assignchap11-18-25

assignchap11-18-25 - PROBLEM 11.18 KNOWN: Concentric tube...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: PROBLEM 11.18 KNOWN: Concentric tube heat exchanger with area of 50 In; with operating conditions as shown on the schematic. FIND: (a) Outlet temperature of the hot fluid; (b) Whether the exchanger is operating in counterflow or parallel flow; or can’t tell from information provided; (c) Overall heat transfer coefficient; (d) Effectiveness of the exchanger; and (e) Effectiveness of the exchanger if its length is made very long SCHEMATIC: Operating conditions Concentric tube HXer, A = 50 m2 Hot fluid (h) Cold fluid (:2) Capacity rate, kaK 6 3 Inlet temperature, “C 60 30 Outlet temperature, °C -- 54 ASSUMPTIONS: (1) Negligible heat loss to sun‘oundings. (2} Negligible kinetic and potential energy changes, and (3) Constant properties. ANALYSIS: From overall energy balances on the hot and cold fluids, find the hot fluid outlet temperature q:cc (Tc:o_Tc,i):Ch (Thn _Th.o) (l) 3000 WIK (54—30)K : 6000 (ISO—Thy) Th,o = 48°C < (h) HXer must be operating in counter-flow (CF) since Tim < Tm. See schematic for temperature distribution. (c) From the rate equation with A : 50 mg, with liq. (1} for q, q 2 CC (Tao ’ch ) = UAATEm (2) a M 60—54 K— 48—3 K arfm— “1 M3 J l ( 0) —l().9°C (3) Fm(AT1fAT2) E’n{6/18) 3000W/K(54—30]K:U><50 m2x10.9K U=132Wlm21K < {d} The effectiveness, from liq. 11.20. with the cold fluid as the minimum fluid. CC = Cmin. 8_ q _ Came—Tm) :(54—3[))K qmax Cmin (Th‘i—chi) (60—30)K 20.8 < (e) For a very long CF HXer. the outlet ofthe minimum fluid. Cmm = CC. will approach TM. That is. q *9 Cmin (Too _Tc.i ) "9 iilmax E :1 < PROBLEM 11.25 KNOWN: Cooling milk from a dairy operation to a safe—to—store temperature. Tim : 13°C. using ground water in a countcrflow concentric tube heat exchanger with a 50—min diameter inner pipe and '3 overall heat transfer coefficient of 1000 Wim‘K. FIND: (a) The UA product required for the chilling process and the length I. of the exchanger. (h) The outlet temperature ofthe ground water. and (c) the milk outlet temperatures for the cases when the water flow rate is halved and doubled,- using the UA product found in part (a) SCHEMATIC: , I U = 1000 WImE-K Vh = 250 IiterJh D : 50 mm Thlo S. 13°C 9: = 0.72 m3fh To = 10°C ASSUMPTIONS: (1} Steady—state conditions, (2) Negligible heat loss to surroundings and kinetic and potential energy changes, and (3) Constant properties. PROPERTIES: Table A—6, Water (To = 287' K, assume T00 : 18°C): ,0 : 1000 kg 1' 1113. CD = 4187 J/kg-K; Milk(given): 921030 kg/m3, ep =386011kg-K. ANALYSIS: (:1) Using the effectiveness-NTU method. determine the capacity rates and the minimum fluid. Horfluia’, milk: mh = ,9th = 1030 kglrn3 x250liter/hxltj'3m3fliterx1h13600 5:0.0715 kgfs Ch : r'nhch =0.n715 kg/sx38601/kg-K=276 wa Coldfluid, water: cC = r'ncec =1000 kg2'1113x(0.72f3600 m3/s]><41871/kg-K = 337 er It follows that (3mm = Ch. The effectiveness ofthe exchanger from Eqs. 11.19 and l 1.21 is 8: q =Mzt38-6—Im: qmax ijn[Thai-‘Tcgi) (38.6-10)K 0.895 (1) The NTU can be calculated from Eq. 10.30b. where CT = Cminfcnmx = 0.330, NTU= ] time—‘1 (2) CT—l cCr—l' NTU : in [ 0.895 H-l “flu—J = 2.842 0.895 X0.330—l 0.330—1 PROBLEM 11.25 (Cont) From Eq. 11.25. find UA [1M]: NTU'Cmin :2.842x276W/K:785 er < and the exchanger tttbe length with A = rt 1)Lis t.=[UA]/:zou = 785 W/K/rrODSU m>< 1000 w; m2 -K :50 m < (b) The water outlet temperature, ’1‘“). can be calculated from the heat rates. Ch (Thi "1119) : Cc (Too ‘Tc.i) ' (3) 276 W! K (38.6—13JK : 837 W/ K (rtO — t0ji< TuU =18.4°C < (c) Using the foregoing Eqs. (1 — 3) in the MIT workspace. the hot fluid (milk) outlet temperatures are evaluated with UA : 785 W/K for different water flow rates. The results. including the hot fluid outlet temperatures. are compared to the base case. part (a). Case Cc. (W/K) Tm (0C) Ti“, (0C) 1, halved flow rate 419 14.9 25.6 Base, part (a) 837 13 18.4 2, doubled flow rate 1675 12.3 14.3 COMMENTS: {1) Frotn the results table in p‘dl'1{C), note that ifthe water flow rate is halved. the milk will not be properly chilled. since T“, = 149°C > 13°C. Doubling the water flow rate reduces the outlet milk temperature by less than 1°C. (2) From the results table, note that the water outlet temperature changes are substantially larger than those ol‘the milk with changes in the water flow rate. Why is this so‘? What operational advantage is achieved using the heat exchanger under the present conditions? (3) The water thermophysical properties were evaluated at the average cold fluid temperature, Tc : (Tm +1139]! 2. We assumed an outlet temperature of 18°C, which as the results show. was a good choice. Because the water properties are not highly temperature dependent, it was acceptable to use the same values for the calculations of part {c}. You could, of course. use the properties function in lHTthat will automatically use the appropriate values. ...
View Full Document

Page1 / 3

assignchap11-18-25 - PROBLEM 11.18 KNOWN: Concentric tube...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online