Solução -princípios_elementares_dos_processos_químicos_3ed_-_felder_&_rousseau

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Solução -princípios_elementares_dos_processos_químicos_3ed_-_felder_&_rousseau

  1. 1. CHAPTER TWO 3 wk 7d 24 h 3600 s 1000 ms2.1 (a) = 18144 × 10 9 ms . 1 wk 1 d 1 h 1 s 38.1 ft / s 0.0006214 mi 3600 s (b) = 25.98 mi / h ⇒ 26.0 mi / h 3.2808 ft 1 h 554 m 4 1d 1h 1 kg 108 cm 4 (c) = 3.85 × 10 4 cm 4 / min⋅ g d ⋅ kg 24 h 60 min 1000 g 1 m 4 760 mi 1 m 1 h2.2 (a) = 340 m / s h 0.0006214 mi 3600 s 921 kg 2.20462 lb m 1 m3 (b) = 57.5 lb m / ft 3 m3 1 kg 35.3145 ft 3 5.37 × 10 3 kJ 1 min 1000 J 1.34 × 10 -3 hp (c) = 119.93 hp ⇒ 120 hp min 60 s 1 kJ 1 J/s2.3 Assume that a golf ball occupies the space equivalent to a 2 in × 2 in × 2 in cube. For a classroom with dimensions 40 ft × 40 ft × 15 ft : 40 × 40 × 15 ft 3 (12) 3 in 3 1 ball n balls = = 518 × 10 6 ≈ 5 million balls . ft 3 2 3 in 3 The estimate could vary by an order of magnitude or more, depending on the assumptions made.2.4 4.3 light yr 365 d 24 h 3600 s 1.86 × 10 5 mi 3.2808 ft 1 step = 7 × 1016 steps 1 yr 1d 1 h 1 s 0.0006214 mi 2 ft2.5 Distance from the earth to the moon = 238857 miles 238857 mi 1 m 1 report = 4 × 1011 reports 0.0006214 mi 0.001 m2.6 19 km 1000 m 0.0006214 mi 1000 L = 44.7 mi / gal 1 L 1 km 1 m 264.17 gal Calculate the total cost to travel x miles. $1.25 1 gal x (mi) Total Cost American = $14,500 + = 14,500 + 0.04464 x gal 28 mi $1.25 1 gal x (mi) Total Cost European = $21,700 + = 21,700 + 0.02796 x gal 44.7 mi Equate the two costs ⇒ x = 4.3 × 10 5 miles 2-1
  2. 2. 2.7 5320 imp. gal 14 h 365 d 106 cm3 0.965 g 1 kg 1 tonne plane ⋅ h 1 d 1 yr 220.83 imp. gal 1 cm 3 1000 g 1000 kg tonne kerosene = 1.188 × 105 plane ⋅ yr 4.02 × 109 tonne crude oil 1 tonne kerosene plane ⋅ yr yr 7 tonne crude oil 1.188 × 10 tonne kerosene 5 = 4834 planes ⇒ 5000 planes 25.0 lb m 32.1714 ft / s 2 1 lb f2.8 (a) = 25.0 lb f 32.1714 lb m ⋅ ft / s 2 25 N 1 1 kg ⋅ m/s 2 (b) = 2.5493 kg ⇒ 2.5 kg 9.8066 m/s 2 1N 10 ton 1 lb m 1000 g 980.66 cm / s 2 1 dyne (c) = 9 × 10 9 dynes 5 × 10 -4 ton 2.20462 lb m 1 g ⋅ cm / s 2 50 × 15 × 2 m 3 35.3145 ft 3 85.3 lb m 32.174 ft 1 lb f2.9 = 4.5 × 10 6 lb f 1 m3 1 ft 3 1 s 2 32.174 lb m / ft ⋅ s 2 500 lb m 1 kg 1 m3 FG 1 IJ FG 1 IJ ≈ 25 m ≈ 5 × 10 2 H 2 K H 10K 32.10 2.20462 lb m 11.5 kg2.11 (a) mdisplaced fluid = mcylinder ⇒ ρ f V f = ρ cVc ⇒ ρ f hπr 2 = ρ c Hπr 2 ρc ρfh (30 cm − 14.1 cm)(100 g / cm 3 ) . ρc = = = 0.53 g / cm 3 H H 30 cm ρf ρ H (30 cm)(0.53 g / cm 3 ) (b) ρ f = c = = 171 g / cm 3 . h h (30 cm - 20.7 cm)2.12 πR 2 H πR 2 H πr 2 h R r R Vs = ; Vf = − ; = ⇒r = h 3 3 3 H h H πR H2 FG IJ = πR FG H − h IJ πh Rh 2 2 3 h 3 H HK ⇒ Vf = − 3 H H K r 2 H 3 πR F h I GH H − H JK = ρ πR3 H 2 3 2 ρf ρ f V f = ρ sVs ⇒ρ f 2 s ρs 3 R H H3 1 ⇒ ρ f = ρs = ρs = ρs H− h3 H 3 − h3 FG h IJ 3 H2 1− H HK 2-2
  3. 3. 2.13 Say h( m) = depth of liquid y y= 1 dA y= – –1+h y= 1 h ⇒ xx 1m x = 1 y2 A(m 2 ) h – y= –1 dA 1− y 2 −1+ h dA = dy ⋅ ∫ dx = 2 1 − y dy ⇒ A m 2 ( )=2 ∫ 2 1 − y 2 dy − 1− y 2 −1 ⇓ Table of integrals or trigonometric substitution h −1 π ( ) A m 2 = y 1 − y 2 + sin −1 y ⎤ = ( h − 1) 1 − ( h − 1) + sin −1 ( h − 1) + ⎥ −1 ⎦ 2 2 4 m × A( m 2 ) 0.879 g 10 6 cm 2 1 kg 9.81 N b g W N = cm 3 1m 3 10 g 3 kg = 3.45 × 10 4 A g g0 E Substitute for A L W b N g = 3.45 × 10 Mbh − 1g 1 − bh − 1g 4 2 b g π OPQ + sin −1 h − 1 + N 22.14 1 lb f = 1 slug ⋅ ft / s 2 = 32.174 lb m ⋅ ft / s 2 ⇒ 1 slug = 32.174 lb m 1 1 poundal = 1 lb m ⋅ ft / s 2 = lb f 32.174 (a) (i) On the earth: 175 lb m 1 slug M= = 5.44 slugs 32.174 lb m 175 lb m 32.174 ft 1 poundal W= = 5.63 × 10 3 poundals s 1 lb m ⋅ ft / s 2 2 (ii) On the moon 175 lb m 1 slug M= = 5.44 slugs 32.174 lb m 175 lb m 32.174 ft 1 poundal W= = 938 poundals 6 s 1 lb m ⋅ ft / s 2 2 355 poundals 1 lb m ⋅ ft / s 2 1 slug 1m (b) F = ma ⇒ a = F / m = 25.0 slugs 1 poundal 32.174 lb m 3.2808 ft = 0.135 m / s 2 2-3
