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Basics of Fluid Mechanics, 2014a

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13.9. COUNTER–CURRENT FLOW 561<br />

For this shear stress, the critical upward interface velocity is<br />

( 2 3 − 1 2)<br />

{}}{ (<br />

1 ρL gh 2 )<br />

U critical | interface<br />

=<br />

(13.71)<br />

6 μ L<br />

The wall shear stress is the last thing that will be done on the liquid side. The wall<br />

shear stress is<br />

⎛<br />

(<br />

dU<br />

τ L | @wall<br />

= μ L dx ∣ = μ L<br />

⎜<br />

ρ L g<br />

⎝ ✟ ✟✯ 0 )<br />

2 x − h +<br />

x=0<br />

μ L<br />

τ i<br />

{ }} {<br />

2 ghρ L<br />

3<br />

⎞<br />

1<br />

⎟<br />

μ L<br />

⎠<br />

x=0<br />

(13.72)<br />

Simplifying equation (13.72) 12 becomes (notice the change <strong>of</strong> the sign accounting for<br />

the direction)<br />

τ L | @wall<br />

= ghρ L<br />

3<br />

(13.73)<br />

Again, the gas is assumed to be in a laminar flow as well. The shear stress on gas<br />

side is balanced by the pressure gradient in the y direction. The momentum balance on<br />

element in the gas side is<br />

dτ xyG<br />

dx<br />

= dP<br />

dy<br />

(13.74)<br />

The pressure gradient is a function <strong>of</strong> the gas compressibility. For simplicity, it is<br />

assumed that pressure gradient is linear. This assumption means or implies that the<br />

gas is incompressible flow. If the gas was compressible with an ideal gas equation <strong>of</strong><br />

state then the pressure gradient is logarithmic. Here, for simplicity reasons, the linear<br />

equation is used. In reality the logarithmic equation should be used ( a discussion can be<br />

found in “Fundamentals <strong>of</strong> Compressible Flow” a Potto project book). Thus, equation<br />

(13.74) can be rewritten as<br />

dτ xyG<br />

dx<br />

= ΔP<br />

Δy = ΔP<br />

L<br />

(13.75)<br />

Where Δy = L is the entire length <strong>of</strong> the flow and ΔP is the pressure difference <strong>of</strong> the<br />

entire length. Utilizing the Newtonian relationship, the differential equation is<br />

d 2 U G<br />

dx 2<br />

= ΔP<br />

μ G L<br />

(13.76)<br />

12 Also noticing that equation (13.70) has to be equal ghρ L to support the weight <strong>of</strong> the liquid.

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