Question 8.15: At sea level, water from the outlet of a river is diverted t...

At sea level, water from the outlet of a river is diverted to a power plant that operates on the principle of osmosis. A turbine is installed in the pipe that brings the river water to an osmotic membrane separating the clear water from the salt water of the sea. The sea water at the location of the membrane is assumed to have a constant low salt concentration c, i.e. c\ll 1. The pure water pressure in the river and in the sea is p_0. Because of osmosis, water is driven from the  river through the turbine and then across the osmotic membrane into the sea. Just after the turbine and before the membrane the pressure is p_1 = p_0 − Δ_p .
Calculate the mechanical power of the water flowing through the turbine,

P_W = Δ p \dot{N} ν

where ν is the molar volume of water and \dot{N} are the number of moles per unit time flowing through the osmotic membrane. The hydrodynamics of the turbine is such that we can assume,

Δ p = \frac{R_H \dot{N}}{ν}

so that P_W = R_H \dot{N}^ 2 , i.e. it is similar to the form of the Joule power for electrical heating.
Use the ideal mixture relation (8.68) to determine the mechanical power P_W . Since the salt concentration c is low enough, we can assume at ambient temperature that Δμ \gg R T c .

μ_A (T, p, c_A) = μ_A (T, p) + R T \ln (c_A)                 (8.68)
Show that the mechanical power is given by,

P_W = Δ μ \dot{N}

where Δμ is the chemical potential drop between the river and the sea water.

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The water chemical potential difference between the two sides of the membrane is given by,

Δ μ = μ (p_0, 1 − c) − μ (p_1)

Here, we use the notation of § 8.6. In particular, when a chemical potential does not depend on a concentration, it means that we are referring to the pure substance. According to relation (8.85), we have for pure water,

\mu (T,p) – \mu (T,p_{ext}) = \int_{p_{ext}}^{p}{\frac{\partial \mu }{\partial p} dp} = \nu \int_{p_{ext}}^{p}{dp} = \nu (p-p_{ext})        (8.85)

μ (p_1) = μ (p_0) + ν (p_1 − p_0) = μ (p_0) − ν Δ p

For the salt water, assuming c \ll 1 , we use relation (8.84) and write,

μ (T, p_{ ext}) = μ (T, p) − c R T        (chemical equilibrium)     (8.84)

μ (p_0, 1 − c) = μ (p_0) − R T c

Hence, since Δμ \gg R T c, we find that,

Δ μ = νΔ p − R T c \simeq νΔ p .

Thus, using the previous relation and the assumption ν Δ p = R_H \dot{N} ,

Δμ = R_H \dot{N}

Therefore, the mechanical power is given by,

P_W = R_H \dot{N}^ 2 = Δ μ\dot{N}

This expression is analogous to the electrical power, expressed as the product of a current (i.e. \dot{N} ) and potential difference (i.e. Δμ).

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