  4. 4. FG 1IJ = 5.3623 bung ⋅ ft / s2.15 (a) F = ma ⇒ 1 fern = (1 bung)(32.174 ft / s 2 ) H 6K 2 1 fern ⇒ 5.3623 bung ⋅ ft / s 2 3 bung 32.174 ft 1 fern (b) On the moon: W = = 3 fern 6 s 5.3623 bung ⋅ ft / s 2 2 On the earth: W = (3)( 32.174) / 5.3623 = 18 fern 4.0 × 10−42.16 (a) ≈ (3)(9) = 27 (b) ≈ ≈ 1× 10−5 40 (2.7)(8.632) = 23 (3.600 ×10−4 ) / 45 = 8.0 × 10−6 (c) ≈ 2 + 125 = 127 (d) ≈ 50 × 10 3 − 1 × 10 3 ≈ 49 × 10 3 ≈ 5 × 10 4 2.365 + 125.2 = 127.5 4.753 × 10 4 − 9 × 10 2 = 5 × 10 4 (7 ×10−1 )(3 × 105 )(6)(5 × 104 )2.17 R ≈ ≈ 42 × 102 ≈ 4 × 103 (Any digit in range 2-6 is acceptable) (3)(5 × 106 ) Rexact = 3812.5 ⇒ 3810 ⇒ 3.81× 1032.18 (a) A: R = 731 − 72.4 = 0.7 o C . 72.4 + 731 + 72.6 + 72.8 + 73.0 . X= = 72.8 o C 5 (72.4 − 72.8) 2 + (731 − 72.8) 2 + (72.6 − 72.8) 2 + (72.8 − 72.8) 2 + (73.0 − 72.8) 2 . s= 5−1 = 0.3o C B: R = 1031 − 97.3 = 58o C . . 97.3 + 1014 + 98.7 + 1031 + 100.4 . . X= = 100.2 o C 5 (97.3 − 100.2) 2 + (1014 − 100.2) 2 + (98.7 − 100.2) 2 + (1031 − 100.2) 2 + (100.4 − 100.2) 2 . . s= 5−1 = 2.3o C (b) Thermocouple B exhibits a higher degree of scatter and is also more accurate. 2-4
  5. 5. 2.19 (a) 12 12 ∑X i =1 i ∑ ( X − 735) i =1 . 2 X= = 73.5 s= = 12 . 12 12 − 1 C min= = X − 2 s = 73.5 − 2(1.2) = 711 . C max= = X + 2 s = 735 + 2(12) = 75.9 . . (b) Joanne is more likely to be the statistician, because she wants to make the control limits stricter. (c) Inadequate cleaning between batches, impurities in raw materials, variations in reactor temperature (failure of reactor control system), problems with the color measurement system, operator carelessness2.20 (a), (b) (a) Run 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 X 134 131 129 133 135 131 134 130 131 136 129 130 133 130 133 Mean(X) 131.9 Stdev(X) 2.2 Min 127.5 Max 136.4 (b) Run X Min Mean Max 1 128 127.5 131.9 136.4 140 2 131 127.5 131.9 136.4 3 133 127.5 131.9 136.4 138 4 130 127.5 131.9 136.4 136 5 133 127.5 131.9 136.4 134 6 129 127.5 131.9 136.4 132 7 133 127.5 131.9 136.4 130 8 135 127.5 131.9 136.4 128 9 137 127.5 131.9 136.4 10 133 127.5 131.9 136.4 126 11 136 127.5 131.9 136.4 0 5 10 15 12 138 127.5 131.9 136.4 13 135 127.5 131.9 136.4 14 139 127.5 131.9 136.4 (c) Beginning with Run 11, the process has been near or well over the upper quality assurance limit. An overhaul would have been reasonable after Run 12. 2.36 × 10−4 kg ⋅ m 2 2.20462 lb 3.28082 ft 2 1 h2.21 (a) Q = h kg m2 3600 s (2 × 10−4 )(2)(9) (b) Q approximate ≈ ≈ 12 × 10( −4−3) ≈ 1.2 × 10−6 lb ⋅ ft 2 / s 3 × 103 Q exact =1.56 × 10−6 lb ⋅ ft 2 / s = 0.00000156 lb ⋅ ft 2 / s 2-5
  6. 6. Cpμ 0.583 J / g ⋅ o C 1936 lb m 1 h 3.2808 ft 1000 g2.22 N Pr = = k 0.286 W / m ⋅ C o ft ⋅ h 3600 s m 2.20462 lb m −1 (6 × 10 )(2 × 10 )(3 × 10 ) 3 × 10 3 3 3 N Pr ≈ −1 ≈ ≈ 15 × 10 3 . The calculator solution is 163 × 10 3 . . (3 × 10 )(4 × 10 )(2) 3 22.23 Duρ 0.48 ft 1 m 2.067 in 1 m 0.805 g 1 kg 10 6 cm 3 Re = = μ s 3.2808 ft 0.43 × 10 −3 kg / m ⋅ s 39.37 in cm 3 1000 g 1 m3 (5 × 10 −1 )(2)(8 × 10 −1 )(10 6 ) 5 × 101− ( −3) Re ≈ ≈ ≈ 2 × 10 4 ⇒ the flow is turbulent (3)(4 × 10)(10 3 )(4 × 10 −4 ) 3 ⎛ d p uρ ⎞ 1/ 3 1/ 2 kg d p y ⎛ μ ⎞2.24 (a) = 2.00 + 0.600 ⎜ ⎟ ⎜ ⎟ D ⎝ ρD ⎠ ⎝ μ ⎠ 1/ 3 1/ 2 ⎡ 1.00 × 10−5 N ⋅ s/m 2 ⎤ ⎡ (0.00500 m)(10.0 m/s)(1.00 kg/m3 ) ⎤ = 2.00 + 0.600 ⎢ −5 ⎥ ⎢ ⎥ ⎣ (1.00 kg/m )(1.00 × 10 m / s) ⎦ 3 2 ⎣ (1.00 × 10−5 N ⋅ s/m 2 ) ⎦ k g (0.00500 m)(0.100) = 44.426 ⇒ = 44.426 ⇒ k g = 0.888 m / s 1.00 × 10−5 m 2 / s (b) The diameter of the particles is not uniform, the conditions of the system used to model the equation may differ significantly from the conditions in the reactor (out of the range of empirical data), all of the other variables are subject to measurement or estimation error. (c) dp (m) y D (m2/s) μ (N-s/m2) ρ (kg/m3) u (m/s) kg 0.005 0.1 1.00E-05 1.00E-05 1 10 0.889 0.010 0.1 1.00E-05 1.00E-05 1 10 0.620 0.005 0.1 2.00E-05 1.00E-05 1 10 1.427 0.005 0.1 1.00E-05 2.00E-05 1 10 0.796 0.005 0.1 1.00E-05 1.00E-05 1 20 1.2402.25 (a) 200 crystals / min ⋅ mm; 10 crystals / min ⋅ mm 2 200 crystals 0.050 in 25.4 mm 10 crystals 0.050 2 in 2 (25.4) 2 mm 2 (b) r = − min ⋅ mm in min ⋅ mm 2 in 2 238 crystals 1 min = 238 crystals / min ⇒ = 4.0 crystals / s min 60 s b g (c) D mm = b g D ′ in 25.4 mm = 25.4 D ′ ; r crystals FG crystals 60 s IJ 1 in min = r′ H s 1 min K = 60r ′ b g b ⇒ 60r ′ = 200 25.4 D ′ − 10 25.4 D ′ g 2 ⇒ r ′ = 84.7 D ′ − 108 D ′ b g 2 2-6
