Question

6 Problem 4 (25%) A 2-shell passes and 4-tube passes heat exchanger is used to heat glycerin (Cp.gly = 2447) entering at 15 °C by hot water (Cp water = 4180 f.), which enters at 90°C. The thin-walled inner tube has a 4 cm diameter and a total length of 15 m. The hot water flows through the tube at a total rate of 0.265 kg/s, and the glycerin through the shell at a rate of 0.6 kg/s. The convection heat transfer coefficient is h; = 700 W/m².C on the inner surface of the tube and ho = 250 W/m2.C on its outer surface. Cold glycerin me = 0.6 kg/s 15°C în = 0.265 kg/s Hot water 90°C a. (20%) Determine the wtal leat transfer rate in the heat exchanger. The Figures needed are attached in the next pages. W b. (5%) What are the outlet temperatures of cold and hot fluids? Tcout °C Th.out °C

Answer 1

U n = 2 pat ㅗ L hi ho 250 ( 700 U - 2447 to Tcp = 15°C fund ( water) A = nxPX HDL 2X4 X X. 04.X S. AE 15.08 m² Overall heat Transfer Transfer coefficient tot [t 184 .211 W/m²°C for coldey fluid (glycerin) =1460.2W mg = 0.6kg/s E = mg cpg 68.2W CPg / kegic hot 1107.74 1/19°C (Cwl < Cc Grald facto Cmin 7545 Cmag Rate the maximum heat transfer Emin (Trg - Tes) 1107.7 x(90-15) = 83077.5 W 03.08 KW Cwsmw Coa 0.265 Kg/s 4100 لا mw = Coo 90°c cc) C for that fund is minimum . c amax

NTV of of this heat exchanger INTU Net Transfer unit) u Asurface c min 184.211 X 15:00 i107.7. NTV = 2:5078 find & Coffective we E for tid it 2 may [crsuster en de econeguru as we know (€, NTU) we can to find a H.EU J HE can find (1) By chart ( i used foom Heat of mass transfer cengel Fig 11.27 (2) By formula off effectiveness you counter yow & counter |-e-(1-curo 1-ceci-CNTU we get [E = 0.776.] - We know effectives ivers (E) of Qactual LE Q maximum Qactual E amoximum 0.776 X 83:08 KW 64.47 kW Heat exchangerdhe a = mq (pa (Tc, -Tcz) = mw Cow (Thi Tha) -

д- How we know heat transfer rate. mg Cpg (Tez Te ) 0 •6X24+7 (T2-15 Те, = sü: 4 °C , +7 — x 10 Aus в — ThCru (1», – Th) = 3-74 °C Ina Aus

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