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47

The efficiency of a thermoelectric generator is around 5 - 8%. The efficiency of a large steam turbine power plant aproaches 40%. In fact the thermodynamic efficiency of a large steam turbine power plant is over 90%, so it's about as efficient as anything could be. The maximum possible efficiency of a steam driven engine is given by the idealised model ...


25

I'm not sure what all you've read on them, but I'll try to clarify at least a few things. I would certainly disagree with several of your assertions. For starters, you say "...they don't produce anything you could feasibly use as a source of material for nuclear weapons." Thorium reactors use Thorium as a fertile fuel that transmutes into fissile U233. ...


23

Suppose you are using a waterwheel to do some form of work (e.g. grind corn). You need a head of water to make the wheel move, and you could use either 1kg of water at a height of a million metres or you could use a million kg of water at a height of one metre. In both cases the water would do the same amount of work as it flowed through your wheel. The ...


23

Let's work this out from the Stephan Boltzmann law. What color is a fire? If you look at color charts for black body radiators at various temperatures, I estimate it to be about 1000K. (Be careful: some flames are colored by strong emission spectra, making their light very different from a blackbody radiator's color). Glancing around the web from various ...


22

I was going to comment on other people's answers, but this was going to become too long. Almost everyone fails to separate Thorium (which is a fuel type) and reactor type. Safety is a function of the reactor type, and molten salt in particular for this question. Does the fuel choice impact ultimate reactor safety? Yes, but to a limited extent. So how ...


22

It's all a question of if they need it. Most that are staying within a couple AU of the sun can get sufficient power from solar panels. It's when they start getting further away that they use an RTG. For example, New Horizons, which launched in 2006 (which is considered to be 'modern' when you only launch a few probes per year) is going to Pluto, so it ...


18

If you throw a bunch a uranium ore in one blob, nothing happens. If you chemically purify the ore so that the only element present is uranium, still nothing happens. The runaway chain reaction needed for a uranium-powered bomb involves U-235, an isotope having three fewer neutrons than the most common natural isotope U-238. According to Wikipedia, The ...


18

my question is about whether it's possible in principle The answer is yes. and whether anyone tried it. The answer is by all chances, no. So, how come? The effect The thermoelectric effect for electricity generation (called the Seebeck effect) is the phenomenon that a voltage is generated at a temperature different across the ends of a ...


16

The conductor material (copper, aluminum, whatever) expands when heated. When the temperature increases, the length of the power line between two towers increases due to thermal expansion, and the line sags because of the increased slack.


15

edit: I originally had some points about the inefficiency of RTGs, but after some more research prompted by @Jeremy I found that it's not really a valid point when they're used appropriately for the spacecraft's mission. The RTGs used by Galileo at Jupiter generated 300W of power, whereas the solar panels that will be used by Juno at Jupiter will generate ...


14

Dear Thomas, the diameter of the beams of these HeNe lasers is between 0.5 and 1 millimeter, so the power 1 mW is coming to $10^{-6}$ squared meters or so. The ratio of power and area is $10^{-3}/10^{-6} = 10^{+3}$ Watts per squared meter. On the other hand, when a 3W LED is watched from the distance 0.1 meters, the power of 3 W is divided to $4\pi R^2 = ...


12

Within power systems such as regional or national electricity grids, $\frac{\mathrm{d}^2E}{\mathrm{d}t^2}$ is called the slew rate: it's used to denote the rate of change of power demanded from, or supplied to, electricity grids. It's typically either expressed as MW/s or GW/h, being two time periods of interest in balancing electricity grids. ...


12

The real problem with RTGs is that the US stopped making Pu238 in the 80s and has been very slow to start up production again, purchasing all our spacecraft Pu238 from the Russians (who have now also run out). I don't know about the byproducts from the breeder reactors, but Pu238 itself is actually not that dangerous to handle, and only toxic if ingested.


11

Any reasonably flat piece of sort-of reflective metal will function perfectly well as a heat collector, but would not be terribly suited to do astronomical observations with. In principle, you could probably pull it off. But it would require a lot more accurately shaped mirrors, with a lot better quality reflective surface. Also, there's good reasons ...


11

What is the size/scale of a wood fire that is producing 1kW? Based on what my understandings re energy content of fuels and combustion processes, about 0.5 to 1 cubic inch per minute of typically dry wood in an open fire. Based on an utterly superb 80 page Wood Fuels handbook which I discovered along the way - about the same, rather to my surprise. ...


