Question
Asked Dec 12, 2019
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A man attaches a divider to an outdoor faucet so that water flows through a single pipe of radius 9.00 mm into two pipes, each with a radius of 6.00 mm. If water flows through the single pipe at 1.25 m/s, calculate the speed (in m/s) of the water in the narrower pipes.

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Expert Answer

Step 1

Given:

Radius of the original pipe = 9 mm

Radius of the other two pipes = 6 mm

Velocity of flow in original pipe = 1.25 m/s

Step 2

Calculating the velocity of flow in other pipes:

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Area of cross-section of original pipe, A = nr =(3.14)(9 ×10-³m)² = 2.54x10-“m² Velocity of flow, v, =1.25m/s Rate of flow from original pipe, R= A,v, =(2.54×10“m²)×(1.25m /s) = 3.175x10ʻm³ Area of narrower pipes, A, = A, =(3.14)(6×10³m)² =1.13×10“m² As we know that, Amount of water entering into the original pipe per second is equal to the sum of amount of water leaving at the end of narrower pipes per second.

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Step 3

Calculating the speed of the flow ...

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Or A,v, = A,v, +A, v, ...(1) As both the narrower pipes are similar, hence, v, =v, =v(let 's say) Substituting the values in the equation 1, 3.175x 10 m = (1.13x10 m² +1.13x10*m²)v 3.175x10-m Rearranging for velocity of flow in narrower pipes, v= =1.4m/s 2.26x10-m Speed of the water through narrower pipes is 1.4 m/s

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Physics

Fluid Mechanics

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