Question

heat transfer

answer is given

A cannonball of diameter d -20 cm is cast by pouring molten metal into a mold. When the ball has solidified and cooled to 150 °C, it is removed from the mold and set on a rack to cool, surrounded by still air at T = 20 °C. Can the cannonball (which is a sphere) be approximated as an isothermal lump as it cools? If so, calculate how long until it cools to T, = 40 °C. If not, describe a possible solution technique to solve for the cooling time. Properties that you might need: Air properties (the fluid): ky = 33.8x103 w v= 26.4 x 106 as = 38.3x106m 7 Pr=0. 0 0000 - 38.3x10 Cannonball material properties (the solid): Ry - 60 = 7800G = 430

Answer 1

niven diameter of Cannonbau (d) - 20cm=0.2 m volume of Cannon bau (v) 24/2583 25/d3 =5%* (0.213 Curtale area of common the area of Cannonbau (A 2 ñda 5 x 10.2.2 2 Thermal conductivity of Cannon bay. Ka 60/wank density, 82 7800kglan? specifichear of Cannon bas. cp = 4 305 legk casing Biot number Bir hb = NGX (0.23 for sphere, b = xconn ? 0.2 lif Bison they cannonbau isothermal eumped body. Canbe treated as for lymped systery en (3-Too) - Axt nep To-too on viiven, Ts = 486 Tooa 20c initial (To) 21500 Temperature

as Heat Transfer Coefficient is not given Hence we have to calculate heat transfer coeff от Сладkok'o Слт) 2 2 g.hxo.orgx 230x0.23 (26.4x10or рація = 4.0987x107 селу - их хот = 4.098 7x107x0.4 - 2.87X 107 nas Musselt no. MYR 0.54 (Rg 77/4 nd 2 39.54 i 23 hao.2 -39.524 33.8x103 - ha 6. 6796 whick How Bioh no. Ric nb = 6.629 6x 0.2 20.00371 sou 68 60 Hence luuped analysis is possible. - TC How en (20) = - 6.67968 of 570.22 77600 23x90x7500 a) t= 31329. 34 see