Again, I originally assumed black bodies because they're simpler: black bodies don't reflect any radiation. That means "power in" depends on the chamber walls and "power out" through that boundary only depends on the heat source.
I have it on record where you insisted that we assume gray bodies so that we could include a term for emissivity. Seriously. You insisted. I'm not going to look it up this late at night, because you are getting completely ridiculous. But I am sure as hell going to include it in my publication. [Jane Q. Public, 2014-10-08]
Remember to include that part where I originally assumed black bodies because they're simpler. And the part where Jane insisted that "we should use real materials with real emissivities and absorptivies. Just to keep everybody honest."
But black bodies aren't "dishonest". Also, Jane should make sure to include the part where Jane said I was "lying again" for considering a black body source.
But if Jane wants to work on the simpler black body problem that I originally proposed months ago, that's fine with me. It's simpler, and easier to learn from.
... the equation for radiant power emittance at steady state does NOT say "X + ( (epsilon * sigma) * T^4) - X". It simply says (epsilon * sigma) * T^4. Because it's already known that X cancels out!!!... [Jane Q. Public, 2014-10-08]
No, that's because the equation for radiant power emittance doesn't have anything to do with conservation of energy, so those extra terms wouldn't be in the Stefan-Boltzmann equation in the first place.
That's what I've been trying to tell you, Jane. The Stefan-Boltzmann equation can give you "radiative power out" but only a completely different principle called "conservation of energy" can give you a totally different quantity known as "electrical heating power".
My energy conservation equation is this: electrical power in = (epsilon * sigma) * T^4 * area = radiant power out [Jane Q. Public, 2014-10-08]
So you're saying electrical heating power is the same as radiative power out? What did that energy conservation equation look like before you cancelled terms? It's important.
Jane still hasn't written down an energy conservation equation for a boundary around the heated source which links "electrical heating power" to "radiative power out".
Yes, I have. I have done it at least several times before, and I just did it again. Not only did I give you the equations, I showed you my exact calculations. Why are you lying again, and trying to claim I did not do something that I very clearly and provably did do? In fact, since you seem to be so obsessed with archiving other people's comments, I am sure you have several instances of where I've showed this to you before. [Jane Q. Public, 2014-10-08]
Jane, just two days ago you claimed that you didn't say radiative power out was the same as electrical heating power, and that they don't need to be the same. Today you're saying they are the same.
Again, why does Jane think if something doesn't affect the power out, it can't affect the power in? For example, black body "power in" depends on the chamber walls even though "power out" through that boundary doesn't depend on the chamber walls.
Not according to my thermodynamics textbooks. Simply stating this, and linking to yourself stating it again elsewhere, isn't any kind of argument. In analyzing Spencer's challenge, we could have assumed black bodies. The only reason we didn't was because YOU insisted that you wanted to include an emissivity figure. But it still doesn't change the general principle. [Jane Q. Public, 2014-10-08]
Again, I originally assumed black bodies because they're simpler: black bodies don't reflect any radiation. That means "power in" depends on the chamber walls and "power out" through that boundary only depends on the heat source.
Again, why does Jane think if something doesn't affect the power out, it can't affect the power in?
Conservation of energy. Your own idea of power in through a boundary = power out through that boundary. If your boundary is around JUST the heat source, the only NET power in is electricity, and the only NET power out is radiation. I see absolutely no reason (if we assume 100% efficiency) that these should not be equal. [Jane Q. Public, 2014-10-08]
The crucial assumption isn't 100% efficiency, it's that nothing inside the boundary is changing. If not, power in != power out. Either way, conservation of energy doesn't imply that "if something doesn't affect the power out, it can't affect the power in." Otherwise it would apply to black bodies, and that isn't true. Otherwise it would apply even if that source is warming, so power in > power out, but that isn't true either.
Your continued assertion that, at steady-state, the presence of a nearby cooler body somehow affects the power output of a warmer body at known temperature is a bizarre violation of the Second Law of Thermodynamics. The power output is what it is. It depends only on emissivity and temperature. [Jane Q. Public, 2014-10-08]
Once again, Jane confuses "radiative power out" which depends only on emissivity and temperature, with "electrical heating power" which goes to zero if the chamber walls are also at 150F.
This does not even remotely resemble my equation. The textbook thermodynamic answer is: radiant power out at steady-state, per unit area, equals (emissivity * Stefan-Boltzmann constant) * T^4. That's all. The end. [Jane Q. Public, 2014-10-08]
Jane coyly says that my attempt to understand Jane's energy conservation equation doesn't even remotely resemble Jane's super-secret energy conservation equation. Which he still refuses to write down.
Then, once again, Jane writes down the Stefan-Boltzmann equation which only determines "radiative power out" without even trying to write down an energy conservation equation to show how it relates to "electrical heating power". This means Jane either doesn't understand that "radiative power out" is different than "electrical heating power", or Jane doesn't understand that conservation of energy is necessary to l
Yes, liquid water's vapor pressure is a problem in vacuum. Your solution seems like a good idea: loosely glue a patch over the hole before spraying. After the water freezes, its vapor pressure also decreases. Then the patch could be removed and reused if the cost of dealing with the remaining sublimation loss is less than the cost of replacing bladders hit by micrometeorites.
The remaining sublimation loss could be minimized by keeping the glazed shield as cold as possible. At first it seemed like this would be easier in the outer solar system, but then I realized that the side of the ship facing the sun would probably be covered with solar panels anyway. Moving farther from the sun either requires larger solar panels, or a large cheap mirror to collect more sunlight.
