Upcoming SlideShare
×

Thanks for flagging this SlideShare!

Oops! An error has occurred.

×
Saving this for later? Get the SlideShare app to save on your phone or tablet. Read anywhere, anytime – even offline.
Standard text messaging rates apply

# Chen204solutions

82,637

Published on

Published in: Education
18 Likes
Statistics
Notes
• Full Name
Comment goes here.

Are you sure you want to Yes No
• Be the first to comment

Views
Total Views
82,637
On Slideshare
0
From Embeds
0
Number of Embeds
0
Actions
Shares
0
522
0
Likes
18
Embeds 0
No embeds

No notes for slide

### Transcript

• 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. 2.7 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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
• 26. 0.09 mol CH 4 16.04 g 0.91 mol Air 29.0 g Air3.23 M = + = 27.83 g mol mol mol 700 kg kmol 0.090 kmol CH 4 = 2.264 kmol CH 4 h h 27.83 kg 1.00 kmol mix 2.264 kmol CH 4 0.91 kmol air = 22.89 kmol air h h 0.09 kmol CH 4 2.264 kmol CH 4 0.95 kmol air 5% CH 4 ⇒ = 43.01 kmol air h h 0.05 kmol CH 4 Dilution air required: b43.01 - 22.89g kmol air 1000 mol = 20200 mol air h h 1 kmol 20.20 kmol Air 29 kg Air Product gas: 700 kg + = 1286 kg h h h kmol Air 43.01 kmol Air 0.21 kmol O2 32.00 kg O2 h kg O2 = 0.225 h 1.00 kmol Air 1 kmol O2 1286 kg total kg mi m M3.24 xi = , ρi = i , ρ = M Vi V mi mi 1 mi2 A: ∑ xi ρi = ∑ M Vi = M ∑V ≠ρ Not helpful. i xi mi Vi 1 V 1 B: ∑ρ = ∑M mi = M ∑ Vi = M = ρ Correct. i 1 xi 0.60 0.25 0.15 ρ = ∑ = + + ρ i 0.791 1.049 1.595 = 1.091 ⇒ ρ = 0.917 g / cm 3 R20 × 80 = 64 mol CO | 25 ⇒S 23.25 (a) Basis: 100 mol N 2 ⇒ 20 mol CH 4 | 20 × 40 = 32 mol CO T 25 N total = 100 + 20 + 64 + 32 = 216 mol 32 64 xCO = = 0.15 m ol C O / m ol , x C O 2 = = 0.30 m ol C O 2 / m ol 216 216 20 100 x CH 4 = = 0.09 mol CH 4 / mol , x N 2 = = 0.46 mol N 2 / mol 216 216 (b) M = ∑ yi M i = 015 × 28 + 0.30 × 44 + 0.09 × 16 + 0.46 × 28 = 32 g / mol . 3-10
• 27. 3.26 (a) Samples Species MW k Peak Mole Mass moles mass Area Fraction Fraction 1 CH4 16.04 0.150 3.6 0.156 0.062 0.540 8.662 C2H6 30.07 0.287 2.8 0.233 0.173 0.804 24.164 C3H8 44.09 0.467 2.4 0.324 0.353 1.121 49.416 C4H10 58.12 0.583 1.7 0.287 0.412 0.991 57.603 2 CH4 16.04 0.150 7.8 0.249 0.111 1.170 18.767 C2H6 30.07 0.287 2.4 0.146 0.123 0.689 20.712 C3H8 44.09 0.467 5.6 0.556 0.685 2.615 115.304 C4H10 58.12 0.583 0.4 0.050 0.081 0.233 13.554 3 CH4 16.04 0.150 3.4 0.146 0.064 0.510 8.180 C2H6 30.07 0.287 4.5 0.371 0.304 1.292 38.835 C3H8 44.09 0.467 2.6 0.349 0.419 1.214 53.534 C4H10 58.12 0.583 0.8 0.134 0.212 0.466 27.107 4 CH4 16.04 0.150 4.8 0.333 0.173 0.720 11.549 C2H6 30.07 0.287 2.5 0.332 0.324 0.718 21.575 C3H8 44.09 0.467 1.3 0.281 0.401 0.607 26.767 C4H10 58.12 0.583 0.2 0.054 0.102 0.117 6.777 5 CH4 16.04 0.150 6.4 0.141 0.059 0.960 15.398 C2H6 30.07 0.287 7.9 0.333 0.262 2.267 68.178 C3H8 44.09 0.467 4.8 0.329 0.380 2.242 98.832 C4H10 58.12 0.583 2.3 0.197 0.299 1.341 77.933 (b) REAL A(10), MW(10), K(10), MOL(10), MASS(10), MOLT, MASST INTEGER N, ND, ID, J READ (5, *) N CN-NUMBER OF SPECIES READ (5, *) (MW(J), K(J), J = 1, N) READ (5, *) ND DO 20 ID = 1 , ND READ (5, *)(A(J), J = 1, N) MOLT = 0. 0 MASST = 0. 0 DO 10 J = 1, N MOL(J) = MASS(J) = MOL(J) * MW(J) MOLT = MOLT + MOL(J) MASST = MASST + MASS(J) 10 CONTINUE DO 15 J = 1, N MOL(J) = MOL(J)/MOLT MASS(J) = MASS(J)/MASST 15 CONTINUE WRITE (6, 1) ID, (J, MOL(J), MASS (J), J = 1, N) 20 CONTINUE 1 FORMAT ( SAMPLE: `, I3, /, ∗ SPECIES MOLE FR. MASS FR., /, 3-11
• 28. 3.26 (cont’d) ∗ 10(3X, I3, 2(5X, F5.3), /), /) END \$DATA ∗ 4 16. 04 0. 150 30. 07 0. 287 44 . 09 0. 467 58. 12 0. 583 5 3. 6 2. 8 2. 4 1. 7 7 . 8 2. 4 5. 6 0. 4 3. 4 4 . 5 2. 6 0. 8 4 . 8 2. 5 1. 3 0. 2 6 . 4 7. 9 4 . 8 2. 3 [OUTPUT] SAMPLE: 1 SPECIES MOLE FR MASS FR 1 0.156 0.062 2 0.233 0.173 3 0.324 0.353 4 0.287 0.412 SAMPLE: 2 (ETC.) (8.7 × 10 6 × 0.40) kg C 44 kg CO 23.27 (a) = 1.28 × 10 7 kg CO 2 ⇒ 2.9 × 105 kmol CO 2 12 kg C (11 × 10 6 × 0.26) kg C 28 kg CO . = 6.67 × 10 5 kg CO ⇒ 2.38 × 10 4 kmol CO 12 kg C ( 3.8 × 10 5 × 0.10) kg C 16 kg CH 4 = 5.07 × 10 4 kg CH 4 ⇒ 3.17 × 10 3 kmol CH 4 12 kg C (1.28 × 10 7 + 6.67 × 10 5 + 5.07 × 10 4 ) kg 1 metric ton metric tons m= = 13,500 1000 kg yr M = ∑y i M i = 0.915 × 44 + 0.075 × 28 + 0.01 × 16 = 42.5 g / mol3.28 (a) Basis: 1 liter of solution 1000 mL 1.03 g 5 g H 2 SO 4 mol H 2 SO 4 = 0.525 mol / L ⇒ 0.525 molar solution mL 100 g 98.08 g H 2 SO 4 3-12
• 29. 3.28 (cont’d) 55 gal 3.7854 L min 60 s (b) t = V = = 144 s V gal 87 L min 55 gal 3.7854 L 10 3 mL 1.03 g 0.0500 g H 2 SO 4 1 lbm = 23.6 lb m H 2 SO 4 gal 1L mL g 453.59 g V 87 L m 3 1 min (c) u = = = 0.513 m / s A min 1000 L 60 s (π × 0.06 2 / 4 ) m 2 L 45 m t= = = 88 s u 0.513 m / s3.29 (a) n1 (mol/min) n2 (mol/min) 0.180 mol C6H14/mol 0.050 mol C6H14/mol 0.820 mol N2/mol 0.950 mol N2/mol 1.50 L C6H14(l)/min n3 (mol C6H14(l)/min) . 150 L 0.659 kg 1000 mol n3 = = 1147 mol / min . min L 86.17 kg Hexane balance: 0.180n1 = 0050n2 + 1147 (mol C6 H14 / min) . . U ⇒ R n = 838 mol / min V |n = 72.3 mol / min . W S solve 1 Nitrogen balance: 0.820n1 = 0950n2 (mol N2 / min) . | T 2 n3 1147 . (b) Hexane recovery = × 100% = × 100% = 76% n1 0180 838 . . b g 30 mL 1L 0.030 mol 172 g3.30 = 0155 g Nauseum . 103 mL lL 1 mol 3-13
• 30. 3.31 (a) kt is dimensionless ⇒ k (min -1 ) (b) A semilog plot of CA vs. t is a straight line ⇒ ln CA = ln CAO − kt 1 y = -0.4137x + 0.2512 0 -1 R2 = 0.9996 ln(CA) -2 -3 -4 -5 0.0 5.0 10.0 t (m in) k = 0.414 min −1 ln CAO = 02512 ⇒ CAO = 1286 lb - moles ft 3 . . FG 1b - molesIJ = C′ mol 28.317 liter 2.26462 lb - moles (c) C A H ft K liter 1 ft 3 A 3 1000 mol = 0.06243C A ′ t ′bsg 1 min t bming = = t ′ 60 60 s C A = C A 0 exp(− kt ) b g b g b g drop primes 0.06243C A = 1334 exp −0.419t ′ 60 ′ . ⇒ C A mol / L = 214 exp −0.00693t . t = 200 s ⇒ C A = 5.30 mol / L 2600 mm Hg 14.696 psi3.32 (a) = 50.3 psi 760 mm Hg 275 ft H 2 O 101.325 kPa (b) = 822.0 kPa 33.9 ft H 2 O 3.00 atm 101325 × 105 N m2 . 12 m2 (c) 2 2 = 30.4 N cm2 1 atm 100 cm 280 cm Hg 10 mm 101325 × 106 dynes cm2 1002 cm2 . dynes (d) = 3.733 × 1010 1 cm 760 mm Hg 2 1 m 2 m2 20 cm Hg 10 mm 1 atm (e) 1 atm − = 0.737 atm 1 cm 760 mm Hg 3-14
• 31. 3.32 (cont’d) (f) 25.0 psig 760 mm Hg gauge b g = 1293 mm Hg bgaugeg 14.696 psig (g) b25.0 + 14.696gpsi 760 mm Hg = 2053 mm Hg abs b g 14.696 psi (h) 325 mm Hg − 760 mm Hg = −435 mm Hg gauge b g P 35.0 lbf 144 in 2 ft 3 s2 32.174 lbm ⋅ ft 100 cm Eq. (3.4-2) ⇒ h = = (i) ρg in2 1 ft 2 1.595x62.43 lbm 32.174 ft s ⋅ lbf 2 3.2808 ft = 1540 cm CCl4 0.92 × 1000 kg 9.81 m / s2 h (m) 1N 1 kPa3.33 (a) Pg = ρgh = m 3 1 kg ⋅ m / s 2 103 N / m2 ⇒ h (m) = 0111Pg (kPa) . h Pg Pg = 68 kPa ⇒ h = 0111 × 68 = 7.55 m . FG moil = ρV = 0.92 × 1000 kg IJ FG × 7.55 × π × 16 2 3 IJ H m3 K H 4 m = 14 × 10 6 kg . K (b) Pg + Patm = Ptop + ρgh b g b g 68 + 101 = 115 + 0.92 × 1000 × 9.81 / 103 h ⇒ h = 5.98 m3.34 (a) Weight of block = Sum of weights of displaced liquids ρ h + ρ 2 h2 (h1 + h2 ) Aρ b g = h1 Aρ 1 g + h2 Aρ 2 g ⇒ ρ b = 1 1 h1 + h2 (b) Ptop = P + ρ1gh0 , P atm bottom = P + ρ1g(h0 + h ) + ρ2 gh2 , W = ρb (h + h2 ) A atm 1 b 1 ⇒Fdown = ( P + ρ1gh0 ) A + ρb (h1 + h2 ) A , Fup = [ P + ρ1g(h0 + h1) + ρ2 gh2 ]A atm atm Fdown = Fup ⇒ ρb (h1 + h2 ) A = ρ1gh1 A + ρ2 gh2 A ⇒ Wblock = W displaced liquid 3-15
• 32. 3.35 b g Δ P = Patm + ρgh − Pinside = 1 atm − 1 atm + b105g1000 kg . 9.8066 m 150 m 12 m2 1N m3 s2 1002 cm2 1 kg ⋅ m / s2 F= 154 N 65 cm2 = 100 × 104 N × . FG 022481 lb f = 2250 lb f IJ cm2 . 1N H K 14 × 62.43 lb m . 1 ft 3 2.3 × 106 gal3.36 m = ρV = 3 = 2.69 × 107 lb m ft 7.481 gal P = P0 + ρgh lb f 14 × 62.43 lb m 32.174 ft 30 ft . 1 lb f 12 ft 2 = 14.7 2 + in ft 3 s2 32.174 lb m ⋅ ft / s2 12 2 in 2 = 32.9 psi — Structural flaw in the tank. — Tank strength inadequate for that much force. — Molasses corroded tank wall π × 24 2 × 3 in 3 1 ft 3 8.0 × 62.43 lb m3.37 (a) mhead = 3 3 = 392 lb m 4 12 in ft 3 392 lb m 32.174 ft / s 2 1 lb f W = mhead g = = 392 lb f 32.174 lb m ⋅ ft / s 2 ⎡( 30 + 14.7 ) ⎤ lb f π × 202 in 2 ⎣ ⎦ Fnet = Fgas − Fatm − W = in 2 4 14.7 lbf π × 242 in 2 − 2 − 392 lb f = 7.00 × 103 lbf in 4 The head would blow off. F 7.000 × 10 lbf 3 32.174 lb m ⋅ ft/s 2 Initial acceleration: a = net = = 576 ft/s 2 mhead 392 lb m 1 lb f (b) Vent the reactor through a valve to the outside or a hood before removing the head. 3-16
• 33. 3.38 (a) Pa = ρgh + Patm , Pb = Patm If the inside pressure on the door equaled Pa , the force on the door would be F = Adoor ( Pa − Pb ) = ρghAdoor a Since the pressure at every point on the door is greater than 2m b Pa , Since the pressure at every point on the door is greater 1m than Pa , F >ρghAdoor (b) Assume an average bathtub 5 ft long, 2.5 ft wide, and 2 ft high takes about 10 min to fill. V 5 × 25 × 2 ft 3 . Vtub = ≈ = 25 ft 3 / min ⇒ V = 5 × 25 = 125 ft 3 / min . . . t 10 min (i) For a full room, h = 7 m 1000 kg 9.81 m 1N 7 m 2 m2 ⇒F > ⇒ F > 1.4 ×105 N m 3 s2 1 kg ⋅ m/s 2 The door will break before the room fills (ii) If the door holds, it will take V t fill = room = b 5 × 15 × 10 m3 g 35.3145 ft 3 1h = 31 h V 12.5 ft 3 / min 1 m3 60 min He will not have enough time.3.39 (a) Pgd i tap = 25 m H 2 O 101.3 kPa 10.33 m H 2 O = 245 kPa dP i g = b g 25 + 5 m H 2 O 101.3 kPa = 294 kPa junction 10.33 m H 2 O (b) Air in the line. (lowers average density of the water.) (c) The line could be clogged, or there could be a leak between the junction and the tap.3.40 Pabs = 800 mm Hg Pgauge = 25 mm Hg Patm = 800 − 25 = 775 mm Hg 3-17
• 34. b g3.41 (a) P1 + ρ A g h1 + h2 = P2 + ρ B gh1 + ρ C gh2 b g ⇒ P1 − P2 = ρ B − ρ A gh1 + ρ C − ρ A gh2b g (b) P1 = 121 kPa + LMb10 − 0.792g g 981 cm 30.0 cm + b137 − 0.792g g . . 981 cm 24.0 cmOP N cm 3 s 2 cm 3 s2 Q F 1 dyne I F ×G I H 1 g ⋅ cm / s JK GH 1.01325101325dynes / cm JK = 123.0 kPa 2 . × 10 kPa 6 23.42 (a) Say ρt (g/cm3) = density of toluene, ρm (g/cm3) = density of manometer fluid 500 − h ρ t g (500 − h + R ) = ρ m gR ⇒ R = ρm −1 ρt (i) Hg: ρ t = 0.866, ρ m = 13.6, h = 150 cm ⇒ R = 238 cm . (ii) H 2 O: ρ t = 0.866, ρ m = 100, h = 150 cm ⇒ R = 2260 cm . Use mercury, because the water manometer would have to be too tall. (b) If the manometer were simply filled with toluene, the level in the glass tube would be at the level in the tank. Advantages of using mercury: smaller manometer; less evaporation. (c) The nitrogen blanket is used to avoid contact between toluene and atmospheric oxygen, minimizing the risk of combustion.3.43 gb Patm = ρ f g 7.23 m ⇒ ρ f = P 7.23 g atm F P − ρ gIJ b26 cmg P − P = d ρ − ρ i gb26 cmg = G H 7.23 m K atm a b f w w F756 mmHg 1 m −1000 kg 9.81 m/s 2 Ib g = GH 7.23 m 100 cm m 3 1N 760 mmHg 1m 1 kg⋅ m/s2 1.01325×105 N m2 100 cm JK 26 cm ⇒ Pa − Pb = 81 mm Hg . 75 psi 760 mm Hg .3.44 (a) Δh = 900 − hl = = 388 mm Hg ⇒ hl = 900 − 388=512 mm 14.696 psi 338 mm Hg 14.696 psi (b) Δh = 388 − 25 × 2 = 338 mm ⇒ Pg = = 6.54 psig 760 mm Hg 3-18
• 35. 3.45 (a) h = L sin θ b g b g (b) h = 8.7 cm sin 15° = 2.3 cm H 2 O = 23 mm H 2 O3.46 (a) P = Patm − Poil − PHg 920 kg 9.81 m / s2 0.10 m 1N 760 mm Hg = 765 − 365 − m3 1 kg ⋅ m / s 1.01325 × 105 N / m2 2 = 393 mm Hg (b) — Nonreactive with the vapor in the apparatus. — Lighter than and immiscible with mercury. — Low rate of evaporation (low volatility). c h3.47 (a) Let ρ f = manometer fluid density 110 g cm 3 , ρ ac = acetone density . c0.791 g cm h 3 Differential manometer formula: ΔP = ρ f − ρ ac gh d i b ΔP mm Hg =g b110− 0791gg 981 cm h (mm) 10 cm 1 g⋅dynes . . cm s 3 1 2 mm 1 cm/ 2 760 mm Hg 1.01325×106 dyne/ cm2 = 0.02274 hb mmg V b mL sg 62 87 107 123 138 151 hb mmg 5 10 15 20 25 30 ΔPb mm Hgg 0.114 0.227 0.341 0.455 0.568 0.682 b g (b) lnV = n ln ΔP + ln K 6 5.5 y = 0.4979x + 5.2068 ln(V) 5 4.5 4 -2.5 -2 -1.5 -1 -0.5 0 ln( P) From the plot above, ln V = 04979 ln ΔP + 52068 . . b g ml s ⇒ n = 04979 ≈ 05 , ln K = 5.2068 ⇒ K = 183 . . mm Hg b g 0.5 3-19
• 36. 3.47 (cont’d) b gb g (c) h = 23 ⇒ ΔP = 0.02274 23 = 0.523 mm Hg ⇒ V = 183 0.523 b g 0.5 = 132 mL s 132 mL 0.791 g 104 g 1 mol = 104 g s = 180 mol s . s mL s 58.08 g3.48 (a) T = 85° F + 4597 = 544° R / 18 = 303 K − 273 = 30° C . . (b) T = −10° C + 273 = 263 K × 18 = 474° R − 460 = 14° F . 85° C 10° K . 85° C 18° F . 85° C 1.8° R (c) ΔT = = 85° K; = 153° F; = 153° R 10° C . 1° C 10° C . 150° R 1° F 150° R 1.0 K 150° R 1.0° C (d) = 150° F; = 83.3° K; = 83.3° C 1° R 1.8° R 1.8° R3.49 (a) T = 0.0940 × 1000 FB + 4.00 = 98.0 C ⇒ T = 98.0 × 1.8 + 32 = 208 F (b) ΔT ( C) = 0.0940ΔT ( FB) = 0.94 C ⇒ ΔT (K) = 0.94 K 0.94 C 1.8 F ΔT ( F) = = 169 F ⇒ ΔT ( R) = 169 R . . 1.0 C (c) T1 = 15 C ⇒ 100 L ; T2 = 43 C ⇒1000 L T ( C) = aT ( L) + b a= b43 − 15g C = 0.0311FG C IJ ; b = 15 − 0.0311 × 100 = 119 C . b1000 - 100g L H LK T ( C) = 0.0311T ( L) + 11.9 and ⇒ 1 T ( L) = ⎡0.0940T (o FB)+4.00-11.9 ⎤ = 3.023T (o FB)-254 0.0311 ⎣ ⎦ (d) Tbp = −88.6 C ⇒ 184.6 K ⇒ 332.3 R ⇒ -127.4 F ⇒ −9851 FB ⇒ −3232 L . (e) ΔT = 50.0 L ⇒ 1.56 C ⇒ 16.6 FB ⇒ 156 K ⇒ 2.8 F ⇒ 2.8 R . 3-20
• 37. 3.50 bT g = 100° C bT g b H 2O m AgCl = 455° C (a) V b mVg = aT b° Cg + b 5.27 = 100a + b a = 0.05524 mV ° C ⇒ 24.88 = 455a + b b = −0.2539 mV b g V mV = 0.05524T ° C − 0.2539 b g ⇓ b g T ° C = 1810V mV + 4.596 . b g (b) 100 mV →136 mV ⇒1856° C →2508° C ⇒ . . . . = b . . g dT 2508 − 1856 ° C = 326 ° C / s . dt 20 s3.51 (a) ln T = ln K + n ln R T = KR n n= b ln 250.0 110.0 g = 1184 b ln 40.0 20.0 g . ln K = ln 1100 − 1184(ln200) = 1154 ⇒ K = 3169 ⇒T = 3169R1184 . . . . . . . F 320 IJ (b) R = G 1/1.184 H 3169 K . = 49.3 (c) Extrapolation error, thermocouple reading wrong.3.52 (a) PV = 0.08206nT Pbatmg = b g P′ psig + 14696 . bg d i , V L = V ′ ft × 28317 ft 3 . 3 . 14696 L T ′( F) − 32 b g b n mol = n ′ lb - moles × g 453.59 mol lb − moles , T( K) = 1.8 + 27315 . b P′ + 14.696g × V ′ × 28317 = 0.08206 × n′ × 453.59 × L(T ′ − 32) + 27315O ⇒ 14.696 . 1 MN 1.8 . P Q 0.08206 × 14.696 × 453.59 b ⇒ P ′ + 14.696 × V ′ = g 28.317 × 18. b × n ′ × T ′ + 459.7 g b g ⇒ P′ + 14.696 V ′ = 1073n′ T ′ + 459.7 . b g 3-21
• 38. 3.52 (cont’d) (b) ntot = ′ b500 + 14.696g × 35 = 0.308 lb - mole . 10.73 × b85 + 459.7g 0308 lb - mole 0.30 lb - mole CO 28 lb m CO . mCO = = 2.6 lbm CO lb - mole lb - mole CO (c) T ′ = b3000 + 14.696g × 35 − 459.7 = 2733 F . 10.73 × 0.308 b g b3.53 (a) T ° C = a × r ohms + b g 0 = 23624a + b . U⇒ V a = 10634 . b g b g ⇒ T ° C = 10634r ohms − 25122 . . 100 = 33028a + bW . b = −25122 . FG kmol IJ = n′ (kmol) 1 min = n′ (b) n H s K min 60 s 60 P′bmm Hgg P′ Pbatmg = bg b g 1 atm = , T K = T ′ ° C + 27316 . 760 mm Hg 760 Fm I m 3 VG J =V ′ 3 1 min V ′ = H s K min 60 s 60 n′ 12186 P′ = . V′ ⇒ n′ = . b g d 0016034 P′ mm Hg V ′ m3 min i 60 760 T ′ + 27316 60 . T ′ ° C + 27316 . b g (c) T = 10.634r − 25122 . r1 = 26159 ⇒ T1 = 26.95° C . ⇒ r2 = 26157 ⇒ T2 = 26.93° C . r3 = 44.789 ⇒ T3 = 2251° C . P (mm Hg) = h + Patm = h + (29.76 in Hg) FG 760 mm Hg IJ = h + 755.9 H 29.92 in Hg K h1 = 232 mm ⇒ P = 987.9 mm Hg 1 ⇒ h2 = 156 mm ⇒ P2 = 9119 mm Hg . h3 = 74 mm ⇒ P3 = 829.9 mm Hg 3-22
• 39. 3.53 (cont’d) (d) n1 = b0.016034gb987.9gb947 60g = 0.8331 kmol CH min 26.95 + 27316 4 . n2 = b0.016034gb9119gb195g = 9.501 kmol air min . 26.93 + 27316 . n3 = n1 + n2 = 10.33 kmol min (e) V3 = b n3 T2 + 27316 . g b = gb 10.33 2251 + 27316 . . g = 387 m3 min 0.016034 P3 b gb 0.016034 829.9 g 0.8331 kmol CH 4 16.04 kg CH 4 kg CH 4 (f) = 13.36 min kmol min 0.21× 9.501 kmol O2 32.0 kg O2 0.79 × 9.501 kmol N2 28.0 kg N2 kg air + = 274 min kmol O2 min kmol N2 min 13.36 kg CH 4 min xCH4 = = 0.0465 kg CH 4 kg (13.36 + 274) kg / min3.54 REAL, MW, T, SLOPE, INTCPT, KO, E REAL TIME (100), CA (100), TK (100), X (100), Y(100) INTEGER IT, N, NT, J b g READ 5, ∗ MW, NT DO 10 IT=1, NT b g READ 5, ∗ TC, N TK(IT) = TC + 273.15 b g READ 5, ∗ (TIME (J), CA (J), J = 1 , N) DO 1 J=1, N bg bg CA J = CA J / MW XbJg = TIMEbJg YbJg = 1./CAbJg 1 CONTINUE CALL LS (X, Y, N, SLOPE, INTCPT) b g K IT = SLOPE WRITE (E, 2) TK (IT), (TIME (J), CA (J), J = 1 , N) WRITE (6, 3) K (IT) 10 CONTINUE DO 4 J=1, NT bg bg X J = 1./TK J YbJg = LOGcKbJgh 3-23
• 40. 3.54 (cont’d) 4 CONTINUE CALL LS (X, Y, NT, SLOPE, INTCPT) b KO = EXP INTCPT g E = −8.314 = SLOPE WRITE (6, 5) KO, E 2 FORMAT ( TEMPERATURE (K): , F6.2, / * TIME CA, /, * (MIN) (MOLES), / * 100 (IX, F5.2, 3X, F7.4, /)) 3 FORMAT ( K (L/MOL – MIN): , F5.3, //) 5 FORMAT (/, KO (L/MOL – MIN) : , E 12.4, /, E (J/MOL): , E 12.4) END SUBROUTINE LS (X, Y, N, SLOPE, INTCPT) REAL X(100), Y(100), SLOPE, INTCPT, SX, SY, SXX, SXY, AN INTEGER N, J SX=0 SY=0 SXX=0 SXY=0 DO 10 J=1,N SX = SX + X(J) SY = SY + Y(J) SXX = SXX + X(J)**2 SXY = SXY + X(J)*Y(J) 10 CONTINUE AN = N SX = SX/AN SY = SY/AN SXX = SXX/AN SXY = SXY/AN SLOPE = (SXY – SX*SY)/(SXX – SX**2) INTCPT = SY – SLOPE*SX RETURN END \$ DATA [OUTPUT] 65.0 4 TEMPERATURE (K): 367.15 94.0 6 TIME CA 10.0 8.1 (MIN) (MOLS/L) 20.0 4.3 10.00 0.1246 30.0 3.0 20.00 0.0662 40.0 2.2 30.00 0.0462 50.0 1.8 40.00 0.0338 3-24
• 41. 3.54 (cont’d) 60.0 1.5 50.00 0.0277 60.00 0.0231 b g K L / MOL ⋅ MIN : 0.707 bat 94°Cg 110. 6 10.0 3.5 20.0 1.8 TEMPERATURE (K): 383.15 30.0 1.2 40.0 0.92 b g K L / MOL ⋅ MIN : 1.758 50.0 0.73 60.0 0.61 127. 6 b g K0 L / MOL − MIN : 0.2329E + 10 ETC E bJ / MOLg: 0.6690E + 05 3-25
• 42. CHAPTER FOUR4.1 a. Continuous, Transient b. Input – Output = Accumulation No reactions ⇒ Generation = 0, Consumption = 0 kg kg dn dn kg 6.00 − 3.00 = ⇒ = 3.00 s s dt dt s c. 100 m3 1000 kg 1 s . t= = 333 s 1 m3 3.00 kg4.2 a. Continuous, Steady State b. k = 0 ⇒ C A = C A0 k = ∞ ⇒ CA = 0 c. Input – Output – Consumption = 0 Steady state ⇒ Accumulation = 0 A is a reactant ⇒ Generation = 0 FG m IJ C FG mol IJ = V FG m IJ C FG mol IJ + kVC FG mol IJ ⇒ C 3 3 CA0 V H sK Hm K H sK Hm K A0 3 A 3 HsK A A = 1+ kV V4.3 a. b mv kg / h g 100 kg / h 0.850 kg B / kg 0.550 kg B / kg Input – Output = 0 0.150 kg T / kg Steady state ⇒ Accumulation = 0 0.450 kg T / kg b ml kg / h g No reaction ⇒ Generation = 0, Consumption = 0 0.106 kg B / kg 0.894 kg T / kg (1) Total Mass Balance: 100.0 kg / h = mv + ml (2) Benzene Balance: 0.550 × 100.0 kg B / h = 0.850mv + 0106ml . Solve (1) & (2) simultaneously ⇒ mv = 59.7 kg h, ml = 40.3 kg h b. The flow chart is identical to that of (a), except that mass flow rates (kg/h) are replaced by masses (kg). The balance equations are also identical (initial input = final output). c. Possible explanations ⇒ a chemical reaction is taking place, the process is not at steady state, the feed composition is incorrect, the flow rates are not what they are supposed to be, other species are in the feed stream, measurement errors. 4- 1
