Tugas TRK Ii CH 11 [PDF]

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TUGAS TRK II CHAPTER 11 Gading Bagus M



02211740000100



Samuel Bagas W



022117400000113



Aura Selena Kinayomi



022117400000114



11.6 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?



Given : ID = 10 cm H = 19,1 m vgas = 60.000 cm3/s vliquid = 300 cm3/s



Answer : 𝜋



𝜋



Volume total = 4 𝐷2 𝐻 = 4 (10 . 10−2 )2 (19,1) = 0,15 m3 V gas = 60.000 cm3/s x 2 s = 120.000 cm3 = 0,12 m3 0,12



%v volume liquid = 0,15 𝑥 100 % = 80 % V liquid = 300 cm3/s x 100 s = 30.000 cm3 = 0,03 m3 0,03



%v volume liquid = 0,15 𝑥 100 % = 20 %



A liquid macro fluid reacts according to A + R as it flows through a vessel. Find the conversion of A for the flow patterns of Figs. P11.7 to P1l.ll and kinetics as shown. 11. 7



𝑥𝑦 = 1 2𝑦 = 1



(



(



𝑦=



1 2



𝐸=



1 2



∞ 𝐶𝐴 )= ∫ 𝐶𝐴𝑂 0



∞ 𝐶𝐴 )= ∫ 𝐶𝐴𝑂 0



2 𝐶𝐴 ( )= ∫ 𝐶𝐴𝑂 0



(



𝐶𝐴 𝐸 𝑑𝑡 ) 𝐶𝐴𝑂 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 1



[1 + (𝑛 − 1)𝐶𝐴𝑂



(𝑛−1)



(0,5−1)



[1 + (0,5 − 1)1



2 𝐶𝐴 ( )= ∫ 𝐶𝐴𝑂 0



𝑘𝑡]1−𝑛 𝐸 𝑑𝑡



𝑘𝑡]



1 1−0,5



[1 − 𝑡]2 . 0,5 𝑑𝑡



2



1 − 𝑥𝐴 = ∫



[1 − 𝑡]2 . 0,5 𝑑𝑡



0



1 − 𝑥𝐴 = 𝑥𝐴 =



2 3



1 3



.



1 𝑑𝑡 2



11. 8



1 + 𝑦2 × 0.5 = 1 2 (1 + 𝑦2 ) × 0.5 = 2 (1 + 𝑦2 ) = 4 𝑦2 = 3 𝑦 − 𝑦1 𝑥 − 𝑥1 = 𝑦2 − 𝑦1 𝑥 − 𝑥1 𝑦−1 𝑥−0 = 3 − 1 0.5 − 0 𝑦−1 𝑥 = 2 0.5 𝑦 = 4𝑥 + 1 𝐸 = 4𝑡 + 1 (