So the most important variable is how much power the ship needs. I'm working on a simple design which has enough garden space to feed 4 people, and I calculated the power needed to light the garden as it would be lit on Earth. Assuming blackbody radiators at 0C (which puts a lower bound on the attainable interior temperature), dissipating that power requires that ~30% of the ship's surface not covered with solar panels would need to be covered with radiators.
And that's just the power needed to light the garden. So you're probably right: it would be better to cover the rest of the surface with bladders to reduce sublimation loss. Those bladders could be covered with radiators and individually connected or disconnected to the ship's interior via insulated loop heat pipes.
Also, I liked your pykrete link. Wood pulp is likely to be expensive in space, but glass fiber made from moon dust could probably be cheap.
... don't try to tell me you're calculating the TOTAL electrical power needed to both heat the source and cool the walls, because that would be a different experiment. Spencer stipulated "electrical power" to the heat source. He left power to the walls unstated, except to say that they are maintained at 0 degrees F. He did not say the power to the heat source AND to the walls was constant.... [Jane Q. Public, 2014-10-07]
Again, I've repeatedly explained that the power needed to cool the walls is irrelevant, and that it isn't required to be constant.
The problem with your theory is that you have failed to show that electrical power in = anything BUT power out. It isn't heat transfer, as you have several times asserted. Heat transfer to a cooler body has NO relevance to the radiated power output of a warmer body at known temperature. And since it does not affect the power out, it does not affect the power in. QED. [Jane Q. Public, 2014-10-07]
Again, why does Jane think if something doesn't affect the power out, it can't affect the power in? For example, black body "power in" depends on the chamber walls even though "power out" through that boundary doesn't depend on the chamber walls.
Since we agree that "electrical heating power" goes to zero when the chamber walls are also at 150F, has Jane also noticed that "net heat transfer" also goes to zero when the chamber walls are also at 150F?
Isn't that a weird coincidence? So why does Jane keep using an equation that depends on "electrical heating power = radiative power out" without even writing down an energy conservation equation to try to justify that claim? Has Jane even considered the possibility that if he applied conservation of energy, he'd find that electrical heating power really is determined by net heat transfer, rather than "radiative power out" which stays constant even if the chamber walls are also at 150F?
If you draw your boundary around just the heat source itself, since there is NO NET RADIATIVE POWER COMING IN (which doesn't then just go right back OUT, yielding a net of 0)... [Jane Q. Public, 2014-10-07]
If there's no net radiative power coming in, that must mean all the "power in" from the chamber walls just goes back out. That would yield a net of zero. But as usual Jane didn't write down the power in = power out equation showing these terms before they supposedly cancel. Is this what you mean, Jane?
Draw a boundary around the heat source:
Jane's power in = electrical heating power + radiative power in from chamber walls
Jane's power out = radiative power out from source + radiative power from chamber walls, re-emitted back out
At steady state, Jane's power in = Jane's power out:
electrical heating power + radiative power in from chamber walls = radiative power out from source + radiative power from chamber walls, re-emitted back out (Jane's equation?)
Jane, is that your equation for required electrical heating power? By "NO NET RADIATIVE POWER COMING IN", are you saying "radiative power in from the chamber walls" = "radiative power from chamber walls, re-emitted back out"?
... don't try to tell me you're calculating the TOTAL electrical power needed to both heat the source and cool the walls, because that would be a different experiment. Spencer stipulated "electrical power" to the heat source. He left power to the walls unstated, except to say that they are maintained at 0 degrees F. He did not say the power to the heat source AND to the walls was constant.... [Jane Q. Public, 2014-10-07]
Again, I've repeatedly explained that the power needed to cool the walls is irrelevant, and that it isn't required to be constant.
The problem with your theory is that you have failed to show that electrical power in = anything BUT power out. It isn't heat transfer, as you have several times asserted. Heat transfer to a cooler body has NO relevance to the radiated power output of a warmer body at known temperature. And since it does not affect the power out, it does not affect the power in. QED. [Jane Q. Public, 2014-10-07]
Again, why does Jane think if something doesn't affect the power out, it can't affect the power in? For example, black body "power in" depends on the chamber walls even though "power out" through that boundary doesn't depend on the chamber walls.
Since we agree that "electrical heating power" goes to zero when the chamber walls are also at 150F, has Jane also noticed that "net heat transfer" also goes to zero when the chamber walls are also at 150F?
Isn't that a weird coincidence? So why does Jane keep using an equation that depends on "electrical heating power = radiative power out" without even writing down an energy conservation equation to try to justify that claim? Has Jane even considered the possibility that if he applied conservation of energy, he'd find that electrical heating power really is determined by net heat transfer, rather than "radiative power out" which stays constant even if the chamber walls are also at 150F?
If you draw your boundary around just the heat source itself, since there is NO NET RADIATIVE POWER COMING IN (which doesn't then just go right back OUT, yielding a net of 0)... [Jane Q. Public, 2014-10-07]
If there's no net radiative power coming in, that must mean all the "power in" from the chamber walls just goes back out. That would yield a net of zero. But as usual Jane didn't write down the power in = power out equation showing these terms before they supposedly cancel. Is this what you mean, Jane?
Draw a boundary around the heat source:
Jane's power in = electrical heating power + radiative power in from chamber walls
Jane's power out = radiative power out from source + radiative power from chamber walls, re-emitted back out
At steady state, Jane's power in = Jane's power out:
electrical heating power + radiative power in from chamber walls = radiative power out from source + radiative power from chamber walls, re-emitted back out (Jane's equation?)
Jane, is that your equation for required electrical heating power? By "NO NET RADIATIVE POWER COMING IN", are you saying "radiative power in from the chamber walls" = "radiative power from chamber walls, re-emitted back out"?
