Linearizing Pouseuilles' Law

[ggb123]

Homework Statement

The viscous passage of gas through a capillary from a region (Region 1) to another (Region 2) is described by Pouseuilles' Law. I have to find a way to linearize Pouseuilles' law such that I can have a linear equation where the slope is dependent on the viscosity of a gas η.

Homework Equations

$\frac{ d (P_1(t) V_1 )}{dt} = - \frac{\pi D^4 \bar{P} }{128 \eta L} (P_1(t) - P_2(t))$

Where D is the diameter of the capillary, P-bar is the average pressure of the system, P1(t) is the pressure of region 1, P2(t) is the pressure of region 2.

The Attempt at a Solution

The function for pressure is exponential, so I'm assuming I'm going to have to take the log of this at some point, but I'm not entirely sure how to get there or where to go from there. Any help would be greatly appreciated!

Thanks!

 

[Mute]

I'm not exactly clear on what you want to do. Do you

1) want to linearize the differential equation? It's looks linear to me so unless something depends on $P_1(t)$ that you haven't mentioned I'm guessing this isn't want you want to do.

2) want to linearize the solution of the differential equation about some point in time? It would help if you write down what you got for your solution.

3) want to plot your solution in such a way that the plot is linear? i.e., do you want to find a function y = f(P_1(t)) and plot it against some x = g(t), for some functions f an g, such that the plot of y vs. x is linear? In this case it would again be helpful if you write down the solution you get for the differential equation.

 

[ggb123]

The solution I got was a decaying exponential of the form

$P(t) = P_0 e^{-a t}$.

This is experimental data, so I didn't solve the equation. I'm trying to modify this equation so I can use the experimentally found exponential relationship in this equation to find a linear equation with its slope dependent on η. Then, by finding the slope of this line, I should be able to find a value for η.

The function P1(t) describes gas leaving region 1 and entering region 2. So I guess P2(t) could be of the form P2(t) = P_0 - P1(t). That just leaves the left hand side of the equation, and I get stuck on how to handle the V1 term. I'm not sure, if in this case, it is to be treated as a constant or as the rate of the volumetric flow of the fluid from one region to another.

 

[Mute]

The solution I got was a decaying exponential of the form

$P(t) = P_0 e^{-a t}$.

This is experimental data, so I didn't solve the equation. I'm trying to modify this equation so I can use the experimentally found exponential relationship in this equation to find a linear equation with its slope dependent on η. Then, by finding the slope of this line, I should be able to find a value for η.

So, let me see if I understand you: You fit an exponential curve to some experimental data, and you wish to extract $\eta$ from the fit parameters? Is that correct?

The function P1(t) describes gas leaving region 1 and entering region 2. So I guess P2(t) could be of the form P2(t) = P_0 - P1(t). That just leaves the left hand side of the equation, and I get stuck on how to handle the V1 term. I'm not sure, if in this case, it is to be treated as a constant or as the rate of the volumetric flow of the fluid from one region to another.

Your differential equation is not hard to solve for arbitrary $V_1(t)$ or $P_2(t)$, as long as they do not depend on $P_1(t)$. However, you will not get a solution of the form $P_1(t) = P_0\exp(-at)$ unless $V_1$ is at least approximately constant. A non-zero $P_2(t)$ will also generate some extra terms in the solution for $P_1(t)$. If you set $V_1$ to a constant and $P_2$ to zero you can get a solution which is exponential, and you can relate $P_0$ and $a$ to the parameters in the differential equation.

Or did you need to do something else? Sorry if I'm still not clear on what you want to do.

 

[paranormal]

One possible approach to linearizing Pouseuilles' Law is by taking the natural logarithm of both sides of the equation. This will allow us to convert the exponential function into a linear function, which can then be solved for the slope.

\ln{\frac{ d (P_1(t) V_1 )}{dt}} = \ln{\left(- \frac{\pi D^4 \bar{P} }{128 \eta L} (P_1(t) - P_2(t))\right)}

Using the properties of logarithms, we can simplify this equation to:

\ln{\frac{ d (P_1(t) V_1 )}{dt}} = \ln{\left(- \frac{\pi D^4 \bar{P} }{128 \eta L}\right)} + \ln{(P_1(t) - P_2(t))}

Now, we can define a new variable, y, as the natural logarithm of the left side of the equation:

y = \ln{\frac{ d (P_1(t) V_1 )}{dt}}

And another variable, x, as the natural logarithm of the right side of the equation:

x = \ln{(P_1(t) - P_2(t))}

Substituting these values into the original equation, we get:

y = \ln{\left(- \frac{\pi D^4 \bar{P} }{128 \eta L}\right)} + x

This is now in the form of a linear equation, y = mx + b, where m is the slope and b is the y-intercept. In this case, m = 1 and b = \ln{\left(- \frac{\pi D^4 \bar{P} }{128 \eta L}\right)}.

Therefore, the linearized form of Pouseuilles' Law is:

y = x + \ln{\left(- \frac{\pi D^4 \bar{P} }{128 \eta L}\right)}

And the slope, m, is equal to 1. This means that the slope of the graph will depend on the viscosity of the gas, η, which is what we wanted to achieve.

I hope this helps!