Question

A pulse tracer is introduced in an ideal CSTR (with a mean residence time T) at time = 0. The time taken for the exit concentration of the tracer to reach half of its initial value will be:

• A. \( 2 \, \text{T} \)
• B. \( 0.5 \, \text{T} \)
• C. \( \frac{\text{T}}{0.693} \)
• D. \( 0.693 \, \text{T} \)

Hint:

An ideal CSTR tracer balance behaves exactly like a standard first-order radioactive decay process with a decay constant \(\lambda = 1/T\). The half-life equation for any first-order decay process is given by: \[ t_{1/2} = \frac{\ln(2)}{\lambda} = \frac{0.693}{1/T} = 0.693 \cdot T \]

Answer 1

The Correct Option is D

Concept: The residence time distribution (RTD) of an ideal Continuous Stirred-Tank Reactor (CSTR) can be determined experimentally by injecting a pulse tracer input at the reactor inlet and measuring the tracer concentration profile at the outlet over time. When a tracer pulse of total mass $M$ is injected instantly at time $t = 0$ into an ideal CSTR of volume $V$ operating at a constant volumetric flow rate $v$, the tracer mixes uniformly throughout the vessel immediately. The initial concentration of the tracer inside the reactor and at the exit at time $t = 0^+$ is given by: \[ C_{\text{init}} = C_0 = \frac{M}{V} \]

Step 1: Setting up the transient tracer material balance equation.
Let us perform a transient mass balance for the tracer inside the CSTR for times $t \gt 0$, noting that no additional tracer enters the reactor after the initial pulse ($\text{In} = 0$): \[ \text{In} - \text{Out} = \text{Accumulation} \] \[ 0 - v \cdot C = V \cdot \frac{dC}{dt} \] Let us rearrange this differential equation to separate the variables $C$ and $t$: \[ \frac{dC}{C} = -\left(\frac{v}{V}\right) \cdot dt \] Recall that the mean residence time of an ideal CSTR is defined as $T = \frac{V}{v}$. Substituting this definition into the differential equation yields: \[ \frac{dC}{C} = -\frac{1}{T} \cdot dt \]

Step 2: Integrating the differential equation to find the concentration profile.
Let us integrate both sides of the separated differential equation from the initial state ($t = 0, C = C_0$) to an arbitrary future state ($t, C$): \[ \int_{C_0}^{C} \frac{dC}{C} = -\frac{1}{T} \cdot \int_{0}^{t} dt \] \[ \ln\left(\frac{C}{C_0}\right) = -\frac{t}{T} \] Taking the exponential of both sides gives the classic first-order exponential decay equation for the tracer exit concentration profile: \[ C(t) = C_0 \cdot e^{-t/T} \]

Step 3: Calculating the time required to reach half of the initial concentration.
The problem asks for the time ($t_{1/2}$) required for the exit concentration to drop to exactly half of its initial value ($C(t_{1/2}) = 0.5 \cdot C_0$): \[ 0.5 \cdot C_0 = C_0 \cdot e^{-t_{1/2}/T} \] Dividing both sides by the non-zero initial concentration $C_0$: \[ 0.5 = e^{-t_{1/2}/T} \quad \Rightarrow \quad \frac{1}{2} = e^{-t_{1/2}/T} \] Taking the natural logarithm ($\ln$) of both sides of the equation: \[ \ln\left(\frac{1}{2}\right) = -\frac{t_{1/2}}{T} \] \[ -\ln(2) = -\frac{t_{1/2}}{T} \] Eliminating the negative signs from both sides yields: \[ t_{1/2} = T \cdot \ln(2) \] Substituting the known numerical value for the natural logarithm of 2 ($\ln(2) \approx 0.69315$): \[ t_{1/2} = 0.693 \cdot T \] Therefore, the time required for the tracer concentration to reach half of its initial value is equal to $0.693 \, \text{T}$.