# Heat transfer Problem

A cylindrical rod of stainless steel is insulated on its exterior
surface except for the ends. The steady-state temperature
distribution is T(x) = a-bx/L, where a = 305 K
and b =10 K. The diameter and length of the rod are
D = 20 mm and L =100 mm, respectively. Determine
the heat flux along the rod, Hint: The mass of the rod
is M = 0.248 kg.

T(x) = 305 - 100x.
dT(x)/dx = -100.
I know that heat flux is modeled by Fourier's Law for conduction: qx'' = -k*dT/dx. What I'm having trouble finding is the conductivity k and I'm wondering whether it has something to do with the mass of the rod, although I am not seeing the connection here.
Thanks.

Chestermiller
Mentor
A cylindrical rod of stainless steel is insulated on its exterior
surface except for the ends. The steady-state temperature
distribution is T(x) = a-bx/L, where a = 305 K
and b =10 K. The diameter and length of the rod are
D = 20 mm and L =100 mm, respectively. Determine
the heat flux along the rod, Hint: The mass of the rod
is M = 0.248 kg.

T(x) = 305 - 100x.
dT(x)/dx = -100.
I know that heat flux is modeled by Fourier's Law for conduction: qx'' = -k*dT/dx. What I'm having trouble finding is the conductivity k and I'm wondering whether it has something to do with the mass of the rod, although I am not seeing the connection here.
Thanks.

You need to have more confidence in yourself. So far, what you have done is correct. You need to look up the thermal conductivity of stainless steel on google. Just multiply it by minus the temperature gradient, and you're done. The diameter is extraneous, as is the mass (unless the thermal conductivity of steel is given as a function of density, which I doubt).

rude man
Homework Helper
Gold Member
. The diameter is extraneous, as is the mass (unless the thermal conductivity of steel is given as a function of density, which I doubt).

Oh?
dQ/dt = -kA dT/dx
k = thermal conductivity
A = cross-sectional area
dT/dx = temperature gradient = -b/L

Am I missing something?

Chestermiller
Mentor
Oh?
dQ/dt = -kA dT/dx
k = thermal conductivity
A = cross-sectional area
dT/dx = temperature gradient = -b/L

Am I missing something?

Yes. The problem asked for the heat flux (heat flow per unit area), not the total rate of heat flow.

rude man
Homework Helper
Gold Member
Yes. The problem asked for the heat flux (heat flow per unit area), not the total rate of heat flow.

I would question that.

For example, in magnetics, flux is the B field times the area.

Chestermiller
Mentor
That may very well be the case in magnetics, but I've been doing engineering heat transfer for almost 50 years, and heat flux has always been heat flow per unit area. I've never seen the term used in any other way.

rude man
Homework Helper
Gold Member
That may very well be the case in magnetics, but I've been doing engineering heat transfer for almost 50 years, and heat flux has always been heat flow per unit area. I've never seen the term used in any other way.

I checked and you are right.