Question

The front surface of the glass cover of a flat plate solar collector of dimensions 1 m x 1 m has a temperature of 40°C. The collector loses heat by convection to atmospheric air at temperature 14°C. For a wind speed of 5 m/s blowing along and parallel to the glass cover, determine the rate of heat loss by convection. Calculate the net heat loss from the glass cover due to radiation exchange with blackbody surroundings at temperature 7°C if the glass emissivity is 0.9.

Answer 1

Sadicative heat loss, ad recta Black body temperaten, Tb = fat f7 273 280k missivity of glaus, € = og Temperation of glau flat plate, Is = 40 c= 40 +24 Ts=313k Q Rad & ECTS TH stoforo BoH3 ms.com Const, T-S-6770w pad = 5.64x10x 0.9 ( 3137_2807) lorod - 176.12 W

pirmensions of s of that plate- I mim Teroperatore q tal plate Pocets Teropperalame of almosphecic air=14 codo Speed of aim, u sol sec. Heat loss by Convection. Q=hACTS-Too) Heat toonster Coefficient, h=2 Nie = 0.664 Re, Pr - IL At T = Tet To 40+14 24c, Propertica of air ale Cp=100049 kJ l bgt e = 10184 kgl . v = 1.568x10 sec. Pr = 0.1296.ca 2 nolsee x = 21141x10 UL = 5xx = 26.24 milimki = 318 877.55 e is 6exios (esxior) laminas flow. so 2 113 bxl .26.24x10 'Nü = 0.664(31884+55) Co.#2963+hx! [ = 8 , 86 look Convective heat booss for allea, A = 1x/= lon a comment hacts-Too ) = 8.86 x 1 x 40-14) Rconv = 230.3 k O convective - ecc