar