Jawaban Matek Nomor 12.docx

Diketahui : KA = 500 L KB = 400 L kA = 40 gr/L kB = 5 gr/L Vin = 10 L/min + 2 L/min = 12 L/min Vout = 12 L/min IN – OUT = Akumulasi IN = kA × Vin = 40 gr/L × 12 L/min = 480 gr/min 𝑄

X gr

12𝑋

OUT = 𝐾 × V out = 400 L × 12 = 400 𝑔𝑟/𝑚𝑖𝑛 IN – OUT = 480 gr/min -

12𝑋 400

𝑔𝑟/𝑚𝑖𝑛 𝑑𝑥 12𝑥 = 480 − 𝑑𝑡 400 𝑑𝑥 480(400) 12𝑥 = − 𝑑𝑡 400 400 ∫

𝑑𝑥 𝑑𝑥 = ∫ 400 192000 − 12𝑥

𝑡 𝑑𝑥 = ∫ 400 192000 − 12𝑥 *𝐦𝐢𝐬𝐚𝐥, 𝐮 = 𝟏𝟗𝟐𝟎𝟎𝟎 − 𝟏𝟐𝐱 *𝐝𝐮 = −𝟏𝟐 𝐝𝐱 𝟏

*𝐝𝐱 = − 𝟏𝟐 𝐝𝐮 1 − 12 𝑡 = ∫ 𝑑𝑢 400 𝑢 𝑡 1 = − ln(𝑢) + 𝑐 400 12 𝑡 1 = − ln(192000 − 12𝑥) + 𝑐 400 12 1 𝑡 = ln(192000 − 12𝑥)−12 + 𝑐 400

t = 0, x = kB × KB = 5 gr/L × 400 L = 2000 gr

1 0 = ln(192000 − 12(2000))−12 + 𝑐 400

𝑐 = ln

(168000) −12

𝑐=

12,0 −12

𝑐 = −1

Ditanya : a.) Konsentrasi keluar tangki B pada waktu 1 jam ? Jawab : 1 𝑡 − 1 = ln(192000 − 12𝑥)−12 400 1 60 − 1 = ln(192000 − 12𝑥)−12 400 1 60 − 400 = ln(192000 − 12𝑥)−12 400 1 340 = ln(192000 − 12𝑥)−12 400 34

𝑒 −30 =

1

(192000 − 12𝑥)−12

26903,18 = 192000 − 12𝑥 12𝑥 = 165096,8 𝑥 = 13758,0678 Konsentrasi =

13758,07 𝑔𝑟 400 𝐿

= 34,3951 𝑔𝑟/𝐿

Ditanya : b.) Jika tangki tidak mempunyai kapasitas maksimum, tentukan konsentrasi larutan untuk jangka waktu tak terbatas! Jawab : X = kA × KB = 40 gr/L × 400 L = 16000 gr

1 𝑡 − 1 = ln(192000 − 12𝑥)−12 400 1 𝑡 − 400 = ln(192000 − 12(16000))−12 400 1 𝑡 − 400 = ln(192000 − 192000)−12 400 1

= ln(0)−12 𝑡 − 400 = [ −

1 ln(0)] 400 12

lim 𝑙𝑛 (𝑥) = −∞

𝑥→0

𝑡 − 400 = [−

1 × (−∞)] 12

𝑡=∞