Question

Figure A series of ultrafiltration units is used to concentrate a dye-water solution from an input fresh feed of 10% dye (by mass) to an entire process output of 32% dye (by mass). The fresh feed flow rate to the process is 50.0 kg/min. The figure is shown above. The system can be considered to have a dye in solution with water and the dye remains soluble over the range of concentrations in the process. The ultrafiltration units produce streams C and D with dye concentrations of 1.0% and 2.5%, respectively. Streams E and G are more concentrated in the dye. Design information on ultrafilter #1 states that 1.0 kg of solution C is produced for every 4.0 kg of stream B entering the filter, Information on ultrafilter #2 is not known and is not necessarily the same as #1). A flowmeter in the stream connecting ultrafilter #1 and #2 (that is, stream E) reads 45 kg/min. Assume that no reaction occurs and that the system is continuous and steady-state. a) Determine the total flow rate (kg/min) of stream C. b) Determine the dye concentration (mass fraction) and total flow rate (kg/min) of stream B. c) Determine the total flow rate (kg/min) of stream H exiting the system.

Answer 1

soli Given F xd=0.37 E of tr A = 50kg UF# 45 kg/min BLUF#IL D X = 0.37 Xj= 0.015 Hy=0.0) * 1 kg of solution a produced for query 4 kg of stream B. Note: F 8H have same dye concentration because splitter I won't change the concentrations, around of #1 :- mass balance [Given, & = 4 B = C+ E Bac +45 B - B/4 +45 B(34) = 45 - 45x4 Goky/min B - around mixing mass balance point A + F - B 50 +F = 60 F = 60 -50 = 10kg/min flow rate of stream I (a) Total c= 3/4 = Go = 15 kg /min

E (6) B Total How rate of stream B = 60kylmin dye balance around mixing - point 60 X Xd = 50X0.1 + 0x0.3 5 + 3y = 60 x 2d 8.7 13,6% 0.136 = Xd - around the system mass balance A = (+ D + H > 0 50 = 15+ D + H around the system dye balance D X 0.025 + # X 0.3 15X0.01 + y (2) 50 x 0.1 = from (1) HOD=-H+ 25 (3) substitut eqn (3) h (2) 50X0.1 = 15x0.01 + (35-H) X0.075+ #x0.37 5 = 0,15 + 0,875- 0.02H +0.34 H 5 = 1.025 + 0. 304 o 3.975 0.34 : 13.25 Kalmin H = 3.975 0.3 .