zdas04: To answer your question on the "conservative approximation," your first reading is the way that I meant it. If the insulative effect of the ice is considered, the rate of freezing will slow. If this was for an icemaker, then you are right--ignoring the ice formation would be...
doug:
I agree with everything ione has posted except for the last equation. This assumes a linear cooling rate.
Rather, the cooling will be exponential (e^-t/tau) while the water is still liquid, and then essentially constant* while the water is freezing. The time constant for the system...
jparrish, sailoday28:
Heat flow in the radial direction will be governed by fourier's law, so the heat rate will be proportional to the temperature gradient in the various directions. At the very center of the cylinder (assuming symmetry of boundary conditions and construction of the...
jparrish:
For the amount of "current" to add at each node, the amount is determined by the volume of the ring and the volumetric generation rate (the q''' from the way back).
I would take the radius of the cylindrical capacitor and divide it into, say, five "rings" (like rings in a tree) and...
jparrish:
To answer your earlier question, the radial heat flow near the center of the cylinder will be essentially zero (at the very center, it will be zero, due to the symmetry). To this end, I would draw the thermal circuit so that "resistor" in the radial direction was "open circuit" as...
zekeman:
You originally wrote:
"Not exactly. You can get the equivalent K =Keq from the well known equation for radial flow(L1+L2)/Keq=L1/K1+L2/K2Thus for the long cylinder , you can use Keq as the homogeneous value of conductivity.Also, given the thicknesses of each layer in microns you...
OK. That is much clearer. A few comments on your terminology and your computations:
1) "K" is typically reserved for thermal conductivity (W/mK or W/mC--they are the same since the K or C int he denominator is actually a temperature difference), while "h" is used for a convection coefficient...
jparrish:
I have to admit I don't really understand what you're doing above. For starters, the formula you have has no equals sign...what does it produce?
I would caution against trying different equations until you find one that seems to be close. "A stopped clock is right twice a day" as...
jparrish:
To answer your question, yes, q''' is simply heat (in watts) divided by volume (which is what you have).
I am not so sure that the Bessel function route is the way to go. The assumption in such a solution is that the cylinder is of a homogeneous material of constant properties. If...
Your derivation is correct. q''' (triple prime) is the volumetric heat generation rate (W/m^3). If you are thinking of heat generation in a wire (due to current passing through), this will be the watts of heat generated per unit length divided by the cross sectional area.
Some texts (e.g...
It is unclear to me exactly what is vibrating. The original post seems to suggest that it is the engine itself that is torsonally vibrating about the crankshaft, not torsonal vibrations within the crankshaft itself (as is evidenced by the damage to the intake manifold...the latter would damage...
joedvo:
If the tubing is in fact between 45 and 53 F, it will lose heat to the 40F ambient environment.
This assumes, of course, that the tubing is not within the boundary layer that will form on the tank surface. I have not done any computations on this, but prex seems to be on the right...
I write this assuming that I interpreted your drawing properly.
First, a question: What's the temperature of the tubing? If these too are at 40F, then there is no heat lost to the tubing, since there would be no temperature difference between them and the ambient environment.
I suspect you...
Stoveman:
Modern freestanding stoves are available with circulation fans. They don't actually increase the combustion efficiency of the stove--they just increase the convection coefficient from the stove to the air and help circulate the air around the room a little...reducing the temperature...
To answer your question about minimum temperatures, I think 250 F is considered the point at which creosote begins to condense. I suggest purchasing a magnetic wood stove thermometer (it sticks to the outside) and keeping your wood stove burning at least this hot (they are usually labelled with...
