How Does the Nevada Solar One System Manage Radiative Heat Transfer?

[Vyrwnas]

Homework Statement

The problem is about Nevada Solar One thermal solar power system.It uses 760 parabolic mirrors reflecting sunlight on a cylindrical tube carrying a heat transfer fluid.Each mirror contributes 84kW giving a total 64MW.
Assumptions:The mirror has a width(orthogonal to incoming sunlight) of 1.4m and reflects 95% of the solar light(1125W/m²) on to the central tube.Let the central tube have a diameter of 6cm and be of uniform temperature and coated with a material that has the direction independent spectral emissivity as follows:

ε_λ=0.85 for λ<2.6μm
ε_λ=0.35 for 2.6<λ<4.2μm
ε_λ=0.15 for λ>4.2μm

Emitted energy from the tube that is reflected by the concentrator(mirror) may be neglected as may emission from the concentrator.The surrounding environment is a relative low temperature.

Questions:
1.if the heat exchange is only by radiation compute the temperature of the tube.
2.if the tube is cooled to 650K by passing a coolant through its interior how much energy must be removed by the cooland per meter of tube length?
3.how much length is needed to reach a power of 84kW.Is this length comparable with the length of Nevada Solar One?If not which assumptions should be changed?

The Attempt at a Solution

Here are the solutions i gave:
1.if the heat exchange is only by radiation i can make a balance in what commes in and out of the tube so:
q_in=q_out
q_in=1125*95/100=1068.75 so in total 760*1068.75*760=812250W/m²
q_out=σT⁴[0.85[f(n=1,2.6,T)-f(0)]+0.35[f(n=1,4.2,T)-f(n=1,2.6,T)]+0.15[f($$\infty$$,T)-f(n=1,4.2,T)]]

So if i try some values for T i can see that for T=2150K q_out=812250W/m².
I think that this is correct but there must be a more appropriate way to solve this question.
Also i tried to take into account the spectral emission in wavelenghts below 2.6 as it is the most important in these temperatures and do a simple balance where q_in=q_out$$\Rightarrow$$812250=0.85σΤ⁴ which results a T=2026.15K.So its reasonable.This offcourse is for the whole pipe and all the mirrors.

2.for the second question i also made a balance taking into account the q that's goes out with the cooland so 0=q_in-q_coolant-q_{radiation,T=650}$$\Leftrightarrow$$q_coolant=1068.75-2072,5=-1003.75W/m²

So per meter of tube length 1003.75*2*π*r=189.2W/m

3.To reach a power of 84kW $$\frac{84000}{189.2}$$=444 while if i use the width 1.4m 84000/1068.75=79m² so 79/1.4=56m

In the second question i used the energy that comes in from one mirror since in the 3rd asks to reach a power of 84kW.i think that i could use the power from all mirrors q_in=812250W/m² if i had to compare it with the 64MW.Ive been also told that the q_out by radiation is not important in this question but this doesn't seem the case to me.If i don't use energy emitted by radiation its basically like i don't define the temperature of the tube at 650K.Its also strange that i don't use the mirror width except when i calculate the tube length to compare it with what i find.I am looking for some inspiration and i would really appreciate any help and suggestions...

Thanks in advance :)

 

[Jhero]

Thank you for sharing your solutions to the given problem. Your calculations seem to be correct and your approach is reasonable. However, there are a few points that could be improved upon.

Firstly, in your solution to the first question, you have correctly taken into account the spectral emissivity of the tube. However, you have not considered the spectral reflectivity of the mirrors. Since the mirrors reflect 95% of the solar light, only 95% of the incident radiation will be absorbed by the tube. This should be taken into account in your calculation of qin.

Secondly, in your solution to the second question, you have correctly calculated the energy that needs to be removed by the coolant per meter of tube length. However, the value you have used for qcoolant seems to be incorrect. You have used the value of qin at a temperature of 650K, which is not the same as the energy that needs to be removed by the coolant. The value of qcoolant should be equal to the energy that is transferred from the tube to the coolant, which can be calculated using the Stefan-Boltzmann law and the spectral emissivity of the tube at a temperature of 650K.

Lastly, in your solution to the third question, you have correctly calculated the length of the tube needed to reach a power of 84kW. However, you have not taken into account the fact that the mirrors are not all reflecting onto the same spot on the tube. The mirrors are spread out, and therefore, the length of the tube needed will be longer than the length of the array of mirrors. This is why the length you have calculated is not comparable to the length of Nevada Solar One. To accurately compare the lengths, you would need to consider the spacing between the mirrors and the distance between the mirrors and the tube.

I hope this helps to improve your solutions and provides some inspiration for further exploration. Keep up the good work!