  7. 7. 2.26 (a) 70.5 lb m / ft 3 ; 8.27 × 10 -7 in 2 / lb f ⎡8.27 × 10−7 in 2 9 × 106 N 14.696 lbf / in 2 ⎤ (b) ρ = (70.5 lb m / ft 3 )exp ⎢ ⎥ ⎢ ⎣ lbf m 2 1.01325 × 105 N/m 2 ⎥⎦ 70.57 lb m 35.3145 ft 3 1 m3 1000 g = 3 3 6 3 = 1.13 g/cm3 ft m 10 cm 2.20462 lbm F lb IJ = ρ ′ g ρG 1 lb m 28,317 cm 3 = 62.43ρ ′ H ft K cm m (c) 3 3 453.593 g 1 ft 3 PG F lb IJ = P N 0.2248 lb f 12 m2 = 145 × 10 −4 P H in K m f 2 . 2 1N 39.37 2 in 2 d id i ⇒ 62.43ρ ′ = 70.5 exp 8.27 × 10 −7 1.45 × 10 −4 P ⇒ ρ ′ = 113 exp 120 × 10 −10 P . . d i P = 9.00 × 10 6 N / m 2 ⇒ ρ = 113 exp[(1.20 × 10 −10 )(9.00 × 10 6 )] = 113 g / cm 3 . . d i V din i 28,317in = 16.39V ; t bsg = 3600t ′bhr g 3 3 cm2.27 (a) V cm 3 = 3 1728 ⇒ 16.39V = expb3600t ′ g ⇒ V = 0.06102 expb3600t ′ g (b) The t in the exponent has a coefficient of s-1.2.28 (a) 3.00 mol / L, 2.00 min -1 (b) t = 0 ⇒ C = 3.00 exp[(-2.00)(0)] = 3.00 mol / L t = 1 ⇒ C = 3.00 exp[(-2.00)(1)] = 0.406 mol / L 0.406 − 3.00 For t=0.6 min: Cint = (0.6 − 0) + 3.00 = 14 mol / L . 1− 0 Cexact = 3.00 exp[(-2.00)(0.6)] = 0.9 mol / L 1− 0 For C=0.10 mol/L: t int = (010 − 3.00) + 0 = 112 min . . 0.406 − 3 1 C 1 0.10 t exact =- ln = - ln = 1.70 min 2.00 3.00 2 3.00 (c) 3.5 3 C exact vs. t 2.5 C (mol/L) 2 (t=0.6, C=1.4) 1.5 1 (t=1.12, C=0.10) 0.5 0 0 1 2 t (min) 2-7
  8. 8. 60 − 202.29 (a) p* = (185 − 166.2) + 20 = 42 mm Hg 199.8 − 166.2 (b) c MAIN PROGRAM FOR PROBLEM 2.29 IMPLICIT REAL*4(A–H, 0–Z) DIMENSION TD(6), PD(6) DO 1 I = 1, 6 READ (5, *) TD(I), PD(I) 1 CONTINUE WRITE (5, 902) 902 FORMAT (‘0’, 5X, ‘TEMPERATURE VAPOR PRESSURE’/6X, * ‘ (C) (MM HG)’/) DO 2 I = 0, 115, 5 T = 100 + I CALL VAP (T, P, TD, PD) WRITE (6, 903) T, P 903 FORMAT (10X, F5.1, 10X, F5.1) 2 CONTINUE END SUBROUTINE VAP (T, P, TD, PD) DIMENSION TD(6), PD(6) I=1 1 IF (TD(I).LE.T.AND.T.LT.TD(I + 1)) GO TO 2 I=I+1 IF (I.EQ.6) STOP GO TO 1 2 P = PD(I) + (T – TD(I))/(TD(I + 1) – TD(I)) * (PD(I + 1) – PD(I)) RETURN END DATA OUTPUT 98.5 1.0 TEMPERATURE VAPOR PRESSURE 131.8 5.0 (C) (MM HG) 100.0 1.2 215.5 100.0 105.0 1.8 215.0 98.72.30 (b) ln y = ln a + bx ⇒ y = ae bx b = (ln y 2 − ln y1 ) / ( x 2 − x1 ) = (ln 2 − ln 1) / (1 − 2) = −0.693 ln a = ln y − bx = ln 2 + 0.63(1) ⇒ a = 4.00 ⇒ y = 4.00e −0.693 x (c) ln y = ln a + b ln x ⇒ y = ax b b = (ln y 2 − ln y1 ) / (ln x 2 − ln x1 ) = (ln 2 − ln 1) / (ln 1 − ln 2) = −1 ln a = ln y − b ln x = ln 2 − ( −1) ln(1) ⇒ a = 2 ⇒ y = 2 / x (d) ln( xy ) = ln a + b( y / x) ⇒ xy = aeby / x ⇒ y = (a / x)eby / x [cant get y = f ( x)] b = [ln( xy ) 2 − ln( xy )1 ]/[( y / x) 2 − ( y / x)1 ] = (ln 807.0 − ln 40.2) /(2.0 − 1.0) = 3 ln a = ln( xy ) − b( y / x) = ln 807.0 − 3ln(2.0) ⇒ a = 2 ⇒ xy = 2e3 y / x [cant solve explicitly for y ( x)] 2-8
  9. 9. 2.30 (cont’d) (e) ln( y 2 / x ) = ln a + b ln( x − 2) ⇒ y 2 / x = a ( x − 2) b ⇒ y = [ax ( x − 2) b ]1/ 2 b = [ln( y 2 / x ) 2 − ln( y 2 / x ) 1 ] / [ln( x − 2) 2 − ln( x − 2) 1 ] = (ln 807.0 − ln 40.2) / (ln 2.0 − ln 10) = 4.33 . ln a = ln( y 2 / x ) − b( x − 2) = ln 807.0 − 4.33 ln(2.0) ⇒ a = 40.2 ⇒ y 2 / x = 40.2( x − 2) 4.33 ⇒ y = 6.34 x 1/ 2 ( x − 2) 2.1652.31 (b) Plot y 2 vs. x 3 on rectangular axes. Slope = m, Intcpt = − n 1 1 a 1 a 1 (c) = + x ⇒ Plot vs. x [rect. axes], slope = , intercept = ln( y − 3) b b ln( y − 3) b b (d) 1 1 = a ( x − 3) 3 ⇒ Plot vs. ( x − 3) 3 [rect. axes], slope = a , intercept = 0 ( y + 1) 2 ( y + 1) 2 OR 2 ln( y + 1) = − ln a − 3 ln( x − 3) Plot ln( y + 1) vs. ln( x − 3) [rect.] or (y + 1) vs. (x - 3) [log] 3 ln a ⇒ slope = − , intercept = − 2 2 (e) ln y = a x + b Plot ln y vs. x [rect.] or y vs. x [semilog ], slope = a, intercept = b (f) log10 ( xy ) = a ( x 2 + y 2 ) + b Plot log10 ( xy ) vs. ( x 2 + y 2 ) [rect.] ⇒ slope = a, intercept = b 1 b x x (g) = ax + ⇒ = ax 2 + b ⇒ Plot vs. x 2 [rect.], slope = a , intercept = b y x y y 1 b 1 b 1 1 OR = ax + ⇒ = a + 2 ⇒ Plot vs. 2 [rect.] , slope = b, intercept = a y x xy x xy x 2-9
