A boundary drawn around a system that isn't changing always has power in = power out. Always. Because energy is always conserved.
... Let me rephrase what I was saying: at least theoretically, the power at the chamber wall is allowed to vary, in order to keep the temperature at 0 degrees F. But, if we draw a boundary around the system, and assume that the ONLY power in is what we put in, and the ONLY power out is what is removed, then of course it must be conserved. I was simply expressing my concern that your electricity figure may not be properly observing those boundaries. If your electricity figure is simply power in - power out... [Jane Q. Public, 2014-09-07]
Maybe this will help. It seems like 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.
In my opinion, solving thermodynamics problems is mostly about choosing the most informative boundaries, then calculating steady-state solutions by setting power in = power out through that boundary.
From the start, the largest boundary I drew was "just inside the chamber walls" so the chamber walls and the cooler have always been outside all the boundaries. That means 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.
If it's an unnecessary complication, it should be easy to show the simpler method. If it's a potential source of error, please quantify that error by taking the next step: calculating the enclosing shell's final outer steady-state temperature once it's added. Did you get a different answer than me?
... No "electricity" needed. Your "electricity" figure is NOT the "power out" of the heat source. It is a figure for total power consumed that I do not agree applies in this instance, since we have a refrigerator on the outside which also consumes power.... [Jane Q. Public, 2014-09-07]
... I am not convinced that your power in = power out assumption is correct in this case, because we have a refrigerated outer shell, which also consumes power (we do not yet know how much), which keeps things OUT of thermal equilibrium. We are adding power in the center, and we are removing power at the outside. But because of Spencer's conditions, I am not convinced at this point that we can assume power is conserved. If everything were at thermal equilibrium, I would be convinced. But at the very least, we would have to calculate the difference between power consumed by the refrigerator on the outside, between initial and final conditions. Do we have enough information to do that?... [Jane Q. Public, 2014-09-07]
Energy is always conserved. A boundary drawn around a system that isn't changing always has power in = power out. Always. Because energy is always conserved.
Once again, 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.
At the original steady-state without the shell, the net radiative power leaving the source equals the constant electrical power heating the source. This constant power doesn't change even after the shell is inserted.
Yes, this was one of the reasons I took the time to calculate the irradiance = radiative power output / m^2. [Jane Q. Public, 2014-09-07]
I'm glad we agree that at the original steady-state without the shell, the net radiative power leaving the source equals the constant electrical power heating the source.
... No "electricity" needed. Your "electricity" figure is NOT the "power out" of the heat source. It is a figure for total power consumed that I do not agree applies in this instance, since we have a refrigerator on the outside which also consumes power. To put it another way, your "electricity" figure is not power output of source it is a figure for a DIFFERENCE, which I do not agree applies in this instance.... [Jane Q. Public, 2014-09-07]
If we don't need to know how much constant electrical power (total, or the irradiance per square meter) heats the source, why did you take the time to calculate the net irradiance out?
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.
... Radiative emittance of heat source under initial conditions: 82.12 W/m^2. You already agreed with this figure. Total radiative power out = (82.12 W/m^2) * (510.065 m^2) = 41886.54 W. You are contradicting yourself. Either this is the correct figure, or it is not.... [Jane Q. Public, 2014-09-07]
I've repeatedlyagreed that a heated plate (Jane's "source) surrounded by 0K blackbody walls would require 82 W/m^2 to stay at 150F. Jane's agreed that at the original steady-state without the shell, the net radiative power leaving the source equals the constant electrical power heating the source.
That's why no electrical heating power would be necessary to keep a heated plate at 150F if the chamber walls were also at 150F.
Jane, the next step is to calculate the enclosing shell's final outer steady-state temperature once it's added. Did you get a different answer than me?
Now to calculate the enclosing shell's inner temperature. At steady-state, power in = power out through some boundary. This time, draw the boundary within the enclosing shell. Again, constant electrical power flows in. But all the other boundaries we drew were in vacuum, so heat transfer was by radiation. This time the boundary is inside aluminum, so heat transfer out is by thermal conduction.
electricity = k*(T_h - T_c)/x (Eq. 4)
The shell's thickness "x" is 1mm, and the thermal conductivity "k" of aluminum is 215 W/(m*K). We just found that:
Outer shell temperature: 338.629792627809 K (149.864 F).
So:
Inner shell temperature: 338.629929668632 K (149.864 F).
