Question

Steam Generator air (inlet) - 320°C P. = 100 kPa th =0.5 kg/s heat exchanger - water (inlet) T = 20°C Pw = 100 kPa m = 0.025 kg/s Problem sketch Solve with EES. Document all necessary balances The problem sketch illustrates a heat exchanger in which hot air is used to generate steam. Air enters the heat exchanger at 1a, in = 320C, Pa = 100 kPa, and ma -0.5 kg/s. Model air as an ideal gas with constant specific heat capacity (Ra = 287 J/kg-K, Cpva = 1005 J/kg K). Water enters the heat exchanger at Tuan - 20°C and P -100 kPa. The mass flow rate of the water is my = 0.025 kg/s. The pinch point associated with the heat exchanger is on the hot side (i.e., where the air enters and water leaves). The approach temperature differences AT H 160 K. You may neglect pressure drop on the air and the water sides of the heat exchanger a. Determine the rate of heat transfer from the air to the water in the heat exchanger. b. Determine the total rate of entropy generation in the heat exchanger c. Determine the effectiveness of the heat exchanger d. If the approach temperature difference of the heat exchanger were reduced to ATH OR the heat exchanger became "perfect') then would the entropy generation in the heat exchanger be reduced to zero? Justify your answer.

Answer 1

Pr Inle (3) 4 Ta,aut Ta, in F 320'c Pa- 0KPS watey (in let) 2 T in Pw 20'c - -100)epg mw- Ga025 lg apprmach mp dit 160K 160 c Ta,aut Twsin 160 ta,out 2o -tha Ta aut 7 Hrome fes Rat Meal (ATu Tain Ta ast (320-180 0.5 X 100Sn (-Ta, out t Tain) O'SX (005 x (-180 1 E SC 320) 703 S0 wat 7o 35 Ku Lat enhely choye for Censt ma Cp Ln Tout Tin (AS)air 1804273 320+273 -135.324 Wwatt K

enheby Chay fo loake Touk, w Tin w Tout o 0.025x 4180 XIn SCCD r balomee m, a(AT)ar (320-184) (Taut-20) &7.32°C Tout, w 87-3L+ 273 (As x 4180 xn 2 001 0.2 5 20t 273 216.13 W CAS)w 2 A)(03) (as)nE 2)6. 12 00.806 enhepy geetin AS)ME XTO atmepkne lemp C8 300K 1 80 806x 300 Sgen Syen 24.242 Kw

S02.5 S O SX1005 forcosky m C O.25x 418oT m)smtt S02.5 J - effechicnes CPr Fo 35x103 (mcp)smll x (AT)mon 70 35x10 AT merr Tin a- Tin w 502.5X (320- 20) O467 E 46.67 / Thappeak d Naw Touta= Tin 20'c 293 K 1 O257 4180x (Tort-23) O.5K1005x (3 20-20) Tout164.26 (ASa (AS) 2an(sy maC In Tout a out w mwcf n Tin a Tin w O.25x4180 x ln 164 23 +223 20123) 20t233 O.5X 1005 xn 320t273

-354.223 ys 3 AS)ME CAs)WE Loctt Gy.027 To Sgen 64.027 x 30 19.208 KW |Sgen a Ende Py genehin wll hot loceoleced fo but 9t wll b Redluree Zero
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