exam3extraprob - PROBLEM 8.70 KNOW N Inner and outer radii...

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Unformatted text preview: PROBLEM 8.70 KNOW N: Inner and outer radii and thermal conductivity of a teflon tube. I-‘lowratc and temperature of confined water. Heat flux at outer surface and temperature and convection coefficient of ambient air. FIND: Fraction of heat transfer to water 11 . . . . . 911 and temperature of tube outer surface. T m SCHEMATIC: 010103" _9.p E0 EHé'Q/rf) ZfiLk r0 =13mm (hiAil‘I Teflon, 9:! 7;” k=0.55WlmK m ’ ASSUMPTIONS: (I) Steady—state conditions. (2) Fully~devclopcd flow, (3) One—dimensional conduction, [4) Negligible tape contact and conduction resistances. —( PROPERTIES: Table A43, Water (Tm = 290K}: it I 1080 x 10 ) kgrls-m, k = 0.598 Whit-K. Pr = 7.56. ANALYSIS: The outer surface temperature follows from a surface energy balance T 4‘... r. - r (2?! equ”=L_l+# (1102111101.) (111(10 Mimi: L-k]+(11’21r 11m.) Tso * Tin ”:1 '1‘. 40., —_ q 10( 5‘0 )+(r0fk]ln(rbfri)+(1‘01’r'1]1"'hi With ReD =4 rr11"[71' Dy) :4[0.2kgfs)f[rr[0.02in)1080xl(}_6 kgr's -m] :1 1.789 the flow is turbulent and liq. 8.60 yields r1]- : (lo-Di )0023ReD’S P104 = (0.598 W.-"m K1002 m)(0.023 )(I 1,?09)4"5(756)U'4 : 2m: wrm2 -K. l-lcnce T SI 0 —290 K 2000 Via-«1113 = 25 was-1112 -K (TM, 300K )+ 7 (00131111035 wrm -K)111 (1.3}+(1.3).-‘[2792 \\-'.-'111-1K) and solving for T510.- Tsfi. = 308.3 K. < The heat flux to the air is L113: h11lTs.11— Too] 2 25 vii/1112 -K (3003—300) K : 207.5 w..--'1113. Hence. q"; 111’: (2000- 207.5) w1-‘11121'2000 won? 2 0.00. < COMMENTS: The resistance to heat transfer by convection to the air substantially exceeds that due to conduction in the tellon and convection in the water. Hence. most or the heat is transferred to the water. PROBLEM 3.76 KNOWN: Fluid enters a thinrwalled tube of SAmm diameter and 27m length with a flow rate of 0.04 kgfs and temperature of Tm‘i = 85°C.; tube surface temperature is maintained at T8 = 25°C; and, for this base operating condition, the outlet temperature is Tm0 = 311°C. FIND: The outlet temperature it'the flow rate is doubled? SCHEMATIC: T5 = 25°C Tmib = 31.300 ' ”Tr”? 7 Tmli = 85°C M)” " ' m1, = 0.04 kgls Thin-walled tube _ or _ _ -———-——-"i D = 5 mm rnn = kaJ ie—————-——_ 1- ‘ 2 m ASSUMPTIONS: [1) Flow is fully developed and turbulent, (2) Fluid properties are independent of temperature, and (3) Constant surface temperature cooling conditions. ANALYSIS: For the bro-e operating condition [b]. the rate equation. Eq. 8.42b. with C = rhcp. the capacity rate, is TS _Tm._.0)b = exp _ PLEb Ts ’de Cb (1) Substituting numerical values, with P = TED, find the ratio, hb J'Cb, 25 v 31.1 — m _ exp[n' x0005 rn><2 m(hb rcb )] Eb rcb = 72.77 m_2 For the new operating condition (n), the flow rate is doubled, C” = 2Cb. and the convection coefficient scales; according to the Dittus—Boelter relation. Eq. 8.60, hi Rc%8|l rho-8 En 2 203m, and (En Jen): 20-8 r2(Eb/Cb) (2) Using the rate equation for the new Operating condition. find mzexp[_PLHn ]:exp[-PLXO.87l(hb/Cb)] (3) 15 ‘TmJ C11 25 -—Tm!0) —” : cxp[—jr><0.005 m><2 m><0.87l X7277 III—2] 25—85 PROBLEM 11.33 KNOWN: Concentric tube heat exchanger operating in paratlel flow (PF) conditions with a thin-walled separator tube of lOO-ntm diameter; fluid conditions as specified. FIND: (a) Required length for the exchanger; (b) Convection