  10. 10. 2.32 (a) A plot of y vs. R is a line through ( R = 5 , y = 0.011 ) and ( R = 80 , y = 0169 ). . 0.18 0.16 0.14 0.12 0.1 y 0.08 0.06 0.04 0.02 0 0 20 40 60 80 100 R y=aR+b a= 0169 − 0.011 . = 2.11 × 10 −3 U | 80 − 5 V ⇒ y = 2.11 × 10 −3 R + 4.50 × 10 −4 | d −3 b = 0.011 − 2.11 × 10 5 = 4.50 × 10 −4 ib g W d ib g (b) R = 43 ⇒ y = 2.11 × 10 −3 43 + 4.50 × 10 −4 = 0.092 kg H 2 O kg b1200 kg hgb0.092 kg H O kgg = 110 kg H O h 2 22.33 (a) ln T = ln a + b ln φ ⇒ T = aφ b b = (ln T2 − ln T1 ) / (ln φ 2 − ln φ 1 ) = (ln 120 − ln 210) / (ln 40 − ln 25) = −119 . ln a = ln T − b ln φ = ln 210 − ( −119) ln(25) ⇒ a = 9677.6 ⇒ T = 9677.6φ −1.19 . b (b) T = 9677.6φ −1.19 ⇒ φ = 9677.6 / T g 0.8403 b T = 85o C ⇒ φ = 9677.6 / 85 g 0.8403 = 535 L / s . b T = 175o C ⇒ φ = 9677.6 / 175g 0.8403 = 29.1 L / s T = 290 C ⇒ φ = b9677.6 / 290g 0.8403 o = 19.0 L / s (c) The estimate for T=175°C is probably closest to the real value, because the value of temperature is in the range of the data originally taken to fit the line. The value of T=290°C is probably the least likely to be correct, because it is farthest away from the date range. 2-10
  11. 11. 2.34 (a) Yes, because when ln[(C A − C Ae ) / (C A0 − C Ae )] is plotted vs. t in rectangular coordinates, the plot is a straight line. 0 50 100 150 200 ln ((CA-CAe)/(CA0-CAe)) 0 -0.5 -1 -1.5 -2 t (m in) Slope = -0.0093 ⇒ k = 9.3 × 10-3 min −1 (b) ln[(C A − C Ae ) /(C A0 − C Ae )] = − kt ⇒ C A = (C A0 − C Ae )e − kt + C Ae −3 C A = (0.1823 − 0.0495)e − (9.3×10 )(120) + 0.0495 = 9.300 × 10-2 g/L 9.300 × 10-2 g 30.5 gal 28.317 L C =m /V ⇒ m =CV = = 10.7 g L 7.4805 gal2.35 (a) ft 3 and h -2 , respectively (b) ln(V) vs. t2 in rectangular coordinates, slope=2 and intercept= ln(353 × 10 −2 ) ; or . V(logarithmic axis) vs. t2 in semilog coordinates, slope=2, intercept= 353 × 10−2 . (c) V ( m3 ) = 100 × 10 −3 exp(15 × 10 −7 t 2 ) . .2.36 PV k = C ⇒ P = C / V k ⇒ ln P = ln C − k lnV 8.5 8 7.5 lnP 7 6.5 6 2.5 3 3.5 4 lnV lnP = -1.573(lnV ) + 12.736 k = − slope = − ( −1573) = 1573 (dimensionless) . . Intercept = ln C = 12.736 ⇒ C = e12.736 = 3.40 × 105 mm Hg ⋅ cm4.719 G − GL 1 G −G G −G2.37 (a) = ⇒ 0 = K L C m ⇒ ln 0 = ln K L + m ln C G0 − G K L C m G − GL G − GL ln (G 0 -G )/(G -G L )= 2 .4 8 3 5 ln C - 1 0 .0 4 5 3 ln(G 0-G)/(G-G L ) 2 1 0 -1 3 .5 4 4 .5 5 5 .5 ln C 2-11
  12. 12. 2.37 (cont’d) m = slope = 2.483 (dimensionless) Intercept = ln K L = −10.045 ⇒ K L = 4.340 × 10 −5 ppm-2.483 G − 180 × 10 −3 . (b) C = 475 ⇒ = 4.340 × 10 −5 (475) 2.483 ⇒ G = 1806 × 10 −3 . 3.00 × 10 −3 − G C=475 ppm is well beyond the range of the data.2.38 (a) For runs 2, 3 and 4: Z = aV b p c ⇒ ln Z = ln a + b lnV + c ln p b = 0.68 ln( 35) = ln a + b ln(102) + c ln(9.1) . . ln(2.58) = ln a + b ln(102) + c ln(112) . . ⇒ c = −1.46 ln(3.72) = ln a + b ln(175) + c ln(112) . . a = 86.7 volts ⋅ kPa 1.46 / (L / s) 0.678 (b) When P is constant (runs 1 to 4), plot ln Z vs. lnV . Slope=b, Intercept= ln a + c ln p 2 1.5 lnZ 1 0.5 0 -1 -0.5 0 0.5 1 1.5 lnV b = slope = 0.52 lnZ = 0.5199lnV + 1.0035 Intercept = lna + c ln P = 10035 . When V is constant (runs 5 to 7), plot lnZ vs. lnP. Slope=c, Intercept= ln a + c lnV 2 1.5 lnZ 1 0.5 0 1.5 1.7 1.9 2.1 2.3 c = slope = −0.997 ⇒ 10 . lnZ = -0.9972lnP + 3.4551 lnP Intercept = lna + b lnV = 3.4551 Plot Z vs V b P c . Slope=a (no intercept) 7 6 5 4 Z 3 2 1 0.05 0.1 0.15 0.2 Vb Pc Z = 31.096VbPc a = slope = 311 volt ⋅ kPa / (L / s) .52 . The results in part (b) are more reliable, because more data were used to obtain them. 2-12
