assignments11and12

assignments11and12 - PROBLEM 17.50 KNOWN Flow rate inlet...

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Unformatted text preview: PROBLEM 17.50 KNOWN: Flow rate, inlet temperature of atmospheric air passing through a duct of prescribed diameter and length maintained with a constant heat flux; average convection coefficient. FIND: (a) The heat transfer rate, q, and the mean temperature of outlet air Tmp; (b) Surface temperature at the tube inlet TSJ and outlet T53; (0) Sketch the axial variations of TS and Tm with distance from the inlet x; also sketch their variations for the more realistic case in which the local convection coefficient varies with x. SCHEMATIC: 53E) q§=1000 WIm2 Tm,Ts Tm,i=20°C HHH HHHl p = 1 atm l rh = 0.005 kg/s -—° F = 25 W/m2-K Ts,i Constant h |—>x ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible kinetic and potential energy effects; no shaft work, (3) Fully developed flow conditions, and (4) Constant properties. D=50mm L=3m PROPERTIES: Table HT —3, Air (assume Tm =340 K, 1 atm): cp = 1009 J/kg-K. ANALYSIS: (a) From an overall energy balance on the tube, the total heat transfer provided by the applied surface heat flux is q=q§AS =qgnDL=1OOOW/m2(nxo.050 m)x3m=471.2W < From an overall energy balance on the fluid, Eq. 17.48, q = mop (Tm, 4m) T =T -+q/mc =20°C+47l.2W/0.005kg/sx1009J/kg-K m,o m,1 p Tm, =113.4°C < (b) Since the convection coefficient is uniform within the tube, the difference between the tube surface and mean fluid temperatures will be constant q; = 3(Ts 'Tm) (TS — Tm)=q's'/h = 1000 W/m2 /25 W/m2 -K = 40°C Hence the tube surface temperatures at the inlet and outlet are T5,, = 60°C Tsyo =153.4°c < (c) The axial variations for Tm and Ts shown in the sketch above are both linear with distance x. However, if h is no longer constant because of entrance region effects, the surface temperature distribution is no longer linear so that TS — Tm is smaller near the inlet and increases with distance until fiilly developed conditions are achieved. PROBLEM 17.51 KNOWN: Flow rate, inlet and outlet temperatures of atmospheric air passing through a duct of prescribed diameter, length, and surface temperature. FIND: (a) The heat transfer rate, (b) Log mean temperature difference, ATlm; (0) Average convection coefficient for the air flow; and ((1) Sketch axial variations of TS and Tm with distance from the inlet; comment on key features. SCHEMATIC: D=50mm (LEE) rh = 0.04 kg/s Tm'i = 60°C p = 1 atm ASSUMPTIONS: (l) Steady-state conditions, (2) Negligible kinetic and potential energy effects; no shaft work, (3) Fully developed flow conditions, and (4) Constant properties. PROPERTIES: Table HT-3, Air (Tm = (Tm; + Tm, )/2 = 318 K, 1 atm): cp = 1008 J/kg-K ANALYSIS: (a) From the overall energy balance on the duct, Eq. 17.48, the heat transfer rate from the air to the tube is q = mop (Tm, — Tm,i)=0.04 kg/s x1008 J/kg-K(30 — 60)°C = —1210 w < (b) The log mean temperature difference, Eq. 17.58, is _ . _ O _ _1 O ATim = ATO ATl = (30 15) C (60 5) C =27-30C < 1n(ATo/ATi) ln(15/45) (c) From the rate equation, Eq. 17.57, and knowledge of the heat rate and ATlm, find the average convection coefficient, Clconv = hAs AT1m With AS = 1rDL, and substituting numerical values, 3— qconv _ 1210w —-———————=9.4W/m2-K < 7rDL ATlm ”(0.150 m)10m x 27.3 K (d) The