HW4solution

HW4solution - In order to increase the rate of heat...

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Unformatted text preview: In order to increase the rate of heat dissipation from one side of an electronic device 1 m wide and 1 m tall the addition of aluminum fins has been suggested. The fins are to be rectangular in cross section, 2.5 cm long and 0.25 cm thick. There are to be 100 fins per meter. The convective heat transfer coefficient, both for the wall and the fins, is estimated at 35 Wz’m2 K. With this information determine the percent increase in the rate of heat transfer of the finned wall compared to the bare wall. GIVEN - Aluminum fins with a rectangular cross section e Dimensions: 2.5 cm long and 0.25 mm thick - Number of fins per meter = 100 __ - The convective heat transfer coefficient (11:) = 35 W/m2 K FIND The percent increase in the rate of heat transfer of the finned wall compared to the bare wall. ASSUMPTIONS - Steady State heat transfer SKETCH PROPERTIES AND CONSTANTS From Appendix 2, Table 12: The thermal conductivity of aluminum (k) = 240 W/m K at 127°C SOLUTION Since the fins are of uniform cross section, Table 2.1 can be used to calculate the heat transfer rate from a single tin with convection at the tip: M sinh (mL) + (Hg/mk) cosh (mL) q‘ ' cosh (mu—T { Jka sin—h {mL} where M = ‘Ih‘cPkA' as = ,{fic 2{t+w) k (tw)‘ a, BS : T u T S u For a l m width [w='l m): M = H35 (1.0025 m)(240 (.0025 m?) 05 = 6.49 0‘ ¥ '5 P E 2[t+w) {35 %)2(1.0025m) L k‘_A :1. _° =0.025m —'“ mL 1‘ (tw) (240 m_“‘l'<)(.0025m1) Lm = 0.025m[10.81 %) = 0.270 F 35 4 __°_ = _—M = 0.0135 rnk (10.31 #240 Therefore the rate of heat transfer from one fin, 1 meter wide is: W sinhf0.27} + 0.0135 cosh(0.27} = 6.493 _ ____._.___._.._.___ qf " K C0511 (0.27} + 0.0135 Sinhw27) W = 1.7929 — qf 5 K In I rn2 of wall area there are 100 fins covering “100 tw = 100 (.0025 m)(1m) z 0.25 rrt2 of wall area leaving 0.75 rn2 of bare wall. The total rate of heat transfer from the wall with fins is the sum of the heat transfer from the bare wall and the heat transfer from 100 fins. qlot ; qbare + qun = FAlure as + qfin _ W a w _ W q“ _ (35fi)(0.75m )0, + 100 (1.792) 0‘? - 205.3 0‘? The rate of heat transfer from the wall without fins is: gm = Rae, = [35—1] (1:112) 0' = 35.011; The percent increase due to the addition of fins is: 205.3 — 35 x 100 = 486% 3:: % increase = COMMENTS This problem illustrates the dramatic increase in the rate of heat transfer that can be achieved with properly designed fins. The assumption that the convective heat transfer coefficient is the same for the fins and the wall is an oversimplification of the real situation, but does not affect the final results appreciably. In later chapters we will learn how to evaluate the heat transfer coefficient from physical parameters and the geometry of the system. The tip of a soldering iron consists of a 0.6 cm 01') copper rod, 7.6 cm long. if the tip must be 204°C, what is the required minimum temperature of the base and the heat flow, in Btu’s per hour and in watts, into the base? Assume that h = 22.7 W'sz K and 13,, = 21°_c. GIVEN - Tip of soldering iron consists of a copper rod - Outside diameter (D) = 0.6 cm = .006 m - Length (L) = 7.6 cm = ‘076 rn - Temperature of the tip (TL) = 204°C - Heat transfer coefficient (5} x 22.7 W/m2 K - Ambient