D&D Engineering Problem

Factual Questions
1

There is a magic item in D&D called the Decanter of Endless Water which can produce an unlimited gush at such a prodigious rate that it occurred to me that it could easily be set to run a turbine. But based on its description, I’m not sure how to calculate how much energy it would put out in its three modes:

*  "Stream" pours out 1 gallon per round.
* "Fountain" produces a 5-foot-long stream at 5 gallons per round.
* "Geyser" produces a 20-foot-long, 1-foot-wide stream at 30 gallons per round.

Any measurement of power or energy seems to require a distance that an amount of mass is moved, which is not indicated in these specifications. The “fountain” and “geyser” settings specify the length of the stream, but this isn’t really a measure of how far the total mass of water is moved, only how far in game terms you can effectively hope the stream to reach. Is there enough data here to indicate how much energy is involved in moving this volume of water?

2

I’m guessing that you would have to build some sort of reservoir and a chute or pipe for it to go into in order to build up the necessary pressure to turn a turbine usefully.

3

If it was done with a sort of millwheel configuration, I suppose it would be possible to supply the distance missing from the data. But then the volume being shot out per second would be less important than the acceleration of gravity.

4

You need to know the diameter of the opening. From this you can calculate the speed of the water as it is leaving the Decanter, and therefore its kinetic energy.

That would give an exact answer, but the velocity could also be estimated from the distance the water travels.

If the distance they give is the maximum the stream will travel, we will make the assumption that there is little air drag making the optimum angle for distance 45 degrees.

If the initial speed is V, then now the horizontal and vertical speeds are each V/sqrt(2).

The equation of motion in the vertical direction is:

a[sub]y[/sub]=-g

where a[sub]y[/sub] is the acceleration of the water and g is the acceleration due to gravity. Integrating that gives:

v[sub]y[/sub]=v[sub]y0[/sub]-g*t

where v[sub]y[/sub] is the vertical speed and v[sub]y0[/sub] is the initial vertical speed. When v[sub]y[/sub]=0 it has reached the top of its arc. That will occur at:

t[sub]peak height[/sub]=v[sub]y0[/sub]/g

The total hang time for the water will be twice this value (it takes just as long to come down as it does to go up), or 2*v[sub]y0[/sub]/g.

Now lets look in the horizontal direction. Since there is no acceleration in that direction (remember we are neglecting air resistance because otherwise this becomes very difficult) the equation is quite simple:

x=v[sub]x0[/sub]*t

Here x is horizontal distance traveled and v[sub]x0[/sub] is initial horizontal speed. We know the time it is traveling from above, so we combine those two and get:

x=2*v[sub]x0[/sub]*v[sub]y0[/sub]/g

Remember that both of those speeds are the same and equal to V/sqrt(2). With this substitution we get:

x=V[sup]2[/sup]/g

Rearrange that since we know the distance traveled:

V=sqrt(x*g)

Great. Now we know the speed of the water. Lets combine that with the known flow rater, f, to get a power. Power is energy per unit time:

P=.5mV[sup]2[/sup]/t

Lets put that in terms of flow rate where flow rate is volume per unit time and density is mass per unit volume. So, mass per unit time is f*rho where rho is density:

P=.5frho*V[sup]2[/sup]

And now we add the velocity we already calculated:

P=.5frhoxg

With the max flow rate of 30 gallons per 6-second round (that is still the way they do it, right?), a density for water of 62.42 lb/ft[sup]3[/sup], a distance of 20 feet, and a gravity of 32.2 ft/s[sup]2[/sup] I get:

P=0.391 hp

Assuming I got the conversions right you are not exactly burning things up there. Note that that is on the low end as they probably assume a more horizontal stream of water. You will have major efficiency losses though if you try to convert this to usable power using any medieval technology.

In fact, you would probably do a lot better to just set the thing up high and use the potential energy from the water rather than its kinetic energy as you essentially have as much free potential energy as you want by elevating the thing as high as you want.

5

For each force, there is an equal and opposite force. It requires a strength of 12 to avoid being knocked over; given the non-fractioning system of D&D, we can say that the geyser effect produces a force matchable by a person with 12 strength but not by someone with 11.

