Asked Dec 12, 2019

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.


Expert Answer

Step 1


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.

Step 3

Calculating the speed of the flow ...


Image Transcriptionclose

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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Fluid Mechanics

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