( ) = 9.03 lb m. mass of H lb m. O = lb m. = lb m. Determine density of aqueous KOH solu5on in lb m /gal. =

Transcription

1 To detere the ass frac5ons of KOH and HO we need the ass of each coponent and the total ass of the solu5on. We are given the ass of pure KOH in 1 gallon of solu5on is lb of KOH. We are also given the Specific Gravity of the en5re solu5on (KOH + HO) To be Plan 1. calculate total ass of solu5on using specific gravity.. Detere ass of water by subtrac5ng KOH ass fro solu5on ass 3. Calculate ass frac5ons Detere density of aqueous KOH solu5on in /gal SG ρ ρ = = ρ ρref ρ aquesous KOH = SGρ water = lb water ft 3 = ft 3 Detere total ass of aqueous KOH solu5on in ρ = ass volue ass = ρv = lb ft 3 ft 3 1 gal gal ( ) = 9.03 ass fraction KOH = KOH 9.03 = KOH ass of H O = = 8.17 ass fraction H O = 8.17 H O 9.03 = H O

2 Extra Credit Coponent ass KOH KOH lb 56 KOH/ ole = lb - oles 8.17 H O H O 8.17 lb 18 H O/ ole = lb - oles Total oles = ( ) - oles = 0.47 ole fraction KOH = 0.015lb - oles 0.47 oles = ole KOH ole ole fraction H O = lb - oles 0.47 oles = lb - ole H O ole

3 The F in the unit of T is a teperature. The calculation is best done in two steps. Step 1: Substitute for T( F) and siplify the resulting equation: t(f) 3 t( o C) = 1.8 t( o F) 3 t( o F) 3 ρ = ρ = ( 0.556t( o F) 17.78) ( 0.556t( o F) 17.78) ρ = t t Step : Define a new variable ρ in ters of /ft 3 ρ = ρ' ft 3 1kg ft 3 =16.0ρ' ρ'= ( ) t t ρ'= 6.8x t t

4 Makeup Proble The F in the unit of T is a teperature. The calculation is best done in two steps. Step 1: Substitute for T( F) and siplify the resulting equation: t(f) 3 t( o C) = 1.8 t( o F) 3 t( o F) 3 ρ = ρ = ( 0.556t( o F) 17.78) ( 0.556t( o F) 17.78) ρ = t t Step : Define a new variable ρ in ters of /ft 3 ρ = ρ' ft 3 1kg ft 3 =16.0ρ' ρ'= ( ) t t ρ'= 6.8x t t

5 P 1 = P P 3 = P 4 P 4 = P at + ρ oil gh oil P 1 P P 3 = P + ρ Hg gh hg + ρ air gh air = P 1 + ρ water gh water P 3 P 4 = P 1 + ρ water gh water P = P 3 ρ Hg gh hg ρ air gh air = ( P 3 ρ Hg gh hg ρ air gh ) air + ρ water gh water = ( P at + ρ oil gh oil ρ Hg gh hg ρ air gh ) air + ρ water gh water = P at kg 9.81 s ( ) kg kg 9.81 s s 0.15 ( ) ( ) kg = P at + ( Pa) ( Pa) ( 3.8 Pa) + ( Pa) =101,35 Pa 1,66 Pa = 89,059 Pa Lower then atosphere pressure s 0.45 ( )

6 This proble cae fro Module 4, Prac5ce Proble # Inlet Wet pulp 40 wt% water 60 wt% pulp W = 00 kg/ Dryer Outlet O = kg Dried pulp/ x wt% water y wt% pulp E = kg Evaporated H 0/ 70% of original water Plan: Calculate the Aount of water reoved by drying Perfor overall aterial balance Assue: the exi5ng dried pulp contains soe water since not all of the water is evaporated H O evaporated = 0.70 kg H O evaporated 00 kg H O entering dryer H O evaporated = 56 kg H O evaporated 60 kg wet pulp 0.40 kg H O kg wet pulp = 3,360 kg H O evaporated Overall Mass Balance Input Output + Genera5on Consup5on = Accuula5on Input = Output W = E + O 00 kg wet pulp = 56 kg H O evaporated + O O = = 8,640

7 Makeup Exa Inlet Wet pulp 50 wt% water 50 wt% pulp W = 00 kg/ Dryer Outlet O = kg Dried pulp/ x wt% water y wt% pulp E = kg Evaporated H 0/ 60% of original water Plan: Calculate the Aount of water reoved by drying Perfor overall aterial balance Assue: the exi5ng dried pulp contains soe water since not all of the water is evaporated H O evaporated = 0.60 kg H O evaporated 00 kg H O entering dryer H O evaporated = 60 kg H O evaporated 60 kg wet pulp 0.50 kg H O kg wet pulp = 3,600 kg H O evaporated Overall Mass Balance Input Output + Genera5on Consup5on = Accuula5on Input = Output W = E + O 00 kg wet pulp = 60 kg H O evaporated + O O = = 8,400

8 No, not with these units Need to ul5ply by g c f = D ( ΔP ) f = D ( ΔP ) ρv z = ft 3 ft lb f ft ft s = ft lb f ft s = lb f s ft ρv z = lb f s ft * ft /s = diensionless lb f This proble cae fro Module, Prac5ce Proble #9

9 Input Output + Genera3on Consup3on = Accuula3on Input aterial (ass) that enters tough the syste boundaries Genera5on aterial produced by reac5on within the syste Output Material that leaves tough the syste boundaries Consup5on Material that is consued by reac5on within the syste Accuula5on aterial that builds up in the syste Input Output + Genera5on = Accuula5on IOGA Genera5on this ter is posi5ve if aterial is produced within the syste and nega5ve if aterial is being consued within the syste.