Solving Solar Heating System Problem

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SUMMARY

The discussion focuses on designing a solar heating system utilizing an aluminum pipe to transfer heat to water. The key equation derived is Q=(alpha)(area of pipe)(solar Heat flux) = mdot(Cp)(T'out-T'in), where alpha represents absorptivity and mdot is the mass flow rate. The user grapples with transient heat transfer and the complexities of calculating the outlet temperature of the fluid, considering both convective and radiative heat losses. The consensus is to assume an average temperature for calculations, which simplifies the process while still providing a reasonable estimate of the exit temperature.

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  • Basic proficiency in solar energy systems design
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amish99
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Bit Stumped at the moment...

I am attempting to design a system to take advantage of the solar energy available this summer, but am struggle to get the equation correct.

Broken down, i have an aluminium pipe with water running thorugh it, which will gain energy from solar radiation, but will also loose energy to the air around via convection and to the atmosphere in thermal radiation. The pipe will also be covered in black absorber coating.

My target is to find the temperature of the fluid as is leaves the exposed pipe,
However my unknown at this point is that of the fluid, obviosuly i know the inlet temperature, but obviosuly this would change as we go down the length of the pipe,

I think right now I am looking at tranient heat trasnfer, am i making this more complex than it is?

My solution so far is,

Q=(alpha)(area of pipe)(solar Heat flux) = mdot(Cp)(T'out-T'in)
alpha being absobitivty
mdot is mass flow rate

I then assume the temperature of the pipe is assumed to be that of T'out and use it in the following;

Q/L=2pie(T'pipe-T'air) / [(1/h1r1)+(1/k1)(lnr2/r1)+(1/r2h2)]

r1 being the inner pipe radia
r2 being the outer pipe radia
h1 being convection of water
h2 being convetion of air
k1 is conductivity of pipe


This doesn't make sense for me, unless i am expected to assume that the water is maintained at the pre calcuated outlet temperature, or do i take an average instead?
 
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I Forget to mention the thermal radiation lost

Q/L= (planks constant)(emistivity)(2pie)(r)(T^4-T^4)
 
You should assume an average temperature. That will give you an average heat loss per unit length. From there, you can apply that heat loss and mass flow to get a good estimate of the temperature at the exit.

I suppose the proper way would be to integrate over the length of the pipe, but unless the pipe is extremely large, your method should get you in the ballpark. Particularly being that convection coefficients will vary significantly throughout the day (wind, etc).
 

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