Oops, I actually plotted inverse density versus half-value layer thicknesses. This doesn't affect the conclusion, but here's a plot with a corrected file name.
Oops, I actually plotted inverse density versus half-value layer thicknesses. This doesn't affect the conclusion, but here's a plot with a corrected file name.
I suspect that glazing a ship in ice would always be cheaper than building a network of chambers to act as a radiation shield. That's because it seems like improvements in robotic construction (prefabrication, etc.) could also be applied to the glazing process.
It seems even more likely that the repair costs of a "glazed" shield would be lower than a shield made out of water chambers. If a "large" micrometeorite blasts a chunk of a glazed shield away, you just send a robot with a water tank out to the hole and let it spray more water on the shield. If that micrometeorite hits a shield made out of water chambers, you have to repair or replace whatever chambers and pumps were damaged in the explosion.
Again, metric tons per square meter = thickness * density. That means the half-value layer should be inversely proportional to the shield's density. So if metric tons per square meter are relevant to the half-value layer, the half-value layer should be inversely proportional to the shield's density.
Yes, on a first glance it should. But in fact it does not. It highly depends on the material you use.
Again, to a good approximation, those half-value layers are inversely proportional to the shield's density.
I plotted those half-value layers against the inverse densities of concrete, steel, lead, tungsten and uranium. The blue squares are for the iridium source, and the red circles are for the cobalt source. Since those points lie close to a straight line, radiation absorption is determined primarily by density. So metric tons per square meter is a good first order approximation, at least for those materials.
What do you mean by that? What are "people like me"? "Laymen"?
Obviously...
Why are you making assumptions about who I am? If you clicked on my homepage, it would only take a few seconds to realize that you're wrong. But more importantly, it's not necessary or productive to accuse someone of being a layman. There's no reason to be nasty. Just discuss the science, and leave your assumptions about who the other person is out of it.
I linked you the 'half value layer' articles... metric tons per square meter are irrelevant.
No. Metric tons per square meter = thickness * density, so if density is relevant then metric tons per square meter is also relevant.
Relevant is how dense the material is and what its actual properties are to 'break' or capture cosmic rays. A ton of water simply does not equal a ton of lead, even if you believe so after you got missleaded by that NASA article:)
The NASA article I showed you explicitly calculated the required shielding using silicon dioxide (Moon dust) as I've failed to explain. They're not saying a ton of water exactly equals a ton of lead, and neither am I.
no one uses tons per m^2 to describe radiation absobtion.
A measure like that would imply the material used is irrelevant, which it is not. The correct material is the prime shielding factor.
No, density is the prime shielding factor. That means metric tons per square meter is a good first order approximation.
That is a laymen explanation for people like you. I linked you the 'half value layer' articles... metric tons per square meter are irrelevant. Relevant is how dense the material is and what its actual properties are to 'break' or capture cosmic rays. A ton of water simply does not equal a ton of lead, even if you believe so after you got missleaded by that NASA article:)
Again, metric tons per square meter = thickness * density. That means the half-value layer should be inversely proportional to the shield's density. So if metric tons per square meter are relevant to the half-value layer, the half-value layer should be inversely proportional to the shield's density.
Did you try plotting those half-value layers against the inverse densities for concrete, steel, lead, tungsten and uranium? If you did, you'd notice that they're all close to a straight line. So metric tons of shielding per square meter is a good first order approximation.
Also, you claimed I mixed up the travel time, but you still haven't shown that my 3.5 day travel time to Mars at 0.25g is somehow wrong. What travel time did you get?
What "I propose" is the textbook answer to this question. It's not even "my" idea, as I clearly showed you just yesterday. YOU are the one going against "established" physics here. So I daresay it's up to you to prove your point, rather than arguing with me about it. Which you will never do, because you're wrong. If you could actually show how the physics textbook idea of heat transfer was wrong, you would be world famous by now. Instead, you're arguing ineffectively with some person on Slashdot, about something every textbook on the subject, as well as other sources, say your are wrong about. [Jane Q. Public, 2014-10-06]
Once again, your textbooks don't say I'm wrong. They just say that "radiative power out per square meter = (e*s)*T^4". Once again, I agree with that statement. So how am I going against "established" physics or arguing with "every textbook on the subject"?
Seriously, "radiative power out" is different than "electrical heating power". For instance, we agree that "radiative power out" stays constant even if the chamber walls are also at 150F, but "electrical heating power" goes to zero. So they can't be the same.
electrical heating power out per square meter = (e*s)*T1^4.
But the Stefan-Boltzmann law in your textbooks actually says:
radiative power out per square meter = (e*s)*T1^4.
Jane, don't you see how your equation for electrical heating power would only be true if "radiative power out = electrical heating power"? If you "didn't say they were the same" then why does your equation depend on them being the same?
Instead of calling me a blathering religious zealot liar who wasn't ever actually a physicist, could you calmly explain why you disagree with this energy conservation equation?
I already did so, several times. What, do you honestly think that If I fail to refute this idea just one more time, it will somehow magically become correct? The heat transfer scenario I presented, and my calculations of temperatures, were correct within a reasonable degree of precision. Yours, on the other hand, were not. By what stretch of your imagination am I obligated to KEEP refuting your same, lame arguments? This is all old news now. You can read about it all again later, when I write this all up and publish it. In the meantime, if you want answers to these questions AGAIN, you can go back and read our prior discussion of the matter. [Jane Q. Public, 2014-10-06]
Jane seemed to try to explain why he disagrees here by saying "A = A" and helpfully asking if I knew what a zero was. But as usual Jane refused to actually write down what Jane considered to be the "correct" energy conservation equation. When I asked what equation Jane meant, Jane said that wasn't it. So Jane's never written down an energy conservation equation around the heated source, which is the first step to calculating the required electrical heating power.
