## Thermal Analysis Of Flatplate Collectors

In this section, the thermal analysis of the collectors is presented. The two major types of collectors, flat plate and concentrating, are examined separately. The basic parameter to consider is the collector thermal efficiency. This is defined as the ratio of the useful energy delivered to the energy incident on the collector aperture. The incident solar flux consists of direct and diffuse radiation. While flat-plate collectors can collect both, concentrating collectors can utilize direct radiation only if the concentration ratio is greater than 10 (Prapas et al., 1987).

Glass cover

Absorber plate

FIGURE 3.23 Radiation transfer between the glass cover and absorber plate.

In this section, the various relations required to determine the useful energy collected and the interaction of the various constructional parameters on the performance of a collector are presented.

The prediction of collector performance requires information on the solar energy absorbed by the collector absorber plate. The solar energy incident on a tilted surface can be found by the methods presented in Chapter 2. As can be seen from Chapter 2, the incident radiation has three special components: beam, diffuse, and ground-reflected radiation. This calculation depends on the radiation model employed. Using the isotropic model on an hourly basis, Eq. (2.97) can be modified to give the absorbed radiation, S, by multiplying each term with the appropriate transmittance-absorptance product as follows:

where the terms [1 + cos(|3)]/2 and [1 - cos(|3)]/2 are the view factors from the collector to the sky and from the collector to the ground, respectively. The same equation can be used to estimate the monthly average absorbed solar radiation, S, by replacing the hourly direct and diffuse radiation values with the appropriate monthly average values, HB and HD, RB with RB, and various (to) values with monthly average values, (ra) in Eq. (3.1). More details on this are given in Chapter 11.

The combination of cover with the absorber plate is shown in Figure 3.23, together with a ray tracing of the radiation. As can be seen, of the incident energy falling on the collector, to is absorbed by the absorber plate and (1 - o)t is reflected back to the glass cover. The reflection from the absorber

plate is assumed to be diffuse, so the fraction (1 - a)T that strikes the glass cover is diffuse radiation and (1 - a)TpD is reflected back to the absorber plate. The multiple reflection of diffuse radiation continues so that the fraction of the incident solar energy ultimately absorbed is

Typical values of (Ta) are 0.7-0.75 for window glass and 0.9-0.85 for low-iron glass. A reasonable approximation of Eq. (3.2) for most practical solar collectors is

The reflectance of the glass cover for diffuse radiation incident from the absorber plate, pD, can be estimated from Eq. (2.57) as the difference between Ta and t at an angle of 60°. For single covers, the following values can be used for pD:

For KL = 0.0125, pD = 0.15. For KL = 0.0370, pD = 0.12. For KL = 0.0524, pD = 0.11.

For a given collector tilt angle, (3, the following empirical relations, derived by Brandemuehl and Beckman (1980), can be used to find the effective incidence angle for diffuse radiation from sky, 9e, D, and ground-reflected radiation, 9e,G:

where 3 = collector slope angle in degrees.

The proper transmittance can then be obtained from Eq. (2.53), whereas the angle dependent absorptance from 0 to 80° can be obtained from (Beckman et al., 1977):

an where

9e = effective incidence angle (degrees).

on = absorptance at normal incident angle, which can be found from the properties of the absorber. Incidence angle, 9 (degrees)

FIGURE 3.24 Typical (ra)/(ra)n curves for one to four glass covers. (Reprinted from Klein (1979), with permission from Elsevier.)

Incidence angle, 9 (degrees)

FIGURE 3.24 Typical (ra)/(ra)n curves for one to four glass covers. (Reprinted from Klein (1979), with permission from Elsevier.)

Subsequently, Eq. (3.2) can be used to find (Ta)fl and (toi)g. The incidence angle, 9, of the beam radiation required to estimate RB can be used to find (to)b.

Alternatively, (to)„ can be found from the properties of the cover and absorber materials, and Figure 3.24 can be used at the appropriate angle of incidence for each radiation component to find the three transmittance-absorp-tance products.

When measurements of incident solar radiation (It) are available, instead of Eq. (3.1), the following relation can be used:

where (To)av can be obtained from

Example 3.1

For a clear winter day, IB = 1.42 MJ/m2 and ID = 0.39 MJ/m2. Ground reflectance is 0.5, incidence angle is 23°, and RB = 2.21. Calculate the absorbed solar radiation by a collector having a glass with KL = 0.037, the absorptance of the plate at normal incidence, an = 0.91, and the refraction index of glass is 1.526. The collector slope is 60°.