• 43. 4.4 b. n (mol) b 0.500 n mol N 2 28 g N 2 1 kg g = 0.014 n kg N 2 b g 0.500 mol N 2 mol mol N 2 1000 g 0.500 mol CH 4 mol c. 100.0 g / s nE = b g 100 x E g C 2 H 6 1 lb m lb - mole C2 H 6 3600 s b g x E g C2 H 6 g s 453593 g 30 lb m C2 H 6 . h xP bg C H gg 3 8 b = 26.45x E lb - mole C 2 H 6 / h g xB bg C H gg 4 10 d. b g n1 lb - mole H 2 O s nO2 = 0.21n2 ( lb-mole O 2 / s ) Rn blb - mole DA sg | 0.21 lb - moleO lb - mole DA U 2 | n1 ⎛ lb-mole H 2 O ⎞ S | 0.79 lb - mole N lb - mole DAV 2 | xH2O = n1 + n2 ⎜ lb-mole ⎟ ⎝ ⎠ T 2 W 0.21n2 ⎛ lb-mole O 2 ⎞ xO2 = n1 + n2 ⎜ lb-mole ⎟ ⎝ ⎠ e. n ( mol ) nN2O4 = n ⎡0.600 − yNO2 ⎤ ( mol N 2 O4 ) ⎣ ⎦ 0.400 mol NO mol yNO2 ( mol NO 2 mol ) 0.600 − yNO2 ( mol N 2 O 4 mol )4.5 a. Basis: 1000 lbm C3H8 / h fresh feed (Could also take 1 h operation as basis - flow chart would be as below except that all / h would be deleted.) b n6 lb m / h g 0.02 lb m C3H8 / lb m 1000 lb m C3H 8 / h b n7 lb m / h g 0.98 lb m C3H 6 / lb m 0.97 lb m C3H8 / lb m Still 0.03 lb m C3H 6 / lb m Compressor b n1 lb m C3H 8 / h g b n1 lb m C3H 8 / h g Reactor n blb 2 m C H / hg 3 6 b n2 lb m C3H 6 / h g n blb CH / h g 3 n blb m H / hg 4 b n3 lb m CH 4 / h g 4 m 2 b n4 lb m H 2 / h g Stripper b n5 lb m / h g Note: the compressor and the off gas from Absorber the absorber are not mentioned explicitly in the process description, but their presence should be inferred. b n1 lb m C3H 8 / h g b n2 lb m C3H 6 / h g n blb 5 m oil / h g 4- 2
• 44. 4.5 (cont’d) b. Overall objective: To produce C3H6 from C3H8. Preheater function: Raise temperature of the reactants to raise the reaction rate. Reactor function: Convert C3H8 to C3H6. Absorption tower function: Separate the C3H8 and C3H6 in the reactor effluent from the other components. Stripping tower function: Recover the C3H8 and C3H6 from the solvent. Distillation column function: Separate the C3H5 from the C3H8.4.6 a. 3 independent balances (one for each species) b. 7 unknowns ( m1 , m3 , m5 , x2 , y2 , y4 , z4 ) – 3 balances – 2 mole fraction summations 2 unknowns must be specified c. y2 = 1 − x2 FG kg A IJ = m + b1200gb0.70g FG kg A IJ A Balance: 5300 x2 H hK 3 H hK F kg I Overall Balance: m + 5300 G J = m + 1200 + m G J F kg I 1 H hK 3 H hK 5 B Balance: 0.03m + 5300 x G F kg BIJ = 1200 y + 0.60m FG kg BIJ 1 H hK 2 H hK 4 5 z4 = 1 − 0.70 − y44.7 a. 3 independent balances (one for each species) b. bg 400 g 0.885 g H 2O mR g 0.995 g H 2O = ⇒ mR = 356 g min Water Balance: min g min gb g Acetic Acid Balance: b400gb0115g FGH g CH OOH IJK = 0.005m . min 3 R + 0.096mE FG g CH OOH IJ H min K 3 ⇒ mE = 461g min FG g IJ = m + m FG g IJ ⇒ m = 417 g min Overall Balance: mC + 400 H min K H min KR E C c. b0115gb400g − b0.005gb356g FGH min IJK = b0.096gb461g FGH min IJK ⇒ 44 g min = 44 g min . g g 4- 3
• 45. 4.7 (cont’d) d. H 2O CH 3COOH some CH3COOH CH3COOH H 2O Extractor C4 H9OH Distillation C 4 H9OH CH3COOH Column C4 H 9OH4.8 a. X-large: 25 broken eggs/min 35 45 unbroken eggs/min 120 eggs/min 0.30 broken egg/egg Large: n 1 broken eggs/min 0.70 unbroken egg/egg n 2 unbroken eggs/min b. 120 = 25 + 45 + n1 + n2 ( eggs min ) ⇒ n1 + n2 = 50 ⎫ n1 = 11 ⎪ ⎬ ⇒ ( 0.30 )(120 ) = 25 + n1 ⎪ n2 = 39 ⎭ c. n1 + n2 = 50 large eggs min b n1 large eggs broken/50 large eggs = 11 50 = 0.22 g d. b g 22% of the large eggs (right hand) and 25 70 ⇒ 36% of the extra-large eggs (left hand) are broken. Since it does not require much strength to break an egg, the left hand is probably poorly controlled (rather than strong) relative to the right. Therefore, Fred is right-handed.4.9 a. b m1 lb m strawberries g b m3 lb m W evaporated g 015 lb m S / lb m . 0.85 lb m W / lb m 1.00 lb m jam c m2 lb m S sugar h 0.667 lb m S / lb m 0.333 lb m W / lb m b. 3 unknowns ( m1 , m2 , m3 ) – 2 balances – 1 feed ratio 0 DF c. Feed ratio: m1 / m 2 = 45 / 55 (1) S balance: 0.15 m1 + m 2 = 0.667 (2) Solve simultaneously ⇒ m1 = 0.49 lb m strawberries, m 2 = 0.59 lb m sugar 4- 4
• 46. 4.10 a. 300 gal b g m1 lb m 4 unknowns ( m1 , m2 ,V40 , m3 ) 0.750 lb m C 2 H 5OH / lb m 0.250 lb m H 2O / lb m b g m3 lb m – 2 balances 0.600 lb m C 2 H 5OH / lb m – 2 specific gravities 0.400 lb m H 2O / lb m 0 DF b g V40 gal m b lb g 2 m 0.400 lb m C 2 H 5OH / lb m 0.600 lb m H 2 O / lb m b. 300 gal 1ft 3 0.877 × 62.4 lb m m1 = = 2195 lb m 7.4805 gal ft 3 Overall balance: m1 + m2 = m3 (1) C2H5OH balance: 0.750m1 + 0.400m2 = 0.600m3 (2) Solve (1) & (2) simultaneously ⇒ m2 = 1646 lb m, , m3 = 3841 lb m 1646 lb m ft 3 7.4805 gal V40 = = 207 gal 0.952 × 62.4lb m 1ft 34.11 a. 3 unknowns ( n1 , n2 , n3 ) b n1 mol / s g – 2 balances 0.0403 mol C3H 8 / mol 1 DF 0.9597 mol air / mol b n3 mol / s g 0.0205 mol C3H 8 / mol b n2 mol air / s g 0.9795 mol air / mol 0.21 mol O 2 / mol 0.79 mol N 2 / mol b. Propane feed rate: 0.0403n1 = 150 ⇒ n1 = 3722 mol / s b g Propane balance: 0.0403n1 = 0.0205n3 ⇒ n3 = 7317 mol / s b g Overall balance: 3722 + n2 = 7317 ⇒ n2 = 3600 mol / s b g c. > . The dilution rate should be greater than the value calculated to ensure that ignition is not possible even if the fuel feed rate increases slightly. 4- 5
• 47. 4.12 a. b m kg / h g 1000 kg / h 0.960 kg CH3OH / kg 2 unknowns ( m, x ) 0.500 kg CH 3OH / kg 0.040 kg H 2O / kg – 2 balances 0.500 kg H 2O / kg 0 DF 673 kg / h b x kg CH 3OH / kg g 1 − x b kg H O / kgg 2 b. Overall balance: 1000 = m + 673 ⇒ m = 327 kg / h b g b g b g Methanol balance: 0.500 1000 = 0.960 327 + x 673 ⇒ x = 0.276 kg CH 3OH / kg Molar flow rates of methanol and water: 673 kg 0.276 kg CH3OH 1000 g mol CH3OH = 5.80 × 103 mol CH3OH / h h kg kg 32.0 g CH3OH 673 kg 0.724 kg H 2O 1000 g mol H 2O = 2.71 × 104 mol H 2O / h h kg kg 18 g H 2O Mole fraction of Methanol: 5.80 × 103 = 0176 mol CH 3OH / mol . 5.80 × 103 + 2.71 × 104 c. Analyzer is wrong, flow rates are wrong, impurities in the feed, a reaction is taking place, the system is not at steady state.4.13 a. Product Reactor effluent 1239 kg Feed Reactor Purifier R = 583 2253 kg 2253 kg W aste R = 388 b g m w kg R = 140 Analyzer Calibration Data 1 1.364546 x p = 0.000145R xp 0.1 0.01 100 R 1000 4- 6
• 48. 4.13 (cont’d) b g b. Effluent: x = 0.000145 388 1.3645 = 0.494 kg P / kg p Product: x = 0.000145b583g 1.3645 p = 0.861 kg P / kg Waste: x = 0.000145b140g 1.3645 p = 0123 kg P / kg . 0.861b1239g Efficiency = × 100% = 95.8% 0.494b2253g c. Mass balance on purifier: 2253 = 1239 + mw ⇒ mw = 1014 kg P balance on purifier: b gb g Input: 0.494 kg P / kg 2253 kg = 1113 kg P Output: b0.861 kg P / kggb1239 kgg + b0123 kg P / kggb1014 kgg = 1192 kg P . The P balance does not close . Analyzer readings are wrong; impure feed; extrapolation beyond analyzer calibration data is risky -- recalibrate; get data for R > 583; not at steady state; additional reaction occurs in purifier; normal data scatter.4.14 a. b n1 lb- mole/ h g . 00100 lb- mole H2O/ lb- mole . 09900 lb- mole DA/ lb- mole b g n3 lb- mole/ h . 0100 lb- mole H2O/ lb- mole b n2 lb- mole HO/ h 2 g . 0900 lb- mole DA/ lb- mole v dft / hi 2 3 4 unknowns ( n1 , n2 , n3 , v ) – 2 balances – 1 density – 1 meter reading = 0 DF Assume linear relationship: v = aR + b v2 − v1 96.9 − 40.0 Slope: a = = = 1.626 R2 − R1 50 − 15 Intercept: b = v a − aR1 = 40.0 − 1.626 15 = 15.61 b g b g v2 = 1.626 95 + 15.61 = 170 ft / h c 3 h n2 = 170 ft 3 62 .4 lb m lb - mol h ft 3 18.0 lb m b = 589 lb - moles H 2 O / h g DA balance: 0.9900n1 = 0.900n3 (1) Overall balance: n1 + n2 = n3 (2) Solve (1) & (2) simultaneously ⇒ n1 = 5890 lb - moles / h, n3 = 6480 lb - moles / h b. Bad calibration data, not at steady state, leaks, 7% value is wrong, v − R relationship is not linear, extrapolation of analyzer correlation leads to error. 4- 7
• 49. 4.15 a. b m kg / s g 100 kg / s 0.900 kg E / kg 0.600 kg E / kg 0100 kg H 2 O / kg . 0.050 kg S / kg 3 unknowns ( m, xE , xS ) 0.350 kg H 2 O / kg – 3 balances b m kg / s g 0 DF b x E kg E / kg g x b kg S / kg g S 1 − x − x b kg H O / kg g E S 2 b. Overall balance: 100 = 2m ⇒ m = 50.0 kg / s b g b g b g S balance: 0.050 100 = xS 50 ⇒ xS = 0100 b kgS / kgg . b g b g b g E balance: 0.600 100 = 0.900 50 + x E 50 ⇒ x E = 0.300 kg E / kg kg E in bottom stream 0.300b50g kg E in bottom stream = = 0.25 kg E in feed 0.600b100g kg E in feed c. bg bg bg x = aRb ⇒ ln x = ln a + b ln R lnb x / x g lnb0.400 / 0100g . b= 2 1 = = 1491 lnb R / R g lnb38 / 15g . 2 1 lnbag = lnb x g − b lnb R g = lnb0100g − 1491lnb15g = −6.340 ⇒ a = 1764 × 10 1 1 . . . −3 x = 1764 × 10−3 R1.491 . F x I F 0.900 IJ 1 1 R=G J =G b 1.491 H a K H 1764 × 10 K . −3 = 655 . d. Device not calibrated – recalibrate. Calibration curve deviates from linearity at high mass fractions – measure against known standard. Impurities in the stream – analyze a sample. Mixture is not all liquid – check sample. Calibration data are temperature dependent – check calibration at various temperatures. System is not at steady state – take more measurements. Scatter in data – take more measurements. 4- 8
• 50. 4.16 a. 4.00 mol H 2SO 4 0.098 kg H 2SO 4 L of solution L of solution mol H 2SO 4 1213 kg solution . b = 0.323 kg H 2SO 4 / kg solution g b. v1 L bg 5 unknowns ( v1 , v2 , v3 , m2 , m3 ) – 2 balances 100 kg bg v3 L – 3 specific gravities m b kg g 0.200 kg H 2SO 4 / kg 0 DF 3 0.800 kg H 2 O / kg 0.323 kg H 2SO 4 / kg SG = 1139 . 0.677 kg H 2 O / kg bg v2 L SG = 1.213 m b kg g 2 0.600 kg H 2SO 4 / kg 0.400 kg H 2 O / kg SG = 1.498 Overall mass balance: 100 + m2 = m3 U ⇒ m = 44.4 kg V m = 144 kg 2 Water balance: 0.800b100g + 0.400m = 0.677m W 2 3 3 100 kg L v1 = = 87.80 L20%solution 1139 kg . 44.4 kg L v2 = = 29.64 L 60% solution 1498 kg . v1 87.80 L 20%solution = = 2.96 v2 29.64 L 60% solution c. 1250 kg P 44.4 kg 60%solution L = 257 L / h h 144 kg P 1498 kgsolution .4.17 b g m1 kg @\$18 / kg 0.25 kg P / kg 0.75 kg H2O / kg 100 kg . b g m2 kg @\$10 / kg 017 kg P/ kg . 012 kg P / kg . 0.83 kg H2O / kg 0.88 kg H 2O / kg Overall balance: m1 + m2 = 100 . (1) Pigment balance: 0.25m1 + 012m2 = 017 100 . . . b g (2) Solve (1) and (2) simultaneously ⇒ m = 0.385 kg 25% paint, m = 0.615 kg12% paint Cost of blend: 0.385b\$18.00g + 0.615b\$10.00g = \$13.08 per kg 1 2 Selling price: 110b\$13.08g = \$14.39 per kg . 4- 9
• 51. 4.18 a. b gb m1 kg H 2O 85% of entering water g 100 kg 0.800 kgS / kg 0.200 kg H 2O / kg b g m2 kgS m b kg H Og 3 2 gb g b 85% drying: m1 = 0.850 0.200 100 = 17.0 kg H 2O Sugar balance: m = 0.800b100g = 80.0 kgS 2 Overall balance: 100 = 17 + 80 + m3 ⇒ m3 = 3 kg H 2O 3 kg H 2O xw = = 0.0361 kg H 2O / kg b 3 + 80 kg g m1 17 kg H 2O = = 0.205 kg H 2O / kg wet sugar m2 + m3 80 + 3 kg b g b. 1000 tons wet sugar 3 tons H 2 O = 30 tons H 2 O / day day 100 tons wet sugar 1000 tons WS 0.800 tons DS 2000 lb m \$0.15 365 days = \$8.8 × 107 per year day ton WS ton lb m year c. xw = 1 10 b g x w1 + x w 2 +...+ x w10 = 0.0504 kg H 2 O / kg SD = 1 9 b 2 g x w1 − x w +...+ x w10 − x w b 2 g = 0.00181 kg H 2 O / kg Endpoints = 0.0504 ± 3 0.00181 b g Lower limit = 0.0450, Upper limit = 0.0558 d. The evaporator is probably not working according to design specifications since x w = 0.0361 < 0.0450 .4.19 a. c h v1 m 3 m b kg H O g 5 unknowns ( v1 , v2 , v3 , m1 , m3 ) d i 1 2 SG = 1.00 v3 m 3 – 1 mass balance m b kg suspension g 3 – 1 volume balance v2 m d i 3 SG = 1.48 – 3 specific gravities 0 DF 400 kg galena SG = 7.44 Total mass balance: m1 + 400 = m3 (1) 4- 10
• 52. 4.19 (cont’d) Assume volume additivity: b g m1 kg m3 + 400 kg m 3 m kg = 3 m3 (2) b g 1000 kg 7440 kg 1480 kg Solve (1) and (2) simultaneously ⇒ m1 = 668 kg H 2O, m3 = 1068 kg suspension 668 kg m3 v1 = = 0.668 m 3 water fed to tank 1000 kg b. Specific gravity of coal < 1.48 < Specific gravity of slate c. The suspension begins to settle. Stir the suspension. 1.00 < Specific gravity of coal < 1.484.20 a. b n1 mol / h g b n2 mol / h g 0.040 mol H 2O / mol b x mol H 2O / mol g 0.960 mol DA / mol b 1 − x mol DA / mol g b n3 mol H 2O adsorbed / h g 97% of H 2O in feed Adsorption rate: n3 = b354 − 3.40g kg . mol H 2O = 1556 mol H 2O / h . 5h 0.0180 kg H 2O b g 97% adsorbed: 156 = 0.97 0.04n1 ⇒ n1 = 401 mol / h . . Total mole balance: n1 = n2 + n3 ⇒ n2 = 401 − 1556 = 38.54 mol / h . . Water balance: 0.040 ( 40.1) = 1.556 + x ( 38.54 ) ⇒ x = 1.2 × 10−3 ( mol H 2 O/mol ) b. The calcium chloride pellets have reached their saturation limit. Eventually the mole fraction will reach that of the inlet stream, i.e. 4%.4.21 a. 300 lb m / h 0.55 lb m H 2SO 4 / lb m 0.45 lb m H 2O / lb m b mC lb m / h g b mB lb m / h g 0.75 lb m H 2SO 4 / lb m 0.90 lb m H 2SO 4 / lb m 0.25 lb m H 2O / lb m 010 lb m H 2O / lb m . Overall balance: 300 + mB = mC (1) b g H2SO4 balance: 0.55 300 + 0.90mB = 0.75mC (2) Solve (1) and (2) simultaneously ⇒ mB = 400 lb m / h, mC = 700 lb m / h 4- 11
• 53. 4.21 (cont’d) 500 − 150 b. mA − 150 = 70 − 25 b g RA − 25 ⇒ mA = 7.78 RA − 44.4 800 − 200 mB − 200 = 60 − 20 b g RB − 20 ⇒ mB = 15.0 RB − 100 ln 100 − ln 20 ln x − ln 20 = 10 − 4 b g Rx − 4 ⇒ ln x = 0.2682 Rx + 1923 ⇒ x = 6.841e 0.2682 Rx . 300 + 44.4 400 + 100 mA = 300 ⇒ RA = = 44.3, mB = 400 ⇒ RB = = 333, . 7.78 15.0 1 FG 55 IJ x = 55% ⇒ Rx = 0.268 ln H 6.841 = 7.78 K c. Overall balance: mA + mB = mC H2SO4 balance: 0.01xmA + 0.90mB = 0.75mC = 0.75 mA + mB ⇒ mB = b g b0.75 − 0.01xgm A 015 . ⇒ 15.0 RB − 100 = d 0.75 − 0.01 6.841e 0.2682 Rx i b7.78 R A − 44.4 g 015 . d ⇒ RB = 2.59 − 0.236e 0.2682 Rx iR A + 135e0.2682 Rx − 813 . . b g b g − 813 = 333 Check: RA = 44.3, Rx = 7.78 ⇒ RB = 2.59 − 0.236e e 0.2682 7.78 j44.3 + 135e . 0.2682 7.78 . .4.22 a. b n A kmol / h g 010 kmol H 2 / kmol . 100 kg / h 0.90 kmol N 2 / kmol b nP kmol / h g 0.20 kmol H 2 / kmol b nB kmol / h g 0.80 kmol N 2 / kmol 0.50 kmol H 2 / kmol 0.50 kmol N 2 / kmol b g b MW = 0.20 2.016 + 0.80 28.012 = 22.813 kg / kmol g 100 kg kmol ⇒ nP = = 4.38 kmol / h h 22.813 kg Overall balance: n A + nB = 4.38 (1) H2 balance: 010n A + 0.50 nB = 0.20 4.38 . b g (2) Solve (1) and (2) simultaneously ⇒ n A = 3.29 kmol / h, n B = 110 kmol / h . 4- 12
• 54. 4.22 (cont’d) b. mP nP = 22.813 mP Overall balance: n A + nB = 22.813 x m H2 balance: x A n A + x B nB = P P 22.813 ⇒ nA = b mP x B − x P g nB = b mP x P − x A g b 22.813 x B − x A g b 22.813 x B − x A g c. Trial XA XB XP mP nA nB 1 0.10 0.50 0.10 100 4.38 0.00 2 0.10 0.50 0.20 100 3.29 1.10 3 0.10 0.50 0.30 100 2.19 2.19 4 0.10 0.50 0.40 100 1.10 3.29 5 0.10 0.50 0.50 100 0.00 4.38 6 0.10 0.50 0.60 100 -1.10 5.48 7 0.10 0.50 0.10 250 10.96 0.00 8 0.10 0.50 0.20 250 8.22 2.74 9 0.10 0.50 0.30 250 5.48 5.48 10 0.10 0.50 0.40 250 2.74 8.22 11 0.10 0.50 0.50 250 0.00 10.96 12 0.10 0.50 0.60 250 -2.74 13.70 The results of trials 6 and 12 are impossible since the flow rates are negative. You cannot blend a 10% H2 mixture with a 50% H2 mixture and obtain a 60% H2 mixture. d. Results are the same as in part c.4.23 Venous blood Arterial blood 195.0 ml / min 200.0 ml / min 1.75 mg urea / ml 1.90 mg urea / ml Dialysate Dialyzing fluid b v ml / min g c b mg urea / mlg 1500 ml / min a. Water removal rate: 200.0 − 195.0 = 5.0 ml / min . b . g b g Urea removal rate: 190 200.0 − 175 195.0 = 38.8 mg urea / min b. v = 1500 + 5.0 = 1505ml / min 38.8 mg urea/min c= = 0.0258 mg urea/ml 1505 ml/min c. b2.7 − 11g mg removed . 1 min 10 3 ml 5.0 L = 206 min (3.4 h) ml 38.8 mg removed 1L 4- 13
• 55. 4.24 a. b n1 kmol / min g 20.0 kg CO 2 / min b n3 kmol / min g b n2 kmol / min g 0.023 kmol CO 2 / kmol 0.015 kmol CO 2 / kmol 20.0 kg CO 2 kmol n1 = = 0.455 kmol CO 2 / min min 44.0 kg CO 2 Overall balance: 0.455 + n2 = n3 (1) CO2 balance: 0.455 + 0.015n2 = 0.023n3 (2) Solve (1) and (2) simultaneously ⇒ n2 = 55.6 kmol / min, n3 = 561 kmol / min . b. 150 m u= = 8.33 m / s 18 s 1 561 kmol . m3 1 min s A = πD 2 = ⇒ D = 108 m . 4 min 0123 kmol 60 s 8.33 m .4.25 b Spectrophotometer calibration: C = kA ====> C μg / L = 3.333 A A = 0.9 g C =3 . . . b gb g Dye concentration: A = 018 ⇒ C = 3333 018 = 0.600 μg / L 0.60 cm 3 1L 5.0 mg 103 μ g Dye injected = = 3.0 μ g 103 cm 3 1L 1 mg b g bg ⇒ 3.0 μ g V L = 0.600 μ g / L ⇒ V = 5.0 L4.26 a. 1000 L B / min b n3 kmol / ming b m2 kg B / min g y b kmol SO / kmolg 3 2 1 − y b kmol A / kmolg V d m / min i 3 3 m b kg / min g 1 n b kmol / min g 4 x b kg SO / kgg 1 y b kmol SO / kmolg 4 2 1 − x b kg B / kgg 1 2 1 − y b kmol A / kmolg 1 4 4- 14
• 56. 4.26 (cont’d) 8 unknowns ( n1 , n3 , v1 , m2 , m4 , x4 , y1 , y3 ) – 3 material balances – 2 analyzer readings – 1 meter reading – 1 gas density formula – 1 specific gravity 0 DF b. Orifice meter calibration: d i A log plot of V vs. h is a line through the points h1 = 100, V1 = 142 and h2 = 400, V2 = 290 . d i ln V = b ln h + ln a ⇒ V = ah b b= d h = lnb290 142g = 0.515 ln V2 V1 lnbh h g lnb400 100g 2 1 ln a = ln V − b ln h = lnb142g − 0.515 ln 100 = 2.58 ⇒ a = e 1 1 2 .58 = 13.2 ⇒ V = 13.2h 0.515 Analyzer calibration: ln y = bR + ln a ⇒ y = aebR b= b ln y 2 y1 g = lnb01107 0.00166g = 0.0600 . U | R2 − R1 90 − 20 | | ln a = ln y1 − bR1 = lnb0.00166g − 0.0600b20g = −7.60V ⇒ y = 5.00 × 10 −4 e 0.0600 R E | | a = 5.00 × 10 −4 | W c. h1 = 210 mm ⇒ V1 = 13.2 210 b g = 207.3 m min 0.515 3 ρ feed gas = b12.2g b150 + 14.7g 14.7 batmg = 0.460 mol / L = 0.460 kmol / m 3 b75 + 460g 18 bKg . E 207.3 m 3 0.460 kmol n1 = = 95.34 kmol min min m3 b g R1 = 82.4 ⇒ y1 = 500 × 10−4 exp 0.0600 × 82.4 = 0.0702 kmol SO2 kmol . R3 = 116 ⇒ y3 = 500 × 10−4 . . expb0.0600 × 116g = 0.00100 kmol SO . 2 kmol 1000 L B 130 kg . m2 = = 1300 kg / min min LB 4- 15