Here's how to use the principle of conservation of energy. Draw a boundary around the heat source:
power in = electrical heating power + radiative power in from the chamber walls
power out = radiative power out from the heat source
You're just re-hashing old arguments that I've already shot down. Why are you doing that, if your purpose is not dishonest? [Jane Q. Public, 2014-10-06]
Dishonest? Shot down? Have you even considered the possibility that radiative power out might actually be different than electrical heating power?
For instance, we agree that "radiative power out" stays constant even if the chamber walls are also at 150F, but "electrical heating power" goes to zero. So they can't be the same. Is saying that dishonest?
This doesn't cause "radiative power out" to depend on anything but its emissivity and temperature. Is saying that dishonest?
If you want to propose some relationship between "radiative power out" and "electrical heating power" then you need to use conservation of energy. Is saying that dishonest?
Here's how to use the principle of conservation of energy. Draw a boundary around the heat source:
power in = electrical heating power + radiative power in from the chamber walls
power out = radiative power out from the heat source
Since power in = power out through any boundary where nothing inside is changing:
electrical heating power + radiative power in from the chamber walls = radiative power out from the heat source
Instead of calling me dishonest, could you calmly explain why you disagree with this energy conservation equation?
Seriously, "radiative power out" is different than "electrical heating power". For instance, we agree that "radiative power out" stays constant even if the chamber walls are also at 150F, but "electrical heating power" goes to zero. So they can't be the same.
This doesn't cause "radiative power out" to depend on anything but its emissivity and temperature.
If you want to propose some relationship between "radiative power out" and "electrical heating power" then you need to use conservation of energy.
Here's how to use the principle of conservation of energy. Draw a boundary around the heat source:
power in = electrical heating power + radiative power in from the chamber walls
power out = radiative power out from the heat source
Since power in = power out through any boundary where nothing inside is changing:
electrical heating power + radiative power in from the chamber walls = radiative power out from the heat source
Instead of calling me a blathering religious zealot liar who wasn't ever actually a physicist, could you calmly explain why you disagree with this energy conservation equation?
I've repeatedly failed to communicate that I agree radiative power out is a function of emissivity and temperature only:
"Again, radiative power out is dependent only on emissivity and thermodynamic temperature. We don't disagree about that, despite your repetitive claims to the contrary."
Once again, I'm just saying that "radiative power out" is different than "electrical heating power".
... But note that the equation for power out clearly implies it is independent of transfer to cooler bodies.... [Jane Q. Public, 2014-10-05]
I've repeatedly failed to communicate that I agree radiative power out is independent of transfer to cooler bodies:
"Once again, I agree that "power out" through a boundary drawn around the heat source is given by the Stefan-Boltzmann law."
Once again, I'm just saying that "radiative power out" is different than "electrical heating power".
... In this case, note that it gives the equation for power output as distinct from radiation "loss" (heat transfer). BECAUSE THEY ARE TWO DIFFERENT THINGS. One is the power output of a SINGLE gray body at a given temperature. The other is radiative transfer to another body. One requires ONLY emissivity and temperature to calculate. The other involves 2 bodies.... [Jane Q. Public, 2014-10-05]
Exactly. Radiative power out is different than electrical heating power, because only electrical heating power goes to zero when the chamber walls are also at 150F. So electrical heating power involves 2 bodies, but radiative power out requires ONLY emissivity and temperature to calculate.
... are you going to continue to erroneously claim that radiative POWER output is dependent on the presence of cooler bodies?... [Jane Q. Public, 2014-10-05]
I've never claimed that radiative power out is dependent on the presence of cooler bodies. Once again, I've repeatedly agreed that radiative power output doesn't depend on the presence of cooler bodies:
Once again, I'm claiming that "radiative power out" is different than "electrical heating power". For instance, we agree that "radiative power out" stays constant even if the chamber walls are also at 150F, but "electrical heating power" goes to zero. So they can't be the same.
... Your misguided attempt to include the power used to cool the chamber walls does not change that.... It is neither necessary nor called for to calculate the power used to cool the chamber walls in order to find the temperatures of the other bodies.... [Jane Q. Public, 2014-10-05]
Good grief, Jane. Once again, I never attempted to include the power used to cool the chamber walls! In fact, I've repeatedly told you it's irrelevant. Once again, that's not what "power out" means. Months ago, after I asked Jane if he agreed that power in = power out, Jane misunderstood my question and responded:
... As long as the power used by the source and the power used by the cooler are constant as required, any relationship between them has no bearing on the experiment. [Jane Q. Public, 2014-08-02]
So I explained that "I've never even mentioned the power used by the cooler of the chamber walls... none of these equations has anything to do with the power used by the cooler.... Jane's also wrong to claim that the power used by the cooler is required to be constant...."
I tried again a month later: "I've repeatedlyfailed to explain that the power consumed by the refrigerator on the outside is irrelevant. So obviously we'll have to agree to disagree about that."
I tried once again: "... Jane might think I meant power in = electrical heating power, and power out = cooling power of the chamber walls. If so, that's not what I meant, and I'm sorry for not being more clear. I take full responsibility. Just to be clear, power in = power flowing into the boundary in question, and power in = power flowing out of that boundary.... any power used by the cooler is simply being moved from some point outside the boundary to another point which is also outside the boundary. Because that power never crosses the boundary, it's irrelevant."
I tried yet again: "I've explained why the power used to set the chamber wall temperature is irrelevant. Any power used is simply being moved from some point outside the boundary to another point which is also outside the boundary. Because that power never crosses the boundary, it's irrelevant."
