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HW 10 solution ME 114 S08

HW 10 solution ME 114 S08 - PROBLEM 11.45 50(01fw V19 KNOWN...

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Unformatted text preview: PROBLEM 11.45, 50(01fw V19 KNOWN: Single-pass, cross-flow heat exchanger with both fluids (water) unmixed; hot water enters at 90°C and at 10,000 kg/h while cold water enters at 10°C and at 20,000 kg/h; effectiveness is 60%. ' ' FIND: Cold water exit temperature, Tc'o. \i -= -90°C SCHEMATIC: J.“ 72,,- =10°c. vac MOW/<9”? 7,1,- : 90°C, —l> ‘ 7- mc=20, 000kg/h mh —D [,0 72'] :lonc é 8:0.60 Arr'angemenf‘ ASSUMPTIONS: (1) Negligible heat loss to surroundings, (2) Negligible kinetic and potential energy changes, (3) Constant properties. PROPERTIES. Table A- 6, Water (Tc =(10—1—40)°C/2~ 300K): cc: —4179 J/kgK; Table A—6, Water (Th (90+60)°C/2= 350K). ch=4195 J/kgK. ANALYSIS: From an energy balance on the cold fluid, Eq. 11 7, the outlet temperature can be expressed as Tao = Tc.i + Q/IIJC Cc The heat rate can be written in terms of the effectiveness and qmax. Using Eqs. 11.20 and 11.19, q: eclmax=ecmincrh,i_ TIc,i)- By inspection, it can be noted that the hot fluid 15 the minimum capacity fluid. Substituting numerical values, q = 8 (filh Ch) (Thi_Tc,i) , q = 0.6000,000kg/h/36OOS/h)419SJ/kg-K (90—10)°C = 559.3XIO3 W . The exit temperature of the coldwater is then 20,000 ' . _ ‘0 3600 kg/sx4179J/kgK—34.1C. <1 T“, = 10°C + 559.3x103 W/ COMMENTS: (1) The properties of the cold fluid should be evaluated at =(T H0+Tc1 /2= (34. 1+10)°C/2= 295K. Note the analysis assumed Tc 2,300K hence little error is incurred. For best precision, one should check Th and Ch. (2) From Fig. 11.18, the value of NTU could be determined. First evaluate the term _ . . _ 10,000x4195 _ Cmm/Cm‘“ —thn/mc C° ' 20,000x4179 ‘0'50 and with 8:0.60, find NTU: 1.2. 7'". Hm,“ litilhfililrllll 1 WWW 114115111011“ PROBLEM 11.46 KNOWN: HXer consisting of 32 tubes in 0.6m square duct. Hot water enters tubes at 150°C with mean velocity 0.5 m/s. Atmospheric air at 10°C enters exchanger with volumetric flow rate of 1 m3 /s. Heat transfer coefficient on tube outer surfaces is 400 W/m2 -K. FIND: Outlet temperatures of the fluids, Tao and The- SCHEMATIC: Tube.- 1:067", N=32 h0=400 w/mZ-K 7;],- =150°C 1.1,,7 = 0.5777/5 —'D , ,1 T=10°C fig D,'=10.2mm . 7;] _ ‘ 7; 00-125mm ASSUMPTIONS: (1) Negligible heat loss to surroundings, (2) Negligible potential and kinetic energy changes, (3) Constant properties, (4) HXer is a single-pass, cross-flow type with one fluid mixed (air) and the other unmixed (water), (5) Tube water flow is fully developed, (6) Negligible thermal resistance due to tube wall. PROPERTIES: Table A-4, Air (Tc, = 10°C: 238K, 1 atm): p = 1.2407 kg/ms; Table A-4, Air (assume Tc", 240°C, Tc =(lO+40)°C/2 =298K, 1 atm): cP = 1007 J/kg-K; Table A-6, Water (Assume T1,, = 140°C, Th = (140+150)°C/2 =418K): p = 1/vf = 