Question

Oil flow in a journal bearing can be treated as parallel flow between two large isothermal plates with the lower plate moving at a constant velocity of V_∞ =12 m/s and the other stationery. Consider such a flow with a uniform spacing of d=0.5mm between the plates. The temperatures of the upper and lower plates are T₁=50 °C and T₂=15°C, respectively. Use thermal conductivity k=0.145 W/m.K and dynamic viscosity µ =0.8374 kg/m.s for the oil.

(a) Simplify the mass continuity, momentum, and energy equations. (5 points)

(b) Determine the velocity and temperature distribution in the oil. (5 points)

(c) Find the maximum temperature and where it occurs. (5 points)

Answer 1

Average temperature at bearing as Te Title Ti= 80445 2 To= 32.5 given Ka) 8.145 wink o dx 2 20 dx Miss 0.8374 Ns/m² Consider flow drection in x , and hormals as y so Continuity egpalien du dv dy flow is Partiel vzo a dy so idu S24 - So M= Mly) OP → gradient (presure) momentum equation du de Pludu dy dy² dx Plux 2 y = malu dyz a Zu dyz zo ax u + Odu ر = دل x o -O 20

On integration Од – С. ५ argain integrate (9) - G9 +12 Екофа» Now to) -o 4(L) y clearance of beely O=Cҳо +12 2 -0 V z Cıxl to Ciz До uly) = ( 9+02 щ9) 2) yy to асу) 2) v2 Уу on differentiation Ә4. ay 0= кат + ( 2 ) ду. ат + Oе к ду? От д че (*

(32= To-liter? ОТ ay integration MY Z + c3 again integration & Tly=-uy ²1 %) + (zyrly- boundary conditions I (0) = Ti 2K T(L) = 72 Ti= u(83? (7)+ +(3(0) teu 21 Toz otottu Cu= Ti ak t Te-T, = - UL?? 2k1² +(₂2 ak tur? ZKL so (3= T2 - 1 1 + TCY)= My ? (2 + czytly ak T2-T Y+T, + 2kL Tyz-MY

7 t & F. 2 272 키 료? a-ti)y E-)y tury y² ) 시? ㅠ 쐞는 + 쟈 + (27) J velocity distribution is July) - 보 temperature distribution is Ty) 랴 빠 - F) 10 T 삎는 2y 2 리 림) 十 하 랴 May () temperature dI zo (12) - (1) dy O Mv² 랴나 ) 2Y L 1. 2(72) u2 느-UTE) 2r2 ye 늘 Y= L kL(T-Z) 2. urz Now put the vales 0. DOOS y 0-146 / OoOGO5 x (IS-5) 00837Yx 122 + 2104 85 Ye Do0S 12 2-5xid t2.10455 Yz271x10

so location of tmav 0.27 mm ya (max temps 2 +TI (I-IOy+ le2 Tmax ₂ 3 ) (분 Vy T 2k Y= 0.271 0.000271 7 to OaQba27) 0.0005 0.8374 x 12 2x 0.145 0.0005 + 50-150) X0:00027) ordoos + 56 + 18.97 +15 415.81 (0.542 -0.293) 103+18097+15 o2 137.50 Tmar