temperature distributions for TS and Tm(x) as a function of distance from the inlet are shown in the Schematic above. The tube wall temperature is constant, however, the fluid temperature changes exponentially. The log-mean temperature represents the mean difference between the wall and fluid temperatures. PROBLEM 17.52 KNOWN: Ethylene glycol flowing through a coiled, thin walled tube submerged in a well-stirred water bath maintained at a constant temperature. FIND: Heat rate and required tube length for prescribed conditions. SCHEMATIC: J71h=0.01kg/s T = 55°C T Tm,,-=85°c c. ’"I" "’6" Thin-walled Well-sflrred T 'I‘ube, D=5mm wafer ba‘l‘h, 5 7;, =25 °C ASSUMPTIONS: (l) Steady-state conditions, (2) Tube wall thermal resistance negligible, (3) Convection coefficient on water side infinite; cooling process approximates constant wall surface temperature distribution, (4) Negligible kinetic and potential energy effects; no shaft work, (5) Constant properties, (6) Negligible heat transfer enhancement associated with the coiling. PROPERTIES: Table HT -4, Ethylene glycol (Tm = (85 + 35)°C/2 = 60°C = 333 K): cp = 2562 J/kg-K, u = 0.522 X 10-2 N‘s/m2, k = 0.260 W/m‘K, Pr = 51.3. ANALYSIS: From an overall energy balance on the tube, qconv = m cp (Tmp ‘Tm,i ) = 0.01 kg/sx2562 J/kg(35 —85)° C = —1281 W. (1) < For the constant surface temperature condition, from the rate equation, As = (Iconv / HATflm (2) AT 0 o 35 — 25 0 AT = AT -AT- M 0=[35—25 C— 85—25 C]/€ =27.9 C. 3 6m ( o 1) nATi ( ) ( ) 1185—25 () Find the Reynolds number to determine flow conditions, ' x . k ReD= 4m _ 4 001 g/s 2813. (4) 7r Dfl _ 7r><0.003 mX0.522x10'2N-s/m2 Hence, the flow is laminar and, assuming the flow is fully developed, the appropriate correlation is fip=h—D=3.66, h=Nu£=3.66x0.260i/0.003m=317W/m2-K. (5) k D m-K From Eq. (2), the required area, AS, and tube length, L, are As = 1281 W/317 W/m2 ~Kx27.9°C = 0.1448 m2 L=AS/7r D=0.1448m2/7r(0.003m)=15.4m. < COMMENTS: Note that for fully developed laminar flow conditions, the requirement of Eq. 17.41 is satisfied: (L/D) / ReD Pr = (153/0003) / (813 X 51.3) = 0.122 > 0.05. Note also the sign ofthe heat rate ‘lconv when using Eqs. (1) and (2). PROBLEM 17.65 KNOWN: Diameter, length and surface temperature of condenser tubes. Water velocity and inlet temperature. FIND: Water outlet temperature evaluating properties at Tm = 300 K. SCHEMATIC: L = s m “fir-”’9' D = 25.4 mm Tmp 1* H Um = 1 m/s Tm,i = 290 K a '- T5 = 350 K ASSUMPTIONS: (1) Negligible tube wall conduction resistance, (2) Negligible kinetic and potential energy effects; no shaft work. PROPERTIES: Table HT-5, Water (assume Tm = 300 K): p = 997 kg/m3, cp = 4179 J/kg-K, u = 855 x 10'6 kg/s-m, k = 0.613 W/m-K, Pr = 5.83. ANALYSIS: From the energy balance relation, Eq. 17.55b, Tm, = Ts — (TS —Tm’i )exp [—(nDL/r'ncp )h]. Evaluating properties at Tm = 300 K, find the Reynolds number as _ pumD _ 997kg/m3(1m/s)0.0254m _ RCD H 855x10‘6 kg/s~m 29,618 The flow is turbulent, and since L/D = 197, it is reasonable to assume fully developed flow throughout the tube. From the Dittus-Boelter correlation, Eq. 17.64 with n = 0.4 since Ts > Tm, NuD =o.023 Re4D/ 5 Pro‘4 = 0.023(29,618)4/5 (5.83)0‘4 = 176 E = NuD (k/D)=176(0.613W/m-K/0.0254m)= 4248 w/m2 -K. The mass flow rate is . _ 2 / _ 3 2 _ m — pum ED 4 — (7r/4)997 kg m (lm/s)(0.0254m) — 