Radiative Heat Transfer

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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:

ελ=0.85 for λ<2.6μm
ελ=0.35 for 2.6<λ<4.2μm
ελ=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 enviroment 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 lenght 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:
qin=qout
qin=1125*95/100=1068.75 so in total 760*1068.75*760=812250W/m2
qout=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([tex]\infty[/tex],T)-f(n=1,4.2,T)]]

So if i try some values for T i can see that for T=2150K qout=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 qin=qout[tex]\Rightarrow[/tex]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 thats goes out with the cooland so 0=qin-qcoolant-qradiation,T=650[tex]\Leftrightarrow[/tex]qcoolant=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 [tex]\frac{84000}{189.2}[/tex]=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 qin=812250W/m2 if i had to compare it with the 64MW.Ive been also told that the qout by radiation is not important in this question but this doesnt seem the case to me.If i dont use energy emitted by radiation its basically like i dont define the temperature of the tube at 650K.Its also strange that i dont 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 :)
 

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