Heat transfer- Solar energy collector problem (took me 3 days, unsolve

1. Aug 23, 2014

jkim6881

1. The problem statement, all variables and given/known data

I have attached the problem as an image and they are same as below.

The collector-receiver combination tested above is chosen for use in a solar thermal power plant. plant consists of 100 rows of collector-receiver modules. For the receiver efficiency use the value you calculated in Part A for the no wind case. For the collector optical efficiency use the value you calculated in Part B. Assume a plant thermal efficiency of 35% and no thermal storage. Use a simple model account for variation in atmospheric absorption through the day in which the solar irradiation GS at the earth’s surface (on a plane perpendicular to the sun’s rays) is a function of solar altitude angle a,
Gs = GBx( 1- Γ cosa)
GB is baseline solar irradiation and Γ is the coefficient of absorption. Assume GB = 820W/m2 and Γ = 0.32
Using a spreadsheet or Matlab, consider two cases – collector modules aligned east-west and north south. For each case plot the electrical power output of the plant against solar time at 5 minute intervals. over a 24 hour period on the following days: 21 December, 21 March, 21 June and 21 September in 2014. Hence estimate the total electrical energy (in kWh) and the average power output that would be generated by the plant on those days.

Based on these results give a rough estimate of the total annual electrical energy (in kWh), annual average power and capacity factor of the plant for the two cases. Assume a constant annual average cloudiness factor of 17%. Cloudiness factor is the reduction in daily average solar flux at the surface due to cloudiness.

collector-receiver model: parabolic collector, 5m wide aperture width, 30m long, receiver diameter is 55mm
Location is Hobart

2. Relevant equations

Where;
X = (360*(N-1))/365.24

EoT = (0.1236*SIN((PI()*X/180))-0.0043*COS((PI()*X /180))+0.1538*SIN(2*(PI()*X /180))+0.0608*COS(2*(PI()*X /180)))

Solar time = Time – EoT +(1.21*4/60) <<< -1 for day light saving>>>

(solar hours) hs =15*(Solar time -12)

Latitude angle (from web)= - 42.8806

(sun declination angle) Delta =23.45*SIN((PI()/180)*(360/365.24*(N+284)))

Elevation =ASIN(COS((PI()/180)*L)*COS((PI()/180)*Delta)*COS((PI()/180)* hs)+SIN((PI()
/180)*Delta)*SIN((PI()/180)*L))*180/PI()

Azimuth =ATAN(SIN((PI()/180)* hs)/(SIN((PI()/180)*L)*COS((PI()/180)* hs)-COS((PI()/180)
*L)*TAN((PI()/180)*Delta)))*180/PI()

Incidence angle =(ACOS(SQRT(1-(COS((PI()/180)*Elevation))^2)*((SIN((PI()/180)
*Azimuth))^2))))*180/PI()

From previous problem,
Optical efficiency = 88.7%
N = number of day from 1st of January.

All equations are checked (from a reference book) except following equation as it is not given..

Thermal efficiency*20*4*30/1000

3. The attempt at a solution

Previous problem I got all correct answer, up to calculating incident angle column..

I considered the day light saving.
I have attached my excel file.

I double checked all the equations but can not find anything wrong..
I am pretty sure I got all right upto Gs part...

I guess something is wrong in calculating power output in terms of unit.

If I can get a right answer for a total electrical energy I can do other parts.

Can anyone find my problem?

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2. Aug 27, 2014