b08a2ba6-2d78-491e-8e05-62f35ca147fa.jpg - 2AT X 10 2/24S T = 20"C T1 = 400"C-D= 0.003 m Soldering L 0.01 m Iron qe = hco A AT = he/D/4 7DL(T-T = q q =

b08a2ba6-2d78-491e-8e05-62f35ca147fa.jpg - 2AT X 10 2/24S T...

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Unformatted text preview: 2AT X 10 -% 2)/24S T_. = 20"C T1 = 400"C -D= 0.003 m Soldering L- 0.01 m Iron qe = hco A AT = he (/D /4+ 7DL) (T,-T.) = q q = 20 W/(m. K) 7 (0.003 m)? 4 -+ /(0.003m) (0.01m) (400.C - 20C) q = 0.77 W qr = A1 61 6(T)* - T.') - (TD + R DL ) eo(; - T.." 7 (0.003 m )? 4 -+ 7(0.003m) (0.01m) | (0.8) [5.67 x 10-$ W/(m' K')] [(673 K)* - (293 K)*] qr = 0.91 W The power requirement of the soldering iron, q, is equal to the total rate of heat loss from the tip. The total heat loss is equal to the sum of the convectiond radiative losses q = qe + qr = 0.77 W + 0.91 W = 1.68 W (40 pnt) The cross-section of a spherical fuel element (Thorium) in a nuclear reactor is shown. Energy generation occurs uniformly in the thorium fuel rod (k-60 W/m.K), which is of diameter D-20 mm. It is proposed that, under steady-state conditions, the system operates with a generation rate of e gen =5x10* W/m' and cooling system characteristics of T.-20"C and h=10 kW/m K. If the melting temperature of the Thorium is 2023 K, will the fuel melt inside the nuclear reactor? Coolant T_, h fuel rod dr 1 1 B. C. Ther tell Symmetry two dTo =>4-0 T()--3+ca dr 64 end gc. T() - 9 TE (."-t) +30. +To k dT(rero) . h. (T - Tea) 3h dr T()~Slo? (01013 2, S.1053.0) -t.(-s.co. ) . h . (T(.)-Too + (204231) =138, 83 - 1 282.885, 9cz 41666, 614 293 C2 - 36 40 (T( ) - 2038 , 56 -4385 804,5,2 may ~ T(r =0) = 2098, 56 k. 2 2023 1 14 Melts! Time: 90 min. Good Luck......
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