coefficient for water flow. assumed to be fully developed; to) Compute and plot the heat transfer rate. q, and fluid inlet temperatures, TM and '1" as a function of the tube length for 60 S L S 400 m with the PF arrangement and overall coefficient [1.1: 200w/m2 - K). inlet temperatures (rm = ’22an and 'rm- = 30°C). and fluid flow rates from Problem 1123; ((1) Reduction in required length relative to the value found in part (a) ifthe exchanger were operated in the counterflow (CF) arrangement; and (e) Compute and plot the effectiveness and fluid outlet temperatures as a function of tube length for 60 S L S 400 tn for the CF arrangement of part (c). (2.0! SCHEMATIC: o E u=2oowrm2-K r :225 0 Q) T "" hrmzmot’c \Ll ;.(s;;;;p ASSUMPTIONS: (1) No losses to surroundings, (2) Negligible kinetic and potential energy changes. (3) Separation tube has negligible thermal resistance, (4) Water flow is fully developed, (5) Constant properties, (6) Exhaust gas prepetties are those of atmospheric air. PROPERTIES: Table A-4, Hot fluid. Air(l attn. T = (325 +l[}0)°C I2 = 436 K): on 2 1019 J/kg‘K: Table A-6. Cold fluid. Water T: (30 + 80)°C l2 2: 328 K): p = l/vr: 9854 kg/m". cl, : 4183 Jfkg‘K‘ k = 0.648 W/m-K. u = 505 x 10"” N‘s/ml. Pr = 3.58. ANALYSIS: (a) From the rate equation. E-q. 11.14, with A = TIDL. the length of the exchanger is l. =q/U-erarrn‘PF. (l) The heat rate follows from an energy balance on the cold fluid. using Eq. 11.7. find q = mccc (Tm —'1‘L.,i ) : 3kg/sx41831/kg-K(80—30)K = 627.5xtn3w. Using an energyr balance on the hot fluid. find n1“ for later use. mh = q/Ch ("on firm ) = 627.5x103 w 10191;"ikg . K (225 —-100)K : 4.93. kg/s (2) For parallel How. Eqs. 11.15 and 11.10. an — er; _ (3:5 — 30)“ C — (loo —: 80)° t'.‘ A'nm H. — . _ — _ J _ 768°C. ' math-at: 111(225 — 30),..(100—30) Substituting numerical values into Eq. ( l ). find .= 027.5 x103 xiv/”gooair-”m2 -K(rr><().1m)7o.8K : 130m. < Continued... PROBLEM 11.33 (Cont.) (b) Considering the water flow within the separator tube, from Eq‘ 8.6, ReD = 4zn/7rDtt : 4x3kgf’s/(rr XOmeSOleO—6 N/s- n12 ) : 75.638. Since Rm > 2300. the flow is turbulent and since flow is assumed to be fully developed. use the Dittus- Boelter correlation with n = 0.4 for heating, NuD = 0023113958 Pr0‘4 = 0.023(75.638)0'8 (358)0'4 = 306.4 k . h : Nun — = 306.4x0.648W/tn<K/(0.lm) =1985W/m2 ‘K. < D (c) Using the [HT Heat Exchanger loaf, Concentric Tube. Parallel Flow, Effa‘tivcncss relation, and the Properties Tool for Water and Air. :1 model was developed for the PF arrangement. With U = 200 \Wln“-K and prescribed inlet temperatures. '1".,_,- = 225°C and Tc.i = 30°C, the outlet temperatures. Th0 and Tm and heat rate, q. were computed as a function oftuhe length I... Parallel llow arrangement i; 140 1 a: ' l a | = | .— . . 1....___I.__+. .. a l l l .l ! 3 Its : i ; . . . E l I l a; l : E . . —"'————"'."—' E i l i l 6 90 - _ .“i Q i ' i a: i g 65 H) D. E | .3 | . 4t)- . I ———¢; ' . 50 120 130 240 300 350 Tune length, L {m} ‘-’*— Cold outlet temperature. Tco (CJ “9-— Hot outlet temperature The {C} Heat rate. q‘tD‘N-d (W) As the tube length increases, the outlet temperatures approach one another and eventually reach Tm, : TN, = 816°C. (d) lithe exchanger as for part (a) is operated in T Counterflow (rather than parallel flow). the log mean temperature difference is Tm: 225 DC AT — A'l‘ ATl“ m C1: '— l ‘7'";' Too = 80 DC ‘ {HATI \12 r—\ Ix.) I | 2 80)—(|00—30) . —--.