The basic radiation transfer equation is given by equation (6.5),
Qab = TabqoaFabAa W/nm (8.2)
where Qab is the radiation power incident on area Ab that originates at area Aa. Also, xab is the transmittance of the media between Aa and Ab, qoa is radiation power per unit area per wavelength leaving Aa and Fab is the view-factor from area Aa to area Ab. Equation (8.2) assumes qoa is uniform across Aa. In addition, assume the radiation fluxes arriving at an area, qi, are also uniform across the area. In that case, Qia = qiaAa, where qia is the total incident flux on area Aa. Also, the view-factor relation given by equation (6.7)
applies. Thus, applying equation (8.2) and equation (6.7) to the radiation leaving all areas that view area Ab the following is obtained.
qib = ^bqoaFba (8.3)
Also, the flux leaving a surface is the sum of the emitted and reflected fluxes [equation (6.30)]. Therefore, equation (8.3) becomes the following,
qob =Pbqib +£beb (^,Tb ) = Pb ^TA, aFba +^beb (^,Tb ) (8.4)
where Єь is the spectral emittance of Aa and eb(X, Tb) is the blackbody emissive power at the temperature of Ab, which is given by equation (1.120).
Now apply equation (8.4) to each of the surfaces in Figure 8.1, noting that tab = 1 since a vacuum exists in the optical cavity.
qoE _Pe [FEC1qoa +FEC2qoC2 + FEC3qoC3 + FEbqob ] _ ЄЬЄЬ (^,TE ) (8.5)
qob (-1 _PbFbb ) _Pb [FbEqoE + FbaqoC1 + FbC2qoC2 + FbC3qoC3 ] _ ЄЬЄЬ (^,Tb ) (8.6)
qoC1 _Pc1 [FC1EqoE + FC1bqob ] _ £C1eb (^,TC1 ) (8.7)
qoC2 Pc2 [FC2EqoE + FC2bqob ] _ £C2eb (^,TC2 ) (8.8)
qo3 _Pc3 [FC3EqoE + FC3bqob ] _ £C3eb (^, TC3 ) (8.9)
Notice the pbFbb term in equation (8.6). It occurs because the reflector has a view of itself.
Equations (8.5) through (8.9) form a set of five linear algebraic equations for the five unknown qo’s. The following solutions for qoE and qob are obtained after a tedious amount of algebraic manipulation,
a = SEeb (^,TE ) + PE [FEC1SС1ЄЬ (^,TC1 ) + FEC2SС2ЄЬ, TC2 ) + FEC3SC3 ,TC3 )]
b = Sbeb (^,Tb ) + Pb [FbC1SC1eb (^,ТИ ) + FbC2SC2eb (^,TC2 ) + FbC3eb (^,TC3 )] B1 _ Pb [FbE + PaFC1EFbC1 +PC2FC2EFbC2 + PC3FC3EFbC3 ]
B2 _ 1 _ Pb [Fbb + pC1FC1bFbC1 +PC2FC2bFbC2 + PC3FC3bFbC3 ]
E1 _ 1 _Pe [pC1FC1EFEC1 +Pc2FC2EFEC2 +Pc3FC3EFEC3 ]
E2 _ Pe [FEb + PaFC1bFEC1 +PC2FC2bFEC2 +PC3FC3bFEC3 ]
DEN = E1B2 – E2B1
Equations (8.10) and (8.11) are solutions for qoE(X) and qob(X) in terms of the view – factors, optical properties, and temperatures of each of the five areas. Knowing qoE and qob, solutions for qoC1, qoC2, and qoC3 are obtained using equations (8.7) through (8.9).
The view-factors depend upon the geometry of the optical cavity. For the model shown in Figure 8.1, the areas AE, AC1, AC2, and AC3 can be either circular or square with a corresponding cylindrical or square prism shaped reflector. View-factor expressions for parallel circular disks, identical, parallel opposed rectangles, and unequal, parallel, coaxial squares are given in Figure 6.3. These three view-factors are all that are required to calculate the necessary view-factors for either the circular or square geometry as shown in Figure 8.1. Consider FEC1; this view-factor is given in Figure 6.3a for a circular geometry and Figure 6.3c for a square geometry. Thus, using the reciprocity relation [equation (6.7)] the view-factor FC1E can be calculated.
FC1E =T^FEC1 (8.19)
From view-factor algebra,
Obtaining FC2b and FC3b requires the following view-factor algebra.
parameters required to obtain the solutions for qoE, qob, qoC1, qoC2, and qoC3. Knowing the fluxes allows the calculation of the cavity and photovoltaic efficiencies.
The cavity efficiency is the following,
where Qin = Qe + Qb is the power input to the emitter and reflector. Qe is given by equation (6.85).
