Assignment 1-solar thermalsystem

ASSOLA SUBMITT

SSIGAR THER

TED TO: Dr. A

MENRMAL SY

ABDUL WAHEE

NT‐YSTEMS

ED BADAR

IMR

1S

RAN SAJID12SP

D SHAHIDP‐MS‐ME‐002

10/15/201

D2 2

2 IMRAN SAJID SHAHID | 12SP‐MS‐ME‐002

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QUESTION-1:

Using a spreadsheet program and the relations presented in this chapter, try to create a

program that estimates all solar angles according to the latitude, day of year, hour, and slope of

surface.

Answer:

A MATLAB program was written to calculate the solar angles with latitude, day of year, hour

and slope of surface as input. Copy the code lines in m‐file to run the program. M‐file is also

attached with email.

clc clear close all % This program Calculates the Solar angles as an output, includes; % Declination angle(delta), Altitude Angle(alpha_s), % Max. Altitude angle(alpha_smax), Hour Angle(omega), Day Length(N), % Zenith Angle(theta_z), Solar Azimuth angle(Gamma), Incident Angle(theta), % Surface Azimuth(Gamma_s), Solar-Surface Azimuth(Gamma_ss) % Solar_angles(40,6,15,14,0,0,90) disp('~INTRODUCTION:') disp('--------------') disp('This program Calculates the Solar angles as an output') disp('~INPUT FORMAT:') disp('-------------') disp('LATITUTDE, MONTH, DATE, SOLAR TIME HR., SOLAR TIME MIN., MERADIAN, SRF. TILT, SRF. AZIMUTH') disp('E.G; 35, 2, 13, 10, 30, am, 15, 45') disp('~OUTPUT:') disp('--------') disp('DELINATION ANGLE, ALTITUDE ANGLE, MAX. ALTITUDE ANGLE,') disp('HOUR ANGLE, DAY LENGTH, ZENITH ANGLE, SOLAR AZIMUTH,') disp('INCIDENT ANGLE, SURFACE AZIMUTH, SOLAR SRF.AZIMUTH,') disp('===================================================================') disp(' WRITTEN BY: IMRAN SAJID SHAHID ') disp(' REGISTRATION # 12SP-MS-ME-002 ') disp(' DATED:14th OCT 2012 ') disp('===================================================================') % !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! % ---------------------------------INPUT DECK------------------------------ disp('*******************************************************************') disp('************************** INPUT ******************************') disp('*******************************************************************') pHI = input('Enter the your Latitutde position:---> '); Month =input('Enter the desired month in number:--->'); Date=input('Enter the date of the desired month:--->'); Solar_Time_hr=input('Enter the Solar time hour,(10 from 10:30 time):--->'); Solar_Time_min=input('Enter the Solar time minute,(30 from 10:30 time ):--->'); mer=input('Enter am for a.m. & pm for p.m. in lowercase:--->','s');


