The concentration will change on the feed side from inlet (feed) to outlet (retentate) because water is continuosly permeating, see figure
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1 SOLUTION TO EXAM SIK2010 H2002 / M-B Hägg PROBLEM Revers Osmosis The concentration will change on the feed side from inlet (feed) to outlet (retentate) because water is continuosly permeating, see figure a) Checking the specification: c1 c2 R c2 c 1 = ) / 2 = ppm NaCl (average log mean value could also have been used) = where c1 is the average concentration from inlet to outlet, c 1 c f c r = 2 Calculates c 2 : c.995 = c2 = c p = 2ppm. NaCl < 00ppm OK! b) Calculating the osmotic pressure over the membrane at inlet and outlet: nrt π = V m where n = (c 1 -c 2 )2 / M NaCl and V m is volume of water (Note we multiply by 2 because NaCl is completely dissociated; Na + and Cl - Further; the density given at 25 C should read: density of water, not seawater. This makes only a minor difference for the calculations, and both ways are accepted.) Also remember ppm = parts per million on weight basis, hence.5 w% NaCl = ppm 0.2 w% NaCl = 2 ppm Inserting into Vant Hoff s equation above: 6 ( ) π innløp = N / m = 29. 6bar 58.5 (1/ 997) 6 ( ) π utløp = = N / m = 8. 8bar 58.5 (1/ 997) Average osmostic pressure over the module: ( ) / 2 = 9.2 bar c) Calculating the membrane area based on information about test module: General equation: N w = Aw( p π ) where A w = water flux (m /h. m 2. bar) (A w for test module = A w real module) or: J = N A where A M = membrane area (m 2 ) w w M hence: testmodule 60 l/h = Aw (25-0). A 1 given: A 1 = 1m 2 real module 1000 l/h = Aw (55 9.2).A 2 Solving for A 2 gives membrane area:a 2 = 26. m 2
2 PROBLEM Humidity and drying a) efinitions of the variables below are found on pages in Geankoplis: -dry bulb temperature -wet bulb temperature -adiabatic saturation temperature -percentage humidity -percentage relative humidity b) Given for an air stream: temperature 85 C (=dry bulb temperature) dew point 27.C C 5% relative humidity; H R = 100 (p A / p AS ) total pressure of humid air is 1 bar Vapor pressure: From table A.2-9 saturation pressure, at 85 C is found; p AS = 57.8 kpa vapor pressure is calculated: p A = (H R. p AS ) / 100 p A = (5/100) = 2.87 kpa From the diagram, the remaining variables can be found: At 27.5 C a vertical line is drawn up to 100% humidity (dew point) Then going horisontal to the humidity axis we read: Absolute humidity at 85 C: H = kg H 2 O = / kg dry air We read precentage humidity for our gas stream where the vertical line from 85 C is crossing this horisontal line found above: hence % humidity, Hp = 100 (H / Hs) = % (app.) If this air stream is cooled adiabatic to 100% saturation (follow the adiabatic saturation curves in the diagram), we find the wet bulb temperature T w = 8 C ( adiabatic saturation temperature) (Note: H s at this temperature is 0.05 kg H2O / kg dry air, but this is NOT the same as H s will be for our air stream at 85 C)
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