temperature (T_) = 21°C FIND (a) Minimum temperature of the base (T) (b) Heat flow into the base (:1) in Btu/h and W ASSUMPTIONS - The tip is in stead)»r state - The thermal conductivity of copper is uniform and constant, i.e. not a function of temperature. - The copper tip can be treated as a fin SKETCH T i wait Tw=21°0 PROPERTIES AND CONSTANTS From Appendix 2, Table 12: The thermal conductivity of copper (K)= 388 W/m K at 227%: SOLUTION (a). From Table 2.1, the temperature distribution for a fin with a uniform cross section and convection from the tip is: e cosh[m(l,ex)]+(_)sinh[rn[L—x)] _ mk 8’ cosh (m L} + sinh [m L) where 6 = T—T_ and 6s = 8(0) = T'-T_ 4 (22.71) rn’K EP F100 45 Lm=L ..._=L = _.=0.G76m kA kEDZ kD Lm = 0.076 m[6.25_1.] = 0.475 (388.3%) (.006 m) m a 227—? _k = _._""i_ = 0.00935 :11 (6.25 Ln] [388.3%] Evaluating the temperature at x = L: 9L = TL-T- cosh(0) + .0093651nhw) _ - .___...._.—.... = 0.8932 0 T‘——T_ ‘ 005 H.475) + .00936 50010475) Solving for the base temperature: T -T o _ :- T=T+ L '=21°C+M ’ " .8932 .8932 = 226” C (b). To maintain steady state conditions, the rate of heat transfer into the base must be equal to the rate of heat loss from the rod. From Table 2.1, the rate of heat loss is: sinh(m L) + c05h(m L} where M =‘lfi PkA 0‘ =1lE k‘T'EP (Ts-T) q! = M —.-. cosh (m L) + sinh (m L} 2 M: ‘02.? W )(3ss£)“_(.006m)3 (226°C—21°C)=14.045W m2K mK 4 sinh (.475) + .00936 cosh {.475} = 6 3 W ‘cosh _(.475')—+ .00'936" sinh (.475) ' 3.412 Btu/h c1I = 14.045W 6.3WE ) = 21.513111!!! COMMENTS A small soldering iron such as this will typically be rated at 30W to allow for radiation heat losses and more rapid heat-up. One end of a 0.3 m iong steel rod is connected to a wall at 204°C. The other end is connected to a wall which is maintained at 93°C. Air is blown across the rod so that a heat transfer coefficient of 1‘? W I m2 K is maintained over the entire surface. If the diameter of the rod is 5 cm and the temperature of the air is 38°C, what is the net rate of heat loss to the air? GIVEN - A steel rod connected to walls at both ends - Length of rod (L) = 0.3 m - Diameter of the rod (D) = 5 cm = .05m - Wall temperatures: Ts= 204°C TL: 93°C - Heat transfer coefficient (11—0) = '17 W/m2 K - Air temperature (T__) = 38°C FIND The net rate of heat loss to the air (q.) ASSUMPTIONS - The wall temperatures are constant - The system is in steady state - The rod is 1% carbon steel - The thermal conductivity of the rod is uniform and not dependent on temperature - One dimensional conduction along the rod SKETCH Tl- ‘- €300 PROPERTIES AND CONSTANTS Pram Appendix 2, Table 10.- The thermal conductivity of 1% carbon steel (k) : 43 W/m K [at 20°C) SOLUTION The rod can be idealized as a tin of uniform cross section with fixed temperatures at both ends. From Table 2.1 the rate of heat loss is: cosh (m L) - (elf 95) = M q‘ sinh(m L) where: 8L =TL -T_ = 93°C - 38°C = 55°C and 3' = "I" - T“ = 204°C - 38°C =166°C E P ficflD Lm=L 31.4 = le‘ 4 1 2 M=JHPkA es= hfifnakef \(17.;K)14.