• 57. 4.26 (cont’d) b gb g b g A balance: 1 − 0.0702 95.34 = 1 − 0.00100 n3 ⇒ n3 = 88.7 kmol min SO2 balance: b0.0702gb9534g( 64.0 kg / kmol) = b0.00100gb88.7g( 64) + m x . 4 4 (1) B balance: 1300 = m4 (1 − x4 ) (2) Solve (1) and (2) simultaneously ⇒ m4 = 1723 kg / min, x4 = 0.245 kg SO2 absorbed / kg SO2 removed = m4 x4 = 422 kg SO 2 / min d. Decreasing the bubble size increases the bubble surface-to-volume ratio, which results in a higher rate of transfer of SO2 from the gas to the liquid phase.4.27 a. d V2 m 3 / min i b n3 kmol / min g m b kg B / min g y b kmolSO / kmol g 3 2 1 − y b kmol A / kmol g 2 3 d V1 m / min 3 i R3 n b kmol / min g 1 b m4 kg / min g y b kmolSO / kmol g 1 2 x b kgSO kg g 4 2 1 − y b kmol A / kmol g 1 1 − x b kg B / kg g 4 P , T1 , R1 , h1 1 b. 14 unknowns ( n1 ,V1 , y1 , P , T1 , R1 , h1 ,V2 , m2 , n3 , y3 , R3 , m4 , x4 ) 1 – 3 material balances – 3 analyzer and orifice meter readings – 1 gas density formula (relates n1 and V1 ) – 1 specific gravity (relates m2 and V2 ) 6 DF b g b A balance: 1 − y1 n1 = 1 − y3 n3 g (1) x4 m4 SO2 balance: y1n1 = y3n3 + (2) 64 kgSO 2 / kmol b B balance: m2 = 1 − x4 m4 g (3) Calibration formulas: y1 = 5.00 × 10−4 e0.060 R1 (4) −4 0.060 R3 y3 = 5.00 × 10 e (5) V1 = 13.2h10.515 (6) Gas density formula: n1 = b 12.2 P + 14.7 / 14.7 1 g bT + 460g / 18 1 . V1 (7) Liquid specific gravity: SG = 130 ⇒ V2 = . m2 kgb gm3 (8) h 1300 kg 4- 16
• 58. 4.27 (cont’d) c. T1 75 °F y1 0.07 kmol SO2/kmol P1 150 psig V1 207 m3/h h1 210 torr n1 95.26 kmol/h R1 82.4 Trial x4 (kg SO2/kg) y3 (kmol SO2/kmol) V2 (m3/h) n3 (kmol/h) m4 (kg/h) m2 (kg/h) 1 0.10 0.050 0.89 93.25 1283.45 1155.11 2 0.10 0.025 1.95 90.86 2813.72 2532.35 3 0.10 0.010 2.56 89.48 3694.78 3325.31 4 0.10 0.005 2.76 89.03 3982.57 3584.31 5 0.10 0.001 2.92 88.68 4210.72 3789.65 6 0.20 0.050 0.39 93.25 641.73 513.38 7 0.20 0.025 0.87 90.86 1406.86 1125.49 8 0.20 0.010 1.14 89.48 1847.39 1477.91 9 0.20 0.005 1.23 89.03 1991.28 1593.03 10 0.20 0.001 1.30 88.68 2105.36 1684.29 V2 vs. y3 3 .5 0 3 .0 0 2 .5 0 V 2 (m /h) 2 .0 0 3 1 .5 0 1 .0 0 0 .5 0 0 .0 0 0 .0 0 0 0.0 2 0 0 .0 4 0 0 .06 0 y 3 ( k m o l S O 2 /k m o l) x4 = 0 .1 0 x4 = 0 .20 For a given SO2 feed rate removing more SO2 (lower y3) requires a higher solvent feed rate ( V2 ). For a given SO2 removal rate (y3), a higher solvent feed rate ( V2 ) tends to a more dilute SO2 solution at the outlet (lower x4). d. Answers are the same as in part c.4.28 Maximum balances: Overall - 3, Unit 1 - 2; Unit 2 - 3; Mixing point - 3 Overall mass balance ⇒ m3 Mass balance - Unit 1 ⇒ m1 A balance - Unit 1 ⇒ x1 Mass balance - mixing point ⇒ m2 A balance - mixing point ⇒ x2 C balance - mixing point ⇒ y2 4- 17
• 59. 4.29 a. 100 mol / h b n2 mol / h g b n4 mol / h g 0.300 mol B / mol b g x B 2 mol B / mol 0.940 mol B / mol 0.250 mol T / mol Column 1 x b mol T / molg Column 2 0.060 mol T / mol T2 0.450 mol X / mol 1 − x − x b mol X / molg B2 T2 n b mol / h g 3 b n5 mol / h g 0.020 mol T / mol b g x B5 mol B / mol 0.980 mol X / mol x b mol T / molg T5 1 − x − x b mol X / molg B5 T5 Column 1 Column 2: 4 unknowns ( n2 , n3 , x B 2 , xT 2 ) 4 unknowns ( n3 , n4 , n5 , y x ) –3 balances – 3 balances – 1 recovery of X in bot. (96%) – 1 recovery of B in top (97%) 0 DF 0 DF Column 1 b gb g 96% X recovery: 0.96 0.450 100 = 0.98n3 (1) Total mole balance: 100 = n2 + n3 (2) b g B balance: 0.300 100 = x B 2 n2 (3) T balance: 0.250b100g = x T 2 n2 + 0.020n3 (4) Column 2 97% B recovery: 0.97 x B 2 n2 = 0.940n4 (5) Total mole balance: n2 = n4 + n5 (6) B balance: x B 2 n2 = 0.940 n4 + x B5 n5 (7) T balance: xT 2 n2 = 0.060n4 + xT 5 n5 (8) b. (1) ⇒ n3 = 44.1 mol / h (2) ⇒ n2 = 55.9 mol / h (3) ⇒ x B 2 = 0.536 mol B / mol (4) ⇒ x T 2 = 0.431 mol T / mol (5) ⇒ n4 = 30.95 mol / h ( 6) ⇒ n5 = 24.96 mol / h (7) ⇒ x B5 = 0.036 mol B / mol (8) ⇒ x T 5 = 0.892 mol T / mol b 0.940 30.95 g × 100% = 97% Overall benzene recovery: b g 0.300 100 0.892b24.96g × 100 = 89% 0.250b100g Overall toluene recovery: 4- 18
• 60. 4.30 a. 100 kg / h b m3 kg / h g b g m4 kg / h b m10 kg / h g 0.035 kg S / kg b x3 kg S / kg g 4 x b kg S / kgg 10 0.050 kg S / kg 0.965 kg H 2O / kg 1 b 1 − x3 kg H 2O / kg g 4 1 − x b kg H O / kgg 4 2 0.950 kg H 2O / kg b 0100mw kg H 2O / h . g b 0100mw kg H 2O / h . g b 0100mw kg H 2O / h . g b mw kg H 2O / h g b. Overall process 100 kg/h m10 ( kg / h) 0.035 kg S/kg 0.050 kg S/kg 0.965 kg H2O/kg 0.950 kg H2O/kg mw ( kg H 2 O / h) b g Salt balance: 0.035 100 = 0.050m10 Overall balance: 100 = mw + m10 H2O yield: Yw = b mw kg H 2O recovered g b 96.5 kg H 2O in fresh feed g First 4 evaporators 100 kg/ h b m4 kg / h g 0.035 kg S/ kg b x 4 kg S/ kg g 0.965 kg H2 O / kg b g 1 − x4 kg H2 O / kg 4 × 0100m b kg H O / hg . w 2 b Overall balance: 100 = 4 0100 mw + m4 . g b g Salt balance: 0.035 100 = x4 m4 c. Yw = 0.31 x 4 = 0.0398 4- 19
• 61. 4.31 a. 2n1 mol b g 0.97 mol B / mol Condenser 0.03 mol T / mol b g n1 mol b g n1 mol (89.2% of Bin feed ) 0.97 mol B / mol 0.97 mol B / mol 100 mol 0.03 mol T / mol 0.03 mol T / mol 0.50 mol B / mol Still 0.50 mol T / mol b gb n4 mol 45% of feed to reboiler g y b mol B / molg B 1 − y b mol T / molg B b g n3 mol b g n2 mol Reboiler x b mol B / molg z b mol B / molg B B 1 − x b mol T / molg 1 − z b mol T / molg B B Overall process: 3 unknowns ( n1 , n3 , x B ) Still: 5 unknowns ( n1 , n2 , n4 , y B , z B ) – 2 balances – 2 balances – 1 relationship (89.2% recovery) 3 DF 0 DF Condenser: 1 unknown ( n1 ) Reboiler: 6 unknowns ( n2 , n3 , n4 , x B , y B , z B ) – 0 balances – 2 balances 1 DF – 2 relationships (2.25 ratio & 45% vapor) 3 DF Begin with overall process. b. Overall process b gb g 89.2% recovery: 0.892 0.50 100 = 0.97 n1 Overall balance: 100 = n1 + n3 b g B balance: 0.50 100 = 0.97 n1 + x B n3 Reboiler e j = 2.25 yB / 1 − yB / b1 − x g Composition relationship: xB B Percent vaporized: n4 = 0.45n2 (1) Mole balance: n2 = n3 + n4 (2) (Solve (1) and (2) simultaneously.) B balance: z B n2 = x B n3 + y B n4 4- 20
• 62. 4.31 (cont’d) c. B fraction in bottoms: x B = 0100 mol B / mol . Moles of overhead: n1 = 46.0 mol Moles of bottoms: n3 = 54.0 mol b1 − x gn × 100% = b1 − 010gb54.02g × 100% = 97% B 3 . 0.50b100g 0.50b100g Recovery of toluene:4.32 a. b m3 kg H 2O g Bypass Mixing point Basis: 100 kg 100 kg b g m1 kg Evaporator b g m4 kg b g m5 kg 0.12 kg S / kg 012 kg S / kg . 0.58 kg S / kg 0.42 kg S / kg 0.88 kg H2O / kg 0.88 kg H2O / kg 0.42 kg H 2O / kg 0.58 kg H2O / kg b g m2 kg 012 kg S / kg . 0.88 kg H2O / kg Overall process: 2 unknowns ( m3 , m5 ) Bypass: 2 unknowns ( m1 , m2 ) – 2 balances – 1 independent balance 0 DF 1 DF Evaporator: 3 unknowns ( m1 , m3 , m4 ) Mixing point: 3 unknowns ( m2 , m4 , m5 ) – 2 balances – 2 balances 1 DF 1 DF b g Overall S balance: 012 100 = 0.42m5 . Overall mass balance: 100 = m3 + m5 Mixing point mass balance: m4 + m2 = m5 (1) Mixing point S balance: 0.58m4 + 012m2 = 0.42m5 . (2) Solve (1) and (2) simultaneously Bypass mass balance: 100 = m1 + m2 b. m1 = 90.05 kg, m2 = 9.95 kg, m3 = 714 kg, m4 = 18.65 kg, m5 = 28.6 kg product . m2 Bypass fraction: = 0.095 100 c. Over-evaporating could degrade the juice; additional evaporation could be uneconomical; a stream consisting of 90% solids could be hard to transport. 4- 21
• 63. 4.33 a. b m4 kg Cr / h g b m1 kg / h g b m2 kg / h g b g m5 kg / h b g m6 kg / h 0.0515 kg Cr / kg 0.0515 kg Cr / kg Treatment x b kg Cr / kgg 5 x b kg Cr / kgg 6 0.9485 kg W / kg 0.9485 kg W / kg Unit 1 − x b kg W / kgg 5 1 − x b kg W / kgg 6 b m3 kg / h g 0.0515 kg Cr / kg 0.9485 kg W / kg b. b m1 = 6000 kg / h ⇒ m2 = 4500 kg / h maximum allowed value g Bypass point mass balance: m3 = 6000 − 4500 = 1500 kg / h b gb g 95% Cr removal: m4 = 0.95 0.0515 4500 = 220.2 kg Cr / h Mass balance on treatment unit: m5 = 4500 − 220.2 = 4279.8 kg / h Cr balance on treatment unit: x5 = b 0.0515 4500 − 220.2 g = 0.002707 kg Cr / kg 4779.8 Mixing point mass balance: m6 = 1500 + 4279.8 = 5779.8 kg / h Mixing point Cr balance: x6 = b g 0.0515 1500 + 0.0002707 4279.8 b = 0.0154 kg Cr / kg g 5779.8 c. m 1 (kg/h) m 2 (kg/h) m 3 (kg/h) m 4 (kg/h) m 5 (kg/h) x5 m 6 (kg/h) x6 1000 1000 0 48.9 951 0.00271 951 0.00271 2000 2000 0 97.9 1902 0.00271 1902 0.00271 3000 3000 0 147 2853 0.00271 2853 0.00271 4000 4000 0 196 3804 0.00271 3804 0.00271 5000 4500 500 220 4280 0.00271 4780 0.00781 6000 4500 1500 220 4280 0.00271 5780 0.0154 7000 4500 2500 220 4280 0.00271 6780 0.0207 8000 4500 3500 220 4280 0.00271 7780 0.0247 9000 4500 4500 220 4280 0.00271 8780 0.0277 10000 4500 5500 220 4280 0.00271 9780 0.0301 4- 22
• 64. 4.33 (cont’d) m 1 vs. x 6 0.03500 0.03000 x 6 (kg Cr/kg) 0.02500 0.02000 0.01500 0.01000 0.00500 0.00000 0 2000 4000 6000 8000 10000 12000 m 1 (kg/h) d. Cost of additional capacity – installation and maintenance, revenue from additional recovered Cr, anticipated wastewater production in coming years, capacity of waste lagoon, regulatory limits on Cr emissions.4.34 a. b 175 kg H 2O / s 45% of water fed to evaporator g b m1 kg / s g b m4 kg K 2SO 4 / s g b m6 kg K 2SO 4 / s g Filter cake 0196 kg K 2SO 4 / kg . b m5 kg H 2O / s g Evaporator b m7 kg H 2O / s g Crystallizer Filter b 10 m2 kg K 2SO 4 / s g 0.804 kg H 2O / kg b g m2 kgsoln / s R0.400 kg K SO / kg U S0.600 kg H O / kg V 2 4 Filtrate T 2 W b m3 kg / s g 0.400 kg K 2SO 4 / kg 0.600 kg H 2 O / kg Let K = K2SO4, W = H2 Basis: 175 kg W evaporated/s Overall process: 2 unknowns ( m1 , m2 ) Mixing point: 4 unknowns ( m1 , m3 , m4 , m5 ) - 2 balances - 2 balances 0 DF 2 DF Evaporator: 4 unknowns ( m4 , m5 , m6 , m7 ) Crystallizer: 4 unknowns ( m2 , m3 , m6 , m7 ) – 2 balances – 2 balances – 1 percent evaporation 2 DF 1 DF Strategy: Overall balances ⇒ m1 , m2 U verify that each | chosen subsystem involves % evaporation ⇒ m5 | V Balances around mixing point ⇒ m , m | no more than two Balances around evaporator ⇒ m , m | unknown variables 3 4 W 6 7 4- 23
• 65. 4.34 (cont’d) Overall mass balance: m1 = 175 + 10m2 + m2 U | 0196m1 = 10m2 + 0.400m2 V | Overall K balance: . W Production rate of crystals = 10m2 45% evaporation: 175 kg evaporated min = 0.450m5 W balance around mixing point: 0.804m1 + 0.600m3 = m5 Mass balance around mixing point: m1 + m3 = m4 + m5 K balance around evaporator: m6 = m4 W balance around evaporator: m5 = 175 + m7 m4 Mole fraction of K in stream entering evaporator = m4 + m5 b. Fresh feed rate: m1 = 221 kg / s Production rate of crystals = 10m2 = 416 kg K s s . bg b m3 kg recycle s g= 352.3 = 160 kg recycle Recycle ratio: b m1 kg fresh feed s g 220.8 . kg fresh feed c. Scale to 75% of capacity. Flow rate of stream entering evaporator = 0.75(398 kg / s) = 299 kg / s 46.3% K, 53.7% W d. Drying . Principal costs are likely to be the heating cost for the evaporator and the dryer and the cooling cost for the crystallizer. 4- 24
• 66. 4.35 a. Overall objective: Separate components of a CH4-CO2 mixture, recover CH4, and discharge CO2 to the atmosphere. Absorber function: Separates CO2 from CH4. Stripper function: Removes dissolved CO2 from CH3OH so that the latter can be reused. b. The top streams are liquids while the bottom streams are gases. The liquids are heavier than the gases so the liquids fall through the columns and the gases rise. c. b n1 mol / h g b g n5 mol N 2 / h 0.010 mol CO 2 / mol n b mol CO / h g 6 2 0.990 mol CH 4 / mol 100 mol / h Absorber b n2 mol / h g Stripper 0.300 mol CO 2 / mol 0.005 mol CO 2 / mol b n5 mol N 2 / h g 0.700 mol CH 4 / mol 0.995 mol CH 3OH / mol b g n3 mol CO 2 / h n b mol CH OH / h g 4 3 Overall: 3 unknowns ( n1 , n5 , n6 ) Absorber: 4 unknowns ( n1 , n2 , n3 , n4 ) – 2 balances – 3 balances 1 DF 1 DF Stripper: 4 unknowns ( n2 , n3 , n4 , n5 ) – 2 balances – 1 percent removal (90%) 1 DF Overall CH4 balance: b0.700gb100g bmol CH / hg = 0.990n 4 1 Overall mole balance: 100b mol / h g = n + n 1 6 Percent CO2 stripped: 0.90 n3 = n6 Stripper CO2 balance: n3 = n6 + 0.005n2 Stripper CH3OH balance: n4 = 0.995n2 d. n1 = 70.71 mol / h , n2 = 6510 mol / h, n3 = 32.55 mol CO 2 / h, n4 = 647.7 mol CH3OH / h, . n6 = 29.29 mol CO 2 / h 30.0 − 0.010n1 Fractional CO2 absorption: f CO 2 = = 0.976 mol CO 2 absorbed / mol fed 30.0 4- 25
• 67. 4.35 (cont’d) Total molar flow rate of liquid feed to stripper and mole fraction of CO2: n3 n3 + n4 = 680 mol / h, x3 = = 0.0478 mol CO 2 / mol n3 + n4 e. Scale up to 1000 kg/h (=106 g/h) of product gas: b g b MW1 = 0.01 44 g CO 2 / mol + 0.99 16 g CH 4 / mol = 16.28 g / mol g bn g = d10 × 10 g / hib16.28 g / molg = 6142 × 10 mol / h 1 new. 6 . 4 bn g = b100 mol / hg (6142 × 10 mol / h) / (70.71 mol / h) = 8.69 × 10 feed new . 4 4 mol / h f. Ta < Ts The higher temperature in the stripper will help drive off the gas. Pa > Ps The higher pressure in the absorber will help dissolve the gas in the liquid. g. The methanol must have a high solubility for CO2, a low solubility for CH4, and a low volatility at the stripper temperature.4.36 a. Basis: 100 kg beans fed e m kg C H 5 6 14 j Condenser e m kg C H j 300 kg C 6 H14 b g m2 kg b g m4 kg b m6 kg oil g 1 6 14 Ex b x2 kg S / kg g F b g y4 kg oil / kg Ev y b kg oil / kgg 2 1 − y b kg C H 4 6 14 / kg g 1 − x − y b kg C H 2 2 6 14 / kg g 13.0 kg oil 87.0 kg S b g m3 kg 0.75 kg S / kg b y3 kg oil / kg g b 0.25 − y3 kg C 6 H14 / kg g Overall: 4 unknowns ( m1 , m3 , m6 , y3 ) Extractor: 3 unknowns ( m2 , x2 , y2 ) – 3 balances – 3 balances 1 DF 0 DF Mixing Pt: 2 unknowns ( m1 , m5 ) Evaporator: 4 unknowns ( m4 , m5 , m6 , y4 ) – 1 balance – 2 balances 1 DF 2 DF Filter: 7 unknowns ( m2 , m3 , m4 , x2 , y2 , y3 , y4 ) – 3 balances – 1 oil/hexane ratio 3 DF Start with extractor (0 degrees of freedom) Extractor mass balance: 300 + 87.0 + 13.0 kg = m2 4- 26
• 68. 4.36 (cont’d) Extractor S balance: 87.0 kg S = x2 m2 Extractor oil balance: 13.0 kg oil = y2 m2 Filter S balance: 87.0 kg S = 0.75m3 b g Filter mass balance: m2 kg = m3 + m4 Oil / hexane ratio in filter cake: y3 y2 = 0.25 − y3 1 − x2 − y2 Filter oil balance: 13.0 kg oil = y3m3 + y4 m4 Evaporator hexane balance: 1 − y4 m4 = m5 b g Mixing pt. Hexane balance: m1 + m5 = 300 kg C6 H14 Evaporator oil balance: y4 m4 = m6 b. Yield = m6 = 118 kg oil . 100 100 kg beans fed = 0118 kg oil / kg beans fed . b g m Fresh hexane feed = 1 = 28 kg C6 H14 100 100 kg beans fed = 0.28 kg C 6 H14 / kg beans fed b g m Recycle ratio = 5 = m1 272 kg C 6 H14 recycled 28 kg C6 H14 fed b = 9.71 kg C 6 H14 recycled / kg C 6 H14 fed g c. Lower heating cost for the evaporator and lower cooling cost for the condenser.4.37 b m lb m dirt 1 g 98 lb m dry shirts 3 lb m Whizzo 100 lbm 2 lbm dirt 98 lbm dry shirts Filter Tub b m lb m Whizzo 2 g b g m lb m 3 b g m lb m 4 5 b g m lb m 0.92 lb m dirt / lb m 0.03 lb m dirt / lb m 013 lb m dirt / lb m . 0.08 lb m Whizzo / lb m 0.97 lb m Whizzo / lb m 0.87 lb m Whizzo / lb m b g m6 lbm b g 1 − x lb m dirt / lbm x blb Whizzo/ lb g m m Strategy 95% dirt removal ⇒ m1 ( = 5% of the dirt entering) Overall balances: 2 allowed (we have implicitly used a clean shirt balance in labeling the chart) ⇒ m2 , m5 (solves Part (a)) 4- 27
• 69. 4.37 (cont’d) b g Balances around the mixing point involve 3 unknowns m3 , m6 , x , as do balances around the filter b m , m , x g , but the tub only involves 2 b m , m g and 2 balances are 4 6 3 4 allowed for each subsystem. Balances around tub ⇒ m3 , m4 Balances around mixing point ⇒ m6 , x (solves Part (b)) a. b gb g 95% dirt removal: m1 = 0.05 2.0 = 010 lb m dirt . Overall dirt balance: 2.0 = 010 + b0.92gm ⇒ m = 2.065 lb dirt . 5 5 m Overall Whizzo balance: m = 3 + b0.08gb2.065g blb Whizzog = 317 lb 2 . m m Whizzo b. Tub dirt balance: 2 + 0.03m3 = 010 + 013m4 . . (1) Tub Whizzo balance: 0.97m3 = 3 + 0.87m4 (2) Solve (1) & (2) simultaneously ⇒ m3 = 20.4 lb m , m4 = 19.3 lb m Mixing pt. mass balance: 317 + m6 = 20.4 lb m ⇒ m6 = 17.3 lb m . Mixing pt. Whizzo balance: 3.17 + x (17.3) = ( 0.97 )( 20.4 ) ⇒ x = 0.961 lb m Whizzo/lb m ⇒ 96% Whizzo, 4% dirt4.38 a. 2720 kg S mixer 3 C 2L kg L Discarded Filter 3 C 2S kg S C 3L kg L C 3S kg S F 3L kg L 3300 kg S F 3S kg S 620 kg L mixer 1 mixer 2 C 1L kg L Filter 1 Filter 2 C 1S kg S F 1L kg L F 2L kg L F 1S kg S F 2S kg S To holding tank mixer filter 1: b g 0.01 620 = F1L ⇒ F1L = 6.2 kg L balance: 620 = 6.2 + C ⇒ C1L = 6138 kg L . 0.01b6138 + F g = F U 1L F2 L = 6.2 kg L mixer filter 2: balance: . 6138 + F = F + C V ⇒ . | 3L 2L C2 L = 613.7 kg L mixer filter 3: 0.01C = F2L 3L | W 3L 2L 3L F3L = 61 kg L . balance: 613.7 = 6.1+ C3L ⇒ C3L = 607.6 kg L 4- 28
• 70. 4.38 (cont’d) Solvent m f 1: .b g 015 3300 = C1S ⇒ C1S = 495 kg S balance: 3300 = 495 + F ⇒ F1S = 2805 kg S 015b495 + F g = C U 1S C2 S = 482.6 kg S m f 2: . 495 + F = C + F | 3S | 2S F2 S = 2734.6 kg S balance: m f 3: 3S 2S 015b2720 + C g = C | . V⇒ 2S C3S = 480.4 kg S 2720 + C = F + C | 2S 3S balance: 2S W 3S 3S F3S = 2722.2 kg S Holding Tank Contents 6.2 + 6.2 = 12.4 kg leaf 2805 + 2734.6 = 5540 kg solvent b. 5540 kgS QR kg b g Q0 kg b g 0165 kg E / kg . Extraction 013 kg E / kg . Steam 0.200 kg E / kg 0.835 kg W / kg Unit 0.15kg F / kg Stripper 0.026 kg F / kg b g QD kg D 0.855kg W / kg 0.774 kg W / kg Q b kg Fg F b g QE kg E QB kg b g Q b kg D g 0.013 kg E / kg D Q b kg Fg 0.987 kg W / kg F b Q3 kg steam g 1 kg D 620 kg leaf Mass of D in Product: = 0.62 kg D = QD 1000 kg leaf Water balance around extraction unit: 0.835 5540 = 0.855QR ⇒ QR = 5410 kg b g Ethanol balance around extraction unit: b g b g 0165 5540 = 013 5410 + QE ⇒ QE = 211 kg ethanol in extract . . b g c. F balance around stripper b g b 0.015 5410 = 0.026Q0 ⇒ Q0 = 3121 kg mass of stripper overhead product g E balance around stripper b g b g 013 5410 = 0.200 3121 + 0.013QB ⇒ QB = 6085 kg mass of stripper bottom product . b g W balance around stripper b g b g b 0.855 5410 + QS = 0.774 3121 + 0.987 6085 ⇒ QS = 3796 kg steam fed to stripperg4.39 a. C2 H 2 + 2 H 2 → C2 H 6 2 mol H 2 react / mol C 2 H 2 react 0.5 kmol C 2 H 6 formed / kmol H 2 react 4- 29
• 71. 4.39 (cont’d) b. nH 2 = 15 < 2.0 ⇒ H 2 is limiting reactant . nC2 H2 15 mol H 2 fed ⇒ 10 mol C 2 H 2 fed ⇒ 0.75 mol C 2 H 2 required (theoretical) . . 10 mol fed − 0.75 mol required . % excess C 2 H 2 = × 100% = 333%. 0.75 mol required c. 4 × 106 tonnes C 2 H 6 1 yr 1 day 1 h 1000 kg 1 kmol C 2 H 6 2 kmol H 2 2.00 kg H 2 yr 300 days 24 h 3600 s tonne 30.0 kg C 2 H 6 1 kmol C 2 H 6 1 kmol H 2 = 20.6 kg H 2 / s d. The extra cost will be involved in separating the product from the excess reactant.4.40 a. 4 NH 3 + 5 O 2 → 4 NO + 6 H 2O 5 lb - mole O 2 react = 125 lb - mole O 2 react / lb - mole NO formed . 4 lb - mole NO formed b. dn i O 2 theoretical = 100 kmol NH3 5 kmol O 2 h 4 kmol NH 3 = 125 kmol O 2 40% excess O 2 ⇒ nO 2 d i fed b g = 140 125 kmol O 2 = 175 kmol O 2 . c. b50.0 kg NH gb1 kmol NH / 17 kg NH g = 2.94 kmol NH 3 3 3 3 b100.0 kg O gb1 kmol O / 32 kg O g = 3125 kmol O 2 2 . 2 2 F n I = 3125 = 106 < F n I = 5 = 125 GH n JK 2..94 . GH n JK 4 . O2 NH 3 O2 NH3 fed stoich ⇒ O 2 is the limiting reactant 3125 kmol O 2 4 kmol NH 3 . Required NH3: = 2.50 kmol NH 3 5 kmol O 2 2.94 − 2.50 % excess NH 3 = × 100% = 17.6% excess NH 3 2.50 Extent of reaction: nO 2 = nO2 d i −v 0 O2 . b g ξ ⇒ 0 = 3125 − −5 ξ ⇒ ξ = 0.625 kmol = 625 mol 3125 kmol O 2 4 kmol NO 30.0 kg NO . Mass of NO: = 75.0 kg NO 5 kmol O 2 1 kmol NO4.41 a. By adding the feeds in stoichometric proportion, all of the H2S and SO2 would be consumed. Automation provides for faster and more accurate response to fluctuations in the feed stream, reducing the risk of release of H2S and SO2. It also may reduce labor costs. 4- 30
