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How to boil water faster

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coppice:

--- Quote from: PlainName on January 09, 2024, 12:03:56 am ---Warm water from the hot tank contains contaminants (hot water + metal = corrision). Hot water systems fur up where cold water doesn't - worth thinking about.

--- End quote ---
Pipes fur up more rapidly with hot water, but if you live is an area with very hard water the cold pipes, taps, etc, fur up very well. All pipes and tanks are susceptible to corrosion and contaminants. There is nothing special about a hot water tank there.

Hydro:
I didn't read the whole topic, but I remembered another task.
How to use 10 liters of water at a temperature of 100 degrees to heat another 10 liters of water at a temperature of 0 degrees to 80 degrees.
Someone will immediately claim that this violates the laws of thermodynamics.

If you know the answer, don't tell yet.

Zero999:

--- Quote from: coppice on January 09, 2024, 02:14:37 am ---
--- Quote from: PlainName on January 09, 2024, 12:03:56 am ---Warm water from the hot tank contains contaminants (hot water + metal = corrision). Hot water systems fur up where cold water doesn't - worth thinking about.

--- End quote ---
Pipes fur up more rapidly with hot water, but if you live is an area with very hard water the cold pipes, taps, etc, fur up very well. All pipes and tanks are susceptible to corrosion and contaminants. There is nothing special about a hot water tank there.

--- End quote ---
If that's true, then hot water would be more pure, because the minerals have come out of the solution and deposited in the pipes.

hans:
The added energy needed to boil 1x10L of water, or 10x1L of water, is obviously the same.
I assume that the question asserts that both water heaters have equal power, and therefore surface area, and the water is at an identical temperature at the start. Then only changes can be made from the boiling process and the way we treat losses.

So lets assume 2 sources of heat loss: evaporation from the hot/boiling water, and heat convection into the kettle walls.
I'm not going to speculate whatever those rates exactly are. However, if we assume they scale linearily with surface area, then we can design a kettle that has an optimal trade-off ratio between radius and height. Assuming here that a larger radius will result in more evaporation losses, and higher walls will result in more convection losses with the walls. We can find optimums for both a 1L and 10L container.

The losses of evaporation is presumably going to scale with R^2, since thats the surface area of the water body on the top of the container.
The losses in the walls is presumably going to scale with 2*R*H, as thats the surface area of the walls of the kettle.

Therefore, to balance these ratios, e.g. alpha *R^2 = beta * 2*R*H, where alpha is evaporation loss coefficient and beta is the convection loss coefficient, we can see that we must solve for R and H giving alpha and beta. Since the volume of the container is also constrained by V=R^2*H, we have a system of equation(s) that can be solved.

It means that if we go from a 1L => 10L container, we can do that either by making H 10x bigger, or R 3.16x larger. Or a mix.. whichever has the more favourable coefficient..

"Solving" the problem by having 10 containers of 1L, and boiling them one at a time (with assuming 0s switching time), is not going to help much. We have a 10x larger body of water, however, we also have 10x the losses from that smaller container.
Lets assume some extreme edge cases for such case:

Assume that alpha=0 and thus we have no evaporation losses. Having 10x 1L containers means effectively we have cut up a 10x H container into sections of 1x H. The losses into the walls should be identical, and thus it should have no influence if we boil each pot sequentially. However, since the evaporation losses is zero, we ideally would want to make R -> infinity and thus H -> 0, which makes for a non-numerical solution.

Assume that beta=0, and thus we have no losses in the walls. In this case we would want to have R -> 0, and thus H -> infinity (again non-numerical case). However, since we chop our container up in 10 sections we can reason that we also have 10x the evaporation losses. (Even where R -> 0). Ratiometrically speaking, the losses in boiling 10 separate containers is 10x worse.

The practice will probably be somewhere in between.

So my bet is on a 10L pot being more efficient, as one can set a new optimal ratio between radius and height to balance the losses of its volume.

.RC.:
If you put the 10l in a pressure cooker, then put a vacuum pump on the pressure cooker, it will boil quicker.   

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