... Your misguided attempt to include the power used to cool the chamber walls does not change that.... It is neither necessary nor called for to calculate the power used to cool the chamber walls in order to find the temperatures of the other bodies.... [Jane Q. Public, 2014-10-05]
After I repeatedly explained that the power used to cool the chamber walls is irrelevant, it's bewildering that Jane accuses me of trying to include it.
... The problem is that an analysis of this kind, based on the assumption that power-in = power-out, is doomed to fail except in coincidental cases. Even conservation of energy can give very misleading results.... power-in = power-out is not necessarily true, and i
no one uses tons per m^2 to describe radiation absobtion.
Except NASA: "Passive shielding is known to work. The Earth's atmosphere supplies about 10 t/m^2 of mass shielding and is very effective. Only half this much is needed to bring the dosage level of cosmic rays down to 0.5 rem/yr. In fact when calculations are made in the context of particular geometries, it is found that because many of the incident particles pass through walls at slanting angles a thickness of shield of 4.5 t/m^2 is sufficient."
If you think I mixed up the travel time, try calculating the travel time at 0.25g from Earth to Mars when it's closest to Earth at 55 million kilometers. Again, that only takes 3.5 days.
Just no. That is not even remotely what I meant, and I explained this to you clearly at least several times already. I have no reason to continue to re-explain it just because you keep asking. Instead I'm going to repeat something else I have stated several times: pick up a textbook on heat transfer, and see what the accepted, textbook, "consensus" science says about it. Hint: they don't agree with you. [Jane Q. Public, 2014-10-05]
Jane, mainstream physics is based on conservation of energy. That means power in = power out through any boundary where nothing inside is changing. If your textbook doesn't agree with that principle, it's either wrong or you're misinterpreting what it says. For instance:
I will do you a favor here, and say: don't bother to go calculating the energy, either. The problem is that an analysis of this kind, based on the assumption that power-in = power-out, is doomed to fail except in coincidental cases. Even conservation of energy can give very misleading results. The black body example I gave shows why your "energy conservation just inside the surface" won't work. Aside from just "view factor" and a few other things, a certain amount of the power in (often a very significant amount) just ends up going right back out, but you often don't see that in the formulas. Quote from one of my references, "Fundamentals of Heat and Mass Transfer", by Inropera, et al., 6th edition, 2006, p13. I have to type this in by hand from the book so any typographical errors are mine. Emphasized words have been capitalized.
Relationship to Thermodynamics
At this point it is appropriate to note the fundamental differences between heat transfer and thermodynamics. Although thermodynamics is concerned with the heat interaction and the vital role it plays in the first and second laws, it considers neither the mechanisms that profide for heat exchange nor the methods that exist for computing the RATE of heat exchange. Thermodynamics is concerned with EQUILIBRIUM states of matter, where an equilibrium state necessarily precludes the existence of a temperature gradient. Although thermodynamics may be used to determine the amount of energy required in the form of heat to pass from one equilibrium state to another, it does not acknowledge that HEAT TRANSFER IS INHERENTLY A NONEQUILIBRIUM PROCESS. For heat transfer to occur, there must be a temperature gradient and, hence, thermodynamic nonequilibrium. The discipline of heat transfer therefore seeks to do what thermodynamics is inherently unable to do, namely, to quantify the RATE at which heat transfer occurs in terms of the degree of thermal nonequilibrium. This is done via the rate equations for the three modes...
Heat transfer requires a temperature gradient, and therefore thermodynamic non-equilibrium (as we established early on). I was hoping you would catch on that this also implies that power-in = power-out is not necessarily true, and in fact that is probably a very rare exception. Therefore, you aren't going to prove anything with this approach. I wanted to stop you before you wasted more of your time. [Jane Q. Public, 2014-09-07]
No Jane, you've misinterpreted your textbook. Energy is always conserved, so power in = power out through any boundary where nothing inside is changing. This isn't a "very rare exception". It's a fundamental law called "conservation of energy". Does Jane seriously think his textbook says that using a fundamental law like "conservation of energy" is "doomed to fail"?
Again, it really sounds like Jane opened a textbook and found "radiative power out per square meter = (e*s)*T^4" and simply assumed that "radiative power out" is just a f
That makes sense. I was thinking in terms of the lighthuggers in Revelation Space which are literally "glazed" in water ice because that's cheaper than setting up a system of tanks and pumps. That shield would be very easy to repair even after a collision with a "large" micrometeorite because there would be no infrastructure. Just melt more water and apply.
The system you describe would also be useful as an alternate (albeit temporary) way to dump heat in case the external radiators were damaged. I've been thinking about a similar setup, but using loop heat pipes instead of steam pumps because heat pipes don't have moving parts.
If a VASIMR drive could sustain 0.25g acceleration, its fuel tanks would be enormous. It would also use a lot of power, requiring either a nuclear reactor or huge solar panels capable of supporting themselves at 0.25g.
But if it could be done, continuously accelerating at 0.25g to the midpoint then decelerating at 0.25g would result in an Earth-Mars travel time much shorter than 3 months. When Mars is closest to Earth, the travel time would only be 3.5 days. Even when Mars is on the other side of the Sun, the travel time would only be 9.4 days.
Slashdot Mirror
Answered here.
Remember to include that part where I originally assumed black bodies because they're simpler. And the part where Jane insisted that "we should use real materials with real emissivities and absorptivies. Just to keep everybody honest."
But black bodies aren't "dishonest". Also, Jane should make sure to include the part where Jane said I was "lying again" for considering a black body source.
But if Jane wants to work on the simpler black body problem that I originally proposed months ago, that's fine with me. It's simpler, and easier to learn from.
