ME327Soln5A2024 (1)
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University of Saskatchewan**We aren't endorsed by this school
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ME 327
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Mechanical Engineering
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Dec 17, 2024
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Name: ID #: University of Saskatchewan Department of Mechanical Engineering ME 327 Heat Transfer Fall, 2024 Assignment #5A (10 points total) 1. Problem 5.9 in 8thed. of Text (Problem 5.8 in 7thed.)* 2. Problem 5.57(a) in 8thed. of Text (Problem 5.74(a) in 7thed.) 3. Problem 5.69(a) in 8thed. of Text (Problem 5.89(a) in 7thed.) 4. Problem 5.72(a) in 8thed. Of Text (Problem 5.92(a) in 7thed.) 5. Question 5, 2010 Midterm Exam A 1.5 in. (38 mm) thick piece of yellow pine is tested using the electric heater in a cone calorimeter. The specimen is initially at a uniform temperature of 20ºC. The heat flux from the electric heater to the surface of the specimen is 10 kW/m2. a) Determine the temperature at a depth of 10 mm after 5 minutes of exposure to the electric heater. b) Estimate the length of exposure required for the surface temperature of the specimen to reach 300ºC. denotes that problem has been revised for 8thedition (e.g., new values for parameters)
PROBLEM 5.9KNOWN:The temperature-time history of a pure copper sphere in a hydrogen stream. FIND:The heat transfer coefficient between the sphere and the hydrogen stream. SCHEMATIC:D = 20 mmT(0) = 75°CT(97s) = 55°CASSUMPTIONS:(1) Temperature of sphere is spatially uniform, (2) Negligible radiation exchange, (3) Constant properties. PROPERTIES:Table A-1, Pure copper (338 K): r= 8933 kg/m3, cp= 384 J/kg⋅K, k = 399 W/m⋅K. ANALYSIS:The time-temperature history is given by Eq. 5.6 with Eq. 5.7. ( )2tsitts3tptt1expwhere RADR ChADCVc V6θpθpr=−====TT .θ∞=−Recognize that when t = 97 s, ( )()()ittt5527Ct97 s0.583expexp7527Cθθtt−===−=−−and solving for ttfind t180 s.t=Hence, ()333p22s t8933 kg/m0.02m/ 6384 J/kgKVchA0.02m180 sprtp⋅==×2h63.5 W/mK.=<COMMENTS:Note that with Lc= Do/6, 2-4chL0.02Bi63.5 W/mKm/399 W/mK5.3 10.k6==⋅×⋅=×Hence, Bi < 0.1 and the spatially isothermal assumption is reasonable.
PROBLEM 5.69 KNOWN:Irreversible thermal injury (cell damage) occurs in living tissue maintained at T ≥48°C for a duration ∆t ≥10s. FIND:(a) Extent of damage for 10 seconds of contact with machinery in the temperature range 50 to 100°C, (b) Temperature histories at selected locations in tissue (x = 0.5, 1, 5 mm) for a machinery temperature of 100°C. SCHEMATIC: ASSUMPTIONS:(1) Portion of worker’s body modeled as semi-infinite medium, initially at a uniform temperature, 37°C, (2) Tissue properties are constant and equivalent to those of water at 37°C, (3) Negligible contact resistance. PROPERTIES:Table A-6, Water, liquid (T = 37°C = 310 K): ρ= 1/vf= 993.1 kg/m3, c = 4178 J/kg⋅K, k = 0.628 W/m⋅K, α= k/ρc = 1.513 ×10-7m2/s. ANALYSIS:(a) For a given surface temperature -- suddenly applied -- the analysis is directed toward finding the skin depth xbfor which the tissue will be at Tb≥48°C for more than 10s? From Eq. 5.60, ()()[ ]1/2bsbisT x ,tTerfx2terfTTαη−==−. For the two values of Ts, the left-hand side of the equation is ()()s48100CT100 C:0.82537100C−==−()()s4850CT50 C:0.1543750C−==−The burn depth is []()[]()[]1/ 21/ 27241/ 2bxw 2tw 2 1.513 10mst7.77910w tα−−==××=×. Continued...
PROBLEM 5.69 (Cont.) Using Table B.2 to evaluate the error function and letting t = 10s, find xbas Ts= 100°C: xb= 7.779 ×10-4[0.96](10s)1/2= 2.362 ×103m = 2.36 mm <Ts= 50°C: xb= 7.779 ×10-4[0.137](10s)1/2= 3.37 ×103m = 0.34 mm <Recognize that tissue at this depth, xb, has not been damaged, but will become so if Tsis maintained for the next 10s. We conclude that, for Ts= 50°C, only superficial damage will occur for a contact period of 20s. (b) Temperature histories at the prescribed locations are as follows. 01530Time, t(s)37475767778797Temperature, T(C)x = 0.5 mmx = 1.0 mmx = 2.0 mmThe critical temperature of 48°C is reached within approximately 1s at x = 0.5 mm and within 7s at x = 2 mm. COMMENTS:Note that the burn depth xbincreases as t1/2.
PROBLEM 5.72 KNOWN:Thick oak wall, initially at a uniform temperature of 25°C, is suddenly exposed to combustion products at 800°C with a convection coefficient of 20 W/m2⋅K. FIND:(a) Time of exposure required for the surface to reach an ignition temperature of 400°C, (b) Temperature distribution at time t = 325s. SCHEMATIC: ASSUMPTIONS:(1) Oak wall can be treated as semi-infinite solid, (2) One-dimensional conduction, (3) Constant properties, (4) Negligible radiation. PROPERTIES:Table A-3, Oak, cross grain (300 K): ρ= 545 kg/m3, c = 2385 J/kg⋅K, k = 0.17 W/m⋅K, α= k/ρc = 0.17 W/m⋅K/545 kg/m3 ×2385 J/kg⋅K = 1.31 ×10-7m2/s. ANALYSIS:(a) This situation corresponds to Case 3 of Figure 5.7. The temperature distribution is given by Eq. 5.63 or by Figure 5.8. Using the figure with ()iiT 0,tT400250.48TT80025∞−−=−−and ()1/ 2x02tα=we obtain h(αt)1/2/k ≈0.75, in which case t ≈(0.75k/hα1/2)2. Hence, ()21/ 2272t0.750.17 W m K20W mK 1.31 10ms310s−≈×⋅⋅×=<(b) Using the IHT Transient Conduction Modelfor a Semi-infinite Solid, the following temperature distribution was generated for t = 325s. 00.0050.010.0150.020.0250.03Distance from the surface, x(m)25100175250325400Temperature, T(C)The temperature decay would become more pronounced with decreasing α(decreasing k, increasing ρcp) and in this case the penetration depth of the heating process corresponds to x ≈0.025 m at 325s. COMMENTS:The result of part (a) indicates that, after approximately 5 minutes, the surface of the wall will ignite and combustion will ensue. Once combustion has started, the present model is no longer appropriate.