Elementary Principles 241 - 6.98. a. 1.50 L / min 25o C,...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
6- 75 6.98 . a. ± ± n PV RT 0 == ⋅⋅ = (1atm)(1.50 L / min) (0.08206 L atm / mol K)(298 K) 0.06134 mol / min r.h.=25% p p HO *o 2 2 C) ( . 25 025 = Silica gel saturation condition : X p p * * .. * . . === 12 5 12 5 0 25 3125 2 2 2 gH Oads 100 g silica gel Water feed rate : y pC p o 0 0 25 23756 760 0 00781 = .( ) .(. ) . * 2 2 mm Hg mm Hg mol H O mol ± m H 2 O 0.06134 mol 0.00781mol H O 18.01g H O min mol mol H O 0.00863 g H O / min 22 2 2 Adsorption in 2 hours = = (. 0 00863 g H O / min)(120min) 1.035 g H O Saturation condition : 1.035g H O (g silica gel) 3.125g H O 100 g silica gel 33.1g silica gel M M =⇒ = Assume that all entering water vapor is adsorbed throughout the 2 hours and that P and T are constant. b. Humid air is dehumidified by being passed through a column of silica gel, which absorbs a significant fraction of the water in the entering air and relatively little oxygen and nitrogen. The capacity of the gel to absorb water, while large, is not infinite, and eventually the gel
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 02/27/2012 for the course CHEMICAL E 312 taught by Professor Cheung during the Fall '11 term at The University of Akron.

Ask a homework question - tutors are online