Resistance to flow by membrane

In calculating the osmotic pressure , we found a relationship between the pressure across a membrane and the flux of liquid across it, V ˙ w = A L p Δ p o s m \dot{V}_w = AL_p\Delta p_{osm} .

We can flip this relation around and expose a powerful perspective:

Δ p o s m = V ˙ w × 1 A L p \Delta p_{osm} = \dot{V}_w\times\frac{1}{AL_p}

Does this look familiar?

In fact, this has the same form as Ohm's circuit law, V = I R V=IR .

Here, our rate of volume flow is akin to current, the osmotic pressure takes the place of voltage, and the resistance is given by the inverse of hydraulic conductivity times surface area of the membrane. I.e.

R m e m = Δ p V ˙ w = 1 A L p \mathcal{R}_{mem} = \frac{\Delta p}{\dot{V}_w} = \frac{1}{AL_p}

In any situation where we have a flow resulting from a pressure, we can characterize the circuit element (membrane, pipe, etc.) according to this ratio ( Δ p / V ˙ w \Delta p/\dot{V}_w ), which we call the hydraulic resistance R \mathcal{R} .

Calculate the hydraulic resistance (in Pascal seconds per meter 3 ^3 ) of the semipermeable membrane connecting xylem and phloem, R m e m \mathcal{R}_{mem}

Assumptions

  • The length of the xylem:phloem section is 0.1 m
  • The radius of the phloem is 20 × 1 0 6 \times 10^{-6} m
  • The hydraulic permeability of the membrane is 5 × 1 0 14 \times 10^{-14} m/s/Pa


The answer is 1.59E+18.

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1 solution

Tom Engelsman
Jan 10, 2016

If one models the membrane as a cylinder, then all we have is simple algebra:

R = 1 2 π r h L p = 1 2 π × 20 × 1 0 6 0.1 × 5 × 1 0 14 = 1.59 × 1 0 18 Pa s/m 3 R = \frac{1}{2\pi rh L_p} = \frac{1}{2\pi \times 20\times 10^{-6} 0.1 \times 5\times 10^{-14}} = 1.59\times 10^{18}\textrm{ Pa s/m}^3 .

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