Top-Rated Free Essay
Preview

FlatPlate

Good Essays
1662 Words
Grammar
Grammar
Plagiarism
Plagiarism
Writing
Writing
Score
Score
FlatPlate
Flat-Plate Solar Collectors

The Absorption of Solar Radiation
A flat-plate solar collector usually has a non-selective or a selective black plate with one or two glass covers a few centimeters above the black plate, and a well insulated back. The length of the plate is typically about 2m. Edge effects are usually small.
The transmittance () of a glass cover for solar radiation depends on the angle of incidence . Typical values for clear glass are given in Table 1.
Table 1. Transmittance of a Glass Cover.
:

60°
70°
80°
90°
():
0.9
0.8
0.65
0.35
0
The absorptance () of the black plate for solar radiation also depends on the angle of incidence . Table 2 shows typical values for () and the product ().().
Table 2. Absorptance of a Black Plate.
:

60°
70°
80°
90°
():
0.92
0.85
0.75
0.60
0
().():
0.83
0.68
0.49
0.21
0
The solar irradiance Iin incident on the cover glass is given by
Iin = Ib cos + Id, where Ib is the beam solar irradiance, is the angle of incidence, and Id is the diffuse irradiance.
If there is one glass cover the solar irradiance on the black plate is
().Ib cos + mId, where m is the mean value of (). The solar radiation flux qabs absorbed by the black plate is given by qabs = ().() Ib cos + (.)mId, where (.)m is the mean value of ().(). The mean value of ().() can be found by means of integrals over the hemispherical sky as follows:
(.)m = [0/2 ().().sin .cos .d()] / [0/2 sin .cos .d()].
For one glass cover the result is approximately (.)m = 0.70.
Heat Losses
The glass cover behaves nearly as a black body for long-wave radiation. We can assume that the emittance c of the glass cover is 0.95.
The emittance b of the black plate for long-wave radiation depends on whether the surface is non-selective or selective. Typically we have b = 0.92 for a non-selective surface, b = 0.10 for a selective surface.
We shall consider a collector with one glass cover. Let
Ta = ambient temperature,
Tb = black plate temperature,
Tc = glass cover temperature, where absolute temperatures must be used for radiation calculations.
Heat is lost by conduction through the back insulation. It can be reduced to a low rate by inexpensive insulation materials. Typically the back loss might be given by the formula hba(Tb Ta), where the heat transfer coefficient is hba = 0.3W/m2K.
Heat is lost from the black plate to the glass cover by convection and radiation. Experience has shown that, for free convection the Nusselt number Nu in air spaces between parallel plates with Grashof numbers Gr in the range 104 to 107, we have
Nu = 0.152 Gr0.281 for horizontal plates,
Nu = 0.093 Gr0.310 for plates tilted at an angle 45°.
Here
Gr = g.(Tb Tc)L3/2, where we assume as a typical example for air: g = acceleration of gravity == 9.8m/s2, = coefficient of thermal expansion = 1/T, T = 60°C = 333K,
Tb Tc = 80°C 40°C = 40K,
L = spacing = 50mm = 0.05m, = kinematic viscosity = 0.194×104m2/s.
This gives Gr = 3.91×105, which is within the range 104 to 107 mentioned above.
Assume a tilt angle 15°. Then we estimate, by interpolation, Nu = 0.132 Gr0.291 = 5.572.
Also since
Nu = hL/k, where h = heat transfer coefficient,
L = 0.05m, k = thermal conductivity of air = 0.02750W/mK, we have h = 3.06W/m2K. This calculation shows that in general the heat transfer coefficient h is a function of Tb and Tc.
For the heat loss by radiation between the black plate and the glass cover we have the expression
[.(Tb4 Tc4)] / [b1 + c1 1] = bc.(Tb4 Tc4), where is the Stefan-Boltzman constant 56.7×109W/m2K4.
Thus the total heat loss qba from the black plate can be written qba = hba(Tb Ta) + hbc(Tb Tc) + bc..(Tb4 Tc4), .....(1) where hbc depends on Tb and Tc, and on the angle of tilt of the collector.
The heat loss from the glass cover to the surroundings must be the same, in the steady state, as the heat loss from the black plate to the glass cover. We have for the heat loss from the glass cover qca = hca.(Tc Ta) + c..Tc4 c.L, .....(2) where the convection heat transfer coefficient hca is difficult to estimate because it is partly due to free convection and partly due to forced convection by wind blowing over the collector. The following formula is recommended: hca = 2.8 + 3.0V W/m2K, where V is the wind speed in meters per second. The second term c..Tc4 in equation (2) is the long-wave radiation from the glass cover, and the third term c.L is the long-wave radiation absorbed by the glass cover from the sky.
Collector Efficiency in the Steady State
We must calculate the temperatures Tb and Tc from the radiation fluxes and the ambient temperature Ta.
First assume a value for qca and solve equation (2) for Tc. Now equation (1) shows that qca = hbc(Tb Tc) + bc(Tb4 Tc4), .....(3) from the heat balance of the glass cover in the steady state. This equation is solved for Tb. Finally qba is found from equation (1).
Repeat the calculation for different values of qca to obtain qba as a function of Tb. This is easy to do numerically with a programmable calculator.
Let qout be the heating power output of the collector per unit area. It can be varied within the feasible limits by controlling the operating conditions of the collector. The heat balance for the black plate gives qabs = qba + qout, where qabs depends on Ib, , and Id; and the heat loss rate qba is known as a function of Tb. The overall efficiency = qout/qin is therefore known as a function of the radiation fluxes and the black plate temperature Tb.
Simplifications in the Theory
In practical flat-plate solar collectors the temperature of the flat plate is not uniform. In tube-in-sheet designs the temperature of the sheet between the tubes is higher than the temperature of the tubes. Furthermore, the temperature of the tubes is higher at the outlet ends than at the inlet ends. The use of a single black plate temperature Tb is therefore a simplification.
Another simplification is to put qabs = .qin, where is the optical efficiency, which is equal to the mean transmittance-absorptance product (.)m.
The heat balance equation is then written qout = .qin U(Tb Ta), where U is the overall heat loss coefficient between the black plate and the surroundings. The efficiency is now given by = qout/qin = U(Tb Ta)/qin.
Thus is the efficiency when Tb = Ta. Often U is assumed to be constant, and the stagnation temperature Tmax obtained when no heat is extracted from the collector is estimated to be
Tmax = Ta + .qin/U.
In reality, because of the non-linear relations between the heat losses and the temperature differences, U varies with Tb Ta. With good approximation we can write qout = .qin U1(Tb Ta) U2(Tb Ta)2 and = U1(Tb Ta)/qin U2(Tb Ta)2/qin, where U2 is small compared with U1. The graph of versus (Tb Ta) is slightly curved. (See Fig. 1.)