• 72. 4.41 (cont’d) b. 3.00 × 10 2 kmol 0.85 kmol H 2 S 1 kmol SO 2 nc = = 127.5 kmol SO 2 / h h kmol 2 kmol H 2 S c. C a lib r a t io n C u r v e 1 .2 0 1 .0 0 X (mol H 2 S/mol) 0 .8 0 0 .6 0 0 .4 0 0 .2 0 0 .0 0 0 .0 2 0 .0 4 0 .0 6 0 .0 8 0 .0 1 0 0 .0 R a (m V ) X = 0.0199 Ra − 0.0605 d. b nc kmol SO 2 / h g b n f kmol / h g b x kmol H 2S / kmol g Blender Flowmeter calibration: n f = aR f Un V = 20 Rf n f = 100 kmol / h , R f = 15 mV W f 3 Control valve calibration: nc = 25.0 kmol / h, R c = 10.0 mV Un V = 7 Rc + 5 nc = 60.0 kmol / h, R = 25.0 mV W c c 3 3 1 7 5 1 20 FG IJ b g Stoichiometric feed: nc = n f x ⇒ Rc + = 2 3 3 2 3 R f 0.0119 Ra − 0.0605 H K ⇒ Rc = 10 7 R f 0.0119 Ra − 0.0605 − 5 7 b g 3 n f = 3.00 × 10 2 kmol / h ⇒ R f = n f = 45 mV 20 4- 31
• 73. 4.41 (cont’d) Rc = 10 7 b gb gb g 5 45 0.0119 76.5 − 0.0605 − = 53.9 mV 7 7 b g 5 ⇒ nc = 53.9 + = 127.4 kmol / h 3 3 e. Faulty sensors, computer problems, analyzer calibration not linear, extrapolation beyond range of calibration data, system had not reached steady state yet.4.42 165 mol / s b n mol / s g b x mol C 2 H 4 / mol g 0.310 mol C 2 H 4 / mol b 1 − x mol HBr / mol g 0173 mol HBr / mol . 0.517 mol C 2 H 5 Br / mol C 2 H 4 + HBr → C 2 H 5 Br C balance: b g 165 mol x mol C 2 H 4 2 mol C b gb g b = n 0.310 2 + n 0.517 2 gb g (1) s mol mol C2 H 4 Br balance: 165 (1 − x )(1) = n ( 0.173)(1) + n ( 0.517 )(1) (2) (Note: An atomic H balance can be obtained as 2*(Eq. 2) + (Eq. 1) and so is not independent) Solve (1) and (2) simultaneously ⇒ n = 108.77 mol / s, x = 0.545 mol C 2 H 4 / mol b g ⇒ 1 − x = 0.455 mol HBr / mol Since the C2H4/HBr feed ratio (0.545/0.455) is greater than the stoichiometric ration (=1), HBr is the limiting reactant . bn g = b165 mol / sgb0.455 mol HBr / molg = 75.08 mol HBr HBr fed 75.08 − ( 0.173)(108.8 ) Fractional conversion of HBr = = 0.749 mol HBr react/molfed 75.08 (n )C2 H 4 stoich = 75.08molC 2 H 4 (n )C2 H 4 fed = (165mol/s )( 0.545molC2 H 4 /mol ) = 89.93molC2 H 4 89.93 − 75.08 % excess of C2 H 4 = = 19.8% 75.08 ( Extent of reaction: nC2 H5 Br = nC2 H5Br ) 0 + vC2 H5 Brξ ⇒ (108.8 )( 0.517 ) = 0 + (1) ξ ⇒ ξ = 56.2 mol/s 4- 32
• 74. 4.43 a. 1 2HCl + O 2 → Cl 2 + H 2 O Basis: 100 mol HCl fed to reactor 2 100 mol HCl b n2 mol HCl g n b mol O g 3 2 b n1 mol air g n b mol N g 4 2 0.21 mol O 2 / mol n b mol Cl g 5 2 0.79 mol N 2 / mol n b mol H Og 6 2 35% excess bO gstoic = 100 mol HCl 0.5mol HCl = 25 mol O 2 2 mol O 2 2 35% excess air: 0.21n b mol O fed g = 135 × 25 ⇒ n = 160.7 mol air fed 1 . 2 1 85% conversion ⇒ 85 mol HCl react ⇒ n2 = 15 mol HCl 85 mol HCl react 1 mol Cl 2 n5 = = 42.5 mol Cl 2 2 mol HCl b gb g n6 = 85 1 2 = 42.5 mol H 2O N 2 balance: b160.7gb0.79g = n 4 ⇒ n4 = 127 mol N 2 O balance: b160.7gb0.21g mol O 2 2 mol O = 2n3 + 42.5 mol H 2O 1 mol O ⇒ n3 = 12.5 mol O 2 1 mol O 2 1 mol H 2O Total moles: 5 15 mol HCl mol HCl mol O 2 mol N 2 ∑nj = 239.5 mol ⇒ 239.5 mol = 0.063 mol , 0.052 mol , 0.530 mol , j= 2 mol Cl 2 mol H 2 O 0177 . , 0177 . mol mol b. As before, n1 = 160.7 mol air fed , n2 = 15 mol HCl 1 2HCl + O 2 → Cl 2 + H 2O 2 ni = ni b g 0 + vi ξ E HCl: 15 = 100 − 2ξ ⇒ ξ = 42.5 mol 4- 33
• 75. 4.43 (cont’d) b g 2 ξ = 12.5 mol O O 2 : n3 = 0.21 160.7 − 1 2 N 2 : n4 = 0.79b160.7g = 127 mol N 2 Cl 2 : n5 = ξ = 42.5 mol Cl 2 H 2 O: n6 = ξ = 42.5 mol H 2 O These molar quantities are the same as in part (a), so the mole fractions would also be the same. c. Use of pure O2 would eliminate the need for an extra process to remove the N2 from the product gas, but O2 costs much more than air. The cheaper process will be the process of choice.4.44 b g FeTiO3 + 2H 2SO 4 → TiO SO 4 + FeSO 4 + 2H 2O Fe O + 3H SO → Fe bSO g + 3H O 2 3 2 4 2 4 3 2 bTiOgSO + 2H O → H TiO bsg + H SO 4 2 2 3 2 4 H TiO bsg → TiO bsg + H O 2 3 2 2 Basis: 1000 kg TiO2 produced 1000 kg TiO 2 kmol TiO 2 1 kmol FeTiO 3 = 12.52 kmol FeTiO 3 decomposes 79.90 kg TiO 2 1 kmol TiO 2 12.52 kmol FeTiO 3 dec. 1 kmol FeTiO 3 feed = 14.06 kmol FeTiO 3 fed 0.89 kmol FeTiO 3 dec. 14.06 kmol FeTiO3 1 kmol Ti 47.90 kg Ti = 6735 kg Ti fed . 1 kmol FeTiO 3 kmol Ti b g 673.5 kg Ti / M kg ore = 0.243 ⇒ M = 2772 kg ore fed b Ore is made up entirely of 14.06 kmol FeTiO3 + n kmol Fe 2O 3 (Assumption!) g 14.06 kmol FeTiO3 151.74 kg FeTiO3 n = 2772 kg ore − = 6381 kg Fe 2O 3 . kmol FeTiO3 638.1 kg Fe 2O 3 kmol Fe 2O 3 = 4.00 kmol Fe 2O 3 159.69 kg Fe 2O 3 14.06 kmol FeTiO3 2 kmol H2SO4 4.00 kmol FeTiO3 3 kmol H2SO4 + = 4012 kmol H2SO4 . 1 kmol FeTiO3 1 kmol Fe2O3 . . b . g 50% excess: 15 4012 kmol H 2SO 4 = 6018 kmol H 2SO 4 fed 60.18 kmol H 2SO 4 98.08 kg H 2SO 4 Mass of 80% solution: = 5902.4 kg H 2SO 4 1 kmol H 2SO 4 5902.4 kg H 2 SO 4 / M a bkg solng = 0.80 ⇒ M a = 7380 kg 80% H SO 2 4 feed 4- 34
• 76. 4.45 a. Plot C (log scale) vs. R (linear scale) on semilog paper, get straight line through d R = 10, C 1 1 = 0.30 g m 3 and i FH R2 = 48, C2 = 2.67 g m 3 IK ln C = bR + ln a ⇔ C = ae br b= b ln 2.67 0.30 g = 0.0575 , ln a = lnb2.67g − 0.0575b48g = −1.78 ⇒ a = e −1.78 = 0169 . 48 − 10 ⇒ C = 0169e 0.0575 R . d C g m3 = i C ′(ftlb 3 m) 453.6 g 35.31 ft 3 1 lb m 1 m3 = 16,020C ′ E d i 16,020C = 0169e 0.0575 R ⇒ C ′ lb m SO 2 ft 3 = 1055 × 10 −5 e 0.0575 R . . b. d2867 ft sib60 s ming = 138 ft 3 3 lb m coal 1250 lb m min d . i R = 37 ⇒ C ′ lb m SO 2 ft 3 = 1055 × 10−5 eb gb g = 8.86 × 10 0.0575 37 −5 lb m SO 2 ft 3 8.86 × 10−5 lb m SO 2 138 ft 3 lb SO 2 3 = 0.012 < 0.018 m compliance achieved ft 1 lb m coal lb m coal c. S + O 2 → SO 2 1250 lb m coal 0.05 lb m S 64.06 lb m SO 2 = 124.9 lb m SO 2 generated min min 1 lb m coal 32.06 lb m S 2867 ft 3 60 s 886 × 10−5 lbm SO2 . = 152 lbm SO2 min in scrubbed gas . s 1 min ft3 air scrubbing fluid 1250 lb m coal/min furnace stack gas scrubber scrubbed gas 62.5 lb m S/min 124.9 lbm SO2 /min 15.2 lb m SO2 /min ash liquid effluent (124.9 – 15.2) lbm SO2 (absorbed)/min % removal = b124.9 − 15.2g lb SO 2 scrubbed min m × 100% = 88% 124.9 lb m SO 2 fed to scrubber min d. The regulation was avoided by diluting the stack gas with fresh air before it exited from the stack. The new regulation prevents this since the mass of SO2 emitted per mass of coal burned is independent of the flow rate of air in the stack. 4- 35
• 77. 4.46 a. A + B ===== C + D nA = nA − ξ 0 nB = nB − ξ 0 e y A = n A − ξ nT 0 j nC = n C + ξ yB = en B0 − ξj nT 0 nD = nD + ξ yC = en C0 + ξj n T 0 nI = nI yD = en D0 + ξj n T 0 Total nT = ∑ ni yC y D = b gb nC0 + ξ c nD0 + ξ c g = 4.87 ( nT ’s cancel) At equilibrium: y A yB b gb n A0 − ξ c nB0 − ξ c g c b gh b 387ξ 2 − nC0 + nD0 + 487 nA0 + nB0 ξ c − nC0nD0 − 487nA0nB0 = 0 . c . . g [aξ 2 + bξ c + c = 0] c a = 387 . ∴ξ c = 1 2a e j −b ± b2 − 4ac where b = − nC0 + nD0 + 487 nA0 + nB0 . b g c = − nC0nD0 − 487nA0nB0 . b. Basis: 1 mol A feed nA0 = 1 nB0 = 1 nC0 = nD0 = nI 0 = 0 Constants: a = 3.87 b = −9.74 c = 4.87 ξe = 1 2 ( 3.87 ) ( 9.74 ± ( 9.74 ) 2 ) − 4 ( 3.87 )( 4.87 ) ⇒ ξ e1 = 0.688 (ξ e 2 = 1.83 is also a solution but leads to a negative conversion ) nA0 − nA ξ e1 Fractional conversion: X A ( = X B ) = = = 0.688 nA0 nA0 c. nB0 = 80, nC0 = nD0 = nJ 0 = 0 nC 0 = 0 nC = 70 = nC 0 + ξ c =======> ξ c = 70 mol n A = n A0 − ξ c = n A0 − 70 mol n B = n B 0 − ξ c = 80 − 70 = 10 mol nC = nC 0 + ξ c = 70 mol n D = n D0 + ξ c = 70 mol 4.87 = y C y D nC n D = ⇒ b gb g 70 70 = 4.87 ⇒ n A0 = 170.6 mol methanol fed y A y B n A nB b gb g n A0 − 70 10 4- 36
• 78. 4.46 (cont’d) Product gas n A = 170.6 − 70 = 100.6 mol U y | y A = 0.401 mol CH3OH mol nB = 10 mol |⇒ V y B = 0.040 mol CH3COOH mol nC = 70 mol | y | C = 0.279 mol CH3COOCH 3 mol nD = 70 mol W D = 0.279 mol H 2O mol ntotal = 250.6 mol d. Cost of reactants, selling price for product, market for product, rate of reaction, need for heating or cooling, and many other items.4.47 a. CO + H 2O ← → CO 2 + H 2 ⎯ (A) (B) (C) (D) 100 mol . b n A mol CO g 0.20 mol CO / mol n b mol H O g B 2 n b mol CO g 010 mol CO 2 / mol . C 2 n b mol H g 0.40 mol H 2O / mol D 2 0.30 mol I / mol n b mol Ig I Degree of freedom analysis: 6 unknowns ( n A , nB , nC , nD , n I , ξ ) – 4 expressions for ni ξ bg – 1 balance on I – 1 equilibrium relationship 0 DF b. Since two moles are prodcued for every two moles that react, b g b g b g ntotal out = ntotal in = 100 mol . n A = 0.20 − ξ (1) nB = 0.40 − ξ (2) nC = 010 + ξ . (3) nD = ξ (4) n I = 0.30 (5) ntot = 100 mol . yC y D nC nD b 010 + ξ ξ . gb g 4020 FG ⇒ ξ = 0110 mol IJ At equilibrium: = y A y B n A nB = b gb 0.20 − ξ 0.40 − ξ = 0.0247 exp g 1123 . H K b y D = nD = ξ = 0110 mol H 2 . / molg c. The reaction has not reached equilibrium yet. 4- 37
• 79. 4.47 (cont’d) d. T (K) x (CO) x (H2O) x (CO2) Keq Keq (Goal Seek) Extent of Reaction y (H2) 1223 0.5 0.5 0 0.6610 0.6610 0.2242 0.224 1123 0.5 0.5 0 0.8858 0.8856 0.2424 0.242 1023 0.5 0.5 0 1.2569 1.2569 0.2643 0.264 923 0.5 0.5 0 1.9240 1.9242 0.2905 0.291 823 0.5 0.5 0 3.2662 3.2661 0.3219 0.322 723 0.5 0.5 0 6.4187 6.4188 0.3585 0.358 623 0.5 0.5 0 15.6692 15.6692 0.3992 0.399 673 0.5 0.5 0 9.7017 9.7011 0.3785 0.378 698 0.5 0.5 0 7.8331 7.8331 0.3684 0.368 688 0.5 0.5 0 8.5171 8.5177 0.3724 0.372 1123 0.2 0.4 0.1 0.8858 0.8863 0.1101 0.110 1123 0.4 0.2 0.1 0.8858 0.8857 0.1100 0.110 1123 0.3 0.3 0 0.8858 0.8856 0.1454 0.145 1123 0.5 0.4 0 0.8858 0.8867 0.2156 0.216 The lower the temperature, the higher the extent of reaction. An equimolar feed ratio of carbon monoxide and water also maximizes the extent of reaction.4.48 a. A + 2B → C ln K e = ln A0 + E T K b g E= b ln K e1 / K e 2 g = lnd10.5 / 2.316 × 10 i = 11458 −4 1 T1 − 1 T2 1 373 − 1 573 ln A0 = ln K e1 − 11458 T1 = ln 10.5 − 11458 373 = −28.37 ⇒ A0 = 4.79 × 10−13 c b gh Ke = 4.79 × 10 −13 exp 11458 T K atm−2 ⇒ Ke (450K ) = 0.0548 atm−1 b. n A = n A0 − ξ U | b gb y A = n A0 − ξ nT 0 − 2ξ g nB = nB 0 − 2ξ | V ⇒ b gb y B = nB 0 − 2ξ nT 0 − 2ξ g nC = nC 0 + ξ | b gb yC = nC 0 + ξ nT 0 − 2ξ g nT = nT 0 − 2ξ | W b nT 0 = n A0 + nB 0 + nC 0 g At equilibrium, yC 1 = C0 b n + ξ e nT 0 − 2ξ e gb g 2 1 bg bg = Ke T (substitute for K T from Part a.) 2 y A yB P 2 b n A0 − ξ e nB 0 − 2ξ egb g 2 P2 e c. Basis: 1 mol A (CO) n A0 = 1 nB 0 = 1 nC 0 = 0 ⇒ nT 0 = 2 , P = 2 atm , T = 423K b ξ e 2 − 2ξ e g 2 1 b g = K e 423 = 0.278 atm -2 ⇒ ξ 2 − ξ e + 01317 = 0 b1 − ξ gb1 − 2ξ g e e 2 4 atm 2 e . 4- 38
• 80. 4.48 (cont’d) (For this particular set of initial conditions, we get a quadratic equation. In general, the equation will be cubic.) ξ e = 0156 , 0.844 . Reject the second solution, since it leads to a negative nB . b y A = 1 − 0156 . g c2 − 2b0156gh ⇒ y = 0.500 . A y = c1 − 2b0156gh c2 − 2b0156gh ⇒ y = 0.408 B . . B y = b0 + 0156g c2 − 2b0156gh ⇒ y = 0.092 C . . C n −n ξ Fractional Conversion of CO b Ag = A0 = A = 0156 mol A reacted / mol A feed . n nA0 A0 d. Use the equations from part b. i) Fractional conversion decreases with increasing fraction of CO. ii) Fractional conversion decreases with increasing fraction of CH3OH. iii) Fractional conversion decreases with increasing temperature. iv) Fractional conversion increases with increasing pressure. REAL TRU, A, E, YA0, YC0, T, P, KE, P2KE, C0, C1, C2, C3, EK, EKPI, * FN, FDN, NT, CON, YA, YB, YC INTEGER NIT, INMAX TAU = 0.0001 INMAX = 10 A = 4.79E–13 E = 11458. READ (5, *) YA0, YB0, YC0, T, P KE = A * EXP(E/T) P2KE = P*P*KE C0 = YC0 – P2KE * YA0 * YB0 * YB0 C1 = 1. – 4. * YC0 + P2KE * YB0 * (YB0 + 4. * YA0) C2 = 4. * (YC0 –1. – P2KE * (YA0 + YB0)) C3 = 4. * (1. + P2KE) EK = 0.0 (Assume an initial value ξe = 0. 0 ) NIT = 0 1 FN = C0 + EK * (C1 + EK * (C2 + EK * C3)) FDN = C1 + EK * (2. * C2 + EK * 3. * C3) EKPI = EK - FN/FDN NIT = NIT + 1 IF (NIT.EQ.INMAX) GOTO 4 IF (ABS((EKPI – EK)/EKPI).LT.TAU) GOTO 2 EK = EKPI GOTO 1 2 NT = 1. – 2. * EKPI YA = (YA0 – EKPI)/NT YB = (YB0 – 2. + EKPI)/NT YC = (YC0 + EKPI)/NT 4- 39
• 81. 4.48 (cont’d) CON = EKPI/YA0 WRITE (6, 3) YA, YB, YC, CON STOP 4 WRITE (6, 5) INMAX, EKPI 3 FORMAT ( YA YB YC CON, 1, 4(F6.3, 1X)) FORMAT (DID NOT CONVERGE IN, I3, ITERATIONS,/, * CURRENT VALUE = , F6.3) END \$ DATA 0.5 0.5 0.0 423. 2. RESULTS: YA = 0.500, YB = 0.408, YC = 0.092, CON = 0.156 Note: This will only find one root — there are two others that can only be found by choosing different initial values of ξa4.49 a. CH 4 + O 2 ⎯ → HCHO + H 2O ⎯ (1) CH 4 + 2O 2 ⎯ → CO 2 + 2H 2O ⎯ (2) 100 mol / s 0.50 mol CH 4 / mol b n1 mol CH 4 / s g 0.50 mol O 2 / mol n b mol O / s g 2 2 n b mol HCHO / sg 3 n b mol H O / s g 4 2 n b molCO / g n (mol CO s) 5 5 2 2 7 unknowns ( n1 , n2 , n3 , n4 , n5 , ξ1 , ξ 2 ) e – 5 equations for ni ξ 1 , ξ 2 j 2 DF b. n1 = 50 − ξ1 − ξ 2 (1) n2 = 50 − ξ1 − 2ξ 2 (2) n3 = ξ1 (3) n4 = ξ1 + 2ξ 2 (4) n5 = ξ 2 (5) c. Fractional conversion: b50 − n g = 0.900 ⇒ n 1 1 = 5.00 mol CH / s 50 4 Fractional yield: n 3 = 0.855 ⇒ n3 = 42.75 mol HCHO / s 50 Equation 3 ⇒ ξ 1 = 42.75 4 U y CH = 0.0500 mol CH 4 / mol | y O = 0.0275 mol O 2 / mol Equation 1 ⇒ ξ 2 = 2.25 2 | | Equation 2 ⇒ n2 = 2.75 ⇒ V y HCHO = 0.4275 mol HCHO / mol | y H O = 0.4725 mol H 2 O / mol Equation 4 ⇒ n4 = 47.25 2 | | Equation 5 ⇒ n5 = 2.25 2 W y CO = 0.0225 mol CO 2 / mol Selectivity: [(42.75 mol HCHO/s)/(2.25 mol CO 2 /s) = 19.0 mol HCHO/mol CO 2 4- 40
• 82. 4.50 a. Design for low conversion and feed ethane in excess. Low conversion and excess ethane make the second reaction unlikely. b. C2H6 + Cl2 → C2H5Cl + HCl, C2H5Cl + Cl2 → C2H4Cl2 + HCl Basis: 100 mol C2H5Cl produced n1 (mol C2H6) 100 mol C2H5Cl 5 unknowns n2 (mol Cl2) n3 (mol C2H6) –3 atomic balances n4 (mol HCl) 2 D.F. n5 (mol C2H5Cl2) c. Selectivity: 100 mol C 2 H 5 Cl = 14n5 (mol C 2 H 4 Cl 2 ) ⇒ n5 = 7.143 mol C 2 H 4 Cl 2 b g 15% conversion: 1 − 0.15 n1 = n3 U ⇒ n = 714.3 mol C H in | V 2n = 2b100g + 2n + 2b7.143g| n = 114.3 mol C H out 1 2 6 C balance: 1 W 3 3 2 6 H balance: 6b714.3g = 5b100g + 6b114.3g + n + 4b7.143g ⇒ n = 607.1 mol HCl 4 4 Cl balance: 2n = 100 + 607.1 + 2b7.143g ⇒ n = 114.3 mol Cl 2 2 2 Feed Ratio: 114.3 mol Cl 2 / 714.3 mol C2 H 6 = 016 mol Cl 2 / mol C2 H 6 . Maximum possible amount of C2H5Cl: 114.3 mol Cl 2 1 mol C 2 H 5 Cl n max = = 114.3 mol C 2 H 5 Cl 1 mol Cl 2 nC2 H5Cl 100 mol Fractional yield of C2H5Cl: = = 0.875 n max 114.3 mol d. Some of the C2H4Cl2 is further chlorinated in an undesired side reaction: C2H5Cl2 + Cl2 → C2H4Cl3 + HCl4.51 a. C2H4 + H2O → C2H5OH, 2 C2H5OH → (C2H5)2O + H2O Basis: 100 mol effluent gas 100 mol 0.433 mol C 2 H 4 / mol n1 (mol C 2 H 4 ) 3 unknowns 0.025 mol C 2 H 5OH / mol -2 independent atomic balances n2 [mol H 2 O (v)] 0.0014 mol (C H ) O / mol -1 I balance 2 5 2 n 3 (mol I) 0.093 mol I / mol 0 D. F. 0.4476 mol H 2 O (v) / mol b (1) C balance: 2n1 = 100 2∗0.433 + 2∗0.025 + 4∗0.0014 g b (2) H balance: 4n1 + 2n2 = 100 4∗0.433 + 6∗0.025 + 10∗0.0014 + 2∗0.4476 g b (3) O balance: n2 = 100 0.025 + 0.0014 + 0.4476 g Note; Eq. (1)∗2 + Eq. ( 3)∗2 = Eq. (2) ⇒2 independent atomic balances (4) I balance: n3 = 9.3 4-41
• 83. 4.51 (contd) b. (1) ⇒ n1 = 46.08 mol C 2 H 6 U | ⇒ Reactor feed contains 44.8% C H , 46.1% H O, 9.1% I (3) ⇒ n2 = 47.4 mol H 2 O V | 2 6 2 (4) ⇒ n3 = 9.3 mol I W 46.08 − 43.3 % conversion of C2H4: × 100% = 6.0% 46.08 If all C2H4 were converted and the second reaction did not occur, nC2 H5OH d i max = 46.08 mol ⇒ Fractional Yield of C2H5OH: nC2 H5OH / nC2 H5OH d i max b g = 2.5 / 46.08 = 0.054 Selectivity of C2H5OH to (C2H5)2O: 2.5 mol C 2 H 5 OH = 17.9 mol C 2 H 5 OH / mol (C 2 H 5 ) 2 O 0.14 mol (C 2 H 5 ) 2 O c. Keep conversion low to prevent C2H5OH from being in reactor long enough to form significant amounts of (C2H5)2O. Separate and recycle unreacted C2H4.4.52 bg bg CaF2 s + H 2 SO 4 l → CaSO 4 s + 2HF gbg bg 1 metric ton acid 1000 kg acid 0.60 kg HF = 600 kg HF 1 metric ton acid 1 kg acid Basis: 100 kg Ore dissolved (not fed) 100 kg Ore d issolved nn1 (kg CaSO44) 1 (kg CaSO ) 0.96 kg CaF 2 /kg n2 (kg HF) n2 (kg HF) 0.04 kg SiO 2/ kg nn3 (kg H2SiF66 ) 3 (kg H 4 ) SiF n (kg 93% H2 SO4 ) A nn4 (kg HH 2SO 4 4 (kg 2SO4) ) 0.93 H2 SO4 kg/ kg 0.07 H2 O kg/ kg nn5 (kg HH2 O) 5 (kg 2 O) Atomic balance - Si: 0.04 (100 ) kg SiO 2 28.1 kg Si n3 (kg H 2SiF6 ) 28.1 kg Si = ⇒ n3 = 9.59 kg H 2SiF6 60.1 kg SiO 2 144.1 kg H 2SiF6 Atomic balance - F: 0.96 (100 ) kg CaF2 38.0 kg F n2 (kg HF) 19.0 kg F = 78.1 kg CaF2 20.0 kg HF 9.59 kg H 2SiF6 114.0 kg F + ⇒ n2 = 41.2 kg HF 144.1 kg H 2SiF6 600 kg HF 100 kg ore diss. 1 kg ore feed = 1533 kg ore 41.2 kg HF 0.95 kg ore diss. 4-42
• 84. 4.53 a. C 6 H 6 + Cl 2 → C 6 H 5 Cl + HCl C 6 H 5 Cl + Cl 2 → C 6 H 4 Cl 2 + HCl C 6 H 4 Cl 2 + Cl 2 → C 6 H 3 Cl 3 + HCl Convert output wt% to mol%: Basis 100 g output species g Mol. Wt. mol mol % C6 H 6 65.0 78.11 0.832 73.2 C 6 H 5 Cl 32.0 112.56 0.284 25.0 C 6 H 4 Cl 2 2.5 147.01 0.017 1.5 C 6 H 3 Cl 3 0.5 181.46 0.003 0.3 total 1.136 Basis: 100 mol output n 4 (mol HCl(g )) n1 (mol C6 H6 ) n 3 (mol I) 4 unknowns -3 atomic balances 65.0 mo l C6 H6 -1 wt% Cl 2 in feed n2 (mol Cl 2) 73.2 mol C6H6 0 D.F. 32.0 mo l C6 H 5 Cl n3 (mol I) 25.0 mol C6H5Cl 2.5 mo l C 6 H 4 Cl2 1.5 mol C6H4Cl2 0.5 mol C6 H 3 Cl 3 0.3 mo l C H Cl 6 3 3 b. C balance: 6n1 = 6 ( 73.2 + 25.0 + 1.5 + 0.3) ⇒ n1 = 100 mol C 6 H 6 H balance: 6 (100 ) = 6 ( 73.2 ) + 5 ( 25.0 ) + 4 (1.5 ) + 3 ( 0.3) + n4 ⇒ n4 = 28.9 mol HCl Cl balance: 2n2 = 28.9 + 25.0 + 2 (1.5 ) + 3 ( 0.3) ⇒ n2 = 28.9 mol Cl2 Theoretical C 6 H 6 = 28.9 mol Cl 2 (1 mol C6 H 6 1 mol Cl2 ) = 28.9 mol C6 H 6 (100 − 28.9 ) 28.9 ×100% = 246% excess C6 H 6 Excess C 6 H 6 : Fractional Conversion: (100 − 73.2 ) 100 = 0.268 mol C 6 H 6 react/mol fed Yield: (25.0 mol C6 H 5 Cl) (28.9 mol C6 H 5 Cl maximum)=0.865 28.9 mol Cl2 70.91 g Cl2 1 g gas ⎫ Gas feed: = 2091 g gas ⎪ mole Cl2 0.98 g Cl2 ⎪ g gas ⎬ ⇒ 0.268 ⎛ 78.11 g C6 H 6 ⎞ ⎪ g liquid Liquid feed: (100 mol C6 H 6 ) ⎜ ⎟ = 7811 g liquid ⎪ ⎝ mol C6 H 6 ⎠ ⎭ c. Low conversion ⇒ low residence time in reactor ⇒ lower chance of 2nd and 3rd reactions occurring. Large excess of C 6 H 6 ⇒ Cl 2 much more likely to encounter C 6 H 6 than substituted C 6 H 6 ⇒ higher selectivity. d. Dissolve in water to produce hydrochloric acid. e. Reagent grade costs much more. Use only if impurities in technical grade mixture affect the reaction rate or desired product yield. 4-43