Solution

Using Eq. (3.5) for the beam radiation at 9 = 23°,

— = 1 + 2.0345 X 10-3 9e - 1.99 X 10-4 92 n + 5.324 X 10-6 93 - 4.799 X 10-8 9¡ = 1 + 2.0345 X 10-3 X 23 - 1.99 X 10-4 X 232 + 5.324 X 10-6 X 233 - 4.799 X 10-8 X 234 = 0.993

For the transmittance we need to calculate t„ and Tr. For the former, Eq. (2.51) can be used. From Eq. (2.44), 02 = 14.8°.Therefore,

0.037 I

0.962

From Eqs. (2.45) and (2.46) r± = 0.054 and r = 0.034. Therefore, from Eq. (2.50a),

Finally, from Eq. (2.53), t = TaTr = 0.962 X 0.916 = 0.881.

Alternatively, Eq. (2.52a) could be used with the above r values to obtain t directly.

(Ta)B = 1.01t (a/an )an = 1.01 X 0.881 X 0.91 X 0.993 = 0.804 w 0.80 From Eq. (3.4a), the effective incidence angle for diffuse radiation is

D = 59.68 - 0.1388(3 + 0.001497p2 = 59.68 - 0.1388 X 60 + 0.001497 X 602 = 57°

From Eq. (3.5), for the diffuse radiation at 9 = 57°, a/a„ = 0.949. From Eq. (2.44), for 9X = 57°, 92 = 33°. From Eqs. (2.45) and (2.46), r± = 0.165 and r = 0.

From Eq. (2.50a), Tr = 0.858, and from Eq. (2.51), to = 0.957. From Eq. (2.53), t = 0.957 X 0.858 = 0.821

and from Eq. (3.3), (t«)d = 1.01T(a/an )an = 1.01 X 0.821 X 0.949 X 0.91 = 0.716 w 0.72

From Eq. (3.4b), the effective incidence angle for ground reflected radiation is

From Eq. (3.5), for the ground reflected radiation at 9 = 65°, o/on = 0.897. From Eq. (2.44), for 9i = 65°, 92 = 36°. From Eqs. (2.45) and (2.46), r± = 0.244 and r = 0.012.

From Eq. (2.50a), Tr = 0.792, and from Eq. (2.51), to = 0.955. From Eq. (2.53), t = 0.792 X 0.955 = 0.756

And from Eq. (3.3), (Ta)G = 1.01T(a/an )an = 1.01 X 0.756 X 0.897 X 0.91 = 0.623 « 0.62 In a different way, from Eq. (3.3),

(Ta)n = 1.01 X 0.884 X 0.91 = 0.812 (note that for the transmittance the above value for normal incidence is used, i.e., Tn)

From Figure 3.24, for beam radiation at 9 = 23°, (Ta)/(Ta)n = 0.98. Therefore,

From Figure 3.24, for diffuse radiation at 9 = 57°, (Ta)/(Ta)n = 0.89. Therefore,

From Figure 3.24, for ground-reflected radiation at 9 = 65°, (Ta)/(Ta)n = 0.76. Therefore,

All these values are very similar to the previously found values, but the effort required is much less.

Finally, the absorbed solar radiation is obtained from Eq. (3.1):

2.72 MJ/m2

### 3.3.2 Collector Energy Losses

When a certain amount of solar radiation falls on the surface of a collector, most of it is absorbed and delivered to the transport fluid, and it is carried away as useful energy. However, as in all thermal systems, heat losses to the environment by various modes of heat transfer are inevitable. The thermal network for a single-cover, flat-plate collector in terms of conduction, convection, and radiation is shown in Figure 3.25a and in terms of the resistance between plates in Figure 3.25b. The temperature of the plate is Tp, the collector back temperature is Tb, and the absorbed solar radiation is S. In a simplified way, the various thermal losses from the collector can be combined into a simple resistance, RL, as shown in Figure 3.25c, so that the energy losses from the collector can be written as

where

UL = overall heat loss coefficient based on collector area Ac (W/m2-K). Tp = plate temperature (°C).

The overall heat loss coefficient is a complicated function of the collector construction and its operating conditions, given by the following expression:

where

Ub = bottom heat loss coefficient (W/m2-K).