—= 103.0 (7. m [-225 — 801.5100 — 30 [_Jsing Lit]. (I). the required length is L:(:27.5><103WfEOOW/n12-K xn x0.lm><103.0l< 2 97 m. The reduction in required length 01' CF relative to FF operation is PROBLEM 11.33 (Cont.) AL = (LH; —L(_;I.-)/LPF 2(103797)/103= 5.8% < (C) Using the IHT Heat hitclronger Tool, Concentric Tube. Cotmterflow, hflec'tiveness relation. and the Properties Tool for Water and Air, 21 model was developed for the CF arrangement. For the same conditions as part (c). but CF rather than PF. the effectiveness and fluid outlet temperatures were computed as a function of tube length L. Counterflow arrangement 140 120 100 80 50 Tho, Too (0] or eps'too 40 20 -. . . 60 120 180 240 300 360 Tube length. L (m) #7 Cold oullet temperature. Too [8} —9— Hot outlel temperature. The (C! — Ettectrveness, eps‘t 00 Note that as the length increases. the effectiveness tends toward unity, and the hot fluid outlet temperature tends toward TW = 300C. Remember the heat rate for an infinitely long CF heat exchanger is (1...... and the minimum Fluid {hot in our case) experiences the temperature change. Tm - '1}... COMMENTS: (1) As anticipated. the required length for CF operations was less than for PE operation. (2) Note that U is substantially less than hl implying that the gas—side coefficient must be the controlling thermal resistance. PROBLEM 11.69 KNOWN: Inlet temperatures and flow rates of water (c) and ethylene glycol (h) in a shell—and—tube heat exchanger (one shell pass and two tube passes) of prescribed area and overall heat transfer coefficient. FIND: (:1) Heat transfer rate and fluid outlet temperatures and (b) Compute and plot the effectiveness, E and fluid outlet temperatures. Tm and Tim as a function ofthe flow rate of ethylene glycol. tin1 . for the range 0.5 S iiih S 5 kg/s. s SCHEMATIC: l 1;”,th = 2 kgts ' r One Shell Too pass and two tube passes Tam n'qc : 5 kgls Ethylene glycol ‘— Tw = 10 ”C. water l Tim A =15 m2. U = 300 wrmZ-K ASSUMPTIONS: (l) Negligible heat loss to surroundings. (2)I Negligible kinetic and potential energy changes, (3) Constant properties, and (4) Overall coefficient remains unchanged. PROPERTIES: Table 45, Ethylene glycol {Tm = 40°C.): c. = 2474 J/kg-K; Tabch-é, Water (Tm = 15°C): c. = 41 so J/kgK. ANALYSIS: (3) Using the erNTU method we first obtain Ch = (mhcp’h ) = (2kg/sx2474J/kg - K) : 4948 w/K cc =(meow):(5kg/sx41361/kg- K): 20,930w/K. Hence with Cmm = Ch : 4948 W/K and Cr = Cum/Cm. : 0.236, UA (sonwr’in2 -K)15m2 N'ru : =—-——=2.43. Cmin 4948 W] K From Fig. [1.16. 8 = 0.81 and from Eq. 11.23 q 2 ecmin (TM :13. ) = (181(4948 W/K)(60—10]K : 2x105 W. < From qu. 11.6 and 11.7, energy balances on the fluids, 5 0 2x10 W O Th.,:’1*hi—i=60 01,:1943 c < ' ' ch 494swn< S ., 2x10 w o rc(,=rci+i=10C++=19.ec. < 1 ‘ cC 20,930w,ri<; (b) Using the IHT Heal? Exchanger Tool, Shell and Tube. and the Properties Tool for Water and .8: 100 Ethylene Glycol. Tm. T1,... and E as a function of g so 4 m}. were computed and plotted. g :3 O r: At very low Cm-m. (low l‘hh ) note that E ——> 1 while E 2: _ Tim—Sr '1‘“... As mh increases. both fluid outlet F temperatures increase and the effectiveness Hot fluid llow rate. rndoln (kgis) decreases. + TeolC}, max fluid + Tho (C), min lltlld —~— Elfectlveness.eps‘100 ...
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