Elementary Principles 176 - 6.19 Liquid H 2 O initially...

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

View Full Document Right Arrow Icon
6- 10 6.19 Liquid H O initially present: L kg 1 kmol L k g kmol H O l 2 2 25 100 1802 1387 . . . = bg Saturation at outlet: C mm Hg mol H O mol air HO 2 2 2 y p P = ° = × = * . . . 25 2376 15 760 mm Hg 00208 1 0 0208 0 0212 . . . = mol H O mol dry air 2 Flow rate of dry air: 15 L STP 1 mol min 22.4 L STP mol dry air min = 0670 . Evaporation Rate: mol dry air mol H O min mol dry air mol H O min 2 2 00212 00142 .. . = Complete Evaporation: 1.387 kmol mol 1 h kmol mol 60 min h d a y s 3 10 1628 678 min . . = 6.20 a. Daily rate of octane use = 4 30 (18 8) 7.069 10 ft 7.481 gal day ft 5.288 10 gal / day 2 33 3 4 π ⋅⋅− = × () . SG CH 81 8 = 0703 5288 10 3 4 . / ×× gal 1 ft 0.703 62.43 lb day 7.481 gal ft .10 10 lb C H day 3 m 3 5 m81 8 b. m 2 2 mf lb ft 32 2 2 f s 0.703 62.43 lb 32.174 ft 1 lb (18-8) ft 1 ft 29.921 in Hg 6.21 in Hg ft s 32.174 144 in 14.696 lb /in p × Δ= = c. Table B.4: pp y P o f 2 octane octane 8 F) 20.74 mm Hg 14.696 psi
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