## Cggb

10,904

Therefore, from Eq. (3.11), we have the following.

1708

5830

0.0288

0.04

1708

132,648 X cos(35)

5830

1708

5830

0.0267

0.02

1708

5830

0.333

The convection heat transfer coefficient from glass to ambient is the wind loss coefficient given by Eq. (3.28). In this equation, the characteristic length is the length of the collector, equal to 2 m. Therefore, hcgi-a = hw = 8.6(2.5)0'6/20'4 = 11.294W/m2-K

To check whether the assumed values of Tg1 and Tg2 are correct, the heat transfer coefficients are substituted into Eqs. (3.15), (3.17), and (3.22):

Qt/Ac = (hc,p-2 + hr,p-g2)(Tp - Tg2) = (2.918 + 0.835)(353 - 314.7) = 143.7 W/m2

Qt A = (hc,g2-g1 + Kgg2-g1)(Tg2 - Tg1) = (2.852 + 5.098)(314.7 - 296.8) = 142.3W/m2

QtA = (hc,g1-a + K,g1-a)(Tg1 - Ta) = (11.294 + 4.991)(296.8 - 288) = 143.3 W/m2

Since these three answers are not exactly equal, further trials should be made by assuming different values for Tg1 and Tg2. This is a laborious process which, however, can be made easier by the use of a computer and artificial intelligence techniques, such as a genetic algorithm (see Chapter 11). Following these techniques, the values that solve the problem are Tg1 = 296.80 K and Tg2 = 314.81 K. These two values give Q/Ac = 143.3 W/m2 for all cases. If we assume that the values Tg1 = 296.8 K and Tg2 = 314.7 K are correct (remember, they were chosen to be almost correct from the beginning), Ut can be calculated from

hc,p-g2 + hr,p-g2 hc,g2-g1 + hr,g2-g1 hc,g1-a + hr,g1-a

Example 3.3

Repeat Example 3.2 using the empirical Eq. (3.25) and compare the results. Solution

First, the constant parameters are estimated. The value of hw is already estimated in Example 3.2 and is equal to 11.294 W/m2-K. From Eq. (3.26), f = (1 - 0.04hw + 0.0005hW)(1 + 0.091Ng)

f = (1 - 0.04 X 11.294 + 0.0005 X 11.2942)(1 + 0.091 X 2) = 0.723 From Eq. (3.27),

C = 365.9(1 - 0.00883(3 + 0.0001298(2) C = 365.9(1 - 0.00883 X 35 + 0.0001298 X 352) = 311

0.33

11.294

The difference between this value and the one obtained in Example 3.2 is only 4.6%, but the latter was obtained with much less effort.

3.3.3 Temperature Distribution Between the Tubes and Collector Efficiency Factor

Under steady-state conditions, the rate of useful heat delivered by a solar collector is equal to the rate of energy absorbed by the heat transfer fluid minus the direct or indirect heat losses from the surface to the surroundings (see Figure 3.26). As shown in Figure 3.26, the absorbed solar radiation is equal to Gt(ra), which is similar to Eq. (3.6). The thermal energy lost from the collector to the surroundings by conduction, convection, and infrared radiation is represented by the product of the overall heat loss coefficient, UL, times the difference between the plate temperature, Tp, and the ambient temperature, Ta. Therefore, in a steady state, the rate of useful energy collected from a collector of area Ac can be obtained from

Qu = Ac[Gt(Ta) - UL(Tp - Ta)] = mCp[T0 - T ] (3.31)

Equation (3.31) can also be used to give the amount of useful energy delivered in joules (not rate in watts), if the irradiance Gt (W/m2) is replaced with irradiation It (J/m2) and we multiply UL, which is given in watts per square meter in degrees Centigrade (W/m2-°C), by 3600 to convert to joules per square meter in degrees Centigrade (J/m2-°C) for estimations with step of 1 h.

