0 o C. H vap. H evap

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Jack Mosley
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1 Solution. Energy P (00 ) Pν x 0 5 ρ 850,4 J kg - J kg Power kg s 000, W Solution. 00 o C H evap H vap 0 o C H liq 00 t H liq (4. x0 t ) dt 4.t x kj kg - H evap 40,68 J mol - (From Appendix D) kj kg - 8 From Appendix O, the specific heat of the vapour is given by: C p x x x 0-9 Where C p is in J mol - K - and is in K. Now 00 o C 7.5K and 00 o C 7.5K H vap ( x x0.596 x0 ) d 7.5

2 ( x x0.596 x0 ) 4 7.5, kj kmol kj kg - herefore, specific enthalpy: H liq 40 H evap 60 H vap kj kg - From Steam ables: 876 kj kg -. Error 98 kj kg - (.5 %). Solution. Calculation of the enthalpy of reactions:. CO + ½O CO H F (kj mol - ) H R (-0.6) -8.5 kj mol - CO. H + ½O H O H R kj mol - H. CH 4 + O CO + H O H R [ (-4.00)] (-74.86) kj mol - CH 4 4. C H 6 + ½O CO + H O H R [(-9.77) + (-4.00)] (-84.74) kj mol - C H 6

3 5. C H 4 + 6O CO + H O H R [(-9.77) + (-4.00)] kj mol - C H 4 6. C 6 H 6 + 7½O 6CO + H O H R [6(-9.77) + (-4.00)] kj mol - C 6 H 6 Composition (mol %): CO : 4, CO: 5, H : 50, CH 4 :, C H 6 :, C H 4 : 4, C 6 H 6 :, N :. Basis: 00 mol Component Quantity -( H R ) H (kj) CO 4 -- CO H CH C H C H C 6 H N (kj/00 mol) herefore, H ( )(0) kj kmol - Gross CV (kj m - ) 404, ,07 kj m - ( 485 BU ft - ).4 o calculate the Net CV, subtract the heat of vapourisation of the H O burned.

4 Solution.4 Heat ransfer Fluid 0bar NB, 0 o C 500 kg h - H, 0 o C 66 kg h - Molecular weight of nitrobenzene and H 500 Molar flow of nitrobenzene ()(600) Molar flow of H 66 ()(600) 50.8 x 0 - kmol s x 0 kmol s x0 Partial pressure of nitrobenzene 0 [ ].0 bar + Using the Antoine Equation: B ln P A + C he Antoine constants are obtained from Appendix D. ( bar 500 mm Hg) 40.6 ln (500) K 55 o C he boiling point of nitrobenzene at atm 0.6 o C (Appendix D) evap 55 o C 0.6 o C H 0 o C NB 0 o C

5 he specific heat capacity of the nitrobenzene liquid can be estimated using Chueh and Swanson s method. CH C N O (8.4 x 5) (5.7 x ) otal 9 kj kmol - C - he specific heat capacity of the nitrobenzene gas: a b x 0 c x 0 4 d x 0 6 HC (x 5) C NO Nitrobenzene: H liq (5.646 x 0 - )(9)(0.6 0) 08 kw H gas kw ( ) d H evap H : 44,0 kj kmol kmol 48.5 kw s H gas ( kw herefore: otal H kw Note: It is not worth correcting the heat capacities for pressure ) d

6 Solution kg h - NB H Inerts mol % NB 0.45 AN 0.7 H O.68 Cycl. 0. Inerts.66 H 6.67 Nitrogen Balance: Molar flow of nitrobenzene 500 ()(600) herefore, katoms N x 0 - s - Let the total mass out be x, then: x x 00 x kmol s x 0 - kmol s - H reacted 0.7 Aniline produced (0.05) Cyclo-hexylamine produced (0.05) Unreacted H (0.05) kmol s So, total H In kmol s - Now, H reaction 55,000 kj kmol - (Appendix G8) From H f (Appendix D) NB kj mol - AN 86.9 H O -4.00

7 H reaction Σ products Σ reactants [ (-4.00)] (-67.49) kj mol - 9,590 kj kmol - Reactions: C 6 H 5 NO + H C 6 H 5 NH + H O C 6 H 5 NH + H C 6 H NH he second reaction can be ignored since it represents a small fraction of the total. he problem can be solved using the ENRGYBAL program. Heat capacities can be found in Appendix D and calculated values for nitrobenzene obtained from Solution.4. Solution.6 A straight-forward energy balance problem. Best to use the energy balance programs: ENERGY, page 9 or ENRGYBAL, Appendix I, to avoid tedious calculations. Data on specific heats and heats of reaction can be found in Appendix D. What follows is an outline solution to this problem. 00 o C 95 % H 5 % N Cl sat CW 5 o C.5bar 50 o C 0,000 yr - HCl H + Cl HCl Mass balance ( % excess) gives feed. Solution:. sat for Cl from Antoine Equation (Appendix D),. H reaction from the HCl heat of formation,. C p s from Appendix D,

