A viscous fluid is flowing through a small tube at a rate of 0.200 m3/s. If the pressure difference across the tube is doubled, and the length of the tube is halved, and the radius is also doubled, then what is the new rate of flow of the fluid? A) 9.2 m³/s B) 10.2 m³/s C) 12.8 m³/s D) 13.4 m³/s E) 14.6 m³/s

Question

Quizplus · Accepted Answer

First, we can use the continuity equation $A_1v_1=A_2v_2$ to relate the initial flow rate to the final flow rate. Here, $A$ is the cross-sectional area of the tube and $v$ is the velocity of the fluid.

$\Phi = A_1v_1 = A_2v_2$

Next, we can use the fact that the pressure difference across the tube is doubled to find the new velocity.

$\Delta P \propto v^2 \Rightarrow v_2=\sqrt{2}\cdot v_1$

Then, we use the fact that the length of the tube is halved and the radius is doubled to find the new cross-sectional area.

$A_2 = \dfrac{1}{2}\cdot A_1 \cdot \left(\dfrac{r_2}{r_1}ight)^2 = 2\cdot A_1$

Now we can solve for the new flow rate:

$\Phi' = A_2v_2 = 2A_1 \cdot \sqrt{2}\cdot v_1 = 2\sqrt{2}\cdot \Phi\approx12.8 	ext{ m}^3/	ext{s}$

Therefore, the answer is C.

A Viscous Fluid Is Flowing Through a Small Tube at a Rate