Qe = Ae f^b (*,Te)-qoE]& (8.37)
Qb = Ab [eb (X, Te)-qob]dX (8.38)
In obtaining equations (8.37) and (8.38), the relation є = 1 – p is used since the emitter and reflector are opaque, x = 0 [see equation (1.229)]. If areas AC1, AC2, and AC3 contain PV arrays, then the incident, useful power, Qc is the following,
where xC1, xC2, and xC3 are the filter transmittances and Xg is the wavelength corresponding to the PV array bandgap energy, Eg.
1 24 x!03
A. = ————- nm Eg in eV (8.40)
g E g
If there is no filter, then xC1 = xC2 = xC3 = 1 in equation (8.39). Since pq;=qo – seb(A, T), equations (8.7)-(8.9) can be used to replace qic2 and qic3 in equation (8.39).
Qc = AC1 J^CI [FC1EqoE + FC1bqob + AC2 ^ТС2 [FC2EqoE + FC2bqob
+AC3 ^ТС3 [FC3EqoE + FC3bqob
Using the view-factor reciprocity relation, this can be written as follows.
QC = AE ji’t:C1FEC1 +TC2FEC2 + TC3FEC3 JfoE^^ o
+Ab ^[ТС1^Ьс1 +TC2FbC2 +TC3FbC3 }fobd^
If only Ac1 contains a PV array, then the Tc2Fec2, Tc3Fec3, ^Fbc2, and Tc3Fbc3 terms must be removed from equation (8.41b). As can be seen from equations (8.37), (8.38), and (8.41b), once qoE(X) and qob(X) are known, the cavity efficiency can be calculated. The PV efficiency is the following [equation (6.88)],
qPV = Q – (6.88)
where the electrical power output, PEL, depends upon which of the three areas, AC1, AC2, and AC3 have PV arrays. If all three areas have PV arrays, then PEL is the following,
PEL = PELC1 + PELC2 + PELC3 (6.89)
where the electrical power from each array is given by equation (5.228).
Pel = N (VM – RsIlM )Ilm (5.228)
Here Nj is the number of series connected junctions in the PV array, VM is the junction voltage for maximum PEL, and is the root of equation (5.223). In addition, Rs is the series resistance of each junction, and Ilm is the load current for maximum PEL and is given as a function of VM by equation (5.229).
The electrical power output of an array can also be given in terms of the fill factor, FF, short circuit current, Isc, and open circuit voltage, Voc [equation (5.239)].
Pel = FFV0CISC (5.239)
Where Isc is the root of equation (5.217a), Voc is the root of equation (5.218a), and FF is given by equation (5.240).
All the quantities that appear in equations (5.228) and (5.239) for Pel are functions of the PV array series resistance, R^, and shunt resistance, Rsh, the photon generated current, Iph, and the dark saturation current, Is. The resistance and dark saturation current can be determined experimentally for each PV array. And the photon generated current is given as a function of the internal quantum efficiency, Pq, by equation (5.249),
where AJa is the active area of a single junction in the PV array, RC is the spectral reflectivity at the filter-PV array interface or the vacuum-PV interface if no filter is used, and qiC is the incident radiation flux. For the no filter case, xC = 1. The incident flux, qiC, for each of the three areas, AC1, AC2, and AC3, are obtained using equations (8.7) through (8.9) and the result pq; = qo – eeb(X, T). For example, if area AC1 contains a PV array, then IphC1 is the following.
If PV arrays exist on Aa and AC3, then IphC2 and IphC3 are obtained from equation (8.42) by replacing the C1 subscripts by C2 and C3.
The spectral reflectivity, RC(X), and transmittance, xC(X), and the emitter spectral emittance, eE(X), are the important spectral control parameters. Obviously, the ideal situation would have eE ^ 1 and xC ^ 1 (pC = 0) for X < Xg and eE ^ 0 and pC ^ 1 (xC = 0) for X > Xg. In that case, pc ^ 1. The use of a selective emitter, such as the rare earths, discussed in Chapter 3, will produce large eE for X > Xg but has the disadvantage of large eE for X > 5 pm. Tandem plasma-interference and the resonant array filters discussed in Chapter 4 can be fabricated that have pC ^ 1 for X > Xg. Therefore, a combination of a selective emitter and a filter may be a viable solution for effective spectral control. However, absorptance in the filter is a problem, as is shown in Section 8.4.2. An alternative to a filter is the backside reflector (BSR) on the PV array as
discussed in Chapter 4. In that case, by using an anti-reflective coating on the PV array, xC ^ 1 and pC ^ 0 for X < Xg. Also, there is no filter absorptance loss.