• 85. 4.54 a. 2CO 2 ⇔ 2CO + O 2 2A ⇔ 2B + C O 2 + N 2 ⇔ 2NO C + D ⇔ 2E n A = n A 0 − 2ξ e1 yA = bn A0 − 2ξ e1 g bn T0 + ξ e1 g nB = n B 0 + 2ξ e 2 yB = bn B0 + 2ξ e1 g bn T0 + ξ e1 g nC = nC 0 + ξ e1 − ξ e 2 ⇒ y C = bn C0 gb + ξ e1 − ξ e 2 nT 0 + ξ e1 g nD = n D0 − ξ e2 yD = bn D0 gb − 1ξ e 2 nT 0 + ξ e1 g nE = n E 0 + 2ξ e 2 yE = bn E0 gb + 2ξ e 2 nT 0 + ξ e1 g ntotal = nT 0 + ξ e1 bn T0 = n A0 + n B 0 + nC 0 + n D 0 + n E 0 g Equilibrium at 3000K and 1 atm 2 y B yC = bn B0 g bn + ξ − ξ g = 01071 + 2ξ e1 2 C0 . e1 e2 y2 A bn − 2ξ g bn + ξ g A0 e1 2 T0 e1 ( nE 0 + 2ξ e 2 ) 2 2 yE = = 0.01493 yC yD ( nA0 + ξ e1 − ξ e 2 )( nD 0 − ξ e 2 ) E b f 1 = 01071 n A 0 − 2ξ e1 . g bn 2 g b + ξ e1 − n B 0 + 2ξ e1 g bn + ξ − ξ g = 0U Defines functions 2 | f bξ , ξ g and g − bn + 2ξ g = 0 V f bξ , ξ g T0 C0 e1 e2 b f 2 = 0.01493 nC 0 + ξ e1 − ξ e 2 gbn D0 − ξ e2 E0 e2 | W 2 1 2 1 1 2 2 b. Given all nio’s, solve above equations for ξe1 and ξe2 ⇒ nA, nB, nC, nD, nE ⇒ yA, yB, yC, yD, yE c. nA0 = nC0 = nD0 = 0.333, nB0 = nE0 = 0 ⇒ ξe1 =0.0593, ξe2 = 0.0208 ⇒ yA = 0.2027, yB = 0.1120, yC = 0.3510, yD = 0.2950, yE = 0.0393 d. a11d 1 + a12 d 2 = − f 1 a 21d 1 + a 22 d 2 = − f 2 a12 f 2 − a 22 f 1 a 21 f 1 − a11 f 2 d1 = d2 = a11a 22 − a12 a 21 a11a 22 − a12 a 21 bξ g e1 new = ξ e1 + d 1 bξ g e 2 new = ξ e1 + d 2 (Solution given following program listing.) . IMPLICIT REAL * 4(N) WRITE (6, 1) 1 FORMAT(1, 30X, SOLUTION TO PROBLEM 4.57///) 30 READ (5, *) NA0, NB0, NC0, ND0, NE0 IF (NA0.LT.0.0)STOP WRITE (6, 2) NA0, NB0, NC0, ND0, NE0 4-44
• 86. 4.54 (cont’d) 2 FORMAT(0, 15X, NA0, NB0, NC0, ND0, NE0 *, 5F6.2/) NTO = NA0 + NB0 + NC0 + ND0 + NE0 NMAX = 10 X1 = 0.1 X2 = 0.1 DO 100 J = 1, NMAX NA = NA0 – X1 – X1 NB = NB0 + X1 + X1 NC = NC0 + X1 – X2 ND = ND0 – X2 NE = NE0 + X2 + X2 NAS = NA ** 2 NBS = NB ** 2 NES = NE ** 2 NT = NT0 + X1 F1 = 0.1071 * NAS * NT – NBS * NC F2 = 0.01493 * NC * ND – NES A11 = –0.4284 * NA * NT * 0.1071 * NAS – 4.0 * NB * NC – NBS A12 = NBS A21 = 0.01493 * ND A22 = –0.01493 * (NC + ND) – 4.0 * NE DEN = A11 * A22 – A12 * A21 D1 = (A12 * F2 – A22 * F1)/DEN D2 = (A21 * F1 – A11 * F2)/DEN X1C = X1 + D1 X2C = X2 + D2 WRITE (6, 3) J, X1, X2, X1C, X2C 3 FORMAT(20X, ITER *, I3, 3X, X1A, X2A =, 2F10.5, 6X, X1C, X2C =, * 2F10.5) IF (ABS(D1/X1C).LT.1.0E–5.AND.ABS(D2/X2C).LT.1.0E–5) GOTO 120 X1 = X1C X2 = X2C 100 CONTINUE WRITE (6, 4) NMAX 4 FORMAT(0, 10X, PROGRAM DID NOT CONVERGE IN, I2, ITERATIONS/) STOP 120 YA = NA/NT YB = NB/NT YC = NC/NT YD = ND/NT YE = NE/NT WRITE (6, 5) YA, YB, YC, YD, YE 5 FORMAT (0, 15X, YA, YB, YC, YD, YE =, 1P5E14.4///) GOTO 30 END \$DATA 0.3333 0.00 0.3333 0.3333 0.0 0.50 0.0 0.0 0.50 0.0 0.20 0.20 0.20 0.20 0.20 SOLUTION TO PROBLEM 4.54 NA0, NB0, NC0, ND0, NE0 = 0.33 0.00 0.33 0.33 0.00 ITER = 1 X1A, X2A = 0.10000 0.10000 X1C, X2C = 0.06418 0.05181 ITER = 2 X1A, X2A = 0.06418 0.05181 X1C, X2C = 0.05969 0.02986 ITER = 3 X1A, X2A = 0.05969 0.02486 X1C, X2C = 0.05937 0.02213 4-45
• 87. 4.54 (cont’d) ITER = 4 X1A, X2A = 0.05437 0.02213 X1C, X2C = 0.05931 0.02086 ITER = 5 X1A, X2A = 0.05931 0.02086 X1C, X2C = 0.05930 0.02083 ITER = 6 X1A, X2A = 0.05930 0.02083 X1C, X2C = 0.05930 0.02083 YA, YB, YC, YD, YE = 2.0270E − 01 1.1197 E − 01 3.5100E − 01 2.9501E − 01 3.9319 E − 02 NA0, NB0, NC0, ND0, NE0 = 0.20 0.20 0.20 0.20 0.20 ITER = 1 X1A, X2A = 0.10000 0.10000 X1C, X2C = 0.00012 0.00037 ↓ ITER = 7 X1A, X2A = –0.02244 –0.08339 X1C, X2C = –0.02244 –0.08339 YA, YB, YC, YD, YE= 2.5051E − 01 1.5868E − 01 2.6693E − 01 2.8989E − 01 3.3991E − 024.55 a. (B) (1 − f ) m0 (kg/h) xR = 0 Reactor: 99% (A) conv. of R (P) m0 (kg/h) (1 − f ) m0 (kg/h) m1 (kg/h) mP (kg/h) xRA (kg R/kg) xRA (kg R/kg) xR1 (kg R/kg) 0.0075 kg R/kg fm0 (kg/h) xRA (kg R/kg) Mass balance on reactor: 2(1 − f )m0 = m1 (1) 99% conversion of R: m1 xR1 = 0.01(1 − f )m0 xRA (2) Mass balance on mixing point: m1 + fm0 = mP (3) R balance on mixing point: m1 xR1 + fm0 xRA = 0.0075mP (4) The system has 6 unknowns (m0 , xRA , f , m1 , xR1 , mP ) and four independent equations relating them, so there must be two degrees of freedom. b. 2(1 − f )m0 = m1 m1 xR1 = 0.01(1 − f )m0 xRA m1 + fm0 = mP m0 = 2780 kg/h ⎯⎯⎯⎯ E-Z Solve → m1 xR1 + fm0 xRA = 0.0075mP f = 0.254 kg bypassed/kg fresh feed mP = 4850 xRA = 0.0500 4-46
• 88. 4.55 (cont’d) c. mP xRA mA0 mB0 f 4850 0.02 3327 1523 0.54 4850 0.03 3022 1828 0.40 4850 0.04 2870 1980 0.31 4850 0.05 2778 2072 0.25 4850 0.06 2717 2133 0.21 4850 0.07 2674 2176 0.19 4850 0.08 2641 2209 0.16 4850 0.09 2616 2234 0.15 4850 0.10 2596 2254 0.13 mP xRA mA0 mB0 f 2450 0.02 1663 762 0.54 2450 0.03 1511 914 0.40 2450 0.04 1435 990 0.31 2450 0.05 1389 1036 0.25 2450 0.06 1359 1066 0.22 2450 0.07 1337 1088 0.19 2450 0.08 1321 1104 0.16 2450 0.09 1308 1117 0.15 2450 0.10 1298 1127 0.13 f v s . x RA f (kg bypass/kg fresh feed) 0.60 0.50 0.40 0.30 0.20 0.10 0.00 0.00 0.02 0.04 0.06 0.08 0.10 0.12 x R A (k g R / k g A ) 4-47
• 89. 4.56 a. 900 kg HCHO 1 kmol HCHO = 30.0 kmol HCHO / h h 30.03 kg HCHO n (kmol CH OH / h) 1 3 30.0 kmol HCHO / h n2 (kmol H 2 / h) n3 (kmol CH 3OH / h) 30.0 % conversion: = 0.60 ⇒ n1 = 50.0 kmol CH 3 OH / h n1 b. n (kmol CH OH / h) 30.0 kmol HCHO / h 30.0 kmol HCHO / h 1 3 n2 (kmol H 2 / h) n2 (kmol H 2 / h) n3 (kmol CH 3OH / h) n (kmol CH OH / h) 3 3 Overall C balance: n1 (1) = 30.0 (1) ⇒ n1 = 30.0 kmol CH3OH/h (fresh feed) 30.0 Single pass conversion: = 0.60 ⇒ n3 = 20.0 kmol CH 3OH / h n1 + n3 n1 + n3 = 50.0 kmol CH3OH fed to reactor/h c. Increased xsp will (1) require a larger reactor and so will increase the cost of the reactor and (2) lower the quantities of unreacted methanol and so will decrease the cost of the separation. The plot would resemble a concave upward parabola with a minimum around xsp = 60%.4.57 a. Convert effluent composition to molar basis. Basis: 100 g effluent: 10.6 g H 2 1 mol H 2 = 5.25 mol H 2 2.01 g H 2 64.0 g CO 1 mol CO H : 0.631 mol H / mol = 2.28 mol CO 2 2 28.01 g CO ⇒ CO: 0.274 mol CO / mol 25.4 g CH 3OH 1 mol CH 3 OH CH OH: 0.0953 mol CH OH / mol 3 3 32.04 g CH 3OH = 0.793 mol CH 3OH 4-48
• 90. 4.57 (cont’d) n4 (mol/min) 0.004 mol CH3OH(v)/mol x (mol CO/mol) (0.896 - x ) (mol H 2 / mol) 350 mol/ min Cond. n3 (mol CH 3OH(l)/min) Reactor n1 (mol CO/min) 0.631 mol CH 3OH(v)/ mol n2 (mol H2 / min) 0.274 mol CO/ mol CO + H 2 → CH 3OH 0.0953 mol H / mol 2 Condenser Overall process 3 unknowns (n3 , n4 , x) 2 unknowns (n1 , n2 ) –3 balances –2 independent atomic balances 0 degrees of freedom 0 degrees of freedom Balances around condenser ⎫ n3 = 32.1 mol CH 3OH(l)/min CO: 350 ∗ 0.274 = n4 ∗ x ⎪ ⎪ H : 350 ∗ 0.631 = n4 ∗ (0.996 − x) ⎬ ⇒ n4 = 318.7 mol recycle/min 2 ⎪ x = .301 molCO/mol CH OH : 350 ∗ 0.0953 = n3 + 0.004 ∗ n4 ⎪ 3 ⎭ Overall balances C: n1 =n3 ⎫ n1 = 32.08 mol/min CO in feed ⎬⇒ H : 2n2 =4n3 ⎭ n2 = 64.16 mol/min H 2 in feed Single pass conversion of CO: (32.08 + 318.72 ∗ 0.3009) − 350 ∗ 0.274 × 100% = 25.07% (32.08 + 318.72 ∗ 0.3009) 32.08 − 0 Overall conversion of CO: × 100% = 100% 32.08 b. – Reactor conditions or feed rates drifting. (Recalibrate measurement instruments.) – Impurities in feed. (Re-analyze feed.) – Leak in methanol outlet pipe before flowmeter. (Check for it.) 4-49
• 91. 4.58 a. Basis: 100 kmol reactor feed/hr n3 (kmol CH 4 /h) 100 kmol /h Solvent Reactor Cond. Absorb n3 (kmol CH 4 /h) n3 (kmol CH 4 /h) n1 (kmol CH 4 /h) 80 kmol CH4 /h n4 (kmol HCl /h) n4 (kmol HCl /h) n4 (kmol HCl /h) n2 (kmol Cl 2 /h) 20 kmol Cl2 /h 5n5 (kmol CH 3Cl /h) n5 (kmol CH 2Cl 2 /h) 5n5 (kmol CH 3Cl /h) Still 5n5 (kmol CH 3Cl /h) n5 (kmol CH 2Cl 2 /h) n5 (kmol CH 2Cl 2 /h) Overall process: 4 unknowns (n1, n2, n4, n5) -3 balances = 1 D.F. Mixing Point: 3 unknowns (n1, n2, n3) -2 balances = 1 D.F. Reactor: 3 unknowns (n3, n4, n5) -3 balances = 0 D.F. Condenser: 3 unknowns (n3, n4, n5) -0 balances = 3 D.F. Absorption column: 2 unknowns (n3, n4) -0 balances = 2 D.F. Distillation Column: 2 unknowns (n4, n5) -0 balances = 2 D.F. Atomic balances around reactor: 1) C balance : 80 = n 3 + 5n 5 + n 5 ⎫ ⎪ 2) H balance : 320 = 4n 3 + n 4 + 15n 5 + 2n 5 ⎬ ⇒ Solve for n 3 , n 4 , n 5 3) Cl balance : 40 = n 4 + 5n 5 + 2n 5 ⎪ ⎭ CH4 balance around mixing point: n1 = (80 – n3) Solve for n1 Cl2 balance: n2 = 20 b. For a basis of 100 kmol/h into reactor n1 = 17.1 kmol CH4/h n4 = 20.0 kmol HCl/h n2 = 20.0 kmol Cl2/h 5n5 = 14.5 kmol CH3Cl/h n3 = 62.9 kmol CH4/h c. (1000 kg CH3Cl/h)(1 kmol/50.49 kg) = 19.81 kmol CH3Cl/h 19.81 kmol CH 3Cl/h Scale factor = = 1.366 14.5 kmol CH 3Cl/h n = (17.1)(1.366) = 23.3 kmol CH 4 /h ⎫ n tot = 50.6 kmol/h Fresh feed: 1 ⎬ ⇒ n 2 = ( 20.0)(1.366) = 27.3 kmol Cl 2 /h ⎭ 46.0 mol% CH 4 , 54.0 mole% Cl 2 Recycle: n3 = (62.9)(1.366) = 85.9 kmol CH4 recycled/h 4-50
• 92. 4.59 a. Basis: 100 mol fed to reactor/h ⇒ 25 mol O2/h, 75 mol C2H4/h n1 (mol C 2H 4 //h) n2 (mol O 2 /h) n3 (mol C 2H 4O /h) reactor Seperator separator 75 mol C 2H 4 //h n1 (mol C 2H 4 //h) nC2H4 ( mol C 2H 4 /h) 25 mol O 2 /h n2 (mol O 2 /h) nO2 (mol O 2 /h) n3 (mol C 2H 4O /h) n4 (mol CO 2 /h) n4 (mol CO 2 /h) n5 (mol H 2O /h) n5 (mol H 2O /h) Reactor 5 unknowns (n1 - n5) -3 atomic balances -1 - % yield -1 - % conversion 0 D.F. Strategy: 1. Solve balances around reactor to find n1- n5 2. Solve balances around mixing point to find nO2, nC2H4 (1) % Conversion ⇒ n1 = .800 * 75 90 mol C 2 H 4 O (2) % yield: (.200)(75) mol C 2 H 4 × = n 3 (production rate of C 2 H 4 O) 100 mol C 2 H 4 (3) C balance (reactor): 150 = 2 n1 + 2 n3 + n4 (4) H balance (reactor): 300 = 4 n1 + 4 n3 + 2 n5 (5) O balance (reactor): 50 = 2 n2 + n3 + 2 n4 + n5 (6) O2 balance (mix pt): nO2 = 25 – n2 (7) C2H4 balance (mix pt): nC2H4 = 75 – n1 Overall conversion of C2H4: 100% b. n1 = 60.0 mol C2H4/h n5 = 3.00 mol H2O/h n2 = 13.75 mol O2 /h nO2 = 11.25 mol O2/h n3 = 13.5 mol C2H4O/h nC2H4 = 15.0 mol C2H4/h n4 = 3.00 mol CO2/h 100% conversion of C2H4 c. 2000 lbm C 2 H 4 O 1 lb - mole C 2 H 4 O h lb − mol / h Scale factor = = 3.363 h 44.05 lbm C 2 H 4 O 13.5 mol C 2 H 4 O mol / h nC2H4 = (3.363)(15.0) = 50.4 lb-mol C2H4/h nO2 = (3.363)(11.25) = 37.8 lb-mol O2/h 4-51
• 93. 4.60 a. Basis: 100 mol feed/h. Put dots above all n’s in flow chart. 100 mol/h n1 (mol /h) n2 (mol CH3OH/h) n3 (mol CH 3 OH / h) reactor cond. 32 mol CO/h .13 mol N 2 /mol 64 mol H 2 / h 4 mol N 2 / h 500 mol / h x1 (mol N 2 /mol) n3 (mol / h) x2 (mol CO / mol) x1 (mol N 2 /mol) 1-x1-x2 (mol H 2 / h) x2 (mol CO / mol) 1-x1-x2 (mol H 2 / h) Purge Mixing point balances: total: (100) + 500 = n1 ⇒ n1 = 600 mol/h N2: 4 + x1 * 500 = .13 * 600 ⇒ x1 = 0.148 mol N2/mol Overall system balances: N2: 4 = .148 * n3 ⇒ n3 = 27 mol/h Atomic C: 32 = n2 + x2*27 n2 = 24.3 mol CH3OH/h ⇒ Atomic H: 2 * 64 = 4*24.3 + 2*(1-0.148-x2)*27 x2 = 0.284 mol CO/mol Overall CO conversion: 100*[32-0.284(27)]/32 = 76% Single pass CO conversion: 24.3/ (32+.284*500) = 14% b. Recycle: To recover unconsumed CO and H2 and get a better overall conversion. Purge: to prevent buildup of N2.4.61 a. N2 + 3H2 2 + 3H2 -> NH3 2N 2NH3 (1-yp) (1-fsp) n1 (mol N2) yp (1-fsp) n1 (mol N2) (1-yp) (1-fsp) 3n1 (mol H2) yp (1-fsp) 3n1 (mol H2) (1-yp) n2 (mol I) yp n2 (mol I) (1-fsp) n1 (mol N2) (1-fsp) 3n1 (mol H2) n2 (mol I) 1 mol nr (mol) nr (mol) np (mol) Reactor Condenser (1-XI0)/4 (mol N2 / mol) n1 (mol N2) (1-fsp) n1 (mol N2) 2 fsp n1 (mol NH3) 3/4 (1-XI0) (mol H2 / mol) 3n1 (mol H2) (1-fsp) 3n1 (mol H2) XI0 (mol I / mol) n2 (mol I) n2 (mol I) 2 fsp n1 (mol NH3) 4-52
• 94. 4.61 (cont’d) At mixing point: N2: (1-XI0)/4 + (1-yp)(1-fsp) n1 = n1 I: XI0 + (1-yp) n2 = n2 Total moles fed to reactor: nr = 4n1 + n2 Moles of NH3 produced: np = 2fspn1 (1 − X I0 ) / 4 − y p (1 − f sp )n 1 Overall N2 conversion: × 100% (1 − X I0 ) / 4 b. XI0 = 0.01 fsp = 0.20 yp = 0.10 n1 = 0.884 mol N2 nr = 3.636 mol fed n2 = 0.1 mol I np = 0.3536 mol NH3 produced N2 conversion = 71.4% c. Recycle: recover and reuse unconsumed reactants. Purge: avoid accumulation of I in the system. d. Increasing XI0 results in increasing nr, decreasing np, and has no effect on fov. Increasing fsp results in decreasing nr, increasing np, and increasing fov. Increasing yp results in decreasing nr, decreasing np, and decreasing fov. Optimal values would result in a low value of nr and fsp, and a high value of np, this would give the highest profit. XI0 fsp yp nr np fov 0.01 0.20 0.10 3.636 0.354 71.4% 0.05 0.20 0.10 3.893 0.339 71.4% 0.10 0.20 0.10 4.214 0.321 71.4% 0.01 0.30 0.10 2.776 0.401 81.1% 0.01 0.40 0.10 2.252 0.430 87.0% 0.01 0.50 0.10 1.900 0.450 90.9% 0.10 0.20 0.20 3.000 0.250 55.6% 0.10 0.20 0.30 2.379 0.205 45.5% 0.10 0.20 0.40 1.981 0.173 38.5% 4-53
• 95. 4.62 a. i - C 4 H 10 + C 4 H 8 = C 8 H 18 Basis: 1-hour operation n 2 (n-C 4 H10 ) n 1 (C 8 H18 ) n 3 (i-C 4 H 10) D P n 2 (n-C 4 H10 ) n 1 (C 8 H18 ) m4 (91% H 2 SO4 ) F decanter still E n 1 (C 8 H18 ) Units of n: kmol n 2 (n-C 4 H10 ) Units of m: kg n 3 (i-C 4 H 10) reactor n 5 (n-C 4 H10 ) n 6 (i-C 4 H 10) n 7 (C 8 H18 ) m8 (91% H 2 SO 4 ) C B m4 (kg 91% H 2 SO4 ) 40000 kg A n 0 kmol n 3 (i-C 4 H 10) 0.25 i-C4 H10 0.50 n-C4 H10 0.25 C4 H 8 Calculate moles of feed b gb M = 0.25 M L − C4 H10 + 0.50 M n − C4 H10 + 0.25 M C4 H 8 = 0.75 5812 + 0.25 5610 . . g b gb g = 57.6 kg kmol b gb n0 = 40000 kg 1 kmol 57.6 kg = 694 kmol g b gb g Overall n - C 4 H 10 balance: n2 = 0.50 694 = 347 kmol n - C 4 H 10 in product C 8 H 18 balance: n1 = b0.25gb694g kmol C H 4 8 react 1 mol C 8 H 18 = 1735 kmol C 8 H 8 in product . 1 mol C 4 H 8 b At (A), 5 mol i - C 4 H 10 1 mole C 4 H 8 ⇒ n mol i - C 4 H 10 g = b5gb0.25gb694g = 867.5 kmol A i -C 4 H 10 at moles C 4 H 8 at b A g and b Bg A=173.5 b g Note: n mol C 4 H 8 = 173.5 at (A), (B) and (C) and in feed b gb g i - C 4 H 10 balance around first mixing point ⇒ 0.25 694 + n3 = 867.5 ⇒ n3 = 694 kmol i - C 4 H 10 recycled from still At C, 200 mol i - C4 H10 mol C4 H 8 b ⇒ n mol i - C4 H10 g = b200gb1735g = 34,700 kmol i - C H C . 4 10 4-54
• 96. 4.62 (cont’d) i - C 4 H 10 balance around second mixing point ⇒ 867.5 + n6 = 34,700 ⇒ n6 = 33,800 kmol C 4 H 10 in recycle E Recycle E: Since Streams (D) and (E) have the same composition, b g = n bmoles i - C H g ⇒ n n5 moles n - C 4 H 10 E 6 4 10 E = 16,900 kmol n - C 4 H 10 n2 bmoles n - C H g n bmoles i - C H g 4 10 D 3 4 10 D 5 n7 bmoles C H g = n ⇒ n = 8460 kmol C H 8 18 E 6 n1 bmoles C H g n 8 18 D 3 7 4 18 Hydrocarbons entering reactor: b347 + 16900gbkmol n - C H g FGH 58.12 kmol IJK kg 4 10 F . kg IJ + 1735 kmol C H FG 5610 kg IJ + b867.5 + 33800gb kmol i - C H g G 5812 H kmol K . 4 H . kmol K 10 4 8 F + 8460 kmol C H G 114.22 kg I J = 4.00 × 10 kg . H kmol K 6 8 18 H SO solution entering reactor 4.00 × 10 kg HC 2 kg H SO baq g 2 4 6 band leaving reactor g = 2 4 1 kg HC = 8.00 × 10 kg H SO baq g 6 2 4 m b H SO in recycleg n b n - C H in recycleg = 8 2 4 5 4 10 8.00 × 10 b H SO leaving reactor g n + n b n - C H leaving reactor g 6 2 4 2 5 4 10 ⇒ m = 7.84 × 10 kg H SO baq g in recycle E 8 6 2 4 m4 = H 2 SO 4 entering reactor − H 2 SO 4 in E b g = 16 × 10 5 kg H 2 SO 4 aq recycled from decanter . d ib g b g ⇒ 16 × 10 5 0.91 kg H 2 SO 4 1 kmol 98.08 kg = 1480 kmol H 2 SO 4 in recycle . d16 × 10 ib0.09gkg H O b1 kmol 18.02 kgg = 799 kmol H O from decanter . 5 2 2 Summary: (Change amounts to flow rates) Product: 173.5 kmol C 8 H 18 h , 347 kmol n - C 4 H 10 h Recycle from still: 694 kmol i - C 4 H 10 h Acid recycle: 1480 kmol H 2 SO 4 h , 799 kmol H 2 O h Recycle E: 16,900 kmol n - C 4 H 10 h , 33,800 kmol L - C 4 H 10 h , 8460 kmol C 8 H 18 h, 7.84 × 10 6 kg h 91% H 2 SO 4 ⇒ 72,740 kmol H 2 SO 4 h , 39,150 kmol H 2 O h 4-55
• 97. 4.63 a. A balance on ith tank (input = output + consumption) b g b g b v L min C A, i −1 mol L = vC Ai + kC Ai C Bi mol liter ⋅ min V L gbg E ÷ v, note V / v = τ C A , i −1 = C Ai + kτ C Ai C Bi B balance. By analogy, C B , i −1 = C Bi + kτ C Ai C Bi Subtract equations ⇒ C Bi − C Ai = C B , i −1 − C A, i −1 = C B , i − 2 − C A, i − 2 =… = C B 0 − C A0 A from balances on bi −1g tank st b. C Bi − C Ai = C B 0 − C A 0 ⇒ C Bi = C Ai + C B 0 − C A 0 . Substitute in A balance from part (a). b g C A, i −1 = C Ai + kτ C Ai C Ai + C B 0 − C A0 . Collect terms in C Ai , C 1 , C Ai . 2 Ai 0 2 C Ai b g kτ + C AL 1 + kτ C B 0 − C A0 − C A , i −1 = 0 ⇒α 2 C AL b + β C AL + γ = 0 where α = kτ , β = 1 + kτ C B 0 − C A 0 , γ = − C A, i −1 g − β + β 2 − 4αγ Solution: C Ai = (Only + rather than ±: since αγ is negative and the 2α negative solution would yield a negative concentration.) c. k= 36.2 N gamma CA(N) xA(N) v= 5000 1 -5.670E-02 2.791E-02 0.5077 V= 2000 2 -2.791E-02 1.512E-02 0.7333 CA0 = 0.0567 3 -1.512E-02 8.631E-03 0.8478 CB0 = 0.1000 4 -8.631E-03 5.076E-03 0.9105 alpha = 14.48 5 -5.076E-03 3.038E-03 0.9464 beta = 1.6270 6 -3.038E-03 1.837E-03 0.9676 7 -1.837E-03 1.118E-03 0.9803 8 -1.118E-03 6.830E-04 0.9880 9 -6.830E-04 4.182E-04 0.9926 10 -4.182E-04 2.565E-04 0.9955 11 -2.565E-04 1.574E-04 0.9972 12 -1.574E-04 9.667E-05 0.9983 13 -9.667E-05 5.939E-05 0.9990 14 -5.939E-05 3.649E-05 0.9994 (xmin = 0.50, N = 1), (xmin = 0.80, N = 3), (xmin = 0.90, N = 4), (xmin = 0.95, N = 6), (xmin = 0.99, N = 9), (xmin = 0.999, N = 13). As xmin → 1, the required number of tanks and hence the process cost becomes infinite. d. (i) k increases ⇒ N decreases (faster reaction ⇒ fewer tanks) (ii) v increases ⇒ N increases (faster throughput ⇒ less time spent in reactor ⇒ lower conversion per reactor) (iii) V increases ⇒ N decreases (larger reactor ⇒ more time spent in reactor ⇒ higher conversion per reactor) 4-56