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bETA=input('Enter the tilt of surface in degrees:--->'); gAMMA=input('Enter the surface azimuth in degrees:--->'); % !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! % DAY OF YEAR, n % n for ith Day of Month i=Date; n_i = [i 31+i 59+i 90+i 120+i 151+i 181+i 212+i 243+i 273+i 304+i 334+i]; n = n_i(1,Month); % DECLINATION ANGLE, dELTA FROM EQ. OFCOPPER(1969) dELTA = 23.45 * sind( 360* ((284 + n) / 365) ); % HOUR ANGLE, oMEGA Solar_Time_min= Solar_Time_min/60; Solar_Time=Solar_Time_hr+Solar_Time_min; if ischar(mer) == 1; if strcmp(mer,'am') == 1; oMEGA = 15 * (Solar_Time - 12); else strcmp(mer,'pm') == 1; if Solar_Time_hr == 12; oMEGA = 15 * (Solar_Time - 12); else oMEGA = 15 * (Solar_Time); end end else break; fprintf(1,'PLEASE CHECK YOUR INPUT!!!.\n') end % ALTITUDE ANGLE,aLPHA_s aLPHA_s = asind([cosd(pHI)*cosd(oMEGA)*cosd(dELTA)]+[sind(pHI)*sind(dELTA)]); % MAX. ALTITUDE ANGLE,aLPHA_smax if Solar_Time_hr == 12 && Solar_Time_min == 0.000 aLPHA_smax = aLPHA_s else aLPHA_smax = 90 - abs(pHI-dELTA); end % INCIDENT ANGLE, tHETA and ZENITH ANGLE, tHETA_s % for Horizontal Surface incident angle becomes equal to Zenith Angle %if bETA > 0 c_tHETA = (sind(dELTA) * sind(pHI) * cosd(bETA)) - ... (sind(dELTA)* cosd(pHI)* sind(bETA)* cosd(gAMMA)) + ... (cosd(dELTA)* cosd(pHI)* cosd(bETA)* cosd(oMEGA)) + ... (cosd(dELTA)* sind(pHI)* sind(bETA)* cosd(gAMMA)* cosd(oMEGA)) + ... (cosd(dELTA)* sind(bETA)* sind(gAMMA)* sind(oMEGA)); tHETA = acosd(c_tHETA); % SOLAR AZIMUTH ANGLE, gAMMA_s gAMMA_s = asind([cosd(dELTA)*sind(oMEGA)]/cosd(aLPHA_s)); % SURFACE-SOLAR AZIMUTH ANGLE, gAMMA_s gAMMA_ss = gAMMA + gAMMA_s; % SUNRISE/SUNSET HOUR ANGLE, oMEGA_s & DAY LENGTH oMEGA_s = acosd(-tand(pHI)*tand(dELTA)); N = (2/15)*acosd(-tand(pHI)*tand(dELTA)); % !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! % ---------------------------------OUTPUT---------------------------------- disp('*******************************************************************') disp('************************** OUTPUT *****************************') disp('*******************************************************************') fprintf(1,'Declination Angle = %6.2f deg\n',dELTA) fprintf(1,'Hour Angle = %6.2f deg\n',oMEGA)


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fprintf(1,'Altitude Angle = %6.2f deg\n',aLPHA_s) fprintf(1,'Incident Angle = %6.2f deg\n',tHETA) fprintf(1,'Max. Altitude Angle = %6.2f deg\n',aLPHA_smax) fprintf(1,'Solar Azimuth Angle = %6.2f deg\n',gAMMA_s) fprintf(1,'Surface-Solar Azimuth Angle = %6.2f deg\n',gAMMA_ss) fprintf(1,'Sunset hour angle = %6.2f deg\n',oMEGA_s) fprintf(1,'Day Length = %6.2f hours\n',N) fprintf(1,'NOTE: Read incident angle as zeinth angle for 0.0deg Surface tilt\n') % !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Test Run: Calculate the solar angles for Latitude is 43° at 9:30am on February 13?

Run the solar_angles.m program in MATLAB. Input and output of the program is shown below:

!*****************************************************************************

~INTRODUCTION: -------------- This program Calculates the Solar angles as an output ~INPUT FORMAT: ------------- LATITUTDE, MONTH, DATE, SOLAR TIME HR., SOLAR TIME MIN., MERADIAN, SRF. TILT, SRF. AZIMUTH E.G; 35, 2, 13, 10, 30, am, 15, 45 ~OUTPUT: -------- DELINATION ANGLE, ALTITUDE ANGLE, MAX. ALTITUDE ANGLE, HOUR ANGLE, DAY LENGTH, ZENITH ANGLE, SOLAR AZIMUTH, INCIDENT ANGLE, SURFACE AZIMUTH, SOLAR SRF.AZIMUTH, =================================================================== WRITTEN BY: IMRAN SAJID SHAHID REGISTRATION # 12SP-MS-ME-002 DATED:14th OCT 2012 =================================================================== ******************************************************************* ************************** INPUT ****************************** ******************************************************************* Enter the your Latitutde position:---> 43 Enter the desired month in number:--->2 Enter the date of the desired month:--->13 Enter the Solar time hour,(10 from 10:30 time):--->9 Enter the Solar time minute,(30 from 10:30 time ):--->30 Enter am for a.m. & pm for p.m. in lowercase:--->am Enter the tilt of surface in degrees:--->0 Enter the surface azimuth in degrees:--->90 ******************************************************************* ************************** OUTPUT ***************************** ******************************************************************* Declination Angle = -13.95 deg