No, that's because the equation for radiant power emittance doesn't have anything to do with conservation of energy, so those extra terms wouldn't be in the Stefan-Boltzmann equation in the first place.
That's what I've been trying to tell you, Jane. The Stefan-Boltzmann equation can give you "radiative power out" but only a completely different principle called "conservation of energy" can give you a totally different quantity known as "electrical heating power".
So you're saying electrical heating power is the same as radiative power out? What did that energy conservation equation look like before you cancelled terms? It's important.
Jane, just two days ago you claimed that you didn't say radiative power out was the same as electrical heating power, and that they don't need to be the same. Today you're saying they are the same.
Jane responds.
Again, I originally assumed black bodies because they're simpler: black bodies don't reflect any radiation. That means "power in" depends on the chamber walls and "power out" through that boundary only depends on the heat source.
The crucial assumption isn't 100% efficiency, it's that nothing inside the boundary is changing. If not, power in != power out. Either way, conservation of energy doesn't imply that "if something doesn't affect the power out, it can't affect the power in." Otherwise it would apply to black bodies, and that isn't true. Otherwise it would apply even if that source is warming, so power in > power out, but that isn't true either.
Once again, Jane confuses "radiative power out" which depends only on emissivity and temperature, with "electrical heating power" which goes to zero if the chamber walls are also at 150F.
Jane coyly says that my attempt to understand Jane's energy conservation equation doesn't even remotely resemble Jane's super-secret energy conservation equation. Which he still refuses to write down.
Then, once again, Jane writes down the Stefan-Boltzmann equation which only determines "radiative power out" without even trying to write down an energy conservation equation to show how it relates to "electrical heating power". This means Jane either doesn't understand that "radiative power out" is different than "electrical heating power", or Jane doesn't understand that conservation of energy is necessary to l
Yes, liquid water's vapor pressure is a problem in vacuum. Your solution seems like a good idea: loosely glue a patch over the hole before spraying. After the water freezes, its vapor pressure also decreases. Then the patch could be removed and reused if the cost of dealing with the remaining sublimation loss is less than the cost of replacing bladders hit by micrometeorites.
The remaining sublimation loss could be minimized by keeping the glazed shield as cold as possible. At first it seemed like this would be easier in the outer solar system, but then I realized that the side of the ship facing the sun would probably be covered with solar panels anyway. Moving farther from the sun either requires larger solar panels, or a large cheap mirror to collect more sunlight.
So the most important variable is how much power the ship needs. I'm working on a simple design which has enough garden space to feed 4 people, and I calculated the power needed to light the garden as it would be lit on Earth. Assuming blackbody radiators at 0C (which puts a lower bound on the attainable interior temperature), dissipating that power requires that ~30% of the ship's surface not covered with solar panels would need to be covered with radiators.
And that's just the power needed to light the garden. So you're probably right: it would be better to cover the rest of the surface with bladders to reduce sublimation loss. Those bladders could be covered with radiators and individually connected or disconnected to the ship's interior via insulated loop heat pipes.
Also, I liked your pykrete link. Wood pulp is likely to be expensive in space, but glass fiber made from moon dust could probably be cheap.
Again, I've repeatedly explained that the power needed to cool the walls is irrelevant, and that it isn't required to be constant.
Again, why does Jane think if something doesn't affect the power out, it can't affect the power in? For example, black body "power in" depends on the chamber walls even though "power out" through that boundary doesn't depend on the chamber walls.
Since we agree that "electrical heating power" goes to zero when the chamber walls are also at 150F, has Jane also noticed that "net heat transfer" also goes to zero when the chamber walls are also at 150F?
Isn't that a weird coincidence? So why does Jane keep using an equation that depends on "electrical heating power = radiative power out" without even writing down an energy conservation equation to try to justify that claim? Has Jane even considered the possibility that if he applied conservation of energy, he'd find that electrical heating power really is determined by net heat transfer, rather than "radiative power out" which stays constant even if the chamber walls are also at 150F?
If there's no net radiative power coming in, that must mean all the "power in" from the chamber walls just goes back out. That would yield a net of zero. But as usual Jane didn't write down the power in = power out equation showing these terms before they supposedly cancel. Is this what you mean, Jane?
Draw a boundary around the heat source:
Jane's power in = electrical heating power + radiative power in from chamber walls
Jane's power out = radiative power out from source + radiative power from chamber walls, re-emitted back out
At steady state, Jane's power in = Jane's power out:
electrical heating power + radiative power in from chamber walls = radiative power out from source + radiative power from chamber walls, re-emitted back out (Jane's equation?)
Jane, is that your equation for required electrical heating power? By "NO NET RADIATIVE POWER COMING IN", are you saying "radiative power in from the chamber walls" = "radiative power from chamber walls, re-emitted back out"?
Again, I've repeatedly explained that the power needed to cool the walls is irrelevant, and that it isn't required to be constant.
Again, why does Jane think if something doesn't affect the power out, it can't affect the power in? For example, black body "power in" depends on the chamber walls even though "power out" through that boundary doesn't depend on the chamber walls.
Since we agree that "electrical heating power" goes to zero when the chamber walls are also at 150F, has Jane also noticed that "net heat transfer" also goes to zero when the chamber walls are also at 150F?
Isn't that a weird coincidence? So why does Jane keep using an equation that depends on "electrical heating power = radiative power out" without even writing down an energy conservation equation to try to justify that claim? Has Jane even considered the possibility that if he applied conservation of energy, he'd find that electrical heating power really is determined by net heat transfer, rather than "radiative power out" which stays constant even if the chamber walls are also at 150F?