∞ 𝐶𝐴 )= ∫ 𝐶𝐴𝑂 0



(



∞ 𝐶𝐴 )= ∫ 𝐶𝐴𝑂 0



1 𝐸 𝑑𝑡 1 + 𝑘. 𝐶𝐴𝑂 . 𝑡



(



(



0,5 𝐶𝐴 )= ∫ 𝐶𝐴𝑂 0



𝐶𝐴 𝐸 𝑑𝑡 ) 𝐶𝐴𝑂 𝑒𝑙𝑒𝑚𝑒𝑛𝑡



1 (4𝑡 + 1) 𝑑𝑡 1 + 2.2. 𝑡 0,5



1 − 𝑥𝐴 = ∫ 0



4𝑡 + 1 𝑑𝑡 1 + 4𝑡 0,5



1 − 𝑥𝐴 = ∫ dt 0



1 − 𝑥𝐴 = 0.5



𝑥𝐴 = 0,5



11.9



𝑥𝑦 = 1 3𝑦 = 1



(



(



∞ 𝐶𝐴 )= ∫ 𝐶𝐴𝑂 0



𝑦=



1 3



𝐸=



1 3



(



𝐶𝐴 𝐸 𝑑𝑡 ) 𝐶𝐴𝑂 𝑒𝑙𝑒𝑚𝑒𝑛𝑡



∞ 𝐶𝐴 −1 ) = ∫ [1 − (CA0 kt)] × 𝐸 𝑑𝑡 𝐶𝐴𝑂 0



(



3 𝐶𝐴 1 1 𝑑𝑡 ) = ∫ [1 − 3t)] × 𝐶𝐴𝑂 6 3 0



(



31 𝐶𝐴 1 ) = ∫ − t 𝑑𝑡 𝐶𝐴𝑂 6 0 3



1 − 𝑥𝐴 = 0.25 𝑥𝐴 = 0.75



11.10



𝐶𝐴0 𝑥𝐴 = 𝑘𝑡 𝐶𝐴0 (1 − (1 − 𝑓(𝑡) = (



𝐶𝐴 ) = 𝑘𝑡 𝐶𝐴0



𝐶𝐴 𝑘𝑡 ) = 𝐶𝐴0 𝐶𝐴0 𝐶𝐴 𝑘𝑡 ) =1− 𝐶𝐴0 𝐶𝐴0



∞ 𝐶𝐴 = ∫ 𝛿(𝑡 − 𝑡0 )𝑓(𝑡)𝑑𝑡 = 𝑓(𝑡0 ) 𝐶𝐴0 0



𝑘𝑡 3 𝐶𝐴 𝛿(𝑡 − 3) (1 − ) 𝑑𝑡 =∫ 𝐶𝐴0 𝐶𝐴0 0 𝑘𝑡



1 − 𝑥𝐴 = 1 − 𝐶



𝐴0



1 − 𝑥𝐴 = 1 − 𝑥𝐴 =



11.11



3 4



saat t=3



1×3 4



Hydrogen sulfide is removed from coal gas by contact with a moving bed of iron oxide particles which convert to the sulfide as follows: Fe203 -+ FeS



In our reactor the fraction of oxide converted in any particle is determined by its residence time t and the time needed for complete conversion of



11.12



1𝑦 = 1 𝑦=1 𝐸=1 (



∞ 𝐶𝐴 )= ∫ 𝐶𝐴𝑂 0



(



∞ 𝐶𝐴 )= ∫ 𝐶𝐴𝑂 0



(



𝐶𝐴 𝐸 𝑑𝑡 ) 𝐶𝐴𝑂 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝑡 3 (1 − ) 𝐸 𝑑𝑡 𝜏



(



1 𝐶𝐴 )= ∫ 𝐶𝐴𝑂 0



(1 − 𝑡)3 .1 𝑑𝑡



1



1 − 𝑥𝐴 = ∫



(1 − 3𝑡 + 3𝑡 2 − 𝑡 3 ) .1 𝑑𝑡



0



1−𝑥𝐴 = 0,25 𝑥𝐴 = 1 − 0,25 = 0,75 = 75%



11.13



𝑡 3 𝑓(𝑡) = (1 − ) 𝜏 With 𝜏 = 1 ℎ𝑟 ∞ 𝐶𝐴 = ∫ 𝛿(𝑡 − 𝑡0 )𝑓(𝑡)𝑑𝑡 = 𝑓(𝑡0 ) 𝐶𝐴0 0 0.5 𝐶𝐴 𝛿(𝑡 − 0.5)(1 − 𝑡)3 𝑑𝑡 =∫ 𝐶𝐴0 0



1 − 𝑥𝐴 = (1 − 𝑡)3 saat t=0.5 1 − 𝑥𝐴 = (1 − 0.5)0.5 𝑥𝐴 = 0.293



11.14



∞



∫



𝐸 𝑑𝑡 = 1



0 1,5



∫



𝐸 𝑑𝑡 = 1



0,5



𝐸(𝑡 − 𝑡𝑜) = 1 𝐸(1,5 − 0,5) = 1 𝐸=1 ∞ 𝐶𝐴 ( )= ∫ 𝐶𝐴𝑂 0



(



∞ 𝐶𝐴 )= ∫ 𝐶𝐴𝑂 0



(



𝐶𝐴 𝐸 𝑑𝑡 ) 𝐶𝐴𝑂 𝑒𝑙𝑒𝑚𝑒𝑛𝑡



𝛿(𝑡 − 𝑡0 )𝑓(𝑡) 𝑑𝑡



11.15 Cold solids flow continuously into a fluidized bed where they disperse rapidly enough so that they can be taken as well mixed. They then heat up, they devolatilize slowly, and they leave. Devolatilization releases gaseous A which then decomposes by first-order kinetics as it passes through the bed. When the gas leaves the bed decomposition of gaseous A stops. From the following information determine the fraction of gaseous A which has decomposed. Data: Since this is a large-particle fluidized bed containing cloudless bubbles, assume plug flow of gas through the unit. Also assume that the volume of gases released by the solids is small compared to the volume of carrier gas passing through the bed. Mean residence time in the bed: t = 15 min, t = 2 s for carrier gas For the reaction: A →products, -rA = kcA, k = 1 s-1



∞



∫



𝐸 𝑑𝑡 = 1



0 2



∫



𝐸 𝑑𝑡 = 1



0



𝐸(𝑡 − 𝑡𝑜) = 1 𝐸(2 − 0) = 1 𝐸 = 0.5 𝐶



∞



(𝐶 𝐴 ) = ∫0 𝐴𝑂



𝐶



(𝐶 𝐴 ) 𝐴𝑂



𝑒𝑙𝑒𝑚𝑒𝑛𝑡



(



(



𝐸 𝑑𝑡 , karena merupakan reaksi orde 1 maka :



∞ 𝐶𝐴 )= ∫ 𝐶𝐴𝑂 0



2 𝐶𝐴 )= ∫ 𝐶𝐴𝑂 0



(



𝑒−𝑘𝑡 . 𝐸 𝑑𝑡 𝑒−1.𝑡 . 0,5 𝑑𝑡



2 𝐶𝐴 ) = 0,5 . ∫ 𝐶𝐴𝑂 0



𝑒−𝑡 𝑑𝑡



1 − 𝑥𝐴 = 0,4323 𝑥𝐴 = 0,5677 = 56,77 %