Tugas mattek 2.docx

1. Initially, a tank contain 500 kg of salt solution containing 10 % salt. A stream enters at constant flow rate of 10 kg/h containing 20 % salt. A stream leaves at a constant rate 5 kg/h. The tank is well stirred. Derive an equation relating the weight fraction wA of the salt in the tank at any time t in hours.

A 10 kg/ min

ρ, C1

10% 500 kg

ρ2, C2 5 kg/ min

Neraca Massa Total Acc = input – output t = 0  C = 0,2 ; m = 500 kg

dm dt =10−5=∫ 5 dt ¿ ∫¿ m = 5t + K t=0 m = 500 kg m = 5t + K 500 = 5(0) + K1 K1 = 500 m = 5t + 500

dm.c  10C1  5C 2 dt

C

d ( m) dC m  2  5C dt dt

C(5) + (5t + 500)

dC dt

= 2 – 5C

dC dt  2  10C 5t  500

dt 5 t+500 ¿ dC 2−10 C =∫ ¿ ¿ ∫¿ −1 10

ln (2 – 10C) = ln (5t + 500) + K2

t = 0  C = 0,1

−1 10

ln 1 = ln (5(0) + 500) + K2

K2 = - ln 500

−1 10

ln (2 – 10C) = ln (5t + 500) + (-ln 500)

Ln (2 – 10C)-1/10= ln

5t  500 500

2 – 10C = (0,01t +1)-10

2  (0,01t  1) 10

-10

=C

2. A well stirred storage vessel contains 10000 kg of solution of a dilute methanol solution ( wA = 0.05 mass fraction alcohol). A constant flow of 500 kg/min of pure water is suddenly introduced into the tank, and a constant rate of withdrawal of 500 kg/min of solution is started. These two flow are continued & remain constant. Assuming that the densities of the solutions are the same and that the total contents of the tank remain constant at 10000 kg of solution, calculate the time for the alcohol to drop to 1.0 wt%.

A 500 kg/ min

ρ, C1

10000 kg XA 0,05

500 kg/ min

ρ2, C2

Neraca Massa Total Acc = input – output



d ( m)  V1  V 2  0 dt

d ( V )  500 1  500  2 dt



dV 0 dt

Neraca Massa Alkohol Acc = input – output

d (VC )  500C1  500C 2 dt dC 2 dC 2  500C 2  0,05C 2 10000 dt dt

dC 2 dt =∫ −0,05 dt ¿ ∫¿ ln C2 = ln – 0,05 t + ln K ln C2 = ln K C2 = K

e−0,05 t

e−0,05 t

t= 0 ; C2 = 0,05 0,05 = K

e−0,05(0)

0,05 = K. 1 K = 0,05 C2 = 0,05

−0,05 t

e

t saat C2 0,01 0,01

= 0,05 .

−0,05 t

e

e-0,05 t . 5 = 1 e0,05 t = 5 ln ( e

0,05 t

¿ = ln 5

0,05 t = ln 5

ln 5

t 0,05

= 32,18 min

1. Initially, a tank contain 500 kg of salt solution containing 10 % salt. A stream enters at constant flow rate of 10 kg/h containing 20 % salt. A stream leaves at a constant rate 5 kg/h. The tank is well stirred. Derive an equation relating the weight fraction wA of the salt in the tank at any time t in hours

A 500 kg/ min

ρ, C1

10000 kg XA 0,05

ρ2, C2 500 kg/ min

3. A storage vessel is well stirred and contains 500 kg of total solution with concentration 5% salt. A constant flow rate of 900 kg/h of salt solution containing 16.67 % salt is suddenly introduced into the tank and a constant withdrawal rate 600 kg/h is also started. These two flows remain constant thereafter. Derive an equation relating the outlet withdrawal concentration as function of time. Also, calculate the concentration after 2.0 h. Neraca Massa Total

dm dm1 dm0   dt dt dt dm  900kg / h  600kg / h dt

dm  300kg / h dt m

t

500

0

.............................. (1)

 dm   300dt

m-500 = 300t + k1 m = 300t + 500 k1 ............................ ..............(2) t=0 k1 = 0 m = 500 Neraca Massa Garam

dmC  0,1667 x900  600C dt

dmC Cdm   150,03  600C dt dt (300  500)

dC  C 300  150,03  600C dt

dC 150,03  900C  dt 300t  500 m

t

dC dt 500150,03  900C  0 300t  500 

1 1 ln( 150,03  900C )  ln( 300  500) 900 300

t = 0 ; C= 0,015

k2  

1 1 ln( 150,03  900C )  ln( 300  500) 900 300

k2  

1 1 ln( 105,03)  ln( 800) 900 300



1 105,03  900 1 300  500 ln( ) ln( ) 900 105,03 300 500

ln ( 1,428 – 8,56 C) = ln (0,6t + 1)-3 1,428 – 8,56 C = (0,6t + 1)-3 8,56 C = 1,428 – (0,6t + 1)-3

C  0,1667 

1 8,56(0,6  1)  3

t = 2 jam

C  0,1667 

1 8,56(0,62  1)3

C = 0,1667 – 0,0109 C = 0,1557