Two hoses are connected to the same outlet using a Y-connector, as the drawing s

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 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 = 1.44. Find the ratio of the speed of the water in hose B to the speed in hose A.

Explanation / Answer

?P = 8µLQ/(pi*r^4) The pressure difference is equal in both instances, also the length L, the viscosity µ , radius 1 is 1, radius 2 = 1.21, so 8µL*Q1/(pi*1^4) = 8µL*Q2/(pi*1.21)^4 Q1 = Q2/1.21^4 Q2 = 2.1434*Q2 ----- Volume1/s = cross section area 1*length of tube/s with L/s = speed of water in tube V/s = Ø*L/s: V1 = Ø1*L1 v2 = Ø2*L2 with V2 = 2.1434*V1 and Ø2 = pi*1.21^2/pi*Ø2: L1 = V1/Ø1 L2 = V2/Ø2 L1 = V1/Ø1 L2 = 2.1434*V1/(Ø1*1.21^2) L1/L2 = 1.21^2/2.1434 L1/L2 = 0.683 or L2/L1 = speed in B/speed in A = 1/0.683 = 1.464

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