Chapter-3
Solar Energy Collectors
Thermal Analysis of flat plate collector

Example-3.6.1_Page No: 100_NCES_GDRai

Q) Data for a flat plate collector used for heating the building are given below

1. Location and latitude : Baroda, 22 deg N
   
2. Day and time: January 1, 11:30-12:30(IST)
   
3. Annual average intensity, of solar radiation : 0.5 langley per minute
   
4. Collector tilt : latitude + 15 deg
   
5. No. of glass covers : 2
   
6. Heat removal factor for collector : 0.810
   
7. Transmittance of the glass :0.88
   
8. Absorptance of the glassv: 0.90
   
9. Top loss coefficient: 7.88 W/sq.m dec C(6.80 kcal/hr m2 dec C)
   
10. Collector fluid temperature : 60 dec C
    
11. Ambient temperature : 15 deg C
    Calculate: (i)solar altitude angle (ii) Incident angle (iii)Collector efficiency

Solution:

Given data:
1) Location and latitude(\(\phi\)) : Baroda, 22 deg N
2) Day(n) and time(LST): January 1, 11:30-12:30(IST)
3) Annual average intensity of solar radiation(H) : 0.5 langley per minute
4) Collector tilt(or slope,S) : latitude + 15 deg : \(\phi\) + 15⁰
5) No. of glass covers : 2
6) Heat removal factor for collector,F_{R} : 0.810 7) Transmittance of the glass(\(\tau\)) :0.88
8) Absorptance of the glass(\(\alpha\)): 0.90
9) Top loss coefficient,\(U_L\): 7.88 W/m^{2 0}C(6.80 kcal/hr m^{2 0}C)
10) Collector fluid temperaturem \(T_{fi}\) : 60⁰C
11) Ambient temperature, \(T_a\) : 15 deg

(i) Solar altitude angle (\(\alpha\))

• we know that
  Solar altitude angle \(\alpha\) = asin \((cos\alpha cos\delta cos\omega_s + sin\phi_0 sin\delta)\) ——>(1) As per the equation one, we need to calculate the declination angle \(\delta\) and hour angle \(\omega_s\), the other data is given in the question

## [1] "The number of the day is 1 "

## [1] "The declination anlge is -23.01 deg"

## [1] "The hour angle is 0 deg "

## [1] "The solar altitude angle is 44.99 deg "

(ii) Incident angle (\(\theta\)) (dependent on altitude angle)

• we know that incident anlge \(\theta = \frac{\pi}{2}-\alpha\)

# incident angle(theta)
theta<-((pi/2)*(180/pi))-alpha
theta

## [1] 45.01

(iii) Collector Efficiency (\(\eta_c\))

• we know the formula for collector efficiency, \[\eta_c = \frac{q_u}{H_b.R_b}\]
• where,
  1. \(q_{u} = F_{R}[S-U_{L}(T_{fi}-T_{a})]\) is the useful gain
  2. \(U_{L}\) = overall heat loss coeficient : difference between average collector surface temp and the surrounding air temperature in W/m^{2 0}C
  3. \(S = H_b.R_b <\tau.\alpha>\)
     1. \(H_b\) = the rate of incident beam radiation on a unit area of surface of any orientation
        (ii)\(R_b\) = factor to convert beam or diffuse radiation to that on the plane of collector
        (iii)\(\tau.\alpha\) = effective transmittance absorptance product of cover system for beam and diffuse radiation \[<\tau.\alpha> = \frac{\tau.\alpha} {1-(1-\alpha)\rho_d}\] \(\rho_d\) = diffuse reflectance for two glasses = 0.24
  4. \(T_{fi}\) = the average temperature of the upper surface of the absorber plate or collector fluid,⁰C
  5. \(T_{a}\) = atmospheric temperature, ⁰C
  6. \(F_{R}\) = heat removal factor

# Collector efficiency
#given
#tau.alpha calculation
tau<- 0.88
alpha<- 0.90
rho_d<-0.24 #to ensure given or assumed
tau.alpha <-(tau * alpha)/(1-(1-alpha)* rho_d)
tau.alpha

## [1] 0.8114754

#calculation of R_b
phi<-22
s<-phi+15
omega_deg              # hour angle calculated above

## [1] 0

delta                  # declination anlge calculated above

## [1] -23.01

R_b<-round(((cos((phi-s)*(pi/180))*cos(delta*(pi/180))*cos(omega_deg*(pi/180)))+(sin((phi-s)*(pi/180))*sin(delta*(pi/180))))/((cos(phi*(pi/180))*cos(delta*(pi/180))*cos(omega_deg*(pi/180)))+(sin(phi*(pi/180))*sin(delta*(pi/180)))),digits=2)
sprintf("Factor to convert beam radiation to that on the plane of the collector is %s deg", R_b)

## [1] "Factor to convert beam radiation to that on the plane of the collector is 1.4 deg"

# average solar radiation given(considering only beam),H_b=0.5 langlet/min
H_b<-300 #(kcal-hr per sq.meter)
S<-round(H_b*R_b*tau.alpha,digits=3)
sprintf(" The value of S is calulated as %s kcal/hr sq.m",S)

## [1] " The value of S is calulated as 340.82 kcal/hr sq.m"

#useful gain calculation
F_r<-0.810
S

## [1] 340.82

U_L<-6.80
T_fi<-60
T_a<-15
q_u<-round(F_r*(S-(U_L*(T_fi-T_a))),digits=3)
q_u 

## [1] 28.204

sprintf(" The useful gain q_u is calculated as %s kcal/hr s.m",q_u)

## [1] " The useful gain q_u is calculated as 28.204 kcal/hr s.m"

#Final calculation of collector efficiency
collector_efficiency<-round(q_u/(H_b*R_b),digits=3)
sprintf("The collector efficiency is found to be %s percent", collector_efficiency)

## [1] "The collector efficiency is found to be 0.067 percent"