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...
...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 ("tau") is simply m*c/(h*A), where m is the mass of the system, c is the specific...
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...
...inner radius is greater than the wall thickness, the error drops. For thin-walled vessels, the difference if essentially nil (even at r_outer = 2*r_inner, the error between flat wall and radial is only about 4%). The flat wall approximation uses a slightly simpler expression for thermal...
...This is parallel flow through thermal resistors, and the appropriate equivalent thermal resistance is:
L1*L2/(K1*K2*A1*A2)
Req = R1*R2/(R1+R2) = -----------------------
L1/(K1*A1) + L2/(K2*A2)
In the special case where...
...increased size of each ring, which will increase linearly with radius from the center. The area over which the heat is passing is increasing (A=2*pi*r*L, where L is the length of the capacitor).
c) At each node, add in a "current source" that reflects the amount of heat added at that layer...
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...
For the "ramp" part of the heating cycle (where the oven is warming up at a steady rate), it may be helpful to note that the response of a first order system (i.e. lumped capacitance thermal system) is itself a ramp function that simply lags behind the ambient temperature by a time period equal...
I'm with IRStuff. I don't follow your statement that " if node 1 has computed 7.4 and the boundary value of node 1 has been set at 50 then the value of node 1 is re-set to 50 before the next time step"
You should NOT be computing node 1. If the temperature of node 1 is in fact specified, then...
Free convection correlations are for "quiescent" conditions, which are likely hard to achieve in your (apparently) simple experimental setup. The real coefficient will typically be higher, due to mixed free/forced effects. Like the others mentioned, add radiation to the mix as well.
Dave
microhydro:
Thanks for the link. I can see now (and admire) what you are trying to do.
The fundamental problem you will encounter is the freezing of the water on the evaporator coil. Once ice starts forming on the surface of the pond, the water below the ice will be liquid water at 32F...
...lumped-parameter approximation (also called "lumped capacitance"). In this case, the tank will cool according to the relation T=Tinf-(T(0)-Tinf)*e^(-t/Tau), where "T" is the temperature as a function of time, T(0) is its starting temp, Tinf is the ambient temp, and "Tau" is the time constant...
