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College of Engineering and Computer Science Mechanical Engineering Department Mechanical Engineering 375 Heat Transfer Spring 2007 Number 17629 Instructor: Larry Caretto

Exercise Eleven – Fundamentals of Radiation

1. The variation of spectral transmissivity of a 0.6-cm thick glass window is shown in the figure at the right. Determine the average transmissivity for this window for solar radiation (T = 5,800 K) and radiation coming from surfaces in the room (t = 300 K). Also, determine the amount of solar radiation transmitted through the window for incident solar radiation of 650 W/m2. (Problem and figure P12.51 from Çengel, Heat and Mass Transfer.) Here we have to evaluate the average transmissivity, , which is given by the general equation (which is the first integral on the right) that is subsequently applied to the profile in this problem. 1  1 0.3 m 1 3 m 1     E d  0E d  0.92E d  0E d 4   b, 4  b, 4  b, 4  b, T 0 T 0 T 0.3 m T 3 m The non-zero part of this equation is simply 0.92 times the fraction of total radiation between the

wavelengths of 1 = 0.3 m and 2 = 3 m. We can find this from the Table 12-2 on page 672 of

the text which gives the fraction f as a function of the product T. 0.92 3 m   E d  0.92 f  T  f  T  4  b,   2  1  T 0.3 m

For T = 5800 K, 1T = (0.3 m)(5800 K) = 1740mK and 2T = (3 m)(5800 K) = 17400mK. Interpolation in Table 12-2 gives .039341.019718 f  T   f 1740 m  K   0.019718  1740 1600  0.033454  1  1800 1600 .980860 .973814 f  T   f 17400 m  K   0.019718  17400 160  0  0.978746  2  18000 16000 Thus the total transmissivity of the surface for radiation with a source temperature of 5800 K is 0.978746 – 0.033454 = 0.870.

If the temperature is 300 K, 1T = (0.3 m)(300 K) = 90mK and 2T = (3 m)(300 K) = 900mK. Interpolation in Table 12-2 gives .0.0  0.0 f  T   f 90 m K   0.0  90  0  0.0  1  200  0

Jacaranda (Engineering) 3333 Mail Code Phone: 818.677.6448 E-mail: [email protected] 8348 Fax: 818.677.7062 0.000321  0.000016 f  T   f 900 m  K   0.000016  900  800  0.00016  1  1000  800 Thus the total transmissivity of the surface for radiation with a source temperature of 5800 K is 0.00016 – 0.000000 = 0.00016. The amount of incoming solar radiation transmitted through the glass uses the transmissivity for a radiation source at 5,800 K. So, if the incident radiation is 650 W/m2, then the transmitted radiation is 0.870 times this or 566 W/m2.

2. The variations of spectral emissivity of two surfaces are shown in the figure at the right. Determine the average emissivity of each surface at T = 3000 K. Also determine the average absorptivity and reflectivity of each surface for radiation coming from a source at 3000 K. Which surface is more suitable to serve as a solar absorber? (Problem and figure P12.47 from Çengel, Heat and Mass Transfer.) For both surfaces the average emissivity, , is given by the general equation (which is the first integral on the right) that is subsequently applied to the profile in this problem. 1  1 3 m 1     E d   E d   E d  4   b, 4  1 b, 4  2 b, T 0 T 0 T 3 m

1 f 3 mT   2 1 f 3 mT   2  1   2  f 3 mT 

For T = 3000 K, (3 m)T = (3 m)(3000 K) = 9000 mK; at this value of T, f = 0.890029 in Table 12-2 on page 672 of the text. Thus we have the following computational equation.

   2  1   2  f 3 mT    2  0.8900291  2 

For the surface, with 1 = 0.2 and 2 = 0.9

   2  0.8900291   2   0.9  0.8900290.2  0.9 = 0.28 By Kirchoff’s law  =  so  = 0.28 for this surface. To determine the reflectivity for this surface we use the relationship that  +  +  = 1. We do not have any data on the transmissivity so we cannot really apply this equation. We can get an estimate of the maximum reflectivity by assuming that the surface is opaque; that is  = 0. In this case  +  = 1, so  = 1 –  = 1 – 0.28 or  = 0.72.

For the surface, with 1 = 0.8 and 2 = 0.1

   2  0.8900291   2   0.1 0.8900290.8  0.1= 0.72 By Kirchoff’s law  =  so  = 0.72 for this surface. To determine the reflectivity for this surface we again assume that  = 0.to get a maximum value for . For this surface,  = 1 –  = 1 – 0.72 or  = 0.28.

Jacaranda (Engineering) 3333 Mail Code Phone: 818.677.6448 E-mail: [email protected] 8348 Fax: 818.677.7062 The surface with 1 = 0.8 and 2 = 0.1 is a better surface for a solar collector. It has a high absorptivity in the low wavelength region where solar energy is concentrated and a low emissivity at longer wavelengths where most of the radiation emitted by the solar collector would occur. 3. A solar collector has a surface whose solar absorptivity is 0.87 and whose emissivity at terrestrial temperatures is 0.09. At a time when the air temperature is 25oC and the effective sky temperature is 15oC, the collector is shut off so there is no heat transfer out of the back of the collector. Determine the equilibrium temperature of the collector surface if the incident solar radiation is 600 W/m2 and the convective heat transfer coefficient is 10 W/m2oC. If there is no heat transfer out of the back of the collector the net heat transfer into the front of the collector must be zero. This consists of the solar irradiation absorbed by the collector minus the heat lost by convection and radiation to the environment. Setting this heat balance gives the following relationship among heat fluxes entering the collector.

4 4  sGsolar  hT Ts   Tsky Ts  0 Substituting the given problem data into this equation, using absolute temperatures for the radiation calculations, gives

4 4  sGsolar  hT Ts   Tsky Ts  0 8 600W 10W 5.670x10 W 4 4 0.87  298.15 K Ts  0.09 288.15 K  Ts  0 m2 m2  K m2  K 4

This equation for Ts cannot be solved explicitly. Using numerical software such as the goal seek method of Excel gives Ts = 346 K.

Jacaranda (Engineering) 3333 Mail Code Phone: 818.677.6448 E-mail: [email protected] 8348 Fax: 818.677.7062