... 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 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.
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 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.
... 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 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.
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.
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.
After the last of 16 mounting screws has been removed from an access cover, it will be discovered that the wrong access cover has been removed.
