The effect of a cover

Подпись:Many flat plate solar absorbers have a cover to reduce heat loss via the greenhouse effect; a schematic is shown in Fig. 8.7. This complicates the analysis for two reasons. Firstly, the amount of radiation that reaches and is absorbed by the absorber plate must be determined. This is written || ат || Pd (в), where the symbol || • || represents the result of a thorough calculation and is not merely the product of the absorber plate absorbance and glass cover transmittance. It is normally called the absorbance-transmittance product though. The full calculation results in an infinite series that accounts for light transmission through the cover, then absorption by the absorber plate and then reflection by the absorber plate. The reflected radiation can then be transmitted back through the cover, absorbed or reflected back to the absorber plate. This process occurs multiple times and the result in summing the terms

Подпись: Fig. 8.8 Schematic of a cover used in a flat plate solar energy absorber. The light is refracted according to Snell’s law to an angle 6T after entering at 6 S . Подпись: Table 8.2 Refractive indices for various materials. PMMA is Polymethymethacrylate and is sometimes called Plexiglass or Perspex and PET is Polyethylene terephthalate and is sometimes called Mylar. Material Refractive index Air 1.00 Water 1.33 PMMA 1.49 Glass 1.53 PET 1.64 is that |ат || и ат, where a is the absorbance of the absorber plate and т is the transmittance of the cover; this approximate relation will be used here after reviewing the more detailed calculation.

The second reason that the analysis is complicated is that there are two heat transfer rates which occur in series: those from the absorber plate to the cover (qpc) and from the cover to the air (qca). Each, though, has two heat transfer rates that occur in parallel: radiative (qrad) and convective (qconv). This certainly makes the calculation complicated since an iterative solution must be pursued, yet, not impossible. This is not a challenge for present day equation solvers.

We will first determine how much radiation reaches the absorber plate and consider transmission of the beam component; different procedures must be used for the diffuse and albedo radiation components that are considered later. Reference to Fig. 8.8 shows that the beam component Bter makes an angle 0S with the normal to the glass cover surface. The radiation is refracted at an angle 0T which is given by Snell’s Law to be

nr = sin(^S) Snell’s Law (8.21)

nS sin(dT)

where nS is the refractive index of the surrounding medium (air which has nS и 1.0) and nr is the refractive index of the cover material. Many if not most covers are made of low-iron glass to reduce the absorption of radiation; there are some that use other materials and their refractive indices are listed in Table 8.2.

The beam component is unpolarized radiation (light) and must be considered to have two polarization components. Before this is dis­cussed a brief description of waves is presented. Electromagnetic (light) waves are transverse waves whose electric field oscillates transversely, or perpendicular to, their direction of propagation (x-axis), as seen in Fig. 8.9. The amplitude of the wave is given by the magnitude of the electric vector (E). Unpolarized light has the direction of the electric vector in any arbitrary direction perpendicular to the x-axis (i. e. in the y-z plane) for any given wave. Polarized light has the all the waves’ electric vectors oscillating in only one direction, say the y-direction. A polarizer is used to filter out all other directions. There are other types of waves, sound waves for example are longitudinal waves, as shown in Fig. 8.9. A longitudinal wave has its oscillation in the direction of propagation and for sound waves the amplitude is given by the pressure. High pressure at the apex of the wave is called condensation, while low pressure is called rarefaction and is at the wave nadir. Because of this sound waves can­not be polarized. You may have noted that sound waves are longitudinal waves upon viewing a speaker as it oscillates to produce regions of high and low pressure, thereby making sound.

Now consider the beam component of solar radiation and the two polarization directions required to analyze light propagation through the cover. One polarization direction is perpendicular to the plane made by the incident beam and the surface normal while the other will be

Подпись: □ □ P = P° Подпись:Подпись: (8.22a) (8.22b) Подпись: waves.image520Подпись:Подпись: т □ = TdПодпись: 1 - pg 1 - pg Td Подпись: aparallel to this plane. The symbols 1 and || will be used as superscripts to the variables below for the perpendicular and parallel components, respectively. If the variable has the superscript □, such as pg, the same equation can be used for either the 1 or || component and the □’s within the equation are changed to either 1’s or ||’s accordingly.