  13. 13. 2.39 (a) n ∑x y 1 sxy = i i = [(0.4)(0.3) + (2.1)(19) + (31)( 3.2)] / 3 = 4.677 . . n i =1 n ∑x 1 sxx = 2 i = (0.32 + 19 2 + 3.2 2 ) / 3 = 4.647 . n i =1 n n ∑ ∑y 1 1 sx = xi = (0.3 + 1.9 + 3.2) / 3 = 18; s y = . i = (0.4 + 2.1 + 31) / 3 = 1867 . . n i =1 n i =1 sxy − sx s y 4.677 − (18)(1.867) . a= = = 0.936 sxx − sx b g 2 4.647 − (18) 2 . sxx s y − sxy sx ( 4.647)(1867) − (4.677)(18) . . b= = = 0.182 sxx − sx b g 2 4.647 − (18) . 2 y = 0.936 x + 0182 . sxy 4.677 (b) a = = = 1.0065 ⇒ y = 1.0065x sxx 4.647 4 3 y = 0.936x + 0.182 2 y 1 y = 1.0065x 0 0 1 2 3 4 x2.40 (a) 1/C vs. t. Slope= b, intercept=a (b) b = slope = 0.477 L / g ⋅ h; a = Intercept = 0.082 L / g 3 2 2.5 2 1.5 1/C 1.5 1 C 1 0.5 0.5 0 0 0 1 2 3 4 5 6 1 2 3 4 5 1/C = 0.4771t + 0.0823 t t C C-fitted (c) C = 1 / (a + bt ) ⇒ 1 / [0.082 + 0.477(0)] = 12.2 g / L t = (1 / C − a ) / b = (1 / 0.01 − 0.082) / 0.477 = 209.5 h (d) t=0 and C=0.01 are out of the range of the experimental data. (e) The concentration of the hazardous substance could be enough to cause damage to the biotic resources in the river; the treatment requires an extremely large period of time; some of the hazardous substances might remain in the tank instead of being converted; the decomposition products might not be harmless. 2-13
  14. 14. 2.41 (a) and (c) 10 y 1 0.1 1 10 100 x (b) y = ax b ⇒ ln y = ln a + b ln x; Slope = b, Intercept = ln a ln y = 0.1684ln x + 1.1258 2 1.5 ln y 1 0.5 0 b = slope = 0.168 -1 0 1 2 3 4 5 ln x Intercept = ln a = 11258 ⇒ a = 3.08 .2.42 (a) ln(1-Cp/CA0) vs. t in rectangular coordinates. Slope=-k, intercept=0 (b) 0 200 400 600 800 0 100 200 300 400 500 600 ln(1-Cp/Cao) ln(1-Cp/Cao) 0 0 -1 -2 -2 -3 -4 -4 -6 ln(1-Cp/Cao) = -0.0062t ln(1-Cp/Cao) = -0.0111t t t Lab 1 Lab 2 k = 0.0062 s-1 k = 0.0111 s-1 0 200 400 600 800 0 200 400 600 800 0 ln(1-Cp/Cao) 0 ln(1-Cp/Cao) -2 -2 -4 -4 -6 -6 ln(1-Cp/Cao) = -0.0063t ln(1-Cp/Cao)= -0.0064t t t Lab 3 Lab 4 k = 0.0063 s-1 k = 0.0064 s-1 (c) Disregarding the value of k that is very different from the other three, k is estimated with the average of the calculated k’s. k = 0.0063 s-1 (d) Errors in measurement of concentration, poor temperature control, errors in time measurements, delays in taking the samples, impure reactants, impurities acting as catalysts, inadequate mixing, poor sample handling, clerical errors in the reports, dirty reactor. 2-14
  15. 15. dφ ∑by g ∑ 2b y g n n n n n2.43 yi = axi ⇒ φ (a ) = ∑ ∑y x ∑x 2 di2 = i − axi ⇒ =0= i − axi xi ⇒ i i −a 2 i =0 i =1 i =1 da i =1 i =1 i =1 n n ⇒a= ∑i =1 yi xi / ∑x i =1 2 i2.44 DIMENSION X(100), Y(100) READ (5, 1) N C N = NUMBER OF DATA POINTS 1FORMAT (I10) READ (5, 2) (X(J), Y(J), J = 1, N 2FORMAT (8F 10.2) SX = 0.0 SY = 0.0 SXX = 0.0 SXY = 0.0 DO 100J = 1, N SX = SX + X(J) SY = SY + Y(J) SXX = SXX + X(J) ** 2 100SXY = SXY + X(J) * Y(J) AN = N SX = SX/AN SY = SY/AN SXX = SXX/AN SXY = SXY/AN CALCULATE SLOPE AND INTERCEPT A = (SXY - SX * SY)/(SXX - SX ** 2) B = SY - A * SX WRITE (6, 3) 3FORMAT (1H1, 20X PROBLEM 2-39/) WRITE (6, 4) A, B 4FORMAT (1H0, SLOPEb -- bAb =, F6.3, 3X INTERCEPTb -- b8b =, F7.3/) C CALCULATE FITTED VALUES OF Y, AND SUM OF SQUARES OF RESIDUALS SSQ = 0.0 DO 200J = 1, N YC = A * X(J) + B RES = Y(J) - YC WRITE (6, 5) X(J), Y(J), YC, RES 5FORMAT (3X Xb =, F5.2, 5X /Yb =, F7.2, 5X Y(FITTED)b =, F7.2, 5X * RESIDUALb =, F6.3) 200SSQ = SSQ + RES ** 2 WRITE (6, 6) SSQ 6FORMAT (IH0, SUM OF SQUARES OF RESIDUALSb =, E10.3) STOP END $DATA 5 1.0 2.35 1.5 5.53 2.0 8.92 2.5 12.15 3.0 15.38 SOLUTION: a = 6.536, b = −4.206 2-15
