# Step-by-Step Solution ## Step 1: Write Down Known Values - Inlet concentration of glutamine in blood, \(C_{B,in}\): 1.05 g/L - Blood flow rate, \(Q_B\): .85 L/min - Dialysate flow rate, \(Q_D\): .95 L/min - Membrane area, \(A\): 500 cm² - Overall (membrane) mass transfer coefficient, \(K_m\): \(2.86 \times 10^{-2}\) cm/s - Blood side mass transfer coefficient, \(k_B\): \(8.77 \times 10^{-2}\) cm/s - Dialysate side mass transfer coefficient, \(k_D\): \(3.44 \times 10^{-2}\) cm/s ## Step 2: Convert Flow Rates to Consistent Units Convert all flow rates to cm³/s (since 1 L = 100 cm³, 1 min = 60 s): \[ Q_B = .85 \times 100 \div 60 = 14.17\ \text{cm}^3/\text{s} \] \[ Q_D = .95 \times 100 \div 60 = 15.83\ \text{cm}^3/\text{s} \] ## Step 3: Calculate the Overall Mass Transfer Coefficient (\(K\)) The overall mass transfer coefficient combines resistances in series: \[ \frac{1}{K} = \frac{1}{k_B} + \frac{1}{K_m} + \frac{1}{k_D} \] Plug in values: \[ \frac{1}{K} = \frac{1}{8.77 \times 10^{-2}} + \frac{1}{2.86 \times 10^{-2}} + \frac{1}{3.44 \times 10^{-2}} \] \[ \frac{1}{K} = 11.4 + 34.97 + 29.07 = 75.44 \] \[ K = \frac{1}{75.44} = .01326\ \text{cm/s} \] ## Step 4: Calculate Number of Transfer Units (NTU) For counter-current flow, the number of transfer units (NTU) is: \[ NTU = \frac{K \cdot A}{Q_B} \] Plug in the values: \[ NTU = \frac{.01326 \times 500}{14.17} = \frac{6.63}{14.17} = .468 \] ## Step 5: Calculate the Extraction Factor (S) \[ S = \frac{Q_B}{Q_D} = \frac{14.17}{15.83} = .895 \] ## Step 6: Apply Counter-Current Dialyzer Equation For a membrane dialyzer with no net fluid flow across the membrane, the outlet blood concentration (\(C_{B,out}\)) can be found using: \[ \frac{C_{B,out} - C_{D,in}}{C_{B,in} - C_{D,out}} = e^{-NTU \cdot (1 - S)} \] Assuming the dialysate inlet is pure water (\(C_{D,in} = \)), and the dialysate outlet concentration is negligible, so \(C_{D,out} \approx \). This simplifies to: \[ \frac{C_{B,out}}{C_{B,in}} = e^{-NTU \cdot (1 - S)} \] Plug in the values: \[ NTU \cdot (1 - S) = .468 \times (1 - .895) = .468 \times .105 = .0491 \] \[ \frac{C_{B,out}}{1.05} = e^{-.0491} = .952 \] \[ C_{B,out} = 1.05 \times .952 = .999\ \text{g/L} \] ## Final Answer **The residual concentration of glutamine in the blood after dialysis is approximately \(.999\ \text{g/L}\).** --- **Summary Table** | Quantity | Value | |------------------------|------------| | \(C_{B,in}\) | 1.05 g/L | | \(C_{B,out}\) | .999 g/L | | \(Q_B\) | 14.17 cm³/s| | \(Q_D\) | 15.83 cm³/s| | \(A\) | 500 cm² | | \(K\) | .0133 cm/s| | \(NTU\) | .468 | | \(S\) | .895 |