Question

A liquid \(L\) containing a dissolved gas \(S\) is stripped in a countercurrent operation using a pure carrier gas \(V\). The liquid phase inlet and outlet mole fractions of \(S\) are \(0.1\) and \(0.01\), respectively. The equilibrium distribution of \(S\) between \(V\) and \(L\) is governed by \(y_e = x_e\), where \(y_e\) and \(x_e\) are the mole fractions of \(S\) in \(V\) and \(L\), respectively. The molar feed rate of the carrier gas stream is twice that of the liquid stream. Under dilute solution conditions, the minimum number of ideal stages required is \(\underline{\hspace{2cm}}\) (in integer).

Hint:

For dilute absorption/stripping with linear equilibrium \(y^\ast=mx\), use the Kremser factors: \(A=\dfrac{L}{mV}\) (absorption) and \(S=\dfrac{V}{mL}\) (stripping).
With pure entering gas for stripping, \(x_1 = \dfrac{x_{N+1}}{\sum_{k=0}^{N-1} S^k}\). A geometric sum makes computations quick.
A larger \(S\) (more carrier gas) or more stages both lower the exit liquid concentration.

Answer 1

Correct Answer: 3

Step 1: Under dilute conditions with a linear equilibrium \(y^\ast = m x\) and \(m=1\), define the stripping factor: \[ S = \frac{V}{mL} = \frac{V}{L} = 2 \] since the carrier gas molar flow rate is twice the liquid flow rate.

Step 2: For countercurrent stripping with solute-free entering gas (\(y_{N+1}=0\)) and linear equilibrium, the Kremser relation for the exiting liquid composition after \(N\) ideal stages is: \[ x_1 = \frac{x_{N+1}}{\,1 + S + S^2 + \cdots + S^{N-1}\,}. \] Here: \[ x_{N+1} = 0.1, \quad x_1 = 0.01. \] Hence: \[ \frac{x_1}{x_{N+1}} = \frac{0.01}{0.1} = \frac{1}{10} = \frac{1}{\,1 + S + S^2 + \cdots + S^{N-1}\,}. \]

Step 3: With \(S=2\), \[ 1 + 2 + 2^2 + \cdots + 2^{N-1} = 2^N - 1 = 10. \] Thus: \[ 2^N = 11 \quad \Rightarrow \quad N = \log_2 11 \approx 3.46. \]

Step 4: Since \(N\) must be an integer number of ideal stages, the minimum integer satisfying the required reduction is: \[ N_{\min} = 3. \]

Final Answer: \[ \boxed{N_{\min} = 3 \;\; \text{ideal stages}} \]