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Hour Angle = -37.50 deg Altitude Angle = 23.50 deg Incident Angle = 66.50 deg Max. Altitude Angle = 33.05 deg Solar Azimuth Angle = -40.11 deg Surface-Solar Azimuth Angle = 49.89 deg Sunset hour angle = 76.61 deg Day Length = 10.21 hours NOTE: Read incident angle as zeinth angle for 0.0deg Surface tilt !*****************************************************************************


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QUESTION-2:

Calculate the sunrise and sunset times and altitude and azimuth angles for London, England, on

March 15 and September 15 at 10:00 am and 3:30 pm solar times?

Given:

Latitude of London, England = 51°30’28.23’’ [1] = 51.5078°

Day of Year = (i) March 15 (ii) September 15

Solar Time = (i) 10:00am (ii) 3:30pm

Find:

(a) Sunrise Time, Sunset time, Altitude angle, Azimuth Angle on 15th March at 10:00am =?

(b) Sunrise Time, Sunset time, Altitude angle, Azimuth Angle on 15th March at 3:30am =?

(c) Sunrise Time, Sunset time, Altitude angle, Azimuth Angle on 15th Sept at 10:00am =?

(d) Sunrise Time, Sunset time, Altitude angle, Azimuth Angle on 15th Sept at 3:30am =?

Solution:

(a) On 15th March at 10:00am, Day of the year can be obtained with the help of Table

1.6.1[2].

59 15 74 The Declination angle can be found from the equation of Copper (1969)

23.45 sin 360284

365

After substituting the value of n, will give us . °.

Under given conditions hour angle is

15° 2 . ° .

By substituting the value of latitude angle ( ), declination ( ), and hour angle ( ), in

following relation, altitude angle can be calculated:

sin


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sin cos 51.5078 cos 30 cos 2.82 sin 51.5078 sin 2.82°

Solar azimuth angle can be calculated form following equation:

sin

sincos 2.82 sin 30

cos 29.99 . °

Using a latitude and declination in following equation for calculating sunset hour

angle:

cos

cos tan 51.5078 tan 2.82 . °

It also follows that number of daylight hours is given by:

2

152

1587.45 .

Sunrise time = 12 6.235 :

Sunset time = 12 17.765 :

(b) On 15th March at 3:30pm, Day of the year can be obtained with the help of Table

1.6.1[2].


23.45 sin 360284

365


Under given conditions hour angle is

15° 3.5 . ° .



sin


8

sin cos 51.5078 cos 52.5 cos 2.82 sin 51.5078 sin 2.82. °


sin

sincos 2.82 sin 30

cos 19.87 . °


angle:

cos

cos tan 51.5078 tan 2.82 . °


2

152

1586.45 .



(c) On 15th Sept at 10:0am, Day of the year can be obtained with the help of Table

1.6.1[2].


23.45 sin 360284

365


At 10:0am hour angle is;

15° 2.5 . ° .




9

sin

sin cos 51.5078 cos 30 cos 2.22 sin 51.5078 sin 2.22. °


sin

sincos 2.22 sin 30

cos 34.67 . °


angle:

cos

cos tan 51.5078 tan 2.22 . °


2

152

1592.79 .



