Liquid in and on pipes

Posted in math on Monday, April 21 2014

One of the nice things about setting up the math for simple fluid flow problems is that you can recycle the initial bits for various other uses. If you set up a balance based on a particular geometry of a differential volume then a wide variety of possible flow cases can be solved by just substituting different boundary conditions.

I'm going to work through some flow problems for an incompressible fluid assuming laminar flow in cylindrical coordinates.

The first one is a falling film along a cylinder (from problem 2B.6 in Transport Phenomena). Imagine you stuck a straw through the top of a bottle of water and slowly squeezed the bottle. Water flows up the straw and flows out and down the straw. If you do this gently enough and the water wets the surface of the straw well enough you might get a nice cylinder of water running down the straw.

problem sketch

First I'm going to set up a basic momentum balance for a cylindrical shell with inner radius r and thickness Δr, then go into the boundary conditions for the problem (hopefully why I do this makes more sense later).

The steady state shell momentum balance is: (rate of momentum in) - (rate of momentum out) + gravity = 0

The rate of momentum entering in the z direction is: $$ \left(2 \pi r \Delta r \right) \phi_{zz} \vert_{z=0} $$

The rate of momentum entering in the r direction is: $$ \left(2 \pi r L \phi_{rz} \right)\vert_{r} $$

The rate of momentum exiting in the z direction is: $$ \left(2 \pi r \Delta r \right) \phi_{zz} \vert_{z=L} $$

The rate of momentum exiting in the r direction is: $$ \left(2 \pi r L \phi_{rz} \right)\vert_{r + \Delta r} $$

The force of gravity acting on the whole volume is: $$ \left( 2 \pi r \left(\Delta r\right) L \right)\rho g $$

Putting it all together the shell momentum balance is: $$ 2 \pi L \left[ \left(r \phi_{rz} \right)\vert_{r} - \left( r \phi_{rz} \right)\vert_{r + \Delta r} \right] + 2 \pi r \Delta r \left[ \phi_{zz} \vert_{z=0} - \phi_{zz} \vert_{z=L} + L \rho g \right] = 0$$

If we separate out the [ ... ]s and cancel out the s we get:

$$ { {\left[ \left(r \phi_{rz} \right)\vert_{r + \Delta r} - \left( r \phi_{rz} \right)\vert_{r} \right]} \over {\Delta r}} = r {{ \left[ \phi_{zz} \vert_{z=0} - \phi_{zz} \vert_{z=L} + L \rho g \right]} \over {L}} $$

If we take the limit Δr → 0 the left hand side is the definition of the partial derivative: $$ {{\partial} \over {\partial r}} r \phi_{rz} = r {{ \left[ \phi_{zz} \vert_{z=0} - \phi_{zz} \vert_{z=L} + L \rho g \right]} \over {L}} $$

Conveniently we can find tables of what the elements of the total momentum flux tensor are in cylindrical coordinates (assuming an incompressible fluid): $$ \phi_{rz} = -\mu \left[ \frac{\partial v_r}{\partial z} + \frac{\partial v_z}{\partial r} \right] + \rho v_r v_z $$

$$ \phi_{zz} = p + -2\mu \frac{\partial v_z}{\partial z} + \rho v_z^2 $$

If we assume vz = vz(r) and vr = 0 , and substitute:

$$ \frac{\partial}{\partial r} \left( -\mu r \frac{\partial v_z}{\partial r} \right) = r \left( { {p_0 + \rho v_z^2 - p_L - \rho v_z^2 + L\rho g} \over {L} } \right)$$

Lets mark this as equation (*)

For the fluid falling down the straw problem let the straw have radius R and the liquid layer to have outer radius aR. The pressure is constant as is the velocity of the liquid along the length we are considering so we get:

$$ \frac{\partial}{\partial r} \left( -\mu r \frac{\partial v_z}{\partial r} \right) = \rho g r$$

Integrating once w.r.t r:

$$ -\mu r \frac{\partial v_z}{\partial r} = \frac{\rho g}{2} r^2 + C_1 $$

Now at the outside boundary, r = aR, we assume the viscous shear is negligible: $$ \tau_{rz} = 0 \rightarrow \frac{\partial v_z}{\partial r}=0 $$

Subbing in and solving for C1: $$ C_1 = - \frac{\rho g}{2} \left( aR \right)^2 $$

$$ \frac{\partial v_z}{\partial r} = \frac{\rho g}{2 \mu} \left( \frac{\left( aR \right)^2}{r} - r \right) $$

Integrating again w.r.t. r: $$ v_z = \frac{\rho g}{2 \mu} \left( \left( aR \right)^2 \ln r - \frac{1}{2} r^2 + C_2 \right) $$

Now for the second boundary condition, at r = R the no-slip boundary condition holds, vz = 0:

$$ 0 = \frac{\rho g}{2 \mu} \left( \left( aR \right)^2 \ln R - \frac{1}{2} R^2 + C_2 \right) $$

$$ C_2 = - \left( aR \right)^2 \ln R + \frac{1}{2} R^2 $$

Finally: $$ v_z = \frac{\rho g}{2 \mu} \left( \left( aR \right)^2 \ln \left( \frac{r}{R} \right) + \frac{1}{2} \left( R^2 - r^2 \right) \right) $$

Now, we can take that exact same equation (*) and revisit it with an entirely different scenario.

Suppose now I have two reservoirs connected by horizontal tube. A shaft runs through the center of this tube and is moving from one reservoir to another at a constant velocity, v0 (problem 2B.7 Transport Phenomena).

problem sketch 2

The liquid is dragged from one reservoir to the next, through the annulus, by the moving shaft.

Suppose the inner diameter of the tube is R and the diameter of the shaft is kR. Also suppose the pressure in each reservoir is the same, and that we can just drop gravity from our concern. We can simplify equation (*) to:

$$ \frac{\partial}{\partial r} \left( -\mu r \frac{\partial v_z}{\partial r} \right) = 0 $$

Integrating once w.r.t r:

$$ -\mu r \frac{\partial v_z}{\partial r} = C_1 $$

and again:

$$ v_z = -\frac{C_1}{\mu} \ln r + C_2 $$

The two boundary conditions are: (1) r = R $$ 0 = -\frac{C_1}{\mu} \ln R + C_2 $$

(2) r = kR $$ v_0 = -\frac{C_1}{\mu} \ln kR + C_2 $$

Subtracting (1) from (2): $$ v_0 = -\frac{C_1}{\mu} \ln kR + C_2 +\frac{C_1}{\mu} \ln R - C_2$$

$$ C_1 = - \frac{\mu v_0}{\ln k} $$

Substituting back into (1) and solving for C2: $$ C_2 = -v_0 \frac{\ln R}{\ln k} $$

Finally: $$ v_z = \frac{v_0}{\ln k} \ln \left( \frac{r}{R} \right) $$

Basically after the balance and equation (*) producing the velocity distributions for two rather dissimilar flow patterns with cylindrical symmetry was a simple matter of plugging in the boundary conditions.

I could have just as easily solved for regular pipe flow, annular flow, or some other thing.