0.505 kg/s. Substituting numerical values into the energy balance relation, find 7r(0.0254m)5m(4248W/m2 K) = 323 K = 50°C < 0.505kg/s(4179J/kg-K) Tm,o = 350K—(60K)exp — COMMENTS: Finding Tm,0 = 50°C, the average mean fluid temperature is Tm = (290 + 323)K/2 = 307 K vs. the assumed value of 300 K. The analysis using the proper Tm yields Tm,0 = 51.7°C. While the difference is only 1.7°C, it is good practice to use the proper value for Tm at which to evaluate properties. PROBLEM 17.82 KNOWN: Vertical plate maintained at a uniform temperature while suspended in quiescent air. FIND: (a) Heat transfer rate from the plate by free convection, (b) Boundary layer thickness at the trailing edge of the plate, and (0) Compare results for heat transfer rate and boundary layer thickness with those for forced convection with an air stream having a velocity of 5 m/s. SCHEMATIC: L = 0.25 m 0’; qhonv qconv TS - . —° b — \ co — 25°C Plate Too = 2500 x / 1 L = 0.25 m Ts = 70°C u = 5 m/s q°°rw x T on ——>Y Free convection (a,b) Forced convection (c) ASSUMPTIONS: (1) Steady-state conditions, (2) Plate is at uniform temperature, (3) Room air is quiescent for free convection condition, and (4) For forced convection, Rex,C = 5 x 105. PROPERTIES: Table HT—3, Air (Tf= (TS + Tag/2 = 320.5 K, 1 atm): v = 17.95 x 10'6 mZ/s, k = 0.02783 W/m-K, or = 25.53 X 10'6 mz/s, 13 = 1/Tf= 0.00312 K-l, Pr = 0.704. ANALYSIS: (a) The heat transfer by free convection from both sides of the vertical plate is given by the rate equation Clizonvz2 E L(TS_T°0) where the average convection coefficient can be estimated from the Churchill-Chu correlation, Eq. 17.74. M 0.387 mi“ 1% =—= 0.825+ L k 8/27 [1+(O.492/Pr)9/16:l Find first the Rayleigh number, Eq. 17.19 or 17.66, g/3(Ts—TOO)L3 va RaL = _9.8 m/52x0.00312 K‘1(70—25)K(0.250 m)3 7 RaL— =4.69><10 17.95x10_6 m2/sx25.53x10_6 mZ/s Substituting numerical values, find 5, Continued PROBLEM 17.82 (Cont) 7 1/6 2 _ 0.387(4.68x10 ) NuL= 0.825+ 9/16 3/27 248.75 [1+(0.492/0.704) ] E=§ITIEL =Wx48.75=5.4 W/m2 -K L 0.250 m and then the convection heat transfer rate per unit width of the plate, qgonv=2x5.4 W/mz-Kx0.250 m(70—25)°C=122 W/m < (b) Since RaL < 109 (see Eq. 17.66), the boundary layer flow is laminar and the thickness may be estimated from Eq. 17.76 6 —1/4 —=6G /4 L (m ) where GrL = RaL/Pr. Substituting numerical values, find Ra ““4 469x107 4/4 6=6L —L =6x0.250 m '——— =23.5 m < 4 Pr 4x0.704 (c) For the forced convection case shown on the schematic, the boundary layer flow condition is characterized by the Reynolds number, ReL=u°°L= 5 m/5x0.25 m =6.96><104 v 17.95x10_6 mz/s Since ReL < Rex,c = 5 x 105, the flow is laminar over the entire plate and Eq. 17.26 is the appropriate correlation for estimating the average coefficient, —HL 1/2 NuL =T=0.664 RelL/2 Pr1/3 =0.664(6.96x104) 1/3 ) (0.704 =155.9 H=5fi 0.02783 W/m-K = x155.9=17.3 W/mZ-K L L 0.250m Using the rate equation, find qgonv=2x173 W/m2 -Kx0.25 m(70—25)°C=390 W/m < Eqs. 17.21 and 17.24 are appropriate for estimating the hydrodynamic and thermal boundary layers, respectively, —1/2 ) =4.7 m < 5=5L Reil/z =5x0.250 m(6.96x104 5t =5Pr‘1/3 =4.7 mm(0.704)‘“3 :53 m < Continued PROBLEM 17.82 (Cont) COMMENTS: (1) Note that the convection heat rate by forced convection with a modest air stream velocity (1100 = 5 m/s) is more than three times that by