Fig. 1. The efficiency of a flat plate collector.
If the solar radiation falling on the collector changes rapidly, due to the passage of clouds, the collector will take time to change its temperature because of its heat capacity. This may be important and require separate analysis. The theory of non-steady state processes in solar collectors is very complicated, and is ignored in steady state calculations.
Practical Collector Performance Parameters
In practice it is convenient to use the fluid temperature Tf instead of the black plate temperature Tb. The total heat extraction rate Qout from a collector of area A is then written
Qout = AF'[.qin U(Tf Ta)], .....(1) where F' is called the collector efficiency factor. In good designs F' is nearly unity. F'. is the effective transmittance-absorptance product; and F'U is the heat loss coefficient between the fluid and its surroundings.
We may define the thermal resistance R between the black plate and the fluid by the equation qout = (Tb Tf)/R, and derive equation (1) from the heat balance equation between the black plate and the surroundings. It is then found that
F' = 1/(1 + RU).
Consider a fluid with mass flow rate m and specific heat capacity c flowing a total distance L through a collector of area A. The heating of the fluid with respect to distance x through the collector is given by mc(dT(x)/dx) = (A.F'/L)[.qin U(T(x) Ta)].
Assume U is constant. Then this is a first order non-homogeneous linear differential equation. It can be solved to give the basic equation relating the inlet and outlet fluid temperatures Tin and Tout as follows:
[.qin U(Tin Ta)]exp(AF'U/mc) = [.qin U(Tout Tin)]. .....(2)
Often the performance of a collector is written in terms of the fluid inlet temperature Tin and a heat removal factor FR, as follows
Qout =AFR[.qin U(Tin Ta)]. .....(3)
Writing Qout = mc(Tout Tin), and eliminating Tout with the help of (2), shows that
FR = (mc/AU)[1 exp(AF'U/mc)]. .....(4)
For small flow FR is small and the fluid temperature approaches the stagnation temperature Tmax. For large flow FR is large, but the rise in fluid temperature is small.
The Collector as a Heat Exchanger

