Question

A cylinder with diameter D=6mm has a volumetric heat generation of 176.8×10⁵ W/m³. The surface is exposed to air convection with a velocity of V_∞ =11.11m/s and temperature at T_∞=10 °C. You can initially evaluate air thermal properties at T_∞ using Table A.4 or any other available data.

(a)Determine the surface temperature Ts of the cylinder (12 points).

(b)Is it reasonable to use T_∞ rather than the film temperature to evaluate the air properties in this case (3 points)?

Answer 1

Heat generation 73 yerden b, Top=10°C dar da? 1 dT T 11 q 1111 s da -()) 1r T =-qr + Gloger + C₂ (2) proof day Assumption (i) steady b. T = 10°C adr-q ada K dna qd? di -qa + da da d [ada ] -ah da 2 dT dr -qra transfer state heat conditions (1") One dimessional heat conduction equation To f(x) only (1) Unitorom heat gereretion rate å = constand (iv) Thermal conductivity I= constant. BC At Q=R T=Ts Heat genoted Heat conducted radially at the surface qo Gr®2) da -R T&T +C 2k ak 9 - do -arta da Integrate again T= -9 + Cloget LE) -k (22RLAT 1 -Q R (da JAR ak CEO ; Ca= Toase ata=0 Ts Sa - har qka T= -qra 4K +o+T mare 4K Tman Ttar 4k I man = Tst q R² 4k - Tous 2 Tmun Tomar is (3)

For steady state of rod, generated in rod = Heat convected to fluid q (0) Heat hL25RL The S Ts q R To

of ain at To Obtain the thermophysical properties S = 1.2474 kg/m² M = 1.7722x10-5 Kgloos K Oo o 24840 0.71759 Calculate Reynolds numbers (Re) Re PVD 1. 2474 x 11'll XC6x16-3) M 1.7722x10-5 Ph = 17 4692.003 heat transfer Coefficient Convective To calculate costelation for we require require the necessary average Nusselt number for forced for forced convection.

For Re between (4000 - 40000) (1/2) 0.618 Nu = 0.193 Re hD - 0.192 (4692.003616 (014125975 K h 0.02489-0 x 166.286 0.006 h 688.424 W/m²K Sia face temperature (Ts ) ca Calculate the Is a 146.8 x 105 x 10 ) 2x688.424 + 10°C Ts=48.52 C reasonable Cases fluid properties. at An actual case, because (6) No, it is not the empirical correlations for the average Ny form Forced conrection over Circular Cylinders in Cross flow film temperature to evaluate to evaluate Reason The fluid temperature in the thermal bounday from Is surface to about layer Ta at the outer edge of the boundary. So fluid properties also vary with temperature and thus with position across the boundary layer so we have to use the film Homperature tas propertius evaluation If = Tst Te Values