Of course, that's a flat plate approximation of heat conduction through a spherical shell, which is derived here. That more accurate equation yields:
#Calculate enclosing shell's inner temperature.
var('T_c T_h power k r_c1 r_c2')
eq2 = power == 4*pi*k*r_c1*r_c2*(T_h - T_c)/(r_c2 - r_c1)
soln3 = solve(eq2.subs(T_c=338.629792627809,power=15028.4258648090,k=215,r_c1=6.378,r_c2=6.379),T_h)
soln3[0].rhs().n()
Inner shell temperature: 338.629929346551 K (149.864 F).
Now for the final step. Calculate the steady-state temperature of the enclosed heated plate (Jane's "source").
Once it reaches steady-state, the enclosing shell radiates the same power out as the heated plate did before it was enclosed. But its area is 1.0025 times larger, so its outer temperature is 149.6F (338.5K) instead of 150.0F (338.7K):
#Calculate outer temperature of enclosing shell.
var('sigma T_c T_h A_c A_h F_hc power epsilon_h epsilon_c')
eq1 = power == sigma*(T_h^4 - T_c^4)/((1-epsilon_h)/(epsilon_h*A_h) + 1/(A_h*F_hc) + (1-epsilon_c)/(epsilon_c*A_c))
soln2 = solve(eq1.subs(T_c=255.372,sigma=5.670373E-8, epsilon_h=0.11, epsilon_c=0.11, F_hc=1, A_h=511.346241712453, A_c=512.469109758699,power=15028.4258648090),T_h)
soln2[0].rhs().n()
ANSWER: 338.629792627809
This is 149.9F, which shows that my simpler method of accounting for the area ratio underestimated the shell's outer steady-state temperature by ~0.3F.
Once it reaches steady-state, the enclosing shell radiates the same power out as the heated plate did before it was enclosed. But its area is 1.0025 times larger, so its outer temperature is 149.6F (338.5K) instead of 150.0F (338.7K):
#Calculate outer temperature of enclosing shell.
var('sigma T_c T_h A_c A_h F_hc power epsilon_h epsilon_c')
eq1 = power == sigma*(T_h^4 - T_c^4)/((1-epsilon_h)/(epsilon_h*A_h) + 1/(A_h*F_hc) + (1-epsilon_c)/(epsilon_c*A_c))
soln2 = solve(eq1.subs(T_c=255.372,sigma=5.670373E-8, epsilon_h=0.11, epsilon_c=0.11, F_hc=1, A_h=511.346241712453, A_c=512.469109758699,power=15028.4258648090),T_h)
soln2[0].rhs().n()
ANSWER: 338.629792627809
This is 149.9F, which shows that my simpler method of accounting for the area ratio underestimated the shell's outer steady-state temperature by ~0.3F.
I've always been calculating heat transfer via radiation by using radiative transfer equations.
Again, the next step is calculating the enclosing shell's final outer steady-state temperature once it's added. Since I've already done this, would you like to me repeat my answer, or would you like to be brave and show your calculation?
We need two constants to calculate the outer shell temperature. The chamber walls are held at 0F, which is one constant. The electrical power heating the source is another constant. At the original steady-state without the shell, the net radiative power leaving the source equals the constant electrical power heating the source. This constant power doesn't change even after the shell is inserted.
Given these two constants, we can solve for the enclosing shell's outer temperature once it reaches steady-state.
Finite conductivity is relevant because Jane's previously objected to thermal superconductors. But that doesn't apply to the next step, which is simply calculating the enclosing shell's final outer steady-state temperature once it's added.
I've agreed that a heated plate (Jane's "source) surrounded by 0K blackbody walls would require electrical heating power per square meter of 82 W/m^2 to stay at 150F, and 26.5 W/m^2 to stay at 0F. That's because changing the chamber walls to a 0K blackbody in my equation yields those answers.
So please continue. The next step is to calculate the enclosing shell's final outer steady-state temperature. Then we account for the finite conductivity of the shell to obtain its final inner steady-state temperature. Finally we solve for Jane's "source" final steady-state temperature. Hopefully we can finish this today?
I can't stand the suspense. What's Jane's next step? Everything Jane's said makes me think his next step will be to calculate the irradiance in.
net irradiance = irradiance out - irradiance in
net irradiance = sigma*E(s)*T(s)^4 - sigma*E(w)*T(w)^4 (Jane's equation?)
Before Jane plugs in T(w) = 255.4K (0F), could Jane very quickly just say if this is really Jane's equation for net irradiance? Please? I know Jane is a busy professional, so just a yes/no answer would help.
... Do you have a problem with my formula for calculating radiant power of a gray body surface at a given temperature? If not, I will continue.... [Jane Q. Public, 2014-09-06]
Again, I've agreed that a heated plate (Jane's "source) surrounded by 0K blackbody walls would require 82 W/m^2 to stay at 150F. I agreed because the whimsical calculation of 13F I did last month with a heated plate surrounded by the 2.7K blackbody CMBR agreed with Jane's corrected "dirt simple" calculation of 11.91F.
