Question

By what factor will the flow rate of a viscous fluid through a pipe change, if the pipe radius is doubled, the fluid viscosity is doubled, the length is doubled and the pressure drop is reduced to a quarter of its previous value?

• A. $4$
• B. $16$
• C. $\frac{1}{4}$
• D. Doesn't change

Answer 1

The correct option is D Doesn't change

We know,
Rate of flow of a viscous fluid,
$\Rightarrow V=\frac{\pi P{r}^4}{8\eta l}$
where, $P=$ Pressure drop across pipe
$r=$ Internal radius of pipe
$l=$ Length of pipe
$\eta =$Fluid viscosity
Given,
If $P^{\prime }=\frac{P}{4},r^{\prime }=2r,{\eta }^{\prime }=2\eta ,l^{\prime }=2l$
Then, $V^{\prime }=\frac{\pi P^{\prime }(r^{\prime }{)}^4}{8{\eta }^{\prime }{l}^{\prime }}=\frac{\pi \left(\frac{P}{4}\right)(2r{)}^4}{8\left(2\eta \right)\left(2l\right)}$
$\Rightarrow \frac{V^{\prime }}{V}=1$
$\therefore$ Volume flow rate remains same.