If we look at the Weight statistics, a person with a score of 11 in strength can carry a maximum of 115lb. A person with 12 strength may carry a total of 130lbs of equipment. Thus the force exerted by the decanter in geyser mode must be equal to a point in the range of the force exerted by a mass of 116-130lbs under the influence of gravity.

Going with F=ma, we have F = (116 - 130) * 9.8m/s, giving us a low of 1136.8 pound-force, which equals roughly 255.6 Newtons of force, and a high of 1274 pound-force on the dot, which gives us 286.4 Newtons. Thus the force exerted by the geyser itself is between those two figures.

Ok, so, the energy. There’s no bloody way i’m going to get into cos/sin stuff, so i’m going to assume that the decanter is pouring out without the effects of gravity affecting it. Work = FD, so we have (255.6 - 286.4) * 20 feet (which is 6.1 metres) giving us a low of 1559.16 and a low of 1747.94.

The energy worked by the Decanter on the water is thus (roughly) somewhere in the range of 1559.16 -1747.94 joules - absent the effects of gravity.

I make it… say, four minutes before someone comes in and points out what i’m sure are incredibly obvious flaws in that.

6

The amount of energy you can get from it depends a lot of where you put it. Build yourself a tower, and have thirty gallons per minute, (An D&D “round” is a minute.) and you can generate more energy with your turbine. Or, given the need for primitive technology, a cascade of water wheels, and spillways, with a thirty gallon per minute flow along it.

By the way, a one foot diameter stream would hardly move rapidly enough to be a geyser at thirty gallons per minute. A foot diameter cylinder of thirty gallons volume is about 12 feet long. So, if it moves at more than 12 feet per minute, (0.2 MPH), it has to be fairly much a spray to get up to one foot in diameter. In the US, the average gasoline fill pump is rated at 30 gpm. Pressure, not capacity is what will determine the distance it travels.

This represents pretty much standard attention to mathematics, and physical details for D&D.

Tris

7

Don’t have anything constructive to add, except to mention that this has got to be the geekiest thread ever. :wink:

8

Nothing to add, except that with the current rules it should be 10 rounds per minute or 6 seconds per round.

9

It’s easier if you work in metric. 1 gallon = 4.5 litres. 30 gallons per round = 135 litres per round or 22.5 litres per second. 1 litre of water weighs 1 kilo. 20 feet is approx 6 metres, so in 1 second the water goes 1 metre, so we get a force of 22.5 N.

If that doesn’t sound a lot, think about being under an apple tree and being hit by 22.5 falling apples every second.

10

To clarify, the item in question was in the 2nd edition DM’s guide IIRC. And a round is a minute in that system.

Under 3rd edition rules a round is six seconds, so unless they’re rewritten the rules in the new books (I could check when I get home) we should still assume the quoted figures are per minute.

11

The item is being cited from the 3rd Edition SRD in which a round is 6 seconds. I haven’t checked whether it matches the 2nd edition equivalent in output.

It had occurred to me that in an ‘age of steam’ sort of campaign you could generate steam using a steam mephit, but this doesn’t work out because surely the creature is not able to survive in the kind of pressure it takes to put steam to work on mechanisms. Perhaps one or more fire mephits could keep a boiler going with its unlimited breath weapon. Of course, the only energy rating for the breath weapon is its potential hit point damage, and I suspect the adventure of figuring out how much energy is in a hit point will be maddeningly pointless.

Since in theory you’d be better off collecting energy from the Decanter of Endless water as the water falls rather than the energy of the actual gout of liquid, I’m inclined to wonder what else in D&D could be used as an energy source to run machinery that needs to be more portable than, say, a hydroelectric tower. What would run a locomotive or a submarine, say? Traditional locomotives burn coal, which is of course always an option for them, but not so appealing for a submarine if you wanted to go all Nemo. I was hoping the Decanter would be a good source, but it doesn’t appear so by flight’s calculations.

12

No, But I’d love to watch the attempt. Imagine the little bugger squeal.

I’ve seen fire elementals used inside a furnace to keep the whole thing hot. As they’re more fire and less physical than a mephit I’d suggest one or two of those.
As I’m sure you know, hit points are not just a physical stat. They represent luck, strength of mind and more besides. So I agree we could never arrive at a figure.

A boiler run by having lots and lots of continual light cast inside it ?
Daylight spells with Permanancy trapped in a black box ?