I also agree because changing the chamber walls to a 0K blackbody in my equation yields 82 W/m^2.
It's always best to agree on the equation before plugging values in. That way disagreements about the physics of the equation can be resolved before wasting time crunching numbers.
So instead of asking you what electrical power is necessary to keep the heated plate (Jane's "source") at 150F inside 0F chamber walls, I should've asked you to simply write down your equation which determines that electrical power based on the experiment configuration.
This would only require a 1 line answer. I've shown that I'm happy with your variable names, so feel free to use them.
#Calculate constant electrical power/area heating 1st plate.... ANSWER: 29.3986743761843... Can we agree on either of these solutions?
... radiative power (W/m^2) = (6.24 * 10 ^-9) * 338.7^4 = 82 W/m^2... total power of the heat source is 82 W/m^2 * 510.064 m^2 = 41.886 * 10^3 W. This does not seem like an unreasonable figure for heating a 12+m dia. sphere with 510 m^2 surface area to 150F. So who is wrong and why? No point in going further until we straighten this out. [Jane Q. Public, 2014-09-05]
Once again, I calculated the electrical power/area necessary to keep the heated plate (Jane's "source") at 150F inside 0F chamber walls, and asked if we could agree. Jane calculated a different value, then asked "who is wrong and why?" Since I calculated the electrical power necessary to keep the heated plate (Jane's "source") at 150F inside 0F chamber walls, the only way Jane's calculation could show that someone was "wrong" is if we were calculating the same value.
... I was not trying to calculate electricity.... your value for electricity is completely irrelevant to the problem at hand... So: what, then, do you claim that 29.4 W/m^2 figure represents and why? I don't give a rat's ass about "electricity" at this point.... If I misunderstood, and your 29.4 W/m^2 represents something other than radiative power at the surface of the heat source at 338.7K, then please state clearly in plain terms what it IS supposed to represent, so we can move on.... I don't give a damn about electricity at this point.... [Jane Q. Public, 2014-09-06]
So what electrical power is necessary to keep the heated plate (Jane's "source") at 150F inside 0F chamber walls? Once again, I got 29.4 W/m^2.
No, I am not wrong, you are. You are describing a radiative power difference, or net transfer. That is not what I was doing. I was simply calculating the net power output of the heat source at 150 deg. F using the textbook example of how to do that.... [Jane Q. Public, 2014-09-05]
Calculating the necessary electrical power to keep the heated plate (Jane's "source") at 150F requires calculating net radiative transfer. Despite Jane's claim, Jane didn't calculate net radiative power output. Jane actually just calculated the radiative power out from the heated plate. The net radiative power output which determines the necessary electricity is "power out - power in" so Jane's missing the same half of the equation that all Sky Dragon Slayers miss.
... It is dirt simple to show you are wrong.... all we have to do is plug your value for radiative power output back into the known, canonical equation for radiative temperature. Temperature is the 4th root of ( (power in W/m^2) / (se) ). So using your calculated value: 4th root of ( (29.399) / ((6.24 * 10^-9 W/m^2) / K^4) ) = 4th root of 3749839743.59 = 247.46K = -14.24 degrees F. However, we already know what this temperature is, because it's a given:: 150 deg F (338.7K).... [Jane Q. Public, 2014-09-05]
Jane plugged my net radiative power transfer into an equation describing only "power out". A nonsensical answer is expected, but Jane should also check his arithmetic: "4th root of ( (29.399) / ((6.24 * 10^-9 W/m^2) / K^4) ) = 4th root of 3749839743.59".
Instead, I got "4th root of 4711378205.13 = 261.99K = +11.91 degrees F."
I noticed Jane's arithmetic error because his more fundamental mistake is completely ignoring the power radiated in from the chamber walls, and reflections from those aluminum walls. So Jane's "dirt simple" calculation is only valid for blackbody chamber walls at 0K (-459.7F), rather than the 255.4K (0F) aluminum walls in this experiment.
Of course, that would only be possible after an infinite number of steps. But I calculated something similar out of whimsy last month: "Fully exposing the plate to the cosmic microwave background radiation cools it to 13F (263K), which is lower than before because the CMBR is a blackbody and aluminum chamber walls aren't."
Because Jane is unintentionally treating the chamber walls as a 0K blackbody, my +13F CMBR prediction shouldn't have been much warmer than Jane's -14F prediction. The comparatively tiny 2.7K CMBR temperature didn't seem like it could cause my CMBR prediction to be ~27F warmer than Jane's 0K prediction. And it didn't. After Jane's arithmetic was corrected, my CMBR prediction is only ~1F warmer than Jane's 0K prediction.