1/1.0850§<10‘3 m3/kg, cp =4297 J/kg-K, u, =188x10‘6 N-s/mz, kf=0.688 W/m-K, Prf = 1.18. , ANALYSIS: Using the e-NTU method, first find the capacity rates. Ch =Ihh Cp.h =_(P Ac um)h New 1 7E , __3 2 m J ' W c =———-—x— 10.2x10 m) x0.5—x32x4297—-—7=5178-— h 1.0350x10‘3m3/kg 4 ( s kg-h K (20 :11c (3,,c =‘(p V)c op,c = 1.241175% 1m3/sx 1007 J/kg-K= 1249¥ . (1,2) In . Note that the cold fluid is the minimum fluid, Cc =Cnfin. The overall heat transfer coefficient follows from Eq. 11.5, 1 1 '1 U°A° [ h,Ai ~11vo ] < ) where hi must be estimated from an appropriate internal flow correlation. The Reynolds number for water flow is Re _ pum D, = (1/1.0850x10-3m3/kg)x0.5m/sx(10.2x1o-3m) D u - 188x10'6N-s/m2 = 25,002 . (4) Continued ..... PRGBLEM 11.46 (Cont) The flow is turbulent and since L/Di =O.6m/10.2><10"3m=59, fully developed conditions may be assumed. The Dittus-Boelter correlation with n: 0.3 is appropriate, h- D- NuD = —‘k—‘— = 0.023 ReOD'8 Pro-3 = 0.023 (25,002)0-8 (1.18)” = 79.7 hi = 35— NuD = W x 79.7 =_ 5376 W/mZ-K. Di 10.2x10'3 m Substituting numerical values into Eq. (3), find —1 1 ] = 366.6W/m2‘K. U0 = [(2222) _"__1__ + __ 10.2mm 5376 W/mz-K 400 W/mz-K It follows from Eq. 11.25, with A0 =N (7: Do L), that U” A“ =‘366.6 W Cmin mz-K From Fig. 11.19, noting that Cmin = Cc is the mixed fluid (solid curves), Cmixed Cmin Cc 1249 W/K Cum“, cmax ch 5178 W/K , and with N TU = 0.22 find 3.: 0.19. From the definition of effectiveness, Eq. 11.20, V q _ CIc (Tc,o;Tc.i) x(32x7tx12.5x10‘3mx0.6m)/1249—V—Iz— = 0.22. NTU= 8 = _..._—_.__ Clmax Cmin (Th,i"Tc,i) - Tm = T... + e (TM—Ted) = 10°C + 0.19 (150—10)°c = 36.6°C. ‘ 4 Equating the energy balances on both fluids, Cc (Tc.o—Tc.i) = Ch (Th.i—Th.o) 01' cc Th,o = Th,i — E: (Tc,o—Tc,i) 1249W/K (36.6-10)°C=143.5°c. ' ~ V ‘ ~ <1 _ 0 _ Th°‘1.50 C 5178W/K ' COMMENTS: (1) Note that the assumptions of Th'o'and Tc,o used in eValuating properties are reasonable. (2) Note that to calculate [11c from V, the density at 10°C is more appropriate than at Tc. ’QK PROBLEM 11.50 KNOWN: Single-shell," two-tube 2pass heat exchanger with surface area 0. 5 m2 and overall heat transfer coefficient of 2000W/m2 K; saturated steam at 100°C condenses on one side while water at a flow rate of 0. 5 kg/s enters at 15°C. * FIND: (a) Outlet temperature of the water, Tc'o, (b) Rate of condensation of steam, 151111. SCHEMATIC: 0779 shell—> fwa fube pass ASSUMPTIONS: (1) Negligible heat loss to surroundings, (2) Negligible, kinetic and potential energy changes, (3) Constant properties * PROPERTIES: Table A- 6, Steam (100°C, 1 atm.:) hfg= 2257 kJ/kg; Table A- 6, Water (Tc =(15+35)°C/2= 300K): cc=--41791/kgK. ANALYSIS: (a) Using the e-NTU method of analysis, recognize that the minimum capacity fluid is the cold fluid since for the hOt fluid, Ch ——> oo. See Fig. 11.9a. That is, €11,111 = rhc cc = 0.5 kg/s x4179 J/kg-K = 2090 W/K . It follows also that, NTU = AU/Cmin = 0.5m2 x 2000 W/mz-K/2090 W/K =- 0.43 . Using NTU=O.48 and C111i11/C11111X=0, find from Figure 11.16 that 8:0.39. Since an, is the minimum fluid, from Eq. 11.22 8 = (Tc,o—Tc,i)/(Th HR, 1) T111, -==Tc1 +e(T111— T01)= 15°C+0.39(100—15)°C= 48. 