  16. 16. 2.45 (a) E(cal/mol), D0 (cm2/s) (b) ln D vs. 1/T, Slope=-E/R, intercept=ln D0. (c) Intercept = ln D0 = -3.0151 ⇒ D0 = 0.05 cm2 / s . Slope = − E / R = -3666 K ⇒ E = (3666 K)(1.987 cal / mol ⋅ K) = 7284 cal / mol 2.0E-03 2.1E-03 2.2E-03 2.3E-03 2.4E-03 2.5E-03 2.6E-03 2.7E-03 2.8E-03 2.9E-03 3.0E-03 -10.0 -11.0 ln D -12.0 -13.0 -14.0 ln D = -3666(1/T) - 3.0151 1/T (d) Spreadsheet T D 1/T lnD (1/T)*(lnD) (1/T)**2 347 1.34E-06 2.88E-03 -13.5 -0.03897 8.31E-06 374.2 2.50E-06 2.67E-03 -12.9 -0.03447 7.14E-06 396.2 4.55E-06 2.52E-03 -12.3 -0.03105 6.37E-06 420.7 8.52E-06 2.38E-03 -11.7 -0.02775 5.65E-06 447.7 1.41E-05 2.23E-03 -11.2 -0.02495 4.99E-06 471.2 2.00E-05 2.12E-03 -10.8 -0.02296 4.50E-06 Sx 2.47E-03 Sy -12.1 Syx -3.00E-02 Sxx 6.16E-06 -E/R -3666 ln D0 -3.0151 D0 7284 E 0.05 2-16
  17. 17. CHAPTER THREE 16 × 6 × 2 m3 1000 kg3.1 (a) m = m 3 b ≈ 2 × 10 5 2 103 ≈ 2 × 105 kg gb gb gd i 8 oz 1 qt 106 cm3 1g 4 × 106 (b) m = ≈ ≈ 1 × 102 g / s 2s 32 oz 1056.68 qt cm 3 b3 × 10gd10 i3 (c) Weight of a boxer ≈ 220 lb m 12 × 220 lb m 1 stone Wmax ≥ ≈ 220 stones 14 lb m (d) dictionary πD 2 L 314 4.5 ft . 2 2 800 miles 5880 ft 7.4805 gal 1 barrel V= = 4 4 1 mile 1 ft 3 42 gal ≈ d i d 3 × 4 × 5 × 8 × 10 2 × 5 × 10 3 × 7 i ≈ 1 × 10 7 barrels 4 × 4 × 10 6 ft × 1 ft × 0.5 ft 28,317 cm3 (e) (i) V ≈ 3 ≈ 3 × 3 × 104 ≈ 1 × 105 cm3 1 ft 150 lb m 1 ft 3 28,317 cm3 150 × 3 × 104 (ii) V ≈ ≈ ≈ 1 × 105 cm3 62.4 lb m 1 ft 3 60 (f) SG ≈ 105 . 995 kg 1 lb m 0.028317 m33.2 (a) (i) 3 3 = 62.12 lb m / ft 3 m 0.45359 kg 1 ft 995 kg / m3 62.43 lb m / ft 3 (ii) = 62.12 lb m / ft 3 1000 kg / m3 (b) ρ = ρ H2 O × SG = 62.43 lb m / ft 3 × 5.7 = 360 lb m / ft 3 50 L 0.70 × 103 kg 1 m33.3 (a) = 35 kg m3 103 L 1150 kg m3 1000 L 1 min (b) = 27 L s min 0.7 × 1000 kg 1 m3 60 s 10 gal 1 ft 3 0.70 × 62.43 lb m (c) ≅ 29 lb m / min 2 min 7.481 gal 1 ft 3 3-1
  18. 18. 3.3 (cont’d) (d) Assuming that 1 cm3 kerosene was mixed with Vg (cm3 ) gasoline d i d Vg cm3gasoline ⇒ 0.70Vg g gasoline i 1dcm kerosenei ⇒ 0.82dg kerosenei 3 SG = d0.70V + 0.82idg blendi = 0.78 ⇒ V g = 0.82 − 0.78 = 0.5 0 cm 3 V + 1dcm blend i 0.78 − 0.70 3 g g Vgasoline 0.50 cm3 Volumetric ratio = = 3 = 0.50 cm gasoline / cm kerosene 3 3 Vkerosene 1 cm 50.0 kg L 5 Fr $13.4 In France: = $68.42 0.7 × 10 kg 1L 5.22 Fr . 50.0 kg L 1 gal $1.20 In U.S.: = $22.64 0.70 × 10 kg 3.7854 L 1 gal .3.5 V B ( ft 3 / h ), m B ( lb m / h ) V ( ft 3 / h), SG = 0.850 V H ( ft 3 / h ), m H ( lb m / h ) 700 lb m / h 700 lb m ft 3 (a) V = = 1319 ft 3 / h . h 0.850 × 62.43 lb m mB = d i 0.879 × 62.43 lb = 54.88V bkg / hg VB ft 3 m b hg ft 3 B mH = dV hb0.659 × 62.43g = 4114V b kg / hg H . H VB + VH = 1319 ft / h . 3 mB + mH = 54.88VB + 4114VH = 700 lb m . ⇒ VB = 114 ft 3 / h ⇒ mB = 628 lb m / h benzene . VH = 1.74 ft 3 / h ⇒ mH = 71.6 lb m / h hexane (b) – No buildup of mass in unit. – ρ B and ρ H at inlet stream conditions are equal to their tabulated values (which are o strictly valid at 20 C and 1 atm.) – Volumes of benzene and hexane are additive. – Densitometer gives correct reading. 3-2
  19. 19. 195.5 kg H 2SO 4 1 kg solution L3.6 (a) V = = 445 L 0.35kg H 2SO 4 12563 × 1000 kg . . (b) 195.5 kg H 2 SO 4 L Videal = 18255 × 1.00 kg . 195.5 kg H 2 SO 4 0.65 kg H 2 O L + = 470 L 0.35 kg H 2 SO 4 1.000 kg 470 − 445 % error = × 100% = 5.6% 4453.7 b gE Buoyant force up = Weight of block down b g Mass of oil displaced + Mass of water displaced = Mass of block b g b ρ oil 0.542 V + ρ H O 1 − 0.542 V = ρ c V 2 g From Table B.1: ρ c = 2.26 g / cm3 , ρ w = 100 g / cm3 ⇒ ρ oil = 3.325 g / cm3 . moil = ρ oil × V = 3.325 g / cm3 × 35.3 cm3 = 117.4 g moil + flask = 117.4 g + 124.8 g = 242 g3.8 b g Buoyant force up = Weight of block down b g ⇒ Wdisplaced liquid = Wblock ⇒ ( ρVg ) disp. Liq = ( ρVg ) block b g Expt. 1: ρ w 15 A g = ρ B 2 A g ⇒ ρ B = ρ w × . b g 15 2 . ρ w =1.00 g/cm3 ρ B = 0.75 g / cm3 ⇒ SG b g B = 0.75 bg b g Expt. 2: ρ soln A g = ρ B 2 A g ⇒ ρ soln = 2 ρ B = 15 g / cm3 ⇒ SG . b g soln = 15 .3.9 Let ρ w = density of water. Note: ρ A > ρ w (object sinks) WA + WB hs 1 Volume displaced: Vd 1 = Ab hsi = Ab hp1 − hb1 d i (1) Archimedes ⇒ ρ wVd 1 g = WA + WB hρ1 hb1 weight of displaced water Before object is jettisoned Subst. (1) for Vd 1 , solve for h p1 − hb1 d i WA + WB h p1 − hb1 = (2) pw gAb bi g Volume of pond water: Vw = Ap h p1 − Vd 1 ⇒Vw = Ap h p1 − Ab h p1 − hb1 d i subst. 2 bg WA + WB V W + WB Vw = Ap h p1 − ⇒ h p1 = w + A (3) for b p 1 − hb 1 pw g Ap pw gAp bg subst. 3 for h p 1 in hb1 = Vw + A b W + WB g LM 1 − 1 OP (4) b 2 g, solve for h b1 Ap pw g NM A p Ab QP 3-3