Ue = heat loss coefficient form the collector edges (W/m2-K).

Therefore, UL is the heat transfer resistance from the absorber plate to the ambient air. All these coefficients are examined separately. It should be noted that edge losses are not shown in Figure 3.25.

Ambient air

Glass cover g

Collector plate

(c) Simple collector network

Collector back hc h.

Rb-a

Ambient air

(a) Heat transfer network (b) Resistance network FIGURE 3.25 Thermal network for a single cover collector in terms of (a) conduction, convection, and radiation; (b) resistance between plates; and (c) a simple collector network.

In addition to serving as a heat trap by admitting shortwave solar radiation and retaining longwave thermal radiation, the glazing also reduces heat loss by convection. The insulating effect of the glazing is enhanced by the use of several sheets of glass or glass plus plastic.

Under steady-state conditions, the heat transfer from the absorber plate to the glass cover is the same as the energy lost from the glass cover to ambient. As shown in Figure 3.25, the heat transfer upward from the absorber plate at temperature Tp to the glass cover at Tg and from the glass cover at Tg to ambient at Ta is by convection and infrared radiation. For the infrared radiation heat loss, Eq. (2.67) can be used. Therefore, the heat loss from absorber plate to glass is given by

Qt, absorber plate to glass cover Achc,p—g ((Tp Ta) +

hc, p-g = convection heat transfer coefficient between the absorber plate and glass cover (W/m2-K). ep = infrared emissivity of absorber plate. eg = infrared emissivity of glass cover.

For tilt angles up to 60°, the convective heat transfer coefficient, hcp-g, is given by Hollands et al. (1976) for collector inclination (9) in degrees:

Nu = hc, P_gL = 1 + 1446 [1 1708 1+ L 1708[sin(1.89)]1'6

5830

where the plus sign represents positive values only. The Rayleigh value, Ra, is given by

where g = gravitational constant, = 9.81 m2/s.

|3' = volumetric coefficient of expansion; for ideal gas, |3' = 1/T.

Pr = Prandtl number.

L = absorber to glass cover distance (m).

The fluid properties in Eq. (3.12) are evaluated at the mean gap temperature (Tp + Tg)/2.

For vertical collectors, the convection correlation is given by Shewen et al. (1996) as

0.0665Ra0 333

The radiation term in Eq. (3.10) can be linearized by the use of Eq. (2.73) as h _ o(Tp + Tg)(T¡ + T¡) (3 14)

Consequently, Eq. (3.10) becomes

Qt, absorber plate to glass cover Ac (hc,p—g ^ hr,p—g )(Tp Tg )

Rp-g k in which

Similarly, the heat loss from glass cover to ambient is given by

Qt, absorber plate to glass cover Ac (hc,g — a hr,g—a )(Tg ^Ta) D (3.17)

Rg — a where hc,g-a = convection heat transfer coefficient between the glass cover and ambient due to wind (W/m2-K). hr,g-a = radiation heat transfer coefficient between the glass cover and ambient (W/m2-K).

The radiation heat transfer coefficient is now given by Eq. (2.75), noting that, instead of Tsky, Ta is used for convenience, since the sky temperature does not affect the results much:

Since resistances Rp_g and Rg-a are in series, their resultant is given by

Therefore,

In some cases, collectors are constructed with two glass covers in an attempt to lower heat losses. In this case, another resistance is added to the system shown in Figure 3.25 to account for the heat transfer from the lower to upper glass covers. By following a similar analysis, the heat transfer from the lower glass at Tg2 to the upper glass at Tg1 is given by

Qt, lower cover to top cover Ac (hc,g2 — g1 ^ hr,g2—g\)(Tg2 Tg1)

where hc, g2-gi = convection heat transfer coefficient between the two glass covers (W/m2-K).

hr, g2-gi = radiation heat transfer coefficient between the two glass covers (W/m2-K).

The convection heat transfer coefficient can be obtained by Eqs. (3.11)-(3.13). The radiation heat transfer coefficient can be obtained again from Eq. (2.73) and is given by h^gl = ^ + ^ + ^ (3.23)

where eg2 and eg1 are the infrared emissivities of the top and bottom glass covers.

Finally, the resistance Rg2-g1 is given by

In the case of collectors with two covers, Eq. (3.24) is added on the resistance values in Eq. (3.20). The analysis of a two-cover collector is given in Example 3.2.

