Energy Engineering - Heat and Mass Transfer

Radiation Problems - Solutions

Divided by topic

0.1 Radiative heat transfer: exercise 3 0.1.1 Description The components of an electronic package in an orbiting satellite are mounted in a compartment that is well insulated on all but one of its sides. The uninsulated side is inclined to the sun radiation (surface temperature: Ts 5800 K, heat flux at the orbit of the earth: qs 1.35 103W m 2) and consists of an isothermal copper plate (squared, flat, opaque, di↵use, side: l 1.0m, plate temperature: Tp 500 K, inclination to sun radiation: ✓ ⇡ 6rad , spectral emissivity: ✏,1 0.20 for 0.0; 2 .0µm ,✏,2 0.80 for 2.0; µm ) whose outer surface is exposed to the vacuum of outer space and whose inner surface is attached to the components. Assuming steady-state operating conditions: 1. compute the power dissipated by the components. 0.1.2 Questions 1. compute the power dissipated by the components. 0.1.3 Data SUN surface temperature: Ts 5800 .0K 5527 .0 C heat flux at the orbit of the earth: qs 1.35 103W m 2 1.35 kW m 2 COPPER PLATE squared flat opaque di↵use side: l 1.000 m plate temperature: Tp 500 .0K 227 .0 C inclination to sun radiation: ✓ 0.5236 rad spectral emissivity: ✏ ✏,1 0.20 0.0; 2 .0 µm ✏,2 0.80 2.0; µm (1) CONSTANTS Stefan - Boltzmann constant: 5.67 10 8W m 2 K 4 Wien’s displacement law constant: Cw 2.90 103µmK 0.1.4 Solution QUESTION 1 The Wien’s law provides the wavelength corresponding to the peak emission of a blackbody at a given surface temperature. That value, which is not strictly required, is useful, selecting the right temperature, to estimate in advance, for a given spectral hemispherical emissivity distribution: •the total hemispherical absorptivity (the sun surface temperature is required), •the total hemispherical emissivity (the copper surface temperature is required), and check the outcome of the computational procedure. The Wien’s law provides that: •the peak emission of the sun tales place at: max s Cw Ts 0.4996 µm (2) •the peak emission of the copper plate tales place at: max p Cw Tp 5.796 µm (3) According to the chart in fig. 1: •the total hemispherical absorptivity: ↵ 0.2( ↵ 0.2 because of the spectral hemispherical emissivity distribution shape), 1 10 -2 10 -1 10 0 10 1 10 2 10 3 wavelength - [ m] 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 spectral emissivity - [-] 10 0 10 2 10 4 10 6 10 8 10 10 10 12 10 14 Blackbody spectral emissive power - E ,b [W/m 2 m] E ,b(Ts) E ,b(Tp) Figure 1: Spectral hemispherical emissivity ✏, spectral emissive power of the Sun E,s , spectral emissive power of blackbody at the plate temperature E,p •the total hemispherical emissivity: ✏ 0.8( ✏ 0.8 because of the spectral hemispherical emissivity distribution shape). As the copper plate does not behaves like a gray body, di↵erent values for the total hemispherical emissivity and the total hemispherical absorptivity has to be expected. To compute the total hemispherical absorptivity of the copper plate are required: 1. the spectral hemispherical absorptivity, which, according to Kirchho↵’s law, is ↵ ✏ ↵,1 0.20 0.0; 2 .0 µm ↵,2 0.80 2.0; µm (4) 2. the Sun surface temperature, because it is the source of irradiation: Ts 5800 .0K 5527 .0 C (5) The blackbody radiation function: F0 F0 T is: ys T s 1.16 104µm K (6) Fs F0 ys 0.9400 (7) In the end, the total hemispherical absorptivity is: ↵s ↵1Fs ↵21 Fs 0.2360 (8) A very similar procedure has to be employed to compute the total hemispherical emissivity, in this case is required the knowledge of: 1. the spectral hemispherical emissivity, which is given: ✏ ✏,1 0.20 0.0; 2 .0 µm ✏,2 0.80 2.0; µm (9) 2. the copper plate surface temperature, because it is the source of emission: Tp 500 .0K 227 .0 C (10) 2 The blackbody radiation function: F0 fT is: yp T p 1000 .0µm K (11) Fp F0 yp 3.28 10 4 (12) In the end, the total hemispherical emissivity is ✏p ✏1Fp ✏21 Fp 0.7998 (13) The power dissipated by the electronic components can be determined writing the energy balance for a control volume surrounding the copper plate (sing convention for the power: Q thermal power positive if provided to the copper plate). To properly write the energy balance, it is important to make some remarks. •The left hand side is set equal to zero because steady state operating condition. •The procedure adopted to compute the thermal power absorbed Qabecause of Sun irradiation provides a positive value (not an algebraic value), that requires to place a sing (plus or minus) in front of it to state if it enters or leaves the control volume. As the irradiation is absorbed a plus sign (+) is required. •The procedure adopted