A tank contains 90 kg of salt and 2000 L of water. Pure water enters a tank at the rate 6 L/min. The solution is mixed and drains from the tank at the rate 3 L/min.

(a) What is the amount of salt in the tank initially? (in kg)

(b) Find the amount of salt in the tank after 3.5 hours. (in kg)

(c) Find the concentration of salt in the solution in the tank as time approaches infinity. (Assume your tank is large enough to hold all the solution.) ( in kg/L)

Question

contestada

Respuesta :

Giangiglia Giangiglia · Answer 1 · 2020-01-07T14:06:01+01:00

Answer:

a) 90 kg

b) 68.4 kg

c) 0 kg/L

Explanation:

Mass balance:

[tex]-w=\frac{dm}{dt}[/tex]

w is the mass flow

m is the mass of salt

[tex]-v*C=\frac{dm}{dt}[/tex]

v is the volume flow

C is the concentration

[tex]C=\frac{m}{V+(6-3)*L/min*t}[/tex]

[tex]-v*\frac{m}{V+(6-3)*L/min*t}=\frac{dm}{dt}[/tex]

[tex]-3*L/min*\frac{m}{2000L+(3)*L/min*t}=\frac{dm}{dt}[/tex]

[tex]-3*L/min*\frac{dt}{2000L+(3)*L/min*t}=\frac{dm}{m}[/tex]

[tex]-3*L/min*\int_{0}^{t}\frac{dt}{2000L+(3)*L/min*t}=\int_{90kg}^{m}\frac{dm}{m}[/tex]

[tex]-[ln(2000L+3*L/min*t)-ln(2000L)]=ln(m)-ln(90kg)[/tex]

[tex]-ln[(2000L+3*L/min*t)/2000L]=ln(m/90kg)[/tex]

[tex]m=90kg*[2000L/(2000L+3*L/min*t)][/tex]

a) Initially: t=0

[tex]m=90kg*[2000L/(2000L+3*L/min*0)]=90kg[/tex]

b) t=210 min (3.5 hr)

[tex]m=90kg*[2000L/(2000L+3*L/min*210min)]=68.4kg[/tex]

c) If time trends to infinity the division trends to 0 and, therefore, m trends to 0. So, the concentration at infinit time is 0 kg/L.