Flow of a polymeric solution

This problem is adapted from the book of Chemical Engineering Fluid Mechanics, 2nd edition, by Ronald Darby. 

A polymeric solution, which is well represented power law fluid, has the following properties:

• flow index of 0.6,
• apparent viscosity of $420 \;cP$ at a shear rate of $1\; s^{-1}$,
• density of $75 \;lb_m/ft^3$.

This fluid will be transported through a $1\; in$ nominal commercial steel pipe diameter  at a flow rate of $4 \;gpm$. Estimate the following:

1. the pressure gradient in $psi/ft$,
2. the shear rate, evaluated, at the pipe wall and the apparent viscosity of the fluid at this shear rate,
3. the pressure gradient if the fluid were newtonian with a viscosity equal to the apparent viscosity from (2) above,
4. the Reynolds number for the polymeric solution and for the above newtonian fluid.

Hints:

• For this problem a laminar flow regime should be considered
• Attempt an anlytical approach to solve the problem
• This results can be determined numerically (not algebraic). No data is missing

The solution

It is not spedified but for small pipe diameter and low flow rates laminar flow regime can be considered although it should be further demonstrated in question (4). Then, the flow rate can be expressed as,

$Q=\frac{\pi R^3}{1/n+3}\left( \frac{\Delta P \; R}{2mL} \right)^{1/n}$        Eq. (1)

On the other hand, the shear stress evaluated at the pipe wall $r=R$ is given by:

$\tau_w=\frac{\Delta P \; R}{2L}$        Eq. (2)

In the above equations $R$ is the pipe internal diameter and $L$ the total pipe length. Be aware that $R$ is not the nominal diameter. For this case $R=0.957 \; in$, for schedule 80 for example.

Question 1

The pressure gradient can at first be isolated from Eq. (1) as,

$\frac{\Delta P}{L}=\frac{2m}{R} \left[ \frac{Q\left( 1/n+3 \right)}{\pi R^3} \right]^n$        Eq. (3)

From Eq. (3) all parameters are known but $m$. The so called  power law parameter of consistency can be determined from the, reduced, constitutive equation:

$\tau_{rz}=m\; \dot{\gamma}_{rz}^n$        Eq. (4)

for the flow of a fluid through a, circular, pipe with radial and axial coordinates $r$ and $z$. $\dot{\gamma}_{rz}$ is as you may know the shear rate. 

On the other hand, the apparent viscosity $\eta$ is related to $\dot{\gamma}_{rz}$ as follows:

$\eta=\dfrac{\tau_{rz}}{\dot{\gamma}_{rz}}$        Eq. (5)

Note: $\dot{\gamma}_{rz}$ is in fact a variable since it changes with the fluid strain (deformation) un less it is evaluated at a given position (radius).

In this way, $m$ in given in Eq. (4) can isolated and be written in terms of $\eta$ as follows:

$\eta\; \dot{\gamma}_{rz}=m\; \dot{\gamma}_{rz}^n$

$\Rightarrow \;\; m=\dfrac{\eta}{\dot{\gamma}_{rz}^{n-1}}$        Eq. (6)

In this way, $\Delta P/L$, given in Eq. (3), can be expressed as:

$\frac{\Delta P}{L}=\frac{2\eta}{R\; \dot{\gamma}_{rz}^{n-1}} \left[ \frac{Q\left( 1/n+3 \right)}{\pi R^3} \right]^n$        Eq. (7)

From Eq. (7), $\Delta P/L$ can be readily calculated since all data is known. This is:

$\frac{\Delta P}{L}=\frac{2\cdot 420 \cdot 0.001 }{0.957\cdot 0.0254\cdot 1^{-0.4}}\left( \frac{4\cdot 6.309e-5 \cdot 4.66}{\pi \left( 0.957\cdot 0.0254 \right)^3} \right)^{0.6}$ (in SI)

$\frac{\Delta P}{L}=34.55 \cdot \left( \dfrac{11.74}{\pi 1.43e-5} \right)^{0.6}$

$\frac{\Delta P}{L}=61,513.35\; Pa/m$

$\frac{\Delta P}{L}=2.71\; psi/ft$

Question 2

The shear rate at the wall can be easily found from Eq. (4) evaluating at $r=R$. This would give,

$\dot{\gamma}_{rz}\; \rvert_{r=R} = \left( \dfrac{\tau_{rz}}{m}\; \rvert_{r=R} \right)^{1/n}$

$\dot{\gamma}_{rz}\; \rvert_{r=R} = \left( \dfrac{\tau_{w}}{m} \right)^{1/n}$        Eq. (8)

which can also be written in terms of the pressure gradient as:

$\dot{\gamma}_{rz}\; \rvert_{r=R} = \left( \dfrac{\Delta P\; R}{2mL} \right)^{1/n}$        Eq. (9)

Remember that $m$ has units of $lb_f\cdot s^n$ so that there should not be any trouble with the units when evaluating Eq. (9).

However, neither $\tau_w$ nor the power law flow consistency index $m$ have been calculated. $\tau_w$ is found from Eq. (2):

$\tau_w=0.1080\; psi$

Next, knowing that $420\; cP=8.77e-3\; lb_f\; s/ft^2$, $m$ can be easily estimated from Eq. (6) to be,

$m = 6.09e-5 \; psi/s^{0.6}$

Note: $m$ does not change with the position or another variable. $m$ is in fact a fitting parameter.

Thus, from Eq. (9) it follows:

$\dot{\gamma}_{rz}=259,822.90\; s^{-1}$

The corresponding new apparent viscosity $\eta$ is found from Eq. (6):

$\eta=4.15e-7\; psi\; s$

$\eta=2.86\; cP$

Question 3

If the fluid wer newtonian, then $n=1$ and $\eta=m=\mu$ , as suggested, from Eq. (6). Also, Eq. (7) should be reduced to:

$\frac{\Delta P}{L}= \frac{8\mu Q}{\pi R^4}$          Eq. (10)

from which the pressure gradient is easily found.

Question 4

Reynolds numbers for bouth fluids follow from the known formulas:

$N_{Re}=\dfrac{\rho V D}{\mu}$

$N_{Re}=\dfrac{8D^nV^{2-n}}{m\left[ 2\left(3n+1\right)/n \right]^n}$

where the fluid velocity $V$ should be written in terms of the flow rate $Q$. Results are then straight.

Any questtion? Write in the comments and I shall try to help.

=========

Ildebrando.