ch05 - CHAPTER FIVE 5.1 Assume volume additivity Av....

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CHAPTER FIVE 5.1 Assume volume additivity Av. density (Eq. 5.1-1): 10 4 0 0 0 0600 0 0 719 ρ =+⇒ = .. . .703 kg L .730 kg L kg L A O A D a. mm t = kg kg min m mass flow rate of liquid mass of tank at time 0 mass of empty tank A A =+ ⇒ == t ±± min . ± 250 150 10 3 14 28 bg ⇒⇒ = = ± ± ( ± . V(L / min) = m(kg / min) kg / L) V kg 1 L min 0.719 kg Lm in 14 28 19.9 b. ( t ) - m t k 0 150 14 28 3 107 −= ± . g 5.2 void volume of bed: 10 0 2335 184 505 33 cm cm cm −− = 3 porosity: 50 5 184 0 274 cm void cm total cm void cm total = bulk density: 6 00 g 184 326 3 cm g cm 3 = . absolute density: 600 g 184 4 3 . cm .49 g cm 3 5.3 CH 66 () l ± ± m ( k g /m i n ) V = 20.0 L / min B B ± ± m (kg / min) V (L / min) ± ± (kg/m ) V (L/m ) T T 78 ( l ± (. . ./ V= V t Dh t m) 4 m 60 min m 22 3 = ππ 4 55 015 00594 m i n Assume additive volumes ± V V-V L/min=39.4 L/min TB 59 4 20 0 ± . mV V kg L L kg L L kg / min BB TT =⋅+⋅= + = ρρ 0879 20 0 0866 39 4 517 x m m kg / L)(20.0 L / min) kg / min) kg B / kg B B = ± ± . 034 5- 1
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5.4 a. PP g h g h h=h h P=P P g h m 1 N 1 Pa gh 10 s l 1 20s l 2 12 12 s l kg m m s kg m s N m sl 32 2 2 =+ U V | W | ⇒− = F H G I K J F H G I K J = ρ ρρ ej bg 1 1 b. 1 1 c c c l x x check units! 1 kg slurry / L slurry kg crystals / kg slurry kg crystals / L crystals kg liquid / kg slurry kg liquid / L liquid L slurry kg slurry L crystals kg slurry L liquid kg slurry L slurry kg slurry =+= c. i.) 3 P kg / m == = 2775 9 8066 0 200 1415 .. b g ii.) 11 1 c c c l c cl s ll x 1-x x F H G 1 1 I K J =− F H G I K J x kg / m kg / m kg / m kg / m kg crystals / kg slurry c 3 3 33 = F H G G I K J J F H G G I K J J = 1 1415 1 12 1000 1 2 3 1000 1 12 1000 0316 . . di iii.) V m kg 1415 kg / m L m L = 175 1000 1238 . iv.) m x m kg crystals / kg slurry kg slurry kg crystals cc s l = 0 316 175 553 b g v.) m kg CuSO H O kmol 249 kg kmol CuSO kmol CuSO H O kg kmol kg CuSO CuSO 42 4 4 4 = = 5 1 1 15 159 6 1 354 . vi.) m x m kg liquid / kg slurry kg slurry kg liquid solution lc s l = = 1 0 684 175 120 bgb g b g vii.) V m k g 1.2 kg / m L m L l l l 3 3 = 120 1000 1000 100 d. h(m) 0.2 ρ l(kg/m^3) 1200 ρ c(kg/m^3) 2300 P(Pa) 2353.58 2411.24 2471.80 2602.52 2747.84 2772.61 2910.35 3093.28 xc 0 0.05 0.1 0.2 0.3 0.316 0.4 0.5 ρ sl(kg/m^3) 1200.00 1229.40 1260.27 1326.92 1401.02 1413.64 1483.87 1577.14 Effect of Slurry Density on Pressure Measurement 0 0.1 0.2 0.3 0.4 0.5 0.6 2300.00 2500.00 2700.00 2900.00 3100.00 Pressure Difference (Pa) Solids Fraction P = 2775, = 0.316 5- 2
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5.4 (cont’d) e. Basis 1 kg slurry x kg crystals V m crystals x kg crystals kg / m cc 3 c c 3 :, ⇒= bg di ρ 1- x kg liquid V m liquid 1-x kg liquid kg / m kg VVm x x cl 3 c l 3 sl 3 c c c l b g b , = = + = + ρρ 11 1 g 5.5 Assume P atm atm = 1 PV RT V = 0.08206 m atm kmol K K 4.0 atm kmol 10 mol mm o l 3 3 3 ±± . . =⇒ = 3132 1 00064 == 1 0 0064 10 45 3 mol 29.0 g 1 kg m a i r mo l g kg m 3 3 . . 5.6 a. V= nRT P mol L atm mol K 373.2 K 10 atm L = = 100 0 08206 306 .. . b. % . . error = 3.06L - 2.8L L 28 100% 9 3% ×= 5.7 Assume P bar atm = 1013 a. PV nRT n bar m kmol K 25+ 273.2 K .08314 m bar kg N kmol kg N 3 2 2 = +⋅ = 10 20 0 0 28 02 249 3 . b. PV PV nV T T P P n V ss s s s s s s = n m 273K bar 1 kmol 298.2K 1.013 bar 22.415 m STP kg N kmol 3 2 = + = 20 0 10 28 02 249 kg N 3 2 . 5.8 a. R= nT atm 1 kmol m 273 K atm m kmol K 33 × 12 2 4 1 5 821 10 2 .
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This note was uploaded on 06/07/2008 for the course CHE 2171 taught by Professor Wetzel during the Spring '08 term at LSU.

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ch05 - CHAPTER FIVE 5.1 Assume volume additivity Av....

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