8 4. Reactor balance to 00 o C (4 % free Cl ), 5. Datum temperature 5 o C, 6. Ignore pressure effects on C p s. Reactor: IN -. H + N zero (at datum temperature),. Cl at sat (note as gas, H reaction for gases),. H reaction at 5 o C (96 % Cl reacted). OU -. HCl + Cl + H (excess) + N at 00 o C,. Cooling in jacket. Cooler: IN -. Reactor outlet H,. 4 % Cl reacted ( H reaction ). OU -. Sensible heat of HCl, H (excess) and N,. Heat to cooling water. Check on sat : ln(.5 750) sat K 8.95 sat sat sat -4.6 o C (Within the temperature limits) he Cl may need preheating. Solution.7 As P < P critical, the simplified equation can be used. 5 bar N 00m h - o

9 γ.4 for air. n P w P v n P where: n and m n n γ m. γ Compression ratio 0 from Figure.7, E p 86 %. P P m E p.4 m 0., n. 49. (.4)(0.86) 0. 0 ( 0 + 7) K 5 o C In practice the compressor cylinder would be fitted with a cooling jacket. v 00 m h m s - w.49 )(0.078) ( kw (Say 0 kw) ( 0) Solution.8 H or HCl 0,000 kg h - HCl H + Cl HCl Burner operating pressure, 600 kn m - required. ake burner as operating at atm. 0 kn m - g or 600 kn m - g. H is compressed from 0 kn m - to 600 kn m -.

10 600 Pressure ratio 5 0 Intermediate pressure P P (0)(600) 68 kn m - Note: For H the inlet temperature will not be the same as the intercooler outlet so the cool stage should be calculated separately. A material balance gives the H flow. he % excess H is ignored in the HCl compressor calculation. Material balance: HCl produced 0,000 (6.5)(600) H required Cl required kmol s kmol s kmol s - Excess H kmol s - he simplified equations (.6a and.8a) can be used since conditions are far removed from critical. ake γ. 4 since both H and HCl are diatomic gases. (.4 ) m (.6a) (.4)(0.7) n.689 (.8a) n n P n w P v (.) n P H : st Stage: v 0.08 m kg

11 w (0 0 )(0.08) 9,9 J kg nd Stage: v m kg w (68 0 )(0.099) 0,04 J kg Power (9,9 + 0,04)(0.084)() 505 W.505 kw HCl: ake both stages as performing equal work with the same inlet temperature. P i P P (0.)(600) 46.5 kn m - (.9) v.97 m kg w (.0 0 )(.97) 50,7 J kg Power (50,7)(0.076)(6.5),47,08 W 47 kw It is necessary to divide by the efficiency to get the actual power but it is clear that the best choice is to compress the H and operate the burner under pressure. Check: emperature of saturated Cl at 600 kn m ln K 6 o C

12 Solutions.9 and.0. Refer to example.7 and the worked solution to problem. Solution. s t C p Streams: ( o C) ( o C) (kw o C - ) 900 kw Preheater C o C 0 o C 50 kw Condenser H o C 60 o C 00 kw Condenser H o C 55 o C 400 kw Reboiler C o C 85 o C 900 kw Reboiler C o C 75 o C 0 kw Cooler H o C 5 o C For min 0 o C + 5 (cold) int out 5 (hot) int out Stream ype act int C H H C C

13 6 H Ranked Streams ( o C) kw Cascade Add C C H 60 H H C H ( ( C p ) ( C p ) ) Hot Utilities 00 kw Cold Utilities 445 kw C H Pinch 60 8 o C

1. (25 points) C 6 H O 2 6CO 2 + 7H 2 O C 6 H O 2 6CO + 7H 2 O

1. (25 points) C 6 H O 2 6CO 2 + 7H 2 O C 6 H O 2 6CO + 7H 2 O MEEBAL Exam 2 November 2013 Show all work in your blue book. Points will be deducted if steps leading to answers are not shown. No work outside blue books (such as writing on the flow sheets) will be considered.