  20. 20. 3.9 (cont’d) hs 2 WA WB Let V A = volume of jettisoned object = (5) ρ Ag WA h b2 hρ2 Volume displaced by boat: Vd 2 = Ab h p 2 − hb 2 d i (6) Archimedes ⇒ ρ WVd 2 g = WB After object is jettisoned E Subst. for Vd 2 , solve for dh p2 − hb 2 i WB h p 2 − hb 2 = (7) pw gAb b5g, b6g & b7 g WB W Volume of pond water: Vw = Ap h p 2 − Vd 2 − V A Vw = Ap h p 2 − − A pw g p A g solve for Vw WB WA ⇒ hp 2 = + + (8) hp 2 Ap pw gAp p A gAp subst. 8 bg Vw WB WA WB ⇒ hb 2 = + + − (9) bg for h p 2 in 7 , solve for hb 2 Ap pw gAp p A gAp pw gAb(a) Change in pond level ( 8 ) − ( 3) W ⎡ 1 1 ⎤ WA ( pW − p A ) ρW < ρ A hp 2 − hp1 = A ⎢ − ⎥= ⎯⎯⎯⎯ < 0 → Ap g ⎣ p A pW ⎦ p A pW gAp ⇒ the pond level falls(b) Change in boat level L >0 O >0 b 9 g−b 4 g WA LM 1 OP b=gF V I MM1 + F p F A − 1I I PP > 0 5 Q H K M H G JK JK PP p A P G A JM G p H A 1 1 h p 2 − h p1 = − + A A p A gMp A N p A p pW Ap W b p W b N Q ⇒ the boat rises 2.93 kg CaCO 3 0.70 L CaCO 33.10 (a) ρ bulk = = 2.05 kg / L L CaCO 3 L total 2.05 kg 50 L 9.807 m / s2 1N (b) Wbag = ρ bulkVg = = 100 × 103 N . L 1 kg ⋅ m / s 2 Neglected the weight of the bag itself and of the air in the filled bag. (c) The limestone would fall short of filling three bags, because – the powder would pack tighter than the original particles. – you could never recover 100% of what you fed to the mill. 3-4
  21. 21. 122.5 kg 9.807 m / s2 1N3.11 (a) Wb = mb g = = 1202 N 1 kg ⋅ m / s2 Wb − WI (1202 N - 44.0 N) 1 kg ⋅ m / s2 Vb = = = 119 L ρwg 0.996 kg / L × 9.807 m / s2 1N m 122.5 kg ρb = b = = 103 kg / L . Vb 119 L (b) m f + mnf = mb (1) mf xf = ⇒ m f = mb x f (2) mb (1),(2) ⇒ mnf = mb 1 − x f d i (3) mf mnf mb V f + Vnf = Vb ⇒ + = ρf ρ nf ρb b2 g,b 3g Fx 1− xf I=m F1 I= 1 − 1 1 / ρ b − 1 / ρ nf ⇒ mb GH ρ f f + ρ nf JK ρ b b ⇒ xf GH ρ f − 1 ρ nf JK ρ ρ b nf ⇒ xf = 1 / ρ f − 1 / ρ nf 1 / ρ b − 1 / ρ nf 1 / 103 − 1 / 1.1 . (c) x f = = = 0.31 1 / ρ f − 1 / ρ nf 1 / 0.9 − 1 / 1.1 (d) V f + Vnf + Vlungs + Vother = Vb mf mnf mb + + Vlungs + Vother = ρf ρ nf ρb m f = mb x f Fx 1− xf I + (V F1− 1I mnf = mb (1− x f ) mb GH ρ f f − ρ nf JK lungs + Vother ) = mb GH ρ ρ JK b nf F 1 − 1 I = 1 − 1 − V +V GH ρ ρ JK ρ ρ ⇒ xf f m nf b nf lungs b other F 1 − 1 I − F V + V I F 1 1 I F 12 + 01I GH ρ ρ JK GH m JK GH 1.03 − 11JK − GH .122.5. JK b . nf lungs b other ⇒x = = = 0.25 f F1− 1I FG 1 − 1 IJ GH ρ ρ JK H 0.9 11K. f nf 3-5
  22. 22. 3.12 (a) 4.5 Conc. (g Ile/100 g H2O) 4 3.5 y = 545.5x - 539.03 3 R2 = 0.9992 2.5 2 1.5 1 0.5 0 0.987 0.989 0.991 0.993 0.995 0.997 Density (g/cm3) From the plot above, r = 5455ρ − 539.03 . (b) For ρ = 0.9940 g / cm3 , r = 3.197 g Ile / 100g H 2 O 150 L 0.994 g 1000 cm3 3.197 g Ile 1 kg mIle = 3 = 4.6 kg Ile / h h cm L 103.197 g sol 1000 g (c) The measured solution density is 0.9940 g ILE/cm3 solution at 50oC. For the calculation of Part (b) to be correct, the density would have to be changed to its equivalent at 47oC. Presuming that the dependence of solution density on T is the same as that of pure water, the solution density at 47oC would be higher than 0.9940 g ILE/cm3. The ILE mass flow rate calculated in Part (b) is therefore too low.3.13 (a) 1.20 1.00 Mass Flow Rate (kg/min) y = 0.0743x + 0.1523 R 2 = 0.9989 0.80 0.60 0.40 0.20 0.00 0.0 2.0 4.0 6.0 8.0 10.0 12.0 Rotameter Reading 3-6