If there's no net radiative power coming in, that must mean all the "power in" from the chamber walls just goes back out. That would yield a net of zero. But as usual Jane didn't write down the power in = power out equation showing these terms before they supposedly cancel. Is this what you mean, Jane?
Draw a boundary around the heat source:
Jane's power in = electrical heating power + radiative power in from chamber walls
Jane's power out = radiative power out from source + radiative power from chamber walls, re-emitted back out
At steady state, Jane's power in = Jane's power out:
electrical heating power + radiative power in from chamber walls = radiative power out from source + radiative power from chamber walls, re-emitted back out (Jane's equation?)
Jane, is that your equation for required electrical heating power? By "NO NET RADIATIVE POWER COMING IN", are you saying "radiative power in from the chamber walls" = "radiative power from chamber walls, re-emitted back out"?
Oops, I actually plotted inverse density versus half-value layer thicknesses. This doesn't affect the conclusion, but here's a plot with a corrected file name.
Oops, I actually plotted inverse density versus half-value layer thicknesses. This doesn't affect the conclusion, but here's a plot with a corrected file name.
I suspect that glazing a ship in ice would always be cheaper than building a network of chambers to act as a radiation shield. That's because it seems like improvements in robotic construction (prefabrication, etc.) could also be applied to the glazing process.
It seems even more likely that the repair costs of a "glazed" shield would be lower than a shield made out of water chambers. If a "large" micrometeorite blasts a chunk of a glazed shield away, you just send a robot with a water tank out to the hole and let it spray more water on the shield. If that micrometeorite hits a shield made out of water chambers, you have to repair or replace whatever chambers and pumps were damaged in the explosion.
Typo: Jane, you're saying: electrical heating power per square meter = (e*s)*T1^4.
Again, to a good approximation, those half-value layers are inversely proportional to the shield's density.
I plotted those half-value layers against the inverse densities of concrete, steel, lead, tungsten and uranium. The blue squares are for the iridium source, and the red circles are for the cobalt source. Since those points lie close to a straight line, radiation absorption is determined primarily by density. So metric tons per square meter is a good first order approximation, at least for those materials.
Why are you making assumptions about who I am? If you clicked on my homepage, it would only take a few seconds to realize that you're wrong. But more importantly, it's not necessary or productive to accuse someone of being a layman. There's no reason to be nasty. Just discuss the science, and leave your assumptions about who the other person is out of it.
No. Metric tons per square meter = thickness * density, so if density is relevant then metric tons per square meter is also relevant.
The NASA article I showed you explicitly calculated the required shielding using silicon dioxide (Moon dust) as I've failed to explain. They're not saying a ton of water exactly equals a ton of lead, and neither am I.
No, density is the prime shielding factor. That means metric tons per square meter is a good first order approximation.
Again, metric tons per square meter = thickness * density. That means the half-value layer should be inversely proportional to the shield's density. So if metric tons per square meter are relevant to the half-value layer, the half-value layer should be inversely proportional to the shield's density.
Did you try plotting those half-value layers against the inverse densities for concrete, steel, lead, tungsten and uranium? If you did, you'd notice that they're all close to a straight line. So metric tons of shielding per square meter is a good first order approximation.
Also, you claimed I mixed up the travel time, but you still haven't shown that my 3.5 day travel time to Mars at 0.25g is somehow wrong. What travel time did you get?
Once again, your textbooks don't say I'm wrong. They just say that "radiative power out per square meter = (e*s)*T^4". Once again, I agree with that statement. So how am I going against "established" physics or arguing with "every textbook on the subject"?
Jane, you're saying:
electrical heating power out per square meter = (e*s)*T1^4.
But the Stefan-Boltzmann law in your textbooks actually says:
radiative power out per square meter = (e*s)*T1^4.
Jane, don't you see how your equation for electrical heating power would only be true if "radiative power out = electrical heating power"? If you "didn't say they were the same" then why does your equation depend on them being the same?
Jane seemed to try to explain why he disagrees here by saying "A = A" and helpfully asking if I knew what a zero was. But as usual Jane refused to actually write down what Jane considered to be the "correct" energy conservation equation. When I asked what equation Jane meant, Jane said that wasn't it. So Jane's never written down an energy conservation equation around the heated source, which is the first step to calculating the required electrical heating power.
Here's how to use the principle of conservation of energy. Draw a boundary around the heat source:
power in = electrical heating power + radiative power in from the chamber walls
power out = radiative power out from the heat source
Since power in = power out throu
Dishonest? Shot down? Have you even considered the possibility that radiative power out might actually be different than electrical heating power?
For instance, we agree that "radiative power out" stays constant even if the chamber walls are also at 150F, but "electrical heating power" goes to zero. So they can't be the same. Is saying that dishonest?
This doesn't cause "radiative power out" to depend on anything but its emissivity and temperature. Is saying that dishonest?
If you want to propose some relationship between "radiative power out" and "electrical heating power" then you need to use conservation of energy. Is saying that dishonest?
Here's how to use the principle of conservation of energy. Draw a boundary around the heat source:
power in = electrical heating power + radiative power in from the chamber walls
power out = radiative power out from the heat source
Since power in = power out through any boundary where nothing inside is changing:
electrical heating power + radiative power in from the chamber walls = radiative power out from the heat source
Instead of calling me dishonest, could you calmly explain why you disagree with this energy conservation equation?
Seriously, "radiative power out" is different than "electrical heating power". For instance, we agree that "radiative power out" stays constant even if the chamber walls are also at 150F, but "electrical heating power" goes to zero. So they can't be the same.
This doesn't cause "radiative power out" to depend on anything but its emissivity and temperature.
If you want to propose some relationship between "radiative power out" and "electrical heating power" then you need to use conservation of energy.
