Step 1: Understanding concentration polarization
In reverse osmosis (RO), water permeates through the membrane while solutes are rejected. However, solutes accumulate near the membrane surface, creating a concentrated boundary layer — this phenomenon is called concentration polarization.
Step 2: Effect on driving force (water flux)
Due to solute accumulation, the effective osmotic pressure at the membrane surface increases. Since net driving pressure is \[ \Delta P_{\text{net}} = \Delta P - \Delta \pi, \] where \(\Delta \pi\) is osmotic pressure difference, the higher osmotic pressure reduces the net driving force. Thus, water flux decreases.
Step 3: Effect on solute rejection
At higher solute concentration near the membrane surface, more solute tends to diffuse through the membrane. This reduces the overall solute rejection capability of the process.
Conclusion:
Concentration polarization leads to: \[ \text{Reduced water flux and reduced solute rejection.} \]
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Consider a process with transfer function: \[ G_p = \frac{2e^{-s}}{(5s + 1)^2} \] A first-order plus dead time (FOPDT) model is to be fitted to the unit step process reaction curve (PRC) by applying the maximum slope method. Let \( \tau_m \) and \( \theta_m \) denote the time constant and dead time, respectively, of the fitted FOPDT model. The value of \( \frac{\tau_m}{\theta_m} \) is __________ (rounded off to 2 decimal places).
Given: For \( G = \frac{1}{(\tau s + 1)^2} \), the unit step output response is: \[ y(t) = 1 - \left(1 + \frac{t}{\tau}\right)e^{-t/\tau} \] The first and second derivatives of \( y(t) \) are: \[ \frac{dy(t)}{dt} = \frac{t}{\tau^2} e^{-t/\tau} \] \[ \frac{d^2y(t)}{dt^2} = \frac{1}{\tau^2} \left(1 - \frac{t}{\tau}\right) e^{-t/\tau} \]