Fig. 2. The flat-plate collector as a heat exchanger.
A solar collector may be regarded as a conventional heat exchanger transferring heat from solar radiation at a constant temperature Tmax to the collector fluid (see Fig. 2). First we define the log mean temperature difference LMTD by the equation
LMTD = (Tout Tin)/log[(Tmax Tin)/(Tmax Tout)]. .....(5)
Then, putting Tmax = Ta + .qin/U, and Qout = mc(Tout Tin), we obtain from equation (2):
Qout = AF'U(LMTD). .....(6)
Another method uses the heat exchanger effectiveness
E = Qout/Qmax, .....(7) where Qmax is the heat transfer rate when Tout = Tmax.
Putting
Qout = mc(Tout Tin),
Qmax = mc(Tmax Tin),
Tmax = Ta + .qin/U, we obtain from equation (2):
E = 1 exp(AF'U/mc).
Standard theory for this type of heat exchanger gives
E = 1 exp(N), .....(8) where N is the number of heat transfer units. Therefore in the collector
N = AF'U/mc. .....(9)
Note that (4), (8), and (9) give the following relation between the different measures of performance:
FRN = F'E.
By R. H. B. Exell, 2000. King Mongkut's University of Technology Thonburi.

You May Also Find These Documents Helpful

  • Satisfactory Essays

    Home Work 7

    • 826 Words
    • 7 Pages

    A solar cooker, really a concave mirror pointed at the Sun, focuses the Sun's rays 19.3…

    • 826 Words
    • 7 Pages
    Satisfactory Essays
  • Powerful Essays

    B. Indirect sunlight (sunlight that hits the plate at an angle of less than 90°)…

    • 2003 Words
    • 9 Pages
    Powerful Essays
  • Good Essays

    Black body radiation is emitted as a broad spectrum, where the shape of the curve is modeled based off Planck’s law of black body radiation. Stefan’s Law shows the relationship between the absolute temperature (T) and the radiancy (RT¬) with the Stefan-Boltzmann constant being σ = 5.67 × 10−8 Wm−2K−4…

    • 530 Words
    • 3 Pages
    Good Essays
  • Better Essays

    The greater the light intensity (shorter distance between light source and table surface) The higher the rate of photosynthesis in leaf discs.…

    • 3533 Words
    • 15 Pages
    Better Essays
  • Best Essays

    Fdi in Japan

    • 3733 Words
    • 15 Pages

    Torikai, S., 2004, Nipponia: Solar Power Lights the Way to the Future, Issue 28, March Edition, Heibonsha Ltd, Japan…

    • 3733 Words
    • 15 Pages
    Best Essays
  • Good Essays

    Zoology Lab

    • 580 Words
    • 3 Pages

    Visible light ranges between 400-700nm ; NA less than one for dry, about 1.25 for oil.…

