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# hw9 SOL - SOlmX‘PCM MEH4 Fag PROBLEM 14 l KNOWN Flow rate...

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Unformatted text preview: SOlmX‘PCM MEH4 Fag PROBLEM 14:. l KNOWN: Flow rate, inlet temperature and desired outlet temperature of water passing through a tube of prescribed diameter-and surface temperature. FIND: (a) Required tube length, L. for prescribed conditions. (b) Required length using tube diameters over the range 30 S D S 50 mm with ﬂow rates m = 1, 2 and 3 kg/s; represent this design information graphically. and (c) Pressure gradient as a function of tube diameter for the three ﬂow rates assuming the ' tube wall is smooth. SCHEMATIC: // Tm’o = 75 c=c D =0.04m rh = 2 kgls ' /. / Tm: 25 oc/v ASSUMPTIONS: (I) Steady—state conditions, (2) Negligible potential energy, kinetic energy and ﬂow ' work changes. (3) Constant properties. PROPERTIES: Table AIS, Water (Tm = 323 K): c], = 4181 J/kgK, [.1 = 547 x 10'6 N-sfrna, k = 0.643 W/m-K. Pr = 3.56. ' ANALYSIS: (5.) From Eq. 8.6. the Reynolds number is 4m 4x 2kg/s =——u_—=1.15 105. ' ' 1 nDtL r:(0.0r-im)547'><10‘6 N‘s/m1 x . U Hence the flow is turbulent. and assuming fully developed conditions throughout the tube, it follows from the Dittus—Boelter correlation, Eq. 3.60, - ‘ 0.643W/m- K 0.04 m a = )“5 (3.56)“ = 6919 w/m2 - K (2) 0.023Reg'5 Pr“ = 0.023(1.16x105 l: D From Eq. 8.42a. we then obtain .—' enAT AT 2k's'41811k -K£n25‘C 75°C - Lzmw=_w¥_)=mﬁm_ < nDh n(0.04m)6919 w/m-- K . Tube IE gih, L {m} equations, the required length flow rate is computed and pl Tuba diameter. D (mm) —G-- Flow rate. mdot = 1 kgls ““- rndoi = 2 kgls "ﬁ— mdui = 3 kgls Continued... PROBLEM if? 2/ KNOWN: Flow rate and inlet temperature of water passing through a tube of prescribed length, diameter and surface temperature. * “mm. mm '1' sewn, ww- :3; FIND: (a) Outlet water temperature and rate of heat transfer to water for prescribed conditions, and (b) Compute and plot the required tube length L to achieve Tm found in part (a) as a function of the surface temperature for the range 85 S T, S 95°C. SCHEMATIC: rh = 2 kgfs /. / Tm,i= 25 UV . ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Negligible kinetic energy, potential energy and flow work effects, (4) Fully deve10ped flow conditions. '- PROPERTIES: Table A6. Water (Tm = 325 K): cFl = 4182 Mtg-K, u = 528 x 10'Ei N-s/ml, k = 0.645 W/m-K, Pr = 3.42. ' ANALYSIS: (3) From Eq. 8.6, the Reynolds number is Hence the flow is turbulent, and assuming fully developed conditions throughout the tube, it folloivs from the Dittus-Boelter correlation, Eq. 8.60, ' 01023 Ref,” pro-4 = W E )415 D ' ' 0.04m E: 0.023(1.21x 105 (142)"-4 : 7064W/m1- K. From the energy balance relation, Eq. 8.42b, Tm = Ts — (T5 — Tm'i)exp[—- nDL h] me P TEX 0.04mx 4m Tm , = 90c— (90°c— 25°C)exp ———.