free convection. The hydrodynamic boundary layer thickness by forced convection is nearly 5 times smaller than by free convection. For free convection, the hydrodynamic and thermal boundary layer thicknesses are of the same extent. For the case of forced convection with air under laminar flow conditions, they are also of the same extent, but only because Pr is nearly unity. (2) It is important to recognize that radiation exchange between the plate and surroundings could be significant. If 8 = 1 and Tsur = Tm, the linearized radiation coefficient from Eq. 15.9 is hrad =80(Ts +Tsur)(T52 +T521'JI')=7'2 W/m2 'K As such, the heat rate for the free convection situation is more than doubled. For the forced convection situation, the heat rate is increased by only 50%. (3) The foregoing equations for parts (a) and (b) —free convection case - were used in the IHT Workspace to calculate the average convection coefficient, heat rate per unit width of the plate, and the boundary layer thickness. Note the use of the Properties function for air to evaluate the properties at the film temperature. II Free convection case, Parts (a) and (b) /* Results, average convection coefficient correlation, heat rate, boundary layer thickness GrL NuLbar RaL delta_mm hbar q'conv 6.661E7 48.75 4.69E7 23.48 5.424 122 l* Properties, air, Tf = 320.5 K Pr Tf alpha beta k nu 0.7041 320.5 2.553E—5 0.00312 0.02782 1.795E-5 */ ll Convection rate equation q'conv = 2 * hbar * L * (Ts - Tinf) //W/m Il Vertical plate, Churchill-Chu correlation NuLbar = ( 0.825 + 0.387 * RaL" (1/6) / ( 1 + (0.492 / Pr )" (9/16) )“(8/27) )"2 // Eq. 17.74 NuLbar = hbar * L / k // Rayleigh number RaL = g * beta * (Ts- Tinf) * L"3 / ( nu * alpha) 9 = 9.8 // Gravitational constant, m"2 / 3 II Boundary layer thickness, Pr = 0.7, RaL <= 10"!) delta / L = 6 * (GrL / 4 )" (-1/4) // Boundary layer thickness, m delta_mm = delta * 1000 // mm RaL = GrL * Pr // Eq. 17.19, 17.66 II Input variables Tinf = 25 + 273 // K Ts = 70 + 273 // K L = 0.25 // m II Air property functions : From Table HT-3 // Units: T(K); 1 atm pressure nu = nu__T("Air",Tf) // Kinematic viscosity, m"2/s k = k_T("Air",Tf) // Thermal conductivity, W/m-K alpha = alpha_T("Air",Tf) // Thermal diffusivity, m"2/s Pr = Pr_T("Air",Tf) // Prandtl number beta = 1/T f // Volumetric coefficient of expansion, K"(-1); ideal gas Tf = (Ts + Tinf) / 2 // Film temperature, K Continued ..... PROBLEM 17.84 KNOWN: Dimensions of vertical rectangular fins; temperature of fins and quiescent air. FIND: Rate of heat transfer from each fin by free convection. Comment on the effect of boundary layer formation on the spacing between the fins. SCHEMATIC: Quiescent ai Too = 27°C PROPERTIES: Table HT-3, Air (Tf= (TS + Tw)/2 = 325 K, 1 atm): v = 18.41 x 10'6 mZ/s, k = 0.02815 W/m-K, a = 26.20 x 10'6 mz/s, 13 = 1/Tf= 0.003077 K'l, Pr = 0.7035. ASSUMPTIONS: (1) Steady-state conditions, (2) Fin surfaces are isothermal, (3) Negligible heat transfer from fin edges, and (4) Ambient air is quiescent. ANALYSIS: The convection rate equation considering both sides of the fin surface is expressed as Clconv =2h(LXH)(TS —T°0) to estimate h, use the Churchill-Chu correlation, Eq. 17.74, 2 N— 0825 0.387 RalL/6 “L” ' + 8/27 [1+(0.492/Pr)9/16:l where the Raleigh number, Eq. 17.19 or 17.66, is written with the characteristic length as