Homework 6 Solutions - Chemical Engineering Thermodynamics | CHE 21100, Assignments of Chemistry

Download Homework 6 Solutions - Chemical Engineering Thermodynamics | CHE 21100 and more Assignments Chemistry in PDF only on Docsity! CHE211 Spring 2009 CHE211 Homework Problem Set Lucky Last Due: Friday May 1st @ 5.00pm Submit assignments to Jenni Layne (Room 2043) 1. Sandler problem 9.4 2. Sandler problem 9.30(a) 3. Sandler problem 9.38 4. Sandler problem 10.1-1 Parts (b) and (c) only [Note: This problem requires some significant iteration. It may be useful to set out the algebraic equations in Excel or similar. Show details of equations and the first iteration in your working.] 5. Sandler problem 10.2-2 [For this problem, you can use the van-Laar expression for activity coefficients.] 6. For the binary system methanol(1)-methyl acetate(2), the following equations provide a reasonable correlation for the activity coefficients: TAwhereAxAx 00523.0771.2ln;ln 212 2 21   The following Antoine equations provide the vapor pressure: 424.53 54.2665 25326.14ln; 424.33 31.3643 59158.16ln 21     T P T P satsat where T is in Kelvin and P is in kPa. (a) Calculate P and yi for T = 318.15K and x1 = 0.25 (b) Calculate the azeotropic pressure and azeotropic composition for T = 318.15K. HW 6 9.4.a i) partial molar volume iV 0 v, 0 0 ln lni ii i i i i U V S S C R U V = + + Taking partial derivatives of the above expression with respect to iV at constant iU , 1i i iU i S R V V ∂ = ∂ From dU TdS pdV= − , we can get another i i U i S V ∂ ∂ by dividing by dV for the i component. i i U i S V ∂ ∂ = p T Therefore, 1 i p R V T = i RT V p = iV = i V N ∂ ∂ = ( , )i i i N V T P N ∂ ∂ ∑ = i i RT N RTp N p ∂ = ∂ ∑ ii) partial molar internal energy iU 0 v, 0 0 ln lni ii i i i i U V S S C R U V = + + Taking partial derivatives of the above expression with respect to iU at constant iV , 9 9.30 (also available as a Mathcad worksheet) i 0 1 10!! x i 0.1 i! fma i x i exp 1.06 1 x i 2!! 1.126! fm i 1 x i exp 1.06 x i 2!! 0.847! x 0 0 1 2 3 4 5 6 7 8 9 10 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 " fma 0 0 1 2 3 4 5 6 7 8 9 10 0 0.266 0.444 0.568 0.66 0.734 0.8 0.867 0.94 1.024 1.126 " fm 0 0 1 2 3 4 5 6 7 8 9 10 0.847 0.77 0.707 0.652 0.602 0.552 0.496 0.427 0.334 0.2 0 " Hma 1.126 exp 1.06( )! Hma 3.25" Hm 0.847 exp 1.06( )! Hm 2.445" aie, 1 E os 9.38 M , Xo,= 0.99, 25°C, 1 bar Air = ——w 2 bar, 25°C + |_ Ny, Xo, 0.05, 25°C, 1 bar No moving parts > W = 0 Mass balance (basis 1 mole air) 0, 0=0.21+099N,+0.05N, ar O=1-N,+N, (Note that a mass balance on nitrogen would be redundant with the two mass balances above.) 0.214.998, +0.05(1-N, ]=0 0.16 =-(0.99-0.03)N, Nyp=-— 0.1702 N, =-0.8298 Energy balance 0=1+H,j,—0.1702 x H; — 0.8298 H, + Q Entropy balance O=1 +S, — 0.1702 = S; — 0.8298 Ss + os. Q4|-Sat 0.1702 S;+ 0.8298 S, — 5.) T Combined with the energy balance 0 = Gri — 0.1702 Gi — 0.8298 Go — Ts, or TS yn = Gair — 0.1702 Gy - 0.8298 G =O New since pressure is low, assume an ideal gas mixture TS. = (0.21 Go, + 0.79 Gy, + 0.21 RT In 0.21 + 0.79 RE In 0.79 — 0.1702 = 0.99 Go, — 0.1702 x 0.99 RT In 0.99 — 0.1702 « 0.01 Gy, - 0.1702 x 0.01 RT In 0.01 0.