A holding tank is required to reduce the concentration of an isotope in water. T

Question

A holding tank is required to reduce the concentration of an isotope in water. The water flow rate is 10 gal/min and the isotope concentration is 0.01 mg/ft^3. How big should the tank be to reduce the isotope concentration by a factor of 10 at steady state operating conditions. If the tank is initially filled with pure water when the contaminated water first enters, how does the effluent concentration change with time? The isotope half-life is 10 minutes and the decay reaction is first order. [V = 1300 gal; C/C_o = {(Q/V)/(k + Q/V)}{1 - exp-(k + Q/V)t}]

Explanation / Answer

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

Flow rate* concentration =10gal/min* 10mg/ft3 = 10 gal/min* 10*7.48 mg/min=7480 mg/min ( 1ft3= 7.48 Us gallon)

Outlet rate concentration =7480/10 =748 mg/min

Change in concentration = 7480-748 = mg/min =6732 mg/min

This is lost due to chemical reaction = KCAV = (0.693/10)*74.8*V

6732= (0.693/10)* 74.8*V

V= 6732*10/(0.693*74.8) =1296 Gallon

For a first order reaction

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

FAO= FA+rA*V+ dNA/dt

For consant volume system, dividing by the volume , time, t = V/Q

FAO = FA+KCAtQ+ dNA/dt

CAO= CA+KCAt + t*dCA/dt

dCA/dt+CA*(1+kt)/t   = CAO/t

When integrated and noting that C=0 at t= 0

C/CO= CAO/(1+Kt){ 1-exp(-1+kt)/t}

But t=V/Q

C/CO= CAO/(1+KV/Q*(1-exp(-1+K*V/Q)/t}

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