H, =gfl(Ts_Too)L3 va RaH 9.8 m/s2x0003077 K'1(77—27)K(0.150 m)3 7 Ray; =——————————=1.055x10 18.41x10‘6 mz/sx26.20x10_6 mz/s and the average convection coefficient is 7 1/6 0.387(1.055x10 ) N—u = 0825+ L )9/16]8/27 =31.66 [1+ (O.492/0.7035 — _0.02815 W/m-K NuH x31.66=5.94 W/mz-K 0.150m 5:5 H Continued ..... PROBLEM 17.84 (Cont) Substituting numerical values into the rate equation, find the heat transfer rate qconv =2x5.94 W/m2 -K(0.020x0.150)m2 (77-27)K Clconv =1-78 W < If the fins are too close together, the boundary layers on adjoining surfaces will coalesce and the heat transfer rate will decrease. If the fins are too far apart, the finned surface area becomes small, and the heat transfer rate will again decrease. To estimate the spacing, calculate the boundary layer thickness at the leading edge. Since RaH < 109, the boundary layer is laminar, and the thickness at the trailing edge can be estimated from Eq. 17.76, 3:6(Gm/4)—1/4=6(RaH/4 Pr)_1/4 1055 107 W4 5=6x0.150m ——' x =20.5 mm 4x0.7035 In order that the boundary layers not coalesce, we require S = 25 z 40 mm. However, by making S some fraction of 25, the added number of fins per unit area of the base may offset any adverse effect of boundary layer interference. COMMENTS: (1) Radiation exchange with the surroundings may not be important in this situation because the fin surfaces mostly “see” each other. (2) The optimum spacing between fin surfaces is an important consideration in the design of fin arrays. This topic is treated in advanced heat transfer courses. PROBLEM 17.88 KNOWN: Horizontal, circular grill of 0.2 m diameter with emissivity 0.9 is maintained at a uniform surface temperature of 130°C when ambient air and surroundings are at 24°C. FIND: Electrical power required to maintain grill at prescribed surface temperature. SCHEMATIC: Quiescent ai Tm. = 24°C Grill surface, D = 250 mm = o .= qconv Ts 130 ca. 0.9 \ P e ASSUMPTIONS: (1) Room air is quiescent, (2) Surroundings are large compared to grill surface, and (3) Sides and bottom of grill are well insulated. PROPERTIES: Table HT—3, Air (n: (To, + Ts)/2 = (24 + 130)°C/2 = 350K, 1 atm): v = 20.92 x 106 mZ/s, k = 0.030 W/m-K, a = 29.9 x 10*5 mZ/s, 0 = 1/Tf. ANALYSIS: The required electrical power, Fe, is equal to the heat transfer from the grill, which is due to free convection with the ambient air and to radiation exchange with the surroundings. Pezqus[H(TS—Too)+ga(Ts4— 51”” (1) Calculate RaL from Eq. 17.19, RaL :gfl(TS—TOO)L%/va where for a horizontal disc from Eq. 17.77, Lc = As/P = (nD2/4)/1tD = D/4. Substituting numerical values, find R 9.8m/s2(1/350 K)(130—24)K(0.25 m/4)3 aL I 20.92x10“6m2/sx29.9x10‘6m2/s Since the grill is an upper surface heated, Eq. 17.79 (Case C, Table 17.6) is the appropriate correlation, 21.158x106. — — 1/4 61/4 NuL =hLLc/k:O.S4RaL =0.54 1.158x10 =17.72 EL :N—uLk/Lc =17.72x0.03OW/m-K/(0.25 m/4) 28.50W/m2 -K. (2) Substituting from Eq. (2) for E into Eq. (1), the required electrical power is P6 = £(0.25m)2 [8.50 (130— 24)K +0.9x 5.67 x10_8 2W ((130+ 273)4 —(24 + 273)4 )K4:l 4 mZ-K m -K4 Pe :44.2W+46.0W=90.2W. < COMMENTS: Note that for this situation, free convection and radiation modes are of equal importance. If the grill surface were highly polished such that a z 0.1, the required power would be reduced by nearly 50%. , ..._1._.i.._imr____..__...