Calculation of Solar radiation and incident energy

Data taken from NASA database

10th lowest value in 12 months of daily solar radiation (horizontal) is in October monthHg= 4.42 kWh/m2-day = 15912 kJ/ m2-dayThis 10th lowest value is taken because to get the required results from the setup as we decided to operate this system for maximum 10 months.

Month Day n

January 17 17

February 16 47

March 16 75

April 15 105

May 15 135

June 11 165

July 17 198

August 16 228

September 15 258

October 15 288

November 14 318

december 10 344

• Declination angle, δ = 23.45 sin ; where n is the day of the year, for 15th October n = 288• So, δ = 23.45 sin ;• δ = -9.6o

• We know, Latitude ф = 13.083o

• Solar angle s = cos-1 (-tan δ * tan ф)• So, s =87.75o

• Now, Daily extra terrestrial radiation(Ho)• Ho = kJ/m2 –day• Isc=1.367 kW/m2

• Substituting all the values we get, Ho= 34175.83 kJ/m2-day• Now, Monthly average of daily diffuse radiation (Hd) • = 1.411 – 1.696• Substituting all the values we get, Hd = 9887.38 kJ/m2-day

Tilt angle (β)• β -= ф – δ;• We take as 0degree as to get average of the day as the collector is

fixed at a particular angle.• β = 23o

Daily radiation on a tilted surface (Ht)

•

• Where, Rb = • st = = 91.69o

• Substituting all the values we get, Rb =1.1275• Rd = = 0.9603;• Rr = = 0.00795; ρ = 0.2• So, using all the above values we get, Ht = 16414.11 kJ/m2-day

Monthly average hourly global radiation (Io)

•

• a = 0.459 + [0.5016;• a = 0.6926• b = 0.6609 – 0.4767;• b = 0.4389• and, Io = = kJ/m2 –hr• = 60o

• Substituting all the values we get, Io = 0.6097 kJ/m2 – hr

Monthly average hourly diffuse radiation (Id)

• Where,• a` = 0.4922 +{} for 0.1 ≤ ≤ 0.7• = 0.6214;• So, a` = 0.942;• b` = • b` = -0.1140• Substituting all the values we get, Id = 561.97 kJ/m2 – hr• Now, Ig = Id + Ib

• We already calculated Ig and Id

• So, Ib = 370.06 kJ/m2 – hr

• For tilted collector surface incident radiation (It)• It = IbRb + IdRd + (Ib + Id)Rr

• We already know all the values for calculating It

• So, It = 964.312 kJ/m2 – hr• S = IbRb(τα)b + {IdRd + (Ib + Id)Rr}(τα)d

• Also, • γ = 0o

• So, = 0.5144• θ = 59.04o

• θ = θ1 (Incident angle)• Now, = ; n2 =1.42

• ρ1 = and ρ2 = • ρ1 = 0.1406 and ρ2 = 0.0019• τr1 = and τr2 = • For M=1 (Single glazing)• τr1 =0.7535 and τr2 = 0.9962• τr = = 0.87485

Flow analysis and derivations

• Bernoulli’s equation applied• at point 1 and 2

• P1, P2, P3 are pressures at point 1, 2 and 3.• are densities at point 1, 2and 3.• V2, V3 are the outlet and inlet rate of velocity of the thermal storage

tank.• at point 2 and 3