Assume a tank with volume V is completely mixed. The tank has a uniform in- and

Question

Assume a tank with volume V is completely mixed. The tank has a uniform in- and

out-flow Q with a constant source of a chemical through the in-flow at a concentration of CI.

(a) Derive a generalized expression of the chemical concentration (assuming the chemical is

conservative) with time. Designate the transient and steady state parts of the solution.

(Note: integration details are necessary).

(b) Assume the chemical is non-conservative with a decay rate of k day-1. Derive a

generalized expression of the chemical concentration with time. Designate the transient

and steady state parts of the solution. (Note: integration details are necessary).

Explanation / Answer

inflow = Qi

outflow = Qo

volume = V

accumulation = dC/dt

so at tranisent state

input - output = accumulation

Qi - Qo = -VdC/dt ...........................(1)

at steady state accumulation = 0

so

Qis - Qos = 0 ............................(2)

given that input and output is uniform

Qis - Qos = -VdC/dt

(Qis - Qos)/V *dt = -dC

integration from t = (0 to t) conc = (Co to Ci)

(Qis - Qos)/V * t = -(C0 - Ci)

Ci = (Qis - Qos)*t/V - C0

if reaction takes place

(Qis - Qos) = -VdC/dt - KC

dC/dt + KC/V + (Qis - Qos)/V =0

it is of form

dy/dc + Py + Q = 0

IF = e^(int(Pdx)) = e^(int(K/V)dt) = e^Kt/V

so solution is

Ci*e^(Kt/V) = int((Qis - Qos))/V * e^Kt/V)dt

Ci*e^(Kt/V) = (Qis - Qos))/V * e^Kt/V) * (V/K)

Ci*e^(Kt/V) = (Qis - Qos)/K * e^Kt/V + Cons

at T = o Ci = C0

so cons = C0 - (Qis - Qos)/K

so

Ci*e^(Kt/V) = (Qis - Qos)/K (e^Kt/V - 1) + C0)

********Note: the sign of final equation may differ as per assumptions taken

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