(.05m}3(43% (166°C)=78.82W cosh (1.687) — E 166 = 78.82W __.._ = 74.4 w q‘ sinh (1.687) COMMENTS In a real situation the convective heat transfer coefficient will not be uniform over the circumference. It will be higher over the side facing the air stream. But because of the hi h thermal conductivit the tem erature at any given section will be nearly uniform. Both ends of a 0.6 cm copper U-shaped rod, as shown in the accompanying sketch, are rigidly affixed to a vertical wall the temperature of which is maintained at 93°C. The developed length of the rod is 0.6 m and it is exposed to air at 38°C. The combined radiative and convective heat transfer coefficient for this system is 34 W/m2 K. {a} Calculate the temperature of the midpoint'of rbd. (b) What will the rate of heat transfer from the rod be? ' - .- -- -3 '. . _ GIVEN - U-shaped copper rod rigidly affixed to a wall - Diameter (D) = 0.6 Cm = 0.006 m - Developed length (L) = 0.6 rn - Wall temperature is constant at (T) = 93 C - Air temperature (T_) = 38°C - Heat transfer coefficient (h) = 34 W/m2 K FIND (a) Temperature of the midpoint (TU) (b) Rate of heat transfer from the rod (M) ASSUMPTIONS - The system is in steady state - Variation in the thermal conductivity of copper is negligible - The U-shaped rod can be approximated by a straight rod of equal length - Uniform temperature across any section of the rod SKETCH \/A| .. I’D-391 I L L— 0.6m use area = 0 Ar 05 re: D: 0.6 cm Z N )2 END PROPERTIES AND CONSTANTS From Appendix 2, Table 12, thermal conductivity of copper (k) = 396 W/m2 K at 64 C SOLUTION By symmetry, the conduction thrctugh the rod at the center must be zero. Therefore, the rod can be thought of as two pin fins with insulated ends as shown in the sketch above. Continued . . . (a). From Table 2.1, the temperature distribution for a tin of uniform cross section with an adiabatic tip is: 9 coshIm (L.r - x)] _ _ where: 9 :T-T_, 8 =T -T__ and Lt=1ength ofthe fin {ia cosh(m L) ‘ ’ W 4(34—2— m: .11“ = _fi.’iP_= = -w—M__=7.57i 1‘" HE D?) (396_}(.006rn) m 4 rnK Evaluating the temperature of the tip of the pin fin: BU.) cosh [m (Ll-Lg] 1 3s _ cosh(m L) H eoshim L) The length of the fin is half of the wire length {L[= 0.3 m): 9(L T L — T 8'): (TflT~= 1___=o.2os 8 _ - coshi7.57':(0.3m) TH.) = .205 (T5 -T_) + T“ = .205 {93°C -38°C) + 38°C 2 492°C The temperature at the tip of the fin is the temperature at the midpoint of the curved rod (492°C). (b). From Table 2.1, the heat transfer from the fin is: qfln =Mtanh(m L) where: M=JHPkA 9.: He: 13114392) (Tl—T.) M = i “T: (34%)(396 TIT—K) (.006 m)3 (93°C -38°C) = 4.653W q“n = 4.553w tanh (7.57%) (03111) = 4.56W The rate of heat transfer from the curved rod is approximately twice the heat transfer of the pin fin: qmd = 2% = 2 (4.56W) = 9.12 w mmmfixssm :- A circumferential tin of rectangular cross section, 3.7 cm OD and 0.3 cm thick surrounds a 2.5 cm diameter tube. The {in is constructed of mild steel. Air blowing over the fin produces a heat transfer coefficient of 28.4 W/rn2 K. It the temperatures of the base of the fin and the air are 260”C and 38°C, respectively, calculate the heat transfer rate from the fin. GIVEN - A mild steel circumferential tin of rectangular cross section on a tube - Tube diameter (DJ = 2.5 cm = .025 m - Fin outside diameter (DD = 3.7 cm = 0.037 m - Fin thickness (t) = 0.3 cm = 0.003 m - Heat transfer coefficient (hc) = 28.4 W/m’ l< - Fin base temperature (T) : 260°C - Air temperature (T_.) = 38°C FIND The rate of heat transfer from the fin (qr...) ASSUMPTIONS - The system has reached steady state - The mild steel is 1% carbon steel - The thermal conductivity of the steel is uniform - Radial