Question

Heat Exchanger V1 = 8.80 L/s acetone n2 mol/s acetone P2 = 1 atm Saturated acetone vapor enters a heat exchanger at a rate of 8.80 L/s at 1 atm, and emerges from the heat exchanger as a saturated liquid at the same pressure. Water at 1 atm enters the heat exchanger at 11.5°C and at a rate of 8.65 mol/s. Assume that the heat exchanger operates adiabatically with respect to its surroundings. Calculate the temperature of exiting water by following the steps below. P1 = 1 atm Т1 At what temperature does acetone enter the heat exchanger? 14 =8.65 mol/s water P4 = 1 atm n3 = 8.65 mol/s water P3 = 1 atm T3 = 11.5°C Number What is the heat of vaporization of acetone at this temperature? Number kJ/mol Use the ideal gas law to calculate the moles of acetone entering the heat exchanger. Scroll down for more questions. Number mol/s

Answer 1

soli- from the given data, Antoine's Equation is given by logio (Psat (mmHg )) - A-8)(T[{3+C) -0 DI Tole Lacetone 17.613 | 156669 / 273.419 7 put p=latm= 760 mm Hg weget, Tsat = 56.87676 Acetone Enters at 36.3767 for the heat of vaporezation, Vant Holfs Equation is given by In ) At Pi=760mm Hg, To = 56.37678= 56.3767+273 = 329.37671 det T2=60c, from Antoine's Equation P2 = 855.9287 sunity Tz = 60+273= 333k In (454943 Warsztato) Duty - 29910-614 J Mol = 29.91 kalmoj

By Ideal gas law, PV = nRT P=latm= 101325Pa V=8.80 LS = 0.0088 m? Is T: 5637670 - 5637677273 - 329.3767k 101325 80.0088 = nx8.314 * 329.3762 n=0.3256089 molls Heat lost by acetore = nx heat of vapourization = 0.3256089 x 22.91K01 mol = 9.7357 KŠIs Cov) = 97357 kJ/5 (68) kW Assuming there is no heat loss to the surrounding. heat gained by water heat lost by acetone = 1228 9.7357kW nco (Tout-Tin) = 9.7357 865 molls X0.0754 kB imol -k x (Tout-19) = 9.7357 kw lout = 28.08349 Tout = 28.08c