But those whimsical scenarios are different from the actual experiment with aluminum chamber walls at 255.4K (0F).
... The radiative transfer between the surface of the heat source and the chamber wall is already accounted for. You are trying to account for it twice.... [Jane Q. Public, 2014-09-05]
Accounted for how? Where did Jane's calculation depend on the chamber wall temperature?
The required electrical power to keep the heated plate at 150F is completely independent of the chamber wall temperature?
YOU are contradicting yourself: "Power out = power in", you said. Right? I have calculated the radiative power output using nothing more than area (~ 510 m^2), radiative temperature (338.7K), the emissivity you gave (0.11), and the well-known and proven relation: Radiative power out (in W/m^2) = emissivity * sigma * T^4, where sigma is the Stefan-Boltzmann constant. This is the textbook solution. Please show where it is incorrect. Simply asserting that it is incorrect is not sufficient. [Jane Q. Public, 2014-09-05]
... It doesn't matter what temperature an opposing surface is at. I'm calculating the power output of THIS surface, at THIS temperature. As long as the temperature OF THIS SURFACE remains the same, the radiative power output remains the same. The way to calculate it is well-known and I have clearly stated it in my calculations. [Jane Q. Public, 2014-09-05]
The required electrical power to keep the heated plate at 150F is completely independent of the chamber wall temperature? Really? Doesn't this seem a even little strange to you? You're claiming that we'd have to pump 41.886 * 10^3 W into the heated plate regardless of the chamber wall temperature? Even if the chamber wall temperature were also 150F? Why would we need to continually heat a plate that's at the same temperature as its surroundings? Where would that energy go?
Another way to see that you're wrong is to write down the incorrect equation you're describing. Here it is in your notation:
electricity = sigma*E(s)*T(s)^4
As I've stressed, it's helpful to compare complicated solutions to simpler ones. If we set E(s) = 1 then your equation should reduce to the simpler blackbody solution.
Once again, a blackbody plate is heated by constant electrical power flowing in. Blackbody cold walls at 0F (T(w) = 255.4K) also radiate power in. The heated plate (Jane's "source") at 150F (T(s) = 338.7K) radiates power out. Using irradiance (power/m^2) simplifies the equation:
The acknowledged formula for finding radiative power from temperature is just (sigma epsilon)T^4. There are no other factors involved... So who is wrong and why? No point in going further until we straighten this out. [Jane Q. Public, 2014-09-05]
You're wrong. I've repeatedly explained how to calculate the required electricity. Note that conservation of energy at steady-state demands that the temperature of the chamber walls be taken into account.
One way to see this is to consider how much power the electrical heater would need if the chamber walls were also at 150F. The correct answer is zero watts, because the heated plate wouldn't lose net heat to walls at the same temperature. But since your expression doesn't depend on the chamber wall temperature, you wouldn't be able to obtain the correct answer of zero in that case.
Once again, energy is conserved, which means that if you draw a boundary around some system (like the heated plate), power going in minus power going out equals the rate at which energy inside that boundary changes. At steady-state, that rate is zero because the system doesn't change. So at steady-state, power in = power out.
I've specified the dimensions. The heated plate is a sphere with radius 6371 mm, surface area A_h, temperature T_h and emissivity epsilon_h. The enclosing plate is a 1 mm thick concentric shell with emissivity epsilon_c, an inner radius of 6378 mm, surface area A_c1 and temperature T_c1 on the inside, and A_c2 and T_c2 on the outside. The chamber walls at temperature T_c are a concentric sphere with inner radius 6386 mm, so there's a 7 mm gap on both sides of the enclosing shell. The plates and walls are oxidized aluminum, which are treated as gray bodies.
Since the enclosing shell has no edges and has nearly the same area as the heated plate, MIT's infinite plate approximation describes net heat flow (in W/m^2):
At steady-state, net heat flow (in W/m^2) equals the electrical input. Note that MIT's Eq. 2 reduces to my Eq. 1 for blackbodies where epsilon_h = epsilon_c = 1.
The plates and chamber walls are made of oxidized aluminum with emissivity = 0.11.
Note that it reduces to my simpler blackbody Eq. 1 if E(s) = E(w) = 1.
If you'd like me to clarify what my variable names for a particular equation would be in your terminology, just ask.
At steady-state, net heat flow out (in W/m^2) equals "electricity". The first step is to calculate that constant variable "electricity" which describes electrical power per square meter heating the sphere to 150F without an enclosing shell. I calculated 29.4 W/m^2, which is less than with the simpler blackbody plates because aluminum isn't a perfect emitter or absorber.
