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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 counterflow (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 WImEK
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/kgK; Milk(given): 921030 kg/m3, ep =386011kgK. ANALYSIS: (:1) Using the effectivenessNTU 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/kgK=276 wa Coldﬂuid, water: cC = r'ncec =1000 kg2'1113x(0.72f3600 m3/s]><41871/kgK = 337 er It follows that (3mm = Ch. The effectiveness ofthe exchanger from Eqs. 11.19 and l 1.21 is 8: q =Mzt386—Im:
qmax ijn[Thai‘Tcgi) (38.610)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 Hl “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. ...
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 Fall '08
 Rothamer
 Heat Transfer

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