11

If you express power loss in a power line as $V^2/R$, the $V$ in that expression is the voltage difference between the two ends of the power line, not the voltage difference between the power line and ground. To supply a fixed amount of power $P_L$ to a load, if the voltage at the load $V_L$ is larger, the current $I=P_L/V_L$ can be smaller. If the power ...


10

The energy will start dissipating in the atmosphere as corona losses, also the high tension(potential difference) between the wires and the ground will make it dangerous. There maybe electrical breakdowns and electric shocks which will then fry the wires along with causing damage to nearby objects and people it will also breakdown the transmission by ...


10

In principle, the drop in the Gibbs energy when the uranium gets converted to the fission products is available for doing useful work. While a steam engine will not come close to the maximum possible efficiency attainable (which is very close to 100%), a thermoelectric device will have much worse performance, as pointed out in detail in the other answers. ...


10

It's just an artifact of using different units for the same types of quantities in the same equation. Suppose we have an ideal pump that puts a force $F$ on a fluid to have it move at a steady velocity $v$. The power required to do this is $$ P = Fv. $$ Since pressure $p$ is force per unit area, then a flow with cross sectional area $A$ has $F = Ap$. At the ...


10

This is a cost to benefit question and can only be answered by a guess in a physics board. There is a new generation of small compact reactors that could be used for powering apartment buildings The new reactor, which is only 20 feet by 6 feet, Seems compact enough, so it is not size but weight that is important, since this weight has to be lifted. ...


9

There is thermal expansion of metals, and in a power line the length will become greater than the installation one. Gravity takes up the slack and lowers the lowest point of the line. The shape taken by suspended strings/cables with only their own weight is called a catenary, and the length depends on the constants of the solution ( alternative analysis in ...


9

One aspect is the concern for if the spacecraft were to fail to launch correctly and ended up crashing back to earth. In such cases, the nuclear radiation pollution could be severe if it ended up crashing in inhabited areas.


9

This would guarantee a meltdown. They're trying to get heat out of the core because---thought the fission chain reaction has been suppressed---various unstable fission daughters continue to decay. Adding hot lead would add heat to the system and not stop this behavior. Total disaster. -- If the core does slag out, it will probably end in a hot ...


9

The upper bound is easy. The solar constant through the atmosphere is about $970W \over m^2$ so you're gathering at most $\pi\cdot((0.75m)^2 \cdot 970W / m^2 = 1.7kW$ If your focal point is perfect (it isn't) then you'd have infinite power density. You will need to measure the actual power produced (and therefor the efficiency) because it depends on a ...


8

The German THTR-300 Thorium High-Temperature Reactor operated for about 16,000 hours and the IAEA produced a report on its shutdown. So there are no physics barriers to thorium reactors: there is an existence proof for thorium reactors. That ends the relevant answer for this site. There are economic, engineering, social, political, technical, and ...


8

As far as I know you have to worry about how much energy being deposited per surface area. And the area of the "hotspot" of the laser can be very small. So the deposited energy it enough to kill cells on your retina. Actually there is a whole article on Wikipedia about it.


7

Here's my guess: As you know, internal combustion engines burn fuel. The power output of the engine is a function of both the current RPM and the amount of fuel you inject. But there's a catch: the engine can burn a limited amount of fuel per cycle, and therefore the higher the RPM, the more fuel can be combusted. So (as @dmckee stated), at a given gear the ...


7

Short, short version: It's complicated. Slightly longer version: Internal combustion engines have at least two relevant performance characteristics: power and torque. Furthermore the maximum attainable values for both are functions of the current engine speed (RPM). Acceleration will cease if the current requirement for either power or torque equal the ...


7

Nuclear power is the only way to make submarines work underneath the water for long distances without coming up because of the oxygen that all other types require. And batteries for electric engines are too heavy. Historically a submarine had to go up to just below sea level to get air through the snorkel for the engines that charge the batteries. Then ...


7

Hmm, lets see. The melting point of lead is fairly low, 327.46 °C and it is a good absorber of radioactivity. I think the problem with the reactors is not the heat per se, but the exposure of the fuel rods to the air without cooling because of escaping steam not being replenished by cooling water. The remaining steam etc may blow out the thick container ...



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