• 98. 4.64 a. Basis: 1000 g gas Species m (g) MW n (mol) mole % (wet) mole % (dry) C3H8 800 44.09 18.145 77.2% 87.5% C4H10 150 58.12 2.581 11.0% 12.5% H2O 50 18.02 2.775 11.8% Total 1000 23.501 100% 100% Total moles = 23.50 mol, Total moles (dry) = 20.74 mol Ratio: 2.775 / 20.726 = 0.134 mol H 2 O / mol dry gas b. C3H8 + 5 O2 → 3 CO2 + 4 H2O, C4H10 + 13/2 O2 → 4 CO2 + 5 H2O Theoretical O2: 100 kg gas 80 kg C 3 H 8 1 kmol C 3 H 8 5 kmol O 2 C3H8: = 9.07 kmol O 2 / h h 100 kg gas 44.09 kg C 3 H 8 1 kmol C 3 H 8 100 kg gas 15 kg C 4 H 10 1 kmol C 4 H 10 6.5 kmol O 2 C 4 H 10 : = 1.68 kmol O 2 / h h 100 kg gas 58.12 kg C 4 H 10 1 kmol C 4 H 10 Total: (9.07 + 1.68) kmol O2/h = 10.75 kmol O2/h 10.75 kmol O 2 1 kmol Air 1.3 kmol air fed Air feed rate: = 66.5 kmol air / h h .21 kmol O 2 1 kmol air required The answer does not change for incomplete combustion4.65 5 L C 6 H 14 0.659 kg C 6 H 14 1000 mol C 6 H 14 = 38.3 mol C 6 H 14 L C 6 H 14 86 kg C 6 H 14 4 L C 7 H 16 0.684 kg C 7 H 16 1000 mol C 7 H 16 = 27.36 mol C 7 H 16 L C 7 H 16 100 kg C 7 H 16 C6H14 +19/2 O2 → 6 CO2 + 7 H2O C6H14 +13/2 O2 → 6 CO + 7 H2O C7H16 + 11 O2 → 7 CO2 + 8 H2O C7H16 + 15/2 O2 → 7 CO + 8 H2O Theoretical oxygen: 38.3 mol C 6 H 14 9.5 mol O 2 27.36 mol C 7 H 16 11 mol O 2 + = 665 mol O 2 required mol C 6 H 14 mol C 7 H 16 O2 fed: (4000 mol air )(.21 mol O2 / mol air) = 840 mol O2 fed 840 − 665 Percent excess air: × 100% = 26.3% excess air 665 4-57
• 99. 4.66 1 1 CO + O 2 → CO 2 H2 + O2 → H 2O 2 2 175 kmol/h 0.500 kmol N2/kmol x (kmol CO/mol) (0.500–x) (kmol H2/kmol) 20% excess air Note: Since CO and H 2 each require 0 .5 mol O 2 / mol fuel for complete combustion, we can calculate the air feed rate without determining x CO . We include its calculation for illustrative purposes. b A plot of x vs. R on log paper is a straight line through the points R1 = 10.0, x1 = 0.05 and g bR 2 g = 99.7, x 2 = 10 . . ln x = b ln R + ln a b g b g b = ln 10 0.05 ln 99.7 10.0 = 1.303 . @ ln a = lnb10g − 1303 lnb99.7g = −6.00 . . ⇒ x = 2.49 × 10 −3 R1303 . x = a Rb a = expb −6.00g = 2.49 × 10 −3 moles CO R = 38.3 ⇒ x = 0.288 mol Theoretical O : 175 kmol 0.288 kmol CO 0.5 kmol O 2 2 h kmol kmol CO 175 kmol 0.212 kmol H kmol O 0.5 kmol O + 2 2 = 43.75 2 h kmol kmol H h 2 43.75 kmol O required 1 kmol air 1.2 kmol air fed kmol air Air fed: 2 = 250 h 0.21 kmol O 1 kmol air required h 24.67 a. CH 4 + 2O 2 → CO 2 + 2H 2 O 100 kmol/h 0.944 CH4 7 C2 H 6 + O2 → 2CO 2 + 3H 2 O 0.0340 C2H6 2 0.0060 C3H8 C 3 H 8 + 5O 2 → 3CO 2 + 4H 2 O 0.0050 C4H10 13 17% excess air C 4 H 10 + O 2 → 4CO 2 + 5H 2 O 2 na (kmol air/h) 0.21 O2 0.79 N2 Theoretical O 2 : b g 0.944 100 kmol CH 4 b g 2 kmol O 2 0.0340 100 kmol C 2 H 6 + 3.5 kmol O 2 h 1 kmol CH 4 h 1 kmol C 2 H 6 + b g 0.0060 100 kmol C 3 H 8 5 kmol O 2 + b g 0.0050 100 kmol C 4 H 10 6.5 kmol O 2 h 1 kmol C 3 H 3 h 1 kmol C 4 H 10 = 207.0 kmol O 2 h 4-58
• 100. 4.67 (cont’d) 207.0 kmol O 2 1 kmol air 1.17 kmol air fed Air feed rate: n f = = 1153 kmol air h h 0.21 kmol O 2 kmol air req. b. b na = n f 2 x1 + 35x 2 + 5x 3 + 6.5x 4 1 + Pxs 100 1 0.21 . gb gb g c. n f = aR f , (n f = 75.0 kmol / h, R f = 60) ⇒ n f = 125R f . na = bRa , (na = 550 kmol / h, Ra = 25) ⇒ na = 22.0 Ra ∑x ∑A 1 xi = kAi ⇒ =k =1 ⇒ k = ∑A i i i i i i Ai ⇒ xi = , i = CH 4 , C 2 H 4 , C 3 H 8 , C 4 H 10 ∑A i i Run Pxs Rf A1 A2 A3 A4 1 15% 62 248.7 19.74 6.35 1.48 2 15% 83 305.3 14.57 2.56 0.70 3 15% 108 294.2 16.61 4.78 2.11 Run nf x1 x2 x3 x4 na Ra 1 77.5 0.900 0.0715 0.0230 0.0054 934 42.4 2 103.8 0.945 0.0451 0.0079 0.0022 1194 54.3 3 135.0 0.926 0.0523 0.0150 0.0066 1592 72.4 d. Either of the flowmeters could be in error, the fuel gas analyzer could be in error, the flowmeter calibration formulas might not be linear, or the stack gas analysis could be incorrect.4.68 a. C4H10 + 13/2 O2 → 4 CO2 + 5 H2O Basis: 100 mol C4H10 nCO2 (mol CO2) nH2O (mol H2O) nC4H10 (mol C4H10) Pxs (% excess air) nO2 (mol O2) nair (mol air) nN2 (mol N2) 0.21 O2 0.79 N2 D.F. analysis 6 unknowns (n, n1, n2, n3, n4, n5) -3 atomic balances (C, H, O) -1 N2 balance -1 % excess air -1 % conversion 0 D.F. 4-59
• 101. 4.68 (cont’d) b. i) Theoretical oxygen = (100 mol C4H10)(6.5 mol O2/mol C4H10) = 650 mol O2 n air = (650 mol O 2 )(1 mol air / 0.21 mol O 2 ) = 3095 mol air 100% conversion ⇒ n C4H10 = 0 , nO 2 = 0 b gb g n N2 = 0.79 3095 mol = 2445 mol U 73.1% N | 2 nCO2 = b100 mol C H react gb4 mol CO mol C H g = 400 mol CO V12.0% CO = b100 mol C H react gb5 mol H O mol C H g = 500 mol H O |14.9% H O 4 10 2 4 10 2 2 n H2O 4 10 2 4 10 W 2 2 ii) 100% conversion ⇒ nC4H10 = 0 20% excess ⇒ nair = 1.2(3095) = 3714 mol (780 mol O2, 2934 mol N2) Exit gas: 400 mol CO2 10.1% CO2 500 mol H2O 12.6% H2O 130 mol O2 3.3% O2 2934 mol N2 74.0% N 2 iii) 90% conversion ⇒ nC4H10 = 10 mol C4H10 (90 mol C4H10 react, 585 mol O2 consumed) 20% excess: nair = 1.2(3095) = 3714 mol (780 mol O2, 2483 mol N2) Exit gas: 10 mol C4H10 0.3% C4H10 360 mol CO2 9.1% CO2 450 mol H2O (v) 11.4% H2O 195 mol O2 4.9% O2 2934 mol N2 74.3% N 24.69 a. C3H8 + 5 O2 → 3 CO2 + 4 H2O H2 +1/2 O2 → H2O C3H8 + 7/2 O2 → 3 CO + 4 H2O Basis: 100 mol feed gas 100 mol 0.75 mol C3H8 n1 (mol C3H8) 0.25 mol H2 n2 (mol H2) n3 (mol CO2) n4 (mol CO) n0 (mol air) n5 (mol H2O) 0.21 mol O2/mol n6 (mol O2) 0.79 mol N2/mol n7 (mol N2) 75 mol C 3 H 8 5 mol O 2 25 mol H 2 0.50 mol O 2 Theoretical oxygen: + = 387.5 mol O 2 mol C 3 H 8 mol H 2 4-60
• 102. 4.69 (cont’d) 387.5 mol O 2 1 kmol air 1.25 kmol air fed Air feed rate: n0 = = 2306.5 mol air h 0.21 kmol O 2 1 kmol air req d. 90% propane conversion ⇒ n1 = 0100(75 mol C 3 H 8 ) = 7.5 mol C 3 H 8 . (67.5 mol C 3 H 8 reacts) 85% hydrogen conversion ⇒ n2 = 0150(25 mol C 3 H 8 ) = 3.75 mol H 2 . 0.95(67.5 mol C 3 H 8 react) 3 mol CO 2 generated 95% CO 2 selectivity ⇒ n3 = mol C 3 H 8 react = 192.4 mol CO 2 0.05(67.5 mol C 3 H 8 react) 3 mol CO generated 5% CO selectivity ⇒ n3 = = 101 mol CO . mol C 3 H 8 react FG mol H IJ + ( 25 mol H 2 )(2) H balance: (75 mol C 3 H 8 ) 8 H mol C 3 H 8 K = (7.5 mol C 3 H 8 )(8) + (3.75 mol H 2 )(2) + n5 ( mol H 2 O)(2) ⇒ n5 = 2912 mol H 2 O . mol O O balance: (0.21 × 2306.5 mol O 2 )(2 ) = (192.4 mol CO 2 )(2) mol O 2 + (101 mol CO)(1) + (2912 mol H 2 O)(1) + 2n6 ( mol O 2 ) ⇒ n6 = 1413 mol O 2 . . . N 2 balance: n7 = 0.79(2306.5) mol N 2 = 1822 mol N 2 Total moles of exit gas = (7.5 + 3.75 + 192.4 + 10.1 + 291.2 + 141.3 + 1822) mol = 2468 mol 101 mol CO . CO concentration in exit gas = × 10 6 = 4090 ppm 2468 mol b. If more air is fed to the furnace, (i) more gas must be compressed (pumped), leading to a higher cost (possibly a larger pump, and greater utility costs) (ii) The heat released by the combustion is absorbed by a greater quantity of gas, and so the product gas temperature decreases and less steam is produced. 4-61
• 103. 4.70 a. C5H12 + 8 O2 → 5 CO2 + 6 H2O Basis: 100 moles dry product gas n1 (mol C5H12) 100 mol dry product gas (DPG) 0.0027 mol C5H12/mol DPG Excess air 0.053 mol O2/mol DPG n2 (mol O2) 0.091 mol CO2/mol DPG 3.76n2 (mol N2) 0.853 mol N2/mol DPG n3 (mol H2O) 3 unknowns (n1, n2, n3) -3 atomic balances (O, C, H) -1 N2 balance -1 D.F. ⇒ Problem is overspecified b. N2 balance: 3.76 n2 = 0.8533 (100) ⇒ n2 = 22.69 mol O2 C balance: 5 n1 = 5(0.0027)(100) + (0.091)(100) ⇒ n1 = 2.09 mol C5H12 H balance: 12 n1 = 12(0.0027)(100) + 2n3 ⇒ n3 = 10.92 mol H2O O balance: 2n2 = 100[(0.053)(2) + (0.091)(2)] + n3 ⇒ 45.38 mol O = 39.72 mol O Since the 4th balance does not close, the given data cannot be correct. c. n1 (mol C5H12) 100 mol dry product gas (DPG) 0.00304 mol C5H12/mol DPG Excess air 0.059 mol O2/mol DPG n2 (mol O2) 0.102 mol CO2/mol DPG 3.76n2 (mol N2) 0.836 mol N2/mol DPG n3 (mol H2O) N2 balance: 3.76 n2 = 0.836 (100) ⇒ n2 = 22.2 mol O2 C balance: 5 n1 = 100 (5*0.00304 + 0.102) ⇒ n1 = 2.34 mol C5H12 H balance: 12 n1 = 12(0.00304)(100) + 2n3 ⇒ n3 = 12.2 mol H2O O balance: 2n2 = 100[(0.0590)(2) + (0.102)(2)] + n3 ⇒ 44.4 mol O = 44.4 mol O √ 2.344 − 100 × 0.00304 Fractional conversion of C5H12: = 0.870 mol react/mol fed 2.344 Theoretical O2 required: 2.344 mol C5H12 (8 mol O2/mol C5H12) = 18.75 mol O2 22.23 mol O 2 fed - 18.75 mol O 2 required % excess air: × 100% = 18.6% excess air 18.75 mol O 2 required 4-62
• 104. 4.71 a. 12 L CH 3 OH 1000 ml 0.792 g mol = 296.6 mol CH 3 OH / h h L ml 32.04 g CH3OH + 3/2 O2 → CO2 +2 H2O, CH3OH + O2 → CO +2 H2O 296.6 mol CH3OH(l)/h n2 ( mol dry gas / h) 0.0045 mol CH3OH(v)/mol DG 0.0903 mol CO2/mol DG n1 (mol O 2 / h) 0.0181 mol CO/mol DG 3.76n1 (mol N 2 / h) x (mol N2/mol DG) (0.8871–x) (mol O2/mol DG) n3 ( mol H 2 O(v) / h) 4 unknowns (n1 , n2 , n3 , x ) – 4 balances (C, H, O, N2) = 0 D.F. b. Theoretical O2: 296.6 (1.5) = 444.9 mol O2 / h C balance: 296.6 = n 2 (0.0045 + 0.0903 + 0.0181) ⇒ n 2 = 2627 mol/h H balance: 4 (296.6) = n 2 (4*0.0045) + 2 n 3 ⇒ n 3 = 569.6 mol H2O / h O balance : 296.6 + 2n1 = 2627[0.0045 + 2(0.0903) + 0.0181 + 2(0.8871 - x)] + 569.6 N2 balance: 3.76 n 1 = x ( 2627) Solving simultaneously ⇒ n1 = 574.3 mol O 2 / h, x = 0.822 mol N 2 / mol DG 296.6 − 2627(0.0045) Fractional conversion: = 0.960 mol CH 3 OH react/mol fed 296.6 574.3 − 444.9 % excess air: × 100% = 29.1% 444.9 569.6 mol H 2 O Mole fraction of water: = 0.178 mol H 2 O/mol (2627 + 569.6) mol c. Fire, CO toxicity. Vent gas to outside, install CO or hydrocarbon detector in room, trigger alarm if concentrations are too high4.72 a. G.C. Say ns mols fuel gas constitute the sample injected into the G.C. If xCH 4 and xC2 H 6 are the mole fractions of methane and ethane in the fuel, then b g b gb ns mol xC2 H 6 mol C 2 H 2 mol 2 mol C 1 mol C 2 H 6 g = 20 n b molg x b mol CH s CH 4 4 molgb1 mol C 1 mol CH g 85 4 E b xC2 H 6 mol C 2 H 6 mol fuel g = 01176 mole C H xCH 4 bmol CH 4 mol fuel g . 2 6 mole CH 4 in fuel gas 4-63
• 105. 4.72 (cont’d) Condensation measurement: b1.134 g H Ogb1 mol 18.02 gg = 0126 2 . mole H 2 O 0.50 mol product gas mole product gas Basis: 100 mol product gas. Since we have the most information about the product stream composition, we choose this basis now, and would subsequently scale to the given fuel and air flow rates if it were necessary (which it is not). CH 4 + 2O 2 → CO 2 + 2H 2 O 7 C 2 H 6 + O 2 → 2CO 2 + 3H 2 O 2 100 mol dry gas / h n1 (mol CH4 ) 0.1176 n1 (mol C2H6) 0.126 mol H O / mol 2 n2 (mol CO2) 0..874 mol dry gas / mol 0.119 mol CO / mol D.G. 2 n3 (mol O2 / h) x (mol N / mol) 2 376 n3 (mol N2 / h) (0.881-x) (mol O / mol D.G.) 2 Strategy: H balance ⇒ n ; C balance ⇒ n 2 ; N 2 balance U⇒ n , x V 1 O balance W 3 b gb g b gb gb g H balance: 4n1 + 6 01176n1 = 100 0126 2 ⇒ n1 = 5.356 mol CH 4 in fuel . . ⇒ 0.1176(5.356) = 0.630 mol C2H6 in fuel b gb g b gb gb g C balance: 5.356 + 2 0.630 + n2 = 100 0.874 0119 ⇒ n2 = 3.784 mol CO 2 in fuel . Composition of fuel: 5.356 mol CH 4 , 0.630 mol C 2 H 6 , 3.784 mols CO 2 ⇒ 0.548 CH 4 , 0.064 C 2 H 6 , 0.388 CO 2 N 2 balance: 3.76n3 = 100 0.874 xb gb g b gb g b gb g b gb O balance: 2 3.784 + 2n3 = 100 0.126 + 100 0.874 2 0119 + 0.881 − x . gb g b g Solve simultaneously: n3 = 18.86 mols O 2 fed , x = 0.813 5.356 mol CH 4 2 mol O 2 0.630 mol C 2 H 6 3.5 mol O 2 Theoretical O 2 : + 1 mol CH 4 1 mol CH 4 = 12.92 mol O 2 required (5.356 + 0.630 + 3.784) mol fuel 7 mol air 0.21 mol O 2 Desired O2 fed: = 14.36 mol O2 1 mol fuel mol air 14.36 − 12.92 Desired % excess air: × 100% = 11% 12.92 b. 18.86 − 12.92 Actual % excess air: × 100% = 46% 12.92 (18.86 / 0.21) mol air Actual molar feed ratio of air to fuel: = 9 :1 9.77 mol feed 4-64
• 106. 4.73 a. C3H8 +5 O2 → 3 CO2 + 4 H2O, C4H10 + 13/2 O2 → 4 CO2 + 5 H2O Basis 100: mol product gas n1 (mol C3H8) 100 mol n2 (mol C4H10) 0.474 mol H2O/mol x (mol CO2/mol) n3 (mol O2) (0.526–x) (mol O2/mol) x 69.4 Dry product gas contains 69.4% CO2 ⇒ = ⇒ x = 0.365 mol CO 2 /mol 0.526 − x 30.6 3 unknowns (n1, n2, n3) – 3 balances (C, H, O) = 0 D.F. O balance: 2 n3 = 152.6 ⇒ n3 = 76.3 mol O2 C balance : 3 n1 + 4 n 2 = 36.5 ⎫ n1 = 7.1 mol C 3 H 8 ⎬⇒ ⇒ 65.1% C 3 H 8 , 34.9% C 4 H10 H balance : 8 n1 + 10 n 2 = 94.8⎭ n 2 = 3.8 mol C 4 H10 b. nc=100 mol (0.365 mol CO2/mol)(1mol C/mol CO2) = 365 mol C nh = 100 mol (0.474 mol H2O/mol)(2mol H/mol H2O)=94.8 mol H ⇒ 27.8%C, 72.2% H From a: 7.10 mol C 3 H 8 3 mol C 3.80 mol C 4 H10 4 mol C + mol C 3 H 8 mol C 4 H10 × 100% = 27.8% C 7.10 mol C 3 H 8 11 mol (C + H) 3.80 mol C 4 H10 14 mol (C + H) + mol C 3 H 8 mol C 4 H104.74 Basis: 100 kg fuel oil 100 kg 0.85 kg C 1 kmol C Moles of C in fuel: = 7.08 kmol C kg 12.01 kg C 100 kg 0.12 kg H 1 kmol H Moles of H in fuel: = 12.0 kmol H kg 1 kg H 100 kg 0.017 kg S 1 kmol S Moles of S in fuel: = 0.053 kmol S kg 32.064 kg S 1.3 kg non-combustible materials (NC) 4-65
• 107. 4.74 (cont’d) 100 kg fuel oil 7.08 kmol C n2 (kmol N2) 12.0 kmol H n3 (kmol O2) 0.053 kmol S C + O2 → CO2 n4 (kmol CO2) 1.3 kg NC (s) C + 1/2 O2 → CO (8/92) n4 (kmol CO) 2H + 1/2 O2 → H2O n5 (kmol SO2) 20% excess air S + O2 → SO2 n6 (kmol H2O) n1 (kmol O2) 3.76 n1 (kmol N2) Theoretical O2: 7.08 kmol C 1 kmol O 2 12 kmol H .5 kmol O 2 0.053 kmol S 1 kmol O 2 + + = 10.133 kmol O 2 1 kmol C 2 kmol H 1 kmol S 20 % excess air: n1 = 1.2(10.133) = 12.16 kmol O2 fed O balance: 2 (12.16) = 2 (6.5136) + 0.5664 + 2 (0.053) + 6 + 2 n3 ⇒ n3 = 2.3102 kmol O2 C balance: 7.08 = n4+8n4/92 ⇒ n4 = 6.514 mol CO2 ⇒ 8 (6.514)/92 = 0.566 mol CO S balance: n5 = 0.53 kmol SO2 H balance: 12 = 2n6 ⇒ n6 = 6.00 kmol H2O N2 balance: n2 = 3.76(12.16) = 45.72 kmol N2 Total moles of stack gas = (6.514 + 0.566 + 0.053 + 6.00 + 2.310 + 45.72) kmol = 61.16 kmol ⇒ 10.7% CO, 0.92% CO, 0.087% SO 2 , 9.8% H 2 O, 3.8% O 2 , 74.8% N 24.75 a. Basis: 5000 kg coal/h; 50 kmol air min = 3000 kmol air h 5000 kg coal / h 0.75 kg C / kg n1 (kmol O2 / h) 0.17 kg H / kg n2 (kmol N2 / h) 0.02 kg S / kg C + 02 --> CO2 n3 (kmol CO / h) 2 0.06 kg ash / kg 2H + 1/2 O2 -->H2O 0.1 n3 (kmol CO / h) S + O2 --> SO2 n4 (kmol SO / h) 2 3000 kmol air / h C + 1/2 O2 --> CO n5 (kmol H2O / h) 0.21 kmol O / kmol 2 0.79 kmol N / kmol 2 mo kg slag / h Theoretical O 2 : C: b g 0.75 5000 kg C 1 kmol C 1 kmol O 2 = 312.2 kmol O 2 h h 12.01 kg C 1 kmol C 4-66
• 108. 4.75 (cont’d) H: b g 0.17 5000 kg H 1 kmol H 1 kmol H 2 O 1 kmol O 2 = 210.4 kmol O 2 h h 101 kg H . 2 kmol H 2 kmol H 2 O S: b g 0.02 5000 kg S 1 kmol S 1 kmol O 2 = 3.1 kmol O2/h h 32.06 kg S 1 kmol S Total = (312.2+210.4 + 3.1) kmol O2/h = 525.7 kmol O 2 h b g O 2 fed = 0.21 3000 = 630 kmol O 2 h 630 − 525.7 Excess air: × 100% = 19.8% excess air 525.7b. Balances: C: b gb gb g 0.94 0.75 5000 kg C react 1 kmol C = n3 + 0.1n3 h 12.01 kg C ⇒ n3 = 266.8 kmol CO 2 h , 01n3 = 26.7 kmol CO h . H: b017gb5000g kg H . 1 kmol H 1 kmol H 2 O = n5 ⇒ n5 = 420.8 kmol H 2 O h h 101 kg H . 2 kmol H S: (from part a) b 3.1 kmol O 2 for SO 2 g 1 kmol SO 2 = n4 ⇒ n4 = 31 kmol SO 2 h . h 1 kmol O 2 N2 : b0.79gb3000g kmol N h = n ⇒ n = 2370 kmol N h 2 2 2 2 O: b0.21g(3000)b2g = 2n + 2b266.8g + 1b26.68g + 2b31g + b1gb420.8g 1 . ⇒ n1 = 136.4 kmol O 2 / h Stack gas total = 3223 kmol h Mole fractions: x CO = 26.7 3224 = 8.3 × 10 −3 mol CO mol x SO 2 = 31 3224 = 9.6 × 10 −4 mol SO 2 mol . 1c. SO 2 + O 2 → SO 3 2 SO 3 + H 2 O → H 2SO 4 3.1 kmol SO 2 1 kmol SO 3 1 kmol H 2SO 4 98.08 kg H 2SO 4 = 304 kg H 2SO 4 h h 1 kmol SO 2 1 kmol SO 3 kmol H 2SO 4 4-67
• 109. 4.76 a. Basis: 100 g coal as received (c.a.r.). Let a.d.c. denote air-dried coal; v.m. denote volatile matter 100 g c.a. r. 1.147 g a.d.c. = 95.03 g air - dried coal; 4.97 g H 2 O lost by air drying 1.207 g c.a. r. 95.03 g a.d.c b1.234 − 1204g g H O = 2.31 g H O lost in second drying step . 2 2 1.234 g a.d.c. Total H 2 O = 4.97 g + 2.31 g = 7.28 g moisture 95.03 g a.d.c b1347 − 0.811g g bv. m.+ H Og − 2.31 g H O = 3550 g volatile matter . 2 . 2 1.347 g a.d.c. 95.03 g a.d.c 0.111 g ash = 8.98 g ash 1.175 g a.d.c. b g Fixed carbon = 100 − 7.28 − 3550 − 8.98 g = 48.24 g fixed carbon . 7.28 g moisture 7.3% moisture 48.24 g fixed carbon 48.2% fixed carbon 35.50 g volatile matter ⇒ 35.5% volatile matter 8.98 g ash 9.0% ash 100 g coal as received b. Assume volatile matter is all carbon and hydrogen. 1 mol O 2 1 mol C 10 3 g 1 mol air C + CO 2 → CO 2 : = 396.5 mol air kg C 1 mol C 12.01 g C 1 kg 0.21 mol O 2 1 0.5 mol O 2 1 mol H 10 3 g 1 mol air 2H + O2 → H 2O : = 1179 mol air kg H 2 2 mol H 1.01 g H 1 kg 0.21 mol O 2 1000 kg coal 0.482 kg C 396.5 mol air Air required: kg coal kg C 1000 kg 0.355 kg v. m. 6 kg C 396.5 mol air + kg 7 kg v. m. kg C 1000 kg 0.355 kg v. m. 1 kg H 1179 mol air + = 3.72 × 10 5 mol air kg 7 kg v. m. kg H 4-68
• 110. 4.77 a. Basis 100 mol dry fuel gas. Assume no solid or liquid products! n1 (mol C) 100 mol dry gas n2 (mol H) n3 (mol S) C + 02 --> CO2 0.720 mol CO / mol 2 C + 1/2 O2 --> CO 0.0257 mol CO / mol 2H + 1/2 O2 -->H2O 0.000592 mol SO / mol 2 S + O2 --> SO2 0.254 mol O / mol 2 n4 (mol O2) n5 (mol H2O (v)) (20% excess) H balance : n 2 = 2 n 5 ⎫ ⎪ O balance : 2 n 4 = 100 [ 2(0.720) + 0.0257 + 2 (0.000592) + 2 (0.254)] + n 5 ⎬ 20 % excess O 2 : (1.20) (74.57 + 0.0592 + 0.25 n 2 ] = n 4 ⎪ ⎭ ⇒ n2 = 183.6 mol H, n4 = 144.6 mol O2, n5 = 91.8 mol H2O Total moles in feed: 258.4 mol (C+H+S) ⇒ 28.9% C, 71.1% H, 0.023% S4.78 Basis: 100 g oil Stack SO 2 , N 2 , O 2, CO 2, H 2O x n 3 mol SO 2 (612.5 ppm SO 2) (N2 , O2 , CO2 , H 2 O) 0.10 (1 – x ) n 5 mol SO2 100 g oil (N2 , O2 , CO2 , H 2 O) 0.87 g C/g 0.10 g H/g furnace 0.03 g S/g Alkaline solution n 1 mol O2 scrubber 3.76 n 1 mol N2 (1 – x ) n 5 mol SO 2 (25% excess) (N2 , O2 , CO2 , H 2 O) n 2 mol N 2 n 3 mol O 2 0.90 (1 – x ) n 5 mol SO2 n 4 mol CO2 n 5 mol SO 2 n 6 mol H 2 O b g 0.87 100 g C 1 mol C 1 mol CO 2 ⇒ n4 = 7.244 mol CO 2 7.244 mol O 2 FG IJ CO 2 : 12.01 g C 1 mol C consumed H K b g 0.10 100 g H 1 mol H 1 mol H 2 O ⇒ n6 = 4.95 mol H 2 O 2.475 mol O 2 FG IJ H 2 O: 101 g H . 2 mol H consumed H K 4-69