When the radiation strikes the glass cover an amount is initially re­flected from the surface and dictated by p°, see Fig. 8.10, that can be determined by

I = sin(6>T – Os)2

^0 sin(0T + Os )2

and

II tan(0T – Os)2 tan(0T +Os)2

Determining the total amount of radiation reflected and transmitted through the cover is the result of considering multiple reflections from the bottom and top of the cover as shown in Fig. 8.10. The result of the analysis is an infinite series which yields for the reflectance p, transmittance т and absorbance a

Tj [1 – pg]2 ‘

1 – [pg Td]2.

[1 – pg]2 1 – [pg Td]2 1 – pg – T□ = [1 – Td] where d is the distance the radiation must propagate through the cover which equals Lc/ cos(0T), where Lc is the cover thickness. Also, the quantity Td, which is the single pass transmittance, is given by

Td = exp (-a x d) (8.24)

where a is the absorption coefficient; values for various materials are given in Table 8.3. One finds the total amount of light reflected or transmitted by taking the arithmetic average of the perpendicular and parallel components via

p = 1 [p1 + p1 ] (8.25a)

T = 1 [t 1 + T11 ] (8.25b)

Подпись: Fig. 8.10 Multiple reflection occurs when radiation propagates through a transparent material. Подпись:In most cases the cover will absorb very little light. For example, using the absorption coefficient for low-iron glass given in Table 8.3 and the cover thickness given in Table 8.1 the amount of radiation transmitted for a single pass through the cover is given by eqn (8.24) to equal 0.9987, assuming the beam component is normal to the cover. Since Td « 1 one finds ag « 0 in eqn (8.23c) allowing eqn (8.23b) to be simplified to

where the definition of t° is given by comparing the two equations. The quantity t^ is frequently introduced to emphasize that this is the part of the radiation that is available after the initial reflection, while Td is that initially transmitted through the glass. Regardless, this approx­imate relation is fairly accurate and can be used with little worry for applications involving covers for solar applications.

Using the above, either the exact or approximate relation, one can find the amount of beam radiation that is transmitted through the cover ( tb ) from

tb = 1 [t 1 + t 11 ] (8.27)

Detailed discussion of the above and that which follows, concerning transmission and absorption of radiation in a solar energy collector, can be found in the textbook of Duffie and Beckman referenced at the end of this chapter.

Example 8.2

Подпись:

image528 image529
Подпись: (8.26)

Determine the amount of light transmitted through the cover for the stan­dard conditions listed in Table 8.1. Compare the exact and approximate techniques.

The first quantity to determine is 9T using Snell’s law given by eqn

(8.21) which is found to be 12.94°. Now pi and p0 need to be calculated with eqns (8.22a) and (8.22b), respectively, and are 0.05148 and 0.03684. Finally, the single pass transmittance Td can be calculated from eqn (8.24) and is 0.9870.

Now, the perpendicular and parallel components of the transmittance (eqns (8.23b) and (8.26)), as well as the total transmittance (eqn (8.27)), can be calculated and are shown in Table 8.4. As can be seen there is very little difference in the values and one must go to four significant figures to see it. Thus, the approximate technique is more than adequate to determine the transmittance through the glass cover.

The transmittance for the diffuse and albedo components, td and ta, respectively, are quite different since they are scattered radiation and come from all directions (isotropic) unlike the beam component. Simu­lations and modeling have been performed where it is found one should use an apparent angle for the total amount of radiation transmitted through the cover which can be written

td и exp I———– / c I where dD = 60° (8.28a)

cos(Q d )

and

TA * exp I——- ^ I where eA = 90° -0.5778в+0.002693в2 (8.28b)

cos(в a)