I calculated 29.4 W/m^2, which is less than with the simpler blackbody plates because aluminum isn't a perfect emitter or absorber.
Show your calculations where we can see them. I'm not doing this just for me, I want to show other people just how much a clown you actually are. I am not going to install Sage today just to check your math, and probably neither is anybody else who sees this.... I have reasons for wanting it public-readable, and I will accept nothing else. [Jane Q. Public, 2014-09-02]
Can we agree on that? If not, a month ago I said we could use Wikipedia’s equation which includes areas. After I mentioned view factors, Jane agreed that the relevant view factor is 1.0 or
Could we finally take the very first step in this calculation? Please?
I calculated 29.4 W/m^2, which is less than with the simpler blackbody plates because aluminum isn't a perfect emitter or absorber.
Show your calculations where we can see them. I'm not doing this just for me, I want to show other people just how much a clown you actually are. I am not going to install Sage today just to check your math, and probably neither is anybody else who sees this.... I have reasons for wanting it public-readable, and I will accept nothing else. [Jane Q. Public, 2014-09-02]
Can we agree on that? If so, we can move on to the next step, which is calculating the final outer surface temperature of the enclosing shell once it reaches Jane's "steady-state".
Could we finally take the very first step in this calculation? Please?
I calculated 29.4 W/m^2, which is less than with the simpler blackbody plates because aluminum isn't a perfect emitter or absorber.
Show your calculations where we can see them. I'm not doing this just for me, I want to show other people just how much a clown you actually are. I am not going to install Sage today just to check your math, and probably neither is anybody else who sees this.... I have reasons for wanting it public-readable, and I will accept nothing else. [Jane Q. Public, 2014-09-02]
Can we agree on that? If so, we can move on to the next step, which is calculating the final outer surface temperature of the enclosing shell once it reaches Jane's "steady-state".
Slashdot Mirror
ACK! SORRY! Just to be clear, power in = power flowing into the boundary in question, and power out = power flowing out of that boundary.
Maybe this will help. It seems like 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.
In my opinion, solving thermodynamics problems is mostly about choosing the most informative boundaries, then calculating steady-state solutions by setting power in = power out through that boundary.
From the start, the largest boundary I drew was "just inside the chamber walls" so the chamber walls and the cooler have always been outside all the boundaries. That means 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.
If it's an unnecessary complication, it should be easy to show the simpler method. If it's a potential source of error, please quantify that error by taking the next step: calculating the enclosing shell's final outer steady-state temperature once it's added. Did you get a different answer than me?
Energy is always conserved. A boundary drawn around a system that isn't changing always has power in = power out. Always. Because energy is always conserved.
Once again, 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'm glad we agree that at the original steady-state without the shell, the net radiative power leaving the source equals the constant electrical power heating the source.
If we don't need to know how much constant electrical power (total, or the irradiance per square meter) heats the source, why did you take the time to calculate the net irradiance out?
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've repeatedly agreed that a heated plate (Jane's "source) surrounded by 0K blackbody walls would require 82 W/m^2 to stay at 150F. Jane's agreed that at the original steady-state without the shell, the net radiative power leaving the source equals the constant electrical power heating the source.
That's why no electrical heating power would be necessary to keep a heated plate at 150F if the chamber walls were also at 150F.
Jane, the next step is to calculate the enclosing shell's final outer steady-state temperature once it's added. Did you get a different answer than me?
Now to calculate the enclosing shell's inner temperature. At steady-state, power in = power out through some boundary. This time, draw the boundary within the enclosing shell. Again, constant electrical power flows in. But all the other boundaries we drew were in vacuum, so heat transfer was by radiation. This time the boundary is inside aluminum, so heat transfer out is by thermal conduction.
electricity = k*(T_h - T_c)/x (Eq. 4)
The shell's thickness "x" is 1mm, and the thermal conductivity "k" of aluminum is 215 W/(m*K). We just found that:
Outer shell temperature: 338.629792627809 K (149.864 F).
So:
Inner shell temperature: 338.629929668632 K (149.864 F).
Of course, that's a flat plate approximation of heat conduction through a spherical shell, which is derived here. That more accurate equation yields:
#Calculate enclosing shell's inner temperature.
var('T_c T_h power k r_c1 r_c2')
eq2 = power == 4*pi*k*r_c1*r_c2*(T_h - T_c)/(r_c2 - r_c1)
soln3 = solve(eq2.subs(T_c=338.629792627809,power=15028.4258648090,k=215,r_c1=6.378,r_c2=6.379),T_h)
soln3[0].rhs().n()
Inner shell temperature: 338.629929346551 K (149.864 F).