—-—-— ' mtg/ex 41821/kg- K 7064_W/m2 - K] = 47.5"c < From the overall energy balance, Eq. 8.37, . q = mcpfrm, _ m.,)= 2kg/s x 41321/kg- K(47.5—— 25)“c =138’kw. (b) urbulent Flaw, along with the T Correlations Tool, Internal Flow, for fullee energy balance and re 'ons uscd above, the requir gth, L, to achieve Tm = 449°C (see comment 1 below) as a function o to perature is computed and plotted below. Continued... PROBLEM 44: 3 KNOWN: Parallel air ﬂow over a uniform temperature, heated vertical plate; the effect of ﬁee convection on the heat transfer coefﬁcient will be 5% when GrL/Rei = 0.08. FIND: Minimum vertical velocity required cf air ﬂow such that free convection effects will be . less than 5% of the heat rate. SCHEMATIC: ASSUNlPTlONS: (1) Steady-state conditions, (2) Criterion for combined free-forced convection determined from experimental results. - PROPERTIES: Table A-4, Air (T4=(T,+r,.)/2=315K, 1 atm): .v=17.40x10"6 'm’l/s, ANALYSIS: To delineate ﬂow regimes, according to Section 9.9, the general criterion for predominately forced convection is that ' GrL/Rei <8: 1 . (1) From experimental results, when GrL/Rei =0.08, free cenvection will be equal to 5% of the ' total heat rate. . ' For the vertical plate using Eq. 9.12, T ——T L3 2 3 sB< 1 2) z 9.8 m/s x1/315Kx(60—~25)Kx(0.3m) = 9311x104 _ Gr = 2 1‘ ‘ v2 (17.40x10-G Eng/S): ( ) For the vertical plate with forced convection, um. L Li... (0.3m) Re = = = 1.724x104 u, . 3 L V 17.42410"6 1112/5 ' C ) By combining Eqs. (2) and (3), G 7 IL = 9.711x10 =0.08 Rea [1.724x104 11.12 ﬁnd that 1.1.. = 2.02 m/s . ‘ <1 That is, when u, 2 2.02 m/s, free convection effects will not exceed 5% of the total heat rate. PROBLEM 14:41, KNOWN: Initial overall heat transfer coefﬁcient of a ﬁre-tube boiler. Fouling factors following one year’s application. FIND: Whether cleaning should be scheduled. S CHEMATIC : Rgfaoorsm-Z-K/W R£°=aooost-K/W u (#9) =400 WImz-K Products 9F combustion ASSUNIPTIONS: (1) Negligible tube wall conduction resistance, (2) Negligible changes in ha and 11],. ANALYSIS: From Equation 11.1, the overall heat transfer coeﬁicient after one year is 1 l l 3‘E+a Since the ﬁrst two terms on therright-hand side correspond to the reciprocal of the initial ‘overall coefﬁcient, 1 1 n — = —~——~—— + 0.0015 + 0.0005 In“ -K = 0.0045 m2'K/W U 400 W/mg'K ( ) M U = 222 W/rnz'K. The reduction in U, and hence in heat transfer, is signiﬁcant (45%). The tube surfaces, particularly the inner surface, should be cleaned. + Eli + R’fp' COMWIENTS: Periodic cleaning of the tube inner surfaces is essential to maintaining efﬁcient ﬁre~tube boiler operation. ‘ ‘ / PROBLEM ‘075 KNOWN: Type-302 stainless tube with prescribed inner and outer diameters used in a cross-flow heat exchanger. Prescribed fouling factors and internal water flow conditions. FIND: (a) Overall coefficient based upon the outer surface. U“. with air at TD =15°C and velocity Vu- = 20 mis in cross-flow; compare thermal resistances due to convection, tube wall conduction and fouling; (it!) Overall coefﬁcient. Um with water (rather than air) at Ta = 15°C and velocity Va = 1 m/s in cross- flow; compare thermal resistances due to convection. tube wall conduction and fouling; (c) For the water-air conditions of part (a), compute and plot UEl as a function of the air cross-ﬂow velocity for 5 5 Va 5 30 mfs for water mean velocities of am = 0.2, 0.5 and 1.0 mfs; and (cl) For the water-water ‘ ' conditions of part (b). compute and plot U, as a function of the water mean velocity for 0.5 S um s 2.5 mls for air cross-flow velocities of Va = l, 3_and 8 m/s. SCI-IEMATIC: . . "f'a = 0.0004 mlellW TUbB, 83302. kW Fouling facmrs "m: 0.0002 mZ-KIW Tm, = 75 00 “mi = 0.5 m/s ﬁisﬁiiﬁﬁiiﬁé‘ééﬁi‘iiﬂﬁﬁiﬁtf:m’éﬂtwﬂﬁé"hTim—rﬁaai-‘E T i 100%?! mm Va i 20 m/s D! = 22 mm ASSUMPTIONS: (l) Steady-state conditions, (2)-Fully developed internal flow, PROPERTIES: Table A], Stainless steel, A151 302 (300 K): kw _= i5.1 W/m‘K'. Table A6. Water . (Tm = 343 K): p. = 974.8 leg/m3. pi = 3.746 x 10“ N‘s/m2. 10: 0.668 W/m‘K. Pr. = 2.354; Table A4, Air (assume Tm, = 315K, 1 atm): k0 = 0.02737 W/m-K. v.3 = 17.35 x 10'6 rug/s, Pro = 0.705. ' ANALYSIS: (0) For the Water-air condition. the overall coefficient, Eq. i 1.1, based upon the outer area can be expressed as the sum of the thermal resistances clue to convection (cv), tube wall conduction (w) and fouling (f): l/Ua An = RID! = Rcvj + R” +RW + Rf.nl+Rcv.o RCVJ = l/Ei AI Roma 2 I/Eo AB RH = iii/Al . Riﬂe: fin/Au and from Eq. 3.28.; Rw =1n(DD/Di)/(21rLkw) The convection coefﬁcients can be estimated from appropriate correlations. Continued... PROBLEM 11.2 (ContJ Estimating H, : For internal ﬂow. characterize the flow evaluating thermophysicai properties at Tm; with - D . . 2 Rem=um i: Dim/5x09 2m 1:28.625 ' vi 3.746x10 N.s/m-/974.3kg/m- For the turbulent flow. uise the Dittus-Boelter correlation. Eq. 8.60, " " ' m”... €0.023Reg'f‘, Pr,M I 0.1! 0.4 Num = 0.023(28,625) (2.354) = 119.1 E. =0qui kl/D, =119.1x0.668W/m2 -K/0.022m =3616W/m1-K Estimating ho : For external flow. characterize the flow with Tmr’ ' I Rem! 9 Re]: n =ﬂ=w=3lilg4 ' v0 17.35x10 m—/s R _ . tr evaluating thermophysical properties at Tm, = (Tm + Tall when the surface temperature is determined from the thermal circuit analysis R result. . w (Tml "— To )/Rtot = (Tao "- To )/Rr.v.o Rip . ‘ T Assume Tm, = 315 K, and check later. Using the Churchill-Bernstein - 5'0 correlation, Eq. 7.57, ﬁnd RIM“ . ' T MS 0 ._.. 0.62 a ”2 P ”3 R 5’“ ' ' . NUD.5=0.3+—_‘—e%ﬁ 1+[3‘h-J it? - [1 + (DA/Pro) ] ~32'000 ' _ . 062(31 194)”2(0 705)”3 ‘31124 5'“ “5 NUDm =0'3+—_;_—;n—TW|:1+[ ’ J ' [1+(O.4/O.705)“ -] 32,000 ' “rm... =102.6 ‘13., = ﬁn... k0 /DD = 102.6 x 0.02737 W/m _. K/0.027m = 104.0 W/m- K Using the above values for E, and En , and other prescribed values, the thermal resistances and overall coefficient can be evaluated and are tabulated below. Rani Rm Rw R120 Rem Uri. Riot (K/W) (Ii/W) (W) (W) (KIW) (W/m'K) (K/W) 0.00436 0.00578 0.00216 0.00236 0.1 134 92.1 . 0.128 The major thermal resistance is due to outside (air) convection. accounting for 89% of the total resistance. The other thermal resistances are of similar magnitude. nearly 50 times smaller than R CV.O' condition, the method of analysis folio that??? part (a). For the internal flow, the estimated convection coe ’ nt is the same a . For an assumed outer ﬁlm coefﬁcient, T“, = 292 K, the conveCW water flow condition V0 2 1 m/s and '1“.I = 15°C. find Continued... ...
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