Question

Two hoses are connected to the same outlet using a Y-connector, as the drawing shows. The hoses A and B have the same length, but hose B has the larger radius. Each is open to the atmosphere at the end where the water exits. Water flows through both hoses as a viscous fluid, and Poiseuille's law
Q =πR4(P2 − P1)
8ηL applies to each. In this law, P2 is the pressure upstream, P1 is the pressure downstream, and Q is the volume flow rate. The ratio of the radius of hose B to the radius of hose A is RB/RA = 1.55. Find the ratio of the speed of the water in hose B to the speed in hose A.
vB/vA=

Answer 1

Answer:

$\dfrac{V_2}{V_1}=2.402$

Explanation:

Given that

Lets take radius of pipe A = R₁

Lets take radius of pipe B = R₂

R₂/R₁ = 1.55

These pipes are connected in the parallel connection that is why pressure difference will be same

$Q=\dfrac{\pi R^4\Delta P}{8\eta L}$

$\Delta P=\dfrac{8\eta LQ}{\pi R^4}{}$

Given that length of both the pipes is same

So we can say that

$\dfrac{Q_1}{R_1^4}=\dfrac{Q_2}{R_2^4}$

We know that

Q= A V

A= Area

V= velocity

A=π R²

So we can say that

$\dfrac{V_1R_1^2}{R_1^4}=\dfrac{V_2R_2^2}{R_2^4}$

$\dfrac{V_1}{R_1^2}=\dfrac{V_2}{R_2^2}$

$\dfrac{V_1}{V_2}=\dfrac{R_1^2}{R_2^2}$

$\dfrac{V_2}{V_1}=\dfrac{R_2^2}{R_1^2}$

$\dfrac{V_2}{V_1}=\dfrac{1.55R_1^2}{R_1^2}$

$\dfrac{V_2}{V_1}=2.402$