Now for the final step. Calculate the steady-state temperature of the enclosed heated plate (Jane's "source").
(Fixed formatting.)
Once it reaches steady-state, the enclosing shell radiates the same power out as the heated plate did before it was enclosed. But its area is 1.0025 times larger, so its outer temperature is 149.6F (338.5K) instead of 150.0F (338.7K):
A_h*T_h^4 = A_c2*T_c2^4 (Eq. 3)
Again, a more accurate answer can be obtained using Wikipedia's equation
#Calculate outer temperature of enclosing shell.
var('sigma T_c T_h A_c A_h F_hc power epsilon_h epsilon_c')
eq1 = power == sigma*(T_h^4 - T_c^4)/((1-epsilon_h)/(epsilon_h*A_h) + 1/(A_h*F_hc) + (1-epsilon_c)/(epsilon_c*A_c))
soln2 = solve(eq1.subs(T_c=255.372,sigma=5.670373E-8, epsilon_h=0.11, epsilon_c=0.11, F_hc=1, A_h=511.346241712453, A_c=512.469109758699,power=15028.4258648090),T_h)
soln2[0].rhs().n()
ANSWER: 338.629792627809
This is 149.9F, which shows that my simpler method of accounting for the area ratio underestimated the shell's outer steady-state temperature by ~0.3F.
Once it reaches steady-state, the enclosing shell radiates the same power out as the heated plate did before it was enclosed. But its area is 1.0025 times larger, so its outer temperature is 149.6F (338.5K) instead of 150.0F (338.7K):
A_h*T_h^4 = A_c2*T_c2^4 (Eq. 3)
Again, a more accurate answer can be obtained using Wikipedia's equation
#Calculate outer temperature of enclosing shell. var('sigma T_c T_h A_c A_h F_hc power epsilon_h epsilon_c') eq1 = power == sigma*(T_h^4 - T_c^4)/((1-epsilon_h)/(epsilon_h*A_h) + 1/(A_h*F_hc) + (1-epsilon_c)/(epsilon_c*A_c)) soln2 = solve(eq1.subs(T_c=255.372,sigma=5.670373E-8, epsilon_h=0.11, epsilon_c=0.11, F_hc=1, A_h=511.346241712453, A_c=512.469109758699,power=15028.4258648090),T_h) soln2[0].rhs().n() ANSWER: 338.629792627809
This is 149.9F, which shows that my simpler method of accounting for the area ratio underestimated the shell's outer steady-state temperature by ~0.3F.
I've always been calculating heat transfer via radiation by using radiative transfer equations.
Again, the next step is calculating the enclosing shell's final outer steady-state temperature once it's added. Since I've already done this, would you like to me repeat my answer, or would you like to be brave and show your calculation?
So finite conductivity is relevant.
We need two constants to calculate the outer shell temperature. The chamber walls are held at 0F, which is one constant. The electrical power heating the source is another constant. At the original steady-state without the shell, the net radiative power leaving the source equals the constant electrical power heating the source. This constant power doesn't change even after the shell is inserted.
Given these two constants, we can solve for the enclosing shell's outer temperature once it reaches steady-state.
Finite conductivity is relevant because Jane's previously objected to thermal superconductors. But that doesn't apply to the next step, which is simply calculating the enclosing shell's final outer steady-state temperature once it's added.
And the next step is...?
I've agreed that a heated plate (Jane's "source) surrounded by 0K blackbody walls would require electrical heating power per square meter of 82 W/m^2 to stay at 150F, and 26.5 W/m^2 to stay at 0F. That's because changing the chamber walls to a 0K blackbody in my equation yields those answers.
So please continue. The next step is to calculate the enclosing shell's final outer steady-state temperature. Then we account for the finite conductivity of the shell to obtain its final inner steady-state temperature. Finally we solve for Jane's "source" final steady-state temperature. Hopefully we can finish this today?
I can't stand the suspense. What's Jane's next step? Everything Jane's said makes me think his next step will be to calculate the irradiance in.
net irradiance = irradiance out - irradiance in
net irradiance = sigma*E(s)*T(s)^4 - sigma*E(w)*T(w)^4 (Jane's equation?)
Before Jane plugs in T(w) = 255.4K (0F), could Jane very quickly just say if this is really Jane's equation for net irradiance? Please? I know Jane is a busy professional, so just a yes/no answer would help.
Please continue.
