A pressure difference of$\mathbf{6}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}$ is required to maintain a volume flow rate of 0.800 m³/s for a viscous fluid flowing through a section of cylindrical pipe that has radius 0.210 m. What pressure difference is required to maintain the same volume flow rate if the radius of the pipe is decreased to 0.0700 m?

• The pressure difference is$P_1=6\times {10}^{4}Pa$.
• The volume flow rate is$\frac{dV}{dt}=0.800m^3/s$ .
• The radius is$r_1=0.210m$ .
• The decreased radius of the pipe is$r_2=0.0700m$ .

In this problem, the pressure difference can be evaluated by using the relation of volume flow rate from Poiseuille’s equation.Also, the volume flow rate for viscous fluid is similar even if the radius is decreased.

The relation of volume flow rate can be written as:

Here, is the pressure difference, is the radius, is the dynamic viscosity and is the length of the pipe.

$V=\frac{P{\mathrm{\pi r}}^4}{8\eta L}$

Since the volume flow rate remains the same and doesn’t change. So, the relation of volume flow rate can be written as:

$\begin{array}{r}{V}_1=V_2\\ \left(\frac{P_1\pi r_1^4}{8\eta L}\right)=\left(\frac{P_2\pi r_2^4}{8\eta L}\right)\\ \left(P_1{r}_1^4\right)=\left(P_2{r}_2^4\right)\\ P_2=P_1{\left(\frac{r_1}{r_2}\right)}^4\end{array}$

Substitute$6\times {10}^{4}Pa$ for$P_1$ , 0.210 m for$r_1$ and 0.700m for$r_2$ in the above relation.

$\begin{array}{r}{P}_2=\left(6\times {10}^4\mathrm{Pa}\right){\left(\frac{0.210\mathrm{m}}{0.0700\mathrm{m}}\right)}^4\\ P_2=4.86\times {10}^6\mathrm{Pa}\end{array}$

Thus, the required pressure difference is$4.86\times {10}^{6}Pa$ .