2°C. 4 (b) The rate of steam condensation can be expressed as Ifih = Cl/hrg - From Eqs. 11.19 and 11.20 C1 = 3 Qmax = E Cmin (Th.i_Tc,i) q = 0.39 x 2090 W/K (100—15)K = 69,284 W. Hence, the condensation rate is 'r'nh = 69,284 W/2257x103 J/kg = 0.031 kg/s . ' V , <1 COMMENTS: (1) Be sure to recognize why C11 -—>°°. Note also that me > rhh. (2) Note that Tc = (Tc'1+Tc,o)/2= (15+48.2)°C/2 2 305K. This compares favorably with the value 300K at which properties of the cold fluid were evaluated. PROBLEM 12.16 KNOWN: Isothermal enclosure of surface area, As, and small opening, A0, through which 70W emerges. ‘ , FIND: (3.) Temperature of the interior enclosure wall if the surface is black, (b) Temperature of the wall surface having £=O.l5. SCHEMATIC: ASSUMPTIONS: (1) Enclosure is isothermal, (2) A0 << A5. ANALYSIS: A characteristic of an isothermal enclosure, according to Section 12.3, is that the " radiant power emerging through a small aperture will correspond to blackbody conditions. Hence ' qrad = A0 Eb (Ts) = A0 GT? where qmd is the radiant power leaving the enclosure opening. That is, 1/4 1/4 70 . . T5: ‘1‘“ = ——————-2 W_8 2 4 =498K. ' <1 A06 . 0.02m x5.670><10 W/m -K . Recognize that the radiated power will be independent of the emissivity of the wall surface. As 1 long as A0 << As and the enclosure is isothermal, then the radiant power will depend only upon the temperature. COMMENTS: It is important to recognize the unique characteristics of isothermal enclosures. See Fig. 12.12 to identify them. > PROBLEM 12.18 KNOWN: Solar flux at outer edge of earth’s atmosphere, 1353 W/mz. FIND: (a) Emissive power cf sun, (b) Surface temperature of sun, (0) Wavelength of maximum solar emission, ((1) Earth equilibrium temperature. SCHEMATIC: Rs.e=1.5XIOI’I” 9” 3 ‘ Earf'h -s—’ ;.'\D =.12%<107m I, ASSUMPTIONS: (1) Sun and earth emit as blackbodies, (2) No attenuation of solar radiation enroute to earth (3) Earth atmosphere has no effect on earth energy balance. ANALYSIS: (a) Applying conservation of energy to the solar energy crossing two concentric spheres, one having the radius of the sun and the other having the radial distance from the edge of the earth’s atmosphere to the center of the sun 1}" 2 (AL? 0",” E (1:132)=41c R —E q 102% wide) 5 s 2 s_- ’ \‘P/GJKII $65)“) ' /\ ~ «r0 , L Hence l/~ 0V ‘ 11 7 2 ‘ 135— 4(1. 5x10 m— --.0 .65><109 In: x1353 W/m2 _ 6.302x107 W/mz . . 4 (1. 39x10 m) (b) From Eq. 12.28, the temperature of the sun is E ”4 . 6 302x107 W/m2 1-” 1 T5 = ~1- = —~—-_—8—'—2"—4'_- = 5774K . 4 o . 5.67x10 W/m -K (c) From Wien’s law, Eq. 12.27, the wavelength of maximum emission is C3 _.2897 6 pm K 7"“ T ‘ 5774K 0'50 ”m' - r {J (d) From an energy balance on the earth’s surface +9 & ,1 fix ((1 d > . E. (nDE) = q”.