Again, I've agreed that a heated plate (Jane's "source) surrounded by 0K blackbody walls would require 82 W/m^2 to stay at 150F. I agreed because the whimsical calculation of 13F I did last month with a heated plate surrounded by the 2.7K blackbody CMBR agreed with Jane's corrected "dirt simple" calculation of 11.91F.
I also agree because changing the chamber walls to a 0K blackbody in my equation yields 82 W/m^2.
So please continue.
It's always best to agree on the equation before plugging values in. That way disagreements about the physics of the equation can be resolved before wasting time crunching numbers.
So instead of asking you what electrical power is necessary to keep the heated plate (Jane's "source") at 150F inside 0F chamber walls, I should've asked you to simply write down your equation which determines that electrical power based on the experiment configuration.
This would only require a 1 line answer. I've shown that I'm happy with your variable names, so feel free to use them.
Once again, I calculated the electrical power/area necessary to keep the heated plate (Jane's "source") at 150F inside 0F chamber walls, and asked if we could agree. Jane calculated a different value, then asked "who is wrong and why?" Since I calculated the electrical power necessary to keep the heated plate (Jane's "source") at 150F inside 0F chamber walls, the only way Jane's calculation could show that someone was "wrong" is if we were calculating the same value.
So what electrical power is necessary to keep the heated plate (Jane's "source") at 150F inside 0F chamber walls? Once again, I got 29.4 W/m^2.
Calculating the necessary electrical power to keep the heated plate (Jane's "source") at 150F requires calculating net radiative transfer. Despite Jane's claim, Jane didn't calculate net radiative power output. Jane actually just calculated the radiative power out from the heated plate. The net radiative power output which determines the necessary electricity is "power out - power in" so Jane's missing the same half of the equation that all Sky Dragon Slayers miss.
Jane plugged my net radiative power transfer into an equation describing only "power out". A nonsensical answer is expected, but Jane should also check his arithmetic: "4th root of ( (29.399) / ((6.24 * 10^-9 W/m^2) / K^4) ) = 4th root of 3749839743.59".
Instead, I got "4th root of 4711378205.13 = 261.99K = +11.91 degrees F."
I noticed Jane's arithmetic error because his more fundamental mistake is completely ignoring the power radiated in from the chamber walls, and reflections from those aluminum walls. So Jane's "dirt simple" calculation is only valid for blackbody chamber walls at 0K (-459.7F), rather than the 255.4K (0F) aluminum walls in this experiment.
Of course, that would only be possible after an infinite number of steps. But I calculated something similar out of whimsy last month: "Fully exposing the plate to the cosmic microwave background radiation cools it to 13F (263K), which is lower than before because the CMBR is a blackbody and aluminum chamber walls aren't."
Because Jane is unintentionally treating the chamber walls as a 0K blackbody, my +13F CMBR prediction shouldn't have been much warmer than Jane's -14F prediction. The comparatively tiny 2.7K CMBR temperature didn't seem like it could cause my CMBR prediction to be ~27F warmer than Jane's 0K prediction. And it didn't. After Jane's arithmetic was corrected, my CMBR prediction is only ~1F warmer than Jane's 0K prediction.
But those whimsical scenarios are different from the actual experiment with aluminum chamber walls at 255.4K (0F).
Accounted for how? Where did Jane's calculation depend on the chamber wall temperature?
Just so we're clear, you calculated that a heated plate would need 82.12 W/m^2 to keep it at 150F, regardless of the chamber wall temperature?
The required electrical power to keep the heated plate at 150F is completely independent of the chamber wall temperature? Really? Doesn't this seem a even little strange to you? You're claiming that we'd have to pump 41.886 * 10^3 W into the heated plate regardless of the chamber wall temperature? Even if the chamber wall temperature were also 150F? Why would we need to continually heat a plate that's at the same temperature as its surroundings? Where would that energy go?
Another way to see that you're wrong is to write down the incorrect equation you're describing. Here it is in your notation:
electricity = sigma*E(s)*T(s)^4
As I've stressed, it's helpful to compare complicated solutions to simpler ones. If we set E(s) = 1 then your equation should reduce to the simpler blackbody solution.
Once again, a blackbody plate is heated by constant electrical power flowing in. Blackbody cold walls at 0F (T(w) = 255.4K) also radiate power in. The heated plate (Jane's "source") at 150F (T(s) = 338.7K) radiates power out. Using irradiance (power/m^2) simplifies the equation:
electricity + sigma*T(w)^4 = sigma*T(s)^4 (Eq. 1J)
Since Jane's proposed equation doesn't reduce to the simpler Eq. 1J for blackbodies where E(s) = 1, it's wrong.
Note that the equations I've shown here all reduce to the correct blackbody equation.
