MIT10_626S11_lec31

# MIT10_626S11_lec31 - VI Electrokinetics Lecture 31...

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VI. Electrokinetics Lecture 31: Electrokinetic Energy Conversion MIT Student 1 Principles 1.1 General Theory We have the following equation for the linear electrokinetic response of a nanochan­ nel: ± Q ² K P K EO Δ P = ± I K K E ²± Δ V ² The basic idea 1 is to apply Δ P and to try to “harvest” the streaming current I or streaming voltage Δ V . 1.1.1 Open-circuit Potential (Streaming Voltage) I = K Δ P + K E Δ V K I = 0 V Δ O = - Δ P K E Where K Δ P is the streaming current and K E Δ V is the Ohmic current. 1 J.F. Osterle, A uniﬁed treatment of the thermodynamics of steady state energy conversion, Appl. Sci. Res. 12 (1964), pp. 425-434. J.F. Osterle, Electro-kinetic energy conversion, J. Appl. Mech. 31 (1964), pp. 161-164. The same idea was also discovered by Quincke in the 1800s. 1

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Lecture 31: Electrokinetic energy conversion 10.626 (2011) Bazant Δ P = P in - P out Δ V O = V in - V out Then Q = K P Δ P + K EO Δ V K 2 = K P 1 - Δ P K P K E = (1 - α ) K P Δ P K 2 Where α = K P K E . Pressure-driven ﬂow is reduced by electro-osmotic back ﬂow. Net ﬂow is 0 when α = 1, but this is not possible. 1.1.2 Second Law of Thermodynamics Work done on system to drive motion per time (power input): P = Q Δ P + I Δ V > 0 = ± Δ P ( Δ P Δ V ) K > 0 Δ V ² P must be positive since irreversible work produces heat in the system. The conductance tensor K must be positive deﬁnite , since this inequality holds for any Δ P , Δ V . So det ( K ) > 0 and thus α < 1. So, we can’t apply pressure and get a ﬂow in the reverse direction! ( Q = (1 - α ) K P Δ P ). 2   
Lecture 31: Electrokinetic energy conversion 10.626 (2011) Bazant 1.1.3 Streaming Current Harvesting Consider harvesting the streaming current via two electrodes connected by a total load resistance R L ( = R external + R internal from interfaces). Then Δ V = - IR L . Also: I = K EO Δ P + K E Δ V = K Δ - K E R L I ± ² K = Δ P 1 + K E R L Let θ = K E R L = R L external load = R E internal resistance . So: K Δ P I = 1 + θ So, applied pressure leads to a current source via streaming current, which ﬂows through internal and external resistors in parallel. We have the equivalent circuit: Q = K P Δ P + K Δ V K 2 - R L Δ V = K P Δ P 1 + θ αθ = K P Δ P ³ 1 - 1 + θ ´ ± 1 + θ = K P Δ P - αθ 1 + θ ² 3       

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Lecture 31: Electrokinetic energy conversion 10.626 (2011) Bazant E f ciency P out
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MIT10_626S11_lec31 - VI Electrokinetics Lecture 31...

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