This problem was adapteed from the book of Dynamics of Polymeric Liquids V1 by Bird RB et al.
A 14.5%, by weight, solution of polyisoprene in isopentane, which behaves as a power law fluid, has the following parameters at 330 K:
- $m=5.2 \times 10^{-3} Pa \; s^n$
- $n=0.23$
This fluid is being pumped through a pipe that is $12.5\; m $ long and has a pressure drop of $35\; kPa$; the flow regime is known to be laminar. It is desired to build another pipe with a length of $24 \; m$ with the same volume rate of flow and the same pipe internal diameter. What should the new presssure drop be?
The solution
Since the fluid is well represented as power law and the flow regime is laminar, the following definition for flow rate comes out,
$Q=\frac{\pi R^3}{1/n+3}\left( \frac{\Delta P \;R}{2mL} \right)^{1/n}$ Eq. (1)
Now, since this situation implies the scaling between two models the same ideas of dimensional analysis can be used here. In other words, we can take,
$Q_1=Q_2$ Eq. (2)
since this should be valid for both situations. Subscripts 1 and 2 indicate the first pipe and the second, future, pipe to be built. Because some parameters are the same for both situations great simplifications can be made as follows,
$\frac{\pi R^3}{1/n+3}\left( \frac{\Delta P_1\; R}{2mL_1} \right)^{1/n} = \frac{\pi R^3}{1/n+3}\left( \frac{\Delta P_2\; R}{2mL_2} \right)^{1/n}$ Eq. (3)
$\left( \frac{\Delta P_1}{L_1} \right)^{1/n} = \left( \frac{\Delta P_2}{L_2} \right)^{1/n}$ Eq. (4)
Then, the pressure at the second pipe is eassily to find as,
$\Delta P_2= \Delta P_1 \frac{\;L2}{L1}$ Eq. (5)
Further substitution of the provided data give you the new pressure. You should notice that pressure drop do not depend on $n$ or $m$ which is expected since the fluid is not changed. As you can see, pressure drop is increased because the fluid runs a larger distance,
$\Delta P_2=67.2\; kPa$
Any questtion? Write in the comments and I shall try to help.
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Ildeebrando
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