HW 9 solution - P8-5 A lb mole Fio hr Tio(F ~ A B B 10 80...

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P8-5 2 A B C + A B C F io lb mole hr - 10 10 0.0 T io (F) 80 80 - ~ Pio Btu C lb mole F ° 51 44 47.5 , lb MW lb mol 128 94 222 , 3 i lb ft ρ 63 67.2 65 20,000 R Btu H lb mol A = , Energy balance with work term included is: [ ] 0 0 0 10 1, 1, 1 10 ( ) S A R i Pi o A B A B AF A Q W X H C T T F F X F Q UA Ts T θ θ θ - - = - = = = = = = - & & % & Substituting into energy balance, [ ] { } [ ] 0 0 0 0 0 0 0 0 0 ( ) ( ) ( ) 63525 199 S S A R AF A pA pB S S A R A pA pB S S A R A pA pB s UA T T W F H X F C C T T UA T T W F H F C C UA T T UA T T W F H T T F C C UA Btu W hr T F - - - = + - - - - = + + - - - - = + + + - = =
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P8-6 A B C + Since the feed is equimolar, C A0 = C B0 = .1 mols/dm 3 C A = C A0 (1-X) C B = C B0 (1-X) Adiabatic: 0 0 [ ( )] 30 15 15 0 i R i P P P pC pB pA X H T T T C X C C C C C φ -∆ = + + = - - = - - = % % ( ) R C B A H T H H H = - - = -41000-(-15000-(-20000) = -6000 cal/mol A 15 15 30 i i pA B pB cal C C C mol K θ θ = + = + = % 6000 300 300 200 30 X T X = + = + 2 2 2 0 (1 ) .01 (1 ) A A r k C X k X - = - = - P8-6 (a) 0 0 PFR A A A CSTR A dX V F r F X V r = - = - For the PFR, F A0 = C A0 v 0 = (.1)(2) = .2 mols/dm 3 See Polymath program P8-6-a.pol . Calculated values of DEQ variables   Variable Initial value Minimal value Maximal value Final value 1  X  0.85  0.85  2  V  308.2917  308.2917  3  Ca0  0.1  0.1  0.1  0.1  4  Fa0  0.2  0.2  0.2  0.2  5  T  300.  300.  470.  470.  6  k  0.01  0.01  4.150375  4.150375  7  ra  -0.0001  -0.0018941  -0.0001  -0.0009338 
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Differential equations 1  d(V)/d(X) = -Fa0 / ra  Explicit equations 1  Ca0 = .1  2  Fa0 = .2  3  T = 300 + 200 * X  4  k = .01*exp((10000 / 2) * (1 / 300 - 1 / T))  5  ra = -k * (Ca0 ^ 2) * ((1 - X) ^ 2)  V = 308.2917dm 3 For the CSTR, X = .85, T = 300+(200)(85) = 470 K.
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