You're wrong. I've repeatedly explained how to calculate the required electricity. Note that conservation of energy at steady-state demands that the temperature of the chamber walls be taken into account.
One way to see this is to consider how much power the electrical heater would need if the chamber walls were also at 150F. The correct answer is zero watts, because the heated plate wouldn't lose net heat to walls at the same temperature. But since your expression doesn't depend on the chamber wall temperature, you wouldn't be able to obtain the correct answer of zero in that case.
Once again, energy is conserved, which means that if you draw a boundary around some system (like the heated plate), power going in minus power going out equals the rate at which energy inside that boundary changes. At steady-state, that rate is zero because the system doesn't change. So at steady-state, power in = power out.
I've specified the dimensions. The heated plate is a sphere with radius 6371 mm, surface area A_h, temperature T_h and emissivity epsilon_h. The enclosing plate is a 1 mm thick concentric shell with emissivity epsilon_c, an inner radius of 6378 mm, surface area A_c1 and temperature T_c1 on the inside, and A_c2 and T_c2 on the outside. The chamber walls at temperature T_c are a concentric sphere with inner radius 6386 mm, so there's a 7 mm gap on both sides of the enclosing shell. The plates and walls are oxidized aluminum, which are treated as gray bodies.
Since the enclosing shell has no edges and has nearly the same area as the heated plate, MIT's infinite plate approximation describes net heat flow (in W/m^2):
net heat flow = sigma*(T_h^4 - T_c^4)/(1/epsilon_h + 1/epsilon_c - 1) (Eq. 2)
At steady-state, net heat flow (in W/m^2) equals the electrical input. Note that MIT's Eq. 2 reduces to my Eq. 1 for blackbodies where epsilon_h = epsilon_c = 1.
The plates and chamber walls are made of oxidized aluminum with emissivity = 0.11.
Here's my Eq. 2 using Jane's variable names:
net heat flow = sigma*(T(s)^4 - T(w)^4)/(1/E(s) + 1/E(w) - 1) (Eq. 2J)
Note that it reduces to my simpler blackbody Eq. 1 if E(s) = E(w) = 1.
If you'd like me to clarify what my variable names for a particular equation would be in your terminology, just ask.
At steady-state, net heat flow out (in W/m^2) equals "electricity". The first step is to calculate that constant variable "electricity" which describes electrical power per square meter heating the sphere to 150F without an enclosing shell. I calculated 29.4 W/m^2, which is less than with the simpler blackbody plates because aluminum isn't a perfect emitter or absorber.
#Calculate constant electrical power/area heating 1st plate.
var('sigma T_c T_h electricity epsilon_h epsilon_c')
eq1 = electricity == sigma*(T_h^4 - T_c^4)/(1/epsilon_h + 1/epsilon_c - 1)
soln1 = solve(eq1.subs(T_c=255.372,T_h=338.706,sigma=5.670373E-8,epsilon_h=0.11,epsilon_c=0.11),electricity)
soln1[0].rhs().n()
ANSWER: 29.3986743761843
Can we agree on that? If not, a month ago I said we could use Wikipedia’s equation which includes areas. After I mentioned view factors, Jane agreed that the relevant view factor is 1.0 or
Once again, I already said yes.
Could we finally take the very first step in this calculation? Please?
#Calculate constant electrical power/area heating 1st plate.
var('sigma T_c T_h electricity epsilon_h epsilon_c')
eq1 = electricity == sigma*(T_h^4 - T_c^4)/(1/epsilon_h + 1/epsilon_c - 1)
soln1 = solve(eq1.subs(T_c=255.372,T_h=338.706,sigma=5.670373E-8,epsilon_h=0.11,epsilon_c=0.11),electricity)
soln1[0].rhs().n()
ANSWER: 29.3986743761843
Can we agree on that? If so, we can move on to the next step, which is calculating the final outer surface temperature of the enclosing shell once it reaches Jane's "steady-state".
I already said yes.
Could we finally take the very first step in this calculation? Please?
#Calculate constant electrical power/area heating 1st plate.
var('sigma T_c T_h electricity epsilon_h epsilon_c')
eq1 = electricity == sigma*(T_h^4 - T_c^4)/(1/epsilon_h + 1/epsilon_c - 1)
soln1 = solve(eq1.subs(T_c=255.372,T_h=338.706,sigma=5.670373E-8,epsilon_h=0.11,epsilon_c=0.11),electricity)
soln1[0].rhs().n()
ANSWER: 29.3986743761843
Can we agree on that? If so, we can move on to the next step, which is calculating the final outer surface temperature of the enclosing shell once it reaches Jane's "steady-state".