All the quantities necessary to calculate pc and pPV have been written as functions of qoE and qob, which are obtained from equations (8.10) and (8.11). However, before qoE and qob can be determined, the emitter and reflector temperatures, TE and Tb, as well as the, temperatures TC1, TC2, and TC3 of AC1, AC2, and AC3 must be known. In a TPV system, the temperatures, TC1, TC2, and TC3 are fixed by the waste heat radiator. If AC1, AC2, and AC3 all contain PV arrays, then that temperature is much lower than TE and Tb so that the eb(X, TC1), eb(X, TC2), and eb(x, TC3) terms in equations (8.10) and (8.11) can be neglected.
To determine TE and Tb as functions of the input power, Qin, an energy balance must be applied to the optical cavity. Conservation of energy requires the following relation be satisfied,
Qin = Qe (Te ) + Qb (Tb )
where Qe and Qb are given by equations (8.37) and (8.38) and depend upon qoE and qob. Therefore, an iterative process is required to satisfy equation (8.43). Since the emitter and reflector are connected, assume TE = Tb. Also, assume TC1 = TC2 = TC3 is given or that the TC1, TC2, and TC3 emittance terms can be neglected. As a result, equation (8.43) is a function of only a single temperature, TE. To begin the iterative process, a value for TE is assumed and qoE and qob are then determined as functions of X by equations (8.10) and (8.11). Using these results for qoE and qob, values for QE and Qb are calculated using equations (8.37) and (8.38). If the correct value of TE is assumed then the sum of QE plus Qb will equal Qin. If too large a value of TE is assumed, then QE + Qb > Qin. If too small a value of TE is assumed, then QE + Qb < Qin . Therefore, an iteration expression for TE that converges to the proper solution is the following,
where TE+1 is the i + 1 approximation for TE and TE is the ith approximation for TE, and Ci is an iteration constant with the units K/W that must be given. QE and Qb are calculated using TE. Once Te has converged to the value that satisfies equation (8.43) the cavity and PV efficiencies can be calculated.
A Mathematic program that calculates all the radiation fluxes and the efficiencies pc and pPV for a square optical cavity like that shown in Figure 8.1 is described in Appendix F. The enclosed CD-ROM contains the program. In the program, PV arrays exist on all three areas AC1, AC2, and AC3. Also, Ci = 0.5 K/W has been chosen for the iteration constant. In addition, the program allows for a gap to exist between the emitter and the reflector. However, if a gap exists between the emitter and the reflector, then Tb ^TE. In that case, a separate energy balance on the reflector must be used to determine Tb for a given TE. Then the overall energy balance, Qin = QE (TE ) , must be applied to determine TE.
Consider a hypothetical TPV system with the following components.
gray body emitter with eE = 0.6, pE = 1 – eE = 0.4 constant reflectance, pb = 0.7, £b = 1 – pb = 0.3 bandgap energy, Eg = 0.6eV (Xg = 2070nm), constant quantum efficiency, Pq = 0.9, dark saturation current density at 300K, Js = 8 x 10-7 A/cm2, ideality factor, Ao = 1, series resistance, Rs = 0.06^, shunt resistance, Rsh = 2000^, percent active area = 0.9
back surface reflector (BSR), pC = 0.1 for 0 < X < Хг, pC = 0.9
for Xg < X < to
InGaAs is a possible material for making a 0.6eV bandgap energy PV array with characteristics like those presented above.
Assume each of the three areas, AC1, AC2, and AC3 are covered by PV arrays. Referring to Figure 8.1, the dimensions of the square optical cavity are the following: wC1 = 6cm, wC2 = 9cm, wC3 = 10cm, h = 0.2cm, and dC = 0.05cm. Each PV array is assumed to consist of 25 junctions. Area AC1 is covered with four arrays, area AC2 is covered with five arrays, and Ac3 is covered with two arrays. Thus, the area of each
of each array and -—^Ц – is the area of each junction on AC3.
(2 x 25)
With this input data, the Mathematica program in Appendix F produces the results shown in Figure 8.2. Shown are the radiation fluxes incident on and leaving the various
surfaces plus the electrical power generated by each of the PV arrays. The emitter and reflector temperatures are TE = Tb = 1232K, the cavity efficiency is pc = 0.57 while the PV efficiency of each of the PV arrays is pPV = 0.31. As a result, the TPV efficiency is pTPV = pcpPV = 0.18 and the electrical power out is PEL = 44W.
TPV efficiency, TJxpv = Ticr|pv = -18 Electrical power out, PEL = 44 W
Note that 22W of radiation leak out the 0.05cm gap between the reflector and the PV array. This is a significant fraction of the total 250W thermal energy input. Also
notice the large radiation fluxes that are circulating within the optical cavity. The emitter is emitting 1171W and receiving 930W. Most of the 1171W is incident on the PV arrays. Thus, if a filter with even a small absorptance were present, the absorptance loss would be significant.