Water flows through two different pipes A and B of the same circular cross-section but at different flow rates. The length of pipe A is 1.0 m and that of pipe B is 2.0 m. The flow in both the pipes is laminar and fully developed. If the frictional head loss across the length of the pipes is same, the ratio of volume flow rates Q_B/Q_A is ______ (round off to two decimal places).

The correct answer is 0.5.

To find the ratio of the volume flow rates $Q_B/Q_A$ for laminar and fully developed flow through the two pipes, we can use the Hagen-Poiseuille equation.

For a fully developed laminar flow of a viscous, incompressible fluid through a circular pipe of constant cross-section, the frictional pressure drop $\Delta P$ is given by:

$\Delta P=\frac{128\mu LQ}{\pi D^4}$

where:
$\mu$ is the dynamic viscosity of the fluid,
$L$ is the length of the pipe,
$Q$ is the volumetric flow rate, and
$D$ is the pipe diameter.

The frictional head loss $h_f$ is related to the pressure drop $\Delta P$ by the relation:

$h_f=\frac{\Delta P}{\rho g}=\frac{128\mu LQ}{\rho g\pi D^4}$

where $\rho$ is the fluid density and $g$ is the acceleration due to gravity.

Since both pipes A and B carry water (meaning $\rho$ and $\mu$ are identical) and have the same circular cross-section (meaning diameter $D$ is identical), we can write the head loss for each pipe as:

$h_{f,A}=\frac{128\mu L_A{Q}_A}{\rho g\pi D^4}$

and

$h_{f,B}=\frac{128\mu L_B{Q}_B}{\rho g\pi D^4}$

We are given that the frictional head loss is the same in both pipes ($h_{f,A}=h_{f,B}$). Equating the two expressions yields:

$\frac{128\mu L_A{Q}_A}{\rho g\pi D^4}=\frac{128\mu L_B{Q}_B}{\rho g\pi D^4}$

By canceling the common constants from both sides, we get:

$L_A{Q}_A=L_B{Q}_B$

Rearranging the equation to solve for the ratio $Q_B/Q_A$ gives:

$\frac{Q_B}{Q_A}=\frac{L_A}{L_B}$

Substituting the given lengths ($L_A=1.0\text{m}$ and $L_B=2.0\text{m}$):

$\frac{Q_B}{Q_A}=\frac{1.0}{2.0}=0.5$

Thus, the ratio of volume flow rates is 0.5.