chemical reaction engineering 3ed edition作者 octave Levensp(8)

2019-09-01 12:45

(b) If the results are consistent, determine the amount of tracer introduced M, and the E curve.

Solution: (a) A1c?t?2h?(25?16)?92h, t??191916tCdt??2519tCdt?Cmax25C163(t?16)tdt??max196(t?25)tdt?1925?19Cmax16Cdt??19Cdt?163(t?16)dt??25C?20sec max196(t?25)dt16?19?3?Cmax?19?25?6?Cmax or?2222C?20.5sec max2?(25?16)t'?V60??4?15sec?t,

So the results are not consistent.

11.4 A step experiment is made on a reactor. The results are shown in Fig. P11.4. (a) Is the material balance consistent with the tracer curve? (b) If so, determine the vessel volume V,t,the F curve and the E curve.

Solution: (a) If Cm?max???0.54?0.125mol/L, A?C1?3maxt?2Cmax?2Cmax so t?2min (b) F??Cm?C?C, max

35

?0,t?[0,1)?1? then F curse is F??(t?1),t?[1,3]

?2??1,t?(3,?)A?216V??2?0.125??8liters

?m0.5?0,t?[0,1)?(3,?)dF?, so E??1 E?dt,t?[1,3]??2E curse is

0.80.60.40.20E

11.5 A batch of radioactive material is dumped into the Columbia River at Hanford, Washington. At Bonneville Dam, about 400 km downstream the flowing waters (6000 m3/s) are monitored for a particular radioisotope (t1/2>10 yr) and the data of Fig.P11.5 are obtained. (a) How many units of this tracer were introduced into the river?

(b) What is the volume of Columbia River waters between Bonneville Dam and the point of introduction of tracer?

012t,min345

Solution: (a) Ac?t?1rM, ?10?6?105?5.25?10?53?day?2?m?5so we obtain M?5.25?10rm3?day?6000?27216unitsof radioisotope 3sm 36

12510?610?6?20t35?20(t?20)d??35t35?125(t?125)dt?tCdt(b) t?? ?0Cdt1?(125?20)?10?62353.15?10?3?60day ?

52.5?10?6m3?V?t???60day?6000?3.11?1010m3

s11.6 A pipeline (10 cm I.D., 19.1 m long) simultaneously transports gas and liquid from here to there. The volumetric flow rate of gas and liquid are 60 000 cm3/s and 300 cm3/s, respectively. Pulse tracer tests on the fluids flowing through the pipe give results as shown in Fig.P11.6. What fraction of the pipe is occupied by gas and what fraction by liquid?

Solution:

tg?2s, tl?100s, so Vg??gtg?6000?2?1.2?105cm3

Vl??ltl?300?100?3?104cm3

Vtotal??4D2L??4?102?19.1?102?1.5?105cm3

1.2?105?100%?80%, %L?20% So we obtain %G?51.5?10A liquid macrofluid reacts according to A → R as it flows through a vessel. Find the conversion of A for the flow patterns of Fig.P11.7 to P11.11 and kinetics as shown.

0.5

11.7 CA0 = 1mol/liter, -rA = kC0.5 /liter0.5·min A, k = 2 mol

37

Figure 11.7

Solution:

????CA1C?rdC?CCAAoAAo?CA?CAo(1?C) AAoCAC?(1??)2?(1??)2 AoCAoSo we obtain CA?CAL11C?Ao?0(C)batch?Edt?Ao?0(1?t)22dt?6

?X1A?1?6?0.833 11.8

CA0 = 2mol/liter, -rA = kC2 A, k = 2 liter / mol·

min Figure 11.8

Solution:

According to page274 eq.15,

CAC?1kC Ao1?AotA, so E, E???4t?1,t?[0,0.5]E?t?1max?3min?1?0,t?(0.5,??)

So CA?CA0.51C?Ao?0(C)elementE?dt?Ao?01?4t(4t?1)dt?0.5,

XA?1?0.5?0.5

11.9 CA0 = 6mol/liter, -rA = k, k = 3 mol/liter·min

38

Figure 11.9

Solution:

C?A???1?ktC,t?2minC Ao?Ao?0,t?2min??AE?t?1, so E??1?3,t?[0,3]

??0,t?(3,??)So

CAC?Ao??0(CAC)3kt11eEdt?Ao?0(1?C)?dt?, Ao33X1A?1?3?0.667

11.10 CA0 = 4mol/liter, -rA = k, k = 1 mol/liter·min

Figure 11.10

Solution:

C?ktA?1??1?tC??CAo4,t?4min, E????,t?3Ao??0,t?4min?0,t?3So CA?CA3?1C?Ao?0(C)eEdt?(1?)Ao4?0Edt?4,

XA?1?14?0.75

11.11 CA0 = 0.1mol/liter, -rA = k, k =0.03 mol/liter·min

39


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