(1tD§/4> . - ——> 1? “U 61:15 ' [15th- 0W. [4 er «1) Hence, from Eq. 12.28, ' _ ~ GA (3. W! ‘54? 1353 W/m2 T = ______——.—— e 4><5.67x10"8 W/mz-K4 II- 1/4 5L5 = 4 0 COMMENTS: The average earth temperature is higher than 278 K due to the shielding effect of the earth’s atmosphere (transparent to solar radiation but not to longer wavelength earth emission). PROBLEM 12.23 KNOWN: Solar disc behaves as a blackbody at 5800 K. FIND: (a) Fraction of total radiation emitted by the sun that is in the visible spectral region, (b) Plot the percentage of solar emission that is at wavelengths less than 7» as a function of 71., and (c) Plot on the same . coordinates the percentage of emission from a blackbody at 300 K that is at wavelengths less than 7» as a function of 9»; compare the plotted results with the upper abscissa scale of Figure 12.23. ASSUMPTIONS: (l) Visible spectral region has limits 2», = 0.40 pm and 71.2 = 0.70 um. ANALYSIS: (a) Using the blacl<body functions of Table 12.], find F(o—o.). the fraction of radiant flux leaving a black surface in the spectral interval O-—->7\. as a function of the product XT. From the tabulated values for Fw—m with T = 5800 K, A: = 0.70 pm MT = 4060 um-K * Paw”): 0.4914 7», = 0.40 pm MT = 2320 ttm-K 110%) = 0.1245 _ Hence, for the visible spectral region, the fraction of total emitted solar flux is F (Ni->1: ) = 110%) — 11M!) = 0.4914 - 0.1245 = 0.3669 or 37% (b,c) Using the [HT Radiation Tool, Band Emission Factor, Fto-m are evaluated for the solar spectrum (T = 5800 K) and that for a blackbody temperature (T = 300 K) as a function of wavelength and are utted below. ~ HQ -> lam ‘1 da) (Va) ' - "WI-II w? 0.1 0.4 0.13 ‘z a 10 40 00 avelangth, la ~ (mum) —— Sf‘ck§3§$“%l“g 330:1??35115 "°“’ The left—hand curve in the plot represents the percentage of solar flux appr 'mated as the 5800 K- blackbody spectrum in the spezéal region less. than 91.. The right-hand curve rep ms the percentage of 300 K—blackbody flux in t spectral region less than 2». Referring to upper abscissa sc leof Figure 12.23, for the solar flu , 75% of the solar flux is at wavelengths shorter than 1 um. For the blackbody flux (300 K), 75% the blackbody‘ flux is at wavelengths shorter than 20 um. These values are in agreement with oints on the solar and 3OOK—blackbody curves, respectively, in the above plot. PROBLEM 12.2 KNOWN. A diffuse surface of area A: 10'4 m2 emits diffusely with total emissive power E: 5 x 104 W/m' . FIND: (a) Rate this emission is intercepted by small surface of area A2 = 5 x 10‘4 m2 at a prescribed ‘ location and orientation, (b) Irradiation G3 on A2, and (c) Compute and ,plot G2 as a function of the , separation distance r3 for the range 0.25 S r2 S 1.0 m for zenith angles 93 = O, 30 and 60°. SCHEMATIC: . n1 A2 = 5x10‘4 m2 n2 ‘—§;€ E ) - ’/’ 92 = 30° _‘/ ,/ ,/ / r2 = O 5 in A1 = 10-4 m2 E1 = 5x104 WIm2 ASSUMPTIONS: 