Now it is possible to find Цат|| and hence the amount of solar en­ergy transferred to the device with one more approximation. Detailed calculations are required to find | ат | B, | ат | D and | ат | a for each com­ponent of the radiation where we now have to consider reflection from the absorber surface to the cover plate and back. Instead, it is found that each is given approximately by the product of the transmittance through the cover multiplied by the absorbance of the absorber plate as discussed above. This may be expected since the materials used in the device would be optimized. For example, the glass cover material would be made to transmit as much radiation as possible and the absorber material would absorb almost all of the radiation. Certainly, this would not apply in general, yet, using this approximation one can write

|ат | Pd (в) * атв B(@) + атв D(@) + атА А(в) (8.29)

Подпись: Table 8.5 Values of the transmittance for the direct beam, diffuse and albedo components through the glass cover under the standard condi- tions given in Table 8.1. variable value TB 0.9035 TD 0.9747 TA 0.9666 Example 8.3

Compare the transmittance for the beam, diffuse and albedo components for the standard conditions listed in Table 8.1.

The beam component has already been calculated in Example 8.2; now we calculate the transmittance for the diffuse and albedo compo­nents with eqns (8.28a) and (8.28b) for the standard conditions given in Table 8.1. The results of the calculation are given in Table 8.5 and, remarkably, the direct beam’s transmittance is the lowest! However, the beam component is usually the largest component of the total insolation so is most likely the largest contributor.

We are now in a position to model the flat plate solar energy collector since the amount of radiation that can reach the absorber plate and heat the flowing water can be determined. However, this is the amount of energy coming into the flat plate solar collector and there will be heat loss terms that must be considered.

Reference to Fig. 8.7 shows that there are two rates of heat transfer through the top of the system: that from the plate to the cover, qpc, and that from the cover to the surrounding atmosphere, qca. These two rates must be equal, at steady state operation, since they operate in series, allowing us to write

Top = qpc = qca (8.30)

where qtop is the rate of heat transfer from the top of the device. This is an important constraint to consider when designing the system and
allows one to determine the heat flow from the top. There are also two heat transfer rates that operate in parallel within qpc and qca, which are convective and radiative heat transfer, given by

Подпись:h [T T ] . ^s[Tpp – T4]

qpc = hpc[Tp Tc] . і і

—- +—– 1

Подпись: and image534 Подпись: (8.31b)

ep ec

Подпись:where hpc is the convective heat transfer coefficient for the air between the absorber plate and cover, hca, the convective heat transfer coefficient between the cover and air, Tp, the absorber plate temperature, Tc, the cover temperature, Ta, the air or ambient temperature, <ts, the Stefan – Boltzmann constant, ep, the (low energy or long wavelength) emissivity of the absorber plate and ec, the emissivity of the cover. The curious form for radiative heat transfer in eqn (8.31a) comes from an analy­sis involving radiative energy exchange between the absorber plate and cover whose derivation can be found in the references at the end of this chapter.8

The same assumption will be made for heat transfer from the bottom of the system as was made in the previous section: the major resistance will be heat transfer through the insulation. Again, since heat transfer for processes in series is limited to the lowest value, here we assume that heat flow through the insulation rather than from the device to the air is limiting, so, this will represent the heat flow from the bottom (qbot) written as

qbot = у1 [Tp – Ta] (8.32)

Li

where it is assumed that the top of the insulation is at the plate tem­perature and the outside of the device is at the ambient temperature, both reasonable assumptions, as discussed in the previous section.

The First Law of Thermodynamics or FLOT can now be written

mCpW [AT ] = [jarPD (в) – qtop – qbot]AD operating line (8.33)

and is the operating line for the flat plate solar energy collector. The temperature difference AT is Tout – Tin.