To model the collector shown in Figure 3.26, a number of assumptions, which simplify the problem, need to be made. These assumptions are not against the basic physical principles and are as follows:

1. The collector is in a steady state.

2. The collector is of the header and riser type fixed on a sheet with parallel tubes.

3. The headers cover only a small area of the collector and can be neglected.

4. Heaters provide uniform flow to the riser tubes.

5. Flow through the back insulation is one dimensional.

6. The sky is considered as a blackbody for the long-wavelength radiation at an equivalent sky temperature. Since the sky temperature does not affect the results much, this is considered equal to the ambient temperature.

7. Temperature gradients around tubes are neglected.

Gt Cover reflection

Gt Cover reflection

temperature Tp

FIGURE 3.26 Radiation input and heat loss from a flat-plate collector.

temperature Tp

FIGURE 3.26 Radiation input and heat loss from a flat-plate collector.

8. Properties of materials are independent of temperature.

9. No solar energy is absorbed by the cover.

10. Heat flow through the cover is one dimensional.

11. Temperature drop through the cover is negligible.

12. Covers are opaque to infrared radiation.

13. Same ambient temperature exists at the front and back of the collector.

14. Dust effects on the cover are negligible.

### 15. There is no shading of the absorber plate.

The collector efficiency factor can be calculated by considering the temperature distribution between two pipes of the collector absorber and assuming that the temperature gradient in the flow direction is negligible (Duffie and Beckman, 1991). This analysis can be performed by considering the sheet-tube configuration shown in Figure 3.27a, where the distance between the tubes is W, the tube diameter is D, and the sheet thickness is 6. Since the sheet metal is usually made from copper or aluminum, which are good conductors of heat, the temperature gradient through the sheet is negligible; therefore, the region between the center line separating the tubes and the tube base can be considered as a classical fin problem.