to compute the thermal power emitted Qebecause of plate surface temperature provides a positive value (not an algebraic value), that requires to place a sing (plus or minus) in front of it to state if it enters or leaves the control volume. As the emitted radiation leaves the surface a minus sign (-) is required. •As the direction of the power dissipated by the electronic components Q is unknown (even if it could be imagined) the best approach is to assume it as an algebraic quantity and set a plus sing (+) in front of it. It means that: – if its value is positive, the thermal power leaves the electronic components and enters in the control volume; – if its value is negative, the thermal power enters in the electronic components and leaves the control volume. dU dt 0 Qa Qe Q (14) Replacing the expression for Qaand Qethe thermal power dissipated by the electronic components is: Q Qe Qa ✏ pT4pL2 ↵sqscos ✓L 2 2.56 103W 2.56 kW (15) The positive value of the power dissipated by the electronic components Q implies Qe Qa, it means that, to keep steady state at the prescribed operating conditions, thermal power has to be provided to the plate by the electronic components. 3 temperature and the interval from 0 to !, this fraction is determined by the ratio of the shaded section to the total area under the curve of Figure 12.13. Hence (12.34) Since the integrand ( E!,b/"T5) is exclusively a function of the wavelength–temperature product !T, the integral of Equation 12.34 may be evaluated to obtain F(0l!)as a function of only !T. The results are presented in Table 12.2 and Figure 12.14. They may also be used to obtain the fraction of the radiation between any two wavelengths !1and !2, since (12.35) F(!1l!2)! ! !2 0E!,b d!" ! !1 0E!,b d! "T 4 ! F(0l!2)" F(0l!1) F(0l!) " ! ! 0E!,b d! ! ! 0E!,b d! ! ! ! 0E!,b d! "T4 ! ! !T 0 E!,b "T5 d(!T)! f (!T) 786 Chapter 12 ! Radiation: Processes and Properties E!,b (!,T) ! ! "0! E!,b d FIGURE 12.13 Radiation emission from a blackbody in the spectral band 0 to !. FIGURE 12.14 Fraction of the total blackbody emission in the spectral band from 0 to ! as a function of !T. F(0 !) 1.0 0.8 0.6 0.4 0.2 00 4 8 12 16 20 !T # 10 –3 ( m •K) µ TABLE 12.2 Blackbody Radiation Functions 200 0.000000 0.375034 #10"27 0.000000 400 0.000000 0.490335 #10"13 0.000000 600 0.000000 0.104046 #10"8 0.000014 800 0.000016 0.991126 #10"7 0.001372 1,000 0.000321 0.118505 #10"5 0.016406 1,200 0.002134 0.523927 #10"5 0.072534 1,400 0.007790 0.134411 #10"4 0.186082 1,600 0.019718 0.249130 0.344904 1,800 0.039341 0.375568 0.519949 2,000 0.066728 0.493432 0.683123 2,200 0.100888 0.589649 #10"4 0.816329 2,400 0.140256 0.658866 0.912155 2,600 0.183120 0.701292 0.970891 2,800 0.227897 0.720239 0.997123 2,898 0.250108 0.722318 #10"4 1.000000 I!, b(!, T) I!, b(!max , T) I!, b(!, T)/"T5 ("m!K!sr) "1 F(0 l !) !T ("m!K) CH012.qxd 2/21/11 5:26 PM Page 786 12.4 ! Blackbody Radiation 787 TABLE 12.2 Continued 3,000 0.273232 0.720254 #10"4 0.997143 3,200 0.318102 0.705974 0.977373 3,400 0.361735 0.681544 0.943551 3,600 0.403607 0.650396 0.900429 3,800 0.443382 0.615225 #10"4 0.851737 4,000 0.480877 0.578064 0.800291 4,200 0.516014 0.540394 0.748139 4,400 0.548796 0.503253 0.696720 4,600 0.579280 0.467343 0.647004 4,800 0.607559 0.433109 0.599610 5,000 0.633747 0.400813 0.554898 5,200 0.658970 0.370580 #10"4 0.513043 5,400 0.680360 0.342445 0.474092 5,600 0.701046 0.316376 0.438002 5,800 0.720158 0.292301 0.404671 6,000 0.737818 0.270121 0.373965 6,200 0.754140 0.249723 #10"4 0.345724 6,400 0.769234 0.230985 0.319783 6,600 0.783199 0.213786 0.295973 6,800 0.796129 0.198008 0.274128 7,000 0.808109 0.183534 0.254090 7,200 0.819217 0.170256 #10"4 0.235708 7,400 0.829527 0.158073 0.218842 7,600 0.839102 0.146891 0.203360 7,800 0.848005 0.136621 0.189143 8,000 0.856288 0.127185 0.176079 8,500 0.874608 0.106772 #10"4 0.147819 9,000 0.890029 0.901463 #10"5 0.124801 9,500 0.903085 0.765338 0.105956 10,000 0.914199 0.653279 #10"5 0.090442 10,500 0.923710 0.560522 0.077600 11,000 0.931890 0.483321 0.066913 11,500 0.939959 0.418725 0.057970 12,000 0.945098 0.364394 #10"5 0.050448 13,000 0.955139 0.279457 0.038689 14,000 0.962898 0.217641 0.030131 15,000 0.969981 0.171866 #10"5 0.023794 16,000 0.973814 0.137429 0.019026 18,000 0.980860 0.908240 #10"6 0.012574 20,000 0.985602 0.623310 0.008629 25,000 0.992215 0.276474 0.003828 30,000 0.995340 0.140469 #10"6 0.001945 40,000 0.997967 0.473891 #10"7 0.000656 50,000 0.998953 0.201605 0.000279 75,000 0.999713 0.418597 #10"8 0.000058 100,000 0.999905 0.135752 0.000019 I!, b(!, T) I!, b(!max , T) I!, b(!, T)/"T5 ("m!K!sr) "1 F(0 l !) !T ("m!K) CH012.qxd 2/21/11 5:26 PM Page 787