Further details are required to determine the operating line. The heat transfer coefficient for the air between the absorber plate and cover (hpc) will be discussed first and is dictated by the Rayleigh number that is used to describe natural convection under the influence of a temperature difference. If air is confined between two flat plates at different temper­atures, that are placed at an angle from the horizontal, then the air will move up along the hot plate, turn around and fall back along the cool plate. The air circulating motion increases the heat transfer rate and is influenced by the temperature difference, the gap between the two plates
and the material’s physical properties. The Rayleigh number Ra is used to gauge the effect of recirculation and written as

2 C

Ra = АТргЛс = CRaATpcL3pc (8.34)

where g, is the gravitational constant, x, the thermal expansion coeffi­cient, p, the density, Cp, the heat capacity, k, the thermal conductivity, p, the viscosity, ATpc, the temperature difference between the two plates (in our case this will be Tp – Tc), Lpc, the gap between the absorber plate and cover and CRa, the group of physical constants that is given in Appendix B. The smaller the Rayleigh number the less heat transfer occurs.9 Note that the air properties are determined at Tm e 1 [Tp + Tc], which is typical of heat transfer calculations.

Подпись:

image538

The Nusselt number used to find hpc can be determined with this correlation,

(8.35)

with the superscript + meaning that if the quantity in the brackets is negative then the entire term is zero. As before, the heat transfer coefficient between the cover and air hca can be determined with eqn (8.6) or (8.7).

The method of solution to find the operating point is an iterative one. This is because the temperatures within the device, such as the absorber plate temperature, Tp, and cover temperature, Tc, are not know a priori, and this influences the physical properties of air. A suggested method is:

Operating line

(1) Know: All the parameters in Table 8.1

(2) Determine: ||ar||PD(в) using eqn (8.29)

(3) Assume: Tp value

(4) Iterate: Change Tc until qpc = qca (eqns (8.31a) and (8.31b)) which will involve calculating the heat transfer coefficients that will depend on temperature. Use the equations in Appendix B to find the effect of temperature on the properties of air, which should be evaluated at the mean temperature at each position (i. e. Tm = 1 [Tp + Tc] for heat transfer from the plate to cover and Tm = 2 [Tc + Ta ] for heat transfer from the cover to air)

(5) Calculate: qbot from eqn (8.32)

(6) Calculate: AT from eqn (8.33)

(7) Repeat: Go to item 3 and assume another Tp value to calculate the operating line for various values of Tp

The following example demonstrates how to perform the calculation.

Example 8.4

Подпись:Find the operating point of a flat plate solar collector using the standard conditions given in Table 8.1. Assume there is a cover and compare the result to that found with no cover in Example 8.1.

The design line is the same as that found in Example 8.1 and it is reproduced in Fig. 8.11. The operating line is calculated using the stan­dard conditions given in Table 8.1 and the transmittance values found in Example 8.3. However, the standard conditions have only a direct beam component and so only tb is required, which was initially calculated in Example 8.2. This allows one to determine \ат||PD(fl) from eqn (8.29) quite easily using tb and the absorber plate absorbance.

Rather than stating the various equations that have to be considered in the above text, outlined on the previous page, the results of the cal­culation are given in Table 8.6 for the reader to compare to their own calculation. The operating line for the standard conditions, using eqn (8.6) for the heat transfer coefficient hca, is plotted in Fig. 8.11.

Some comments should be made about the results presented in Table

8.6. The very top row of data has both Tp and Tc arbitrarily set to Ta, the ambient air temperature, and represents the condition where the (almost) maximum value of the useful heat qu can be obtained. The next four rows of results, between the double horizontal lines, allow Tc to vary for a given Tp so that qpc = qca, the result that follows our procedure discussed above. Since Tc < Ta there will be convective heat transfer from the air to the cover while there is radiative heat transfer from the cover to the sky, which has an effective temperature lower than ambient. This is why Tc can be smaller than Ta. The net result is that qca e 0 for the data in row 2 and is the lowest temperature that Tc can have. Thus, qpc is also identically zero since the condition of qpc = qca must be followed and so Tp e Tc. Amazingly, since Tp < Ta there is heat transfer from the surroundings through the bottom of the device and qbot < 0! The end result is that qu is slightly higher than that in row 1!