r |
1 |

1 |
ß |

1 |
FIGURE 3.27 Flat-plate sheet and tube configuration. (a) schematic diagram. (b) Energy balance for the fin element. (c) Energy balance for the tube element. The fin, shown in Figure 3.27b, is of length L = (W — D)/2. An elemental region of width, Ax, and unit length in the flow direction are shown in Figure 3.27c. The solar energy absorbed by this small element is SAx and the heat loss from the element is ULAx(Tx — Ta), where Tx is the local plate temperature. Therefore, an energy balance on this element gives x+Ax where S is the absorbed solar energy. Dividing through with Ax and finding the limit as Ax approaches 0 gives d 2T = UL dx2 = kb The two boundary conditions necessary to solve this second-order differential equation are dT For convenience, the following two variables are defined: Therefore, Eq. (3.33) becomes dx which has the boundary conditions d V dx and d ^ o m2 tf = O (3.36) Equation (3.36) is a second-order homogeneous linear differential equation whose general solution is tf = C1 emx + C2 e_mx = C1 sinh(mx) + C2 cosh(mx) (3.37) The first boundary yields Ci = 0, and the second boundary condition yields s f = Tb - Ta - — = C2 cosh(mL) cosh(mL) With C1 and C2 known, Eq. (3.37) becomes This equation gives the temperature distribution in the x direction at any given y. The energy conducted to the region of the tube per unit length in the flow direction can be found by evaluating the Fourier's law at the fin base (Kalogirou, 2004): However, kbm/UL is just 1/m. Equation (3.39) accounts for the energy collected on only one side of the tube; for both sides, the energy collection is or with the help of fin efficiency, qfin = (W - D)F[S - Ul(Tb - Ta)] (3.41) where factor F in Eq. (3.41) is the standard fin efficiency for straight fins with a rectangular profile, obtained from The useful gain of the collector also includes the energy collected above the tube region. This is given by qube = D[S - Ul (Tb - Ta)] (3.43) Accordingly, the useful energy gain per unit length in the direction of the fluid flow is Qu = qfin + qube = [(W - D)F + D][S - UL(Tb - Ta)] (3.44) This energy ultimately must be transferred to the fluid, which can be expressed in terms of two resistances as hfinDi cb where hfi = heat transfer coefficient between the fluid and the tube wall. In Eq. (3.45), Cb is the bond conductance, which can be estimated from knowledge of the bond thermal conductivity, kb, the average bond thickness, 1, and the bond width, b. The bond conductance on a per unit length basis is given by (Kalogirou, 2004) The bond conductance can be very important in accurately describing the collector performance. Generally it is necessary to have good metal-to-metal contact so that the bond conductance is greater that 30 W/m-K, and preferably the tube should be welded to the fin. Solving Eq. (3.45) for Tb, substituting it into Eq. (3.44), and solving the resultant equation for the useful gain, we get qu = WF'[S - UL (Tf - Ta)] (3.47) where F' is the collector efficiency factor, given by 1 11 A physical interpretation of F' is that it represents the ratio of the actual useful energy gain to the useful energy gain that would result if the collector absorbing surface had been at the local fluid temperature. It should be noted that the denominator of Eq. (3.48) is the heat transfer resistance from the fluid to the ambient air. This resistance can be represented as 1/U0. Therefore, another interpretation of F' is The collector efficiency factor is essentially a constant factor for any collector design and fluid flow rate. The ratio of UL to Cb, the ratio of UL to hfi, and the fin efficiency, F, are the only variables appearing in Eq. (3.48) that may be functions of temperature. For most collector designs, F is the most important of these variables in determining F'. The factor F' is a function of UL and hfi, each of which has some temperature dependence, but it is not a strong function of temperature. Additionally, the collector efficiency factor decreases with increased tube center-to-center distances and increases with increase in both material thicknesses and thermal conductivity. Increasing the overall loss coefficient decreases F', while increasing the fluid-tube heat transfer coefficient increases F'. ## Example 3.4For a collector having the following characteristics and ignoring the bond resistance, calculate the fin efficiency and the collector efficiency factor: Overall loss coefficient = 6.9 W/m2-°C. Tube spacing = 120 mm. Tube outside diameter = 15 mm. Tube inside diameter = 13.5 mm. Plate thickness = 0.4 mm. Plate material = copper. Heat transfer coefficient inside the tubes = 320 W/m2-°C. Solution From Appendix 5, Table A5.3, for copper, k = 385 W/m-°C. From Eq. (3.34), m 385 X 0.0004 Finally, from Eq. (3.48) and ignoring bond conductance, 0.12 6.9[0.015 + (0.12 - 0.015) X 0.961 n X 0.0135 X 320 0.912 3.3.4 Heat Removal Factor, Flow Factor, and Thermal Efficiency Consider an infinitesimal length 6y of the tube as shown in Figure 3.28. The useful energy delivered to the fluid is qu 8 y. Under steady-state conditions, an energy balance for n tubes gives Dividing through by 8y, finding the limit as approaches 0, and substituting Eq. (3.47) results in the following differential equation: Separating variables gives dT Assuming variables F', UL, and cp to be constants and performing the integrations gives ln S/UL S/UT The quantity nWL in Eq. (3.53) is the collector area Ac. Therefore, S/UL S/UT It is usually desirable to express the collector total useful energy gain in terms of the fluid inlet temperature. To do this the collector heat removal factor Temperature T => dT Temperature T+ ddySy FIGURE 3.28 Energy flow through an element of riser tube. p p needs to be used. Heat removal factor represents the ratio of the actual useful energy gain that would result if the collector-absorbing surface had been at the local fluid temperature. Expressed symbolically: Actual output Output for plate temperature = Fluid inlet temperature |

## Solar Panel Basics

Global warming is a huge problem which will significantly affect every country in the world. Many people all over the world are trying to do whatever they can to help combat the effects of global warming. One of the ways that people can fight global warming is to reduce their dependence on non-renewable energy sources like oil and petroleum based products.

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