Wt frac x sucrose sucrose ms r p k j t m x c solution

This preview shows page 423 - 429 out of 478 pages.

We have textbook solutions for you!
The document you are viewing contains questions related to this textbook.
C++ Programming: Program Design Including Data Structures
The document you are viewing contains questions related to this textbook.
Chapter 5 / Exercise 27
C++ Programming: Program Design Including Data Structures
Malik
Expert Verified
wt frac x sucrose sucrose ms r p K J t M x c. Solution 1. R 1 M 1 R 1 2.1 1 0.9943 0.98803 Then RT equation is p out out x 1 R x 0.01197 x Operating Equation is (17-23) F p out out x 1 .39 0.022 x x x .61 0.61 p out x 0.6393 x 0.036066 Solve RT & operating equation simultaneously. out x 0.05537 , p x 0.000663
We have textbook solutions for you!
The document you are viewing contains questions related to this textbook.
C++ Programming: Program Design Including Data Structures
The document you are viewing contains questions related to this textbook.
Chapter 5 / Exercise 27
C++ Programming: Program Design Including Data Structures
Malik
Expert Verified
438 Check p out x R 1 0.98803 x 0K x water water r p r p ms K J p p a M x t = 2 0.0706 60 1.1 59.895[ 2.1 0.05537 0.000663] 3.67g/(m s) / (1 ) sucrose water p p J J x x 2 3.67 0.000663 / (1 .000663) 0.00243 g (m s) sucrose J Alternatively, sucrose sucrose r p 2 ms K g J Mx x 0.0226[ 2.1 (0.05537) 0.000663] 0.00261 t m s 6.9% different Solution 2. RT Eq. (17-21), p p r p x x 3.140 59.895 0.997 1 3.140 .997 60 1.1 1 x 2.1 1 3.140 59.895 .997 1 x Simplifies to, p p r p x 186.5 x 185.391 x 2.1 391.66 x Linearize r p x @x 0.0003 0.02509 . Note x p = 0.0003 is an arbitrary point. Slope = p r x 0.0003 0.01196 x 0.02509 Then linear form of RT equation is p r x 0.01196 x or p out x 0.01196 x Solve simultaneouly with Operating Equation F p out out x 1 x x 0.6393 x 0.036066 out x 0.05538 , p x 0.0006623 . Very close to value obtained with retention analysis. 17.E2. (was 16.D11 in 2 nd ed.) Eq. (17-59b): PL,in P in out in P C F T T F , in out T T 85 25 60 Stage 1. Assume P y ~ .05 water, 0.95 ethanol P w E 0.95 0.05 2290.3 kJ kg (See Example 17-9) For Feed PL,in PL,w PL,E C 0.1 C 55 C 0.9 C kJ kJ 0.1 4.1915 0.9 2.7595 2.903 kg K kg K Where average temperature from 25 to 85 is 55ºC and P C values are from Perry’s 7 th , pp. 2- 306 and pp. 2-237. P1 P1 in F 2.903 60 0.0760 and F 0.0760 100 7.60 kg hr F 2290.3 .
439 Op. line intersects P out IN y x x 0.10 (water wt. frac.) 1 1 0.760 Slope 12.16 0.0760 Op. Eq. is, P out out 0.1 y 12.16 x 12.16 x 1.32 0.0760 If P out y 1, x 0.32 12.16 0.0263 Plot Op. Line on Figure 16-17a and find intersection: P1 out1 2,in y 0.66, x x 0.055 (water values) out1 in2 F F 100 7.60 92.40 kg h Stage 1 Trial 2. Since y p ≠ y p, assumed , do a second trial. P y 0.66 water, 0.34 E, P w E 0.66 0.34 0.66 2359 0.34 985 1892 kJ kg P1 1in 2.903 60 F 0.0921 F 1892 1 P1 out F 0.0921 100 9.21 kg h, F 100 9.21 91.79 kg h Slope 1 0.0921 9.86 0.0921 , P out out 0.1 y 9.86 x 9.86 x 1.086 0.0921 If P out 0.086 y 1, x 0.00872 9.86 Plot operating line and determine (from graph) 1 1 2 P out in y 0.64, x x 0.05 water. This value of y p is close to the assumed value of 0.66. Can proceed to stage 2. Stage 2: Estimate P y 0.50 (water), P w E 0.50 0.50 1672 For in2 PL,in x 0.050, C 0.05 4.1915 0.95 2.903 2.967 2 2 2 P 2 in in F 2.967 60 0.1065, F .1065 90.66 9.64 kg h F 1672 2 out F 90.66 9.64 8.10 kg hr , Slope 1 0.1065 8.40 0.1065 P out out 0.050 y 8.40 x 8.40 x 0.4421 0.1065 If out P x 0, y 0.4421 . Draw op. line. Intersection gives P y 0.34 For 2 out x use MB. 2 2 2 2 2 in in P P out 2, out F x F y 90.66 0.050 9.64 0.34 x F 81.00 2 2 out ,w out, ,ETOH x 0.0155 or x 0.9845. This is a close as we can get graphically. Mixed Permeate: p,mix 9.21 0.64 9.64 0.34 y 0.487 9.21 9.64 wt frac water
440 P 2 kg 1000 g F h kg Area J g h m , J from Fig. 16-17b based on out x Stage 1 2 J 0.8333 g h m , 2 1 9.34 1000 g h A 11, 208 m 0.8333 Stage 2 2 J 0.208 g h m , 2 2 9.64 1000 A 46,346 m 0.208 Other flow patterns will reduce area. Area is large because of low flux caused by low ethanol permeation rate. 17.F1. RT eqn., x x y 1 1 x , Benz 16.6 .3 x 0.3, 16.6, y 0.87676 1 15.6 .3 Benz 18.3 .2 x 0.2, 18.3, y 0.8266 1 17.3 .2 Benz 6.66 .1 x 0.1, 6.66, y 0.4253 1 5.66 .1 . Plot RT equation. Operating equation Slope 1 .9 9 .1 . Plot on graph. Find Benz P out,Benz y ~ 0.844, x ~ 0.238
441 benz benz P P benz P iP y 1 y 0.844 94.27 1 0.844 164 105.15 cal g benxw Lbenz L,iP PL,in P P benz,in P C x C 1 x C 0.3 0.423 .7 0.73 0.6379 cal g C
442 P in out PL,in T T C 0.1 105.15 50 C 66.48 C 0.6379 17.H1. (was 16.G1 in 2 nd edition) This is set up for Area being the unknown and cut being known. Problem 17.H1 Fr,in 10000.000000 yin,A 0.2500 cut=Fp/Fin 0.2500 tmem,cm 0.002540 pr,cm Hg 300.0000 pp,cm Hg 30.0000 yin,B 0.5500 P,A

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture