Question

Air of 15 °C and 1.0 atm and is moving with a uniform velocity of 15.2 m/s and this flow is laminar. Then, a thin sheet of naphthalene 0.254 cm thick and 10.2 cm square is placed in the air stream parallel to the direction of flow (air flows above and below the sheet). For how long, in hours, must the sheet be exposed to the air stream before one-tenth of its mass will have sublimed? Assume the upper and lower surfaces remain flat at all times. The molecular diffusion coefficient for the air-naphthalene system is 185 cm2/h and the Schmidt number is 2.57 under the conditions given in the problem. The vapor pressure of naphthalene at 15 °C is 0.0364 mmHg. Neglect sublimation from the edges. The kinematic viscos ity of air can be assumed to be 13.28 X 106 m2/s. The density of naphthalene is 1152 kg/m2 Note: the transition to turbulent flow for this system occurs at Rex = 3.0 X 105 so this should be checked

Answer 1

ANSWER data, given ne om AVL diffurton Nu natural 15-axb.o 319 3 9 1o aXlo 36 555 10.2xlo O'0 3193 floo elo and tubulent state, 1f en rate calcutated by mam -Eransfer are Co ffi' cient NA 0 PAI from ound co-Yellation be Can JD 0 036 36555) 0 036 (Re 4.4026 X03 Eransfe Mam Co-Siellation StSe) 3

als Se -3 Ke 4.4026 Xo ( 4.4096 RIO K15.9 Ke al3 C2.5) O356 RT 5684 10 8314 a93 NO kG (PA) .5684 X10 x G.9866 D RIO molm2.sec a 33 tvapour epha prenure Subli mate fate of NASang -4 - & 지0 a 339 5.4986 A1o

Volxdensi ty napthalene initial maM of volM 2.5908 06 6 a.58o Xoxu52 (kg a.9846 grams naphthalene 57986 X10 grams Sublimated /hour only quater but given one 1.e o.74615 9 ms O 7465 X la8.699 houra 5.3986 Xro