(d) On 15th Sept at 3:30pm, Day of the year can be obtained with the help of Table

1.6.1[2].


23.45 sin 360284

365


At 3:30pm hour angle is;

15° 3.5 . ° .


10



sin



sin

sincos 2.22 sin 52.50

cos 24.14 . °


angle:

cos

cos tan 51.5078 tan 2.22 . °


2

152

1592.79 .




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QUESTION-3:

A flat‐plate collector in Nicosia, Cyprus, is tilted at 40° from horizontal and pointed 10° east of

south. Calculate the solar incidence angle on the collector at 10:30 am and 2:30 pm solar times

on March 10 and September10?

Given:

Latitude of Nicosia, Cyprus = 35° 10’00.66’’ [1] = 35.1668°

Surface Tilt angle = = 40°

Surface Azimuth angle = = ‐10°

Day of Year = (a) March 10 (b) September 10

Solar Time = (a) 10:30am (b) 2:30pm

Find:

(a) Solar Incident Angle on 15th March at 10:30am =?

(b) Solar Incident Angle on 15th March at 2:30am =?

(c) Solar Incident Angle on 15th Sept at 10:30am =?

(d) Solar Incident Angle on 15th Sept at 2:30am =?

Solution:

(a) On 10th March at 10:30am, Day of the year can be obtained with the help of Table

1.6.1[2].


23.45 sin 360284

365. °

At 10:30am, hour angle is:

15° 1.5 . ° .




12

sin



sin


cos 44.77 . °

Using a latitude and declination in following equation for calculating solar incidence

angle:

cos

cos sin 4.81 sin 35.1668 cos 40 4.81 35.1668 40 cos 10

4.81 35.1668 40 22.54.81 35.1668 40 10 cos 22.54.81 40 10 22.5

Implies that; . °

(b) On 10th March at 2:30pm, Day of the year can be obtained with the help of Table

1.6.1[2].


23.45 sin 360284

365. °

At 2:30pm, hour angle is:

15° 2.5 . ° .




13

sin



sin


cos 36.72 . °


angle:

cos


4.81 35.1668 40 37.54.81 35.1668 40 10 cos 37.54.81 40 10 37.5

Implies that; . °

(c) On 10th Sept at 10:30am, Day of the year can be obtained with the help of Table

1.6.1[2].


23.45 sin 360284

365. °

At 10:30am, hour angle is:

15° 1.5 . ° .




14

sin



sin


cos 52.71 . °


angle:

cos 0.8927


4.22 35.1668 40 cos 22.54.22 35.1668 40 10 cos 22.54.22 40 10 22.5

Implies that; . °

(d) On 10th Sept, Day of the year can be obtained with the help of Table 1.6.1[2].


23.45 sin 360284

365. °

At 2:30pm, hour angle is:

15° 2.5 . ° .



sin


15



sin


cos 43.56 . °


angle:

cos 0.8927


4.22 35.1668 40 cos 37.54.22 35.1668 40 10 cos 37.54.22 40 10 37.5

Implies that; . °


16

QUESTION-4:

Assuming that the sun is a blackbody at 5777 K, at what wavelength does the maximum

monochromatic emissive power occur? What fraction of energy from this source is in the visible

part of the spectrum in the range 0.38–0.78 μm?

Answer:

(a) Monochromatic emissive power is defined as the energy emitted per unit surface area

at specific wavelength. The wavelength at maximum monochromatic emissive power is

0.48, from Table 1.3.1(a)[2].

(b) The fractions corresponding to wavelength of 0.38 and 0.78 μm are 0.064 and 0.544

respectively [2]. Thus the fraction in the visible range is 0.544 ‐ 0.065 = 0.479.


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REFERENCES

[1] GoogleEarth Software 6.2.2.6613; kh.google.com

[2] John S. Duffie & William A. Beckman, “Solar Engineering of Thermal Processes”, Second

Edition

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Assignment 1-solar thermalsystem

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