just activity 2. a - g O, 67sa (0.6ssey 0,8 TS Q e e Mixing It Up %,ns b) Find a

Question

just activity 2. a - g

O, 67sa (0.6ssey 0,8 TS Q e e Mixing It Up %,ns b) Find an expression for the amount of salt, Q(t in lb in the tank at time t and plot Q(t) vs. t over time interval [0, 200) min c) Determine when the amount of salt doubles from the original amount in the tank. d) Determine when the amount of salt in the tank is 20 lb. e) Determine when the amount of salt in the tank is 30 lb. Determine the maximum amount of salt in the tank and when it occurs. g) Describe the long term behavior of the amount of salt in the tank using accompanying plots to support your description Activity 3: A 1200 gal tank has 400 gal of water in it containing exactly 20 lb of salt at the start of the process. Salt water is being added to the tank. The concentration of the salt in the water being added is 0,3 lb/gal. This salt water mixture initially is added at a rate of 0.1 gal/min and thi rate is steadily increased by 0.02 gal/min each minute.

Explanation / Answer

The mass balance equation can be written as

Rate of mass accumulation = rate of mass in- rate of mass out + rate of loss due to chemical reaction.

Since there is no chemical reaction,

Rate of mass accmulation = rate of mass in-rate of mass out

Since it is well mixed reactor, the concentration at the exit of reactor is same as that in the tank. Let this be c.

Rate of mass at the inlet = 0.25 lb/galllon*3 gallons/min= 0.75 lb/min

Rate of mass out = c*3.5 lb/min

Let the volume of tank be V.

Hence d(V*C)/dt = 0.75-C*0.35 with the boundary condition at t=0 C= 4/100 lb/gallon =0.04 lb/gallon.

(1/V)*dC/dt= 0.75-C*0.35

dC/(0.75-C*0.35)= Vdt

-ln(0.75-C*0.35)= Vt+K, K is integration constant. At t=0, C=0.04

Hence –ln(0.75-0.04*0.35)= K, K=0.31

Hecne –ln(0.75-C*0.35)= V*t+0.31

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