Here's how to use the principle of conservation of energy. Draw a boundary around the heat source:
power in = electrical heating power + radiative power in from the chamber walls
power out = radiative power out from the heat source
Since power in = power out through any boundary where nothing inside is changing:
electrical heating power + radiative power in from the chamber walls = radiative power out from the heat source
Instead of calling me a blathering religious zealot liar who wasn't ever actually a physicist, could you calmly explain why you disagree with this energy conservation equation?
What do you mean by that? What are "people like me"? "Laymen"?
I've explained that net heat transfer = radiative power out - radiative power in, so of course they're not the same.
I've repeatedly failed to communicate that I agree radiative power out is a function of emissivity and temperature only:
"Again, radiative power out is dependent only on emissivity and thermodynamic temperature. We don't disagree about that, despite your repetitive claims to the contrary."
Once again, I'm just saying that "radiative power out" is different than "electrical heating power".
I've repeatedly failed to communicate that I agree radiative power out is independent of transfer to cooler bodies:
"Once again, I agree that "power out" through a boundary drawn around the heat source is given by the Stefan-Boltzmann law."
Once again, I'm just saying that "radiative power out" is different than "electrical heating power".
Exactly. Radiative power out is different than electrical heating power, because only electrical heating power goes to zero when the chamber walls are also at 150F. So electrical heating power involves 2 bodies, but radiative power out requires ONLY emissivity and temperature to calculate.
I've never claimed that radiative power out is dependent on the presence of cooler bodies. Once again, I've repeatedly agreed that radiative power output doesn't depend on the presence of cooler bodies:
"I've been trying to tell Jane: we don't disagree about the equation for radiative power out."
Once again, I'm claiming that "radiative power out" is different than "electrical heating power". For instance, we agree that "radiative power out" stays constant even if the chamber walls are also at 150F, but "electrical heating power" goes to zero. So they can't be the same.
If you want to propose some relatio
Good grief, Jane. Once again, I never attempted to include the power used to cool the chamber walls! In fact, I've repeatedly told you it's irrelevant. Once again, that's not what "power out" means. Months ago, after I asked Jane if he agreed that power in = power out, Jane misunderstood my question and responded:
So I explained that "I've never even mentioned the power used by the cooler of the chamber walls... none of these equations has anything to do with the power used by the cooler. ... Jane's also wrong to claim that the power used by the cooler is required to be constant. ..."
I tried again a month later: "I've repeatedly failed to explain that the power consumed by the refrigerator on the outside is irrelevant. So obviously we'll have to agree to disagree about that."
I tried once again: "... Jane might think I meant power in = electrical heating power, and power out = cooling power of the chamber walls. If so, that's not what I meant, and I'm sorry for not being more clear. I take full responsibility. Just to be clear, power in = power flowing into the boundary in question, and power in = power flowing out of that boundary. ... any power used by the cooler is simply being moved from some point outside the boundary to another point which is also outside the boundary. Because that power never crosses the boundary, it's irrelevant."
I tried yet again: "I've explained why the power used to set the chamber wall temperature is irrelevant. Any power used is simply being moved from some point outside the boundary to another point which is also outside the boundary. Because that power never crosses the boundary, it's irrelevant."
After I repeatedly explained that the power used to cool the chamber walls is irrelevant, it's bewildering that Jane accuses me of trying to include it.
Except NASA: "Passive shielding is known to work. The Earth's atmosphere supplies about 10 t/m^2 of mass shielding and is very effective. Only half this much is needed to bring the dosage level of cosmic rays down to 0.5 rem/yr. In fact when calculations are made in the context of particular geometries, it is found that because many of the incident particles pass through walls at slanting angles a thickness of shield of 4.5 t/m^2 is sufficient."
If I'm mixing lots of stuff up, just explain how this NASA study was wrong to conclude that 4.41 tons/m^2 would be sufficient shielding.
If you think I mixed up the travel time, try calculating the travel time at 0.25g from Earth to Mars when it's closest to Earth at 55 million kilometers. Again, that only takes 3.5 days.
Jane, mainstream physics is based on conservation of energy. That means power in = power out through any boundary where nothing inside is changing. If your textbook doesn't agree with that principle, it's either wrong or you're misinterpreting what it says. For instance:
No Jane, you've misinterpreted your textbook. Energy is always conserved, so power in = power out through any boundary where nothing inside is changing. This isn't a "very rare exception". It's a fundamental law called "conservation of energy". Does Jane seriously think his textbook says that using a fundamental law like "conservation of energy" is "doomed to fail"?
Again, it really sounds like Jane opened a textbook and found "radiative power out per square meter = (e*s)*T^4" and simply assumed that "radiative power out" is just a f
That makes sense. I was thinking in terms of the lighthuggers in Revelation Space which are literally "glazed" in water ice because that's cheaper than setting up a system of tanks and pumps. That shield would be very easy to repair even after a collision with a "large" micrometeorite because there would be no infrastructure. Just melt more water and apply.
The system you describe would also be useful as an alternate (albeit temporary) way to dump heat in case the external radiators were damaged. I've been thinking about a similar setup, but using loop heat pipes instead of steam pumps because heat pipes don't have moving parts.
If a VASIMR drive could sustain 0.25g acceleration, its fuel tanks would be enormous. It would also use a lot of power, requiring either a nuclear reactor or huge solar panels capable of supporting themselves at 0.25g.
But if it could be done, continuously accelerating at 0.25g to the midpoint then decelerating at 0.25g would result in an Earth-Mars travel time much shorter than 3 months. When Mars is closest to Earth, the travel time would only be 3.5 days. Even when Mars is on the other side of the Sun, the travel time would only be 9.4 days.