Elementary Principles 383 - 9.27 a. Basis: 1200 lb m C 9 H...

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

View Full Document Right Arrow Icon
9- 31 9.27 a. b. Basis : 1200 lb C H 1 lb - mole h 120 lb lb - moles cumene produced h m91 2 m = 10 0 . Overall process : 0.75 C H n 1 (lb-moles/h) 36 0.25 C H 41 0 n 2 (lb-moles C H /h) 66 n 3 10.0 lb-moles C H /h 91 2 (lb-moles C H /h) n 4 (lb-moles C H /h) 0 C H l C H l C H l , F Btu lb - mole 2 r bg b g +→ ° = Δ ± H 77 39520 input consumption 2 2 m 66 m6 6 Benzene balance: 10.0 lb - moles C H produced 1 mole C H consumed h 1 mole C H produced 10.0 lb- moles C H lb C H h1 l b - m o l e lb C H h = = == ² . n 2 781 781 input output consumption 2 2 Propylene balance: 0.75 10.0 lb - moles C H 1 mole C H m o l e C H =+ ²² nn 13 C H unreacted lb - moles h lb - moles C H h 20% 075 10 020 075 16 67 250 31 1 3 ⇒= + U V W = = . ² .. ² ² . ² . n n Mass flow rate of C H / C H feed 0.75 lb - moles C H 42.08 lb C H l b - m o l e 0.25 lb - moles C H 58.12 lb C H l b - m o l e lb h 36 41 0 m 36 0 m 41 0 m = += b g b g 1667 16 67 768 . . Reactor : Benzene feed rate 10.0 lb- moles fresh feed moles fed to reactor h 1 mole fresh feed lb - moles C H h
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.
Ask a homework question - tutors are online