The collector efficiency can be obtained by dividing Qu by (G^AJ. Therefore, n = fr

T - Ta 1 | |

n0 - Ul |
where C = concentration ratio, C = AJAr. For Fr, a relation similar to Eq. (3.58) is used by replacing Ac with Ar and using F', given by Eq. (3.126), which does not include the fin and bond conductance terms, as in flat-plate collectors. ## Example 3.12A 20 m long parabolic trough collector with an aperture width of 3.5 m has a pipe receiver of 50 mm outside diameter and 40 mm inside diameter and a glass cover of 90 mm in diameter. If the space between the receiver and the glass cover is evacuated, estimate the overall collector heat loss coefficient, the useful energy gain, and the exit fluid temperature. The following data are given: Absorbed solar radiation = 500 W/m2. Receiver temperature = 260°C = 533 K. Receiver emissivity, er = 0.92. Glass cover emissivity, eg = 0.87. Circulating fluid, cp = 1350 J/kg-K. Entering fluid temperature = 220°C = 493 K. Heat transfer coefficient inside the pipe = 330 W/m2-K. Tube thermal conductivity, k = 15 W/m-K. Solution The receiver area Ar = -DL = - X 0.05 X 20 = 3.14 m2. The glass cover area Ag = -DgL = - X0.09 X 20 = 5.65 m2. The unshaded collector aperture area Aa = (3.5 - 0.09) X 20 = 68.2 m2. Next, a glass cover temperature, Tg, is assumed in order to evaluate the convection and radiation heat transfer from the glass cover. This is assumed to be equal to 64°C = 337 K. The actual glass cover temperature is obtained by iteration by neglecting the interactions with the reflector. The convective (wind) heat transfer coefficient hc,c-a = hw of the glass cover can be calculated from Eq. (3.115). First, the Reynolds number needs to be estimated at the mean temperature /(25 + 64) = 44.5°C. Therefore, from Table A5.1 in Appendix 5, we get p = 1.11kg/m3 Re = pVDg/| = (1.11 X 5 X 0.09)/2.02 X 10~5 = 24,728 Therefore, Eq. (3.115b) applies, which gives and hc,c-a = hw = (Nu)k/Dg = 129.73 X 0.0276/0.09 = 39.8W/m2-K The radiation heat transfer coefficient, hrc_a, for the glass cover to the ambient is calculated from Eq. (2.75): 0.87(5.67 X 10~8)(337 + 298)(3372 + 2982) = 6.34W/m2-K The radiation heat transfer coefficient, hrr_c, between the receiver tube and the glass cover is estimated from Eq. (3.117): _ a(Tr2 + Tg2)(Tr + Tg) (5.67 X 10—8)(5332 + 3372)(533 + 337) 0.87 Since the space between the receiver and the glass cover is evacuated, there is no convection heat transfer. Therefore, based on the receiver area, the overall collector heat loss coefficient is given by Eq. (3.116): 13.95W/m2-K 0.05 Since UL is based on the assumed Tg value, we need to check if the assumption made was correct. Using Eq. (3.119), we get 3.14 X 16.77 X 260 + 5.65(6.34 + 39.8)25 3.14 X 16.77 + 5.65(6.34 + 39.8) This is about the same as the value assumed earlier. The collector efficiency factor can be calculated from Eq. (3.126): 1/UL 1/13.95 D ln 0.05 |

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