• 111. 4.78 (cont’d) b g 0.03 100 g S 1 mol S 1 mol SO 2 ⇒ n5 = 0.0936 mol SO 2 0.0956 mol O 2 FG IJ SO 2 : 32.06 g S 1 mol S consumed H K b 25% excess O 2 : n1 = 125 7.244 + 2.475 + 0.0936 ⇒ 12.27 mol O 2 . g b O 2 balance: n3 = 12.27 mol O 2 fed − 7.244 + 2.475 + 0.0936 mol O 2 consumed g = 2.46 mol O 2 b N 2 balance: n 2 = 3.76 12.27 mol = 4614 mol N 2 . g b SO 2 in stack SO 2 balance around mixing point : g F I b gb g b g H K x 0.0936 + 010 1 − x 0.0936 = 0.00936 + 0.0842 x mol SO 2 n5 . Total dry gas in stack (Assume no CO2 , O2 , or N 2 is absorbed in the scrubber) b CO g b O g 2 2 bN g 2 b . bSO g g 7.244 + 2.46+ 4614 + 0.00936 + 0.0842 x = 5585 + 0.0842 x mol dry gas . b g 2 b 612.5 ppm SO 2 dry basis in stack gas g 0.00936 + 0.0842 x 612.5 = ⇒ x = 0.295 ⇒ 30% bypassed 5585 + 0.0842 x . 10 × 10 6 .4.79 Basis: 100 mol stack gas n 1 (mol C) C + O2 → CO2 n 2 (mol H) 100 mol 1 n 3 (mol S) 2H + O2 → H 2 O 0.7566 N 2 2 0.1024 CO2 S + O 2 → SO 2 n 4 (mol O2 ) 0.0827 H 2 O 3.76 n 4 (mol O2 ) 0.0575 O 2 0.000825 SO 2 a. . b gb C balance: n1 = 100 01024 = 10.24 mol C ⇒ g 10.24 mol C = 0.62 mol C b gb H balance: n2 = 100 0.0827 2 = 16.54 mol H gb g 16.54 mol H mol H The C/H mole ratio of CH 4 is 0.25, and that of C2 H 6 is 0.333; no mixture of the two could have a C/H ratio of 0.62, so the fuel could not be the natural gas. b. b gb S balance: n 3 = 100 0.000825 = 0.0825 mol S g b10.24 mol Cgb12.0 g 1 molg = 122.88 g CU 122.88 = 7.35 g C g H b16.54 mol Hgb1.01 g 1 molg = 16.71 g H| ⇒ 2.65 V 16.71 ⇒ No. 4 fuel oil b0.0825 mol Sgb32.07 g 1 molg = 2.65 g S | 142.24 × 100% = 1.9% S W 4-70
• 112. 4.80 a. Basis: 1 mol CpHqOr 1 mol CpHqOr no (mol S) n2 (mol CO2) C + 02 --> CO2 n3 (mol SO2) Xs (kg s/ kg fuel) 2H + 1/2 O2 -->H2O n4 (mol O2) S + O2 --> SO2 3.76 n1 (mol N2) P (% excess air) n5 (mol H2O (v)) n1 (mol O2) 3.76 n1 (mol N2) Hydrocarbon mass: p (mol C) ( 12 g / mol) = 12 p (g C) q (mol H) (1 g / mol) = q (g H) ⇒ (12 p + q + 16 r) g fuel r (mol O) (16 g / mol) = 16 r (g O) S in feed: (12 p + q + 16r) g fuel X s (g S) 1 mol S X (12 p + q + 16 r) n o= = s (mol S) (1) (1 - X s ) (g fuel) 32.07 g S 32.07(1 - X s ) p (mol C) 1 mol O 2 q (mol H) 0.5 mol O 2 ( r mol O) 1 mol O 2 Theoretical O2: + − 1 mol C 2 mol H 2 mol O = (p + 1/4 q − 1/2 r) mol O 2 required % excess ⇒ n1 = (1 + P/100) (p +1/4 q – ½ r) mol O2 fed (2) C balance: n2 = p (3) H balance: n5 = q/2 (4) S balance: n3 = n0 (5) O balance: r + 2n1 = 2n2 + 2n3 + 2n4 + n5 ⇒ n4 = ½ (r+2n1-2n2-2n3-n5) (6) Given: p = 0.71, q= 1.1, r = 0.003, Xs = 0.02 P = 18% excess air (1) ⇒ n0 = 0.00616 mol S (5) ⇒ n3 = 0.00616 mol SO2 (2) ⇒ n1 = 1.16 mol O2 fed (6) ⇒ n4 = 0.170 mol O2 (3) ⇒ n2 = 0.71 mol CO2 (4) ⇒ n5 = 0.55 mol H2O (3.76*1.16) mol N2 = 4.36 mol N2 Total moles of dry product gas = n2 + n3 + n4 + 3.76 n1=5.246 mol dry product gas Dry basis composition yCO2 = (0.710 mol CO2/ 5.246 mol dry gas) * 100% = 13.5% CO2 yO2 = (0.170 / 5.246) * 100% = 3.2% O2 yN2 = (4.36 / 5.246) * 100% = 83.1% N2 ySO2 = (0.00616 / 5.246) * 106 = 1174 ppm SO2 4-71
• 113. CHAPTER FIVE5.1 Assume volume additivity 1 0.400 0.600 Av. density (Eq. 5.1-1): = + ⇒ ρ = 0.719 kg L ρ 0.703 kg L 0.730 kg L A A ρO ρD = mt + m0 ⇒ m = b250 − 150gkg = 14.28 kg min bm = mass flow rate of liquidg a. m A mass of tank at time t A mass of b10 − 3g min empty tank m(kg / min) 14.28 kg 1L ⇒ V(L / min) = ⇒ V= = 19.9 L min ρ ( kg / L) min 0.719 kg b. m0 = m(t) - mt = 150 − 14.28 3 = 107 kg bg5.2 b void volume of bed: 100 cm3 − 2335 − 184 cm3 = 50.5 cm3 . g porosity: 50.5 cm3 void 184 cm3 total = 0.274 cm3 void cm3 total bulk density: 600 g 184 cm3 = 3.26 g cm3 b absolute density: 600 g 184 − 50.5 cm3 = 4.49 g cm3 g5.3 C 6 H 6 (l ) m B (kg / min) VB = 20.0 L / min m (kg / min) C 7 H 8 (l ) V (L / min) m T (kg / min) VT (L / min) ΔV πD 2 Δh π (5.5 m) 2 015 m . V= = = = 0.0594 m3 / min Δt 4 Δt 4 60 min Assume additive volumes b g VT = V - VB = 59.4 − 20.0 L / min = 39.4 L / min 0.879 kg 20.0 L 0.866 kg 39.4 L m = ρ B ⋅ VB + ρ T ⋅ VT = + = 517 kg / min . L min L min m (0.879 kg / L)(20.0 L / min) xB = B = = 0.34 kg B / kg m (517 kg / min) . 5-1
• 114. P1 = P0 + ρ sl gh1 U | F IF I5.4 a. P2 = P0 + ρ sl gh 2 ⇒ ΔP = P1 − P2 = ρ sl V | e jge jhbmgGH 11 N JK GH 11 Pa JK = ρ gh kg m3 m s2 kg⋅m N sl h = h1 − h 2 W s2 m2 b. 1 = xc + b1 − x g ⇒ check units! c ρ sl ρc ρl 1 kg crystals / kg slurry kg liquid / kg slurry = + kg slurry / L slurry kg crystals / L crystals kg liquid / L liquid L slurry L crystals L liquid L slurry = + = kg slurry kg slurry kg slurry kg slurry ΔP 2775 c. i.) ρ sl = = = 1415 kg / m3 gh 9.8066 0.200 b gb g 1 = xc + b1- x g ⇒ x FG 1 − 1 IJ = FG 1 c − 1IJ ii.) ρ sl ρc ρ lHρ ρ K Hρ c c l sl ρ K l F 1 I GG 1415 kg / m − 1 J 12d1000 kg / m i J xc = H 3 . K 3 F I = 0.316 kg crystals / kg slurry GG 2.3d10001kg / m i − 12d10001kg / m iJJ H . K 3 3 msl 175 kg 1000 L iii.) Vsl = = = 123.8 L ρ sl 1415 kg / m 3 m3 b iv.) mc = x c msl = 0.316 kg crystals / kg slurry 175 kg slurry = 55.3 kg crystals gb g 55.3 kg CuSO 4 ⋅ 5H 2 O 1 kmol 1 kmol CuSO 4 159.6 kg v.) mCuSO 4 = = 35.4 kg CuSO 4 249 kg 1 kmol CuSO 4 ⋅ 5H 2 O 1 kmol b g b vi.) ml = 1 − x c msl = 0.684 kg liquid / kg slurry 175 kg slurry = 120 kg liquid solution gb g ml 120 kg 1000 L Vl = = = 100 L vii.) ρl b1.2gd1000 kg / m i 3 m3 d. h(m) 0.2 ρ l(kg/m^3) 1200 ρ c(kg/m^3) 2300 ΔP(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 0.6 0.5 Solids Fraction 0.4 0.3 ΔP = 2775, ρ = 0.316 0.2 0.1 0 2300.00 2500.00 2700.00 2900.00 3100.00 Pressure Difference (Pa) 5-2
• 115. 5.4 (cont’d) e. b Basis: 1 kg slurry ⇒ x c kg crystals , Vc m3 crystals = g d i b x c kg crystals g d ρ c kg / m 3 i b1- x gbkg liquid g, V dm c l 3 liquid =i b1-ρx dgbkg/ m i g kg c liquid 3 l 1 kg 1 ρ sl = = bV + V gdm i c l 3 xc + b 1 − xc g ρc ρl5.5 Assume Patm = 1 atm 0.08206 m3 ⋅ atm 313.2 K 1 kmol PV = RT ⇒ V = = 0.0064 m3 mol kmol ⋅ K 4.0 atm 103 mol 1 mol 29.0 g 1 kg ρ= 3 = 4.5 kg m3 0.0064 m air mol 103 g nRT 1.00 mol 0.08206 L ⋅ atm 373.2 K5.6 a. V= = = 3.06 L P mol ⋅ K 10 atm b. % error = b3.06L - 2.8Lg × 100% = 9.3% 2.8L5.7 Assume Patm = 1013 bar . a. PV = nRT ⇒ n = b10 + 1013gbar . 20.0 m3 kmol ⋅ K 28.02 kg N 2 = 249 kg N 2 b g 25 + 273.2 K 0.08314 m3 ⋅ bar kmol PV nRT T P n b. = ⇒ n = V⋅ s ⋅ ⋅ s Ps Vs n s RTs T Ps Vs n= 20.0 m3 273K b10 + 1013gbar . 1 kmol 28.02 kg N 2 = 249 kg N 2 298.2K 1.013 bar 3 22.415 m STP b gkmol Ps Vs 1 atm 22.415 m3 atm ⋅ m35.8 a. R= = = 8.21 × 10 −2 n sTs 1 kmol 273 K kmol ⋅ K Ps Vs 1 atm 760 torr 359.05 ft 3 torr ⋅ ft 3 b. R= = = 555 n sTs 1 lb - mole 1 atm 492 R lb - mole ⋅ R 5-3
• 116. 10 cm H 2 O 1m 1 atm5.9 P = 1 atm + 2 = 101 atm . 10 cm 10.333 m H 2 O 2.0 m3 T = 25 C = 298.2 K , V = = 0.40 m3 min = 400 L min 5 min b g m = n mol / min ⋅ MW g / mol b g L g PV 1.01 atm 400 min 28.02 a. m= ⋅ MW = L⋅atm mol = 458 g min RT 0.08206 mol⋅K 298.2 K L g 400 273 K 1 mol 28.02 = 458 g min b g min mol b. m= 298.2 K 22.4 L STP F mI Vdm si = nRT P ⇒ u uG J = 3 nR T2 P1 D12 H s K Ad m i π D 4 u = ⋅ ⋅ ⋅ 2 25.10 Assume ideal gas behavior: 2 2 1 nR T1 P2 D 2 (1.80 + 1.013) bar ( 7.50 cm ) 2 T P D 2 60.0 m 333.2K u 2 = u1 2 1 1 = = 165 m sec (1.53 + 1.013) bar ( 5.00 cm ) 2 T1P2 D 2 2 sec 300.2K5.11 Assume ideal gas behavior: n = PV 100 + 100 atm 5 L = . . b = 0.406 mol g L⋅atm RT 0.08206 mol⋅K 300 K MW = 13.0 g 0.406 mol = 32.0 g mol ⇒ Oxygen5.12 Assume ideal gas behavior: Say m t = mass of tank, n g = mol of gas in tank N2: b gU ⇒ n = 0.009391 mol | 37.289 g = m t + n g 28.02 g mol V g CO : 37.440 g = m + n b44.1 g molg W m = 37.0256 g 2 t | g t unknown: MW = b37.062 − 37.0256gg = 3.9 g mol ⇒ Helium 0.009391 mol Δ V liters ΔV5.13 a. b g Vstd cm3 STP min = Δ t min 273K 296.2K 760 mm Hg 763 mm Hg 103 cm3 1L = 925.3 Δt φ 3 Vstd cm STP min b g U | 5.0 139 | straight line plot V 9.0 268 |φ = 0.031EV + 0.93 | 12.0 370 W std 0.010 mol N 2 22.4 litersbSTPg 10 cm 3 3 Vstd = = 224 cm 3 / min b. min 1 mole 1L d φ = 0.031 224 cm3 / min + 0.93 = 7.9 i 5-4
• 117. g nbkmolgM(kg / kmol) ====> PM n P = b V RT5.14 Assume ideal gas behavior ρ kg L = Vb Lg RT d V2 cm s = V1 cm 3 i d 3 Fρ I si ⋅ G J 12 = V1 P1M1T2 P2 M 2 T1 12 Hρ K 1 2 a. VH 2 = 350 LM cm3 758 mm Hg 28.02 g mol 323.2K OP 12 = 881 cm3 s N s 1800 mm Hg 2.02 g mol 298.2K Q b. b gb g b gb M = 0.25M CH 4 + 0.75M C3H 8 = 0.25 16.05 + 0.75 44.11 = 37.10 g mol g Vg = 350 cm3 LM b758gb28.02gb323.2g OP 12 = 205 cm3 s s N b1800gb37.10gb298.2g Q5.15 a. Reactor Δh soap b. n CO 2 = PV ⇒V= πR 2 Δh π 0.012 m = 2 d i 2 . 12 m 60 s = 11 × 10 −3 m3 / min . RT Δt 4 7.4 s min 755 mm Hg 1 atm 1.1× 10-3 m3 / min 1000 mol n CO2 = m ⋅atm 3 = 0.044 mol/min 0.08206 kmol⋅K 760 mm Hg 300 K 1 kmol5.16 m air = 10.0 kg / h n air (kmol / h) n (kmol / h) y CO (kmol CO 2 / kmol) 2 VCO 2 = 20.0 m3 / h n CO (kmol / h) 2 150 o C, 1.5 bar Assume ideal gas behavior 10.0 kg 1 kmol n air = = 0.345 kmol air / h h 29.0 kg air PV 15 bar . 100 kPa 20.0 m3 / h n CO 2 = = = 0.853 kmol CO 2 / h RT 8.314 m ⋅kPa 1 bar 3 kmol⋅K 423.2 K 0.853 kmol CO 2 / h y CO 2 × 100% = × 100% = 712% b 0.853 kmol CO 2 / h + 0.345 kmol air / h . g 5-5
• 118. 5.17 Basis: Given flow rates of outlet gas. Assume ideal gas behavior m1 (kg / min) 311 m 3 / min, 83o C, 1 atm 0.70 kg H 2 O / kg n 3 (kmol / min) 0.30 kg S / kg 0.12 kmol H 2 O / kmol 0.88 kmol dry air / kmol n 2 (kmol air / min) V2 (m 3 / min) m 4 (kg S / min) 167 o C, - 40 cm H 2 O gauge 1 atm 311 m3 kmol ⋅ K a. n3 = = 10.64 kmol min 356.2K min 0.08206 m3 ⋅ atm 10.64 kmol 0.12 kmol H 2 O 18.02 kg H 2 O balance : 0.70 m1 = min kmol kmol ⇒ m1 = 32.9 kg min milk b g b g S olids balance: 0.30 32.2 kg min = m4 ⇒ m4 = 9.6 kg S min Dry air balance : n 2 = 0.88 (10.64 kmol min ) ⇒ n 2 = 9.36 kmol min air 9.36 kmol 0.08206 m3 ⋅ atm 440K 1033 cm H 2 O V2 = min kmol ⋅ K (1033 − 40 ) cm H 2O 1 atm = 352 m3 air min Vair (m3 / s) 352 m3 1 min u air (m/min)= = = 0.21 m/s A (m 2 ) min 60 s π 4 ⋅ (6 m) 2 b. If the velocity of the air is too high, the powdered milk would be blown out of the reactor by the air instead of falling to the conveyor belt. ρ CO2 PM CO2 M CO 2 44 kg / kmol5.18 SG CO 2 = = RT = = = 152 . ρ air PM air RT M air 29 kg / kmol5.19 a. x CO 2 = 0.75 x air = 1 − 0.75 = 0.25 Since air is 21% O 2 , x O 2 = (0.25)(0.21) = 0.0525 = 5.25 mole% O 2 b. mCO 2 = n ⋅ x CO 2 ⋅ M CO 2 = 1 atm b g 2 × 1.5 × 3 m3 0.75 kmol CO 2 44.01 kg CO 2 =12 kg m3 ⋅atm 298.2 K kmol kmol CO 2 0.08206 kmol⋅K More needs to escape from the cylinder since the room is not sealed. 5-6
• 119. 5.19 (cont’d) c. With the room closed off all weekend and the valve to the liquid cylinder leaking, if a person entered the room and closed the door, over a period of time the person could die of asphyxiation. Measures that would reduce hazards are: 1. Change the lock so the door can always be opened from the inside without a key. 2. Provide ventilation that keeps air flowing through the room. 3. Install a gas monitor that sets off an alarm once the mole% reaches a certain amount. 4. Install safety valves on the cylinder in case of leaks. 15.7 kg 1 kmol5.20 n CO 2 = = 0.357 kmol CO 2 44.01 kg b Assume ideal gas behavior, negligible temperature change T = 19° C = 292.2 K g P1V n1RT n1 P 102kPa = ⇒ = 1 = a. P2 V b n1 + 0.357 RT gn1 + 0.357 P2 3.27 × 103 kPa ⇒ n1 = 0.0115 kmol air in tank n1RT 0.0115 kmol 292.2 K 8.314 m3 ⋅ kPa 103 L b. Vtank = = = 274 L P1 102kPa kmol ⋅ K m3 15700 g CO 2 + 11.5 mol air ⋅ (29.0 g air / mol) ρf = = 58.5 g / L 274 L c. CO 2 sublimates ⇒ large volume change due to phase change ⇒ rapid pressure rise. Sublimation causes temperature drop; afterwards, T gradually rises back to room temperature, increase in T at constant V ⇒ slow pressure rise. b gb g5.21 At point of entry, P1 = 10 ft H 2 O 29.9 in. Hg 33.9 ft H 2 O + 28.3 in. Hg = 37.1 in. Hg . At surface, P2 = 28.3 in. Hg, V2 = bubble volume at entry 1 x x 0.20 0.80 Mean Slurry Density: = solid + solution = + ρ sl ρ solid ρ solution (12)(100 g / cm ) (100 g / cm3 ) . . 3 . cm 3 1.03 g 2.20 lb 5 × 10 −4 ton 10 6 cm 3 = 0.967 ⇒ ρ sl = = 4.3 × 10 −3 ton / gal g cm 3 1000 g 1 lb 264.17 gal 300 ton gal 40.0 ft 3 (STP) 534.7 o R 29.9 in Hg a. = 2440 ft 3 / hr hr 4.3 × 10 −3 ton 1000 gal 492 o R 37.1 in Hg 3π e j D2 3 ==> D 4 D1 = 2 mm P2 V2 nRT V P 2 37.1 b. = ⇒ 2 = 1⇒ = ⇒ D 2 = 1.31D1 3 3 = 2.2 mm e j 2 P1V1 nRT V1 P2 D1 3 28.3 3π 4 2 % change = b2.2 - 2.0g mm × 100 = 10% 2.0 mm 5-7
• 120. 5.22 Let B = benzene n1 , n 2 , n 3 = moles in the container when the sample is collected, after the helium is added, and after the gas is fed to the GC. n inj = moles of gas injected n B , n air , n He = moles of benzene and air in the container and moles of helium added n BGC , m BGC = moles, g of benzene in the GC y B = mole fraction of benzene in room air a. P1V1 = n1RT1 (1 ≡ condition when sample was taken): P1 = 99 kPa, T1 = 306K 99 kPa 2 L mol ⋅ K n1 = = 0.078 mol = n air + n B 101.3 atm 306 K .08206 L ⋅ atm kPa P2 V2 = n 2 RT2 (2 ≡ condition when charged with He): P2 = 500 kPa, T2 = 306K 500 kPa 2 L mol ⋅ K n2 = = 0.393 mol = n air + n B + n He 101.3 atm 306 K .08206 L ⋅ atm kPa P3 V3 = n 3 RT3 (3 ≡ final condition in lab): P3 = 400 kPa, T3 = 296K 400 kPa 2 L mol ⋅ K n3 = = 0.325 mol = (n air + n B + n He ) − n inj 101.3 atm 296 K .08206 L ⋅ atm kPa n inj = n 2 − n 3 = 0.068 mol n2 0.393 mol m BGC (g B) 1 mol n B = n BGC × = = 0.0741 ⋅ m BGC n inj 0.068 mol 78.0 g nB 0.0741 ⋅ m BGC y B (ppm) = × 106 = × 106 = 0.950 × 106 ⋅ m BGC n1 0.078 9 am: y B = (0.950 × 106 )(0.656 × 10 −6 ) = 0.623 ppm U | 1 pm: y B = (0.950 × 106 )(0.788 × 10 −6 | ) = 0.749 ppm VThe avg. is below the PEL 5 pm: y B = (0.950 × 106 )(0.910 × 10 −6 ) = 0.864 ppm| | W b. Helium is used as a carrier gas for the gas chromatograph, and to pressurize the container so gas will flow into the GC sample chamber. Waiting a day allows the gases to mix sufficiently and to reach thermal equilibrium. c. (i) It is very difficult to have a completely evacuated sample cylinder; the sample may be dilute to begin with. (ii) The sample was taken on Monday after 2 days of inactivity at the plant. A reading should be taken on Friday. (iii) Helium used for the carrier gas is less dense than the benzene and air; therefore, the sample injected in the GC may be He- rich depending on where the sample was taken from the cylinder. (iv) The benzene may not be uniformly distributed in the laboratory. In some areas the benzene concentration could be well above the PEL. 5- 8
• 121. 45.23 Volume of balloon = π 10 m = 4189 m3 3 3 b g Moles of gas in balloon 492° R 3 atm b n kmol = g 4189 m3 535° R 1 atm 22.4 m3 STP 1 kmol b g = 515.9 kmol a. He in balloon: b gb m = 515.9 kmol ⋅ 4.003 kg kmol = 2065 kg He g 2065 kg 9.807 m 1N mg = = 20,250 N s 1 kg ⋅ m / s2 2 dP gas in balloon iV = n RT ⇒ n gas = Pair ⋅ n gas = 1 atm ⋅ 515.9 kmol = 172.0 kmol b. dP air displaced iV = n RT air air Pgas 3 atm Fbuoyant 172.0 kmol 29.0 kg 9.807 m 1N Fbuoyant = Wair displaced = = 48,920 N 1 kmol s 1 kg2⋅m 2 s Since balloon is stationary, ∑F = 0 1 Wtotal Fcable Fcable = Fbuoyant − Wtotal = 48920 N − b2065 + 150gkg 9.807 m 1 N = 27,20 s2 1 kg2⋅m s c. When cable is released, Fnet dAi = 27200 N = M tot a 27200 N 1 kg ⋅ m / s2 ⇒a= = 12.3 m s2 b 2065 + 150 kg g N d. When mass of displaced air equals mass of balloon + helium the balloon stops rising. Need to know how density of air varies with altitude. e. The balloon expands, displacing more air ⇒ buoyant force increases ⇒ balloon rises until decrease in air density at higher altitudes compensates for added volume.5.24 Assume ideal gas behavior, Patm = 1 atm 3 PN VN 5.7 atm 400 m / h a. PN VN = Pc Vc ⇒ Vc = = = 240 m3 h Pc 9.5 atm b. Mass flow rate before diversion: 400 m3 273 K 5.7 atm 1 kmol 44.09 kg kg C3 H 6 = 4043 h 303 K 1 atm 22.4 m ( STP ) 3 kmol h 5- 9
• 122. 5.24 (cont’d) Monthly revenue: ( 4043 kg h )( 24 h day )( 30 days month )( \$0.60 kg ) = \$1,747,000 month c. Mass flow rate at Noxious plant after diversion: 400 m3 273 K 2.8 atm 1 kmol 44.09 kg = 1986 kg hr hr 303 K 1 atm 22.4 m3 kmol Propane diverted = ( 4043 − 1986 ) kg h = 2057 kg h5.25 a. PHe = y He ⋅ P = 0.35 ⋅ (2.00 atm) = 0.70 atm PCH 4 = y CH 4 ⋅ P = 0.20 ⋅ (2.00 atm) = 0.40 atm PN 2 = y N 2 ⋅ P = 0.45 ⋅ (2.00 atm) = 0.90 atm b. Assume 1.00 mole gas 4.004 g FG IJ U | 0.35 mol He mol H = 140 g He . K | FG 16.05 g IJ = 3.21 g CH |17.22 g ⇒ mass fraction CH V 3.21 g 0.20 mol CH H mol K 4 | 4 4 = 17.22 g = 0186 . 0.45 mol N G F 28.02 g IJ = 12.61 g N | | H mol K 2 W 2 g of gas c. MW = = 17.2 g / mol mol d. ρ gas m n MW = = d i d i b = P MW = gb 2.00 atm 17.2 kg / kmol = 115 kg / m3 g jb g . V V RTe m3 ⋅atm 0.08206 kmol⋅K 363.2 K5.26 a. It is safer to release a mixture that is too lean to ignite. If a mixture that is rich is released in the atmosphere, it can diffuse in the air and the C3H8 mole fraction can drop below the UFL, thereby producing a fire hazard. b. fuel-air mixture n1 ( mol / s) n 3 ( mol / s) y C3H 8 = 0.0403 mol C 3 H 8 / mol 0.0205 mol C 3 H 8 / mol n C3H 8 = 150 mol C 3 H 8 / s diluting air n 2 ( mol / s) 150 mol C 3 H 8 mol n1 = = 3722 mol / s s 0.0403 mol C 3 H 8 Propane balance: 150 = 0.0205 ⋅ n 3 ⇒ n 3 = 7317 mol / s 5- 10
• 123. 5.26 (cont’d) Total mole balance: n1 + n 2 = n 3 ⇒ n 2 = 7317 − 3722 = 3595 mol air / s c. . b g n 2 = 13 n 2 min = 4674 mol / s V2 = 4674 mol / s 8.314 m3 ⋅ Pa 398.2 K = 118 m3 / s U | mol ⋅ K 131,000 Pa V2 = 1.41 | m3 diluting air V V1 = 3722 mol 8.314 m3 ⋅ Pa 298.2 K = 83.9 m / s 3 V1 | m3 fuel gas | s mol ⋅ K 110000 Pa W 150 mol / s 150 mol / s y2 = = × 100% = 18% n1 + n 2 b 3722 mol / s + 4674 mol / s g. d. The incoming propane mixture could be higher than 4.03%. b g If n 2 = n 2 min , fluctuations in the air flow rate would lead to temporary explosive conditions.5.27 Basis: (12 breaths min )( 500 mL air inhaled breath ) = 6000 mL inhaled min 24 o C, 1 atm 6000 mL / min lungs 37 o C, 1 atm n out (mol / min) n in (mol / min) 0.151 O 2 0.206 O 2 blood 0.037 CO 2 0.774 N 2 0.750 N 2 0.020 H 2 O 0.062 H 2 O 6000 mL 1L 273K 1 mol n in = a. min 3 10 mL 297K 22.4 L STP b g = 0.246 mol min b gb g N 2 balance: 0.774 0.246 = 0.750n out ⇒ n out = 0.254 mol exhaled min O 2 transferred to blood: b0.246gb0.206g − b0.254gb0.151g bmol O 2 g min 32.0 g mol = 0.394 g O 2 min CO 2 transferred from blood: b0.254gb0.037g bmol CO 2 g min 44.01 g mol = 0.414 g CO 2 min H 2 O transferred from blood: b0.254gb0.062g − b0.246gb0.020g bmol H O ming 18.02 g mol 2 = 0.195 g H 2 O min 5- 11