In reality, achieving the conditions given between the double horizon­tal lines in Table 8.6 will be difficult and a very large mass flow rate of water would have to be used, as will be discussed in the next section. Furthermore, whether these conditions can be achieved at all are ques­tionable. The Earth, and the house upon which the device may sit, will be radiating heat too and since the rates of heat transfer are so small under these conditions these perturbations will most likely overcome the very isolated conditions upon which the calculation is made. Regardless, one can start calculations at the very smallest value that Tc can have, Tc, min, given by

hca [Tc, min – Ta] = aSec{T4,min ~ ] (8.36)

Of course, this value for Tc necessitates that qtop = 0 and so one finds Tp = Tc, min which can be the temperature where the calculation be-

Table 8.6 Temperatures and heat flows for the flat plate solar collector with a cover using the standard conditions given in Table 8.1.

Tp (°C)

AT (°C)

Tc (°C)

qtop (W/m2)

qbot (W/m2)

4u (W/m2)

10.0

18.2

10.0

0.0

0.0

764.2

2.516

18.3

2.516

0.0

-3.9

768.0

10.0

17.9

3.8

14.4

0.0

749.8

20.0

17.2

6.1

38.5

5.2

720.4

30.0

16.4

8.5

65.5

10.4

688.3

35.8

15.9

10.0

82.1

13.4

668.7

38.2

15.8

10.6

89.1

14.7

662.8

40.0

15.6

11.1

94.5

15.6

654.1

50.0

14.7

13.8

125.2

20.8

618.1

60.0

13.8

16.7

157.7

26.0

580.4

70.0

12.9

19.6

191.6

31.2

541.2

80.0

11.9

22.7

227.1

36.4

500.7

90.0

10.9

25.8

263.9

41.6

458.6

gins. This will also make Tp < Ta adding heat to the system through the bottom!

The above discussion seems pedantic, however, when the reader be­gins designing these devices using basic heat transfer calculations and then reads the literature, it becomes very confusing. Doing the above detailed calculations and only considering the information for Tc > Ta as relevant to the design is a good way of reconciling differences in design procedures.

Now, the data beginning in row 6, just below the second double line, represents that where Tc > Ta. The first of these rows has Tc = Ta and one can see at this condition Tp = 35.8 °C, which is quite high. So, if the cover temperature is forced to equal the air then the absorber plate is actually quite warm. The next row, which is shaded, is the operating point and the following rows are the result of further calculation.

The intersection of the design line and operating line is the operating point, which is found to be at Tp = 38.2 °C and AT = 15.8 °C. The operating point is close to that for the system with no cover which was Tp = 36.9 °C and AT = 14.9 °C.

The cover does not seem to improve operation much. However, the amount of useful heat Qu calculated using eqn (8.2) does increase with the cover from 3759 W with no cover to 3977 W with a cover, a 6% increase. Thus, 6% more energy will be delivered to the water during any given time period and is substantial.

The efficiency of the flat plate solar collector can be found from

Q u

Pd (в )Ad

 

m Cpw AT
Bter cos(6(s)Ad

 

(8.37)

 

With the efficiency defined this way, the device that has a cover is more efficient at 70.5% compared to 66.7% without a cover. Yet, the cover allows less radiation to reach the absorber plate since jar|| = 0.8132, compared to the system operating with no cover where one simply has absorbance at a = 0.9. However, the absorbed radiation is treated more efficiently when a cover is present since the heat losses are reduced. The heat loss through the top of the device having no cover at its operat­ing point is 204 W/m2 compared to 82.9 W/m2 when a cover is used. One can now appreciate how the cover reduces heat transfer to promote the greenhouse effect. One should also appreciate that using a cover that minimizes reflective losses and has low emissivity as well as low absorbance is a must to ensure efficient operation.

Finally, to conclude the discussion, we consider the efficiency of these devices, which is quite high and of order 70%. The challenge though is that the quality of the energy it produces, hot water, is not that great. If one could produce high pressure steam or electricity at this efficiency a major technological breakthrough would be had. Nevertheless, the ability to make hot water for consumer and industrial use cannot be denied and these devices should be seriously considered for use.

Updated: August 19, 2015 — 1:44 am