When I input higher solar irradiance into my solar cell, its giving me a lower electrical efficiency; this doesn't seem right?

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Jason Dong
Jason Dong on 5 Dec 2021
Answered: Joel Van Sickel on 22 Dec 2021
I'm having a bit of trouble understanding how the electrical portion of MATLAB's example model of a Photovoltaic Thermal (PVT) system works: https://www.mathworks.com/help/physmod/sps/ug/photovoltaic-thermal-pvt-hybrid-solar-panel.html?searchHighlight=solar&s_tid=doc_srchtitle#d123e7823
I read through the article and I understand most of the other parts but I'm not sure how the electrical model works. I assume that its just based off the solar cell model that they also created: https://www.mathworks.com/help/physmod/sps/ref/solarcell.html
However, what is puzzling me is when I input a solar irradiance higher than the default (1000 W/m^2), the electrical efficiency drops instead of rises. Theoretically, since the solar cell is receiving more energy, shouldn't the electrical efficiency also rise? Is there a parameter in the model that I should also adjust in response to the higher solar irradiance to offset this? Any help would be much appreciated. Here is my parameter as reference:
%% Load parameters ------------------------------------
load.R = 4.5; %Resistance. With default parameters, 4.5 Ohm is approximately optimal for maximum electrical power extraction
%% Solar panel parameters ------------------------------
% Initial temperatures [K]
panel.initial.Tg0 = 295; %Glass cover
panel.initial.Tpv0 = 295; %PV cells
panel.initial.Te0 = 295; %Heat exchanger
panel.initial.Tw0 = 295; %Water in the tank
panel.initial.Tb0 = 295; %Back cover
% Geometry
panel.geometry.Acell = 0.0225; %Area of a cell, [m^2]
panel.geometry.Ncell = 72; %Number of cells
% Optical properties
panel.optical.ng = 1.52; %Refractive index ratio glass/air
panel.optical.absg = 0.2; %Absorption coefficient of glass per unit length [1/m]
panel.optical.dg = 0.01; %Thickness of glass cover [m]
panel.optical.rpv = 0.15; %Reflection factor of PV cell
% Heat transfer properties
panel.heatTransfer.Ta = 295; %Temperature of ambient air [K]
panel.heatTransfer.Tsky = 290;%Temperature of sky (for radiative heat transfer) [K]
panel.heatTransfer.Mg = 4; %Mass of glass cover [kg]
panel.heatTransfer.Mpv = 0.2; %Mass of one PV cell [kg]
panel.heatTransfer.Me = 15; %Mass of heat exchanger [kg]
panel.heatTransfer.Mb = 5; %Mass of back cover [kg]
panel.heatTransfer.Cg = 800; %Specific heat of glass [J/kg/K]
panel.heatTransfer.Cpv = 200; %Specific heat of PV cell [J/kg/K]
panel.heatTransfer.Ce = 460; %Specific heat of heat exchanger [J/kg/K]
panel.heatTransfer.Cb = 400; %Specific heat of back cover [J/kg/K]
panel.heatTransfer.epsg = 0.75; %Emissivity of glass
panel.heatTransfer.epspv = 0.7; %Emissivity of PV cell
panel.heatTransfer.hga = 10;%Free convection coefficient between glass and ambient air [W/m^2/K]
panel.heatTransfer.hgpv = 20; %Free convection coefficient between PV cells and glass [W/m^2/K]
panel.heatTransfer.hba = 10;%Free convection coefficient between back cover and ambient air [W/m^2/K]
panel.heatTransfer.ke = 130;%Thermal conductivity of heat exchanger [W/m/K]
panel.heatTransfer.Le = 0.04;%Thickness of heat exchanger [m]
panel.heatTransfer.kins = 0.1;%Thermal conductivity of insulation layer [W/m/K]
panel.heatTransfer.Lins = 0.03;%Thickness of insulation layer [m]
% PV cell electrical properties