conduction only (temperature is uniform across the CTOSS section of the fin) - The heat transfer from the end of the fin can be accounted for by increasing the length by half the thickness and assuming the end is insulated. SKETCH mt v 0,-3.7“. PROPERTIES AND CONSTANTS Thermal conductivity of 1% carbon steeltk) = 43 W/m K at 20°C SOLUTION The rate of heat transfer for the fin can be calculated using the fin efficiency determined from the efficiency graph for this geometry, Figure 2.17. Continued . . . The length of a fin (L) = (D, - DINZ = 0.006 m The parameters needed are: ri=,[2_)‘=0_‘[25m r°=%+L=O.125m +.006m=.0185m 3 w \1 a I 1. 2(28.4_. '-’ (ra + g — ri)’ 7 =[.0815m + '00?“ -.0125m “'K) * ro " (43%)(003mX0185m —.0125m) = 0.176 (r hi)”- : 0.0185m + .0015m =16 ° 2 ' .0125m From Figure 2.17, the fin efficiency for these parameters is: n{=98% The rate of heat transfer from the fin is: 2 qfin:n[EcAfin(Ts-T.) =ntE¢Zfl to 4"; ‘riz (T.'T.) qfin=(.98) (28.4%) 21t[(0.085m + 0.0015m)2 -{0.0125m)’](260°c —38°C) =9.46 w A turbine blade 6.3 cm long, with cross-sectional area A = 4.6 x 1134 m3 and perimeter P 2 0.12 In, is made of stainless steel (k = 18 W/m K). The temperature of the root, T“ is 482°C. The blade is exposed to a hot gas at 871°C, and the heat transfer coefficient 11 is 454 W/mz K. Determine the temperature of the blade tip and the rate of heat flow at the root of the blade. Assume that the tip is insulated. GIVEN - Stainless steel turbine blade - Length of blade (L) = 6.3 cm = 0.063 m » Cross-sectional area (A) = 4.6 x 10* in2 - Perimeter (P) = 0.12 m - Thermal conductivity (k) = '18 W/m K - Temperature of the root (T) = 482°C - Temperature of the hot gas (T) = B71“C - Heat transfer coefficient (h) = 454 W/rn2 K. FIND (a) The temperature of the blade tip (T L) (b) The rate of heat flow (q) at the root of the blade ASSUMPTIONS - Steady state conditions prevail - The thermal conductivity is uniform - The tip is insulated - The cross-section of the blade is uniform ~ One dimensional conduction SKETCH Ta = 51th A224 (A): 4.6.! .6" M’- ?emasrer. (P)= on». 13-. 452% LII 6.3 on Continued... SOLUTION (a) The temperature distribution in a fin of uniform cross-section with an insulated ti p, from Table 2.1, is: 4541!(o.12m) 3 =_—C°5h [ma‘ "01 where:m = E}: = ________"‘ K =31.1i a, cosh {m L) A islxmsxiwmz) m 3 =1“ - T“ At the blade tip, x = L, therefore: 9:_ = TL“T. = cosh [m(0)l = 1 F; T‘—T_ cosh (m L) cosh (In L} T -T a _ c. = , .. = 8710C + 482 C 871 C = 86an L “+ m L) C05 111. cosh (81.1%)(0063mll (b) The rate of heat transfer from the fin is given by Table 2.1 to be: q = MtanhfinL} where M = ‘lhcPkA 8‘ M = 4S4%(0.12m)(18%)[4.6x104m2) (482°C -371°C) = -261W q = (-261W)tanh[81.1.;.(0.063m)] = —261W (out of the blade) COMMENTS in a real situation the heat transfer coefficient will vary over the surface with the highest value near the leading edge. But because of the hibh thermal conductivity of the blade the temperature at any section will be esentially unifonn. ...
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This note was uploaded on 04/08/2008 for the course ENGR 302 taught by Professor Gad-el-hak during the Spring '08 term at VCU.

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HW4solution - In order to increase the rate of heat...

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