Poiseuille's law [Eq. (9-15)] gives the volume flow rate of a viscous fluid through a pipe. (a) Show that Poiseuille's law can be written in the form \(\Delta P=I R\) where \(I=\Delta V / \Delta t\) represents the volume flow rate and \(R\) is a constant of proportionality called the fluid flow resistance. (b) Find \(R\) in terms of the viscosity of the fluid and the length and radius of the pipe. (c) If two or more pipes are connected in series so that the volume flow rate through them is the same, do the resistances of the pipes add as for electrical resistors \(\left(R_{e q}=R_{1}+R_{2}+\cdots\right) ?\) Explain. (d) If two or more pipes are connected in parallel, so the pressure drop across them is the same, do the reciprocals of the resistances add as for electrical resistors \(\left(1 / R_{\mathrm{eq}}=\right.\) $\left.1 / R_{1}+1 / R_{2}+\cdots\right) ?$ Explain.

Poiseuille's Law states that the volume flow rate \(Q\) of a viscous fluid through a pipe is given by: \(Q = \dfrac{\pi r^4 \Delta P}{8 \eta L}\) Where: \(Q\) = volume flow rate (\(\dfrac{m^3}{s}\)) \(r\) = radius of the pipe (m) \(\Delta P\) = pressure drop (Pa) \(\eta\) = fluid viscosity (Pa \(\cdot\) s) \(L\) = length of the pipe (m)

As given in the question, the volume flow rate can be written as: \(I = \dfrac{\Delta V}{\Delta t}\) We want to show that Poiseuille's law can be written in the form \(\Delta P = IR\), where \(R\) is a constant of proportionality called the fluid flow resistance.

Rearrange the Poiseuille's Law equation to get \(\Delta P\): \(\Delta P = \dfrac{8 \eta L}{\pi r^4} Q\) Now, substitute the volume flow rate \(I\): \(\Delta P = \dfrac{8 \eta L}{\pi r^4} I\) This equation can be written in the form \(\Delta P = IR\), where: \(R = \dfrac{8 \eta L}{\pi r^4}\) Hence, we have shown that Poiseuille's law can be written as \(\Delta P = IR\). #b. Finding \(R\) in terms of fluid viscosity, pipe length, and pipe radius# We have already obtained the formula for \(R\) from the previous part: \(R = \dfrac{8 \eta L}{\pi r^4}\) The fluid flow resistance \(R\) is inversely proportional to the fourth power of the pipe radius and directly proportional to pipe length and fluid viscosity. #c. Verifying if resistances add for pipes in series# For electrical resistors, when they are connected in series, the equivalent resistance is the sum of the individual resistances: \(R_{eq} = R_1 + R_2 + \cdots\). We need to check if the same rule applies to fluid resistances in series.

Let's consider two pipes connected in series with resistances \(R_1\) and \(R_2\). In a series connection, the volume flow rate, \(I\), through both pipes is the same.

The equivalent resistance for the series connection can be calculated by: \(R_{eq} = R_1 + R_2\) We can see that the fluid resistances follow the same rule as electrical resistances for pipes connected in series. #d. Verifying if reciprocals of resistances add for pipes in parallel# For electrical resistors, when they are connected in parallel, the reciprocals of the equivalent resistance are the sum of the reciprocals of the individual resistances: \(\dfrac{1}{R_{eq}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \cdots\). We need to check if the same rule applies to fluid resistances in parallel.

Let's consider two pipes connected in parallel with resistances \(R_1\) and \(R_2\). In a parallel connection, the pressure drop, \(\Delta P\), across both pipes is the same.

The equivalent resistance for the parallel connection can be calculated by: \(\dfrac{1}{R_{eq}} = \dfrac{1}{R_1} + \dfrac{1}{R_2}\) We can see that the fluid resistances follow the same rule as electrical resistances for pipes connected in parallel.