The following schematic diagram shows a chemostat with cell recycle (feed flow rate \(F_0\) and recycle flow rate \(F_r\) in \(\mathrm{L\,h^{-1}}\)). The cell concentrations in the reactor, recycle stream, and product stream are \(X_1\), \(X_0\), and \(X\) (in \(\mathrm{g\,L^{-1}}\)), respectively. If \(\dfrac{X_0}{X_1}=1.5\), \(\dfrac{F_r}{F_0}=0.7\), and \(X_1=7.3~\mathrm{g\,L^{-1}}\), the value of \(X\) (in \(\mathrm{g\,L^{-1}}\), rounded off to one decimal place) is ________________.
Show Hint
For separators/recycle systems, write the mass balance on the splitter: \((\text{inlet flow})\times(\text{inlet conc.}) = \sum (\text{outlet flow}\times\text{outlet conc.})\).
Converting given ratios (e.g., \(F_r/F_0\), \(X_0/X_1\)) into variables (\(r\)) simplifies algebra.
Step 1: Perform a steady-state cell mass balance over the \textit{cell separator}. The separator receives \((F_0+F_r)\) from the reactor at concentration \(X_1\) and splits it into product \(F_0\) with concentration \(X\) and recycle \(F_r\) with concentration \(X_0\):
\[
(F_0+F_r)X_1 = F_0X + F_rX_0.
\]
Step 2: Use the given ratios \(r=\dfrac{F_r}{F_0}=0.7\) and \(\dfrac{X_0}{X_1}=1.5 \Rightarrow X_0=1.5X_1\). Divide the balance by \(F_0\) to get
\[
(1+r)X_1 = X + rX_0 = X + r(1.5X_1).
\]
Hence,
\[
X = (1+r)X_1 - 1.5rX_1 = X_1\!\left[1 + r - 1.5r\right] = X_1\!\left(1 - 0.5r\right).
\]
Step 3: Substitute \(r=0.7\) and \(X_1=7.3~\mathrm{g\,L^{-1}}\):
\[
X = 7.3 \times (1 - 0.5\times 0.7) = 7.3 \times 0.65 = 4.745 \approx 4.7~\mathrm{g\,L^{-1}}.
\]
\(\therefore\) The product-stream cell concentration is \(\boxed{4.7~\mathrm{g\,L^{-1}}}\).
