Problem 3.73 Known: omposite wall with outer surfaces eposed to ection process while the inner wall eperiences uniform heat generation Unnown: Volumetric heat generation and thermal conductivity for material reuired for special conditions Schematic: ssumptions:. Steady-state, one-dimensional heat transfer. Negligible contact resistance at interfaces 3. Uniform generation in, zero in and 4. onstant properties nalysis: From an energy balance on wall, E E E E in out g st " " " " ( / ( 0 T T - 0 To determine the heat flues, construct thermal circuits for and : T T " ( / h / (6 5 /000 0.03/ 5 07,73W / m " ( T T / / h ( 5 0.0 / 50 /000 3,857W / m
Using the values for and in euation (, we find: 3 6 /.00*0 4 m W To determine, use the general form of the temperature and heat flu distributions in wall, ( T " ( d There are 3 unnowns,, and, which can be evaluated using three conditions ( ( T T Where T 6 ( ( T T Where T " " ( ( Where " 0773W/m Solving for these euations simultaneously with 3 6 /.00*0 4 m W 5.3W/m.K
Problem 3.9 Known: ong rod eperiencing uniform volumetric generation encapsulated by a circular sleeve eposed to ection Unnown:. Temperature at the interface between rod and sleeve and on the outer surface. Temperature at center of rod Schematic: ssumptions:. One-dimensional radial conduction in rod and sleeve. Steady state conditions 3. Uniform volumetric generation in rod 4. Negligible contact resistance between rod and sleeve 5. onstant properties nalysis: a onstruct a thermal circuit for the sleeve: Where ' E ' gen π r 4,000 π 0. 754.0W / m ln( r / r ln(0.4 / 0. RS.758 0 m. K / W π S π 4 R 3.83 0 m. K W hπd 5π 0.4 / The rate euation can be written as: T T T T ' R R R S
T T '( R S R 7 754 (.758 0 3.83 0 7.8 T ' R 7 754 3.83 0 5.0 T b The temperature at the center of the rod is For -D conduction in cylinder with uniform heat generation, temperature distribution is given by: r r T ( r T 4 r r at r0, we have r 4,000 0. T (0 T0 T 7.8 9 4r 4 0.5 c To minimize the temperature in the center, since we have: r T ( 0 T0 T 4 r Where, r and r are unchanged, so T must be minimized. Since, T T '( R S R and is unchanged. In order to minimize T, R S R must be minimized. ssuming that h is independent of r, then R S R is minimized when r r cr s /h 4/5 0.6m To minimize T 0, the thicness of the sleeve should be decreased from 0.-0.0.m to 0.6-0.0.06m 0.5 d If h 5.8D, then the epression for r will change ln( r / r R tot 0.5 π 5.8(r π r dr tot 0.5 0.5 dr πr.79r π 0.5 r rcr 0. 03m.79 Since r <r, the sleeve should be removed. 0.5 For r r 0., h 5.8(0. 35.35W/m.K T 7754/(35.35*π*0.60.9 T 0 9-(7.8-60.98.
Problem 3 Known: Dimensions and properties of pin fin, dimensions (ecept length of pin fin and properties of pin fin. Same heat flow rate through the both pin fins, i.e Unnown: ength of pin fin if a 0 d Schematic: ( or b h( T T and h( T T d d c T T 80 T b 00 D 0.005m 0.05m 35W/m.K T T 80 D 0.004m, 80W/m.K T,0 h30w/m.k ssumptions:. Steady state -D conduction along. h independent of D; h from the side surface is the same as h from the tip 3. onstant properties 4. Negligible radiation effects nalysis a Negligible rate of heat transfer from the tip Heat flow from the fin with "insulated tip" is given by E. (3.76: M tanh m f Where m For pin fin P ( hp c πd and c πd 4,
4h Therefore m, M hp c θ b Where θ b T b T D For pin fin : 0.05 m D 0.005 m 35 W/m K For pin fin :? D 0.004 m 80 W/m K For same heat transfer rate from both fins: f M tanh( m f, M Solving for, We have 0.05656m, tanh( m Table shows solving results (This is optional m m M f, m [m - ] [-] [W] [W] [m] [m - ] 6.865.309307.438988.43378 0.05656 9.3649 m M f, / [-] [W] [W] [m ] [m ].095304.556695.43445 0.000785 0.0007.04996 Temperature of fin tip can be written as following: Tb T T T cosh( m T, 60.7 o T, 68.3 o b onvective heat transfer from both tips Heat flow from the fin with "ective tip" is given by E. (3.7 sinh( m ( h m cosh( m f M cosh( m ( h m sinh( m ll parameters are as defined before Solving for 0.056635m Table shows solving results (This is optional m m (h/m M f, m [m - ] [-] [-] [W] [W] [m] [m - ] 6.865.309307 0.03737.43899.5498 0.056635 9.3649 m (h/m M f, / [-] [-] [W] [W] [m ] [m ].096748 0.09365.55669.54963 0.000785 0.0007.03565 Temperature of fin tip,
T T T cosh T, 59.5805 o b T ( m ( h m sinh( m T, 67.33974 o Notes: the results for a and b are almost identical, because the area of the tip is very small compared to the total surface area. c Prescribed temperature of the tip; T, T, 80 o ; 80 o Heat flow from the fin of prescribed temperature is given by E. (3.78 cosh( m θ θ b f M sinh( m Solving for 0.04788m Table shows solving results (This is optional m m (θ /θ b M f, m [m - ] [-] [-] [W] [W] [m] [m - ] 6.865.309307 0.75.43899.03665 0.04788 9.3649 m (θ /θ b M f, / [-] [-] [W] [W] [m ] [m ] 0.93783 0.75.55669.037446 0.000785 0.000593.3450