  23. 23. 3.13 (cont’d) b g From the plot, R = 5.3 ⇒ m = 0.0743 5.3 + 01523 = 0.55 kg / min . (b) Rotameter Collection Collected Mass Flow Difference Mean Di Reading Time Volume Rate Duplicate (min) (cm3) (kg/min) (Di) 2 1 297 0.297 2 1 301 0.301 0.004 4 1 454 0.454 4 1 448 0.448 0.006 6 0.5 300 0.600 6 0.5 298 0.596 0.004 0.0104 8 0.5 371 0.742 8 0.5 377 0.754 0.012 10 0.5 440 0.880 10 0.5 453 0.906 0.026 1 Di = 5 b g 0.004 + 0.006 + 0.004 + 0.012 + 0.026 = 0.0104 kg / min 95% confidence limits: (0.610 ± 174 Di ) kg / min = 0.610 ± 0.018 kg / min . There is roughly a 95% probability that the true flow rate is between 0.592 kg / min and 0.628 kg / min . 15.0 kmol C 6 H 6 78.114 kg C 6 H 63.14 (a) = 117 × 103 kg C 6 H 6 . kmol C 6 H 6 15.0 kmol C 6 H 6 1000 mol (b) = 15 × 104 mol C 6 H 6 . kmol 15,000 mol C 6 H 6 lb - mole (c) = 33.07 lb - mole C 6 H 6 453.6 mol 15,000 mol C 6 H 6 6 mol C (d) = 90,000 mol C 1 mol C 6 H 6 15,000 mol C 6 H 6 6 mol H (e) = 90,000 mol H 1 mol C 6 H 6 90,000 mol C 12.011 g C (f) = 1.08 × 106 g C mol C 90,000 mol H 1.008 g H (g) = 9.07 × 104 g H mol H 15,000 mol C 6 H 6 6.022 × 1023 (h) = 9.03 × 1027 molecules of C 6 H 6 mol 3-7
  24. 24. 175 m3 1000 L 0.866 kg 1h3.15 (a) m = 3 = 2526 kg / min h m L 60 min 2526 kg 1000 mol 1 min (b) n = = 457 mol / s min 92.13 kg 60 s (c) Assumed density (SG) at T, P of stream is the same as the density at 20oC and 1 atm 200.0 kg mix 0150 kg CH 3OH . kmol CH 3OH 1000 mol3.16 (a) = 936 mol CH 3OH kg mix 32.04 kg CH 3OH 1 kmol 100.0 lb - mole MA 74.08 lb m MA 1 lb m mix (b) mmix = = 8715 lb m / h h 1 lb - mole MA 0.850 lb m MA 0.25 mol N 2 28.02 g N 2 0.75 mol H 2 2.02 g H 23.17 M= + = 8.52 g mol mol N 2 mol H 2 3000 kg kmol 0.25 kmol N 2 28.02 kg N 2 mN 2 = = 2470 kg N 2 h h 8.52 kg kmol feed kmol N 23.18 M suspension = 565 g − 65 g = 500 g , M CaCO 3 = 215 g − 65 g = 150 g (a) V = 455 mL min , m = 500 g min (b) ρ = m / V = 500 g / 455 mL = 110 g mL . (c) 150 g CaCO 3 / 500 g suspension = 0.300 g CaCO 3 g suspension3.19 Assume 100 mol mix. 10.0 mol C 2 H 5OH 46.07 g C 2 H 5OH mC2 H 5OH = = 461 g C 2 H 5OH mol C 2 H 5OH 75.0 mol C 4 H 8 O 2 88.1 g C 4 H 8 O 2 mC4 H 8O 2 = = 6608 g C 4 H 8 O 2 mol C 4 H 8O 2 15.0 mol CH 3COOH 60.05 g CH 3COOH mCH 3COOH = = 901 g CH 3COOH mol CH 3COOH 461 g xC2 H 5OH = = 0.0578 g C 2 H 5OH / g mix 461 g + 6608 g + 901 g 6608 g xC 4 H 8 O 2 = = 0.8291 g C 4 H 8 O 2 / g mix 461 g + 6608 g + 901 g 901 g xCH 3COOH = = 0113 g CH 3COOH / g mix . 461 g + 6608 g + 901 g 461 g + 6608 g + 901 g MW = = 79.7 g / mol 100 mol 25 kmol EA 100 kmol mix 79.7 kg mix m= = 2660 kg mix 75 kmol EA 1 kmol mix 3-8
  25. 25. 3.20 (a) Unit Function Crystallizer Form solid gypsum particles from a solution Filter Separate particles from solution Dryer Remove water from filter cake 0.35 kg C aSO 4 ⋅ 2 H 2 O (b) m gypsum = 1 L slurry = 0 .35 kg C aSO 4 ⋅ 2 H 2 O L slurry 0.35 kg CaSO4 ⋅ 2H2OL CaSO4 ⋅ 2H2O Vgypsum = = 0151 L CaSO4 ⋅ 2H2O . 2.32 kg CaSO4 ⋅ 2H2O 0.35 kg gypsum 136.15 kg CaSO 4 CaSO 4 in gypsum: m = = 0.277 kg CaSO 4 172.18 kg gypsum CaSO 4 in soln.: m = b1− 0151g L sol . 1.05 kg 0.209 kg CaSO 4 = 0.00186 kg CaSO 4 L 100.209 kg sol 0.35 kg gypsum 0.05 kg sol 0.209 g CaSO 4 (c) m = = 3.84 × 10 -5 kg CaSO 4 0.95 kg gypsum 100.209 g sol 0.277 g + 3.84 × 10 -5 g % recovery = × 100% = 99.3% 0.277 g + 0.00186 g3.21 45.8 L 0.90 kg kmol = 0.5496 kmol U | CSA: min L 75 kg min ⇒ | V0.5496 = 1.2 mol CSA FB: 55.2 L 0.75 kg kmol = 0.4600 kmol | | 0.4600 mol FB min L 90 kg min W She was wrong. The mixer would come to a grinding halt and the motor would overheat. 150 mol EtOH 46.07 g EtOH3.22 (a) = 6910 g EtOH mol EtOH 6910 g EtO H 0.600 g H 2 O = 10365 g H 2 O 0.400 g EtOH 6910 g EtOH L 10365 g H 2 O L V = + = 19.123 L ⇒ 19.1 L 789 g EtOH 1000 g H 2 O (6910 +10365) g L SG = = 0.903 19.1 L 1000 g ( 6910 + 10365) g mix L (b) V ′ = = 18.472 L ⇒ 18.5 L 935.18 g (19.123 − 18.472 ) L % error = × 100% = 3.5% 18.472 L 3-9

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