-.. ,.._....,T W__W_..T._.__._,W__._..-m Wm...“ PROBLEM 18.3 KNOWN: Plate maintained at 200°C by imbedded electrical heating elements; exposed surface has an emissive power of 1200 W/mz, irradiation of 2500 W/mz, and reflectivity of 30%; exposed surface experiences air flow having a freestream temperature of 20°C with a convection coefficient of 15 2 W/m ~K. FIND: (a) The radiosity, J; (b) The net radiation heat flux leaving the surface, qfad, in terms of the radiosity and irradiation; (c) The combined convection and radiation heat flux leaving the surface; (d) Represent a surface energy balance schematically, and label all the radiation processes; and (e) The . . ,, 2 . . . . electrical power requ1rement, Pelee (W/m ) to malntain the plate under these condltrons. SCHEMATIC: G = 2500 WIm2 Ta, = 20°C h =15 W/mZ-K E = 1200 WIm2 P = 0-3 qgonv \ T5 = 200°C / f T Pas ASSUMPTIONS: (1) Steady-state conditions, (2) Plate is opaque, diffuse; and (3) Plate backside is perfectly insulated. ANALYSIS: (a) The radiosity is the sum of the emissive power, E, and the reflected irradiation, Gref, where p is the reflectivity, J=E + Gref 2E + pG=1200 W/m2 +0.3x2500 W/m2 =1950 W/m2 < (b) The net radiation heat fluxing leaving the surface, Qiad , in terms of the radiosity and irradiation is qlr’ad = q’r’ad,out = [Bout _ Efn :lrad : J — G q;ad=1950W/mZ—2500W/mz=—550W/m2 < (c) The combined convection and radiation heat flux leaving the surface is qcv,rad = qlcv + qlfad qgv =h(TS —T°°)=15 W/m2 -K x (200—20)°C=2700 W/m2 qcv+qrad=2700 W/m2+(—550W/m2)=2150W/m2 < (d, e) The surface energy balance is shown schematically in terms of J and G, q;ad, and qcv+rad- q’r'ad \ f (13v . I The surface energy balance, Efn — Eout = 0, expressed as I I I I I 2 Pelee=J—G+qcv=qrad+qcv=qcv+rad=2150W/m < COMMENTS: Note how the directions for the heat fluxes and radiation quantities in the schematic are consistent with the energy balance relations. PROBLEM 18.4 KNOWN: Evacuated, aluminum sphere (D = 2 m) serving as a radiation test chamber. FIND: Irradiation on a small test object when the inner surface is lined with carbon black and at 600 K. What effect will surface coating have? SCHEMATIC: Aluminum sphere, D=2m Small +651“ surface, A1 Carbon b/ack coaflng, 7;=600K ASSUMPTIONS: (l) Sphere walls are isothermal, (2) Test surface area is small compared to the enclosure surface. ANALYSIS: It follows from the discussion of Section 13.3 that this isothermal sphere is an enclosure behaving as a blackbody. For such a condition, see Fig. 18.12(c) and Eq. 18.6, the irradiation on a small surface within the enclosure is equal to the blackbody emissive power at the temperature of the enclosure. That is G1 =Eb (TS)=O-TS4 G1=5.67x10_8W/m2-K4(600K)4 =7348W/m2. < The irradiation is independent of the nature of the enclosure surface coating properties. COMMENTS: (1) The irradiation depends only upon the enclosure surface temperature and is independent of the enclosure surface properties. (2) Note that the test surface area must be small compared to the enclosure surface area. This allows for inter-reflections to occur such that the radiation field, within the enclosure will be uniform (diffuse) or isotropic. (3) The irradiation level would be the same if the enclosure were not evacuated since, in gen...
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