7(1) Surface A. emits diffusely, (2) A may be approximated as a differential surface * - area and that A / r, << 1 ANALYSIS: (a) The rate at which emission from A; is intercepted by A2 follows from Eq. 12.5 written on a total rather than spectral basis. ql_,2 =Ie1(9,¢)A,cos(-3,dm,_,. (1) Since the surface A1 15 diffuse it follows from Eq. 12.13 that I=e.1(e:¢) 113.1: E 1/7t ' . . (2) The solid angle subtended by A2 with respect to A1 is dm,_, 4A,- cose, /r,2 . . 1 (3) Substituting Eqs. (2) and (3) into Eq. (1) with numerical values gives j,“ '7 e 4 2 1 -4 2 u / E1 .1 lcosel -é;?-S—2=5—Xl-O—W/—m—x(10‘4m-XCOS6O°)X 5X10 m xjcosBO sr (4) r2- “Sl- (0.5m)‘ ql—>2 =15,915W/m2sr><(5 x10‘5m3‘)x1.732x10'3sr=1.378><10'-‘W. < (b) From section 12, 2.3, the irradiation is the rate at which radiation is incident upon the surface per unit surface area, fix, 1.378x10‘32W - ' G, _ = ————.— = 2. 76W 5 < ' A2 5x104m (:1: ~~~~~ ( ) 7:) Usingwffie H workspace withh t hiygmq quations, the G2 was computed as a function of the separation distance o 361 ted ze it angles The results are plotted below. Continued... PROBLEM 12.6 KNOWN: Furnace with prescribed aperture and emissive power. FIND: (a) Position of gauge such that irradiation is G = 1000 W/mz, (b) Irradiation when gauge is tilted 0,, = 20°, and (c) Compute and plot the gage irradiation, G, as a function of the separation distance, L, for the range 100 S L S 300 mm and tilt angles of 0,, = 0, 20, and 60°. ’ SCHEMATIC: I Detector, ‘ ’- ‘4 Ad = 1.6x10-5 m2 Furnace aperture, Af _ ...... ‘_ ____________ 7 E = 3.72x105 WIm2 . Irradiation, D=20 mm . G=1ODOW/m2 ASSUMPTIONS: (l) Furnace aperture emits diffusely, (2) Ad << L2. ANALYSIS: (a) The irradiation on the detector area is defined as the power incident on the surface per unit area of the surface. That is G=qf—)d/Ad qfqd =IeAf 03590qu _ . . . (1’2) where q H d is the radiant power which leaves Afand is intercepted by Ad. From Eqs. 12.2 and 12.5, cod_f is the solid angle subtended by surface Ad with respect to Af, . to“r =Adcosed/L2. (3) Noting thatvsince the aperture emits diffusely, I5 = E/n: (see Eq. 12.14), and hence G = (is/1:)Af cos,e(Ad cosfld /L2 )/Ad ' (4) Solving for L2 and substituting for the condition 9 = 0° and 0,, = 0°, , , L2 = Ecosecosed Af/TEG. . p (5) L = [3.72 x' 105 W/m2 x 27500 ><10'3)2 mZ/Tc ><1000W/m3]”2 = 193 mm. < (b) When Gd = 20°, qu (will be reduced by a factor of cos 0,, since a)“ is reduced by a factor cos 0d. Hence, G = 1000 W/mzx cos ed =1000W/m2><cos 20° =, 940 W/mz. ‘ < (c), » ing the IHT workspace with Eq. (4), G is computed and plotted as a function of L for selected 0d. Note that » decreases inversely as L2. As expected, G decreases with increasing 0,, and in the limit, approaches zer . 0d approaches 90°. Irradiation, G (VV/m"2) i Separation distance, L (mm) —9- thetad = 0 deg —- thetad = 20 deg —E- lhetad = 60 deg ...
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