In the preceding equations, solutions by iterations are required for the calculation of the top heat loss coefficient, Ut, since the air properties are functions of operating temperature. Because the iterations required are tedious and time consuming, especially for the case of multiple-cover systems, straightforward evaluation of Ut is given by the following empirical equation with sufficient accuracy for design purposes (Klein, 1975):

g where f = (1 - 0.04hw + 0.0005hW)(1 + 0.091Ng) (3.26)

8.6V06

It should be noted that, for the wind heat transfer coefficient, no well-established research has been undertaken yet, but until this is done, Eq. (3.28) can be used. The minimum value of hw for still air conditions is 5 W/m2-°C. Therefore, if Eq. (3.28) gives a lower value, this should be used as a minimum.

The energy loss from the bottom of the collector is first conducted through the insulation and then by a combined convection and infrared radiation transfer to the surrounding ambient air. Because the temperature of the bottom part of the casing is low, the radiation term (hrb_a) can be neglected; thus the energy loss is given by

kb hc,b—a where tb = thickness of back insulation (m).

kb = conductivity of back insulation (W/m-K).

hc, b-a = convection heat loss coefficient from back to ambient (W/m2-K).

The conduction resistance of the insulation behind the collector plate governs the heat loss from the collector plate through the back of the collector casing. The heat loss from the back of the plate rarely exceeds 10% of the upward loss. Typical values of the back surface heat loss coefficient are 0.3-0.6 W/m2-K.

In a similar way, the heat transfer coefficient for the heat loss from the collector edges can be obtained from

'V "c,e—a where te = thickness of edge insulation (m).

ke = conductivity of edge insulation (W/m-K).

hc, e-a = convection heat loss coefficient from edge to ambient (W/m2-K).

Typical values of the edge heat loss coefficient are 1.5-2.0 W/m2-K.

Example 3.2

Estimate the top heat loss coefficient of a collector that has the following specifications:

Collector area = 2 m2 (1 X 2 m). Collector slope = 35°. Number of glass covers = 2. Thickness of each glass cover = 4 mm. Thickness of absorbing plate = 0.5 mm.

Space between glass covers = 20 mm.

Space between inner glass cover and absorber = 40 mm.

Thickness of back insulation = 50 mm.

Back insulation thermal conductivity = 0.05 W/m-K.

Ambient air temperature = 15°C = 288 K.

Absorber plate emissivity, ep = 0.10.

Glass emissivity, eg = 0.88.

Solution

To solve this problem, the two glass cover temperatures are guessed and then by iteration are corrected until a satisfactory solution is reached by satisfying the following equations, obtained by combining Eqs. (3.15), (3.17), and (3.22):

(hc,p-g2 + hr,p-g2)(Tp _ Tg2) = (hc,g2-g1 + hr,g2-g1)(Tg2 _ Tg1)

However, to save time in this example, close to correct values are used. Assuming that Tg1 = 23.8°C (296.8 K) and Tg2 = 41.7°C (314.7 K), from Eq. (3.14), h v(Tp + Tg2)(Tp2 + Tg22)

(5.67 X 10~8)(353 + 314.7)(3532 + 314.72) (1/0.10) + (1/0.88) - 1 0.835W/m2-K

Similarly, for the two covers, we have h

= c(Tg2 + Tgl)(T¡2 + Ti) (1/£g2) + (1/£g1) - 1 _ (5.67 X 10-8)(314.7 + 296.8)(314.72 + 296.82)

0.88(5.67 X 10~8)(296.8 + 288)(296.82 + 2882) 4.991W/m2-K

From Table A5.1, in Appendix 5, the following properties of air can be obtained:

For Vi(Tp + Tg2) = V (353 + 314.7) = 333.85 K, v = 19.51 X 10~6 m2/s Pr = 0.701 k = 0.0288 W/m-K

For V(Tg2 + Tg1) = V (314.7 + 296.8) = 305.75 K, v = 17.26 X 10~6 m2/s Pr = 0.707 k = 0.0267 W/m-K

By using these properties, the Rayleigh number, Ra, can be obtained from Eq. (3.12) and by noting that |3' = 1/T.

_ _ gß' Pr _ 3 9.81 X 0.701 X (353 - 314.7) X 0.043 Ra--—dp Tg2)L - -

For h