panel.pv.Isc = 8.88; % Short-circuit current, Isc [A]
panel.pv.Voc = 0.62;% Open-circuit voltage, Voc [V]
panel.pv.Is = 1e-6; % Diode saturation current, Is [A]
panel.pv.Is2 = 0; % Diode saturation current, Is2 [A]
panel.pv.Iph0 = 8.88;% Solar-generated current for measurements, Iph0 [A]
panel.pv.Ir0 = 4000;% Irradiance used for measurements, Ir0 [W/m^2]
panel.pv.ec = 1.5; % Quality factor, N
panel.pv.N2 = 2; % Quality factor, N2
panel.pv.Rs = 0; % Series resistance, Rs [Ohm]
panel.pv.Rp = inf ; % Parallel resistance, Rp [Ohm]
panel.pv.TIPH1 = 0; % First order temperature coefficient for Iph, TIPH1 [1/K]
panel.pv.EG = 1.11;% Energy gap, EG [eV]
panel.pv.TXIS1 = 3;% Temperature exponent for Is, TXIS1
panel.pv.TXIS2 = 3;% Temperature exponent for Is2, TXIS2
panel.pv.TRS1 = 0;% Temperature exponent for Rs, TRS1
panel.pv.TRP1 = 0;% Temperature exponent for Rp, TRP1
panel.pv.Tmeas = 25; % Measurement temperature [degC]
%% Pipe parameters ----------------------------------------
pipe.length = 5; % Pipe length [m]
pipe.area = 0.0007; % Cross-sectional area [m^2]
pipe.Dh = 0.03; % Hydraulic diameter [m]
pipe.length_add = 1; % Aggregate equivalent length of local resistances [m]
pipe.roughness = 15e-6; % Internal surface absolute roughness [m]
pipe.Re_lam = 2000; % Laminar flow upper Reynolds number limit
pipe.Re_tur = 4000; % Turbulent flow lower Reynolds number limit
pipe.shape_factor = 64; % Shape factor for laminar flow viscous friction
pipe.Nu_lam = 3.66; % Nusselt number for laminar flow heat transfer
%% Tank params -------------------------------------------
tank.Volmax = 0.25; % Maximum tank capacity [m^3]
tank.Atank = 0.3; % Tank cross-sectional area [m^2]
tank.Voltank0 = 0.1; % Initial volume in the tank [m^3]
tank.Ttank0 = 295; % Initial temperature in the tank [K]
tank.Lins = 0.05; % Insulating layer thickness [m]
tank.kins = 0.1; % Thermal conductivity of insulation layer [W/m/K]
tank.hta = 10; % Free convection coefficient between tank and ambient air [W/m^2/K]
%% Pump flow input params --------------------------------
pumps.mdot_int = 0.02; % Internal circuit mass flow rate [kg/s]
pumps.mdot_dem = 0.005; % Demand mass flow rate (to the sink) [kg/s]
pumps.mdot_sup = 0.005; % Supply mass flow rate (from the source) [kg/s]
And for my output:
****** Efficiency Calculation *********
Total input energy from the sun in the period: 43.7928 kWh
Average input energy from the sun per day: 14.5976 kWh/day
Total electrical energy supplied to the load: 0.51956 kWh
Average electrical energy supplied per day: 0.17319 kWh/day
Total absolute thermal energy in the water supplied to the user: 34.4859 kWh
Total absolute thermal energy in the water extracted from the source: 16.498 kWh
Total used thermal energy (sink - source): 17.9879 kWh
Average used thermal energy per day (sink - source): 5.996 kWh/day
Electrical efficiency: 0.011864
Thermal efficiency: 0.41075
Total efficiency: 0.42261
***************************************
It just seems that an electrical efficiency of 1.2% is too low for a solar irradiance of 4000 W/m^2. Maybe someone with more electrical expertise can make sense of this code.

Answers (1)

Joel Van Sickel
Joel Van Sickel on 22 Dec 2021
Hello Jason,
this example uses a very simple electrical model that isn't ussing MPPT. Without MPPT, efficiency doesn't necessarily improve with higher irradiance. It also doesn't help that the system wasn't designed for that level of irradiance.
Regards,
Joel

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