Since magic inherantly draws energy from another plane into the prime material, all you need is some means to trap it. Or a magic item that replicates the effects.

and yes, this is the geekist thread ever. Love it.

13

Actually, I just thought of a way to get some good power out of it.

Making the assumption that the geyser cannot be “capped” by normal means (as it is probably a portal to the Elemental Plane of Water or something) we can take advantage of water being incompressible. We could enclose the decanter in another container with a smaller orifice. The flow rate through this orifice must be the same as that of the water exiting the Decanter, so it will be going much faster. You could probably get some significant energy that way.

Another way would be to use the Decanter as the power source of some sort of pump. Unlike the real world where flow rate is determined by relative pressures, we have a fixed flow rate, so by making everything very narrow we could run the pump at some huge rpms. You would need some valving, but the only limit on the power you get out would be the strength of the material you make the pump out of.

Of course if I were DM, I would say that the flow rate slowed once you start building up back pressure.

14

We had a Practical Applications of Magic thread awhile back that had some thoughts on the matter—including a rather novel scheme for using undead horse skeletons “set” to continually walk as a form of motor. (Basically, a horseless carriage, with a horse.)

I wonder, as long as we’re screwing with the genre already (parenthesis smiley-face), what other kind of “elementals” one could get away with using. Could you use a Magnetism-elemental to compress deuterium and tritium into fusion? Or function as a primitive Calutron?

15

Obviously, if it gives off light, it must give off energy, but how much? It’s full daylight, within 60 feet, which translates to 18.22 meters. [the sunlight that reaches the earth is potentially about “between 125 and 375 W/m² (3 to 9 kWh/m²/day).” To maximize meter squareage absorbing the light, we’d need a sphere with a radius of 18.22 m. That would have a surface area of 4(pi)r^2, or 4,171 m^2. Taking an average on the power delivery:

P = 250 W/m^2.

So, assuming efficient collection, we’re looking at 250*4,171 or 1042750 watts. 1 watt = 0.00134102209 horsepower, according to Google, so the sphere could collect 1398 horsepower worth of energy, and [url=“http://www.steamtrain.ca/loco_909_en.asp?onglet=1”]this train](Solar power - Wikipedia"According to Wikipedia,[/url) runs on 1000 horsepower. So, it looks like it’s possible to run a train with the power from a Daylight spell. Of course, a sphere 36 m in diameter would fit awkwardly on top, and it’s not clear how you’d transfer the energy to the turbines. But my point is that the spell does appear to generate enough energy. Unless I’ve made some wildly inaccurate assumption.

I recall that in 2nd edition this was specified to be the case, though it’s not so explained in the d20 SRD.

My knowledge of fluid mechanics is lacking. If pinching off the flow doesn’t by itself cause back-pressure, what would? And is it possible that some dynamic could arise by which water would flow back to the elemental plane?

My focus has been on how the kinds of almost limitless energy sources available with D&D magic could be channeled into ‘age of steam’ type technology, it is also worth considering what the magical nuclear age would look like. Someone once mentioned to me that a certain book had a wizard creating a nuclear chain reaction with a cluster of fireballs. Of course, I don’t know how that would work in 3rd edition since it specifies that fireballs do not explode with a concussive force. But those old school fireballs would fill up 33,000 cubic feet or else something had to give.

16

That is my point, it would create a back pressure. The description as stated does not allow for slower flow due to constriction. I was saying that a judicious DM would account for that.

If you accept a changed rate of flow with pressure, then you could create reverse flow with a high enough pressure, but no device designed to extract energy from the system would do this.

17

Well, I was just thinking about how the Elemental Plane of Water might feel about the Prime Material Plane’s backwash. Also, they never address how they make sure that the water coming from the plane isn’t full of elemental extraplanar microbes. What are the celestial whales eating if not plankton with a celestial template?

18

Thinking about the bottomless decanter of water—is there only supposed to be one of them in existence, or can you find/manufacture a number of them? If the latter, could they be channeled into some kind of manifold to increase the water delivered to a turbine or wheel—or even mounted directly on a turbine itself?

19

By 3rd edition rules, any 9th level caster who can cast Control Water can make the item, and it costs 9,000 gp retail.

20

There are no whales in the Elemental Plane of Water, because there is no “surface” to breach for air to breathe…