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Mechanical Engineering

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Oct 30, 2023

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Raquel Riera Homework 1 The thermal conductivity of a sheet of rigid extruded insulation is reported to be 1 = 0.029 N/m.K. The measured temp. diference across a 20mm-thick sheet of the materials is T1 - Tz = 10 a) What is the heat flox through a 2x2[m) sheet of the insulation ? - I A = 2x2m g = 41 1 11 DT 10 C - - =0.029 N/m.K x4m2 x 10 C L 20mm - 1 1000mm 20mm =58 [N] b) What is the rate of heat transfer through the sheet of insulation ? 9" = 1 A q g" = - = W = 14.5 [ A
D A wall has inner & outer surface imperature of 162 and 62 respec tivel f the interior and exterior air Emperatures are 20C &S2 respecti- rely. Calculate the heat tx from the interior air to the wall, from the wall to the exterior air, from the wall to the interior air. -s the I wall under steady-state conditions ? T, = 12 g," Ty, = 5 -> To = 20C I Tz = 62 convection * n = 5W/m2k h= = 20W/m2K I heat flux from interior air to the wall q" = h,.(T - Ti) = 5x (20-16) =20 mm heat flux from the wall to the exterior air q " = nz (Tz - Toc) 2 - 20(6 - 5) =20 W/m2 heat flox from the wall to the interior air: g" = - G, = - 20w m2 is the wall under steady-state conditions 0 Steady-state Ein- Fort +1 -Est Ig - d = 0 g,-q,"Est Este e e dt 20 -20 = Est = 0 The wall is under steady-state
An overhead 25m long, uninsulated industrial steam pipe of 100mm diameter is routed through a building whose walls and air are at 25C. Pressurized stam maintains a pipe surface temperature of 150 and the coefficient associated w natural convection is n = 10W/m2K The surface emissitivity is E = 0.8 a) what is the rate of heat loss from the steam line ? Tw= 252 T h = to L = 2Sm - = 25C q = fcur + grad a 4 Ts = 130 t = 298k q = n.A.(Ts - 5) + do [s" - TwY) - D = 1m 2 = 0.8 Ts = 423R ~ A = π D.2 = 1047.85(423 - 298) + (0.8x5.67x108*(42322984) q = 18405W b) If the steam is generated is a gas-fired boiler operating at - an efficiency, of f=0.90 a natural gas is priced at Cg= $ 0.02 per M5 what The annual cost of heat loss from the line ? E = Q A t 18405*(3654244360075 = 5.8842x10" Et = = 4x"= 6.449/IX 2 = 2y x Ef = 6.449/IX10"
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