    • 580 Words
    • 3 Pages
    Good Essays
  • Powerful Essays

    Leith, E. N., Kozma, A., Upatnieks, J., Marks, J., and Massey, N., "Holographic Data Storage in Three-Dimensional Media," Appl. Opt., 5, No. 8…

    • 11342 Words
    • 167 Pages
    Powerful Essays
  • Good Essays

    A blackbody is a body that absorbs all electromagnetic radiation and emits a continuous spectrum, such as heated cavern with just a small opening to the outside world. In 1900, German physicist Max Planck derived an expression for this blackbody spectrum. Knowing that oscillators emit radiation, he assumed that oscillators within the cavern were responsible for the emissions. In a discovery that launched quantum physics, Planck discovered that only integer multiples of the oscillators natural frequency v multiplied by a constant (his own – h) are capable of producing electromagnetic radiation (i.e. hv, 2hv, 3hv, etc. ).…

    • 661 Words
    • 3 Pages
    Good Essays
  • Good Essays

    This energy is transferred to a solar receiver which it absorbs and conveys this thermal energy to the Stirling engine. In this system, the parabolic dish uses a two-axis tracking system to maintain a zero incidence angle. The cavity receiver in dish/Stirling systems is responsible for absorbing solar radiation reflected by the concentrator and transferring this energy as heat to the working fluid. The most efficient receivers for this system are cavity receivers with a small aperture through which concentrates entered sunlight. The absorber is placed behind the aperture to reduce the intensity of concentrated solar flux. The isolated cavity between the aperture and absorber reduces the amount of heat loss. Increasing receiver aperture will increase the concentrated sunlight. On the other hand, it also increases radiation and convection losses. Consequently, there is a trade-off between the effect of input heat to the receiver and heat losses.…

    • 1308 Words
    • 6 Pages
    Good Essays
  • Powerful Essays

    Government of India. (2010, January 15). Jawaharlal Nehru National Solar Mission. Retrieved February 03, 2012, from Government of India website: http://india.gov.in/allimpfrms/alldocs/15657.pdf…

    • 2215 Words
    • 9 Pages
    Powerful Essays
  • Good Essays

    • • • • • • • • • • • • • AssumptionsX= % of glass absorption Y=% of the medium So data to the ADC is given by ADC= 100 – (X + Y) Now in case of SNF Y ~ 0% As SNF doesn’t absorb the IR rays And in case of FAT Y>X So we have two unknown so create two equation for that we are using two methods Direct transmission and reflective transmission. In direct…

    • 1252 Words
    • 6 Pages
    Good Essays
  • Powerful Essays

    Solar cells

    • 9259 Words
    • 50 Pages

    and manufacturing considerations for the most promising approaches. The effects of varying solar spectrum and current…

    • 9259 Words
    • 50 Pages
    Powerful Essays
  • Powerful Essays

    Synopsis

    • 5898 Words
    • 24 Pages

    Mechanical Engineering Department, Faculty of Engineering, Taif University, KSA alosaimy@yahoo.com Abstract -- In the present work, novel configuration of solar…

    • 5898 Words
    • 24 Pages
    Powerful Essays
  • Satisfactory Essays

    Physics Term Paper

    • 312 Words
    • 2 Pages

    6. To investigate the dependence, of the angle of deviation, on the angle of incidence, using a hollow prism filled, one by one, with different transparent fluids.…

    • 312 Words
    • 2 Pages
    Satisfactory Essays
  • Satisfactory Essays

    Solar Inverter

    • 352 Words
    • 2 Pages

    To continue with sustainable growth, energy needs are increasing day by day throughout the world. At the same time fossil fuels are depleting and lots of environmental issues are coming into picture for addition of any more fossil fuel based power plants. So many countries in the world including Asia Pacific region have placed emphasis on the development of renewable sources of energy in order to reduce fossil fuel consumption and decrease pollution. In the quest for a greener planet and renewable energy sources, solar energy is one of the most widely used and abundant in nature.…

    • 352 Words
    • 2 Pages
    Satisfactory Essays