€298 » 0.05 Go, 0.8298 « 0.05 RT In 0.05 — 0.8298 x 0.95 Gr, — 0.8298 = 0.95 RT In 0.95 L x x x x x V y y y y yi i ! ! 086667 00225 00858 04258 04659 1000 013333 02289 01921 02326 03464 1000 . . . . . . . . . . . . ET P NB MP ET P NB MP (d) For an adiabatic flush vaporization, shown below, the energy balance must also be satisfied liquid liquid vapor X pressure reducing valve or device This is a (two-phase) Joule-Thomson expansion, so that the energy balance yields H Hin out , or x H T P x L x H T P x V y H T P y i i L i i L i i V ! ! ! " # " # $, , , , , , inlet conditions outlet liquid conditions outlet vapor conditions c h This equation must be satisfied, together with the mass balances and phase equilibrium equations of part c. Thus, we have one new unknown here, the outlet temperature, and an additional equation from which to find that unknown. 10.2 10.2-2 The van Laar equation will be used to fit the data given. Starting from i i i iy P x P! vap 1 2 ! ! H O FURF 2 we obtain, at 10 mole % water, 1 5826! . and 2 1266! . . Using eqn. (7.5-10) we get " ! 85648. and # ! 07901. . Thus, ln . . 1 1 2 2 85648 1 10841 ! $ x xa f and ln . . 2 2 1 2 07901 1 00922 ! $ x xa f which we will assume is valid at all temperatures. At the new temperature we have or x P y Pi i i 1 vap ! x y1 1 110352 1013 % ! %. . and x y2 2 201193 1013 % ! %. . which must be solved together with the activity coefficient expressions above, and the criteria that x x1 2 1$ ! and y y1 2 1$ ! . Solution procedure I used was to guess a value of , compute from , computex1 x2 x2 1! & x1 1 and 2 from the expression above, fromyi y x P Pi i i i! vap for i and 2, and then check to see if . Proceeding this way, the following results were obtained ! 1 yi' ! 1 calculated measured x x y y 1 2 1 2 0 075 020 0 925 080 0867 089 0129 011 . . . . . . . . Note = 0.996 which is not quite equal to 1. The discrepancy between the calculated and experimental results indicates the dangers of using approximate solution models. yi' Prob. 6 a. ( , , ) ( )vapi i i iPy x T P x P Tγ= ------------(1) 1 2 1y y+ = --------------(2) Unknowns P, y1, y2, 3 equations iγ can be obtained from the given correlation. 2 2 1 2exp( ) exp((2.771 0.00523 318.15) 0.75 )Axγ = = − × × =1.864 2 2 2 1exp( ) exp((2.771 0.00523 318.15) 0.25 )Axγ = = − × × =1.072 1 1 1 1 vapPy x Pγ= = 3643.310.25 1.864 exp(16.5918 ) 20.75 318.15 33.424 × × − = − ----------(*) 2 1 2 2 2(1 ) vapPy P y x Pγ= − = = 2665.540.75 1.072 exp(14.25326 ) 52.77 318.15 53.424 × × − = − ---------(**) Divide (*) by (**). 1 1 20.75 0.3932 (1 ) 52.77 y y = = − Therefore, y1=0.282 and y2=0.718 P is from either (*) or (**). P=73.58 kPa b. At azeotropic point, 1 1x y= . Therefore, from 1 1 1 1 vapx P y Pγ = , 1 1 73.58 ( ) 1.65 44.52vap P aze P γ = = =