• 124. 5.27 (cont’d) PVin n RT = in in PVout n out RTout FG IJ FG T IJ = FG 0.254 mol min IJ FG 310K IJ = 1078 mL exhaled ml inhaled Vout n ⇒ H K H T K H 0.246 mol min K H 297K K . = out out Vin n in in b. b0.414 g CO lost ming + b0195 g H O lost ming − b0.394 g O gained ming = 0.215 g min 2 . 2 25.28 STACK Ts (K) Ta (K) 2 (M) L M s (g/mol) M a (g/mol) L(m) Ps (Pa) Pc (Pa) PM Ideal gas: ρ = RT a. D = ρgLb g b g − ρgL = Pa M a PM P gL M a M a gL − a s gL = a − LM OP combust. stack RTa RTs R Ta Ts N Q b. b gb g b gb g b gb g M s = 0.18 44.1 + 0.02 32.0 + 0.80 28.0 = 310 g mol , Ts = 655K , . Pa = 755 mm Hg M a = 29.0 g mol , Ta = 294 K , L = 53 m 755 mm Hg 1 atm 53.0 m 9.807 m kmol - K D= 760 mm Hg s 2 0.08206 m3 − atm × LM 29.0 kg kmol − 31.0 kg kmol OP × FG 1 N IJ = 323 N 1033 cm H O Q H 1 kg ⋅ m / s K m 1.013 × 10 N m 2 N 294K 655K 2 2 5 2 = 3.3 cm H 2 O b g P MW MWCCl2O = 98.91 g / mol ρ CCl2O 98.915.29 a. ρ = RT =======> ρ air = 29.0 = 3.41 Phosgene, which is 3.41 times more dense than air, will displace air near the ground. b. Vtube = b g π D in L π 2 2 b = 0.635 cm - 2 0.0559 cm 15.0 cm = 3.22 cm3 gb g 4 4 3.22 cm3 1L 1 atm 98.91 g / mol m CCl O = Vtube ⋅ ρ CCl O = 3 3 L⋅atm = 0.0131 g 2 2 10 cm 0.08206 mol⋅K 296.2 K 3.22 cm3 137 × 1000 g mol . . c. n CCl 2 O(l) = = 0.0446 mol CCl 2 O cm3 98.91 g 5- 12
• 125. 5.29 (cont’d) PV 1 atm 2200 ft 3 28.317 L mol ⋅ K n air = = = 2563 mol air RT 296.2K ft 3 .08206 L ⋅ atm n CCl 2 O 0.0446 = = 17.4 × 10 −6 = 17.4 ppm n air 2563 The level of phosgene in the room exceeded the safe level by a factor of more than 100. Even if the phosgene were below the safe level, there would be an unsafe level near the floor since phosgene is denser than air, and the concentration would be much higher in the vicinity of the leak. d. Pete’s biggest mistake was working with a highly toxic substance with no supervision or guidance from an experienced safety officer. He also should have been working under a hood and should have worn a gas mask.5.30 CH 4 + 2O 2 → CO 2 + 2 H 2 O 7 C 2 H 6 + O 2 → 2CO 2 + 3H 2 O 2 C 3 H 8 + 5O 2 → 3CO 2 + 4 H 2 O 1450 m 3 / h @ 15o C, 150 kPa n1 (kmol / h) 0.86 CH 4 , 0.08 C 2 H 6 , 0.06 C 3 H 8 n 2 (kmol air / h) 8% excess, 0.21 O 2 , 0.79 N 2 n1 = 1450 m3 273.2K b101.3 + 150gkPa 1 kmol h 288.2K 101.3 kPa 22.4 m3 STP b g = 152 kmol h Theoretical O 2 : LM F IJ FG IJ FG IJ OP = 349.6 kmol h O NM GH 152 kmol 2 kmol O 2 3.5 kmol O 2 5 kmol O 2 + 0.08 + 0.06 h 0.86 kmol CH 4 kmol C 2 H 6 K H kmol C 3 H 8 K H K QP 2 Air flow: Vair = b g 108 349.6 kmol O 2 . 1 kmol Air b g = 4.0 × 10 22.4 m3 STP 4 b g m3 STP h h 0.21 kmol O 2 kmol 5- 13
• 126. 5.31 Calibration formulas bT = 25.0; R = 14g , bT = 35.0, R = 27g ⇒ Tb° Cg = 0.77R + 14.2 T T T dP = 0; R = 0i , dP = 20.0, R = 6i ⇒ P bkPag = 3.33R g p g r gauge p dV = 0; R = 0i , dV = 2.0 × 10 , R = 10i ⇒ V dm hi = 200R F p F 3 F F 3 F dV = 0; R = 0h , dV = 1.0 × 10 , R = 25i ⇒ V dm hi = 4000R A A A 5 A A 3 A VF (m 3 / h), T, Pg n F (kmol / h) CH 4 + 2O 2 → CO 2 + 2 H 2 O x A (mol CH 4 / mol) 7 C 2 H 6 + O 2 → 2CO 2 + 3H 2 O x B (mol C 2 H 6 / mol) 2 x C (mol C 3 H 8 / mol) C 3 H 8 + 5O 2 → 3CO 2 + 4 H 2 O x D (mol n - C 4 H 10 / mol) 13 x E (mol i - C 4 H 10 / mol) C 4 H 10 + O 2 → 4CO 2 + 5H 2 O 2 VA ( m 3 / h) (STP) n A (kmol / h) 0.21 mol O 2 / mol 0.79 mol N 2 / mol nF = d VF m3 h i 273.2K dP + 101.3ikPa g 1 kmol bT + 273.2gK 101.3 kPa b g 22.4 m3 STP 0.12031V d P + 101.3i F kmol I bT + 273g GH h JK F g = Theoretical O 2 : dn i o 2 Th c b = n F 2x A + 3.5x B + 5x C + 6.5 x D + x E kmol O 2 req. h gh Air feed: n A = dn i o 2 Th kmol O 2 req. 1 kmol air b1 + Px g 100 kmol feed h 0.21 kmol O 2 1 kmol req. FG P IJ dn i H 100K = 4.762 1 + x o 2 Th VA = n a bkmol air hg d22.4 m bSTPg kmoli = 22.4n 3 A b g m3 STP h RT T(C) R p P (kP R xa xb xc xd xe P (% nF nO th nA g a) f X ) 2, Vf (m3/h) Va (m3/h) Ra 23.1 32.0 7.5 25.0 7.25 0.81 0.08 0.05 0.04 0.02 15 72.2 183.47 1004.74 1450 22506.2 5.63 7.5 20.0 19.3 64.3 5.8 0.58 0.31 0.06 0.05 0.00 23 78.9 226.4 1325.8 1160 29697.8 7.42 46.5 50.0 15.8 52.6 2.45 0.00 0.00 0.65 0.25 0.10 33 28.1 155.2 983.1 490 22022.3 5.51 21 30.4 3 10.0 6 0.02 0.4 0.35 0.1 0.13 15 53.0 248.1 1358.9 1200 30439.2 7.6 23 31.9 4 13.3 7 0.45 0.12 0.23 0.16 0.04 15 63.3 238.7 1307.3 1400 29283.4 7.3 25 33.5 5 16.7 9 0.5 0.3 0.1 0.04 0.06 15 83.4 266.7 1460.8 1800 32721.2 8.2 27 35.0 6 20.0 10 0.5 0.3 0.1 0.04 0.06 15 94.8 303.2 1660.6 2000 37196.7 9.3 5- 14
• 127. 5.32 NO + 2 O 2 ⇔ NO 2 1 1 mol 0.20 mol NO / mol n1 (mol NO) R | 0.80 mol air / mol U | n 2 (mol O 2 ) S | 0.21 O 2 V | n 3 (mol N 2 ) T 0.79 N 2 P0 = 380 kPa W n 4 (mol NO 2 ) Pf (kPa) a. Basis: 1.0 mol feed 90% NO conversion: n1 = 0.10(0.20) = 0.020 mol NO ⇒ NO reacted = 0.18 mol 018 mol NO 0.5 mol O 2 . O 2 balance: n 2 = 0.80(0.21) − = 0.0780 mol O 2 mol NO N 2 balance: n 3 = 0.80(0.79) = 0.632 mol N 2 018 mol NO 1 mol NO 2 . n4 = = 018 mol NO 2 ⇒ n f = n1 + n 2 + n 3 + n 4 = 0.91 mol . 1 mol NO 0.020 mol NO mol NO y NO = = 0.022 0.91 mol mol mol O 2 mol N 2 mol NO 2 y O 2 = 0.086 y N 2 = 0.695 y NO 2 = 0198 . mol mol mol Pf V n f RT n 0.91 mol FG IJ = P0 V n 0 RT ⇒ Pf = P0 f = 380 kPa n0 1 mol = 346 kPa H K Pf 360 kPa b. n f = n0 = (1 mol) = 0.95 mol P0 380 kPa n i = n i0 + υ iξ E n1 (mol NO) = 0.20 − ξ n 2 ( mol O 2 ) = (0.21)(0.80) − 0.5ξ n 3 (mol N 2 ) = (0.79)(0.80) n 4 ( mol NO 2 ) = ξ n f = 1 − 0.5ξ = 0.95 ⇒ ξ = 010 . ⇒ n1 = 010 mol NO, n 2 = 0118 mol O 2 , n 3 = 0.632 mol N 2 , . . n 4 = 010 mol NO 2 ⇒ y NO = 0105, y O 2 = 0124, y N 2 = 0.665, y NO 2 = 0105 . . . . NO conversion = b0.20 - n g × 100% = 50% 1 0.20 360 kPa P (atm) = = 355 atm . 101.3 kPa atm (y NO 2 P) (y NO 2 ) 0.105 1 Kp = = = = 151 atm 2 . ( y NO P)( y O 2 P) 0.5 0.5 ( y NO )( y O 2 ) P 0.5 b (0105) 0.124 . g b3.55g 0.5 0.5 5- 15
• 128. 5.33 Liquid composition: 49.2 kg M 1 kmol 100 kg liquid ⇒ = 0.437 kmol M 0.481 kmol M / kmol 112.6 kg 29.6 kg D 1 kmol = 0.201 kmol D ⇒ 0.221 kmol D / kmol 147.0 kg 21.2 kg B 1 kmol = 0.271 kmol B 0.298 kmol B / kmol 78.12 kg 0.909 kmol a. Basis: 1 kmol C 6 H 6 fed V1 (m 3 ) @ 40o C, 120 kPa n1 (kmol) 0.920 HCl 0.080 Cl 2 1 kmol C 6 H 6 ( 78.12 kg) n 0 (kmol Cl 2 ) n 2 (kmol) 0.298 C 6 H 6 0.481 C 6 H 5Cl 0.221 C 6 H 4 Cl 2 C 6 H 6 + Cl 2 → C 6 H 5Cl + HCl C 6 H 5Cl + Cl 2 → C 6 H 5Cl 2 + HCl 1 kmol C 6 H 6 6 kmol C C balance: = n 2 0.298 × 6 + 0.481 × 6 + 0.221 × 6 1 kmol C 6 H 6 ⇒ n 2 = 100 kmol . H balance: 1 kmol C 6 H 6 6 kmol H 1 kmol C 6 H 6 b = n1 0.920 (1) g + n 2 0.298 × 6 + 0.481 × 5 + 0.221 × 4 ⇒ n1 = 100 kmol . n1RT 100 kmol 1013 kPa 0.08206 m3 ⋅ atm 313.2 K . . V1 = = = 217 m3 . P 120 kPa 1 atm kmol ⋅ K V1 217 m3 . ⇒ = = 0.278 m3 / kg B m B 7812 kg B . πd 2 4 ⋅ Vgas b. Vgas ( m3 / s) = u(m / s) ⋅ A(m2 ) = u ⋅ ⇒ d2 = 4 π ⋅u 3 4m B0 ( kg B) 0.278 m s 1 min 10 cm2 4 d2 = = 5.90m B0 (cm2 ) min kg B π (10) m 60 s m2 ⇒ d(cm) = 2.43 ⋅ m B0 b g 1 2 c. Decreased use of chlorinated products, especially solvents. 5- 16
• 129. 5.34 3.74 SCMM Vb ( m 3 / min) @900 C, 604 mtorr 60% DCS conversion n a ( mol / min) 0.220 DCS n1 (mol DCS / min) U | 0.780 N 2 O n 2 (mol N 2 O / min) n 3 (mol N 2 / min) V n b (mol / min) | n 4 (mol HCl(g) / min) W SiH 2 Cl 2(g) + 2 N 2 O (g) → SiO 2(s) + 2 N 2(g) + 2 HCl (g) 3.74 m3 (STP) 103 mol a. na = =167 mol / min min 22.4 m3 (STP) 60% conversion: n1 = 1- 0.60gFGH 0.220 mol DCS IJK b167 mol / ming = 14.7 mol DCS / min b mol DCS reacted: b0.60gb0.220gb167g mol DCS = 22.04 mol DCS reacted / min min N 2 O balance: n 2 = 0.780 167 b g mol N O min 2 22.04 mol DCS 2 mol N 2 O − = 8618 mol N 2 O / min . min mol DCS 22.04 mol DCS 2 mol N 2 N 2 balance: n 3 = = 44.08 mol N 2 / min min mol DCS 22.04 mol DCS 2 mol HCl HCl balance: n 4 = = 44.08 mol HCl / min min mol DCS n B = n1 + n 2 + n 3 + n 4 = 189 mol / min n B RT 189 mol 62.36 L ⋅ torr 0.001 m3 1173 K ⇒ VB = = = 2.29 × 104 m3 / min P min mol ⋅ K L 0.604 torr n1 14.7 mol DCS/min b. p DCS = x DCS ⋅ P= P= ⋅ 604 mtorr=47.0 mtorr nB 189 mol/min n 86.2 mol N 2 O/min p N2O = x N 2O ⋅ P= 2 P= ⋅ 604 mtorr=275.5 mtorr nB 189 mol/min mol SiO 2 r=3.16 × 10-8 ⋅ p DCS ⋅ p N 2O 0.65 = 3.16 × 10-8 ( 47.0 )( 275.5 )0.65 = 5.7 × 10−5 m2 ⋅ s MW 5.7 × 10−5 mol SiO 2 60 s 120 min 60.09 g/mol 1010 A c. h(A)=r ⋅ t ⋅ = ρSiO2 m2 ⋅ s min 2.25 × 106 g/m3 1 m (Table B.1) =1.1 × 10 A5 The films will be thicker closer to the entrance where the lower conversion yields higher pDCS and p N 2 O values, which in turn yields a higher deposition rate. 5-17
• 130. 5.35 Basis: 100 kmol dry product gas n1 (kmol C x H y ) m1 (kg C x H y ) V2 (m 3 ) R100 kmol dry gasU |0.105 CO | n 2 (kmol air) 0.21 O 2 S0.053 O |0.842 N 2 V | T W 2 0.79 N 2 2 30 o C, 98 kPa n 3 (kmol H 2 O) a. N 2 balance: 0.79n 2 = 0.842(100) ⇒ n 2 = 106.6 kmol air b g b g b O balance: 2 0.21n 2 = 100 2 0105 + 2 0.053 + n 3 ⇒ n 3 = 1317 kmol H 2 O . . g C balance: d n1 kmol C x H y i x bkmol Cg = 100b0105g ⇒ n x = 10.5 . b1g dkmol C H i x y 1 b2 g n 3 =13.17 H balance: n1y = 2n 3 ====> n1y = 26.34 b g b g y = 26.34 = 2.51 mol H / mol C Divide 2 by 1 ⇒ x 10.5 O2 fed: 0.21b106.6 kmol air g = 22.4 kmol O2 in excess = 5.3 kmol ⇒ Theoretical O = b22.4 - 5.3g kmol = 17.1 kmol 2 5.3 kmol O 2 % excess = × 100% = 31% excess air 17.1 kmol O 2 106.6 kmol N 2 22.4 m3 (STP) 1013 kPa 303 K . b. V2 = = 2740 m3 kmol 98 kPa 273 K m1 = b g b n1x kmol C 12.0 kg n1y kmol H 101 kg n1x=10.5 + . =====> m1 = 152.6 kg g kmol kmol n1y=26.34 V2 2740 m3 air m3 air = = 18.0 m1 152.6 kg fuel kg fuel5.36 3N 2 H 4 → 6xH 2 + (1 + 2 x)N 2 + (4 − 4 x)NH 3 a. 0 ≤ x ≤ 1 50 L 0.82 kg 1 kmol b. n N2H4 = = 128 kmol . L 32.06 kg . LM 6x kmol H + b1 + 2xg kmol N + b4 − 4xg kmol NH OP n product = 128 kmol N 2 H 4 2 2 3 N 3 kmol N H 3 kmol N H 2 3 kmol N H 4 Q 2 4 2 4 = 128 . 3 b6x + 1 + 2x + 4 − 4xg = 1707x + 2.13 kmol . 5-18
• 131. 5.36 (cont’d) x nproduct Vp (L) Volume of Product Gas 0 2.13 15447.92 0.1 2.30 16685.93 30000.00 0.2 2.47 17923.94 0.3 2.64 19161.95 25000.00 0.4 2.81 20399.96 20000.00 0.5 2.98 21637.97 V (L) 0.6 3.15 22875.98 15000.00 0.7 3.32 24113.99 10000.00 0.8 3.50 25352.00 0.9 3.67 26590.01 5000.00 1 3.84 27828.02 0.00 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 x c. Hydrazine is a good propellant because as it decomposes generates a large number of moles and hence a large volume of gas.5.37 mA (g A / h) c Vair m3 / h h C A (g A / m 3 ) a. (i) Cap left off container of liquid A and it evaporates into room, (ii) valve leak in cylinder with A in it, (iii) pill of liquid A which evaporates into room, (iv) waste containing A poured into sink, A used as cleaning solvent. FG kg A IJ FG kg A IJ F m I C FG kg A IJ 3 b. mA H hK H hK in = mA out = Vair GH h JK H m K A 3 C e j⋅ V gA mol A A m3 mA RT yA = = ===================> yA = mol air M e j ⋅ n c. gA A mol air m CA = A ; nair = PV k ⋅ Vair M A P k⋅Vair RT d. yA = 50 ×10−6 mA = 90 g / h . 8.314 mol⋅Pa 3 dV h m m RT 9.0 g / h ⋅K 293 K = A = = 83 m3 / h air min −6 d kyA MA P 0.5 50 × 10 101.3 × 10 Pa 104.14 g / mol 3 i Concentration of styrene could be higher in some areas due to incomplete mixing (high concentrations of A near source); 9.0 g/h may be an underestimate; some individuals might be sensitive to concentrations < PEL. e. Increase in the room temperature could increase the volatility of A and hence the rate of evaporation from the tank. T in the numerator of expression for Vair : At higher T, need a greater air volume throughput for y to be < PEL. 5-19
• 132. 5.38 Basis: 2 mol feed gas C3 H 6 + H 2 ⇔ C3 H 8 n p (mol C 3 H 8 ) U | 1 mol C 3 H 6 V (1 - n p )(mol C 3 H 6 ) n 2 = n p + 2(1 − n p ) = 2 − n p | 1 mol H 2 25 C, 32 atm (1 - n p )(mol H 2 ) W 235 C, P2 a. At completion, n p = 1 mol , n 2 = 2 − 1 = 1 mol P2 V n 2 RT2 n T 1 mol 508K 32.0 atm = ⇒ P2 = 2 2 P1 = = 27.3 atm P1V n1RT1 n1 T1 2 mol 298K b. P2 = 35.1 atm P2 T1 35.1 atm 298K 2 mol n2 = n1 = = 1.29 mol P1 T2 32.0 atm 508K 1.29 = 2 − n p ⇒ n p = 0.71 mol C 3 H 8 produced b g ⇒ 1- 0.71 = 0.29 mol C 3 H 6 unreacted ⇒ 71% conversion of propylene c. P2 (atm) n2 C3H8 prod. %conv. 27.5 1.009 0.99075 99.075 Pressure vs Fraction Conversion 28.0 1.028 0.9724 97.24 29.5 1.083 0.91735 91.735 120 30.0 1.101 0.899 89.9 31.5 1.156 0.84395 84.395 100 32.0 1.174 0.8256 82.56 33.0 1.211 0.7889 78.89 33.5 1.229 0.77055 77.055 80 34.0 1.248 0.7522 75.22 % conversion 34.5 1.266 0.73385 73.385 60 %conv. 35.0 1.285 0.7155 71.55 37.0 1.358 0.6421 64.21 39.0 1.431 0.5687 56.87 40 40.0 1.468 0.532 53.2 20 0 25.0 27.0 29.0 31.0 33.0 35.0 37.0 39.0 41.0 Pressure (atm) 5-20
• 133. 5.39 Convert fuel composition to molar basis Basis: 100 g ⇒ b g 95 g CH 4 1 mol 16.04 g = 5.92 mol CH 4 ⇒ 97.2 mol % CH 4 U V b g 5 g C 2 H 6 1 mol 30.07 g = 017 mol C 2 H 6 . 2.8 mol % C 2 H 6 W 500 m3 / h n 2 (kmol CO 2 / h) n1 (mol / h) n 3 (kmol H 2 O / h) 0.972 CH 4 n 4 (kmol O 2 / h) 0.028 C 2 H 6 n 5 (kmol N 2 / h) 40 C, 1.1 bar Vair (SCMH) 25% excess air P1V1 11 bar 500 m3 . kmol ⋅ K n1 = = = 211 kmol h . RT1 313K h 0.08314 m3 ⋅ bar 7 CH 4 + 2O 2 → CO 2 + 2 H 2 O C 2 H 6 + O 2 → 2CO 2 + 3H 2 O 2 Theoretical O 2 = LM 21.1 kmol 0.972 kmol CH 4 2 kmol O 2 h N kmol 1 kmol CH 4 + 0.028 kmol C 2 H 6 3.5 kmol O 2 OP = 431 kmol O . 2 kmol 1 kmol C 2 H 6 Q h Air Feed: . .b 125 431 kmol O 2 g 1 kmol Air 22.4 m bSTPg 3 = 5700 SCMH h 0.21 kmol O 2 1 kmol5.40 Basis: 1 m3 gas fed @ 205° C, 1.1 bars Ac = acetone 1 m 3 @205 C, 1.1 bar n 3 (kmol), 10 C, 40 bar n1 (kmol) condenser y 3 (kmol Ac / kmol) y1 (kmol Ac / kmol) (1- y 3 )(kmol air / kmol) (1 - y1 )(kmol air / kmol) p AC = 0.379 bar p AC = 0.100 bar n 2 (kmol Ac(l)) 1.00 m3 273K 110 bars . 1 kmol n1 = a. 478K 10132 bars 22.4 m3 STP . b g = 0.0277 kmol 0.100 bar 0.379 bar y1 = = 0.0909 kmol Ac kmol , y 3 = = 9.47 × 10 −3 kmol Ac kmol 1.1 bars 40.0 bars b gb g Air balance: 0.0277 0.910 = (1 − 9.47 × 10 −3 )n 3 ⇒ n 3 = 0.0254 kmol Mole balance: 0.0277 = 0.0254 + n 2 ⇒ n 2 = 0.0023 kmol Ac condensed 0.0023 kmol Ac 58.08 kg Ac Acetone condensed = = 0133 kg acetone condensed . 1 kmol Ac 5-21
• 134. 5.40 (cont’d) Product gas volume = 0.0254 kmol 22.4 m3 STP b g . 283K 10132 bars = 0.0149 m3 273K 40.0 bars 20.0 m3 effluent 0.0277 kmol feed 0.0909 kmol Ac 58.08 kg Ac b. = 196 kg Ac h h 0.0149 m3 effluent kmol feed kmol Ac5.41 Basis: 1.00 × 10 6 gal. wastewater day. Neglect evaporation of water. 1.00 × 10 6 gal / day Effluent gas: 68 F, 21.3 psia(assume) n1 (lb-moles H2O/day) n2 (lb-moles air/day) 0.03n1 (lb-moles NH3 /day) n3 (lb-moles NH3 /day) 300 × 10 6 ft 3 air / day Effluent liquid 68 F, 21.3 psia n1 (lb-moles H2O/day) n2 (lb-moles air/day) n4 (lb-moles NH3 /day) a. Density of wastewater: Assume ρ = 62.4 lb m ft 3 ⎡ n1 lb-moles H 2 O 18.02 lb m 0.03n1 lb m NH3 17.03 lb m ⎤ 1 ft 3 7.4805 gal ⎢ + × ⎢ ⎣ day 1 lb-mole day 1 lb-mole ⎥ 62.4 lb m ⎥ ⎦ 1 ft 3 gal = 1.00 × 106 day ⇒ n1 = 4.50 × 105 lb-moles H 2 O fed day , 0.03n1 = 1.35 × 104 lb-moles NH 3 fed day 300 × 106 ft 3 492 R 21.3 psi 1 lb-mole n2 = = 1.13 × 106 lb-moles air day day 527.7 R 14.7 psi 359 ft 3 ( STP ) 93% stripping: n3 = 0.93 × 13500 lb-moles NH 3 fed day = 12555 lb-moles NH3 day Volumetric flow rate of effluent gas PVout nout RT = ⇒ Vout = Vin nout 300 × 10 ft = 6 3 (1.13 × 10 6 ) + 12555 lb-moles day PVin nin RT nin day 1.13 × 106 lb-moles day = 303 × 106 ft 3 day 12555 lb - moles NH 3 day Partial pressure of NH 3 = y NH 3 P = × 213 psi . d1.129 × 10 6 i + 12555 lb - moles day = 0.234 psi 5-22
• 135. 5.42 Basis: 2 liters fed / min 2.0 L soln 1130 g 0.12 g NaOH 1 mol 0.23 NaOH ads. 1 mol Cl 2 mol Cl ads.= = 0.013 60 min L g soln 40.0 g mol NaOH 2 mol NaOH min 2 L / min @ 23 C, 510 mm H 2 O n 2 (mol air / mol) n1 (mol / min) y (mol Cl 2 / mol) (1 - y)(mol air / mol) 0.013 mol Cl 2 / min Assume Patm = 10.33 m H 2 O ⇒ Pabs b g = b10.33 + 0.510g m H O = 10.84 m H O in 2 2 2L 273K 10.84 m H 2 O 1 mol n1 = min 296K 10.33 m H 2 O 22.4 L STP b g = 0.0864 mol min mol Cl 2 Cl balance: 0.0864y = 0.013 ⇒ y = 0150 . ,∴ specification is wrong mol5.43 125 L / min @ 25o C, 105 kPa V3 (L / min) @ 65o C, 1 atm n1 ( mol / min) n C2 H 6 (mol C 2 H 6 / min) y1 (mol H 2 O / mol) Rb g |1- y1 ( mol dry gas / mol) U | n C2 H 4 (mol C 2 H 4 / min) n air (mol air / min) S | 0.235 mol C 2 H 6 / mol DG V | n H 2 O (mol H 2 O / min) T 0.765 mol C 2 H 4 / mol DG W 355 L / min air @ 75o C, 115 kPa n 2 (mol / min) y 2 ( mol H 2 O / mol) (1- y 2 )( mol dry air / mol) a. Hygrometer Calibration ln y = bR + ln a dy = ae i bR b= b ln y1 y 2 g = lnd0.2 10 i = 0.08942 −4 R 2 − R1 90 − 5 bg ln a = ln y1 − bR1 = ln 10 −4 − 0.08942 5 ⇒ a = 6.395 × 10 −5 ⇒ y = 6.395 × 10 −5 e 0.08942R 125 L 273K 105 kPa 1 mol b. n1 = = 5.315 mol min wet gas min 298K 101 kPa 22.4 L STP b g 355 L 273K 115 kPa 1 mol n2 = = 14.156 mol min wet air min 348K 101 kPa 22.4 L STP b g R 1 = 86.0 → y1 = 0.140 , R 2 = 12.8 → y 2 = 2.00 × 10 −4 mol H 2 O mol 5-23
• 136. 5.43 (cont’d) b C 2 H 6 balance: n C2 H 6 = 5.315 mol min gFGH b1 − 0.140g mol DG IJK FGH 0.235 mol CDG IJK mol mol 2 H 6 = 1.07 mol C 2 H 6 min b gb gb g C 2 H 4 balance: n C2 H 4 = 5.315 0.860 0.765 = 3.50 mol C 2 H 4 min Dry air balance: n = b14.156gd1 − 2.00 × 10 i = 14.15 mol DA min air −4 Water balance: n = b5.315gb0.140g + b14.156gd1.00 × 10 i = 0.746 mol H O min H 2O −4 2 n = b1.07 + 3.50 + 14.15g mol min = 18.72 mol min , dry product gas n = b18.72 + 0.746g = 19.47 mol min total 19.47 mol min 22.4 L bSTPg 338K V =3 = 540 liters min mol 273K FG 1.07 IJ × 100% = 5.7% C H , 18.7% C H , 75% dry air Dry basis composition: H 18.72 K 2 6 2 4 0.746 mol H 2 O c. p H 2O = y H 2Ol ⋅ P = × 1 atm = 0.03832 atm 19.47 mol 1 FG 0.03832 IJ y H 2 O = 0.03832 ⇒ R = 0.08942 ln H 6.395 × 10−5 K = 71.55.44 CaCO 3 → CaO + CO 2 1350 m3 273K 1 kmol n CO 2 = = 12.92 kmol CO 2 h h 1273K 22.4 m3 STP b g 12.92 kmol CO 2 1 kmol CaCO 3 100.09 kg CaCO 3 1 kg limestone = 1362 kg limestone h h 1 kmol CO 2 1 kmol CaCO 3 0.95 kg CaCO 3 1362 kg limestone 0.17 kg clay = 279 kg clay h h 0.83 kg limestone Weight % Fe 2 O 3 b g kg Fe 2 O 3 kg clay 279 0.07 × 100% = 18% Fe 2 O 3 1362 + 279 − 12.92 44.1 kg limestone kg clay . b g kg CO 2 evolved 5-24
• 137. 5.45 R864.7 g C b1 mol 12.01 gg = 72.0 mol C |116.5 g H b1 mol 1.01 gg = 115.3 mol H | 1 kg Oil ⇒ S |13.5 g S b1 mol 32.06 gg = 0.4211 mol S Basis: |5.3 g I T 5.3 g I C + O 2 → CO 2 n 1 (mol CO 2 ) 1 72.0 mol C n 2 (mol CO) C + O 2 → CO 115.3 mol H 2 n 3 (mol H 2 O) 0.4211 mol S n 4 (mol SO 2 ) S + O 2 → SO 2 5.3 g I n 5 (mol O 2 ) 1 n 6 (mol N 2 ) 2H + O 2 → H 2 O 2 n a (mol), 0.21 O 2 , 0.79 N 2 15% excess air 175 C, 180 mm Hg (gauge) a. Theoretical O 2 : 72.0 mol C 1 mol O 115.3 mol H 0.25 mol O 2+ 2 1 mol C 1 mol H 0.4211 mol S 1 mol O + 2 = 101.2 mol O 1 mol S 2 Air Fed: ( 1.15 101.2 mol O 2 ) 1 mol Air = 554 mol Air = n 0.21 mol O a 2 554 mol Air 22.4 liter ( STP ) 1 m3 448K 760 mm Hg 3